Cooking as a Chemical Reaction: Culinary Science with Experiments [2nd ed.] 9781138597129, 1138597120, 9781466554801, 1466554800

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Cooking as a ­Chemical ­Reaction

Culinary Science with Experiments

Cooking as a ­Chemical ­Reaction

Culinary Science with Experiments Second Edition

Sibel Özilgen

CRC Press Taylor & Francis Group 6000 Broken Sound Parkway NW, Suite 300 Boca Raton, FL 33487-2742 © 2020 by Taylor & Francis Group, LLC CRC Press is an imprint of Taylor & Francis Group, an Informa business No claim to original U.S. Government works Printed on acid-free paper International Standard Book Number-13: 978-1-138-59712-9 (Hardback) This book contains information obtained from authentic and highly regarded sources. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, please access www.copyright. com (http://www.copyright.com/) or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. CCC is a not-for-profit organization that provides licenses and registration for a variety of users. For organizations that have been granted a photocopy license by the CCC, a separate system of payment has been arranged. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging‑in‑Publication Data Names: Ozilgen, Z. Sibel, author. Title: Cooking as a chemical reaction : culinary science with experiments / Sibel Ozilgen. Description: Second edition. | Boca Raton : CRC Press, Taylor & Francis Group, 2019. | Includes bibliographical references. Identifiers: LCCN 2019006672 | ISBN 9781138597129 (hardback : alk. paper) | ISBN 9781466554801 (pbk. : alk. paper) Subjects: LCSH: Food—Composition. | Food—Analysis. | Chemistry, Technical—Experiments. | Cooking—Mathematics. | Cooking—Study and teaching. Classification: LCC TX545 .O986 2019 | DDC 664/.07—dc23 LC record available at https://lccn.loc.gov/2019006672 Visit the Taylor & Francis Web site at http://www.taylorandfrancis.com and the CRC Press Web site at http://www.crcpress.com

In memory of Prof. Dr. David S. Reid

Contents

Preface xiii Acknowledgments xv Author xvii 1  Measurements and Units 1 Why Do We Need Mathematics in Culinary Processes? 1 Types of Measurements in the Kitchen 1 Units of Measurements in Culinary Calculations 5 Rule 7 Everyday Calculations in the Kitchen 9 Fractions and Percentages 9 Yield Percent 11 Recipe Yield Conversion 11 Simplified Cost Calculations for Culinary Operations 13 Simplified Statistics for Culinary Operations 16 Experiment 1.1 17 Points to Remember 20 Study Questions 21 Selected References 22 2  Basic Food Chemistry 23 Food Processing Is All about Chemistry 23 Experiment 2.1 27 Experiment 2.2 29 The Science Behind the Results 33 Points to Remember 36 Selected References 37

vii

viii Contents

3  Water in Culinary Transformations 39 Functional Properties of Water in Culinary Processes 39 Structure of Water 41 Experiment 3.1 45 The Science Behind the Results 49 Experiment 3.2 51 The Science Behind the Results 55 Experiment 3.3 57 The Science Behind the Results 61 Experiment 3.4 63 The Science Behind the Results 67 Experiment 3.5 69 The Science Behind the Results 73 Experiment 3.6 75 The Science Behind the Results 79 Experiment 3.7 81 The Science Behind the Results 85 Points to Remember 86 More Ideas to Try 87 Study Questions 87 Selected References 87 4  Carbohydrates in Culinary Transformations 89 Functional Properties of Carbohydrates in Culinary Processes 89 Carbohydrate Structure 90 General Formula for Carbohydrates 90 Experiment 4.1 93 The Science Behind the Results 97 Experiment 4.2 99 The Science Behind the Results 105 Experiment 4.3 107 The Science Behind the Results 109 Experiment 4.4 111 Study Questions 113 The Science Behind the Results 115 Experiment 4.5 119 The Science Behind the Results 123 Experiment 4.6 125 The Science Behind the Results 129

Contents ix

Experiment 4.7 131 Study Questions 132 The Science Behind the Results 133 Experiment 4.8 135 Study Question 136 The Science Behind the Results 137 Experiment 4.9 139 The Science Behind the Results 141 Points to Remember 142 More Ideas to Try 143 Study Questions 143 Selected References 143 5  Proteins in Culinary Transformations 145 Functional Properties of Proteins in Culinary Processes 145 Protein Structure 146 Experiment 5.1 151 The Science Behind the Results 155 Experiment 5.2 157 Study Questions 159 The Science Behind the Results 161 Experiment 5.3 163 The Science Behind the Results 165 Experiment 5.4 167 The Science Behind the Results 169 Experiment 5.5 171 The Science Behind the Results 173 Experiment 5.6 175 Study Questions 177 The Science Behind the Results 179 Experiment 5.7 181 The Science Behind the Results 185 Experiment 5.8 187 The Science Behind the Results 189 Experiment 5.9 191 The Science Behind the Results 195 Experiment 5.10 197 Study Questions 198 The Science Behind the Results 201

x Contents

Experiment 5.11 203 The Science Behind the Results 209 Points to Remember 210 More Ideas to Try 211 Study Questions 211 Selected References 211 6  Fats and Oils in Culinary Transformations 213 Functional Properties of Fats and Oils in Culinary Processes 213 Fat and Oil Structure 214 Experiment 6.1 217 The Science Behind the Results 219 Experiment 6.2 223 Experiment 6.3 225 The Science Behind the Results 229 Experiment 6.4 231 Study Questions 233 The Science Behind the Results 235 Experiment 6.5 237 The Science Behind the Results 241 Experiment 6.6 243 The Science Behind the Results 245 Experiment 6.7 249 The Science Behind the Results 253 More Ideas to Try 254 Study Questions 254 Points to Remember 255 Selected References 256 7  Keys to Developing the Perfect Bite: New Food Product Development and Sensory Evaluation Tests 257 Reasons to Develop a New Food Product 257 Stages in New Food Product Development 258 Idea Development 259 Product Development 259 Commercialization 260 How to Carry Out Sensory Evaluation Tests 260 Points to Remember 269 Selected References 270

Contents xi

8  The Science of Flavor and Flavor Pairing 271 Experiment 8.1 273 The Science Behind the Results 275 Experiment 8.2 277 The Science Behind the Results 281 Experiment 8.3 283 Experiment 8.4 287 The Science Behind the Results 289 Points to Remember 303 More Ideas to Try 304 Selected References 304 9  Food Additives in Culinary Transformations 307 Classification of Food Additives 307 Intentional/Direct Food Additives 307 Unintentional/Indirect Food Additives 308 Experiment 9.1 313 Hydrocolloids in Culinary Transformations 317 Experiment 9.2 319 Experiment 9.3 325 The Science Behind the Results 331 Experiment 9.4 339 The Science Behind the Results 343 Study Question 343 More Ideas to Try 343 Points to Remember 344 Selected References 345 10  Food Safety and Hygiene in Culinary Transformations 347 Consumers Have a Right to Expect That the Foods They Consume Will Be Safe and of High Quality 347 Food Safety Has to Do with Controlling Potential Foodborne Hazards 348 Experiment 10.1 349 The Science Behind the Results 353 Experiment 10.2 355 The Science Behind the Results 361 Experiment 10.3 367 The Science Behind the Results 369 Safe Practices are Important to Avoid Foodborne Illness 369

xii Contents

Experiment 10.4 371 The Science Behind the Results 373 Points to Remember 374 More Ideas to Try 375 Study Questions 375 Selected References 375 Extended Glossary 377 Index 387

Preface

This book is written for undergraduate students in culinary arts, nutrition, dietetics, and gastronomy programs. It is intended for students with limited scientific knowledge who are studying different aspects of food preparation and processing. The text uses experiments and experiences from the kitchen rather than theory as the basic means of explaining the scientific facts and principles behind food preparation and processing. Thorough explanations of important scientific concepts that are required to comprehend the text are provided in the glossary. This textbook is prepared such that students first perform certain experiments and record their observations in tables provided in the book. The science behind their expected observations are then subsequently explained. By conducting experiments and using experiences from the kitchen, this textbook aims to engage students in their own learning process. With this book, students are able to make observations that they will frequently see in the kitchen and will be able to learn the science behind these phenomena. Thus, they will be able to control these phenomena, allowing them to create new food products, improve the quality and safety of their dishes, improve the culinary presentations of their food, and understand what goes wrong in the kitchen. Many concepts throughout the book are marked with the symbol

.

This symbol indicates that the concept is an important one that students will come across frequently, both during the study of this text and in the kitchen. The symbol

precedes a scientific explanation of the observations made dur-

ing experiments in the chapter. At the end of each chapter, students are presented with important points to remember, more ideas to try, and study questions to reinforce concepts that were presented in the chapter. It is important to note that it is necessary for students to fully understand the key concepts of each chapter because they will reoccur in subsequent chapters. Sibel Özilgen

xiii

Acknowledgments

I would like to express my gratitude to the many people who have graciously helped in the preparation of this book. To the Yeditepe University in Istanbul, Turkey, for providing facilities to conduct the experiments. To Günsel Keserci for the incredible chef graphics. To my students and colleagues, Yavuz Efe Başeğmez, Fevzi Can Acar, Sinem Turan, Gülben Meral Bozan, Dilek Çiftçi, Nurece Yılmazel, Evren Galip Altaylar, Baran Yağmurlu, Aylin Doğan, Merve İşeri, Fikret Soner, and Melek Gündoğdu for conducting experiments, of which photos were taken to be added to the book. To Semih Özkan, Özcan Yaman, and Barbaros Erdal Çiftçi for photography. To Emel Karakaya for her contributions to the figures and photo editing. To my students for inspiring me to write this book. To my family for always supporting me. Finally, and most importantly, to Burak Arda Özilgen for his scientific and stylistic suggestions, comments, and contributions.

xv

Author

Sibel Özilgen, Ph.D.,  is the Head of the Gastronomy and Culinary Arts Department at Yeditepe University, where she has been a faculty member since 2005. Dr. Özilgen completed her Ph.D. degree in food engineering at Middle East Technical University in Turkey. She attended the University of California as a concurrent student during her Ph.D. study. Dr. Özilgen taught classes and conducted research at Massey University in New Zealand and is the author or coauthor of numerous refereed ­publications, book chapters, and books. Her research and teaching interests lie in the area of food science, food safety, food product development, and the ­eating habits of different ­consumer groups. She holds a patent.

xvii

Chapter 1 Measurements and Units

WHY DO WE NEED MATHEMATICS IN CULINARY PROCESSES? Measuring the ingredients, calculating the proportions of the ingredients, recipe conversion, resizing a recipe for a number of servings, determination of the cost per meal, and portioning all involve basic math operations, such as addition, subtraction, multiplication, and division. An inaccurate measurement of the ingredients is one of the major factors determining the success or failure of a recipe. In addition, imprecise measurements and conversion calculations increase food costs. Inaccurate portioning causes consumer dissatisfaction. Therefore, it is important that all chefs must understand basic math operations.

TYPES OF MEASUREMENTS IN THE KITCHEN There are two types of measurements in culinary operations: qualitative observations and quantitative measurements. Qualitative observations involve using the five senses to observe and describe the properties of food products (Table 1.1). For example, observing the changes in the color of onions during caramelization, comparing the textures of products

1

2  COOKING AS A CHEMICAL REACTION

TABLE 1.1 The Most Common Terms That May Be Used in Qualitative Observations

Attribute of the food product

The common words that may be used to describe the given attribute

Appearance

Appetizing, attractive, bubbly, shiny, cloudy, greasy, clear, opaque, moist, dry, thick, foamy, crumbly, dark, sticky

Taste

Bitter, sweet, sour, salty, bland, spicy, smoky, sharp, tangy, aromatic, creamy, astringent, nutty, plain, pungent

Smell

Floral, fresh, smoky, spicy, burnt, strong, vanilla, fruity, nutty

Texture

Gritty, smooth, hard, slimy, moist, dry, spreadable, chewy, firm, fizzy, foamy, thick, thin, mushy, spongy, elastic, brittle, crumbly, crunchy, crispy, silky, sandy, soft, wilted, fluffy

Sound

Crunchy, fizzy, sizzling, bubbling

prepared with different types of starches, or describing the sweetness of the fruit juices involves qualitative observations. Actual measurements and numbers are not involved in qualitative measurements. Quantitative measurements involve numbers. Instruments, such as kitchen scales, measuring cups, measuring spoons, rulers, and thermometers, are used for quantitative measurements. Liquids are measured by volumes. The most common volume measuring tools in culinary processes are shown in PIC 1.1 and PIC 1.2. Solids are measured by weight. The most common dry weight measuring tools in culinary processes include those shown in PIC 1.3 and PIC 1.4. Temperature is measured by thermometers. The most common thermometers include those shown in PIC 1.5, PIC 1.6, and PIC 1.7.

Measurements and Units  3

PIC 1.1 GRADUATED CUPS.

PIC 1.2 MEASURING CUPS AND SPOONS.

PIC 1.3 SCALES.

4  COOKING AS A CHEMICAL REACTION

PIC 1.4 MEASURING CUPS AND SPOONS.

PIC 1.5 THERMOMETER.

PIC 1.6 THERMOMETER.

Measurements and Units  5

PIC 1.7 THERMOMETER.

UNITS OF MEASUREMENTS IN CULINARY CALCULATIONS In every quantitative measurement, there is a number followed by a unit. Example: A bag of rice is 5 kg (11 lb).

Number from the measurement

Unit

The metric system is the most widely used units of measurements in the world. International System of Units (SI) is based on the metric system. The U.S. Customary System is widely used as the units of measurements in the United States (Table 1.2). Unit uniformity is important in culinary calculations. Unit uniformity is important to understand and compare quantities. Accurate measurement of quantities and their units provide: • standards and stability to the recipes; • effective inventory and cost control; and • standard portion size. Conversion between units of measures is possible. It is important to know how to convert mass, volume, temperature, and length measured in one unit to the corresponding measure in a different unit. The conversion is an easy mathematical operation because the relationship between different units is constant. Those constants are known as conversion factors, and they are tabulated in Tables 1.3 and 1.4.

6  COOKING AS A CHEMICAL REACTION

TABLE 1.2 The Most Common Units of Measurements Used in Culinary Processes

U.S. Customary System of Units

Measurement

International System of Units, SI

Length

inch (in)

meter (m)

Volume

fluid ounce (fl oz); gallon (gal); quart (qt)

liter (L); meter cube (m3)

Weight

pound (lb), ounce (oz)

gram (g)

Temperature

Fahrenheit, °F

Celsius, °C

TABLE 1.3 The Widely Used Conversion Factors for the Measurements of Length, Volume, and Weight

Measurement

Conversion Factor

Length

1 in. = 2.54 cm 1 m = 100 cm

Volume

1 L = 1,000 mL 1 L = 33.8 fl. oz (U.S.) 1 L = 0.264 gal (U.S.)

Weight

1 kg = 1,000 g 1 lb = 454 g 1 lb = 16 oz (U.S.)

TABLE 1.4 The Widely Used Conversion Factor for the Measurements of Temperature

Measurement °C

Conversion Equation T (°F) = 1.8 × T (°C) + 32

Measurements and Units  7

Rule To convert one unit to the corresponding measure in a different unit, the number in original units is multiplied by a conversion factor to produce a result in the desired units.



  New unit   Number in original units × Conversion factor    Original unit    = New number in new unit

Example 1.1 Convert the length of 5.6 in. to its equivalent in units of meters.

Solution From the equation:

  cm  1  m  5.6 in. × 2.54  ×  = 0.142 m   in.  100  cm   

Example 1.2 Convert the mass of 12 pounds to its equivalent in units of grams.

Solution From the equation:

  g  12 lb ×  454    = 5,448 g  lb   

Example 1.3 Convert the oven temperature 303°F into units of °C (Figure 1.1).

Solution From the equation given in Table 1.4:

T ( °C) =

(303°F − 32) = 150.6°C 1.8

8  COOKING AS A CHEMICAL REACTION

Fahrenheit (°F) Boiling point of water

212°F

Celsius (°C)

100°C

180°F

Freezing point of water

32°F

100°C

0°C

Figure 1.1  Temperature scales.

In addition to unit conversion, sometimes it is necessary to convert from volume measures to weight in order to standardize the recipes, especially when adapting a recipe from another source. For example, although solids are generally measured in weights, they are given in volumes in some recipes, or some measurements are given in weights and some are given in volumes in the same recipe. The most practical volume-to-weight conversion method in the kitchen can be explained as: 1. Weigh the measuring tool (e.g., cup) and record its weight. 2. Fill the measuring tool with the ingredient as directed by the recipe. 3. Weigh the measuring tool filled with the ingredient. 4. Subtract the first weight from the last measurement.

Example 1.4 Calculate the weight of 300 mL ketchup.

Solution 1. Weigh the liquid measuring cup and record it. 2. Fill the measuring cup with 300 mL ketchup. 3. Return the filled measuring cup to the scale and record the weight. 4. Subtract the weight of the liquid measuring cup (from Step 1) from the weight of filled measuring cup (from Step 3) to find the weight of 300 mL ketchup. Note: It is possible to convert the measurements given as spoonful, glass, cup, etc., into weight measurements using the same method.

Measurements and Units  9

EVERYDAY CALCULATIONS IN THE KITCHEN Fractions and Percentages Fraction can be defined as the number of parts taken out of a whole quantity divided into equal parts. It can be calculated as:

Fraction =

Number of identical parts being taken from the whole quantity Total number of equally divided parts in the whole quantity

Percentages (%) can be defined as a fraction multiplied by 100.

Percentages (%) = Fraction × 100

Fractions are widely used with measurements of ingredients or with portioning, such as 2½ cups of flour or in portioning (⅔ of a pizza). Percentages are used most commonly to calculate food costs and kitchen and restaurant space requirements. Fraction and percentage calculation knowledge also are required in converting and yielding the recipes.

Example 1.5 A whole apple is sliced into eight equal pieces. What is the fraction of each slice in the whole apple?

Solution From the equation:

1 Fraction of each slice = 8

Example 1.6 A chef prepared an apple pie and divided it into 16 equal slices. He sold ¾ of the pie in the morning. Calculate the number of slices sold and the number of slices left.

Solution Three quarters (¾) means that the whole apple pie is divided into four equal parts and three parts are sold. Because the number of total slices is given, it is possible to calculate the number of slices in each part.

10  COOKING AS A CHEMICAL REACTION

Number of slices each part has =



16 (slices) = 4 slices/part 4 (parts)

 slices  × 3 (parts sold) = 12 slices sold Number of slices sold = 4   part 



Number of slices left = Total number of slices − Number of slices sold Number of slices left = 16 (slices) − 12 (slices) = 4 slices left

Example 1.7 A chef prepared ice creams in two different flavors. Calculate the number of ­consumers who preferred to have strawberry-flavored ice cream, if 20% of the 500 consumers had the other flavor.

Solution From the equation: Number of consumers preferred the other flavor = 500 (consumers) ×



20  consumers preferred the other flavor    100  consumers

= 100 consumers. Number of consumers preferred the strawberry flavor

= 500 (consumers) − 100 ( consumers )



= 400 consumers.

Example 1.8 Calculate the percentage of the consumers who preferred to have strawberry-­flavored ice cream in the previous example, if 87 of the consumers had the other flavor.

Solution Number of consumers preferred

strawberry flavor = 500 (consumers) − 87 (consumers) = 413 consumers

Measurements and Units  11

From the equation:

% of consumers preferred the strawberry flavor =

413  consumers  × 100 500  consumers 

= 82.6%

Yield Percent Yield percent can be defined as the percentage of a whole food item that is usable in production of the meal after any required operations (i.e., cutting, peeling) have been completed. Yield percent can be calculated as: Yield percent =



Edible portion quantity (EPQ) × 100 As-purchased quantity (APQ)

Method to calculate the yield percent can be explained as: 1. Weigh the ingredients to find the as-purchased quantity (APQ). 2. Weigh the ingredients after cleaning, peeling, etc., to find the edible portion quantity (EPQ). 3. Use the equation to calculate the yield percent.

Example 1.9 A recipe requires 5 kg of cleaned, peeled, and diced cucumbers. What is the ­minimum amount of cucumbers to purchase if the yield percent is 82?

Solution From the equation: As-purchased quantity (APQ) =

Edible portion quantity (EPQ) × 100 Yield percent

5 kg × 100 82 = 6.098 kg cucumber is required

APQ =



Recipe Yield Conversion Recipe yield conversion can be defined as adjusting the recipe to increase or decrease the amount that the recipe actually yields.

12  COOKING AS A CHEMICAL REACTION

The amount of each ingredient in the recipe is multiplied by the recipe conversion factor (RCF) to convert the recipe. The RCF can be calculated as: Recipe Conversion Factor (RCF) =



New yield Original yield

Example 1.10 The following ingredients list is given for a recipe that yields 50 apple rolls. Convert the recipe to yield 160 rolls.

Ingredients • • • • • • •

300 g (10.5 oz) butter 30 g (1.05 oz) egg 300 g (10.5 oz) yogurt 150 g (5.3 oz) corn starch 250 g (8.8 oz) sugar 1.1 kg (2.43 lb) flour 360 g (12.7 oz) apple

Solution From the equation:



160 50 = 3.2

RCF =

Multiply all the ingredients by the RCF to calculate the new amounts (Table 1.5). If the units of the measurements are not uniform, convert all quantities to the same units. Butter: 300 g × 3.2 = 960 g (33.8 oz) Egg: 30 g × 3.2 = 96 g (3.9 oz) Yogurt: 300 g × 3.2 = 960 g (33.8 oz) Corn starch: 150 g × 3.2 = 480 g (16.9 oz) Sugar: 250 g × 3.2 = 800 g (28 oz)    1,000 g  100 g • Flour:  1 kg ×  +  × 3.2 = 3,520 g (7.76 lb) = (18.3 oz)  1 kg   

• • • • •

• Apple: 360 g × 3.2 = 1,152 g (2.5 lb)

Measurements and Units  13

TABLE 1.5 Ingredients List for Recipes That Yield 50 Apple Rolls and 160 Apple Rolls, Respectively

Amounts for 50 Apple Rolls (g/oz)

Amounts for 160 Apple Rolls (g/oz)

Butter

300/10.5

960/33.8

Egg

30/1.05

96/3.4

Yogurt

300/10.5

960/33.8

Corn starch

150/5.3

480/16.9

Sugar

250/8.8

800/28

Flour

1,100/38.8

3,520/124

Apple

360/12.7

1,152/40.6

Ingredients

Note: This method is used to convert the amount of ingredients only. Do not apply the RCF to convert the processing temperature and time.

SIMPLIFIED COST CALCULATIONS FOR CULINARY OPERATIONS Example 1.11 The following ingredients list is given for the Cranberry Puree recipe. The fresh cranberries cost $3 per kg. 3.4 kg (7.5 lb) granulated sugar can be purchased for $9.38. The recipe yields approximately 40 cups. Calculate: a. The total cost b. The cost per cup

Ingredients • 9 kg (19.8 lb) fresh cranberries • 3.4 kg (7.5 lb) granulated sugar • 2.4 L (81.2 fl oz) water

14  COOKING AS A CHEMICAL REACTION

Solution a.



Total cost of each ingredient = Number of units × Cost per unit  $  Total cost of fresh cranberries = 9 kg × 3   = $27  kg 

• The total cost is given as $9.38 for 3.4 kg of granulated sugar in the question. • Cost of water is neglected.

Total cost of the recipe =

∑ Total cost of each ingredient

Total cost of the recipe

= Total cost of the cranberries + Total cost of the granulated sugar = $27 + $9.38



= $36.38 b. Total cost of the recipe for 40 cups is $36.38 Total cost of the recipe = Number of units × Cost per unit $36.38 = 40 cups × Cost per cup

Cost per cup =

$36.38 40 cups



= 0.909$/cup The amount 0.909 $/cup rounds up to 1.0 $/cup Note: ∑: sigma sign is a summation sign. It indicates the sum of all values.

Example 1.12 A whole pizza is sliced into eight equal pieces. Each piece is sold for $4.99. The total production cost is $3.75 per pizza. Calculate the net earnings of the chef if she sells approximately 40 pizzas/day.

Measurements and Units  15

Solution

1 Fraction of each piece = 8



Total price of the pizza = Price per unit × Number of units Total price of the pizza = 4.99



$ × 8 pieces piece

= $39.92

The total price of 40 pizzas =  $39.92 × 40 = $1,596.80



Total cost of 40 pizzas = $3.75 × 40 = $150 Net earnings = Total price – Total cost = $1,596.80 − $150





= $1,446.80

Example 1.13 A chef ordered 3 lb (1 kg 360 g) Granny Smith apples, and 0.63 lb (286 g) of granulated sugar for her apple-filling recipe. Granny Smith apple costs $1.98 per pound (per 0.45 kg) and has a 75% yield. The cost of sugar is $1.25 per pound (per 0.45 kg). Calculate the total cost of the apple-filling recipe.



Edible portion cost (EPC) =

As-purchased cost (APC) Yield percent (YP)

Edible portion cost of Granny Smith apple =  

1.98$/lb 0.75

= 2.64 $/lb Note: $2.64 per lb is greater than $1.98 per lb. The difference between the as-­ purchased cost and the edible portion cost arises from the cost of waste from unused parts of the apples such as seeds and peels.

Total cost of each ingredient = Number of units × Cost per unit

16  COOKING AS A CHEMICAL REACTION

Total cost of Granny Smith apple = 3 lb × 2.64



$ lb

= $7.92 Total cost of granulated sugar = 0.63 lb   ×  1.25



$ lb

= $0.79

Total cost of the recipe =

∑ Total cost of each ingredient 

Total cost of apple filling recipe

= Edible portion cost of the Granny Smith apple + Cost of granulated sugar = $7.92 + $0.79 = $8.71

SIMPLIFIED STATISTICS FOR CULINARY OPERATIONS Is it possible to guess how many of each course will be served during weekdays? What is the average number of consumers that will dine tonight? Does the target consumer group like my new food product? What is the average weight of medium sized eggs? Answers are possible with basic statistics knowledge. Statistics are applied to analyze the quantitative data obtained from the samples of measurements or observations. Culinary professionals use the results of the statistical analysis generally to conclude the observations, to design service operations, to design the menu, to develop new food products, to standardize recipes, and to understand consumer behaviors.

Measurements and Units  17

EXPERIMENT 1.1 OBJECTIVE To explain how to apply basic statistical analysis. Ingredients and Equipment • 5 egg cartons with 30 eggs in each • Kitchen scale Method 1. Open the first egg carton and count the eggs. 2. Weigh 10 eggs separately and write the measurements in Data Table 1.1. 3. Repeat the same procedure for each egg carton. 4. Calculate the statistical parameters given in Data Table 1.1. The MEAN can be defined as the average of all scores obtained from the measurements. It can be calculated as:

Mean =

The sum of all scores Total number of measurements

Example 1.14 Calculate the mean of observations 2, 3, 4, 6, 4, 5, 3.

Solution From the equation:

Mean =

2+3+ 4 +6+ 4 +5+3 = 3.86 7

The MODE can be defined as the most frequently occurring score. It is useful when differences between the scores are insignificant.

Example 1.15 What is the mode of observations 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 6?

Solution The answer is 4, because it is the most repeated number in the observation.

18  COOKING AS A CHEMICAL REACTION

The RANGE can be defined as the difference between the most extreme data values. It is the measure of the spread of the data values. It can be calculated as:

Range = Highest point/value − Lowest point/value

Example 1.16 Calculate the range of observations 2, 10, 4, 6, 3.

Solution From the equation: Range = 10 − 2 = 8

#5

#4

#3

#2

#1

Egg Box

Egg #1

DATA TABLE 1.1

Egg #2

Egg #3

Egg #4

Egg #5

Egg #6

Weight of Eggs (g) Egg #7

Egg #8

Egg #9

Egg # 10

Mean

Mode

Range

Statistical Parameters

Measurements and Units  19

20  COOKING AS A CHEMICAL REACTION

POINTS TO REMEMBER There are two types of measurements in culinary operations: qualitative observations and quantitative measurements. Qualitative observations involve the five senses. Quantitative measurements involve numbers. In every quantitative measurement, there is a number followed by a unit. Conversion between units of measures is possible. Unit uniformity is important in culinary calculations. Statistics are applied to analyze the quantitative data obtained from the ­samples of measurements or observation.

Measurements and Units  21

Study Questions 1. Convert the measuring units as indicated: a. 10 kg = ____ lb b. 160°F = ____ °C c. 8 L = ____ U.S. gal d. 4,360 mL = ____ L e. 678 g = ____ kg 2. You purchased 2,500 g (5.5 lb) of squash from the market. After cleaning and peeling, you are left with 1,870 g (4 lb) of squash. Calculate the yield percent for the squash. 3. The shell loss for walnuts is 35%. What is the yield percent? 4. A recipe for tomato soup makes 90 servings. What is the recipe conversion factor if you will be making 200 servings? 5. Calculate the average (mean) weight of a medium-sized apple from the ­measurements given in Table 1.6. 6. A chef is making meatball sauce and plans on serving two spoonsful of the sauce with freshly fried meatballs. Each spoonful of sauce weighs 32 g (1.1 oz). How many kilograms of meatball sauce are needed if the chef plans to serve 600 meatballs? 7. The bakery sells 175 cupcakes every day. Each cupcake costs ¢0.45. Calculate the selling price of each cupcake for $285 total net earnings. TABLE 1.6 Weight of Apples

Sample Number

Weight of the Apple (g/oz)

1

105/3.7

2

102/3.6

3

104/3.66

4

106/3.73

5

101/3.5

22  COOKING AS A CHEMICAL REACTION

SELECTED REFERENCES Besterfield, D. H. 2009. Quality control, 8th ed. London: Pearson Education, Inc. Blocker, L., and J. Hill. 2007. Culinary math, 3rd revised and expanded ed. Hoboken, NJ: John Wiley & Sons. Boeree, G. 2014. Descriptive statistics. Online at: http://webspace.ship.edu/cgboer/­descstats. html. Child Nutritional Programs (CNP) Manager’s Manual. 2014. Weighing and measuring. Online at: http://cnp.alsde.edu/nslp/manuals/CNPManagersManual/Weighing%20 and%20Measuring.pdf. Geankoplis, C. J. 2008. Transport processes and separation process principles (includes unit operations), 4th ed. London: Pearson Education, Inc. Glendale Union High School District Culinary Math Workbook. 2007. Available as e-book: http://aspdf.com/ebook/culinary-math-workbook-pdf.html. Jones, T. 2008. Culinary calculations: Simplified math for culinary professionals, 2nd ed. Hoboken, NJ: John Wiley & Sons. Office of Mathematics, Science, and Technology Education (MSTE) 2014. Introduction to descriptive statistics. University of Illinois, Urbana-Champaign. Online at: http:// mste.­illinois.edu/hill/dstat/dstat.html.

Chapter 2 Basic Food Chemistry

FOOD PROCESSING IS ALL ABOUT CHEMISTRY Foods are made up of chemical compounds that include water, proteins, lipids, carbohydrates, vitamins, and minerals. The composition of the foods determines both physical and chemical properties of the foods. During preparation and processing, such as cooking and drying, a series of chemical reactions occurs in foods. The structures of the chemical compounds are changed and new food products are formed. These types of irreversible changes are called chemical changes. Basic food chemistry primarily deals with 1. the chemical composition of foods; 2. the chemical structures and the properties of primary food compounds, which are water, proteins, lipids, and carbohydrates; and 3. the chemical changes that the foods undergo during food processing, storage, and transportation. Knowledge of basic chemistry concepts is required to understand basic food chemistry. An atom is the smallest unit of an element. Atoms are made up of three particles: protons, neutrons, and electrons (Figure 2.1). Protons and neutrons are found in the nucleus, while the electrons are arranged in shells around the nucleus. The electrons found in the outer shells are called the valence electrons. The valence

23

24  COOKING AS A CHEMICAL REACTION

Proton

Electron

Neutron

Figure 2.1  Structure of an atom.

electrons can be used to form the chemical bonds. All atoms in a molecule must have eight valence electrons to become stable (the Octet Rule). Atoms whose outer shells are filled with electrons have low energy, and they are said to be stable. For most atoms, the outermost shell is incomplete. The atoms are connected to each other by chemical bonds and form molecules, such as NaCl (sodium chloride) and H2O (water) to achieve lower state of energy and become stable. There are different types of bonds that hold atoms of the molecules together. Ionic bond, covalent bond, and hydrogen bonds are the major bonds that exist in foods. An ionic bond is the complete transfer of electrons from the outer shell of one atom to the outer shell of the other atom. In ionic bonds, one atom is negatively charged due to gain of electrons and the other one is positively charged due to loss of electrons. The difference in electronegativity holds the atoms together (Figure 2.2). NaCl is the most common example of a molecule formed by an ionic bond. A covalent bond is the complete sharing of one or more electrons between two atoms. It does not involve transfer of electrons (Figure 2.3). One pair of electrons is shared to form one covalent bond. The bond between the oxygen atom and hydrogen atoms in a water molecule is the most common example of a covalent bond. A hydrogen bond is a special type of bond in which a hydrogen atom of one molecule is attracted to an electronegative atom. For example, this type of bond exists between water molecules (Figure 2.4). A compound is formed if types of atoms in the molecule are different from each other. For example, NaCl is a compound because it has two different atoms in the structure, Na (sodium) and Cl (chlorine). An element is formed if the types of the atoms are the same, such as oxygen. Thus, all compounds are molecules, and elements are either molecules or atoms. The structure of the molecules, hence the types of the atoms and the chemical bonds that form the molecules, determine how food will behave during preparation and processing.

Basic Food Chemistry  25

Transfer of electron

Na

Cl

Sodium atom

Chlorine atom

Na+

Sodium ion

Cl–

Chloride ion

Sodium chloride (NaCl)

Figure 2.2  Ionic bonding.

Oxygen atom

Oxygen atom

O

O

Sharing of electrons

O

O

Oxygen molecule (O2)

Figure 2.3  Covalent bonding.

O == O

26  COOKING AS A CHEMICAL REACTION

δ+

H

O δ+

H

H

δ–

δ–

δ+

O H

δ+ H

H

δ–

δ+

O H

δ– H

O H

δ+

δ+ δ–

δ+

O H

δ+

Figure 2.4  Hydrogen bonding.

δ+

Basic Food Chemistry  27

EXPERIMENT 2.1 OBJECTIVE To show chemical reactions may occur between ingredients when they are mixed together Ingredients and Equipment • 2 cups flour • 1½ teaspoon baking soda • 1½ teaspoon baking powder • 1 cup egg yolks • 1 tablespoon lemon juice • ½ cup water • Whisk • Large mixing bowl • 2 bowls Method (EXP 2.1) 1. Label two bowls “baking soda” and “baking powder.” 2. In a large bowl, beat the egg yolks until thick. 3. Then add flour into the egg mixture and fold in. 4. Divide the mixture equally into two bowls labeled “baking soda” and “baking powder.” 5. Add baking soda to the mixture in the corresponding bowl.

EXP 2.1

28  COOKING AS A CHEMICAL REACTION

6. Add 4–5 tablespoonsful water on the baking soda. 7. Record your observations in Data Table 2.1. 8. Add baking powder to the mixture in the corresponding bowl. 9. Add 4–5 tablespoonsful water on the baking powder. 10. Record your observations in Data Table 2.1. 11. Add ½ tablespoon lemon juice to the egg mixture with baking soda. 12. Record your observations in Data Table 2.1. 13. Add ½ tablespoon lemon juice to the egg mixture with baking powder. 14. Record your observations in Data Table 2.1. Note: This experiment involves the most common ingredients and the steps from the processes of baked and/or dough-based food products. DATA TABLE 2.1 Observation Baking soda mixture without lemon juice Baking soda mixture with lemon juice Baking powder mixture without lemon juice Baking powder mixture with lemon juice

Basic Food Chemistry  29

EXPERIMENT 2.2 OBJECTIVE To understand the same ingredients may give different chemical reactions depending on processing conditions and the order in which the ingredients are combined. Case 1 Cake 1 Ingredients and Equipment • 226 g (8 oz) sugar • 226 g (8 oz) butter • 226 g (8 oz) flour • 226 g (8 oz) eggs • 1 teaspoon vanilla • Bowl • Mixer • Knife • 23 cm (9 in.) non-stick loaf pan • Conventional oven (no fan) Method 1. Set the oven temperature to 177°C (351°F). 2. Put the butter into the bowl and mix at medium speed for 1–2 min. 3. Add the sugar and continue mixing until the mixture gets fluffy. This step takes 2–3 min. 4. Slowly add the eggs while mixing the batter. After each addition, make sure the egg is incorporated into the batter before adding the next one. 5. Reduce the speed of the mixer and add the flour and vanilla. 6. Mix slowly and gently to incorporate the flour. Do not overmix the mixture. 7. Pour the batter into the pan. 8. Bake for 1 h, until top is golden brown. 9. Let it rest for half an hour. 10. Take the cake out of the pan. 11. Slice and analyze the cake. 12. Record your observations in Data Table 2.2.

30  COOKING AS A CHEMICAL REACTION

Case 2 Cake 2 Ingredients and Equipment • 226 g (8 oz) sugar • 226 g (8 oz) butter • 226 g (8 oz) flour • 226 g (8 oz) eggs • 1 teaspoon vanilla • Bowl • Saucepan • Sifter • Mixer • Knife • 23 cm (9 in.) non-stick loaf pan • Conventional oven (no fan) Method 1. Set the oven temperature to 177°C (350°F). 2. Melt the butter in a saucepan or over the warm water. 3. Combine the eggs and sugar in a bowl mix at medium speed until the mixture has tripled in volume. This step may take a few minutes, be careful not to overmix the mixture. 4. Sift the flour and vanilla into the mixture and fold in gently. 5. Fold in the butter until it is completely incorporated. 6. Pour the batter into the pan. 7. Bake for 40–45 min, until top is golden brown. 8. Let it rest for half an hour. 9. Take the cake out of the pan. 10. Slice and analyze the cake. 11. Record your observations in Data Table 2.2. 12. Compare the results from Cases 1 and 2 (EXP 2.2).

Basic Food Chemistry  31

DATA TABLE 2.2 Texture and Appearancea

Tasteb

Crust colorc

Case 1 Case 2 Use the words soft, hard, spongy, greasy, dry, wet, holes, no holes, more holes, and heavy to ­evaluate the texture and appearance. b Use the words buttery, sweet, tasteless, floury, and rich to evaluate the taste. c Use the words brown, golden, yellowish, and dull to evaluate the crust color. a 

CASE 1

CASE 2 EXP 2.2

32  COOKING AS A CHEMICAL REACTION

Basic Food Chemistry  33

The Science Behind the Results Food processing involves a series of chemical reactions. Each food ingredient in the recipe has its own chemical composition. Therefore, each ingredient serves a purpose in culinary processes. For example, baking soda and baking powder are both leavening agents used in baking. They produce carbon dioxide and cause baked goods to rise. Both require an acid ingredient to give a reaction and to produce carbon dioxide. Baking soda does not contain an acidic ingredient; therefore, it requires an acidic ingredient in the recipe to produce carbon dioxide. When an acidic ingredient, such as lemon juice, yogurt, or vinegar, is added, a chemical reaction occurs between the baking soda and the acidic ingredient, and carbon dioxide gas is produced. Baking powder contains baking soda and an acidic ingredient in the same package. When liquid is added, they come together and give a reaction to produce carbon dioxide. Although they are both leavening agents, the two are certainly not interchangeable and they need certain other ingredients to perform. The processing conditions (such as cooking temperature and pH of the medium), chemical properties of other ingredients, and recipe formulation (such as time for cooking and amounts of ingredients and their order of addition in the recipe) will determine the nature of chemical reactions that will take place during food processing and hence the sensory attributes of the final food products. Food production is applied chemistry, therefore, knowing basic properties of the chemical components of foods will help kitchen professionals to • • • • •

better understand their cooking; create new food products; improve the quality and safety of their dishes; improve culinary presentations of their foods; and understand what goes wrong in the kitchen.

Example 2.1 Following are examples of primary functional properties of food components and additives in baking (PIC 2.1).

Fats 1. Enhance the flavor and mouth feel 2. Develop the texture 3. Shorten the dough

34  COOKING AS A CHEMICAL REACTION

PIC 2.1 PUFF PASTRY WITH CREAM FILLING AND STRAWBERRIES.

4. Form emulsions 5. Transfer heat 6. Develop the appearance

Carbohydrates

1. Give sweet taste 2. Provide structure and texture 3. Lower the freezing point 4. Lower the water activity 5. May fat substitutes 6. Undergo reactions that improve the flavors and colors

Water

1. Good cooking medium 2. Greatly affects the texture and the appearance of the foods 3. Dissolves some flavor compounds 4. Medium for chemical reactions 5. Affects the shelf life

Protein

1. Foam formation 2. Gelation 3. Dough formation 4. Flavor development

Basic Food Chemistry  35

5. Viscosity control 6. Water binding 7. Color formation

Food Additives

1. Maintain and improve the nutritional quality 2. Preserve and improve quality and freshness 3. Help in processing or preparation 4. Make food more appealing

36  COOKING AS A CHEMICAL REACTION

POINTS TO REMEMBER Foods are made up of chemical compounds that include water, proteins, lipids, carbohydrates, vitamins, and minerals. Food production is applied chemistry. Food production involves a series of chemical reactions. During preparation and processing, the structures of the chemical compounds in foods change and new food products are formed. Due to their chemical structures, each food ingredient in the recipe serves a purpose in culinary processes. The processing conditions, chemical properties of the ingredients in the recipe, and recipe formulation affect the sensory attributes of the end products. Basic knowledge of the chemical components of foods will help kitchen professionals to: • better understand their cooking; • create new food products; • improve the quality and safety of their dishes; • improve culinary presentations of their foods; and • understand what goes wrong in the kitchen.

Basic Food Chemistry  37

SELECTED REFERENCES Elmhurst College, N. Y. 2003. Virtual chembook. What are compound and molecules? Online at: http://www.elmhurst.edu/~chm/vchembook/103Acompounds.html. Fennema, O. R. 1996. Food chemistry. Boca Raton, FL: CRC Press. Gaman, P. M., and K. B. Sherrington. 1996. The science of food. 4th ed. London, UK: Elsevier Ltd. Lasztity, R. 2009. Food quality and standards–Chemistry. In Encyclopedia of life support systems. Oxford, UK: NESCO Publishing–Eolss Publishers. Online at: www.eolss. net/sample-chapters/c10/e5-08-07-00.pdf. McGee, H. 2007. On food and cooking: The science and lore of the kitchen. New York: Scribner.

Chapter 3 Water in Culinary Transformations

FUNCTIONAL PROPERTIES OF WATER IN CULINARY PROCESSES Water is the major constituent of foods. Most natural foods contain 60%–90% water (Table 3.1). Food processing, such as drying, cutting, and pressing, decreases the amount of water in foods (Table 3.2). The primary functions of water in foods and food processing can be listed as follows: 1. Water is a good cooking medium: It conducts heat to the food during cooking. Foods are usually boiled, steamed, braised, or simmered in water. 2. Water greatly affects the texture and the appearance of foods: It gives the crisp and moist texture of foods. The appearance of foods can tell consumers what the texture will be when the food is eaten. For example, if a vegetable or fruit, such as lettuce or apple, looks wrinkly, then most probably it has lost its crispy bite. 3. Water is a good solvent: It dissolves hydrophilic substances. This property of water primarily enhances the taste of foods because the hydrophilic fl ­ avors, such as salt, sugar, and alcohols, dissolve in water. This property of water

39

40  COOKING AS A CHEMICAL REACTION

TABLE 3.1 Approximate Water Content of Some Natural Foods

Food

Amount of Water in Foods (%)

Tomato

94

Spinach

92

Zucchini

95

Potato

79

Cucumber

96

Broccoli

91

Apple

84

Grape

81

Orange

87

Whole milk

88

Beef

50–70

also enhances the appearance of foods because some color pigments in foods, such as anthocyanins, are also soluble in water. 4. Water is a medium for chemical reactions: It hydrates the medium and facilitates the movement of the molecules to give reactions. It also can participate in chemical reactions, such as hydrolysis. 5. Water content affects the shelf life of foods: Water is essential for the growth of microorganisms. Most foods contain enough water for the microorganisms to grow. That is primarily why foods with low moisture contents, such as dry beans and some biscuits (cookies), are less perishable than those that have high moisture content. Understanding the chemical structure of water is crucial for chefs because the functional properties of water are primarily related to its structure.

Water in Culinary Transformations  41

TABLE 3.2 Approximate Water Content of Some Processed Foods

Food

Amount of Water in Foods (%)

Tomato paste

74

Cereals (rice, wheat, oats)

10–11

Soft cheese, feta cheese

55

Hard cheese, parmesan cheese

29

Bread

34–37

Butter

15

Oil

0

STRUCTURE OF WATER The water molecule has two hydrogen atoms bonded to one oxygen atom by covalent bonds. The molecular formula is H2O (Figure 3.1). It is a polar molecule where the oxygen atom has a partial negative charge and the hydrogen atoms have a partial positive charge. Each water molecule can bond with as many as four other water molecules by weak attractions called hydrogen bonds (Figure 3.2).

δ+ H O The hydrogen side of the water molecule has a partial positive charge

δ+ H

δ–

The oxygen side of the water molecule has a partial negative charge

Figure 3.1  Water molecule.

42  COOKING AS A CHEMICAL REACTION

δ+ H O Covalent bond

δ–

δ+ H

H δ–

δ+

Hydrogen bond δ+

O H

H

H δ–

δ+

O

δ–

δ+

O H

H

H

δ+

δ+

δ+ δ–

O

H

δ+

Figure 3.2  Structure of water.

Water is an excellent solvent because it dissolves a variety of different substances. Its polar nature allows water to dissociate ionic compounds into their constituent positive and negative ions. For example, when table salt (NaCl) is mixed with water, the positive part of salt (Na+) is attracted to the oxygen side of water, while the negative part (Cl−) is attracted to the hydrogen side; thus, salt dissolves in water. Water is known to exist in three different phases: gas, liquid, or solid (Figure 3.3). In the gas phase (vapor), water molecules are well separated and they do not have a regular arrangement. They move freely at high speeds in the gas phase. Water vapor condenses to liquid water as it cools.

Water in Culinary Transformations  43

H H

H H

H

H

H

H

H H

O H

O H

H

O

O

H

H

H

H

H

O

O

O

H

H

H

O

H

H

O

H

H

O

O

H

H

H

O

H H

O

O

H

H

H

O

H

O

O

O

H

O

O

O

H

H H

H H

O

H

H

O

O

H H

H

H

O

H

O

O

H

H

H

H

H

H

H

O

H

H

O

H

H

O

O

H

H

O

H

H

H

O

H

H

H

O

O

H

O

O

H

H

H

H

H

H

O

H

H

H

Figure 3.3  Arrangement of water molecules in the gas phase, liquid phase, and solid phase, respectively.

In the liquid phase, water molecules are closer to each other compared to the gas phase. Once again they do not have a regular arrangement. They move about at lower speeds and slide over each other. In the solid phase, water molecules are tightly packed in a regular shape. They cannot move from one place to the other.

44  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  45

EXPERIMENT 3.1 OBJECTIVE To explain the types of water in foods. Ingredients and Equipment • 2 apples • 2 carrots • 2 trays • Knife or a slicer • Cutting board • Kitchen scale • Conventional oven • Baking paper Method 1. Label two trays “apple tray” and “carrot tray”. 2. Line the trays with baking/parchment paper. 3. Measure the weight of each tray and record your results in Data Table 3.1. 4. Peel the apples. 5. Cut each apple into 1 cm × 1 cm × 1 cm (0.39 in. × 0.39 in. × 0.39 in.) cubes. 6. Arrange the apple cubes on the tray labeled “apple tray” in a single layer. 7. Measure the weight of the apple tray and record your results in Data Table 3.2. 8. Calculate the initial weight of fruits and record your results in Data Table 3.3. 9. Set the oven temperature to 40°C (104°F). 10. Place the tray in an oven. 11. Weigh the tray every 60 min and record your results in Data Table 3.2. 12. R  epeat Steps 10 and 11 until the difference between three consecutive ­readings is less than 1%. 13. Record the time required to dry the apple cubes in Data Table 3.3. 14. Calculate the final weight of apple cubes and record your results in Data Table 3.3. 15. Calculate the total amount of water removed from the apple cubes and record your findings in Data Table 3.3

46  COOKING AS A CHEMICAL REACTION

DATA TABLE 3.1 Initial Weight of the Apple Tray (g/oz)

Initial Weight of the Carrot Tray (g/oz)

DATA TABLE 3.2 Time (min)

Weight of the Tray with Apples (g/oz)

Weight of the Tray with Carrots (g/oz)

0 60 120 180 240 300 360 420 480 DATA TABLE 3.3

Initial Weight of the Apples (g/oz)

Total Time Final Initial Final Water Water for Weight Weight Weight Removed Removed Drying of the of the of the from the from the the Apples Carrots Carrots Apples Carrots Apples (g/oz) (g/oz) (g/oz) (%) (%) (min)

Total Time for Drying the Carrots (min)

Water in Culinary Transformations  47

Repeat the same procedure with the carrots. Calculations 1. Initial weight of the foods (g/oz) = Initial weight of tray with the foods (g/oz) − Weight of the empty tray (g/oz) 2. Final weight of the foods (g/oz) = Final weight of tray with the foods (g/oz) − Weight of the empty tray (g/oz) 3. Total amount of water removed (g/oz) = Initial weight of the foods (g/oz) − Final weight of the foods (g/oz)



Water removed (%) =

Total amount of water removed (g) × 100 Initial weight of the fruit (g)

48  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  49

The Science Behind the Results Water in foods primarily exists in two forms: 1. Free water 2. Bound water Free water is the form of water that is available to support biological and chemical reactions in foods. Free water can act as a solvent. It is freezable water. Free water can be easily removed from the food by cutting, pressing, and drying. Bound water is not available for chemical and biological reactions. Bound water is not a good solvent. It is difficult to freeze bound water under normal food processing conditions. Bound water cannot be easily removed from the food by cutting, squeezing, pressing, and drying because the other food constituents, such as proteins, polysaccharides, and fats, hold it. Although different foods may have the same initial water content, they may have different final water contents after the application of the same drying treatment (as observed in the experiment). This difference primarily arises from the differences in their free water and bound water contents, hence, from the structural differences between the foods (Table 3.3). Lowering the amount of free water content of foods is one of the basic methods for controlling the spoilage of foods. Drying, freezing, concentration, and addition of hydrophilic substances, such as salts, to bind free water in the food are the most common processes to reduce the amount of free water in food products. Free water content of foods also is very important to decide the proper cooking technique (wet cooking or dry cooking) and to predict the amount of water to be released from foods during cooking.

50  COOKING AS A CHEMICAL REACTION

TABLE 3.3 Approximate Chemical Compositions of Apple without Skin and Carrot

Raw Apple, Measure: 100 g Nutrient

Raw Carrot, Measure: 100 g

Units

Value

Nutrient

Units

Value

Water

g

86.67

Water

g

88.29

Protein

g

0.27

Protein

g

0.93

Total lipid

g

0.13

Total lipid

g

0.24

Ash

g

0.17

Ash

g

0.97

Carbohydrate, by difference

g

12.76

Carbohydrate, by difference

g

9.58

Fiber

g

1.3

Fiber

g

2.8

Sugar, total

g

10.10

Sugar, total

g

4.74

Starch

g

0

Starch

g

1.43

Water in Culinary Transformations  51

EXPERIMENT 3.2 OBJECTIVE To understand the difference between water content and water activity. Case 1 Ingredients and Equipment • 4 equal-sized (approximately 10 cm × 10 cm × 2 cm/3.9 in. × 3.9 in. × 0.79 in.) feta cheese slices. They must come from the same natural (with no added preservatives), unsalted feta cheese. • Salt • Water • 4 glass storage containers with lids • 3 bowls • Spoon • Food brush • Kitchen scale Method 1. Label the containers “control”, “5% salt solution”, “10% salt solution”, and “20% salt solution”. 2. Prepare 5%, 10%, and 20% salt solutions in separate bowls. 3. Place a slice of cheese into each glass storage container. 4. Brush the surface of each slice thoroughly with the salt solution according to the label on the respective glasses; start from low concentration if you are using the same brush for all slices. 5. Cover with lids. 6. Store them at room temperature in a safe place. 7. Check the surface of the samples every day at the same time for signs of mold growth. 8. Record your observations in Data Table 3.4. Note: To prepare the x% of salt solution, add x g of salt in (100 − x) g of water. For example, add 10 g (0.35 oz) of salt in 90 g (3.1 oz) of water to make a 10% solution of salt in water.

52  COOKING AS A CHEMICAL REACTION

DATA TABLE 3.4

Time (day)

Mold Growth on the Surface of the Control Sample (without Salt)a

Mold Growth on the Surface of the Sample Brushed with 5% Salt Solution

Mold Growth on the Surface of the Sample Brushed with 10% Salt Solution

Mold Growth on the Surface of the Sample Brushed with 20% Salt Solution

1 2 3 4 … a

 se the terms no growth, not significant sign of a mold growth, and significant mold growth to evaluate U the samples.

Case 2 Turkish Apricot Dessert Recipe Ingredients and Equipment • 500 g (17.6 oz) apricot • 60 g (2.11 oz) sugar • Water • 1 cup halved walnuts • 8 tablespoons lemon juice • 4 pots • 4 plates • Measuring cup • Stove Method 1. Label the plates “control”, “10 g (0.35 oz) sugar”, “20 g (0.7 oz) sugar”, and “30 g (1.05 oz) sugar”, respectively. 2. Clean the apricots. 3. Put 200 mL (6.7 fl oz) water, X g sugar, 2 tablespoons lemon juice, and 125 g (4.4 oz) apricots into a pot. 4. Place the pot on the stove and bring to a boil over medium heat.

Water in Culinary Transformations  53

5. Put the apricots onto a plate when they get soft. 6. Bring them to room temperature. 7. Place a walnut half in each apricot. 8. Store the plates at room temperature in a safe place. 9. Check the surface of the samples every day at the same time for the signs of mold growth. 10. Record your observations in Data Table 3.5. 11. Continue observation until observing a mold growth on the surfaces of all samples. For the purpose of the experiment: Carry out stages 2–7 with: 1. No sugar (0 g): At stage 5, put the apricots on the plate marked “control”. 2. 10 g (0.35 oz) sugar: At stage 5, put the apricots on the plate marked “10 g (0.35 oz) sugar”. 3. 20 g (0.70 oz) sugar: At stage 5, put the apricots on the plate marked “20 g (0.70 oz) sugar”. 4. 30 g (1.05 oz) sugar: At stage 5, put the apricots on the plate marked “30 g (1.05 oz) sugar”.

DATA TABLE 3.5

Time (day)

Mold Growth on the Surface of the Control Sample (without Sugar)a

Mold Growth on the Surface of the Sample with 10 g (0.35 oz) Sugar

Mold Growth on the Surface of the Sample with 20 g (0.70 oz) Sugar

Mold Growth on the Surface of the Sample with 30 g (1.05 oz) Sugar

1 2 3 4 … a

 se the terms no growth, not significant sign of a mold growth, and significant mold growth to evaluate U the samples.

54  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  55

The Science Behind the Results The total water content of foods does not provide information on the state of water—if it is “free water” or “bound water.” Water activity (aw) expresses the availability of water to support chemical and biological reactions in foods. For easier conceptualization, it is possible to say that water activity is the indication of the free water in a food. Water activity ranges from 0 (no free water) to 1.0 (pure water). Usually the products that contain a lower percentage of moisture, such as crackers and dried fruits, have lower water activities. On the other hand, some food products with high moisture content can have very little water activity because of their chemical compositions. The water activities of some foods are given in Table 3.4. Most bacteria require a minimum water activity of 0.90 to grow. At water activity below 0.80, most molds cannot grow. That means foods with lower water activities are more shelf stable.

TABLE 3.4 Approximate Water Content and Water Activity of Some Foods

Food

Water Content (%)

Water Activity

Pure water

100

1.00

Fresh meat

65

0.97

Eggs

75

0.97

Bread

35

0.96

Aged cheddar

36

0.85

Salami

30

0.83

Honey

18

0.75

Dried fruit

20

0.60–0.70

Wheat flour

12

0.70

56  COOKING AS A CHEMICAL REACTION

The rate of most chemical reactions such as enzymatic reactions decreases rapidly with water activity because there is less free water able to act as a solute for the reaction. Changes in physical properties can also be linked to changes in water activity. For example, baked foods may soften as water activity increases or crispy cereals may soften if packed together with dried fruits. Water activity of foods can be adjusted by 1. Physically removing water from foods, such as squeezing, pressing, and drying; 2. Freezing; or 3. Adding substances that bind water, such as sugar and salt.

Water in Culinary Transformations  57

EXPERIMENT 3.3 OBJECTIVE To understand the boiling point temperature and boiling point elevation. Case 1 Determination of boiling point of pure water. Ingredients and Equipment • 60 g (2.11 oz) sugar • Water • Kitchen scale • Measuring cup • 4 saucepans • Thermometer • Stove Methods 1. Measure 500 mL (1 pt) water and place in a saucepan. 2. Place the thermometer in the saucepan. Thermometer should not touch the sides or the bottom of the saucepan. 3. Heat the water. 4. Record your observations in Data Table 3.6 for every 10°C (50°F) increase in temperature. 5. Record the point that temperature stays constant (does not increase) in Data Table 3.7. Case 2 Determination of boiling point of sugar solutions with different sugar concentrations. Methods 1. Label 3 saucepans: “saucepan 1”, “saucepan 2”, and “saucepan 3”. 2. Place 200 mL (6.7 fl oz) of water in “saucepan 1”. 3. Add 10 g (0.35 oz) of sucrose in water. Mix until all crystals are dissolved. 4. Place 200 mL (6.7 fl oz) of water in “saucepan 2”. 5. Add 20 g (0.70 oz) of sucrose to the water. Mix until all crystals dissolve. 6. Place 200 mL (6.7 fl oz) of water in “saucepan 3”. 7. Add 30 g (1.05 oz) of sucrose to the water. Mix until all crystals dissolve.

58  COOKING AS A CHEMICAL REACTION

8. Place the thermometer in the “saucepan 1”. The thermometer should not touch the sides or the bottom of the saucepan. 9. Heat the solution in “saucepan 1”. 10. Record your observations in Data Table 3.6. 11. Record the point that the temperature stays constant (does not increase ­anymore) in Data Table 3.7. 12. Repeat Steps 8–11 with other sugar solutions in saucepans 2 and 3. 13. Compare the results.

DATA TABLE 3.6 Appearancea

Temperature (˚C)

Pure Water

Saucepan 1 (Solution with 10 g/0.35 oz of Sucrose)

Saucepan 2 (Solution with 20 g/0.70 oz of Sucrose)

Saucepan 3 (Solution with 30 g/1.05 oz of Sucrose)

40 50 60 70 80 90 100 a

 se the terms stagnant, small bubbles, bubbly, and boiling to describe the appearance of the U syrups.

Water in Culinary Transformations  59

DATA TABLE 3.7

Sample Pure water Solution prepared with 10 g (0.35 oz) of sucrose Solution prepared with 20 g (0.70 oz) of sucrose Solution prepared with 30 g (1.05 oz) of sucrose

Boiling Point: Point that Temperature Does Not Increase (˚C/˚F)

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Water in Culinary Transformations  61

The Science Behind the Results During cooking, as the temperature increases, molecules gain more kinetic energy and they start to move faster. The intermolecular attractions that hold the molecules together break due to increased molecular motion. As a result, molecules escape from the surface of the liquid, and the liquid becomes vapor. The vapor molecules that have escaped from the liquid phase to the surface apply a pressure on the surface of the liquid. That pressure is known as the vapor pressure. As the temperature of the liquid increases, its vapor pressure also increases because molecules gain more kinetic energy. Boiling occurs when the vapor pressure of the liquid becomes equal to the environmental pressure surrounding the liquid. The temperature at which a compound changes from liquid to vapor (gas) is called the boiling point temperature. Once the liquid starts to boil, the temperature remains constant. Pure water boils at 100°C (212°F) at standard pressure (1 atmosphere). At lower environmental pressures, such as on top of a mountain, water boils at temperatures below 100°C (212°F). In contrast, at higher environmental pressures, such as the pressure generated in a pressure cooker, water boils at temperatures above 100°C (212°F). When a nonvolatile solute, such as sugar or salt, is added to the liquid, the vapor pressure is reduced. This means that water molecules will need more kinetic energy to escape from the surface of the liquid. Therefore, the solution needs to be heated to a higher temperature to boil. This phenomenon is called boiling point elevation. Degree of elevation depends on the concentration of the solutes in the solution. The higher the concentration of the solutes, the higher the boiling point of the solution.

62  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  63

EXPERIMENT 3.4 OBJECTIVE To understand the freezing point temperature and freezing point depression. Ingredients and Equipment • 3 oranges • Water • Ice • Measuring cup • 3 test tubes or glass containers; each is approximately 2 cm (0.79 in.) in diameter • Thermometer • Big bowl • Knife • Juice squeezer • Strainer Case 1 Determination of the freezing point temperature of pure water. Method 1. Place 50 mL (1.69 fl oz) water into a glass container. 2. Place the glass container in a bowl. 3. Fill the bowl with ice. The level of the ice should be higher than the level of water in the glass. 4. Place the thermometer in the water. Thermometer should not touch the sides or the bottom of the glass container. 5. During the experiment, frequently stir the water gently with the thermometer to avoid surface ice crystallization. 6. Measure the temperature every 2 min. 7. Record your readings in Data Table 3.8. 8. Record the measurements at the point that the temperature stays constant (does not decrease) in Data Table 3.9. Case 2 Determination of the freezing point of orange juice with pulp. Method 1. Squeeze 50 mL (1.69 fl oz) of orange juice and place in a clean glass. 2. Place the glass in a bowl.

64  COOKING AS A CHEMICAL REACTION

3. Fill the bowl with ice. The level of the ice should be higher than the level of juice in the glass. 4. Place the thermometer in the juice. The thermometer should not touch the sides or the bottom of the glass. 5. During the experiment, frequently stir the juice gently with the thermometer to avoid surface ice crystallization. 6. Measure the temperature every 2 min. 7. Record your readings in Data Table 3.8. 8. Record the measurements at the point that the temperature stays constant (does not decrease) in Data Table 3.9. Case 3 Determination of the freezing point of orange juice with no pulp. Method 1. Squeeze the oranges. 2. Drain and separate the pulp from the juice. 3. Place 50 mL (1.69 fl oz) of orange juice in a clean glass. 4. Place the glass in a bowl. 5. Fill the bowl with ice. The level of the ice should be higher than the level of juice in the glass. 6. Place the thermometer in the juice. The thermometer should not touch the sides or the bottom of the glass. 7. During the experiment, frequently stir the juice gently with the thermometer to avoid surface ice crystallization. 8. Measure the temperature every 2 min. 9. Record your readings in Data Table 3.8. 10. Record the measurements at the point that the temperature stays constant (does not decrease) in Data Table 3.9. 11. Compare your results. Note: Add more ice if it melts during the experiment.

Water in Culinary Transformations  65

DATA TABLE 3.8 Temperature (˚C/˚F) Time (min)

Water

Orange Juice with Pulp

Orange Juice with no Pulp

0 2 4 6 8 10 12 14 16 18 … DATA TABLE 3.9

Samples Water Orange juice with pulp Orange juice with no pulp

Time for Freezing (min)

Freezing Temperature: Point that Temperature Does Not Decrease (°C/°F)

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Water in Culinary Transformations  67

The Science Behind the Results As temperature decreases, kinetic energy of the molecules decreases. Therefore, the motion of the molecules slows down and they can get closer. Intermolecular forces let the molecules get into a more ordered structure. At a specific temperature, liquids turn into solid. Once the liquid starts to solidify, the temperature remains constant. Freezing point is the temperature at which a liquid changes to a solid. Pure water freezes at 0°C (32°F). When a solute, such as salt or sugar, is added to water, the solute physically interrupts the intermolecular forces in the solvent. A solution freezes at a lower temperature than the pure solvent. This phenomenon is called freezing point depression. Foods freeze below 0°C (32°F) because they naturally contain solutes, such as fibers, fats, vitamins, and sugar.

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Water in Culinary Transformations  69

EXPERIMENT 3.5 OBJECTIVE To understand how freezing rate affects the ice crystal size. Case 1 Ingredients and Equipment • 1 kg (2.2 lb) spinach • Water • Ice • Pot • Bowl • Colander • Blast chiller • Freezer • 2 large Ziploc® bags • Stove • Kitchen scale Method 1. Label the Ziploc bags “conventional freezer” and “blast freezer.” 2. Clean and wash spinach. 3. Boil a large pot of water. 4. Place the spinach into boiling water. 5. After 2 min, place the boiled spinach into a bowl of ice water immediately. 6. Drain the blanched spinach in a colander. 7. Analyze the texture and record in Data Table 3.10. 8. Weigh and equally divide the blanched vegetables into two bags, and close them. 9. Place the first bag in a conventional freezer and leave for freezing. (This stage may take a few hours.) 10. Place the second bag in a blast freezer, sharp freeze, and move to the conventional freezer after full freezing for frozen storage. (Freezing stage takes minutes.) 11. Next day, evaluate the samples in their frozen states.

EXP 3.1

EXP 3.2

EXP 3.3

EXP 3.4

70  COOKING AS A CHEMICAL REACTION

2. Record your observations in Data Table 3.10. 1 13. Thaw both samples completely on the counter. 14. Evaluate the samples. 15. Record your observations in Data Table 3.10. 16. Compare the results (EXP 3.1–3.6).

EXP 3.5

EXP 3.6

DATA TABLE 3.10 Spinach Frozen in Conventional Freezer Amount of ice crystals on the surfacea Size of ice crystals on the surfaceb Texture: Before freezingc Texture: Frozen Texture: After thawing Use the terms insignificant and numerous to evaluate the amount of ice crystals. Use the terms tiny, small, and coarse to evaluate the size of the crystals. c Use the terms crispy, soft, and wilted to evaluate the texture of the samples. a

b

Spinach Frozen in Blast Freezer

Water in Culinary Transformations  71

Case 2 Ingredients and Equipment • 1,000 g (2.2 lb) minced meat • 2 bowls • Blast chiller • Freezer • 2 large Ziploc© bags • Kitchen scale • 2 Colanders Method 1. Label Ziploc© bags “conventional freezer” and “blast freezer”. 2. Label two bowls “conventional freezer” and “blast freezer”. 3. Weigh and equally divide the minced meat into two bags, and close them. 4. Record your measurements in Data Table 3.11. 5. Place the first bag in a conventional freezer and leave for freezing. (This stage may take a few hours.)

DATA TABLE 3.11 Minced Meat Frozen in Conventional Freezer

Minced Meat Frozen in Blast Freezer

Amount of ice crystals on the surfacea Size of ice crystals on the surfaceb Texture: Before freezingc Texture: After freezingc Weight of juice collected (g) Weight of thawed meat (g) % mass loss of frozen meat (g) Use the terms insignificant and numerous to evaluate the amount of ice crystals. Use the terms tiny, small, and coarse to evaluate the size of the crystals. c Use the terms soft and firm to evaluate the texture of the samples. a

b

72  COOKING AS A CHEMICAL REACTION

6. Place the second bag in a blast freezer, sharp freeze, and move to the conventional freezer after full freezing for frozen storage. (Full freezing in blast chiller stage may take 10–15 min.) 7. Next day, evaluate the samples in their frozen states. 8. Record your observations in Data Table 3.11. 9. Take both samples out from the Ziploc© bags and thaw both samples completely in corresponding bowls. 10. Evaluate the samples and record your observations in Data Table 3.11. 11. Drain and weigh the amount of juice collected in each bowl during thawing. 12. Express as the % mass loss of frozen meat. 13. Record your results in Data Table 3.11.

Water in Culinary Transformations  73

The Science Behind the Results When water freezes, its volume expands because the water molecules are locked into an ordered crystalline structure. As discussed previously, in the solid phase, water molecules are tightly packed in a regular shape. That structure keeps the molecules rigidly apart from one another by hydrogen bonds, which causes large gaps between the molecules. In other words, volume created by the same number of molecules is larger in the solid phase (Figure 3.3) compared with the liquid phase. It is observed as an expansion in the volume of the foods when they are frozen. This fact is important because: 1. Containers for food freezing must be designed to accommodate the volume increase. 2. The ice crystals may damage the physical structure of foods because of cell rupture, which may occur depending on the ice crystal size and location. The size and amount of the ice crystals in foods can contribute to several sensory attributes, such as texture and mouthfeel. Large ice crystals in foods may rupture the food cells and may cause a loss of natural juice found in cells upon defrosting. The foods get dry and wilted. Large ice crystals also may give a very gritty and lumpy structure to some frozen foods, such as ice creams and frozen desserts. Rate of freezing determines the size and the number of ice crystals in frozen foods. Generally, a slow freezing rate results in a small number of ice crystals. These ice crystals are usually large in size. Sharp (rapid) freezing results in a large number of ice crystals. These ice crystals are usually small in size. Therefore, rapid freezing is usually favored for the food freezing process. Density can be defined as mass per unit volume of water. Therefore, because ice has fewer molecules than liquid water in an equal volume, it is less dense compared to liquid water. This is why ice cubes float in water.

74  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  75

EXPERIMENT 3.6 OBJECTIVE To understand the effects of temperature fluctuation and mixing on the ice ­crystal size. Basic Ice Cream Recipe Ingredients and Equipment • • • • • • • • • • • • • •

1 cup heavy cream 3 cups half-and-half cream 8 egg yolks 1 cup white sugar 1 / teaspoon salt 8 Heavy saucepan 3 shallow containers 2 large bowls Ice cream maker Stretch film/plastic wrap (to cover the containers) Stove Whisk Refrigerator Freezer

Case 1 Effect of Temperature Fluctuation 1. Take three shallow containers and label them “sample 1”, “sample 2”, and “sample 3”. 2. Pour the heavy cream and half-and-half into a heavy saucepan. 3. Simmer over medium-low heat, stirring frequently. 4. Turn the heat down to low. 5. Whisk together the egg yolks, sugar, and salt in a large bowl. 6. Slowly pour two cups of hot cream mixture into the egg yolk mixture while whisking constantly and thoroughly. 7. Pour the egg yolk mixture back into the heavy saucepan with the remaining hot cream. 8. Whisk constantly over medium-low heat until the mixture thickens; this stage takes 5–8 min. Do not boil the mixture. 9. Pour the ice cream mixture into a clean bowl, and allow to cool at room temperature for about 20 min.

76  COOKING AS A CHEMICAL REACTION

DATA TABLE 3.12 Response Number Sample Number

1

2

3

4

5

6

7

8

9

10

Average Score of Responses

1 2 3

10. Chill overnight in the refrigerator. 11. Next day, place one-third of the mixture into the shallow container labeled “sample 3” and cover. Save it for Case 2. 12. Pour the rest of the mixture into an ice-cream maker that agitates the mixture continuously while freezing it. 13. Freeze both samples thoroughly. 14. After freezing, immediately divide and place the ice cream into two different shallow containers labeled “sample 1” and “sample 2” and cover them. 15. Place both containers in the freezer (−18°C/0°F). 16. After an hour, place “sample 1” in the refrigerator at 5°C–8°C (41°F–46°F) (melting). 17. After an hour, place “sample 1” back in the freezer and freeze overnight (refreezing). 18. Next day, carry out a tasting session with “sample 1” and “sample 2”. Tasting technique: Ask 10 people to taste a small amount of each ice cream sample and evaluate the texture in a ranking of 1 to 5, where 1 = coarse/icy, 5 = smooth. Collect their responses and prepare Data Table 3.12. 19. Calculate the average scores of the responses. 20. Record the results in Data Table 3.12. Case 2 Effect of Mixing 1. After stage 11, place the container labeled “sample 3” in the freezer (−18°C/0°F). 2. Freeze overnight without mixing. 3. Next day, carry out a tasting session as explained in Case 1; record the results in Data Table 3.12.

Water in Culinary Transformations  77

4. Calculate the average scores of the responses. 5. Record the results in Data Table 3.12. 6. Compare the results of the samples. Hint:

Average score =

response 1 + response 2 + response 3 +  + response 10 10

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Water in Culinary Transformations  79

The Science Behind the Results The smooth and creamy texture is one of the major quality indicators of frozen desserts, and it is primarily affected by the size distribution of the ice crystals in the product. The texture becomes gritty and rough as the number of larger crystals in the frozen dessert increases. Rapid freezing and proper agitation (mixing) of foods during freezing are the major requirements for formation of small ice crystals in frozen desserts. The effects of temperature and the rate of cooling on the number and size of the ice crystals have been discussed previously. Agitation physically reduces the size of the crystals. It creates more dispersed ice nucleation because agitation keeps the ice crystals and also water molecules apart from each other. Because they cannot come in contact with each other, they cannot form larger ice crystals. Continuous agitation is required during the processing of the frozen foods to require a smooth texture, such as in ice cream. Quality loss may occur if the frozen food is not stored properly. Ice crystals have a tendency to come together to form larger crystals over time during storage. Therefore, extended storage time may decrease the quality of frozen foods. Temperature fluctuations also enhance the formation of larger crystals. Ice crystals easily melt if the temperature fluctuates and goes above the freezing point during storage. When the temperature drops to the freezing temperature again, the larger ice crystals will be formed as a result of uncontrolled freezing.

80  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  81

EXPERIMENT 3.7 OBJECTIVE To explain osmosis. Case 1 Ingredients and Equipment • 1 large head of lettuce • Two bowls • 5 g (0.18 oz) salt • Knife • Cutting board • Kitchen scale • 2 colanders Method 1. Label the bowls “fresh” and “salted”. 2. Wash and clean the lettuce. 3. Finely chop the lettuce. 4. Weigh and equally divide the lettuce into two bowls. 5. Record the initial weight of fresh lettuce in Data Table 3.13. 6. Sprinkle salt over the lettuce in the bowl labeled “salted”. 7. Record the initial weight of salted lettuce (weight of the lettuce plus the weight of the salt) in Data Table 3.13. 8. Place both bowls in the refrigerator. 9. Every 30 min, observe the textural changes in each sample and record your observations in Data Table 3.13. 10. Drain both samples in separate colanders after two hours. 11. Weigh the samples and record the final weights in Data Table 3.13. 12. Calculate the amount of water released from the lettuce samples. Total amount of water released (g) = Initial weight of the lettuce (g)

− Final weight of the lettuce (g)

Note: The amount of salt dissolved in the released water is assumed as negligible.

82  COOKING AS A CHEMICAL REACTION

DATA TABLE 3.13 Texturea Fresh Lettuce Time (min) 30 60 90 120 Initial weight (g) Final weight (g) Total amount of water released (g) a

Use the terms crispy and wilted to define the texture of the samples.

Case 2 Strawberry jam Ingredients and Equipment • 1 kg (2.2 lb) strawberries • 1 kg (2.2 lb) sugar • 2 tablespoons lemon juice • Container • Plate • Pot • Stove • Stretch film/plastic wrap Method 1. Wash the strawberries and remove stems and leaves. 2. Place the strawberries in a container. 3. Sprinkle sugar over the fruit and cover the container. 4. Let it stand for six to eight hours. 5. Every hour, observe the release of juice from the fruits. 6. Record your observations in Data Table 3.14. 7. Place the strawberries and the juice into a pot.

Salted Lettuce

Water in Culinary Transformations  83

8. Place the pot over low heat and cook until you see bubbles. 9. Then turn the heat to medium and simmer for 15–20 min. 10. Drop some jam on a plate. If the drop keeps its shape when it cools, add the lemon juice, and stir. 11. Cool and store (EXP 3.7–3.10). DATA TABLE 3.14 Time (min)

Observationsa

0 60 120 180 240 300 360 a

Use your own terms to describe the amount of juice released from the fruit.

EXP 3.7

EXP 3.9

EXP 3.8

EXP 3.10

84  COOKING AS A CHEMICAL REACTION

Water in Culinary Transformations  85

The Science Behind the Results Water naturally moves from a region of lower solute concentration (higher solvent concentration) to a region of higher solute concentration (lower solvent concentration), through a semipermeable membrane. This phenomenon is called osmosis. Osmosis may decrease the quality of foods. For example, salad greens become wilted after they have been in contact with salt for a while. Similarly, fruits release their juice when they are sprinkled with sugar. The reason is the movement of water in foods through the cell walls to a region of higher solute (salt or sugar) concentration. On the other hand, osmosis is commonly used to preserve foods. For example, salting (curing) draws moisture out of the foods through osmosis, which decreases the water activity of the foods and makes them less susceptible to biological and chemical reactions.

86  COOKING AS A CHEMICAL REACTION

POINTS TO REMEMBER





The primary functions of water in foods and food processing include the following: 1. Water is a good cooking medium. 2. Water greatly affects the texture and the appearance of foods. 3. Water is a good solvent. 4. Water is a medium for chemical reactions. 5. Water content affects the shelf life of foods. Understanding the structure of water is crucial for chefs because the ­functional properties of water are primarily related to its polar structure. Polar nature allows water to dissociate ionic compounds into their positive and negative ions. Water is known to exist in three different phases: solid, liquid, or gas. Water in foods exists in two forms as: 1. Free water 2. Bound water Availability of water in foods to support chemical and biological reactions is expressed by water activity. The temperature at which a compound changes from a liquid to a vapor is called the boiling point. Nonvolatile solutes in the liquid increase the boiling point temperature. Freezing point is the temperature at which a liquid changes to a solid. Solutes in the liquid decrease the freezing point temperature. As water freezes, its volume expands. Rapid freezing and agitation are the major requirements for the formation of small crystals in frozen desserts. Extended storage time and temperature fluctuations during storage decrease the quality of frozen food. Water naturally moves from a region of lower solute concentration to a region of higher solute concentration through cell walls of the foods. This phenomenon is called osmosis.

Water in Culinary Transformations  87

More Ideas to Try Repeat Case 2 of Experiment 3.7. In this case, skip stages 2–6 and add 100 mL (3.2 fl oz) of water and sugar before cooking. Compare your results.

Study Questions 1. What is the importance of water in food preparation? 2. Which one boils first: pure water or salty water? 3. Why do frozen meats, fruits, and vegetables release water upon thawing? How can you minimize it? 4. What makes jam and honey less perishable? Explain. 5. Which one is more perishable: watermelon or beef? Explain. 6. Why are dried food products more stable than fresh ones? 7. What is the reason for the workers at the grocery store spraying water on the vegetables? Explain.

SELECTED REFERENCES Bastin, S. 1997. Water content of fruits and vegetables. Online at: www2.ca.uky.edu/enri/ pubs/enri129.pdf. Coultate, T. P. 1996. Food: The chemistry of its components, 3rd ed. London/Cambridge: Royal Society of Chemistry (RSC). Cybulska, E. B., and P. E., Doe. 2006. Water and food quality. In Chemical and functional properties of food components, 3rd ed., ed. A. E. Sikorski. Boca Raton, FL: CRC Press. Gaman, P. M., and K. B. Sherrington. 1996. The science of food, 4th ed. London: Elsevier. McGee, H. 2004. On food and cooking, 1st rev. ed. New York: Scribner. Mudambi, S. R., S. M. Rao, and M. V. Rajagopal. 2006. Food science, rev. 2nd ed. New Delhi: New Age Int. Ltd, Publishers. Sun, D.-W. 2011. Handbook of frozen food processing and packaging, 2nd ed. Boca Raton: CRC Press. USDA Nutrient Data Laboratory, Beltsville, MD. Online at: http://fnic.nal.usda.gov/nal_­ display/index.php?info_center=4&tax_level=2&tax_subject=279&topic_id=1387. Vaclavik, V. A., and E. W. Christian. 2008. Essentials of food science, 3rd ed. Berlin: Springer. Water. Online at: http://class.fst.ohio-state.edu/fst605/605%20pdf/Water.pdf.

Chapter 4 Carbohydrates in Culinary Transformations

FUNCTIONAL PROPERTIES OF CARBOHYDRATES IN CULINARY PROCESSES Carbohydrates are the organic compounds naturally found in many foods, ­primarily in plants. They have major roles in food preparation. The primary functional properties of carbohydrates in culinary processes are listed below. 1. They are the major source of energy. 2. They give foods their sweet taste. 3. They provide structure and texture in food products. 4. They lower the freezing point of food products. 5. They lower the water activity of food products. 6. They are the foods for some microorganisms in fermentation, such as yogurt production. 7. They are used as fat substitutes. 8. They undergo reactions that improve the flavors and colors of certain food products: a. Maillard browning b. Caramelization

89

90  COOKING AS A CHEMICAL REACTION

The chemical structure of carbohydrates in foods, molecular conformation, and the number of monomer units in the structure will influence the way the food behaves under different production, preparation, processing, and storage conditions. For example, cooking times, cooking methods, and the final textural properties of rice dishes differ depending on the type of rice used because different types of rice have different carbohydrate compositions. Therefore, the appropriate rice type should be chosen depending on the type of dish that is to be prepared. Similarly, the major difference between potato types is the amount and the nature of the starch each contains. Thus, the potato type chosen for baking is different from the one chosen for boiling. Understanding the basic structure of carbohydrates is crucial for chefs because the functional properties of carbohydrates are primarily related to their structures.

CARBOHYDRATE STRUCTURE Carbohydrates are made of carbon (C), hydrogen (H), and oxygen (O).

General Formula for Carbohydrates Carbohydrates are classified into three major groups according to the number of simple sugar (monomer) units that they have in their structures. 1. Monosaccharides (C6H12O6 or C6(H2O)6) C6(H2O)6 Carbo-

Hydrate

Monosaccharides contain a single (mono) carbohydrate molecule (Table 4.1). The most common monosaccharides include: • Glucose (G) • Fructose (F) • Galactose (GA) Glucose (G) is the most common monosaccharide found in foods. It is the primary monomer of polysaccharides. Fructose (F) is the sweetest of all the sugars and is known as “fruit sugar.” Galactose (GA) is one of the simple sugars found in the structure of milk sugar.

Carbohydrates in Culinary Transformations  91

TABLE 4.1 Structures of Monosaccharides and Disaccharides

Monosaccharides

Disaccharides

GLUCOSE

SUCROSE CH2OH

G

H HO

O

OH

H OH

H

H

OH

G

H

H

H

O OH HO

OH

CH2OH

HO

O

H OH

OH

HO

CH2OH

H

H

G

CH2OH

HO

CH OH 2 O

H OH

H

H

OH

H H

O

CH OH 2 O

H OH

H

H

OH

OH H

LACTOSE O

H OH

G

H

H

GALACTOSE GA

OH

H

MALTOSE

CH2OH

H

HO

H

FRUCTOSE F

F

CH OH O 2

CH2OH O H OH H

H

H

H

GA

G

OH

OH

H HO

CH OH 2 O

H OH

H

H

OH

H

H O

CH OH 2 O

H OH

H

H

OH

OH H

2. Disaccharides (C12H22O11 or C12(H2O)11) Disaccharides are formed when two (di) monosaccharide molecules are chemically linked (bound) together upon condensation (Table 4.1). C12 ( H2O)12 − H2O = C12 ( H2O)11



The most common disaccharides include: • Sucrose (G–F) • Lactose (G–GA) • Maltose (G–G) Sucrose is made when one molecule of glucose and one molecule of fructose link together. It is obtained from sugarcane and sugar beet. It is known as “table sugar.” Lactose is composed of one molecule of glucose and one molecule of galactose. It is found in milk and is known as “milk sugar.” Maltose

92  COOKING AS A CHEMICAL REACTION

is a disaccharide composed of two molecules of glucose. It is found in malted cereals and sprouted grains. 3. Polysaccharides Polysaccharides are the polymers of the simple sugars. They have more than 10 monomer (monosaccharide) units in their structures. The most abundant polysaccharides in foods can be listed as: • Starch • Dietary fibers • Glycogen (animals)

Carbohydrates in Culinary Transformations  93

EXPERIMENT 4.1 OBJECTIVES • To explain the properties of saturated and unsaturated solutions. • To determine the saturation points of sugar solutions at different temperatures. Ingredients and Equipment • 2 kg (4.41 lb) table sugar (sucrose) • Water at room temperature (25°C/77°F) • Food coloring (optional) • Aroma (optional) • Pots • 5 small jars • Measuring cup • Sugar thermometer • Spoons • Pipe cleaner • Kitchen scale • Stove Method 1. Label the jars “55°C,” “65°C,” “75°C,” “85°C,” and “boiled.” 2. Pour 100 mL (3.3 fl oz) water into a pot. 3. Measure the temperature of the water (it should be around 25°C/77°F). 4. Gradually add small amount of the sugar to the water and stir with a spoon. Always completely dissolve the sugar before adding more. Continue adding until the solute will no longer dissolve. 5. Observe the undissolved sugar crystals at the bottom of the pot. 6. Weigh the amount of sugar remaining to determine how much sugar was added to the solution. 7. Calculate the weight of sugar added per 100 mL (3.3 fl oz) water and record the result in Data Table 4.1. 8. Move the pan to the stove over low heat. 9. Bring the solution to 55°C (131°F) and hold it at that temperature. Stir until the sugar dissolves. 10. Repeat Steps 4–7, but make sure to hold the solution at 55°C (131°F) throughout the experiment. 11. Pour the solution into a jar labeled “55°C” while it is hot. 12. Add some food colorings and an aroma of your choice. 13. Tie the pipe cleaner to the pencil.

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DATA TABLE 4.1 Temperature of the Solution (°C/°F)

Total Weight of Sugar Added per 100 ml (3.3 fl oz) of Water (g)

25/77 55/131 65/149 75/167 85/185 Boiled

DATA TABLE 4.2 Temperature (ºC/°F) 55/131

65/149

75/167

85/185

Boiled

Initial weight of the jar (g/oz) Final weight of the jar (g/oz) Amount of water evaporated (g/oz)

14. Place the pencil across the neck of the jar, and wrap the pipe cleaner around the pencil until the pipe cleaner is hanging about 2 cm (¾ in.) from the ­bottom of the jar. 15. Weigh the jar and record in Data Table 4.2. 16. Cover the jar with paper towel (do not close the jar). 17. Keep it undisturbed at room temperature for crystals to develop on the pipe cleaner. 18. Record your observations every 24 h in Data Table 4.3. 19. When the sugar crystal is fully formed, weigh the jar. 20. Calculate the amount of water removed and record in Data Table 4.2.

a

55/131

65/149

75/167

85/185

Boiled 55/131

65/149

75/167

85/185

Boiled

Temperature (ºC/°F)

Temperature (ºC/°F)

Observe the amount of crystals forming on the side of the jar. Explain the size, shape, and number of crystals on the string.

…

96

72

48

24

Time (h)

Crystal Development around the Pipe Cleaner

Crystal Development on the Surface and Bottom of Jara

DATA TABLE 4.3

Carbohydrates in Culinary Transformations  95

96  COOKING AS A CHEMICAL REACTION

Complete the same procedure for solutions at 65°C (149°F), 75°C (167°F), 85°C (185°F), and boiled solution (from stage 10).



Total weight of sugar added per 100 mL (3.3 fl oz) water

= Initial weight of sugar (g/oz) – Weight of sugar remaining (g/oz)

Amount of water evaporated (g/oz)

= Initial weight of the jar (g/oz) – Final weight of the jar (g/oz)

Carbohydrates in Culinary Transformations  97

The Science Behind the Results Monosaccharides and disaccharides are soluble in polar solvents, such as water. Monosaccharides contain polar hydroxyl groups (OH), which can easily form hydrogen bonds with water. The maximum amount of soluble solute that can be dissolved in a fixed volume of solvent is definite. Any excess solute added will stay as crystals at the bottom of the container. The concentration of solute where no more solute can be dissolved in the solvent is called the saturation point and the solution is called the saturated solution, because it cannot accommodate more solute. The solubility of solids in liquids usually increases with temperature. As the ­temperature increases, the molecules gain more energy and they move from one position to another more easily. Sugar molecules move apart as a result of increased molecular motion. Increased molecular motion also causes more solvent molecules to come in contact with solute molecules and attract them with more force. Therefore, more solute can be dissolved in the solution as the temperature gets higher due to the elevated saturation point. Such a solution is said to be supersaturated, and is unstable because the solution contains more sugar than the water can dissolve at room temperature. Solute precipitation and crystallization occur in supersaturated solutions when they are cooled down to room temperature. This process is called recrystallization. Crystal formation from the solution has to do with solubility, or the largest amount of solute that can be dissolved in solvent. Rock candy is one of the best examples to explain supersaturation and crystal formation in solutions. As in Experiment 4.1, a sugar solution is supersaturated with sugar (sucrose) to make rock candy. More sugar than that in the saturated solution is dissolved in the water by heating the sugar solution. During boiling, the water evaporates and the sugar becomes even more concentrated. As the solution cools down, the sugar molecules eventually crystallize on a suitable surface, such as a string or pipe cleaner, for crystal formation. Sugar crystallization can be a problem in some food products like jam and honey because they (and some desserts) are rich in sugars. Also during storage, as the moisture evaporates from the surface, the amount of sugar per volume increases and the solution gradually gets supersaturated. Eventually sugar molecules come back together as sugar crystals because supersaturated solutions are unstable in nature. Sugar crystals grow and the textures of the sweets or desserts become grainy. The size and amount of sugar crystals primarily depend on 1. The concentration of sugar in solution; 2. The temperature at which agitation of a cooling solution is initiated;

98  COOKING AS A CHEMICAL REACTION

3. Presence of foreign substances with similar crystalline structures, and interfering substances. Foreign substances with similar crystalline structure in the sugar solution may act as suitable surfaces for crystal formation, which will grow to larger crystals. Interfering substances, such as butter, milk, cream, and protein, on the other hand, physically prevent the growth of large crystals by coating the crystals and preventing one crystal growing onto another.

Carbohydrates in Culinary Transformations  99

EXPERIMENT 4.2 OBJECTIVES • To explain the effect of temperature on the sensory properties of sugar syrup. • To explain sugar caramelization. • To explain the effect of mixing the hot solution on sugar crystal formation. Case 1 Almond Brittle Recipe Ingredients and Equipment • 2 cups sugar • 1/2 cup water • 2 cups roasted whole almonds • 2 glasses of cold water (use a big water glass or a gastronome) • Measuring cup • 2 heavy saucepans • Stove • Baking paper • Sugar thermometer • Spoon • Pastry brush • Spatula Method 1. Place ¼ cup water and 1 cup sugar in a heavy saucepan. 2. Place the thermometer in the pan. Be sure the ­thermometer does not touch the bottom of the pan. Note: Measure the temperature and record the appearance and the color of the syrup in Data Table  4.4 throughout the experiment at temperatures given in the Data Table. Note: Drop a little of the syrup in a glass of cold water and as soon as it forms a ball, take it out of the water and record the structure in Data Table 4.4 throughout the experiment at temperatures given in the Data Table (EXP 4.1–4.4). 3. Place the saucepan over low-medium heat.

EXP 4.1

EXP 4.2

EXP 4.3

EXP 4.4

Appearancea Colorb

Structurec

Sugar Solution without Mixing Appearancea

b

a

Colorb

Structurec

Sugar Solution with Mixing

Use the terms still, clear, and bubbly to describe the appearance. Use the terms white, light yellow, dark yellow, brown, and dark brown to describe the color. c Use the terms cannot hold its shape, soft ball, firm ball, hard ball, soft crack, and hard crack to describe the structure.

149/300

135/275

125/257

120/248

115/239

50/122

Temperature (ºC/°F)

DATA TABLE 4.4

100  COOKING AS A CHEMICAL REACTION

Carbohydrates in Culinary Transformations  101

4. Stir slowly with a spoon until the sugar dissolves. Clean any crystals from the sides of the saucepan using a moistened pastry brush throughout the experiment. 5. Heat the solution, stirring occasionally, until the thermometer reaches 149°C (300°F). 6. Remove from heat, stir in 1 cup almonds. 7. Stir the solution slowly to dissolve the sugar crystals formed, and return to heat immediately. 8. Pour the mixture onto a baking sheet, spreading the mixture quickly into an even layer with a spatula. 9. Cool for one hour. Complete the same experiment; at stage 7, mix the sugar solution constantly. Compare the results of experiments. Case 2 Candied Nuts Recipe Materials and Equipment • 1 cup sugar • ¼ cup water • 1 cup mixed roasted nuts (peanut, almond, and hazelnut) • Measuring cup • 1 heavy saucepan • Stove • Baking paper • Sugar thermometer • Spoon • 2 forks • Spatula • Pastry brush Method 1. Place ¼ cup water and sugar in a heavy saucepan. 2. Place the thermometer in the pan. Be sure the thermometer does not touch the bottom of the pan. 3. Place the saucepan over low-medium heat. 4. Stir slowly with a spoon until the sugar dissolves. Clean any crystals from the sides of the saucepan using a moistened pastry brush throughout the experiment. 5. Heat the solution, stirring occasionally, until the thermometer reaches 149°C (300°F).

102  COOKING AS A CHEMICAL REACTION

DATA TABLE 4.5 Colora

Structureb

Candied nuts Compare with Case 1 a

Use the terms white, light yellow, dark yellow, brown, and dark brown to describe the color. Use the terms cannot hold its shape, soft ball, firm ball, hard ball, soft crack, and hard crack to describe the structure.

b 

6. Remove the saucepan from the heat, and add the nuts quickly. Stir gently with a spatula. 7. Return the mixture over low-medium heat. 8. Heat until the thermometer reaches 160°C (320°F). 9. Remove from heat immediately. 10. Drop a little of the syrup in a glass of cold water and as soon as it forms a ball take it out of the water and record the structure in Data Table 4.5. 11. Record the color of the syrup in Data Table 4.5. 12. Pour the mixture onto a baking sheet immediately. 13. Use two forks to separate the nuts and coat them with sugar. 14. Let them cool at room temperature. 15. Compare the results with Case 1. Case 3 Fudge Recipe Materials and Equipment • 2 cups sugar • 2 squares (1 oz size) unsweetened chocolate • 1 cup light cream • 1 tablespoon butter • Measuring cup • 1 heavy saucepan • Stove • Sugar thermometer • Wooden spoon • Platter

Carbohydrates in Culinary Transformations  103

Method 1. Chop the chocolate into small pieces. 2. Combine sugar, chocolate, and cream in a heavy saucepan. 3. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 4. Cook the mixture over low heat. Stir constantly until sugar dissolves and chocolate melts. 5. Cook until the mixture reaches 117°C (242°F). 6. Remove from heat and cool slightly. 7. Beat constantly with a wooden spoon until fudge begins to harden. 8. Transfer to a buttered platter. 9. Cut into diamond-shaped pieces before fudge hardens completely. 10. Evaluate the appearance, color, and the texture; record your observations in Data Table 4.6. Case 4 Fudge (cooked at higher temperature) Materials and Equipment • 2 cups sugar • 2 squares (1 oz size) unsweetened chocolate • 1 cup light cream • 1 tablespoon butter • Measuring cup • 1 heavy saucepan • Stove • Sugar thermometer • Wooden spoon • Platter DATA TABLE 4.6 Fudge

Appearancea

Colorb

Texturec

Case 3 Case 4 Use the terms still, clear, and bubbly to describe the appearance. Use the terms white, light yellow, dark yellow, brown, and dark brown to describe the color. c Use the terms smooth, soft, hard, sticky, liquid, solid to describe the texture. a

b

104  COOKING AS A CHEMICAL REACTION

Method 1. Complete the same stages of Case 3; at stage 5, cook until the mixture reaches 128°C (262°F) instead of 117°C (242°F). 2. Compare the texture of the products of Cases 3 and 4.

Carbohydrates in Culinary Transformations  105

The Science Behind the Results The sensory properties of the sweets and desserts mainly depend on the processing temperature, mixing time and temperature, and the other ingredients used in the recipe. Sugar (sucrose) and water are boiled together to make sugar syrups and most of the sweets and desserts (Table 4.2). Sugar syrup is thicker than water since the amount of free water decreases as sugar dissolves in water. The water evaporates and the sugar concentrates during boiling, so the product becomes firmer. Mixing (creaming) the hot sugar solution promotes the formation of large crystals due to rapid movement of the molecules toward each other, resulting in a grainy texture. If the creaming temperature is very low, the crystal size will be very small and the product, such as a candy, will not have the desired texture. Therefore, allowing the sugar solution to cool down to a certain temperature (38°C–54°C/100°F–129°F) before agitation will yield smaller crystals, and a smoother and creamier texture. When sugar is heated at high temperatures, it decomposes into glucose and fructose. This is followed by dehydration of sugar, in which each sugar molecule loses water and they react with each other. New flavor and color compounds are formed; sugar turns brown in color and nutty in flavor. This reaction is called caramelizatıon of sugar. Caramelization gives the desirable flavor and color of many foods such as coffee, beer, brittles, a variety of candies, and confectionary. Caramelization of sucrose starts at 160°C (320°F). It becomes undesirable if the sugar is heated above the caramelization temperature because it gives a burnt smell and color to the food product.

95

99

132–143/​ 269–289

149–154/​ 300–309

160–182/​ 320–359

Hard ball

92

120–130/​ 248–266

Brittle and breakable threads are formed when syrup is dropped into cold water Color goes from clear to brown as the temperature rises; viscosity increases

Caramelization

Hard, but not brittle, threads are formed when syrup is dropped into cold water

Hard ball is formed when syrup is dropped into cold water

Firm ball is formed and it holds its shape when syrup is dropped into cold water

Soft ball is formed and it does not hold its shape when syrup is dropped into cold water

Properties

Hard crack

Soft crack

Firm ball

87

118–120/​ 244–248

Soft ball

Heating Stage

85

Approximate Sugar Concentration in the Syrup (%)

112–116/​ 233–240

Temperature (°C/°F)

The Changes Occurring in Sugar Syrup during Cooking

TABLE 4.2

Caramels

Lollipops

Taffy

Rock candy

Chewy candies (caramels)

Fudge, candy filling

Example

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Carbohydrates in Culinary Transformations  107

EXPERIMENT 4.3 OBJECTIVE To explain the natural sugar contents of foods. Ingredients and Equipment • 2 medium onions • Cooking spray • 1 skillet • Cutting board • Knife • 2 plates • Spoon • Stove Method 1. Label one plate “raw” and the other, “caramelized.” 2. Chop the onions into uniformly sized pieces. 3. Place approximately half of the onions onto the plate labeled “raw.” 4. Spray the skillet with cooking spray. 5. Place the remaining onion in the skillet, and cook over medium heat while stirring. 6. Continue stirring and watch, as the onion’s color turns darker and darker. 7. Remove them onto the plate labeled “caramelized” when the onions are thoroughly browned. 8. Examine the appearance, smell, flavor, and texture of the samples, and record the responses in Data Table 4.7. (See EXP 4.5–4.8.)

EXP 4.5

EXP 4.6

108  COOKING AS A CHEMICAL REACTION

EXP 4.7

EXP 4.8

DATA TABLE 4.7 Colora

Smellb

Tastec

Textured

Raw onion Caramelized onion Use the terms white, yellow, golden, and golden brown to evaluate the color of the samples. Use the terms strong and nutty to evaluate the smell of the samples. c Use the terms pungent and sweet to evaluate the taste of the samples. d Use the terms crispy and soft to evaluate the texture of the samples. a

b

Carbohydrates in Culinary Transformations  109

The Science Behind the Results Some vegetables, such as onions, are rich in “natural sugars” (Table 4.3). When these vegetables are being cooked, the poly- and disaccharides in their structures are broken down into smaller sugar units because the bonds between the units are disturbed by heat. As heating continues, these sugar units undergo caramelization. This is the primary reason behind why onions turn brown and develop a sweeter flavor upon dry heating. TABLE 4.3 The Natural Sugar Content of Some Vegetables

Vegetables serving size (g/oz)

Sugars (g/oz)

Onion: 1 medium (148 g/5.22 oz)

9/0.31

Carrot: 1 carrot (78 g/2.75 oz)

5/0.17

Sweet corn: 1 medium (90 g/3.17 oz)

5/0.17

Sweet potato: 1 medium (90 g/3.17 oz)

7/0.24

110  COOKING AS A CHEMICAL REACTION

Carbohydrates in Culinary Transformations  111

EXPERIMENT 4.4 OBJECTIVES • To explain the structural properties of starch (polysaccharides). • To observe the effects of starch types in starch gels. Turkish Delight Recipe Ingredients and Equipment • 4 cups table sugar • 1¼ cups corn starch • 1 cup icing sugar • 1 teaspoon cream of tartar • 4¼ cups water • 1 tablespoon lemon juice • 1 cup confectioners’ sugar • Thermometer • 2 saucepans • Measuring cup • Whisk • Pan • Wax paper • Knife • Stove Method 1. In a saucepan, combine the lemon juice, table sugar, and 1½ cup water on medium heat. 2. Stir constantly until sugar dissolves. 3. Allow the mixture to boil. 4. Reduce heat to low and allow to simmer, until the mixture reaches 115.5°C (240°F). 5. Remove from heat and set aside. 6. Combine cream of tartar, 1 cup corn starch, and remaining water in another saucepan over medium heat. 7. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 8. Stir until all lumps are gone and the mixture begins to boil. 9. Stop stirring when the mixture has a glue-like consistency. This is called the gelatinization temperature. Record the gelatinization temperature in Data Table 4.8.

112  COOKING AS A CHEMICAL REACTION

DATA TABLE 4.8 Temperature (°C/°F) Time (min)

Corn Starch

Potato Starch

0 10 20 30 40 50 60 70 … Gelatinization Temperature

10. Continue heating. Record the temperature every 10 min in Data Table 4.8 until the mixture becomes golden in color. 11. Stir in the lemon juice and water/ sugar mixture. 12. Stir constantly for about 5 min. 13. Reduce heat to low; allow to simmer for one hour, stirring frequently (EXP 4.9). 14. Once the mixture has become viscous and opaque, pour the mixture into a previously wax paper- or parchmentlined pan to about 1 cm (0.39 in.) from the top. Add nuts (optional).

EXP 4.9

Wheat Starch

Carbohydrates in Culinary Transformations  113

DATA TABLE 4.9 Starch Type

Transparency

a

Color

b

Texture

c

Change in Height (%)

Maize Potato Wheat Use the terms transparent, translucent, and opaque to describe the transparency. Use the terms white, bright golden, golden, and dark golden to describe the color. c Use the terms soft, semisoft, and firm to describe the texture. a

b

15. Spread evenly. 16. Leave the sample at room temperature overnight. 17. Cut in small rectangles, sprinkle with confectioners’ sugar. 18. Analyze the product and record your observations in Data Table 4.9. Repeat above procedure using potato starch and wheat starch. Note: In cases when the observations or the measurements cannot be done, denote that as n/a (not applicable). To calculate the gel strength: 1. Measure the height of the Turkish Delight using a toothpick while it is still on the tray, (hi). 2. Turn over the baking pan containing Turkish Delight onto a clean counter and remeasure the height, (hf ). Calculate the percentage (%) change in the height:

% Change =

( h i − hf ) × 100 hi

Note: A low percentage (%) change indicates a strong gel.

Study Questions 1. Compare the gelatinization temperatures of starches. 2. Discuss the gel strength of samples prepared using different starches. 3. Which starch would be more suitable to make Turkish Delight? Explain your answer.

114  COOKING AS A CHEMICAL REACTION

Carbohydrates in Culinary Transformations  115

The Science Behind the Results Starch is a polysaccharide, which is a major component of many plants and an important ingredient for the culinary industry. Starches are the long-chain polymers of glucose units. The number of glucose molecules varies from hundreds to several hundred thousand, depending on the type of starch. Seeds, cereals, roots, and the tubers are the main sources of starch. The most common starches in culinary processes are corn (maize), rice, wheat, potato, and tapioca (cassava). The functional properties of starches in foods are almost unlimited and the ­primary functions can be listed as: • • • • •

Thickening Body and texture formation Binding Coating Moisture retention

The functional properties of starch are primarily connected to its chemical ­composition. Starch consists of two fractions: amylose and amylopectin. These are packed into water-insoluble granules (Table 4.4). Amylose is a linear chain of ­glucose units. It has a simpler structure compared to amylopectin because ­amylopectin has a branched structure. The relative amounts of amylose and amylopectin differ among different starch varieties depending on the plants from which they are produced (Table 4.5). Starch is insoluble in water. It undergoes starch gelatinization when heated in liquid. When starch is mixed with water and heated, hydrogen bonds break, allowing water to enter the starch granule. Some amylose goes into the water from the surface of the starch granule; more water enters into the granule and the starch granules swell. Gradually, granules lose their natural structures. Hydrogen bonding between amylose and water are formed. The amount of free water decreases, the viscosity of the starch mixture increases. New gel structure is formed upon cooling, and the mixture thickens. This process is called starch gelatinization (Figure 4.1). The chemical structure of the starch and the size of the starch granule determine the viscosity of the starch gel, the speed of gelatinization, the gel strength, and the gelatinization temperature. That is because using the correct type of starch is important in culinary processes.

116  COOKING AS A CHEMICAL REACTION

TABLE 4.4 Properties of Amylose and Amylopectin

Amylose G

Amylopectin G

G

G

G

G

G

G G

• It has a linear chain structure. • It makes approximately 10%–20% of the total amount of starch in plants based on its source.

G

• It has a highly branched structure. • It makes approximately 80%–90% of the total amount of starch in plants based on its source.

TABLE 4.5 Approximate Amounts of Amylose and Amylopectin of Various Starches

Type of Starch (Source)

Amylose (%)

Amylopectin (%)

Corn (cereal)

26

74

Rice (cereal)

17

83

Potato (tuber)

21

79

Cassava (root)

17

83

Wheat (cereal)

25

75

The amylose-to-amylopectin ratio in the starch determines the characteristic properties of the final food. The amount of amylose molecules in the granule determines the gel strength. The more amylose, the stronger the gel, but the less viscous the product. This is because the amylose has a simple linear structure in which molecules can move closer to each other to form bonds. Amylopectin contributes the viscosity of the final food product. The branched structure of amylopectin molecules keeps them from coming closer to form bonds. It does not contribute to the gel formation, but it gives viscosity to the food products due to

Carbohydrates in Culinary Transformations  117

Figure 4.1  Starch gelatinization.

TABLE 4.6 Gelatinization Temperatures of Some Starch Varieties

Type of Starch (source)

Gelatinization Temperature (°C/°F)

Corn (cereal)

62–70/143–158

Rice (cereal)

68–75/154–167

Potato (tuber)

59–68/138–154

Cassava (root)

65–68/149–154

Wheat (cereal)

52–54/125–129

its bulky structure. Therefore, the more amylopectin, the softer the gel because molecules cannot align as easily and, thus, give weaker hydrogen bonding and gel strength. Different starch types have different gelatinization temperatures because they have different amounts of amylose and amylopectin in their structures (Table 4.6).

118  COOKING AS A CHEMICAL REACTION

Carbohydrates in Culinary Transformations  119

EXPERIMENT 4.5 OBJECTIVE To explain the effects of other ingredients in the recipe on starch gels. Case 1 Sugar effect Ingredients and Equipment • 120 g (4.23 oz) corn starch • 30 g (1 oz) sugar • Water • 6 small bowls • 11 saucepans • Measuring cups • Kitchen scale • Thermometer • Spoon • Whisk • Skewers • Refrigerator • Stove Method 1. Label the bowls “control,” “2 g (0.07 oz) sugar,” “4 g (0.14 oz) sugar,” “6 g (0.21 oz) sugar,” “8 g (0.28 oz) sugar,” and “10 g (0.35 oz) sugar.” 2. Place 20 g (0.7 oz) starch and 300 mL (10 fl oz) water in a saucepan. 3. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 4. Heat the mixture over medium heat stirring constantly. 5. Remove from the heat when gelatinization is complete (when the mixture has a glue-like consistency). 6. Record the gelatinization temperature in Data Table 4.10. 7. Pour the gelatinized starch paste into control bowl to about 2 cm (¾ in.) from the top. 8. Rinse the saucepan to remove the paste residues. 9. Place 20 g (0.7 oz) starch, 2 g (0.07 oz) sugar, and 300 mL (10 fl oz) water in a saucepan. 10. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 11. Heat the mixture over medium heat, stirring constantly. 12. Remove from the heat when gelatinization is complete.

120  COOKING AS A CHEMICAL REACTION

3. Record the gelatinization temperature in Data Table 4.10. 1 14. Pour the gelatinized starch paste into the bowl labeled “2 g (0.07 oz)” sugar. 15. Store all the samples in the refrigerator overnight. 16. Assess and compare the gel strengths next day. 17. Record your observations in Data Table 4.10. Repeat Steps 9–17 with 4 g (0.14 oz) sugar, 6 g (0.21 oz) sugar, 8 g (0.28 oz) sugar, and 10 g (0.35 oz) sugar additions in an order. Store all the samples in the ­refrigerator overnight. DATA TABLE 4.10 Sample

Gelatinization Temperature (°C/°F)

% Change in Height

Visual and Textural Propertiesa

Control Sugar (g/oz) 2 4 6 8 10 Acid (mL/fl oz) 2 4 6 8 10 Use the terms transparent, translucent opaque, white, bright golden, golden, and dark golden to describe the visual properties; soft, semisoft, and firm to describe the textural properties. a

Carbohydrates in Culinary Transformations  121

Case 2 Acidity effect Ingredients and Equipment • 120 g (4.23 oz) corn starch • 30 mL (1 fl oz) lemon juice • Water • 6 small bowls • 1 saucepan • Measuring cups • Spoon • Skewers • Kitchen scale • Thermometer • Refrigerator • Stove Method 1. Label the bowls “control,” “2 mL (0.06 fl oz) lemon juice,” “4 mL (0.13 fl oz) lemon juice,” “6 mL (0.20 fl oz) lemon juice,” “8 mL (0.27 fl oz) lemon juice,” and “10 mL (0.33 fl oz) lemon juice.” 2. Place 20 g (0.7 oz) corn starch and 300 mL (10 fl oz) water in a saucepan. 3. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 4. Heat the mixture over medium heat, stirring constantly. 5. Record the gelatinization temperature in Data Table 4.10. 6. Pour the gelatinized starch paste into “control” bowl to about 2 cm (¾ in.) from the top. 7. Rinse the saucepan to remove the paste residues. 8. Place 20 g (0.7 oz) cornstarch, 2 mL (0.06 fl oz) lemon juice, and 300 mL (10 fl oz) water in a saucepan. 9. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 10. Heat the mixture over medium heat stirring all the time. 11. Remove from the heat when gelatinization is complete. 12. Record the gelatinization temperature in Data Table 4.10. 13. Pour the gelatinized starch paste into previously marked bowl to about 2 cm from the top. 14. Store all the samples in the refrigerator overnight. 15. Assess and compare the gel strengths next day. 16. Record your observations in Data Table 4.10.

122  COOKING AS A CHEMICAL REACTION

Repeat Steps 8–16 with 4 mL (0.13 fl oz) lemon juice, 6 mL (0.2 fl oz) lemon juice, 8 mL (0.27 fl oz) lemon juice, and 10 mL (0.33 fl oz) lemon juice. Method to assess the gel strength: 1. Measure the height of the gel using a skewer while still in the bowl, (hi). 2. Turn over bowl containing starch paste onto a clean counter and measure the height again, (hf ). Calculate the percentage (%) change in the height.

% Change =

( h i − hf ) × 100 hi

Note: A low percentage (%) change indicates a strong gel.

Carbohydrates in Culinary Transformations  123

The Science Behind the Results Although the structure of the starch and the starch granule sizes are the primary factors, the viscosity, the speed of gelatinization, the gel strength, and the gelatinization temperature also are affected by the following factors during cooking: 1. Presence of the acidic ingredients 2. Presence of sugar 3. Presence of fats and protein 4. Presence of enzymes 5. Agitation/mixing 6. Rate of cooling. Acidic ingredients, such as lemon, yogurt, and vinegar, change the net charge in foods. They decrease the gel strength because acid breaks down the bonds between the starch molecules. Adding the acid after gelatinization can minimize the acid effect. Sugar addition, especially sucrose addition, decreases the viscosity and firmness of the starch gel because sugar competes with the starch for water. A higher temperature of gelatinization is obtained. Timing of sugar addition is significant. Increased concentration of fats and protein delays the hydration of starch causing a decreased viscosity of the starch paste and decreased gel strength because they coat the surface of the starch granules. Enzymes decrease gel strength because some enzymes hydrolyze the starch. For example, an egg yolk decreases the strength of the gel because it contains alphaamylase, an amylose-digesting enzyme. Agitation throughout the gelatinization process creates a more uniform mixture without lumps because it separates the starch granules. However, it prevents formation of a network and firm gel if the mixture is excessively agitated, or continued after the gel formation. Rate of cooling is also important; too fast or too slow cooling may affect the gel strength.

124  COOKING AS A CHEMICAL REACTION

Carbohydrates in Culinary Transformations  125

EXPERIMENT 4.6 OBJECTIVES • To explain the changes in the starch gel structure during storage. • To explain the effect of storage temperature on starch gel structure during storage. Case 1 Bread staling Ingredients and Equipment • A loaf of plain bread • 15 medium-sized zipped sandwich bags • Sharp bread knife • Cutting board • Refrigerator • Freezer Method 1. Label sandwich bags “room temperature,” “freezer temperature,” and “refrigeration temperature.” Label five bags for each temperature. 2. Cut 15 equal-sized bread slices from the same bread. 3. Place one slice in each bag. 4. Seal the plastic bags. 5. Place five bags labeled “freezer temperature” in the freezer, five bags labeled “refrigeration temperature” in the refrigerator, and five bags labeled “room temperature” in a safe place in the room. 6. After 24 h, take one bag from each storage temperature and let the freezer and refrigerator samples warm to room temperature. 7. Track the accumulation of water on the inside surface of the plastic bags. 8. Take the slices out of the bag and evaluate the changes in the texture. 9. Compare the textures of the slices. 10. Record your evaluations in Data Table 4.11. 11. Dispose of the slices. 12. Repeat the same process over the next four days with the rest of the samples.

126  COOKING AS A CHEMICAL REACTION

DATA TABLE 4.11

Storage Time Day 1

Storage Temperature

Water Accumulation on Inside Surface of Plastic Baga

Texture of the Slice (Hardness, Softness)b

Room Refrigeration Freezer

Day 2

Room Refrigeration Freezer

Day 3

Room Refrigeration Freezer

Day 4

Room Refrigeration Freezer

Day 5

Room Refrigeration Freezer

a b

Use the terms no accumulation, little accumulation, and severe accumulation. Rate the texture of each sample in a ranking of 1 to 5; 1 = hard, 5 = soft.

Comparison of Texture of Slices

Carbohydrates in Culinary Transformations  127

Case 2 Rice dishes prepared with different rice types Ingredients and Equipment • 100 g (3.5 oz) short grain rice (e.g., Arborio rice) • 100 g (3.5 oz) medium grain rice (e.g., Calrose rice) • 100 g (3.5 oz) long grain rice (e.g., Jasmine rice) • Water • 3 small saucepans with lids • Spoon • Fork • Refrigerator • Stove Method 1. Label the saucepans “short grain,” “medium grain,” and “long grain.” 2. Cook short grain rice as directed on the package. 3. When the rice is done, turn off the heat. 4. Cool down to room temperature. 5. Take off the lid. 6. Evaluate the texture. 7. Record your observations in Data Table 4.12. 8. Complete the same stages for all rice types in separate saucepans. 9. Refrigerate the saucepans overnight. 10. Take the saucepans out of the refrigerator and let come closer to room temperature. 11. Evaluate their textures. 12. Record your observations in Data Table 4.12. 13. Compare the degree of changes in textural structures of different rice dishes after refrigeration, and record in Data Table 4.12.

Describe: Rate:

Long grain rice

b

Describe: Rate:

Describe: Rate:

Describe: Rate:

Texture after Refrigeration

Use the terms sticky, fluffy, dry, and moist to describe the texture. Rate the texture of the samples in a ranking of 1 to 5; 1 = hard, 5 = soft.

Describe: Rate:

Medium grain rice

a

Describe:a Rate:b

Texture before Refrigeration

Short grain rice

DATA TABLE 4.12 Comparison of the Structures of Rice Dishes after Refrigeration

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Carbohydrates in Culinary Transformations  129

The Science Behind the Results Why does bread stale during storage? Why does cooked rice harden during storage? Why does pudding release water during storage? The primary cause is aging gel. Starch molecules in gelatinized starch, particularly the amylose chains, have a tendency to reassociate in an ordered crystalline structure during storage. This process is known as retrogradation, which is directly related to the hardening of cooked rice and the staling of baked products, such as bread. Water may squeeze out from the starch gel network as a result of retrogradation. This process is known as syneresis. The collection of water on the surface of pudding during storage is a good example for syneresis. Rate of retrogradation primarily depends on: • Ratio of amylose and amylopectin in the starch. Due to the simpler structure of amylose, rate of retrogradation increases as the amount of amylose in the starch structure increases. For example, long grain rice contains high amounts of amylose and least amylopectin in its structure than short and medium grain rice varieties. That is why cooked long grain rice is typically more prone to retrogradation, and gets harder during storage compared to the other varieties. • Storage temperature has a significant effect on the rate of retrogradation. Rate of retrogradation is higher at the refrigeration temperature than the rate at the freezing temperature and at room temperature.

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Carbohydrates in Culinary Transformations  131

EXPERIMENT 4.7 OBJECTIVE To explain the dextrinization process. Béchamel Sauce Recipe Ingredients and Equipment • 3 tablespoons butter • 2 tablespoons cake flour • 2 cups milk • 1 teaspoon salt • Saucepan • Stove • Whisk Method 1. In a medium saucepan, melt the butter over medium-low heat. 2. Add the flour and salt and stir until smooth. 3. Cook over medium heat, stirring frequently. 4. Every 2 min, record the changes in color, taste, and the texture of the mixture in Data Table 4.13 until the mixture turns a light golden color and has a slightly nutty aroma. 5. Pour in milk slowly while whisking ­constantly until very smooth. 6. Observe the texture of the mixture and record in Data Table 4.14. 7. Bring to a boil while stirring constantly. 8. Reduce the heat to low, cook the mixture 10 min, stirring occasionally. 9. Remove from heat. 10. Observe the texture of the mixture and record in Data Table 4.14 (EXP 4.10–4.13).

EXP 4.10

EXP 4.11

EXP 4.12

EXP 4.13

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DATA TABLE 4.13 Time (min)

Colora

Tasteb

Texturec

2 4 6 Use the terms white, bright golden, golden, and dark golden to describe the color. Use the terms plain, nutty, light, and strong to describe the taste. c Use the terms silky and sandy to describe the texture. a

b

DATA TABLE 4.14 Texture of the Mixture Just after milk addition At the end of cooking

Use the terms silky, sandy, solid, semisolid, and liquid to describe the texture.

Study Questions 1. Explain the effect of heat on the sensory properties of the mixture before the addition of the milk. 2. Explain the effect of heat on the sensory properties of the mixture after the addition of the milk.

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The Science Behind the Results As discussed previously, heat and liquid are required for starch gelatinization. In other words, starch gels when heated in liquid. When starch is heated to high temperatures (100°C–200°C/212°F–392°F) without liquid (dry heating), the linkages between the glucose units are destroyed. Starch polymers are broken down into smaller, sweeter-tasting sugar molecules (dextrins). This process is known as dextrinization. This process is applied in recipes, which call for heating of flour or starch without liquid, such as white and brown roux, some desserts (e.g., Turkish halva), and some soups. The basic purpose of the dextrinization process in the kitchen is to give a nutty aroma, sweeter taste, and a golden color to the foods. It also reduces its ability to thicken into a gel. Note: Dextrinization may occur at lower temperatures if starch is heated with small amounts of acidic ingredients because acid helps to break down the bonds between the starch molecules.

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Carbohydrates in Culinary Transformations  135

EXPERIMENT 4.8 OBJECTIVE To explain the basic function of pectin in foods. Raspberry Jam Recipe Ingredients and Equipment • 1,000 g (2.2 lb) of fresh raspberries • 28 g (1 oz) powdered pectin • 1,200 g (2.6 lb) table sugar • 4 tablespoons lemon juice • 2 pans • Stove • Potato masher or a wooden spoon Method 1. Place 500 g (1.1 lb) of raspberries in a pan. 2. Add 2 tablespoons lemon juice. 3. Bring to a full boil over high heat, stirring constantly. 4. Using a potato masher or a wooden spoon, lightly crush raspberries. 5. Slowly add 600 g (1.3 lb) sugar and 28 g (1 oz) of pectin while stirring constantly until the sugar is fully dissolved. 6. Return to a boil, and let it boil for 5 min. 7. Remove from the heat and let it cool down. 8. Analyze the texture and record in Data Table 4.15. 9. Repeat the experiment without pectin. 10. Analyze the texture and record in Data Table 4.15. DATA TABLE 4.15 Sample

Texture of the Samplesa

With pectin Without pectin a

Use the terms thin, thick, viscous, and less viscous to evaluate the texture.

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Study Question 1. Compare the textures of the jams, and discuss the effect of pectin on the texture of the jam.

Carbohydrates in Culinary Transformations  137

The Science Behind the Results Pectin and cellulose are the polysaccharides found in plants as the principal structural components of the cell walls. They are the most abundant dietary fibers. Fruits, vegetables, and whole grains are the most common sources of the dietary fibers. Functional properties of fibers in culinary processes can be listed as: • • • •

Thickening and gelling Emulsification Stabilization of emulsions Stabilization of foams

There are two types of fibers: water-soluble fibers and water-insoluble fibers. Legumes, oat bran, peas, citrus fruits, apple pulp, and berries are rich in watersoluble fibers. Pectin is the most common example of the water-soluble fibers. Cauliflower, cabbage, apple skin, wheat bran, and whole grain foods are rich in insoluble fibers. Cellulose is the most common example of the water-insoluble fibers. Pectin is primarily used as a gelling agent, thickening agent, and stabilizer in foods, especially in jams, jellies, and some fruit juices. It is found in plant cell walls and mostly concentrated in the skin and the core of fruits. The amount of pectin varies from fruit to fruit (Table 4.7). The stage of ripeness also determines the amount of pectin in fruits. The ripe fruits contain more pectin compared to the over-ripe ones. Therefore, just-ripe fruits are used to thicken fruit syrups, jams, jellies, and marmalades. In the kitchen, fruits that are low in pectin must be combined with one of the higher pectin fruits to form a gel. Alternatively, commercial pectin can be used with low-pectin fruits to form gels.

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TABLE 4.7 Pectin Contents of Some Fruits

Examples of Fruits Rich in Pectin

Examples of Fruits Low in Pectin

• Sour apples

• Apricots

• Currants

• Blueberries

• Cranberries

• Sweet cherries

• Blackberries

• Peaches

• Gooseberries

• Pineapple

• Lemon

• Strawberries

Carbohydrates in Culinary Transformations  139

EXPERIMENT 4.9 OBJECTIVE To explain the effect of different factors on pectin gel formation. Sour Cherry Jam Ingredients and Equipment • 2,000 g (4.4 lb) sour cherries • Water • 1,500 g (3.3 lb) sugar • 10 mL (0.33 fl oz) lemon juice • Ladle • Measuring cups • Skimmer • Kitchen scale • Large pot • Spoons • 4 heat-stable jelly jars with lids • Thermometer • Stove Method 1. Label the jars “250 g (8.8 oz) sugar,” “500 g (17.6 oz) sugar,” “with acid,” and “without acid.” 2. Add 1 L (2.11 pts) of water and 250 g (8.8 oz) of sugar into the pot. 3. Stir until the sugar is completely dissolved. 4. Wash 500 g (17.6 oz) sour cherries and add them to the pot. 5. Cook over high heat, stirring occasionally. 6. Add 5 mL (0.16 fl oz) lemon juice. 7. Once it comes to a boil, start timing while stirring constantly. 8. Boil for 15 min. 9. Remove from heat. 10. Skim off any foam with a spoon. 11. Pour into the jar labeled “250 g (8.8 oz) sugar” and put the lid on. 12. Store for a week. 13. Record your observations in Data Table 4.16. 14. Repeat the same experiment using 500 g (17.6 oz) of sugar. 15. Repeat the same experiments without acid. 16. Compare the samples.

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DATA TABLE 4.16 Amount of Sugar (g/oz)

Amount of Acid (mL/fl oz)

250/8.8

0

250/8.8

5/0.16

500/17.6

0

500/17.6

5/0.16

a b

Flavor

Viscositya

Textureb

Use the words viscous, less viscous, and not viscous to evaluate the viscosity. Use the words sticky, less sticky, and not sticky to evaluate the texture of the samples.

Note: Increasing the amount of sugar to 500 g (17.6) but keeping the amount of lemon juice constant decreases the percentage of acid in total amount of ingredients.

Carbohydrates in Culinary Transformations  141

The Science Behind the Results As explained previously, thickening and gelling are the primary functional ­properties of pectin in culinary processes. During processing, heat, presence of acidic ingredients, acidity of the fruit, and the amount of sugar affect the formation of a pectin gel. Heat is required for gelling to occur. Upon heating, the pectin in fruit becomes water-soluble because heat disturbs the linkages between the molecules. Too high of a temperature or prolonged cooking can destroy the pectin, resulting in a decrease in gel strength. Acid at a certain concentration contributes to the gel formation because it hydrolyzes pectin, and new cross-links are formed. The gel will not set if there is too little acid. On the other hand, too much acid will cause the gel to lose liquid. Acid in fruit is usually enough to form the gel. In some jam recipes, especially if the fruit is not acidic, additional acid, such as lemon and cream of tartar, are required for a proper set. Sugar has two important functions in formation of the pectin gel. The pectin cannot form a strong gel if too much water is present in the medium, which results in a runny food product. Sugar binds excessive water in fruits; therefore, the pectin can form a gel with desired texture in the jam, jellies, and the fruit syrups. Pectin also binds with sugar. Addition of acidic ingredients promotes this reaction. This is one of the reasons for using lemon juice in jam making. Acid hydrolyzes pectin, pectin binds with sugar; a gel with the proper strength is formed as a result of the synergistic effects of acid and sugar. Note: If you are using commercial pectin, be sure to follow the instructions. Each jam, jelly, and the fruit syrup recipe must have a correct balance between pectin, acid, and sugar for the fruit you are using. Do not make unrecommended adjustments.

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POINTS TO REMEMBER Carbohydrates are the organic compounds naturally found in many foods, primarily in plants. Understanding the basic structure of carbohydrates is crucial for chefs because the functional properties of carbohydrates are primarily related to their structures. Carbohydrates are classified into three major groups according to the ­number of simple sugar (monomer) units: 1. Monosaccharides 2. Disaccharides 3. Polysaccharides Monosaccharides and disaccharides are soluble in polar solvents, such as water. Monosaccharides contain polar hydroxyl groups (OH), which can easily form hydrogen bonds with water. The sensory properties of the sweets and desserts depend on the processing temperature and the other ingredients used in the recipe. Some vegetables, such as onions, are rich in “natural sugars.” Starch is a polysaccharide, which is a major component of many plants and an important ingredient for the culinary industry. Starch is insoluble in water. The chemical structure of the starch and the size of the starch granule determine the viscosity of the starch gel, the speed of gelatinization, the gel strength, and the gelatinization temperature. Starch molecules in gelatinized starch, particularly the amylose chains, have a tendency to reassociate in an ordered crystalline structure during storage. This process is known as retrogradation. There are two types of fibers: water-soluble fibers and water-insoluble fibers. Pectin is a water-soluble fiber that is primarily used as a gelling agent, thickening agent, and stabilizer in foods, especially in jams, jellies, and some fruit juices. Cellulose is the most common example of the water-insoluble fibers.

Carbohydrates in Culinary Transformations  143

More Ideas to Try 1. Repeat Experiment 4.7, but skip stages 3 and 4. Compare the color, taste, and the texture of the product with the final product from Experiment 4.7. 2. Repeat Experiment 4.9 by substituting apricot for sour cherry. Try it: a. With lemon juice b. Without lemon juice Compare the results.

Study Questions 1. Discuss the effect of sugar concentration on starch gel strength. 2. Discuss the effect of acid addition on the starch gel strength. 3. Discuss the effect of sugar concentration on pectin gel formation. 4. Discuss the effect of acid addition on pectin gel formation.

SELECTED REFERENCES Astridottenhof, M., and I. Farhat. 2004. Plant biotechnology. Biotechnology and Genetics Engineering Reviews 21:215–228. Campbell-Platt, G. 2009. Food science and technology. Oxford, UK: Wiley-Blackwell Publishing Ltd. Friberg, B. 2003. The advanced professional chef. New York: John Wiley and Sons. Gaman, P. M., and K. B. Sherrington. 1996. The science of food, 4th ed. Oxford, UK: Elsevier Ltd. Lauriston, R. 1996. Gelatinization temperatures for adjuncts. Online at: www.­brewery. org/brewery/library/GelTemps_RL0796.html. McGee, H. 2004. On food and cooking, 1st revised ed. New York: Scribner. Murphy, P. 2000. Handbook of hydrocolloids. Boca Raton, FL: CRC Press. Ozilgen, S. 2012. Failure mode and effect analysis (FMEA) for confectionery manufacturing in developing countries: Turkish delight production as a case study. Ciência e Tecnologia Alimentos 32(3):505–514. Satin, M. 1998. Functional properties of starches. Rome: FAO Agricultural and Food Engineer­ ing Technologies Service. Online at: www.fao.org/ag/magazine/pdf/starches.pdf. Spies, R. D., and R. C. Hoseney. 1982. Effect of sugar on starch gelatinization. Cereal Chemistry 59(2):128–131. Starch. http://food.oregonstate.edu/learn/starch.html. Starch. Online at: www.enotes.com/starch-reference/starch. U.S. Food and Drug Administration (FDA). 2008. Vegetables: Nutritional facts. Online at: w w w.fda.gov/dow nloads/ Food / LabelingNutrition / FoodLabelingGuidance​ Reg ulator yInfor mation / Infor mationforRestaurantsRetailEstablishments / UCM169237.pdf. Vaclavic, V. A., and E. W. Christian. 2008. Essentials in food science, 3rd ed. Berlin: Springer.

Chapter 5 Proteins in Culinary Transformations

FUNCTIONAL PROPERTIES OF PROTEINS IN CULINARY PROCESSES Proteins are the organic compounds naturally found in animals (e.g., milk and meat) and plants (e.g., wheat, soy, and beans). The primary functional properties of proteins in food processing can be listed as: 1. Foam formation 2. Gelation 3. Dough formation 4. Flavor development 5. Viscosity control 6. Water binding 7. Color formation Egg foams when whisked, but overwhisking collapses the structure. Gelatinbased jellies will not set if they are made with kiwi, but blanching the fruit solves that problem. Egg turns white when cooked, but overcooking makes it rubbery. Meat gets softer when marinated, but different marinades are used for different meat cuts—and, so on.

145

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If we can answer the question of why egg albumen turns white when cooked, we will know when to stop the process before the egg gets rubbery. Similarly, if we know how meat gets softer when marinated, we will know which type of marinades to use for different meat cuts. Understanding the basic structure of food proteins is crucial for chefs because  the  functional properties of proteins are primarily related to their structures.

PROTEIN STRUCTURE Proteins are the polymers of different amino acids joined together by covalent bonds (peptide linkages). Each amino acid has a central carbon atom bonded to a hydrogen atom, a carboxyl group, an amino group, and a unique functional group (R group) (Figure 5.1). The chemical properties of R groups primarily determine the properties of proteins in foods. There are 22 known amino acids, which means 22 different R groups. Therefore, proteins in foods, such as in eggs, meat, legumes, and cereals, differ from each other primarily because their R groups, number of amino acids, and the sequence of amino acids are different. Proteins have four structures: 1. Primary structure: The primary structure of protein is the sequence of amino acids bound together by peptide bonds (linkages) (Figure 5.2). 2. Secondary structure: The secondary structure of proteins is formed by coiling of the primary structure in β-sheets or α-helix shapes by formation of hydrogen bonds between amino acids (Figure 5.3a,b). 3. Tertiary structure: The tertiary structure of proteins refers to the overall shape of protein formed by the folding of the secondary structure by Functional group R

H N H Amino group

C H

O C OH Carboxyl group

Figure 5.1  A general structure of amino acids.

Proteins in Culinary Transformations  147

R1

H N H

C

O C

H

N

+ OH

R2

H H

C

O C

H

R1

H N

OH

H

C

H

N

C

OH

H

C

H

C

+H

O

R2

2O

s­ econdary attractions, i.e., disulfide bonds, hydrophobic interactions, and salt bridges (ionic interactions), between R groups of amino acids (Figure 5.4). 4. Quaternary structure: Two or more polypeptide chains join together to form the quaternary structure of proteins (Figure 5.5).

O Peptide bond

Amino acid 1

Amino acid 2

Figure 5.2  Formation of peptide linkages (bonds).

Figure 5.3  (a) α-Helix shapes. (b) β-Sheet.

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Figure 5.4  Tertiary structure of proteins.

Figure 5.5  Quaternary structure of proteins.

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Different proteins in foods may have one of two different molecular shapes: globular or fibrous. Globular Proteins

Fibrous Proteins

• Polypeptide polymers fold back on themselves as globs or spheres to form globular proteins. • Almost all globular proteins are soluble in water. • Milk protein (casein), myoglobin, and egg white protein (albumin) are good examples for the globular proteins found in foods.

• Polypeptide chains are arranged parallel to one another to form a strong rope-like structure. • Fibrous proteins are insoluble in water, and are very stable. • Collagen, actin, myosin, and elastin in meat are good examples of fibrous proteins found in foods.

The structure of the proteins can be affected by processing conditions. The noncovalent bonding interactions of secondary, tertiary, and quaternary structures can be destroyed or destructed by physical or chemical treatments applied during food processing. The protein structure unfolds and a new structure is formed. This process is called protein denaturation (Figure 5.6). A new product is formed because the chemical change has taken place as a result of protein denaturation. The most common physical and chemical factors affecting the protein structure in foods during food preparation and processing include: • • • • • •

Heat treatment Mechanical treatment (whisking, mixing, pounding) Salt addition pH change Enzymatic activities Sugar addition

Figure 5.6  Denaturation of proteins.

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EXPERIMENT 5.1 OBJECTIVE To explain the effect of heat treatment on proteins in foods. Ingredients and Equipment • 4–5 fresh eggs, at room temperature • 1 tablespoon butter • Pan • Stove Method 1. Heat the pan over medium heat. 2. Add butter. 3. When butter begins to sizzle, crack the eggs. 4. Observe the texture and color changes that occur in the eggs as cooking proceeds and record your observations in Data Table 5.1. 5. Cook until the egg whites have just set, and record your observations in Data Table 5.2. DATA TABLE 5.1 Time (min)

Describe the Changes in Colora and Textureb of the Egg Whites

0 2 4 6 8 10 … a b

Use the terms transparent, translucent, and white to describe the color. Use the terms liquid, semisolid, and solid to describe the texture.

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DATA TABLE 5.2 Color and Texture of the Egg Whitesa,b Properly cooked fried eggs Overcooked fried eggs a b

Use the terms transparent, translucent, and white to describe the color. Use the terms liquid, semisolid, and solid to describe the texture.

6. Continue to cook for five more minutes (overcooking), and record your observations in Data Tables 5.1 and 5.2. 7. Compare the textures and colors of the properly cooked eggs and the overcooked eggs (EXP 5.1–5.7).

EXP 5.1

EXP 5.2

EXP 5.3

EXP 5.4

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EXP 5.5

EXP 5.6

EXP 5.7

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Proteins in Culinary Transformations  155

The Science Behind the Results Understanding the composition of foods will help us understand the changes that occur within foods when we process them. The contents of the egg is shown in PIC 5.1. Temperature is one of the primary causes of protein denaturation during cooking. As temperature increases, the energy of the protein molecules also increases and noncovalent interactions in the protein structure are weakened. At some temperature, the protein structure unfolds. The unfolded protein molecules interact (reassociate) with each other to form a new three-dimensional network. During the reassociation of protein molecules, water is entrapped within the network and a gel-like structure is formed (Figure 5.6). The temperature at which this process occurs is known as the protein denaturation temperature. Semisoft, semisolid precipitates are formed when denatured proteins are processed further. This process is called coagulation. Denaturation is the first stage of coagulation. Denaturation and coagulation of proteins are usually desirable in food processing. Differences between raw and cooked eggs are largely a result of protein ­coagulation that occurs during cooking. The egg white is liquid and translucent in its raw state. Upon cooking, albumen coagulation occurs; the texture gradually changes from liquid to solid. Simultaneously, the color turns white. If heating and denaturation continue, the interactions between proteins become very strong. The network collapses as water is removed from the network. The food becomes rubbery and dry as observed in the experiments when the egg white is overcooked.

ALBUMEN (White) is the largest part of the egg. It makes up 2/3 of the egg mass. The total mass of egg white is 88% water and 12% protein. YOLK makes up 1/3 of the egg mass. The total mass of egg yolk is 48% water, 35% lipid, and 17% protein.

PIC 5.1 PROTEIN CONTENT OF THE EGG.

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Proteins in Culinary Transformations  157

EXPERIMENT 5.2 OBJECTIVES • To show that different proteins have different denaturation temperatures. • To show the effects of heating time and heating temperature on proteins in foods. Ingredients and Equipment • 8 eggs • Water bath (if not available, use a large brazier with water at a constant temperature) • 4 food thermometers • Timer • 16 heat-resistant glass containers Method 1. Label four glass containers for each temperature as “60°C (140°F),” “65°C (149°F),” “70°C (158°F),” and “75°C (167°F).” 2. Take two eggs and separate the whites from the yolks. 3. Place the egg whites into two separate glass containers labeled “60°C (140°F).” 4. Repeat Step 3 for the egg yolks. 5. Place the containers in a water bath (or a brazier). 6. Fill the water bath such that the water level is slightly above the level of the samples in the container. 7. Place a thermometer in each of the glass containers. Be sure that the thermometers do not touch the bottom or sides of the containers. The recorded temperatures will be the center temperature of the egg sample. 8. Heat the water in the water bath to 60°C (140°F), and keep the temperature constant. 9. Start timing when the center temperature of the sample reaches 60°C (140°F). 10. R  emove one of the containers with egg white and one with egg yolk from the water bath as soon as the temperatures of the samples reach 60°C (140°F). 11. Evaluate the samples as described in Data Table 5.3. Record your observations and evaluations. 12. After 20 min, remove the other containers from the water bath. 13. Evaluate the samples as described in Data Table 5.3. Record your observations and evaluations. Repeat the same experiment for the temperatures of 65°C (149°F), 70°C (158°F), and 75°C (167°F).

b

a

Texture of the Egg Whiteb t = 0 min

Appearance of the Egg Yolk t = 0 min Texture of the Egg Yolk t = 0 min

Appearance Texture of Appearance Texture of of the Egg the Egg of the Egg the Egg White White Yolk Yolk t = 20 min t = 20 min t = 20 min t = 20 min

Use the terms transparent, translucent, and opaque to describe the appearance of the samples. Use the terms liquid, semisolid, and solid to describe the texture of the samples both at the surface of the tube and at the center of glass custard cups.

75/167

70/158

65/149

60/140

Appearance of the Egg Temperature Whitea (˚C)/(°F) t = 0 min

DATA TABLE 5.3

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Study Questions 1. How does cooking temperature affect the physical properties of the egg white? 2. How does cooking time affect the physical properties of the egg white? 3. How does temperature affect the physical properties of the egg yolk? 4. How does cooking time affect the physical properties of the egg yolk?

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The Science Behind the Results Different proteins denature at different temperatures because they do not share the same molecular structure. The denaturation temperature of egg whites is different from the denaturation temperature of egg yolks. Egg whites begin to denature at 62°C (143°F) and become solid at 65°C (149°F). Egg yolks begin to denature at 65°C (149°F) and become solid at 70°C (158°F). Thus, controlled cooking is required if the mixture of the egg white and the egg yolk are going to be used in the food preparation, such as a soufflé. A temperature gradient occurs in the food during cooking. The surface of the food gets hotter than the interior at the beginning of cooking. If the temperature of the cooking medium is high (intensive heating), the temperature gradient between the surface of the food and the interior will be greater; the surface cooks rapidly and gets undesirably curdled by the time the interior part gets desirably cooked. Therefore, the rate and the intensity of heating must be optimized by considering the properties of the foods and should be controlled during cooking. For example, slow cooking of the whole egg in a 60°C (140°F) water bath for 45 min will yield a soft, semisolid, and translucent egg white and a liquid egg yolk.

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Proteins in Culinary Transformations  163

EXPERIMENT 5.3 OBJECTIVE To explain the effect of heat treatment on the textural properties of the meat. Ingredients and Equipment • 2 pieces of beef steaks (each 5 cm/1.9 in. thick) • 1 tablespoon olive oil • Grill • 2 plates • Food brush • Knife • 2 food thermometers Method 1. Preheat the grill to medium high heat (above 140°C/284°F). 2. Brush the steaks with olive oil. 3. Place the steaks on the grill. 4. Immediately place a thermometer in the center of each steak. 5. Observe the changes occurring in the texture of the samples during cooking. 6. Record the textural properties of the meats at the internal temperatures of 50°C (122°F), 60°C (140°F), and 65°C (149°F) in Data Table 5.4. DATA TABLE 5.4 Internal Temperature (˚C)/(˚F)

Textural Properties of the Samplesa

50/122 60/140 65/149 70/158 75/167 a

Use the terms soft, hard, juicy, dry, and separable to evaluate the textures of the samples.

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7. Take one of the steaks from the grill and place it on one of the plates while the other piece continues to cook. 8. Keep observing the textural changes taking place with the second sample as it continues to cook. 9. Record the textural properties of the meat at the internal temperatures of 70°C (158°F) and 75°C (167°F) in Data Table 5.4. 10. Take the second steak from the grill and place it on the second plate. 11. Slice and analyze both samples (the sample cooked to the internal temperature of 65°C (149°F) and the sample cooked to the internal temperature of 75°C (167°F)), and answer the following: 1. Which sample is softer? 2. Which sample is juicier? 3. Which sample has more separable muscle fibers? Note: In this experiment, for better observation, cook only one side of the steak (EXP 5.8).

EXP 5.8

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The Science Behind the Results Meat is composed of water, protein, fats, minerals, and trace amount of carbohydrates. Composition of lean, raw meat from different animal sources is given in Table 5.1. Meat has muscle fibers and connective tissues, which are the naturally occurring muscle proteins found in animals. Muscle fibers are bundles of the long and very thin cells surrounded by connective tissues. These bundles can be seen when the cooked meat is cut into pieces. They become dryer, tougher, and tear apart upon cooking. Collagen is the major protein that makes up the connective tissues that surround the muscle fibers. It has a very strong, long structure that makes it difficult to break down. Therefore, if meat has too much collagen, it will have a tougher texture. Textural changes are observed in meat during cooking as a result of denaturation of muscle proteins by heat. Upon cooking, the muscle protein structure unfolds and reassociation of protein molecules occurs. The muscle fibers shrink both in diameter and in length with heat, and the water is squeezed out during association of protein molecules. As cooking progresses, collagen becomes tender, the fibrous proteins get tougher, and more moisture is lost, as is observed in Experiment 5.3. Meat contains different proteins, which have different denaturation temperatures. Myosin, one of the contractile filamentous proteins, shrinks both in diameter and in length at 50°C (122°F). During cooking, myosin molecules bind to each TABLE 5.1 The Approximate Chemical Composition of Lean, Raw Meat from Different Animal Sources

Proportion (%) Beef

Lamb

Chicken

Pork

Water

73.0

71.0

75.0

75.0

Protein

22.0

21.0

22.8

22.8

Fat

3.9

7.0

0.9

1.2

Other

1.1

1.0

1.3

1.0

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other, thus squeezing some of the water out of the protein network. The meat becomes firmer and opaque. Raising the temperature between 62°C and 65°C (143°F and 149°F) causes shrinkage to occur more quickly, the meat suddenly releases lots of juice, and improvement of tenderness occurs due to denaturation of collagen. At temperatures of 70°C (158°F) and above, a more rapid shrinkage of ­collagen occurs. Prolonged heating at this temperature results in dissolution of collagen into gelatin, the muscle fibers become more easily separated, and, eventually, the meat gets more tender. Generally, meat (notably beef) heated to an internal temperature of 60°C (140°F) is considered rare; 62°C–68°C (143°F–154°F), medium rare; 68°C–75°C (154°F–167°F), medium; and above 75°C (167°F), well done.

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EXPERIMENT 5.4 OBJECTIVE To explain how to choose the best cooking method for different cuts of meat. Ingredients and Equipment • 2 pieces of identical tougher meat cuts, such as meat from the neck, flank, rump, or round • 2 pieces of identical tender meat cuts, such as meat from the loin, sirloin, or rib • 1 tablespoon olive oil • Water • Grill • Stove • 4 plates • 2 pots • 2 food thermometers Method 1. Label four plates as “tender cut cooked with dry cooking method,” “tougher cut cooked with dry cooking method,” “tender cut cooked with moist cooking method,” and “tougher cut cooked with moist cooking method.” 2. Preheat the grill to medium-high heat (above 140°C/284°F). 3. Brush one sample from each of the meat cuts with olive oil. 4. Place the meats on the grill. 5. Immediately place a thermometer in the center of each sample. 6. Cook the meat cuts on the grill until their internal temperatures read 65°C (149°F). 7. Remove the meats from the grill. 8. Place them on their respective plates, labeled “tender cut cooked with dry cooking method” and “tougher cut cooked with dry cooking method.” 9. Let the samples sit for 15 min. 10. While the samples are sitting, fill two pots with water, and place them on the stove. 11. When the water comes to a boil, place the remaining two meat cuts into separate pots. 12. Boil the samples for 20 min. 13. Remove the meat cuts from the pots. 14. Place them on their respective plates, labeled “tender cut cooked with moist cooking method” and “tougher cut cooked with moist cooking method.”

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DATA TABLE 5.5

Sample

Texture of the Meat Cut Cooked with Dry Heating Method (Grilled)a

Texture of the Meat Cut Cooked with Moist Heating Method (Boiled)

Tender meat cut Tougher meat cut a

Use the terms soft, hard, juicy, dry, and separable to evaluate the texture of the samples.

5. Let them sit for 15 min. 1 16. Analyze the texture of all the samples and record your observations in Data Table 5.5.

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The Science Behind the Results Muscle fibers get larger in size and muscles get stronger as an animal grows and exercises. As the muscle fibers get thicker and stronger, thicker and stronger connective tissues also are needed to bundle them. These are the major reasons why the meat from older animals, and the more active parts of animals, such as the arms and legs, are tougher compared to the meat cuts from the young and also less moving parts of the animal. The ideal cooking method for the given meat cut is directly related to the inherent tenderness of that cut. There are two basic methods of cooking meat: moist heating and dry heating. Tougher cuts need to be exposed to heat for longer durations because they have more connective tissues. However, as the heat treatment gets longer, more water is removed from the meat, and the meat gets hard and dry. Therefore, tougher cuts of meat require moist and long, slow cooking, such as stewing, braising, or steaming. Tender cuts, such as loin, sirloin, or rib, are usually cooked with a dry heat method, such as roasting, grilling, or broiling.

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EXPERIMENT 5.5 OBJECTIVES • To understand the effect of heat treatment on myoglobin. • To understand the contribution of protein to flavor, color, and textural attributes of foods. Ingredients and Material • 2 pieces of 5 cm (1.9 in.) thick identical beef steaks • 1 tablespoon olive oil • Water • Grill • Pot • Stove • 2 plates • Food brush Method 1. Preheat grill for medium-high heat (above 140°C/284°F). 2. Brush the first steak with olive oil. 3. Place on the grill. 4. Cook for 5–7 min per side. 5. Transfer the steak to one of the plates. 6. Analyze and record the color, texture, smell, and taste in Data EXP 5.9 Table 5.6. 7. Put the second beefsteak in the pot. 8. Set the pot on the stove and fill it with water. The water should completely cover the beef. 9. Bring the water to a boil over medium-high heat, and then turn the heat down. 10. Simmer the beef until tender. 11. Transfer it to the second plate. 12. Analyze and record the color, texture, smell, and taste of the sample in Data Table 5.6 (EXP 5.9).

Use the terms soft, hard, juicy, dry, and separable to evaluate the texture of the samples.

Use the terms fresh and nutty to evaluate the smell of the samples. Use the terms raw, bland, and nutty to evaluate the taste of the samples.

c

d

Use the terms pink, purple, red, and brown to evaluate the color of the samples.

b

Grilled Beef Steak

a

Tasted

Smellc

Texture of interior

Surface textureb

Color of interior

Color of surfacea

DATA TABLE 5.6 Boiled Beef Steak

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The Science Behind the Results Protein denaturation is one of the reasons for changes in meat color during cooking. Myoglobin is a protein that primarily gives the color to the meat muscles. Freshly cut meat has a deep purple color. The color changes to red when the meat is exposed to oxygen. Denaturation of myoglobin occurs when the meat is heated. As temperature is increased, the color of the meat changes from red to pink at 50°C (122°F) and from pink to brown at 60°C (140°F) as a result of myoglobin denaturation. Another reason for changes in meat color during cooking is the Maillard browning reaction. The Maillard browning reaction occurs between sugars and amino groups of proteins in foods, usually at higher temperatures (above 150°C/302°F). This reaction gives the toasty and nutty flavor and smell, and the golden brown color to bread crust, fried potatoes, cookies, cakes, roasted coffee beans, and biscuits. Most consumers appreciate these changes. The rate of the Maillard browning reaction increases as the temperature increases. Moisture slows or stops the Maillard browning reaction because the surface temperature of foods cannot exceed 100°C (212°F), especially in cooking methods where an excessive amount of water is present. That is the reason that foods roasted or grilled turn to golden brown, but foods cooked in boiling water never golden brown.

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EXPERIMENT 5.6 OBJECTIVES • To explain the effects of mechanical force on proteins in foods. • To show the foaming properties of some proteins. • To show the effects of sugar on foam formation. • To show the effects of fat on foam formation. Case 1 Meringue Recipe Ingredients and Equipment • 800 g (28 oz) egg whites • 200 g (7 oz) sugar • Salt • Tray • Parchment paper • Piping bag • Mixer • Oven Method 1. Preheat the oven to 110°C (230°F). 2. Line a tray with parchment paper. 3. Place 200 g (7 oz) of egg whites into a mixing bowl. 4. Add a pinch of salt. 5. Beat the egg whites at high speed for 3–4 min until they form soft peaks. 6. Add 100 g (3.5 oz) of sugar a spoonful at a time with the mixer running. 7. Continue to beat until the meringue forms stiff but moist peaks. 8. Observe the structure and record in both Data Tables 5.7 and 5.8. 9. Place a spoonful of the mixture onto trays (or fill a piping bag with the ­mixture and squeeze out round star shapes onto the tray). 10. Place the tray into the oven and reduce the temperature to 80°C (176°F). 11. Leave the oven on for 1½ h or until the meringues are crisp. 12. Turn the oven off and allow the meringues to cool in the oven for 3–4 h. 13. Observe the structure and record in Data Table 5.8. Case 2 1. Place 200 g (7 oz) of egg whites into a mixing bowl. 2. Add a pinch of salt. 3. Beat the egg whites at high speed.

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DATA TABLE 5.7 Case

Foam Structurea

1 2 3 4 a

Use the terms runny, soft, and stiff to evaluate the foam structure of the samples.

DATA TABLE 5.8 Case 1

Texturea

Before cooking After cooking a

Use the terms runny, soft, and stiff to evaluate the foam structure of the samples.

4. Continue beating for 10 more minutes after they form soft peaks. 5. Observe the structure and record in Data Table 5.7. Case 3 1. Place 200 g (7 oz) of egg whites into a mixing bowl. 2. Add a pinch of salt. 3. Add 100 g (3.5 oz) of sugar at once. 4. Beat the mixture at high speed for 3–4 min. 5. Observe the structure and record in Data Table 5.7. Case 4 1. Place 200 g (7 oz) of egg whites into a mixing bowl. 2. Add a pinch of salt. 3. Add one drop of egg yolk. 4. Beat the egg whites at high speed for 3–4 min. 5. Observe the structure and record in Data Table 5.7 (EXP 5.10–5.12).

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EXP 5.10

EXP 5.11

EXP 5.12

Study Questions 1. Explain what happens when the egg whites are whisked longer? 2. What is the effect of the added sugar on foam structure? 3. What is the effect of the added egg yolk on foam structure?

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The Science Behind the Results When raw egg whites, cream, and milk are beaten, their protein structures unfold and the air bubbles are incorporated into the network formed by reassociation of denatured protein molecules. This process is called foaming. Meringue cookies and sponge cakes are good examples of foamed-structure foods. The temperature of the egg and the environmental temperature during foam formation affect foam volume and stability. Egg whites foam faster at room temperature due to a lower surface tension. Heating or heat generation during mixing can weaken foams because heat affects the structures of proteins. Generally, whipping time increases foaming of proteins, but intensive whipping can reduce foam stability, because the bubbles get smaller with excess mechanical force. Overmixing breaks the protein bonds in the network causing air to escape and resulting in loss of volume in the foam. The presence of acid, salt, fat, and sugar affects the stability and ease of formation of foams. Sugar exerts a protective effect on proteins; therefore, the rate of protein denaturation decreases as the amount of sugar increases. The addition of sugar to egg whites may decrease the volume of the foam due to its protective effect, but, on the other hand, it increases the foam stability because sugar binds the excess water. Sugar should be added after the formation of foam for larger foams. Addition of salt influences the interactions due to ionic charges. A small amount of salt promotes foam formation because it helps unfold proteins during initial foaming, but increased amounts will decrease the foam’s stability by weakening the protein structure. Unlike salt, even a trace amount of fat, such as butter, oil, or even a drop of egg yolk will stop the foaming action of egg white proteins because fats compete with protein for special alignment with gas bubbles. Upon cooking of the egg foam, air and/or gas inside the network heat up and expand. The protein network surrounding the air bubbles solidifies due to heat (denaturation by heat), and the firm structure is set. The denaturation temperature is elevated if an egg mixture is diluted or if it is mixed with solids, such as sugars.

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Proteins in Culinary Transformations  181

EXPERIMENT 5.7 OBJECTIVE To explain effect of acidity (pH) on proteins in foods. Case 1 Yogurt Production Ingredients and Equipment • 250 mL (8.4 fl oz) pasteurized milk • Yogurt culture (you can use 1 tablespoonful existing fresh yogurt) • pH meter • 1 pot • 1 spoon • 1 food thermometer • 1 medium-sized jar or a container with a lid • Stove Method 1. Pour the milk into a pot and heat over medium heat, stirring often until it reaches 65°C (149°F). 2. Take the milk off the stove and set it aside to cool. 3. Cool the milk to 35°C (95°F). Pour the cooled milk into a jar. 4. Measure the pH of the milk and record in Data Table 5.9. 5. Evaluate the color and consistency of the milk and record in Data Table 5.9. 6. Add one spoonful of yogurt to provide yogurt culture. 7. Mix the milk and yogurt thoroughly. 8. Cover the jar tightly and place in an incubator or an environment at 35°C (95°F) and ferment for 24 h. 9. Slowly take the jar from the incubator, measure the pH, and evaluate the color and consistency of the content once every 2 h. 10. Record your observations in Data Table 5.9. Case 2 Protein Denaturation with Vinegar Ingredients and Equipment • 250 mL (8.4 fl oz) pasteurized milk • 30 mL (1 fl oz) white vinegar • pH meter • 1 pot

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DATA TABLE 5.9 Time (h)

pH

Describe the Color and the Consistency of the Mixture/Yogurta

0 2 4 6 8 10 12 14 16 18 20 22 a

• • • • • • • • •

Use the terms liquid, semisolid, and solid to evaluate the consistency of the samples.

1 small bowl 1 small plate 1 spoon A cheesecloth large enough to cover the top and 8 cm (3 in.) down the sides of the jar 1 food thermometer Rubber bands 1 medium-sized jar or a container with a lid Scale Stove

Proteins in Culinary Transformations  183

Method 1. Weigh the plate. Record in Data Table 5.10. 2. Pour the milk into a pot and heat over medium heat, stirring often until it reaches 95°C (203°F). 3. Take the milk off the stove and set it aside to cool. 4. Cool the milk to 25°C (77°F), and pour it into the jar. 5. Measure the pH of the milk and record in Data Table 5.10. 6. Evaluate the color and the consistency of the milk and record in Data Table 5.10. 7. Add vinegar to the warm milk and stir for 3 min. 8. Measure the pH of the mixture. 9. Leave the milk undisturbed for 5–10 min to allow for curd formation. 10. Cover the top of the jar with the cheesecloth and fix it in place using rubber bands. 11. Pour the liquid (whey) into the bowl and collect the curds in the cheesecloth. 12. Gather up the cheesecloth and squeeze out the excess whey into the bowl until almost dry. 13. Spread out the cheesecloth. 14. Place the curds on the plate and weigh the plate. 15. Calculate the approximate weight of the curds and record in Data Table 5.10. 16. Evaluate the color and texture of the curds and record in Data Table 5.10. Approximate weight of the curds (g/oz) = Weight of (plate + curds) (g/oz) – Weight of plate (g/oz)

   DATA TABLE 5.10 pH of the Milk

a

pH of the Milk + Vinegar

Weight of the Plate (g)/(oz)

Weight of the Plate + Curds (g)/(oz)

Use your own terms to evaluate the sample.

Approximate Weight of the Curds (g)/(oz)

Color and Consistency of the Milka

Color and Texture of the Curdsa

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Proteins in Culinary Transformations  185

The Science Behind the Results The approximate composition of dairy milk can be given as 88% water, 3.3% protein, 3.3% fat, 4.7% carbohydrate, and 0.7% minerals (primarily calcium). The primary group of milk proteins is casein. Direct (as in the vinegar experiment) or indirect (as in the yogurt experiment) addition of acids and bases disturbs the structure of caseins, and casein curds are formed. The addition of acids, bases, or certain types of salts into foods changes the net charge of the medium. Generally, changes in the net charge of the medium disturb the bridges (bonds) in the protein structure that are held together by ionic charges, causing protein denaturation. This reaction is the primary reason for casein curd formation in milk during yogurt and cheese manufacturing processes, and also in spoiled milk. Yogurt production is one of the most common examples used to explain the effect of acidity on protein coagulation. It is a fermented dairy product. Milk is inoculated with specific types of bacteria to make yogurt. As the bacteria grow, milk sugar (lactose) is converted into lactic acid; therefore, the pH of the medium decreases (this process is called fermentation). The structure of casein unfolds because the bridges in the protein structure held together by ionic charges are disrupted with the accumulation of positive charges (H+) in the medium. The milk proteins clump together due to the charge effects and the casein curds are formed. This is known as milk protein coagulation.

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Proteins in Culinary Transformations  187

EXPERIMENT 5.8 OBJECTIVE To explain the effect of aging on the chemical and physical properties of eggs. Ingredients and Equipment • 2 half-dozen cartons of eggs • 2 plates • 2 mixers • pH meter (if available) Method 1. Purchase the first carton of eggs. 2. Mark the carton “aged.” 3. Leave at room temperature for a week. 4. Purchase the second carton of eggs (preferably on the day of the experiment). 5. Mark the carton “fresh” and keep refrigerated until time for the experiment. 6. Take two plates and mark the first one “aged” and the second one “fresh.” 7. Take one egg from each carton. 8. Crack the eggs onto their respective plates. 9. Observe the physical states of the egg whites and the egg yolks. 10. Record your observations in Data Table 5.11. 11. Carefully measure and record the pH of the egg white and egg yolk of each egg. Complete the following steps for both the aged and the fresh eggs. 12. Crack and separate three eggs. 13. Place the egg whites into a ­mixing bowl. 14. Beat the egg whites for five minutes at high speed. 15. Record your observations in Data Table 5.12 (EXP 5.13).

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DATA TABLE 5.11 Fresh Egg

Aged Egg

Visual appearance of egg whitea Visual appearance of egg yolka pH of white pH of yolk a

Use the terms runny and viscous to evaluate the structure of the samples.

DATA TABLE 5.12 Fresh Egg Visual appearance before mixinga Visual appearance after mixinga Foam stability a

Use the terms runny, foamed, stiff, weak, and stable to evaluate the samples.

Aged Egg

Proteins in Culinary Transformations  189

The Science Behind the Results As the egg ages: 1. It loses moisture through the semipermeable eggshells. The water is replaced with air; therefore, the air space within the egg gets larger. The egg becomes lighter and easier to peel. 2. It loses carbon dioxide through the semipermeable eggshell. The pH of the albumen increases. The egg white becomes thinner and watery because its protein structure, which is held together by ionic charges, is disrupted with the increase in pH. It is for this reason that fresher eggs give larger and more stable foams compared to the aged eggs. The egg yolk becomes flatter and not centered because it is not protected anymore by the thick albumen. That is because pan-fried egg dishes prepared with fresher eggs have wellcentered egg yolks.

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Proteins in Culinary Transformations  191

EXPERIMENT 5.9 OBJECTIVES • To explain the gel formation properties of proteins. • To explain why some fruits prevent gelatin from solidifying. • To explain the effect of temperature on enzyme action. Ingredients and Equipment • 2 kiwi fruits • 1 envelope powdered animal-based gelatin (dessert gelatin) • Water • Ice • 2 pots • Large bowl • Colander • Skimmer • Paper towels • 3 dessert cups (glass) • Cutting board • Spoon • Knife • Stove • Refrigerator Method 1. Label the dessert cups “control,” “blanched,” and “fresh.” 2. Peel and wash the kiwis. 3. Boil a large pot of water. 4. Slice the kiwi on a cutting board. 5. Place half of the kiwi into boiling water. 6. Blanch for 2 min. 7. Immediately place the blanched kiwi into a bowl of ice water. 8. Drain the blanched fruits using a colander. 9. Place the blanched fruits on a paper towel and pat to dry. 10. Prepare the gelatin with water according to the directions given on the package. 11. Stir well with a spoon until all gelatin is dissolved. 12. Place equal amount of gelatin mixture into each dessert cup. 13. Do not add any fruits into the “control” cup. 14. Add blanched kiwi and fresh kiwi to the corresponding cups.

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DATA TABLE 5.13 Sample

Observationa

Control Gelatin dessert with fresh kiwi fruits Gelatin dessert with blanched kiwi fruits a

Use the terms liquid, semisolid, and solid to describe the texture of the sample.

5. Refrigerate all the samples overnight. 1 16. After 24 h, check the contents of each cup for solidification of the contents. 17. Record your observations in Data Table 5.13 (EXP 5.14–5.20).

EXP 5.14

EXP 5.15

Proteins in Culinary Transformations  193

EXP 5.16

EXP 5.17

EXP 5.18

EXP 5.19

EXP 5.20

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Proteins in Culinary Transformations  195

The Science Behind the Results Gelatin is a water-soluble protein prepared from collagen that is obtained from various animal sources. It is commonly used as a gelling agent in foods. Examples include desserts, ice cream, yogurt, and marshmallows. When collagen is heated in water long enough, its structure unwinds and the chains separate, and it dissolves to form gelatin. When gelatin cools, the protein strands form a network (cross-links) that can trap liquid, turning the liquid into solid gels. The strength of a gel depends on the following: • Concentration of the gelatin (protein): The higher the concentration of the protein, the stronger the gel due to increased protein–protein interactions. • Rate of cooling: The higher the rate of cooling, the weaker the gel. The slower the cooling, the stronger the gel. • Interactions with the other ingredients in the recipe, such as sugar, acidic ingredients, salt, or enzymes. For example, the addition of sugar weakens the gels because sugar decreases the protein–protein association, which is required for the formation of a network. The presence of acidic ingredients and salt affects the bridges in the protein structure held together by ionic charges; therefore, a weaker gel is formed. Proteolytic enzymes (proteases) break down the proteins into their peptides or amino acids by cleaving certain bonds in the protein structure (denaturation). Papaya contains papain, kiwi contains actinidin, pineapple contains bromelain, and the fig contains ficin enzymes. Therefore, using these fruits will prevent gelatin from solidifying. On the other hand, proteases also are proteins themselves. Heating of these fruits before adding them to the gelatin will allow the gelatin to solidify because heat denatures the enzymes in these fruits.

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Proteins in Culinary Transformations  197

EXPERIMENT 5.10 OBJECTIVES • To explain the synergistic effects of acids and enzymes on proteins. • To explain the synergistic effects of acids and heat on proteins. Case 1 Cheese making Ingredients and Equipment • 900 mL (30 fl oz) pasteurized milk • Yogurt culture (you can use 2 tablespoons of existing fresh yogurt) • pH meter • 1 pot • 1 spoon • Colander • 1 food thermometer • Rennet • Cheese-pressing frame or curd knife • Cheesecloth • Stove Method 1. Place the cheesecloth in the colander. 2. Pour 450 mL (15 fl oz) milk into the pot and heat it to 35°C (95°F). 3. Add one spoonful of yogurt to provide yogurt culture. 4. Mix the milk and yogurt thoroughly. 5. Cover the inoculated milk. 6. Let it sit in a warm place while occasionally measuring the pH of the milk. 7. Add rennet when the pH of the medium reaches close to 5 (follow the directions provided by the producers for the amount of rennet). This may take as long as 8 h. 8. Cover and leave it undisturbed for 2–3 h. 9. Put your finger into the curd and lift. If the action of rennet is complete, the curd should break cleanly away and the whey should pool in the hole that has been left behind. If not, wait a little longer. 10. Cut the curds into 2–3 mm (0.7–0.11 in.) squares. First cut parallel straight lines 2–3 mm (0.7–0.11 in.) apart, rotate the pot 90 degrees and cut parallel straight lines 2–3 mm (0.7–0.11 in.) apart perpendicular to the first lines. The knife or the frame must touch the bottom of the pot. 11. Place the pot over a very low heat.

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DATA TABLE 5.14 Sample

Textural Propertiesa

Product prepared from milk pasteurized at 35°C (95°F) Product prepared from milk pasteurized at 78°C (172°F) a

Use the terms runny, semisolid, and solid to describe the texture of the sample.

2. Stir curds gently and cut larger curds. 1 13. When the curds are firm enough, slowly pour them into the colander lined with cheesecloth. 14. Cover the curds with the corners of the cheesecloth and lift and press the curds to remove excess whey. 15. Place it back into the colander and put a heavy weight on top. 16. Let it sit until all excess whey has been removed. Repeat the same experiment, but heat the milk to 78°C (172°F) for 15 min at Step 2.

Study Questions 1. Explain your observations for the changes in the milk with the actions of the starter culture. 2. Explain your observations for the changes in the milk after the addition of rennet. 3. Compare the textural properties of the final products prepared from the milk pasteurized at 35°C (95°F) and the milk pasteurized at 78°C (172°F) (use Data Table 5.14). Case 2 Poached Egg • Ingredients and Equipment • 4 eggs • 1,400 mL (47.4 fl oz) water • 2 tablespoons vinegar • 2 tablespoons salt • Pot • Skimmer • Stove

Proteins in Culinary Transformations  199

Method 1. Fill a medium pot with 700 mL (23.7 fl oz) of water. 2. Add vinegar and salt. 3. Move the pot to the stove over low heat. 4. Bring to a simmer. 5. Using skimmer, stir the water with a clockwise motion (form a vortex). 6. Gently crack two eggs into the center of the vortex, letting the water swirl around eggs. 7. Describe the shape and texture of the egg white and record in Data Table 5.15. 8. Let the eggs cook for 4 min. 9. Use a skimmer to take the eggs out of the water. 10. Feel the white for firmness and record your observations in Data Table 5.15. Repeat the same experiment without vinegar and compare the results.

DATA TABLE 5.15 Texture of the Egg White Cracked into the Watera

Texture of the Cooked Egg Whitea

With vinegar Without vinegar a

Use the terms set, separated, liquid, semisolid, and solid to describe the texture.

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The Science Behind the Results The basic principle behind natural cheese production is coagulation of milk protein with synergistic effects of acids and enzymes. Rennet is a proteolytic enzyme obtained from the mammalian stomach. It coagulates casein in milk by cleaving particular bonds in the protein structure, causing milk to solidify. Rennet requires specific temperatures and pH to coagulate milk. Mild heat speeds up the enzyme reaction. The optimum temperature for milk coagulation with rennet is 35°C–42°C (95°F–107°F). The optimum pH for rennet activity is around 5.8. This is one of the reasons for inoculating specific types of bacteria into milk during natural cheese production. As the bacteria grow, milk sugar (lactose) is converted into lactic acid. Therefore, the pH of the medium decreases (as in Experiment 5.7) and coagulation starts. When the pH of the milk goes down to 5.8, which is optimum for rennet activity, rennet is added in milk for further coagulation. In poached egg process, heat and acid work together to denature the proteins. Acid addition (such as vinegar) in the poaching liquid helps to speed up protein coagulation so that the egg white stays solid and compact. Note: Milk is not supposed to be heated above 65°C (149°F) to make cheese and yogurt because high temperature denatures the milk protein and coagulation will not occur.

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Proteins in Culinary Transformations  203

EXPERIMENT 5.11 OBJECTIVES • To demonstrate the properties of wheat protein. • To demonstrate the functional properties of gluten in baking. Case 1 Gluten washing Ingredients and Equipment • 2 cups hard wheat flour (13% protein) • Water • Large bowl • Sink Method 1. Mix water with the wheat flour. 2. Knead to make the dough structure homogeneous. 3. Wash the dough under slow running cold water while kneading it throughout the process with your fingers. 4. Continue washing until the water is clear and no longer a “milky” white. 5. Squeeze the resultant ball with your fingers to remove excessive water. 6. Stretch to observe the elasticity (EXP 5.21–5.24).

EXP 5.21

EXP 5.22

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EXP 5.23

Case 2 Bread recipe Ingredients and Equipment • 565 g (20 oz) bread flour (13%–13.5% gluten) • 565 g (20 oz) g cake flour (7.5%–8% gluten) • 2 teaspoons baker’s yeast • 680 g (24 oz) water • Tray • 2 bowls • Oven • Stretch film • Baking paper • 2 kitchen towels • Knife • Wire rack Method 1. Line a tray with baking paper. 2. Label the bowls “bread flour” and “cake flour.”

EXP 5.24

Proteins in Culinary Transformations  205

3. In the bowl labeled “bread flour,” mix the bread flour and 1 teaspoon baker’s yeast with your hands. 4. Slowly pour 340 g (12 oz) water while kneading the dough. 5. Sprinkle the counter with flour. 6. Place the dough over the counter, and knead it well until it’s soft and uniform in texture, and not sticky. 7. Form the dough into a ball shape. 8. Place it back into the same bowl and cover with stretch film. 9. Complete the same procedure with the cake flour. Keep the time between two processes at a minimum. 10. Let them rise in a warm place until doubled in size (takes about an hour). 11. Remove the dough prepared with the bread flour from the bowl and knead gently a little more. 12. Form the dough in a loaf shape. 13. Divide the tray into two sections and place the dough on the left half of the tray. The placement of the dough on the baking tray must be recorded to remember. 14. Complete stages 11 and 12 with the dough prepared with the cake flour. 15. Place the dough on the right half of the tray. The placement of the dough on the baking tray must be recorded to remember. 16. Cover them both with kitchen towels and let stand for about 30 min. 17. Meanwhile, set the oven to 200°C (392°F). 18. Bake the breads for about 40 min until the top is golden brown. 19. Cool them down on a wire rack. 20. Cut one slice from each bread, and record your observations in Data Table 5.16. Compare your results. DATA TABLE 5.16 Flour Type

Colora

Bread flour Cake flour a b

Use the terms bright golden, dark golden, and brown to describe the color. Use the terms soft, chewy, crumbly, and hard to describe the chewiness.

Chewinessb

206  COOKING AS A CHEMICAL REACTION

Case 3 Chocolate Chip Cookie recipe Ingredients and Equipment • 226 g (8 oz) cookie flour (with 7.5%–9% gluten) • 226 g (8 oz) bread flour (with 13%–13.5% gluten) • 226 g (8 oz) softened butter • 226 g (8 oz) sugar • 226 g (8 oz) brown sugar • 2 eggs • 1 teaspoon baking soda • 1.5 teaspoon salt • 1 tablespoon vanilla extract • 2 cups semi-sweet chocolate chips • Baking paper • 2 bowls • Scale • Mixer • Tray • Oven • Scoop • Wire rack Method 1. Preheat the oven to 190°C (374°F). 2. Line a tray with baking paper. 3. Label the bowls “bread flour” and “cookie flour.” 4. In the bowl labeled “cookie flour,” mix together 226 g (8oz) flour, ½ teaspoon baking soda, and ¾ teaspoon salt. 5. In a separate bowl, stir cream 113 g (4 oz) butter, 113 g (4 oz) white sugar, and 113 g (4 oz) brown sugar on medium speed, until the mixture is smooth, for about 5 min. 6. Combine 1 egg and ½ tablespoon vanilla extract, add to the butter-sugar mixture, and mix thoroughly. 7. On low speed, stir in the dry ingredients and 1 cup chocolate chips until just incorporated. 8. Divide the tray into two sections. Scoop out 1 tablespoon of dough, and place it on the right half of the tray. The placement of the dough on the baking tray must be recorded to remember. 9. Complete stages from 3 to 7 with the bread flour. 10. Scoop out 1 tablespoon of dough, and place it on the left half of the tray. The placement of the dough on the baking tray must be recorded to remember.

Proteins in Culinary Transformations  207

11. Bake the cookies at 190°C (374°F) for 8–10 min, or until the cookies are set. 2. Cool them down on a wire rack. 1 13. Cut one cookie in half from each batch, and record your observations in Data Table 5.17. Compare your results. DATA TABLE 5.17 Flour Type

Colora

Bread flour Cookie flour a b

Use the terms bright golden, dark golden, and brown to describe the color. Use the terms soft, chewy, crumbly, and hard to describe the chewiness.

Chewinessb

208  COOKING AS A CHEMICAL REACTION

Proteins in Culinary Transformations  209

The Science Behind the Results To understand the results, we need to begin by understanding the structure of the protein in flour. Gluten is the protein that gives the elastic structure of dough. Gliadin and glutenin are the two major protein components in wheat flour. Glutenins are large proteins that provide the elasticity and strength of the dough. Gliadins are smaller compared to glutenins; they ensure the elastic structure of the dough because they are more mobile due to their small size. When water and flour are mixed, gliadin and glutenin combine to form weak gluten networks. Kneading of the dough separates these disorganized weaker protein networks, and makes the protein strands align and bind into smoother, stronger, and more organized gluten networks. Dough becomes stronger and stretchable. Inefficient kneading or excessive kneading may affect the strength of the dough because kneading ensures bond formation between the protein strands. Other ingredients in the recipe may affect gluten development. For example, sugar and fat interfere with gluten development and the end product becomes more tender. Acidic ingredients weaken the gluten network because they change the net charge on the protein strands. On the other hand, addition of a limited amount of salt helps in formation of the gluten network. In baking, carbon dioxide gas is produced by yeast or leavening agents. The carbon dioxide is trapped in the gluten network. During baking, the gas entrapped in this network expands so that the volume of the dough increases with increased temperature. As baking continues, the heat denatures the gluten proteins; a rigid framework of the baked product with small holes inside is formed. Most of the cereal grains, such as wheat, oats, and barley, contain gluten, but the amounts and the structures of their glutens are different. Different kinds of wheat flours are made from different kinds of wheat. Bread flour is milled from hard wheat and has high gluten content (12%–14% gluten content). Cake flour is milled from soft wheat, and its gluten content is less than hard wheat flour (6%–8% gluten content). The more gluten in the flour, the stronger and more elastic the dough becomes. The stronger structure holds more gas and expands more easily. Note: In the experiment 5.11 in Case 1, the milky white liquid that is obtained during washing contains starch granules and other water-soluble constituents.

210  COOKING AS A CHEMICAL REACTION

POINTS TO REMEMBER The primary functional properties of proteins in food processing include: • Foam formation • Gelation • Dough formation • Flavor • Viscosity control • Water binding • Color formation Understanding the basic structure of food proteins is crucial for chefs because the functional properties of proteins in foods are primarily related to their structures. Proteins have four structures held together by different types of bonds. The secondary, tertiary, and quaternary structures of proteins can be disrupted or destroyed (protein denaturation) during food processing by: • Heat • Mechanical force • Salt • pH • Enzymes Different proteins may give different reactions during food processing, because they do not share the same molecular structures. Presence or lack of the other ingredients in the recipe may affect the functional properties of the proteins.

Proteins in Culinary Transformations  211

More Ideas to Try 1. What do you expect to observe if you repeat Experiment 5.6, Case 1, adding cream of tartar (acidic ingredient) at the fifth stage? 2. The most common application of using combined effects of acids, salts, and enzymes on protein denaturation in food preparation is meat marination. Marination enriches the flavor, increases the water retention, and improves the tenderness of the meat. The major ingredients used in marinades include salt solutions (2%); acidic ingredients, such as vinegar; and the enzymes, such as papain, bromelain, or enzyme-containing ingredients, such as onion. Design an experiment to show the effects of marination on the textural properties of the meat. 3. Repeat Experiment 5.11 using cookie flour, and compare your observations.

Study Questions 1. Old eggs produce poor foam stability. What is the reason for this? 2. Papain and bromelain are often used as meat tenderizers. Why? 3. What is the function of acidic ingredients in meat marination? 4. Mechanical methods such as pounding or piercing the meat, are sometimes applied to tenderize meat. What is the purpose for it?

SELECTED REFERENCES Bradley, F. A., and A. J. King. 2004. Egg basics for the consumer: Packaging, storage, and nutritional information. Hollister, CA: University of California, Division of Agriculture and Natural Resources. Online at: http://anrcatalog.ucdavis.edu/pdf/8154.pdf. Chemistry in the meat industry resources. New Zealand Institute of Chemistry. Online at: http://nzic.org.nz/ChemProcesses/animal/5A.pdf. Crosby, G. and editors. 2012. The science of good cooking: Master 50 simple concepts to enjoy a lifetime of success in the kitchen. Brookline, MA: America’s Test Kitchen; Cook’s Illustrated. Denaturation of protein. 2003. Elmhurst College. Online at: www.elmhurst.edu/~chm/ vchembook/568­denaturation.html. Food and Agriculture Organization. Meat, fat and other edible carcass parts resources. Rome: FAO. Online at: www.fao.org/docrep/010/ai407e/AI407E03.htm. Forrest, J. C., E. D. Aberle, H. B. Hedrick, and R. A. Merkel. 1975. Principles of meat­science. New York: W. H. Freeman and Company. Gilbert, P. Baking Basics. Online at: www.extension.iastate.edu/sites/www.extension. iastate.edu/files/guthrie/ExtCon091504.pdf.

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Guerrero-Legarreta, I., and Y. H. Hui. 2010. Handbook of poultry science and technology, secondary processing. Hoboken, NJ: John Wiley & Sons. Gustavo, G. 2009. Meat color and pH. Online at: www.docstoc.com/docs/122703427/ Meat-color-and-pH. Kurt, L., and S. Ozilgen. 2013. Failure mode and effect analysis for dairy product manufacturing: Practical safety improvement action plan with cases from Turkey. Safety Science 55:195–206. Lomakina, K., and K. Mikova. 2006. A study of the factors affecting the foaming properties of egg white—A review. Czech Journal of Food Sciences 24(3):110–118. Martens, H., E. Stabursvik, and M. Martens. 1982. Texture and colour changes in meat during cooking related to thermal denaturation of muscle proteins. Journal of Texture Studies 13(3):291–309. McGee, H. 2004. On food and cooking, 1st rev ed. New York: Scribner. Moore Family Center for Whole Grain Foods, Nutrition and Preventive Health. 2012. Corvallis, OR: Oregon State University, College of Public Health and Human Sciences. Online at: http://food.oregonstate.edu/learn. Raikos, V., L. Campbell, and S. R. Euston. 2007. Effects of sucrose and sodium chloride on foaming properties of egg white proteins. Food Research International 40(3):347–355. Ruhlman, M. 2010. Ratio: The simple codes behind the craft of everyday cooking. New York: Scribner. This, H. 2007. Kitchen mysteries: Revealing the science of cooking. New York: Columbia University Press. Vaclavic, V. A., and E. W. Christian. 2008. Essentials in food science, 3rd ed. Berlin: Springer.

Chapter 6 Fats and Oils in Culinary Transformations

FUNCTIONAL PROPERTIES OF FATS AND OILS IN CULINARY PROCESSES Fats and oils are the organic compounds naturally found in animals and plants (Figure 6.1). The primary functional properties of fats and oils in food processing can be listed as: 1. Enhancement of flavor and mouth feel: Fats and oils provide a rich flavor and smoother mouth feel that most people find very satisfying. 2. Development of texture: Fats and oils make foods softer and easier to chew. 3. Shortening: In baking, fats and oils physically separate water and gluten ­molecules and provide a flaky, tender, or crumbly texture. 4. Emulsification: Fats and oils are the primary components of most emulsions, such as mayonnaise, certain salad dressings, sauces, and gravies. 5. Medium for transferring heat: During cooking, fats and oils transfer heat to the foods. 6. Development of appearance: Fats and oils provide a creamy, moist, fluffy, and shiny appearance to foods.

213

214  COOKING AS A CHEMICAL REACTION

Fats

Oils

Animal originated

Plant originated

Animal originated

Plant originated

The most common examples:

The most common examples:

The most common examples:

The most common examples:

• Butter • Cream • Lard

• Cocoa butter • Palm oil

• Fish oil

• Vegetable oils • Flax seed oil • Rapeseed oil • Olive oil

Figure 6.1  The most common food sources of fats and oils.

Selection of fats and oils in food processes is a critical factor because not every fat or oil is suitable for all food processing operations. For example, as butter gives tender texture in pastry and a smooth mouth feel in breads, sunflower oil is ­primarily used in frying processes. Understanding the basic structure of fats and oils is crucial for chefs because their functional properties are primarily related to their structures.

FAT AND OIL STRUCTURE Fats and oils belong to a group of substances called lipids. They are made up of carbon (C), hydrogen (H), and oxygen (O). Lipids are most commonly the triglycerides, which are composed of one molecule of glycerol bonded to three fatty acid molecules (Figure 6.2). All lipids are hydrophobic. This means that lipids are insoluble in water. In the figure, the R groups of fatty acids represent the long hydrocarbon chains. The hydrocarbon chains of the fatty acids can be either saturated (has no carbon–carbon double bond, C=C), monounsaturated (has one C=C), or polyunsaturated (has more than one C=C) (Examples are given in Figure 6.3). Naturally occurring fats and oils are the mixtures of different fatty acids in varying proportions. The degree of saturation of different lipids depends upon their various fatty acids content (Table 6.1). The physical and chemical properties of any specific lipid primarily depend on the chemical structure of the fatty acids, such as the length of the hydrocarbon chains and degree of unsaturation that constitute it (Table 6.2).

Fats and Oils in Culinary Transformations  215

H H

H

HO C

OH

C

R1

H

C

O C

O

R1

O O

HO H

C

OH

C

R2

H

C

O

R2 + 3H2O

C

O

Water O

HO H

C

C

OH

R3

H

O

H Glycerol

C

C

O

R3

H Triglyceride

Fatty acids

Figure 6.2  Formation of fats and oils. R Group H

H

H

H

H

H

C

C

C

C

C

C

H

H

H

H

H

H

H

H

C

C

H H H Monounsaturated

H

HO C O

H

Saturated

Carboxyl Group

H

H

C

C

H

H

H

H

C

C

H

H

HO C O

C

C

H

H

H

HO C O

C

C

C

C

H H H Polyunsaturated

H

H

H

C

C

Figure 6.3  Examples for R groups of fatty acids.

H

216  COOKING AS A CHEMICAL REACTION

TABLE 6.1 Examples for the Saturated, Monounsaturated, and Polyunsaturated Lipids

Degree of Saturation

Example

Saturated

Butter, lard, coconut oil, tallow

Monounsaturated (MUFA)

Olive oil, rapeseed oil, canola oil, peanut oil

Polyunsaturated (PUFA)

Corn oil, cotton seed, soy bean oil

TABLE 6.2 Physical and Chemical Properties of Saturated and Unsaturated Lipids

Saturated Lipids • Hydrocarbon chains do not have double bonds between carbons. • They are stable, and they have longer shelf life compared to unsaturated lipids. • Their melting point is high compared to unsaturated lipids. • They are solid at room temperature. • They are mostly animal originated. • Generally speaking they are known as fats.

Unsaturated Lipids • One or more double bonds exist in the hydrocarbon chain: • Monounsaturated Fatty Acids have one double bond. • Polyunsaturated Fatty Acids have two or more double bonds. • They are less stable, and easily undergo oxidation compared to saturated lipids. • They have lower melting points compared to saturated lipids. • They are liquid at room temperature. • They are mostly plant originated. • Generally speaking, they are known as oils.

Fats and Oils in Culinary Transformations  217

EXPERIMENT 6.1 OBJECTIVE To explain smoke points of different types of lipids. Ingredients and Equipment • 50 mL (1.7 fl oz) pure olive oil • 50 mL (1.7 fl oz) pure corn oil • 50 mL (1.7 fl oz) pure peanut oil • 50 mL (1.7 fl oz) melted clarified butter • 50 mL (1.7 fl oz) melted lard • 50 mL (1.7 fl oz) soybean oil • 6 frying pans • Liquid measuring cup • Food thermometer • Stove Method 1. Pour the olive oil into a frying pan. 2. Place the thermometer in the frying pan. Position the thermometer so that the bulb is in the center of the oil and not touching the sides or bottom of the pan. 3. Heat the oil over medium-high heat until visible smoke appears from the surface of the oil. 4. Record the temperature, at which the visible smoke appears from the surface of the oil in Data Table 6.1. Complete the same procedure for each of the fats and oils. DATA TABLE 6.1 Type of Lipid Olive oil Corn oil Soybean oil Peanut oil Clarified butter Lard

Smoke Point (°C)

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Fats and Oils in Culinary Transformations  219

The Science Behind the Results In cooking, selecting a correct type of lipid is very important because each lipid performs best within a certain range of temperature. The melting point of lipids can be defined as the temperature at which a solid fat becomes a liquid. Different types of lipids have different melting point temperatures due to the differences in their chemical structures (Table 6.3). As a general rule: • The lipids with longer hydrocarbon chains have higher melting points. In other words, the melting point increases as the molecular weight of lipids increases. • The unsaturated lipids tend to have lower melting points than saturated lipids of the same length. In other words, the melting point of lipids decreases as the degree of unsaturation in fatty acids increases. For example, butter has a higher melting point compared to olive oil (Table 6.3). The structure of saturated fatty acids is relatively linear compared to the structure of unsaturated fatty acids. This structure allows fatty acid molecules to get closely stacked together. Therefore, close intermolecular interactions result in relatively high melting points. On the other hand, double bonds in unsaturated fatty acids result in TABLE 6.3 Examples for R groups of fatty acids

Types of Lipid

Melting Temperature (°C/°F)

Corn oil

−11/12.2

Coconut oil

24/75.2

Olive oil

−6/21.2

Sunflower oil

−17/1.4

Soybean oil

−20/−4

Peanut oil

−2/28.4

Butter

36/96.8

Lard

33/91.4

220  COOKING AS A CHEMICAL REACTION

bends in their molecular structures. Therefore, these molecules do not stack very well. The intermolecular interactions are much weaker than saturated molecules. This is because the melting points are much lower in unsaturated fatty acids compared to the saturated fatty acids of the same length. The smoke point of the lipid is one of the main properties to consider when selecting a lipid for high-temperature cooking. Smoke point is the temperature at which lipids give off a visible bluish smoke. Different types of lipids have different smoke points and smoke point temperatures due to the differences in their chemical structures. It is important to know the smoke point temperature of various lipids because the structure of lipids begins to decompose at that temperature (Table 6.4). The chemical decomposition results in off-flavor development, nutritional loss, and generation of harmful cancer-causing chemical components. As a general rule: • The saturated fatty acids have lower smoke points than unsaturated fatty acids. For example, butter gives off a visible bluish smoke at lower temperatures compared to sunflower oil. Therefore, sunflower oil is a better choice for deep-frying processes than butter. TABLE 6.4 Smoke Points of Selected Fats and Oils

Types of Lipid

Smoke Temperature (°C/°F)

Refined corn oil

232/449.6

Extra virgin olive oil

191/375.8

Refined sunflower oil

227/440.6

Refined soybean oil

238/460.4

Peanut oil

225/437.4

Refined canola oil

204/399.2

Coconut oil Lard Butter

175/347 188/370.4 121–149/249.8–300.2

Fats and Oils in Culinary Transformations  221

• The smoke point decreases as the free fatty acid content increases. Prolonged or repeated heating decomposes the structure and produces free fatty acids. That is one of the reasons not to use the same oil for repeated frying processes. • The smoke point also depends on the purity of the lipids. For example, refined oils have higher smoke points compared to the unrefined ones. The presence of food particles also reduces the smoke point.

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Fats and Oils in Culinary Transformations  223

EXPERIMENT 6.2 OBJECTIVES • To explain the importance of selecting a correct type of lipid for cooking. • To explain that not every fat is suitable for high-temperature processes. Ingredients and Equipment • 1 kg (2.2 lb) peeled potatoes • 1 L (4¼ cups) corn oil • 1 L (4¼ cups) peanut oil • 1 L (4¼ cups) canola • 1 L (4¼ cups) soybean oil • 1 L (4¼ cups) sunflower oil • Water • 5 plates • 5 medium-sized bowls • Knife • Skimmer • Colander • Measuring cup • Paper towels • 5 medium-sized frying pans • Food thermometer • Kitchen scale • Stove Method 1. Label the plates “corn oil,” “peanut oil,” “canola oil, “soybean oil,” and “sunflower oil,” and line them with paper towels. 2. Slice the potatoes. Make sure slices are close to the same thickness. 3. Weigh and portion the potatoes equally into 5 bowls, so that each bowl will have 200 g (7 oz) of potatoes. 4. Cover the potatoes with water. 5. Set the bowls aside for 30 min. 6. Place 1 L (4¼ cups) corn oil into a frying pan; 5–6 cm (2–2.3 in.) of space between the top of the oil and the top of the pan is required because the oil will bubble up when potatoes are added. 7. Drain the potatoes in the first bowl and thoroughly pat dry. 8. Place the pan on the stove. 9. Heat the oil to 200°C (392°F). 10. Put the potatoes into the hot oil.

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DATA TABLE 6.2 Sample

Colora

Crunchinessb

Tasteb

Smellb

Corn oil Peanut oil Canola Soybean oil Sunflower oil a b



Use the terms light yellow, golden, and dark yellow to describe the color. Evaluate for crunchiness, taste, and smell, and rate the sample on a scale of 1–5: 1 = very soft, 5 = very crunchy. 1 = less tasty, 5 = very tasty. 1 = strong odor, 5 = acceptable nutty smell.

11. Fry the potatoes until they are golden brown. 2. Transfer the potatoes onto the plate labeled “corn oil.” 1 13. Wait for one minute and remove the paper towel from the plate. 14. Evaluate the color, appearance, taste, and crunchiness of the fried potatoes and record in Data Table 6.2. Complete the same procedure for the other oil types. Do not forget to use a separate frying pan for each type of oil.

Fats and Oils in Culinary Transformations  225

EXPERIMENT 6.3 OBJECTIVE To explain the factors affecting the sensory properties of fried foods. Ingredients and Equipment • 850 g (1.87 lb) peeled potatoes • 4 L (17 cups) corn oil • Water • Ice • 4 plates • 4 medium-sized bowls • Knife • Colander • Measuring cup • Paper towels • 1 medium-sized frying pan • Food thermometer • Kitchen scale • Stove Method 1. Slice the potatoes. Make sure slices are close to the same thickness. 2. Weigh and portion 600 g (1.3 lb) of the potatoes equally into three bowls so each bowl will have 200 g (7 oz) of potatoes. 3. Cover the potatoes in the first two bowls with water and cover the potatoes in the third bowl with ice. 4. Place 250 g (8.8 oz) of the potatoes into the last bowl and cover them with water. 5. Set the bowls aside for 30 min. Case 1 1. Label one of the plates “control” and line it with a paper towel. 2. Drain the potatoes in the first bowl, and thoroughly pat dry. 3. Place 1 L (4¼ cups) corn oil into a frying pan; 5–6 cm (2–2.3 in.) of space between the top of the

Control EXP 6.1

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oil and the top of the pan is required, because the oil will bubble up when potatoes are added. 4. Place the pan on the stove. 5. Heat the oil to 185°C (365°F). 6. Put the potatoes into the hot oil. 7. Measure the temperature of the oil. 8. Measure the time it takes to heat the oil back to 185°C (365°F). 9. Record your measurements in Data Table 6.3. 10. Fry the potatoes until they are golden brown. 11. Transfer the potatoes onto the plate. 12. Wait for one minute and remove the paper towel from the plate. 13. Evaluate the color, oil absorption, and crunchiness of the fried potatoes and record in Data Table 6.4. DATA TABLE 6.3

Sample

Initial Temperature of Frying Oil (°C/°F)

Control

185/365

Cold oil

Room Temperature

Cold potatoes

185/365

More initial potato load

185/365

Temperature of Oil Just After Placing Potatoes in Oil (°C/°F)

Time It Takes to Heat Oil Back to 185°C (365°F) (min)

Fats and Oils in Culinary Transformations  227

DATA TABLE 6.4 Sample

Colora

Crunchiness

Oil Absorption

Control Cold oil Cold potatoes More initial potato load a



 se the terms light yellow, golden, and dark yellow to describe the color of the samples. Evaluate U for crunchiness and oil absorption, and rate the sample on a scale of 1–5: 1 = very soft, 5 = very crunchy. 1 = very oily, 5 = less oil absorption.

Case 2 1. Label one of the plates “cold oil” and line it with paper towels. 2. Drain the potatoes in the second bowl, and thoroughly pat dry. 3. Place 1 L (4¼ cups) corn oil into a frying pan; 5–6 cm (2–2.3 in.) of space between the top of the oil and the top of the pan is required because the oil will bubble up when potatoes are added. 4. Put the potatoes into the cold oil. 5. Place the pan on the stove. 6. Measure the temperature of the oil. 7. Measure the time it takes to heat the oil to 185°C (365°F). 8. Record your measurements in Data Table 6.3. 9. Fry the potatoes until they are golden brown. 10. Transfer the potatoes onto the plate. 11. Wait for one minute and remove the paper towel from the plate. 12. Evaluate the color, oil absorption, and crunchiness of the fried potatoes and record in Data Table 6.4.

Cold Oil EXP 6.2

Cold Potatoes EXP 6.3

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Case 3 1. Label one of the plates “cold potatoes” and line it with paper towels. 2. Drain the potatoes in the third bowl, which has been covered with ice water, and thoroughly pat dry. 3. Place 1 L (4¼ cups) corn oil into a frying pan; 5–6 cm (2–2.3 in.) of space between the top of the oil and the top of the pan is required because the oil will bubble up when potatoes are added. 4. Place the pan on the stove. 5. Heat the oil to 185°C (365°F). 6. Put the potatoes into the hot oil. 7. Measure the temperature of the oil. 8. Measure the time it takes to heat the oil back to 185°C (365°F). 9. Record your measurements in Data Table 6.3. 10. Fry the potatoes until they are golden brown. 11. Transfer the potatoes onto the plate. 12. Wait for one minute and remove the paper towel from the plate. 13. Evaluate the color, oil absorption, and crunchiness of the fried potatoes and record in Data Table 6.4. Case 4 1. Label one of the plates “more potatoes” and line it with paper towels. 2. Drain the potatoes in the last bowl, and thoroughly pat dry. 3. Place 1 L (4¼ cups) corn oil into a frying pan; 5–6 cm (2–2.3 in.) of space between the top of the oil and the top of the pan is required because the oil will bubble up when potatoes are More added. Potatoes 4. Place the pan on the stove. 5. Heat the oil to 185°C (365°F). EXP 6.4 6. Put potatoes into the hot oil. 7. Measure the temperature of the oil. 8. Measure the time it takes to heat the oil back to 185°C (365°F). 9. Record your measurements in Data Table 6.3. 10. Fry the potatoes until they are golden brown. 11. Transfer the potatoes onto the plate. 12. Wait for one minute and remove the paper towel from the plate. 13. Evaluate the color, oil absorption, and crunchiness of the fried potatoes and record in Data Table 6.4 (EXP 6.1–EXP 6.4).

Fats and Oils in Culinary Transformations  229

The Science Behind the Results Frying is cooking of foods in hot oils or fats. It is a popular method used in the food industry because it is possible to heat fats and oils to very high temperatures, well above the boiling point of water. The optimal temperature for frying most foods lies between 180° and 190°C (356°F and 374°F). During frying, several physical and chemical changes occur both in foods and lipids. These changes affect the sensory properties of the end products. The sensory properties of fried foods are primarily influenced by: • The type of lipids used: During frying, a number of deteriorating chemical changes, such as structural decomposition of lipids, occur. The degree of these deteriorating chemical changes primarily depends on the chemical structure of the lipids that are used in the process. Ideal frying lipids have a high smoke point that is well above the optimal frying temperature (Table 6.4). • The temperature of lipids: The frying process involves simultaneous heat and mass transfer. During food frying, when food is immersed in the hot lipid, heat is immediately transferred from the lipid to the product. As the food heats up, water vapor escapes from the food surface into the hot lipid, which results in the crust formation on the surface of the food. As frying proceeds, the crust becomes thicker and firmer. It acts as a barrier to the release of water vapor from the food, and also decreases the lipid absorption. Food becomes juicy inside and crispier outside. If the temperature of the lipid is below the optimal frying temperature or if it drops below the optimal frying temperature during the process, the crust does not form on the surface of the food. This allows lipid absorption into the food and the food becomes soft and oily. • The amount and the initial temperature of foods cooked: Placing a large amount of foods in hot lipids at one time and/or starting with cold foods will decrease the temperature of the lipids below the optimal frying temperature, hence increasing lipid absorption. • The processing time: Prolonged heating will result in decomposition of lipids. Preheating the lipids any longer than required and repeated heating of the same batch of lipid must be avoided. • The formulation of foods: Increased liquid content of the foods and the use of some ingredients, such as eggs and cookie flours in batters, may increase lipid absorption.

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Fats and Oils in Culinary Transformations  231

EXPERIMENT 6.4 OBJECTIVE To explain the shortening power of different fats and oils. Sugar Cookie Recipe Ingredients and Equipment • 2,000 g (4.41 lb) flour • 480 g (1.05 lb) powdered sugar • 250 g (0.55 lb) butter • 250 g (0.55 lb) olive oil • 250 g (0.55 lb) margarine • 250 g (0.55 lb) shortening • Oven • Tray • Bowl • Wooden skewer • Ruler • Baking/parchment paper Method 1. Preheat the oven to 160°C (320°F). 2. Mix 120 g (0.26 lb) sugar with butter in a bowl. 3. Add 500 g (1.1 lb) flour. Knead thoroughly to form homogeneous dough. 4. Shape the cookie dough into balls, approximately 2 cm (¾ in.) in diameter each. 5. Measure the height of three balls (cookies). 6. Calculate the average height of the balls as explained in Experiment 4.4 and record in Data Table 6.5, hi. (Hint: The result should be close to 2 cm (¾ in.).) 7. Transfer the balls onto a tray lined with baking/parchment paper. 8. Bake them in the oven for 15 min. 9. Cool the cookies at room temperature. 10. Measure the height of three cookies. 11. Calculate the average height of the cookies and record in Data Table 6.5, hf. 12. Calculate the percent changes in the height of the cookies and record in Data Table 6.5. 13. Evaluate the crunchiness of the cookies and record in Data Table 6.5. 14. Evaluate the sensory properties of cookies and record in Data Table 6.5. Complete the same procedure for the other lipid types.

The Average Height of Balls Before Cooking (cm/in.) The Average Height of Balls After Cooking (cm/in.) Crunchinessa

% Change in Height

Sensory Properties of Cookies After Cooking (Taste, Shape, Smell, and Color)b



a

Evaluate for crunchiness of the sample on a scale of 1–5: 1 = very soft, 5 = very crunchy. b Use the terms round, flat, and irregularly shaped to evaluate the shape of the cookies. Use your own terms to evaluate the taste, smell, and the color of the cookies.

Shortening

Margarine

Olive oil

Butter

Lipid

DATA TABLE 6.5

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Fats and Oils in Culinary Transformations  233

Hint:

% change in the height =

( h i − hf ) × 100 hi

Study Questions 1. Which type of lipid is more suitable for making these types of cookies? Explain your answer. 2. Discuss the possible reasons for the differences between the sensory properties of the cookies prepared with different lipids.

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Fats and Oils in Culinary Transformations  235

The Science Behind the Results Tenderness and flakiness are the primary quality attributes of baked food products. In batter or dough, lipids physically prevent a contact between water and gluten (flour protein) by coating the gluten molecules. Therefore, lipids inhibit the formation of long gluten chains in the dough. Final products become more tender and crumbly, the lipid is said to shorten the dough. Without shortening, the dough structure gives the feeling of hardness when chewed. This is known as shortening power of the lipids. The shortening power of lipids primarily depends upon: • Proportion of lipid to flour: The product gets more tender with increasing amounts of lipids. • Water contents of lipids: Water in lipids may enhance the gluten development as it hydrates the medium. Therefore, the lipids with little water content have more shortening power than the lipids with high water content. Because of this, hydrogenated fats and lard have higher shortening power than butter and margarine because the later ones contain more water compared to lard and hydrogenated fats. • The kind of flour used: High-gluten flours produce more elastic and less tender pastry due to extensive gluten development. • The homogeneous distribution of the lipids within the mixture. Flakiness is different from tenderness. It is characterized by the thin dough layers within the pastry. The lipid melts during baking and leaves empty spaces. Steam collected in these empty spaces lifts the layers of the dough. Flakiness of the pastry primarily depends upon: • Degree of saturation of the lipids: More saturated lipids produce greater flakiness than less saturated lipids because they are solid at room temperature, have higher melting points, and produce more empty spaces in the dough upon melting. • Temperature of the lipids: Lipids must be well chilled to ensure that they can withstand mixing, rolling out, and handling without being creamed and or absorbed by the flour. • Size of the fat pieces: Fat is cut into small pieces, but chilled to prevent it from becoming melted before cooking. • The kind of flour used: High-gluten flours produce flakier but tougher baked products. • The homogeneous distribution of the lipids within the mixture.

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Fats and Oils in Culinary Transformations  237

EXPERIMENT 6.5 OBJECTIVES • To explain basic structure of emulsions. • To explain the function of an emulsifier. • To explain the parameters affecting the emulsion structure. Case 1: With Emulsifier Mayonnaise Recipe Ingredients and Equipment • 4 egg yolks • 1 tablespoon of vinegar • 550 mL (2¼ cups) olive oil • A mixing bowl • Whisk Method 1. Place the egg yolks in a bowl. 2. Add the vinegar and whisk to blend. 3. Add a very small amount of the oil and whisk until it’s well blended. 4. Continue adding the oil while whisking thoroughly between each addition. 5. Observe the appearance and texture of the mixture when all the oil has been whisked in and record in Data Table 6.6.

DATA TABLE 6.6 Appearance and Texture of the Mixturea Mixture with emulsifier (with egg yolk) Mixture without emulsifier (without egg yolk) a

Use the words viscous, thick, and thin to evaluate the appearance and texture of the mixtures.

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Case 2: With No Emulsifier Ingredients and Equipment • 1 tablespoon of vinegar • 550 mL (2¼ cups) olive oil • A mixing bowl • Whisk Method 1. Place the vinegar in a bowl. 2. Add a very small amount of the oil and whisk until it’s well blended. 3. Continue adding the oil while whisking thoroughly between each addition. 4. Observe the appearance and texture of the mixture when all the oil has been whisked in, and record in Data Table 6.6. Case 3: Amount of Emulsifier Ingredients and Equipment • 2 egg yolks • 1 tablespoon of vinegar • 550 mL (2¼ cups) olive oil • A mixing bowl • Whisk Method 1. Place the egg yolks in a bowl. 2. Add the vinegar and whisk to blend. 3. Add a very small amount of the oil and whisk until it’s well blended. 4. Continue adding the oil while whisking thoroughly between each addition. 5. Observe the appearance and texture of the mixture when all the oil has been whisked in, and record in Data Table 6.7. Case 4: Rate of Oil Addition Ingredients and Equipment • 4 egg yolks • 1 tablespoon of vinegar • 550 mL (2¼ cups) olive oil • A mixing bowl • Whisk

Fats and Oils in Culinary Transformations  239

Method 1. Place the egg yolks in a bowl. 2. Add the vinegar and whisk to blend. 3. Add the oil all at once and whisk until it is well blended. 4. Observe the appearance and texture of the mixture when all the oil has been whisked in, and record in Data Table 6.7. Case 5: A Mechanical Force Ingredients and Equipment • 4 egg yolks • 1 tablespoon of vinegar • 550 mL (2¼ cups) olive oil • A mixing bowl • Whisk Method 1. Place the egg yolks in a bowl. 2. Add the vinegar and whisk to blend. EXP 6.5 3. Add a very small amount of the oil and whisk until it’s well blended. 4. Continue adding the oil while whisking thoroughly between each addition. 5. Continue whisking 10 more minutes when all the oil has been whisked in. 6. Observe the appearance and texture of the mixture, and record in Data Table 6.7 (EXP 6.5).

DATA TABLE 6.7 Appearance and Texture of the Mixturea Amount of emulsifier Rate of oil addition Mechanical force a

Use the words viscous, thick, and thin to evaluate the appearance and texture of the mixtures.

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Fats and Oils in Culinary Transformations  241

The Science Behind the Results Oil and water are immiscible liquids, meaning they do not mix. They do not mix since oil is nonpolar (hydrophobic) and water is polar. When oil and water are mixed and then left to stand, they will separate into two layers. The oil will float above the water because it is less dense than water. Mixtures of two or more liquids that are normally immiscible are called emulsions. In emulsions, fine droplets of one liquid are dispersed in another liquid. Lipids and water are the common liquids in food emulsions. Emulsions contain three components: 1. A dispersed phase: A liquid that is suspended in the form of fine droplets in a continuous phase. 2. A continuous phase: A liquid phase that contains fine droplets of the other liquid (dispersed phase). The volume of the continuous phase is larger than that of the dispersed phase. 3. The emulsifier: A substance that holds water and oil in place in emulsions. Emulsifiers have both polar (hydrophilic) and nonpolar (hydrophobic) ends. In the mixture, the polar end of the emulsifier attracts the water and the nonpolar end attracts the oil. It forms a film at the boundary of oil and water, which coats the droplets of the dispersed phase. Therefore, the oil droplets stay suspended in the water or the water droplets stay suspended in the oil phase. Emulsifiers are usually produced from natural sources. Some food ingredients are used as emulsifiers. For example, egg yolk contains lecithin, which is known to be a good emulsifier. In the emulsion, if the water is the continuous phase and the oil is the dispersed phase, the emulsion is called an oil-in-water (o/w) emulsion. Milk is a good example of natural oil-in-water emulsions, with fat droplets dispersed in a larger amount of the water phase. Mayonnaise and certain salad dressings are the other common examples of oil-in-water (o/w) emulsions. If the oil is the continuous phase and the water is the dispersed phase, the emulsion is said to be a water-in-oil (w/o) emulsion. Butter and margarine are examples of w/o emulsions. Stable emulsions do not separate under normal handling and storage conditions. The major factors affecting emulsion stability can be listed as: • The average droplet size of the dispersed phase: Larger droplets have a tendency to coalesce and form a separate phase. • The density differences between the continuous and dispersed phases: Emulsions are more stable when density differences between the phases are small. There is a possibility of phase separation when the difference between the densities

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• • •

• •

• •

of the phases is large because the less dense phase has a tendency to rise to the surface through the emulsion. The viscosity of the continuous phase: The mobility of the droplets in the emulsion decreases with increased viscosity. Therefore, emulsions are more stable when the viscosity of the continuous phase is high. Type and amount of the emulsifier: To form stable emulsions, the surface of all the droplets must be coated with an emulsifier. Acidity: Changing the pH of the medium may decrease the stability of emulsions by changing the net charge on the emulsions. As explained previously, the attraction between the emulsifier and the polar and nonpolar phases forms the emulsions. Changes in the charge of the emulsions may break down this attraction. For example, acid addition may break down the emulsion stability because acids provide high amounts of positively charged hydrogen ions (H+). Ionic strength: Addition of salts may decrease the stability of emulsions. Some salts drastically change the net charge of the emulsions when they are dissolved in one of the phases. Temperature: Heating or cooling of emulsions affects the emulsion stability. Heating may cause phase separation because the oil droplets melt and coalesce when heated. Upon freezing, destabilization of emulsions occurs because the ice crystals that are formed during freezing physically disturb the film between the phases. Extended storage time: Over time, droplets may combine to form a larger droplet, so the average droplet size increases and phase separation occurs. Extended storage of the emulsified foods should be avoided. Mechanical force: Violent shaking or extended mixing may physically disturb the interfacial film formed by the emulsifier and break the emulsion.

Fats and Oils in Culinary Transformations  243

EXPERIMENT 6.6 OBJECTIVE To explain the effects of storage temperature and light on the oxidation of lipids. Cornmeal Cookie Recipe Ingredients and Equipment • 2/3 cup cookie flour • 1/4 cup cornmeal • 2 tablespoons corn starch • 1/4 teaspoon salt • 1/2 cup softened butter • 1/3 cup sugar • 1/2 teaspoon pure vanilla extract • Measuring cup • 2 bowls • Cookie cutter • 3 dark-colored jars • 1 transparent cookie jar • Baking/parchment paper • Rolling pin • Mixer • Tray • Oven Method 1. Label dark-colored jars “room temperature,” “refrigeration temperature,” and “oven at 40°C (104°F),” respectively. 2. Label the transparent jar “exposed to light at room temperature.” 3. Preheat the oven to 175°C (347°F). 4. Mix the flour, cornmeal, corn starch, and salt together. 5. Blend the butter, sugar, and vanilla until they are creamy. 6. Add all of the flour mixture into the butter mixture and mix until the dough just begins to come together. 7. Roll out the dough on a well-floured surface to 1 cm (⅓ in.) thickness. 8. Using a cookie cutter, cut out 16 cookies, each with a 5 cm (1.9 in.) diameter. 9. Transfer the cookies to a tray lined with baking/parchment paper. 10. Bake them until they become pale golden in color. 11. Let them cool down at room temperature. 12. Place four cookies in each jar and close the lids.

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3. Store each jar according to the conditions marked on the label. 1 14. Take one cookie from each jar and evaluate the aroma and taste of the cookies. 15. Record the results in Data Table 6.8. 16. Repeat the sensory taste every two days. DATA TABLE 6.8 Storage Conditions

Storage Time (days)

Dark-Colored Jar, Room Temp.

Dark-Colored Jar, Refrigeration Temp.

Dark-Colored Jar, 40°C/104/°F

Transparent Jar, Exposed to Light, Room Temp.

Aromaa/Tastea

Aroma/Taste

Aroma/Taste

Aroma/Taste

2 4 6 8 a

Describe and rate the aroma and taste of the samples on a scale of 1–5: 1 = smells normal, no rancid smell, 5 = strong rancid smell. 1 = no rancid/bitter taste, 5 = strong rancid/bitter taste.

Fats and Oils in Culinary Transformations  245

The Science Behind the Results During storage of lipids and food products rich in lipids, development of off-odor and off-flavors may occur as a result of deteriorating changes in the structures of lipids. This phenomenon is called rancidity or lipid oxidation. Lipid oxidation is generally classified into two broad classes: • Enzymatic rancidity: This type of rancidity occurs when the enzymes naturally present in the foods break down lipids into fatty acids and glycerol. For example, lipoxygenase enzymes in soybeans may cause development of offodor and off-flavors in food products that contain soybeans. • Oxidative rancidity: This type of rancidity occurs when oxygen combines with the fatty acids in the lipids. Oxidative rancidity involves a series of complex reactions (Figure 6.4). When lipids are exposed to various factors, such as heat, their structures are degraded and highly reactive compounds that are called free radicals are formed. Free radicals combine with environmental oxygen and give further chain reactions. Compounds responsible for the characteristic unpleasant smell and flavor associated with oxidative rancidity are formed as a result of these reactions. The various factors that affect the development of oxidative rancidity in foods can be listed as: • The degree of unsaturation of fatty acids: Unsaturated lipids are more prone to oxidative rancidity than saturated lipids. For example, lipids in nuts, peanuts, and whole wheat products are more prone to rancidity compared to butter. • Oxygen, heat, light, metal, and moisture: They increase the rate of oxidative rancidity. Several precautions that can be taken to reduce the rate of oxidative rancidity in foods include: • Storing the sensitive foods (the foods that are prone to oxidative rancidity) in dark places. • Reducing the exposure of sensitive foods to direct light (including sunlight). • Packaging the foods in selectively light-absorbent packaging materials, such as dark-colored bottles. • Keeping sensitive foods away from heat. • Storing highly sensitive foods at refrigeration temperature.

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HO C O

H

H

H

H

H

H

H

C

C

C

C

C

C

C

H

H

Fatty acid

H

Light, metal, heat

HO C O

H

H

H

H

H

H

H

C

C

C

C

C

C

C

H

H

Free radical

O2

HO C O

H

H

H

H

H

H

H

C

C

C

C

C

C

C

H

H

Peroxy radical

O O

Fatty acids

HO C O

H

H

H

H

H

H

H

C

C

C

C

C

C

C

H

H

Degradation

O OH

Figure 6.4  Mechanism of oxidative rancidity.

Compounds that give rancid smell and taste

Fats and Oils in Culinary Transformations  247

• Reducing the environmental oxygen concentration to very low levels, such as vacuum packaging the sensitive food products. • Avoiding direct contact of foods with copper and iron equipment and containers. • Using approved natural (i.e., ascorbic acid, tocopherol) or synthetic (BHA, BHT) antioxidants.

248  COOKING AS A CHEMICAL REACTION

Fats and Oils in Culinary Transformations  249

EXPERIMENT 6.7 OBJECTIVE To demonstrate the effects of fat crystals on sensory attributes of some food products. Ingredients and Equipment • 3 blocks of untempered bitter chocolate couvertures (2.5 kg [5.5 lb] each) • Water • 1 milk chocolate • 1 pot • 4 medium-sized metal containers (they will be placed over the top of the pot; therefore, the size of the containers must be chosen accordingly) • 1 metal bowl • 1 container (this will be placed in the metal bowl; therefore, the size of the container must be chosen accordingly) • Stretch film (plastic wrap) • Thermometer • Spatula • Stove • Refrigerator Method Case 1: Fat Blooming 1. Label four medium containers “Tempered,” “Untempered,” “Temperature,” and “Humidity.” 2. Place one of the couvertures in the container labeled “Untempered.” 3. Store at a constant temperature of 16°C–17°C (60.8°F–62.5°F). 4. Break one of the chocolate couvertures into small pieces. 5. Put them in the container labeled “Tempered.” 6. Place a small amount of water into the pot. 7. Place the container over the top of the pot. The bottom of the container should not touch the water. 8. Heat the water until it is very gently simmering, not boiling. 9. Check the temperature of the chocolate continuously while stirring gently. 10. When the temperature of the chocolate reaches 50°C (122°F) on a thermometer, remove the container from the pot. 11. Stir the chocolate with a spatula to melt it homogeneously.

250  COOKING AS A CHEMICAL REACTION

12. As soon as the temperature cools to 28°C (82.4°F), return the container to the pot and reheat, stirring gently. 13. When the temperature of the chocolate reaches 32°C (89.6°F) on a thermometer, remove the container from the pot. 14. Store with the untempered sample. 15. Break the last chocolate couverture into small pieces. 16. Place the chocolate pieces into the container labeled “Temperature.” 17. Place the container over the top of the pot. 18. Heat the water until it is very gently simmering, not boiling. 19. Remove the container from the pot once the chocolate appears to have melted. 20. Cover the container with paper towel and store with the other two samples. 21. Next day, take the sample from the storage area and place it in a warm place that has a temperature of 30°C–35°C (86°F–95°F), for a few hours. 22. Return the container to its storage area. Case 2: Sugar Blooming Meanwhile, 1. Place the milk chocolate into a container labeled “Humidity.” 2. Put a small amount of water into the metal bowl. 3. Place the container inside the metal bowl. 4. Cover them tightly together with stretch film (plastic wrap). That will create a humid environment. 5. Remove the container from the metal bowl after 48 h. 6. Store it with the other samples. 7. Check the surface of the samples every day. Continue the observation for 5–10  days. Record your observations in Data Table 6.9 (EXP 6.6).

EXP 6.6

Note: In this experiment, a special technique, which is called a bain-marie technique, is applied to melt the chocolate in Case 1.

Fats and Oils in Culinary Transformations  251

DATA TABLE 6.9 Surface Appearancea Storage Time (days)

Tempered Sample

Untempered Sample

Sample Exposed to Fluctuating Temps.

Sample Stored in a Humid Environment

0 1 2 3 4 5 6 7 … a

Use the terms no change, sign of white spots, and cloudy to evaluate the surface of the samples.

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Fats and Oils in Culinary Transformations  253

The Science Behind the Results White, cloudy-looking coating, “bloom,” develops on the surface of chocolates if chocolates are processed, stored, or served improperly. There are two main types of chocolate blooms: 1. Fat bloom 2. Sugar bloom Fat bloom is primarily related to the crystallization of cocoa butter found in ­chocolate products on the surface of chocolate. Cocoa butter is composed of six different fat crystals melting at different temperatures. Some of the crystals are stable, but other crystals are not. Unstable crystals change form over time and form a cloudy-looking coating on the surface of the chocolate. During chocolate manufacturing, a process called tempering is used to ensure that only the stable crystals are formed in the final chocolate products. Tempering is a directed precrystallization process to promote the formation of the stable crystal forms only. Tempering is a three-step process: Step 1: Melting chocolate at: • 50°C–55°C (122°F–131°F) for bitter chocolate. • 45°C–50°C (113°F–122°F) for milk chocolate. • 45°C–50°C (113°F–122°F) for white chocolate. Step 2: Rapid cooling to about: • 28°C–29°C (82.4°F–84.2°F) for bitter chocolate. • 27°C–28°C (80.6°F–82.4°F) for milk chocolate. • 26°C–27°C (78.8°F–80.6°F) for white chocolate. Step 3: Warming to: • 31°C–32°C (87.8°F–89.6°F) for bitter chocolate. • 30°C–31°C (86.0°F–87.8°F) for milk chocolate. • 29°C–30°C (84.2°F–86.0°F) for white chocolate. The primary reasons for fat bloom on the surface of chocolate include the following: • Improper tempering of the chocolate: If chocolate is not tempered or if the tempering process is not carried out properly, the unstable forms of cocoa

254  COOKING AS A CHEMICAL REACTION

butter crystals will form and fat bloom will occur on the surface of the chocolate during storage. • High temperature: The chocolate should be stored below the melting point of cocoa butter to avoid surface fat bloom. • Temperature fluctuation: During storage, temperature fluctuation must be avoided. The chocolate is best when stored in an area of stable temperature (15°C–18°C/59°F– 64.4°F) to avoid fat blooming. • Sugar bloom is caused by moisture condensing on the chocolate. Chocolate is usually composed of ground cocoa beans, sugar, EXP 6.7 and sometimes emulsifiers and milk. When chocolate is exposed to moisture, water dissolves the sugar on the surface of the chocolate. The dissolved sugar crystallizes on the surface of the chocolate as the water dries. The resulting small sugar crystals give the chocolate a cloudy appearance. For example, if the chocolate is placed in the refrigerator and then removed and placed at room temperature, moisture from the air will condense on the cold chocolate, and, as water dries, the sugar bloom will appear on the surface of the chocolate. Sugar bloom also may occur if the chocolate is stored in an environment with very high humidity. Chocolate is best stored in an area of low humidity and stable temperatures of 15°C–18°C (59°F–64.4°F) to avoid fat and sugar blooms (EXP 6.7). Fat bloom and sugar bloom on the surface of chocolate products do not have any adverse health affect, but they make the products unappealing to consumers.

More Ideas to Try Put 50 mL (1.7 fl oz) each of sunflower oil, olive oil, and peanut oil in separate bottles and keep the bottles in the refrigerator overnight. Observe the physical changes in oils that occur during storage. Discuss the results.

Study Questions 1. Explain the differences between saturated and unsaturated lipids. 2. Is coconut oil a fat or oil?

Fats and Oils in Culinary Transformations  255

POINTS TO REMEMBER Fats and oils are the organic compounds naturally obtained from animals and plants. Understanding the basic structure of fats and oils is crucial for chefs because the functional properties and shelf life of fats and oils are primarily related to their structures. Fats and oils belong to a group of substances called lipids. The primary functional properties of lipids in food processing can be listed as: a. Enhancement of flavor and mouthfeel b. Development of texture c. Shortening d. Emulsification e. Medium for transferring heat f. Development of appearance Lipids with longer carbon chains have higher melting points. The saturated fatty acids have higher melting points than unsaturated fatty acids of corresponding size. The saturated lipids have lower smoke points than unsaturated lipids. Ideal frying lipids have a high smoke point that is well above the optimum frying temperature. During storage, lipids and products rich in lipids may undergo lipid oxidation reaction. The unsaturated lipids are more prone to lipid oxidation due to the double bonds in their structures. Oxygen, heat, light, metal, and moisture increase the rate of oxidative rancidity. The shortening power of the lipid determines the tenderness of the pastry. A mixture of two or more liquids that are normally immiscible is called an emulsion. Emulsions contain three components: a dispersed phase, a continuous phase, and the emulsifier. Fat crystallization may develop on the surface of chocolate if it is not processed, stored, or served properly. Tempering is a directed precrystallization process to promote the formation of the stable fat crystal forms in chocolates.

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SELECTED REFERENCES Brown, A. C. 2007. Understanding food: Principles and preparation, 3rd ed. Belmont, CA: Wadsworth Publishing. Costa, R. M., F. A. R. Oliveira, O. Delaney, and V. Gekas. 1999. Analysis of the heat transfer coefficient during potato frying. Journal of Food Engineering 39(3):293–299. Costa, R. M., F. A. R. Oliveira, and G. Boutcheva. 2001. Structural changes and shrinkage of potato during frying. International Journal of Food Science and Technology 36:11–23. Culinary Institute of America. 1995. The new professional chef, 6th ed. New York: John Wiley & Sons. Functional properties of fats and oils. 2009. Cereal Chemistry Journal (AACC Int’l) 86(3):May/June. Online at: http://cerealchemistry.aaccnet.org/doi/ pdf/10.1094/9780913250907.001. Ghotra, B. S., S. D. Dyal, and S. S. Narine. 2002. Lipid shortenings: A review. Food Research International 35(10):1015–1048. Lonchampt, P., and R. W. Hartel. 2006. Surface bloom on improperly tempered chocolate. European Journal of Lipid Science and Technology 108(2):159–168. McGee, H. 2004. On food and cooking. The science and lore of the kitchen, 1st rev. ed. New York: Scribner. Meulenaer, B. D., and J. V. Camp. Factors that affect fat uptake during French fries production. Online at: www.euppa.eu/_files/factors-french-fries.pdf. Ozilgen, S., and M. Ozilgen. 1990. Kinetic model of lipid oxidation in foods. Journal of Food Science 55:498–502. Paul, S. R. 2006. Food frying. In Encyclopedia of life support systems (EOLSS), food engineering theme. Online at: www.eolss.net/Sample-Chapters/C10/E5-10-04-06.pdf. University of California/Davis. What is the smoke point of butter? Online at: http://drinc. ucdavis.edu/dairychem.7htm. Vaclavic, V. A., and E. W. Christian. 2008. Essentials in food science, 3rd ed. Berlin: Springer. Weaver, C., and J. Daniel. 2003. The food chemistry laboratory, 2nd ed. Boca Raton, FL: CRC Press.

Chapter 7 Keys to Developing the Perfect Bite: New Food Product Development and Sensory Evaluation Tests

REASONS TO DEVELOP A NEW FOOD PRODUCT Competition in the food industry grows tremendously as a result of: Fast changing consumer trends. Increased consumer demand for new tastes. Increased consumer demand for safe food products. Increased consumer awareness of health. Increased consumer demand for specific food products, i.e., gluten-free food products, diabetic food products. • Increased capacity of the production units that leads to producing more food. Companies can produce more varieties of similar food products in the same production area. For example, dairy producers can produce different types of cheeses at the same production area with the help of new processes and new equipment. • • • • •

257

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Marketing research and consumer survey

Ingredient purchasing

Food processing methods

Food science

Quality control Cost control Sensory evaluation

Food Product Development

Marketing and sales National/ International food laws and regulations

Preservation and storage techniques

Packaging

Food safety

Figure 7.1  Product development is multidisciplinary team work, and it requires knowledge of all items shown in the figure.

New food products and new processes need to be developed continuously to keep the company competitive in a changing food market. Food providers must work hard through innovations and improved processes to meet the consumer’s expectations. Companies invest money, staff, and equipment in new product development. Developing new food products is a reward for the company when it meets the consumers’ needs and demands. On the other hand, possibility of failures of the new food products in the marketplace is a frightening risk for the companies. To minimize the risks, the companies should adopt a multidisciplinary approach. The new food product development process requires multidisciplinary teamwork (Figure 7.1). In properly formed and managed product formulation teams, the individuals come from different disciplines, their contributions are properly harmonized, and they all work toward a common goal.

STAGES IN NEW FOOD PRODUCT DEVELOPMENT The new food product development process may involve: 1. Formulation of completely new food products. 2. Modification of an existing food product, which is called line extension.

PRODUCT DEVELOPMENT & SENSORY EVALUATION  259

The same processing steps are followed in both cases, but line extension requires less financial resources and less time to develop a new food product. The new food product development process has three major stages: 1. Idea development 2. Product development 3. Commercialization

Idea Development The objective of this stage is to gather as many new ideas as possible. For the success of this stage: • • • •

The mission of the organization must be considered. The goal of the company must be clearly identified. The consumer demands for new food products must be considered. Gaps in the market must be identified.

The team members carry out financial and technical reviews, feasibility studies, and legal analysis at this stage. The ideas that are weak in terms of their chances of market success should be filtered out at this stage. The most feasible ideas are carried to the product development stage.

Product Development At this stage: 1. The expected quality attributes of the food product are identified. 2. The product is formulated. Specifications and the amounts of the ingredients are determined. 3. The processing steps and parameters are set. 4. A prototype is prepared. 5. Sensory evaluation tests are carried out. 6. If required, changes are carried out to the prototypes depending on the feedback from the sensory evaluation tests. 7. A trial production (pilot plant production) is carried out to produce large quantities of the improve prototype. 8. The prototypes from the pilot plant production are tested with a range of consumers through the sensory evaluation tests. 9. The product is either modified or discarded based on feedback gained from consumers through sensory evaluation tests. The information gained at this stage is used to move the production to the commercialization stage.

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Commercialization Once the consumers accept the product, the producers move forward to produce the product on a large scale. Finally, the product is introduced to the consumers in the marketplace (launching).

HOW TO CARRY OUT SENSORY EVALUATION TESTS Sensory evaluation tests scientifically measure and analyze the consumers’ responses to the sensory attributes of the food products, such as appearance, odor, sound, texture, and taste. They involve both qualitative observations and quantitative measurements that were previously explained in Chapter 1. In sensory evaluation tests, five senses are used to observe the characteristics of foods (PIC 7.1). Sensory evaluation tests translate qualitative observations to quantitative data to • improve the quality of food products; • understand which attributes make the food products more acceptable to consumers;

Appearance Odor Taste Sound Touch

color, siz

e, shape

, texture

volatile flavor molecules

, bitter our, salty op le, p b b , bu ckle

sweet, s

fizz,

cra

rough and smooth textures, hot, cold, warm,

PIC 7.1 FIVE SENSES ARE USED TO OBSERVE THE CHARACTERISTICS OF FOODS: SIGHT, SMELL, TASTE, HEARING, AND TOUCH.

PRODUCT DEVELOPMENT & SENSORY EVALUATION  261

• understand whether differences exist between the ranges of food products; • understand how big the differences are in a range of food products; and • understand whether a final food product meets its targeted specifications. Requirements for sensory evaluation tests can be listed as: 1. Test environment: Sensory evaluation tests should be carried out in a quiet area with adequate lighting and ventilation. Separate sections/booths that are free from distractions must be used for each panelist. The walls should be white or off-white in color. 2. Timing: Morning is the best time for sensory evaluation tests. 3. Panelists: Panelists are used to evaluate the food samples. The number and specifications of the panelists depend on the purpose of the study. Untrained (consumers), semitrained (experienced), or trained (highly experienced) panelists are incorporated in the tests depending on the purpose of the study. Willingness to participate, availability, food allergies, and food intolerances are the major factors to be considered when choosing the panelists. The panelists should be clearly informed about the purpose and the procedure of the sensory test. Potential consumers are the best panelists. 4. Samples: All food samples of the same type should be served to the panelists at the same temperature. The size, shape, and types of plates and/or cups should be the same for all samples of the same type. Size or volume served should be equal for all samples. The optimum number of samples to be tested in each session should be decided depending on the properties of the food. The samples must be coded to make sure that the panelists cannot recognize or identify the samples. Random three-digit alphanumerical codes, such as GP4, 534, and 1K6, are usually used to label the samples. To avoid bias among the panelists, samples should not be coded A and Z or 1 and 10, since panelists might feel that samples A and 10 are better than samples Z and 1. Samples should be served in a random order; therefore, not all panelists will get the same samples at the same time. For example, if two samples are served, half of the panelists receive one sample first and the rest receive the other sample first. Terminology is very important in sensory evolution tests. It must be consistent from product to product to minimize the panelist confusion and variability in defining the product attributes. Basic terms that are used in qualitative observations are given in Chapter 1. Statistical techniques are often applied to the interpretation of the sensory test results. Basic statistical methods are given in Chapter 1.

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Sensory evaluation tests are grouped into three categories based on the questions that they are addressing: 1. Discrimination/difference tests 2. Descriptive tests 3. Consumer acceptance/preference tests Discrimination/difference tests analyze differences between food products. These tests are carried out to understand whether differences exist between the food products and/or how the consumers would define the difference between the products. For these tests, 25–50 untrained panelists are required. The most common discrimination/difference tests include: 1. Paired comparison test 2. Triangle test 3. Duo-trio test The paired comparison test is applied to compare two samples. Two coded samples are presented to the panelists. Panelists are asked to identify the sample that has a greater degree of intensity in terms of a specific sensorial attribute. Test scorecards are distributed to the panelists and they are asked to follow directions given on the scorecards. For example, panelists are asked to determine which of two samples of Turkish Delights are sweeter. PAIRED COMPARISION TEST SCORECARD You are provided with two samples. Taste the samples in the order from left to right and circle the sample that is sweeter. (Sample A)

(Sample B)

918

871

The triangle test is applied to understand if the differences between two samples are detectable. Three coded food samples are prepared, two of which are the same and the other is different. Samples are arranged in a triangle on a tray and presented to the panelists. Test scorecards are distributed to the panelists and they are asked to follow directions given on the scorecards. For example, panelists are asked to choose the soup sample that is the most different from the other soup samples.

PRODUCT DEVELOPMENT & SENSORY EVALUATION  263

TRIANGLE TEST SCORECARD You are provided with three samples. Two of the samples are identical. Taste the samples in the order from left to right and circle the sample that is different than the order two samples. (Sample A)

(Sample B)

(Sample A)

871

63

918

The duo-trio test determines whether or not a sensory difference exists between two samples. This method is particularly useful when changes are done in ingredients, processing, packaging, or storage of food products currently available. Three food samples are prepared, two of which are the same. One of the two identical samples is marked as reference, and the other two samples are coded with alphanumeric codes. The samples are presented to the panelists. Test scorecards are distributed to the panelists and they are asked to follow directions given on the scorecards. For example, panelists are asked to determine if there is a difference between an original apple pie recipe and the modified apple pie recipe. DUO-TRIO TEST SCORECARD You are provided with three samples and one is marked as reference. One of the other samples is the same as the reference. Taste the samples in the order from left to right and circle the sample that is the same as the reference. (Reference, B)

(Sample A)

(Sample B)

191

438

Descriptive tests are applied to detect how food products differ in preselected sensory attributes. For these tests, 8–12 trained panelists are required. Trained panelists are able to evaluate each attribute and the range of intensity of similar products. For example, a trained panelist presented a sample of chocolate would be able to rate the level of cocoa, cocoa butter, and the crystal structure similar to any instrument that would give a reading. The most common descriptive tests include: 1. Descriptive rating test 2. Descriptive ranking test

264  COOKING AS A CHEMICAL REACTION

The descriptive rating test is used to rate the intensity of preselected attributes of the food products. Test scorecards are distributed to the panelists and they are asked to follow directions given on the scorecards. The attributes of the food samples are generally rated on line scales or spider/star graphs. For example, panelists are asked to rate the intensity of four attributes: crunchiness, texture, color, and taste of the chocolate bar. DESCRIPTIVE RATING TEST SCORECARD

You are provided with a sample of a chocolate bar. Please evaluate the sample for each attribute. Decide on the intensity for each attribute using a scale from 1 to 5. The higher the number, the greater the intensity (1 = very week; 5 = very strong). Mark the number on the line scale. Crunchiness

1

2

3

4

5

2

3

4

5

2

3

4

5

Texture

1 Color

1

PRODUCT DEVELOPMENT & SENSORY EVALUATION  265

DESCRIPTIVE RATING TEST SCORECARD FOR TWO OR MORE SAMPLES

You are provided with two coded chocolate bar samples. Please evaluate the samples for each attribute. Decide on the intensity for each attribute using a scale from 1 to 5. The higher the  number, the greater the intensity (1 = very week; 5 = very strong). Begin with one sample and evaluate it for all  attribute. Taste the other sample for the same attributes. Crunchiness Sample 431

1 Sample 132

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

1

2

3

4

5

Texture Sample 431

Sample 132

Color Sample 431

Sample 132

Taste Sample 431

Sample 132

266  COOKING AS A CHEMICAL REACTION

Crunchiness 10 8 6 4 2 Taste

Texture

0

Color Sample 1

Sample 2

Figure 7.2  Example of a spider graph chart.

Hint: To draw the spider/star chart, the average score of each attribute is calculated from the data obtained from each panelist and marked on the matching scale of the spider/star diagram. The marks are connected to draw the graph. The average scores of different products can be marked on the same chart to compare the sensory properties of the products. Shown in Figure 7.2 is an example of a spider graph. The descriptive ranking test is applied to rank foods in the order of intensity of a specified attribute. Coded samples are presented to the panelists. Test scorecards are distributed to the panelists, and they are asked to follow directions given on scorecards. For example, panelists are asked to rank milkshake samples depending upon the intensity of sweetness. RANK ORDER TEST SCORECARD

You are provided with fi ve coded samples. Taste the samples and place them into rank order depending upon the intensity of sweetness. Write the code of the sample on the rank scale given below. Not sweet

Slightly sweet

Moderately sweet

Very sweet

Extremely sweet

PRODUCT DEVELOPMENT & SENSORY EVALUATION  267

Consumer acceptance/preference tests are used to determine if the product is liked or disliked by the consumer. These tests also are used to determine the most preferred food product among the sets of similar products. For these tests, 25–50 untrained panelists (consumers) are required. Statistics are used to analyze the data. The most common consumer acceptance tests include: 1. Preference tests a. Paired-preference b. Ranking for preference 2. Hedonic rating scale The paired-preference test determines the preference of the consumer between two products. Two coded samples are presented to the panelists. Test scorecards are distributed to the panelists and they are asked to follow directions given on the scorecards. For example, two lemon cake samples are presented to the panelists and they are asked to choose the one that they prefer. PAIRED PREFERENCE TEST SCORECARD

You are provided with two coded samples.Taste the samples and circle the sample that you prefer. (Sample A)

(Sample B)

345

762

The ranking for preference test ranks the food samples with respect to the preference degree. A number of coded samples are presented to the panelists. Test scorecards are distributed to the panelists, and they are asked to follow directions given on the scorecards. For example, panelists are asked to rank the five coded ice cream samples in order of their preferences. RANKING FOR PREFERENCE TEST SCORECARD

You are provided with five coded samples. Taste the samples and rank them in an order of your preference. Please indicate the code your choices in the boxes given below. Sample code 1 choice st

2nd choice 3rd choice 4th choice 5th choice

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The hedonic rating scale test is applied to measure the level of liking or disliking food products by the consumers. Coded samples are presented to the panelists. Test scorecards are distributed to the panelists, and they are asked to follow directions given on scorecards. For example, panelists are asked to indicate the extent of their liking for each of five meatball samples. HEDONIC RANKING TEST SCORECARD

You are provided with five coded samples. Taste the samples and indicate your degree of liking. Check a box, from 1 to 9, to indicate your preference. Sample code

9 6 7 8 3 4 5 1 2 like like like like dislike dislike neither dislike dislike extremely very moderately slightly like or slightly moderately very extremely much dislike much

PRODUCT DEVELOPMENT & SENSORY EVALUATION  269

POINTS TO REMEMBER





New products and new processes need to be developed continuously to keep the company competitive in a changing food market. A new food product development process requires a multidisciplinary teamwork. A new food product development process has three major stages: a. Idea development b. Product development c. Commercialization Sensory evaluation tests scientifically measure and analyze the consumers’ responses to the sensory attributes of the food products, such as appearance, odor, sound, texture, and taste. Sensory evaluation tests are grouped into three categories based on the questions that they are addressing: a. Discrimination/difference tests b. Descriptive tests c. Consumer acceptance/preference tests Sensory evaluation tests provide excellent information for chefs to evaluate and improve their new dishes.

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SELECTED REFERENCES British Nutrition Foundation. 2010. Sensory evaluation: Teachers’ guide. Online at: www. foodafactoflife.org.uk/attachments/276dbf05-695c-44942bb55825.pdf Crawford, I. M. 1997. New product development. In Marketing and Agribusiness texts. Rome: FAO (Food and Agriculture Organization). Online at: www.fao.org/docrep/004/ w3240e/w3240e04.htm Experimental design and sensory analysis. Pullman, WA: Washington State University. Online at: http://public.wsu.edu/~rasco/fshn4202005/EDSA.pdf. Lawless, H. T., and H. Heymann. 2010. Sensory evaluation of food: Principles and practices, 2nd ed. Berlin: Springer. Mason, R. M., and S. L. Nottingham. 2002. Sensory Evaluation Manual. Alimentos Food. Online at: www.scribd.com/doc/890001/sensory. Naes, T., P. B. Brockhof, and O. Tomic. 2010. Statistics for sensory and consumer science. Hoboken, NJ: John Wiley & Sons. O’Mahony, M. 1986. Sensory evaluation of food: Statistical methods and procedures. Boca Raton, FL: CRC Press. Ozer, M. 2009. What do we know about new product idea selection? NC State University, Center for Innovation Management Studies. Online at: http://cims.ncsu.edu/downloads/Research/69_Ozer-%20New%20Prod.%20Idea%20Selection.pdf. Ozilgen, S. 2011. Factors affecting taste perception and food choice. In The sense of taste, ed. E. J. Lynch and A. P. Petrov (pp. 115–126). Hauppauge, NY: Nova Science Publishers. Rudolph, M. J. 2000. The food product development process. In Developing new food products for a changing marketplace. Boca Raton, FL: CRC Press. Woods, T. W., and A. Demiralay. 1998. An examination of new food product development processes: A comparative case study of two hazelnut candy manufacturers. Lexington, KY: University of Kentucky. Online at: http://ageconsearch.umn.edu/ ­bitstream/31979/1/ sp980384.pdf.

Chapter 8 The Science of Flavor and Flavor Pairing

Why do different varieties of the same food such as Granny Smith, Golden, Jonathan, and McIntosh apples have different flavors? Why do we like strawberry jam on toast? Would you choose the same wine if the fish was dipped in a hot and sour sauce? What should I consider when pairing the foods/drinks? Why do certain foods taste particularly good together? How do we know which foods combine well together? Knowledge of basic flavor chemistry is crucial for chefs to understand the p­ rinciples behind the flavor pairings, and to create new food combinations.

271

272  COOKING AS A CHEMICAL REACTION

The Science of Flavor and Flavor Pairing  273

EXPERIMENT 8.1 OBJECTIVES • To explain the difference between taste and flavor • To explain the importance of odor on food flavor and food perception Case 1 Ingredients and Equipment • 1–2 teaspoonful of cumin or cinnamon • Teaspoon Method 1. Hold your nose (it needs to be fully blocked). 2. Take half a teaspoonful of the spice and place it on your tongue. 3. Chew on the spice while still holding your nose. 4. Record your observations in Data Table 8.1 5. Keep your mouth close and release your nose. 6. Record your observations in Data Table 8.1 DATA TABLE 8.1 Flavor Sensation with Nose Closed



Flavor Sensation after Releasing the Nose

Use your own words to describe the flavor sensation

274  COOKING AS A CHEMICAL REACTION

The Science of Flavor and Flavor Pairing  275

The Science Behind the Results Although people often use the terms taste and flavor interchangeably, they are ­different sensory attributes. Flavor is a complicated combination of mainly taste and odor. Each food has its own chemical structure; hence its own hundreds or even ­thousands of taste and odor molecules that make up its flavor (Table 8.1). For example, aromatic substances such as ketones, alcohols, esters, and terpens give the characteristic flavor of fruits and vegetables. Acids, esters, acetals, and aldehydes are the main flavor components of alcoholic beverages. Heterocyclic compounds such as furans, thiazoles, pyrazines, pyrroles, and oxazoles give the characteristic flavor of roasted, baked, and fried foods such as bread, coffee, and French fries. Taste molecules in foods are detected by specialized sensory cells (chemoreceptors) on the tongue. These specialized sensory cells bundle to form taste buds. As food is chewed and mixed with saliva, its particle size reduces, surface area increases, and taste (sweet, sour, etc.) molecules release. These molecules interact and activate the specialized sensory cells, and the signals are sent to the brain. Therefore, the brain detects the taste. Each taste—sweet, sour, bitter, salty, and umami—has specific corresponding type of sensory cells. Similarly, as surface area of the food increases by chewing and mixing with saliva, more volatile odor molecules release. Specialized cells in the nose receive those molecules from the back of the mouth, and send signals to the brain. The concentration of the odor molecules must be above the thresholds to be able to be detected by the brain. Taste and odor signals work together to create the person’s perception of flavor. That is because people with congested nasal passages cannot detect the flavor of the foods. Although some foods may share the similar flavor compounds, they may have completely different flavors due to the differences in their composıtıons. Food is a complex structure. Interaction between food components such as proteins, ­lipids, and carbohydrates may affect the volatility and solubility of the flavor ­molecules; hence their transfer to the specialized sensory cells, which ultimately influences the perception of food flavor.

276  COOKING AS A CHEMICAL REACTION

TABLE 8.1 The Most Common Food Flavor Molecules and Their Chemical Structures

Major Chemical Compound, Formula

Flavor Examples

Diacetyl, C4H6O2

Buttery, naturally occurs in alcoholic beverages

Cinnamic aldehyde, C9H8O

Cinnamon

Isoamyl acetate, C7H14O2

Banana

Butyl acetate, C6H12O2 Limonene, C10H16

Citrus fruit rinds

Ethyl propionate, C5H10O2

Pineapple, kiwi, and strawberry

Allyl hexanoate, C9H16O2

Pineapple

Hydrogen cyanide, HCN

Bitter almond

Benzaldehyde, C7H6O Methyl anthranilate, C8H9NO2

Grape

Acetic acid, CH3COOH

Vinegar

2-Isobutyl-3-methoxy pyrazine, C9H14N2O

Bell pepper

Eugenol, C10H12O2

Clove

Trimethylamine, C3H9N

Fish

Vanillin, C8H8O3

Vanilla

2-isobutylthiazole, C7H11NS

Tomato

Diallyl disulfide, C6H10S2

Garlic

Anethol, C10H12O

Anise seed

Ethyl decadienoate, C12H20O2

Pear

Ethanol, C2H5OH

Alcoholic beverages

The Science of Flavor and Flavor Pairing  277

EXPERIMENT 8.2 OBJECTIVES • To explain the effects of food/drink temperature on food flavor perception. • To explain the effects of temperatures of the foods/drinks that are served together on overall taste perception. Case 1 Ingredients and Equipment • 1 orange • Warm water • Knife • Drinking water at room temperature Method 1. Cut the orange in half. 2. Place one of the halves in the refrigerator. 3. Keep the second half at room temperature. 4. Wait for 1 h. 5. Wash your mouth with warm water. 6. Take the orange from the refrigerator. 7. Peel and slice it immediately. 8. Taste one of the slices, and record your observations in Data Table 8.2. 9. Peel and slice the second half at room temperature. 10. Wash your mouth with warm water. 11. Taste one of the slices, and record your observations in Data Table 8.2.

DATA TABLE 8.2 Sample

Overall Experiencea

Orange at room temperature Orange at refrigeration temperature a

 se the terms fruity, sweet, acidic, flavorful, rich, tasteless, and aromatic to describe the overall U experience.

278  COOKING AS A CHEMICAL REACTION

Case 2 Ingredients and Equipment • 1,200 mL (40.6 fl oz) milk at refrigeration temperature 4°C–5°C (39°F–41°F) • 3 drinking glasses • Small saucepan • Thermometer • Graduated cup • Stove Method 1. Take the milk from a refrigerator and pour 200 mL (6.76 fl oz) into the first glass. 2. Taste and record your observations in Data Table 8.3. 3. Place the rest of the milk into a small saucepan. 4. Place the saucepan over low heat. 5. Place the thermometer in the pan. Be sure the thermometer does not touch the bottom of the pan. 6. Heat until the thermometer reaches 25°C (77°F). 7. Pour 200 mL (6.76 fl oz) into the second glass. 8. Taste and record your observations in Data Table 8.3. 9. Heat the rest of the milk until reaches 35°C (95°F)

DATA TABLE 8.3 Milk Temperature, °C (°F)

Flavora

Refrigeration 25 (77) 35 (95) 45 (113) 55 (131) 65 (149) a

 se the terms pleasantly sweet, sweet, very sweet, creamy, flat, no after taste, strong after taste, mild U after taste, no odor, and milky odor to describe the flavor.

The Science of Flavor and Flavor Pairing  279

Repeat the experiment for the temperatures of 45°C (113°F), 55°C (131°F), and 65°C (149°F). 10. Compare your results. Case 3 Ingredients and Equipment • 2 pieces of chocolate • 2 tablespoons double cream • Icy cold water • Drinking water at room temperature Method 1. Drink a glass of water at room temperature. 2. Taste one piece of chocolate and record your observations in Data Table 8.4. Do not let it melt in your mouth more than 3–5 min before chewing. 3. Wait for 3–5 min, and clean your mouth with warm water to wash the ­chocolate residues away from your mouth. 4. Drink a glass of icy cold water at once. 5. Immediately taste the second piece of the chocolate and record your ­observations in Data Table 8.4. Repeat the same experiment with double cream. DATA TABLE 8.4 Sample

Overall Experiencea

Chocolate Chocolate/icy cold water Double cream Double cream/icy cold water a

 se the terms creamy, silky, smooth, bulky, melting, dry, aromatic, acidic, mouth U coating, dense, sweet, and sticky to evaluate the overall experience.

280  COOKING AS A CHEMICAL REACTION

The Science of Flavor and Flavor Pairing  281

The Science Behind the Results The same food may taste different at different temperatures. There are two major reasons for this. First, the foods are the combinations of different flavor molecules, and the effect of temperature is not uniform across these molecules. In general, increasing temperature of the food between 15°C and 35°C (59°F and 95°F) intensifies the sweet and bitter tastes, but may not have significant effect on perceptions of sourness and saltiness. Similarly, heating or cooling may activate or deactivate (increase or decrease the kinetic energy of) odor volatile compounds associated with ­certain flavors. For example, red wines are served at higher temperatures than white wines because they usually contain more volatile flavor compounds, and low temperature decreases their mobility. In addition, increase or decrease in temperature may change the textural properties of the foods, which may positively or negatively affect the release of activated flavor molecules from the food structure. Second, hot or cold foods and drinks may adversely affect the specialized sensory cells bundled on the tongue; hence taste recognition. It is especially important if the food is going to be eaten in combination with hot or cold foods/drinks, as this might affect the perception of the product. For example, eating immediately after drinking cold water may decrease the perceived intensities of sweetness, specific flavors, and the texture of the consumed food.

282  COOKING AS A CHEMICAL REACTION

The Science of Flavor and Flavor Pairing  283

EXPERIMENT 8.3 OBJECTIVES • To understand ingredients sharing the key chemical compounds may pair well in the recipes. • To understand the use of ingredients in the recipe to modify/mask the ­sensory effects of other ingredients. Case 1 Cauliflower and chocolate Ingredients and Equipment • 1 medium cauliflower • 4 oz (113 g) block of chocolate • Water • Large saucepan • Colander • Small bowl • Plate Method 1. Bring a large saucepan of water to the boil over high heat. 2. Cut the cauliflower into small florets. 3. Add the florets into the boiling water. 4. Cook for 5–10 min or until cauliflower is tender. 5. Drain, cool, and return them onto a plate. 6. Grade the chocolate, and place them in a small bowl. 7. Take two florets in similar sizes. 8. Taste the first one. 9. Record your observations in Data Table 8.5 10. Dip the second one into the bowl with chocolate. Completely cover the head of the cauliflower floret with chocolate. DATA TABLE 8.5 Taste

Aroma

Plain cauliflower Cauliflower dipped in chocolate

Use your own words to describe the taste, aroma, and flavor of the samples.

Flavor

284  COOKING AS A CHEMICAL REACTION

11. Taste it. 12. Record your observations in Data Table 8.5 13. Compare the results (EXP 8.1)

EXP 8.1

Case 2 Onion and coffee Ingredients and Equipment • 2 medium onions • ½ teaspoonful Turkish coffee • Cooking spray • 1 skillet • Cutting board • Knife • Two plates • Spoon • Stove Method 1. Label one plate “plain” and the other, “onion-coffee.” 2. Chop the onions into uniformly sized pieces. 3. Spray the skillet with cooking spray. 4. Place the half of the onions in the skillet, and cook over medium heat while stirring. 5. Continue stirring and watch, as the onion’s color turns light brown in color. 6. Remove them onto the plate labeled “plain.” 7. Record your observations in Data Table 8.6. 8. Clean and spray the skillet with cooking spray. 9. Place the remaining onion in the skillet, and cook over medium heat while stirring. 10. Continue stirring and watch, as the onion’s color turns light brown in color.

The Science of Flavor and Flavor Pairing  285

DATA TABLE 8.6 Colora

Smellb

Tastec

Plain Onion-coffee Use the words light golden, golden, and brownish to evaluate the color. Use the words strong onion smell, onion smell, and no-onion smell to evaluate the smell. c Use the words strong onion taste, onion taste, and no-onion taste to evaluate the taste. a

b

11. Sprinkle the coffee over the onion and stir to combine. 12. Cook for 1 more minute while stirring. 13. Remove them onto the plate labeled “onion-coffee.” 14. Record your observations in Data Table 8.6. 15. Compare the results. Note: Use the same temperature and time combination to cook the samples.

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The Science of Flavor and Flavor Pairing  287

EXPERIMENT 8.4 OBJECTIVE To understand ingredients sharing the key chemical compounds may pair well in the recipes. Case 1 Eggplant salad Ingredients and Equipment • 2 medium eggplants • 1 teaspoonful chopped fresh cilantro • 1 teaspoonful chopped fresh dill • ¼ cup chopped chives • 4 tablespoonsful olive oil • Salt and pepper • Oven • Tray • Bowl • Knife Method • Preheat the oven to 200°C (392°F). • Cut the eggplants into approximately 3 cm × 3 cm × 3 cm (1.2 in. × 1.2 in. × 1.2 in.) cubes. • Put the eggplants on a greased tray. • Sprinkle with a little salt and pepper. • Drizzle them with 1 tablespoonful olive oil. • Bake in the oven for approximately 30 min, turning once half way through. • Place the eggplant in a bowl. • Sprinkle the chopped cilantro, dill, chives, and the rest of olive oil and then mix them together. • Taste the sample and record your observations in Data Table 8.7. Case 2 Eggplant-Orange salad Ingredients and Equipment • 2 medium eggplants • 2 oranges • 1 teaspoonful chopped fresh cilantro

288  COOKING AS A CHEMICAL REACTION

DATA TABLE 8.7 Color

Smell

Taste

Overall Flavor

Eggplant salad Eggplant-Orange salad

Use your own words to describe the sensory attributes of the samples.

• • • • • • • •

1 teaspoonful chopped fresh dill ¼ cup chopped chives 4 tablespoonsful olive oil Salt and pepper Oven Tray Bowl Knife

Method • Preheat the oven to 200°C (392°F). • Cut the eggplants into approximately 3 cm × 3 cm × 3 cm (1.2 in. × 1.2 in. × 1.2 in.) cubes. • Put the eggplants on a greased tray. • Sprinkle with a little salt and pepper • Drizzle them with 1 tablespoonful olive oil. • Bake in the oven for approximately 30 min, turning once half way through. • Meanwhile, peel the orange and cut into small cubes. • Squeeze the half of the second orange. • Mix it with the rest of the olive oil to prepare the dressing. • Place the eggplant, and the orange in a bowl. • Sprinkle the chopped cilantro, dill, chives, and the dressing, and mix them together. • Taste the sample and record your observations in Data Table 8.7 • Compare the results from Cases 1 and 2.

The Science of Flavor and Flavor Pairing  289

The Science Behind the Results The flavor compounds of foods determine the acceptable food pairing. The common, complementing, and contrasting flavor compounds in different foods may create new flavors when they come together. In general, foods those share common or complementing flavor compounds (especially volatile odor compounds) are more likely to pair well in the recipes (Table 8.2). They may interact with each other so; the brain detects it as a new flavor. Although this might be a good starting point in food pairing (flavor pairing), sharing the common or complementing flavor compounds is not solely enough to produce an acceptable new flavor. Besides that, the amount of flavor compounds in foods must be at the detectable level (threshold) by the brain, and the flavor compounds that may be produced as a result of chemical reactions (such as fermentation, dextrinization, and Maillard reaction) during food processing must also be taken into consideration. Some flavor compounds can suppress the specialized sensory cells on the tongue, which may modify the perceived intensity of the other flavor compounds. For example, miraculin (glycoprotein) in miracle fruit affects the taste receptors, therefore foods eaten after miracle fruits taste much sweeter than usual. Cynarin in artichokes inhibits the sweet-perceiving receptors; when it is washed away with water or another food, perceived sweet taste intensity of the next bite is higher than usual. Similarly, eating and drinking acidic foods (such as lemon) activates the sweet receptors. Therefore, the sequence of foods in tasting sessions (such as cheese-wine) and/or the sequence of food layers (such as canapé, and layered bars) are very important for the success of food-pairing designs.

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TABLE 8.2 Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Tomato Tea (green leaf)

186

Potato

159

Apple

158

Cocoa

149

Passion fruit

141

Melon

139

Celery

132

Plum

136

Olive

125

Coconut

119

Clove

110

Coffee

95

Pasta

93

Feta cheese

57

Green beans

111

Cucumber

109

Red beet (Beetroot)

(Continued)

The Science of Flavor and Flavor Pairing  291

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Peanut

108

Cocoa

104

Strawberry

103

Olive

102

Orange

102

Peppermint

100

Garlic

97

Dates

94

Lemon balm

94

Vanilla

94

Coffee

44

Butter

24

Onion

120

Peanut

108

Carrot

106

Ginger

104

Rice

103

Garlic

(Continued)

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TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Fig

102

Grape

101

Strawberry

100

Orange

100

Cucumber

100

Pineapple

99

Cumin

98

Banana

95

Milk

29

Rum

24

Chicken

19

Yogurt

11

Tea

170

Apple

152

Pepper (spice)

148

Carrot

140

Basil

140

Orange

(Continued)

The Science of Flavor and Flavor Pairing  293

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Grape

138

Potato

132

Mushroom

128

Corn

124

Cocoa

123

Peas

122

Mint

118

Olive

116

Melon

116

Anise

114

Sunflower

109

Cauliflower

109

Eggplant

107

Onion

105

Garlic

100

Cheddar cheese

51

Chicken

35

Pork

32

Beef

28 (Continued)

294  COOKING AS A CHEMICAL REACTION

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Fennel Lemon

124

Black current

120

Tea (green leaf)

118

Fig

110

Papaya

110

Tomato

109

Garlic

102

Cocoa

99

Hazelnut

97

Wasabi

93

Beer

45

Coffee

33

Cheddar

25

Beef

14

Turkey

6

Mushroom

126

Vanilla

(Continued)

The Science of Flavor and Flavor Pairing  295

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Mango

123

Rice

118

Buckwheat

115

Green beans

113

Corn

110

Ginger

108

Capsicum

108

Fig

105

Sour cherry

104

Anise

102

Carrot

100

Avocado

100

Lettuce

98

Cabbage

97

Okra

96

Garlic

94

Swiss cheese

49

Feta cheese

44 (Continued)

296  COOKING AS A CHEMICAL REACTION

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Beef Tea (green leaf)

45

Apple

40

Tomato

40

Coffee

37

Rice

39

Blue cheese

35

Grape

32

Strawberry

28

Apricot

28

Vanilla

23

Coconut

24

Garlic

14

Tea (green leaf)

66

Coffee

57

Peanut

55

Mushroom

51

Chicken

(Continued)

The Science of Flavor and Flavor Pairing  297

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Tomato

51

Mango

46

Blue cheese

43

Lima beans

37

Asparagus

36

Grape

33

Feta cheese

33

Banana

28

Onion

25

Cucumber

23

Almond

20

Mustard

20

Olive

20

Coffee

79

Beer

76

Chicken

69

Tea (green leaf)

64

Pork

(Continued)

298  COOKING AS A CHEMICAL REACTION

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Peanut

57

Butter

56

Apple

48

Parmesan

45

Papaya

42

Cranberry

36

Banana

34

Peas

33

Barley

31

Vanilla

29

Pecan

27

Mustard

21

Garlic

20

Cinnamon

19

Olive

18

Black tea

15

Coffee

42

Egg

(Continued)

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TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Tea (green leaf)

41

Potato

37

Rice

37

Cocoa

34

Milk

33

Tomato

31

Mango

30

Corn

29

Pork

26

Apple

24

Banana

23

Okra

23

Strawberry

22

Vanilla

21

Avocado

21

Beef

20

Lemon

19

Garlic

19 (Continued)

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TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Cattle milk Beer

70

Tea (green leaf)

67

Soybean

57

Cocoa

57

Potato

55

White wine

53

Tomato

53

Grape

50

Lima beans

47

Pork

44

Oats

37

Vanilla

35

Egg

33

Walnut

29

Red meat

26

Honey

22

Eggplant

20

Oyster

18 (Continued)

The Science of Flavor and Flavor Pairing  301

TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Coffee Beer

118

Tomato

95

Processed beef

93

Rum

86

Apple

84

Rice

79

Capsicum

78

Strawberry

75

Parmesan cheese

70

Corn

69

Onion

59

Barley

60

Chicken

57

Olive

57

Cabbage

57

Butter

54

Plum

54

Almond

54 (Continued)

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TABLE 8.2 (Continued) Example for Foods Sharing the Common Flavor Compounds

Selected Ingredient

Pairing

Number of Common Compounds

Avocado

46

Broccoli

41

Garlic

39

Chickpea

36

Apple

50

Pineapple

38

Coffee

33

Tea (green leaf)

33

Cocoa

33

White wine

34

Mushrooms

31

Tomato

31

Parmesan cheese

28

Potato

24

Olive

23

Capsicum

21

Black tea

11

Honey

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POINTS TO REMEMBER Knowledge of basic flavor chemistry is crucial for chefs to understand the principles behind the flavor pairings, and to create new food combinations. Flavor and taste are different sensory attributes. Flavor is a complicated combination of mainly taste and odor. Taste molecules in foods are detected by chemoreceptors on the tongue. Chemoreceptors bundle to form taste buds. Specialized cells in the nose receive the volatile odor molecules in the foods from the back of the mouth, and send signals to the brain. Each food has its own hundreds or even thousands of taste and odor ­molecules that make up its flavor. Some foods may share similar flavor compounds, but may have completely different flavors due to differences in their compositions. The same food may taste different at different temperatures. The flavor compounds of foods determine the success of food pairings. The common, complementing, and contrasting flavor compounds in different foods may create new flavors when they come together. The amounts of flavor compounds and the chemical compounds that are produced during food processing are also the key factors for acceptable food pairings.

304  COOKING AS A CHEMICAL REACTION

More Ideas to Try 1. Repeat Experiment 8.2 Case 3 with a slice of chocolate-cream cake but rub your tongue with an ice cube at Step 4. 2. Repeat Experiment 8.3 Case 2 using 1 teaspoon, 11/2 teaspoon, and 2 ­teaspoons Turkish coffee and compare your results.

SELECTED REFERENCES American Chemical Society National Historic Chemical Landmarks. Flavor Chemistry Research at the USDA Western Regional Research. Center Online at: www.acs.org/ content/acs/en/education/whatischemistry/landmarks/usda-flavor-chemistry.html (accessed Month Day, Year). Bachko, K. 2008. Taste the magic: A tropical berry makes sour foods taste sweet. Online at: http://jscms.jrn.columbia.edu/cns/2008-04-15/bachko-magicfruit.html. Briscione, J., and B. Parkhurst. 2018. The flavor matrix: The art and science of pairing common ingredients to create extraordinary dishes. Boston, MA: Houghton Mifflin Harcourt. Brown, L. 2008. The flavor bible. The Essential Guide to Culinary Creativity, Based on the Wisdom of America’s Most Imaginative Chefs, Little, Brown, and Company, USA. Engelena, L., R. A. de Wijka, J. F. Prinza, A. M. Janssena, H. Weenena, and F. Bosman. 2003. The effect of oral and product temperature on the perception of flavor and texture attributes of semi-solids. Appetite 41:273–281. Flavor DB: a database of flavor molecules. Online at: www.ncbi.nlm.nih.gov/pmc/articles/ PMC5753196/. Flavor DB search: A resource for exploring flavor molecules. Online at: https://cosylab. iiitd.edu.in/flavordb/search. Food Flavors: Chemical, Sensory and Technological Properties. 2011. Chemical and Functional Properties of Food Components. Jelen, H. Ed. CRC Press, USA. Green, B. G. 1993. Heat as a Factor in the Perception of Taste, Smell, and Oral Sensation in Nutritional Needs in Hot Environments: Applications for Military Personnel in Field Operations. The National Academies Press, USA. www.nap.edu/read/2094/ chapter/12. Green, B. G., and S. P. Frankmann. 1987. The effect of cooling the tongue on the perceived intensity of taste. Chemical Senses 12(4):609–619. doi:10.1093/chemse/12.4.609. How do we sense the flavors of food? Online at: https://serendipstudio.org/sci_edu/­ waldron/pdf/SensesProtocol.pdf. Mark T., R. Burke, M. O’Sullivan, J. Hannon, and C. Barry-Ryan. 2013. Sensory and Chemical Interactions of Food Pairings (Basmati Rice, Bacon and Extra Virgin Olive Oil) with Banana.

The Science of Flavor and Flavor Pairing  305

Olga Lipatova, O., and Campolattaro, M. M. 2016. The miracle fruit: An undergraduate laboratory exercise in taste sensation and perception. Journal of Undergrad Neuroscience Education 15(1):A56–A60. Online at: https://arrow.dit.ie/cgi/­ viewcontent.cgi?referer=https://www.google.com.tr/&httpsredir=1&article=1200& context=tfschafart. Physiology of Taste. Online at: www.vivo.colostate.edu/hbooks/pathphys/digestion/­ pregastric/taste.html. Rowe, L. 2015. Taste: The infographic book of food. London: Aurum Press. Segnit, N. 2012. The flavor thesaurus: A compendium of pairings, recipes and ideas for the creative cook. New York: Bloomsbury. Talavera, K., Y. Ninomiya, C. Winkel, T. Voets, and B. Nilius. 2007. Influence of temperature on taste perception. Cellular and Molecular Life Sciences 64:377. Talavera, K., K. Yasumatsu, T. Voets, G. Droogmans, N. Shigemura, Y. Ninomiya, R. F. Margolskee, and B. Nilius. 2005. Heat activation of TRPM5 underlies thermal sensitivity of sweet taste. Nature 438:1022–1025. Taste and Smell. Online at: https://courses.lumenlearning.com/boundless-biology/ chapter/taste-and-smell/. Taste, texture and temperature. Online at: https://maxfacts.uk/help/oral-food/ttt. U.S. Department of Health and Human Service. Taste and Smell Researches. Online at: www.nidcd.nih.gov/about/strategic-plan/2017-2021/taste-and-smell-research.

Chapter 9 Food Additives in Culinary Transformations

CLASSIFICATION OF FOOD ADDITIVES Food additives are any substances usually added to foods during production, preparation, processing, packaging, and transportation. They become a part of the food product. Food additives are generally divided into two groups: intentional/direct food additives and unintentional/indirect food additives.

Intentional/Direct Food Additives They are often added during processing to perform a specific purpose. Intentional food additives are usually added to foods to: 1. Maintain and improve nutritional quality. Certain food additives are used either to replace vitamins and minerals lost in processing (enrichment) or to add nutrients to foods that may not have initially contained that nutrient (fortification). Fruit juices with vitamin C added are the best examples of enriched foods. Margarine is a good example of fortified foods because vitamin A, which is not naturally found in margarine, is added to it during processing.

307

308  COOKING AS A CHEMICAL REACTION

2. Preserve or improve quality and freshness. Preservatives are used to extend the shelf life of food products. They decrease the rate of microbial growth and chemical reactions in foods. Organic acids, nitrites, sulfides, sugar, and salts are the best-known examples of antimicrobial preservatives. Antioxidants, such as vitamin C, BHA, and vitamin E maintain the color and flavor of the food products. They prevent rancidity in lipids and the products that are rich in lipids, such as cooking oils, cookies, and nut spreads. In addition, antioxidants keep peeled or cut fresh fruits and fresh vegetables from turning brown when they are exposed to air. The rate of spoilage and quality loss is usually higher than the rate expected by consumers when these additives are not added to foods during production. 3. Help in processing or preparation. Anticaking agents, humectants, leavening agents, maturing and bleaching agents, pH control agents, thickeners, and stabilizers are the most common food additives used to assist in processing and preparation of foods. For example, emulsifiers, such as lecithin, keep products from separating. Stabilizers and thickeners provide an even consistency or texture. Anticaking agents, such as silicone dioxide, keep foods from absorbing moisture. Humectants, such as glycerol, keep food moist and soft. 4. Make food more appealing. Some additives are added to improve, maintain, or enhance the taste, color, and aroma of foods. They are used in foods to replace the color and aroma lost during processing and storage; to produce a uniform product from raw materials that vary in color intensity; and to enhance the natural colors and aroma of foods. Intentional/direct food additives may be natural, synthetic, or nature identical. Natural food additives are primarily derived from plants and animals. For example, strawberries and beet roots are used to produce red food colorants, and carrots and turmeric are used to produce orange food colorants. Synthetic food additives do not occur in nature, but are made in factories. Petroleumbased chemical compounds are the basic sources of synthetic food additives. Nature-identical additives are factory-made copies of substances that occur naturally.

Unintentional/Indirect Food Additives These are substances that enter foods in trace quantities during their production, processing, storage, or packaging. Indirect additives are commonly known as contaminants.

Food Additives in Culinary Transformations  309

Unintentional/indirect food additives may enter foods through: 1. Polluted air, water, and soil, such as heavy metals. 2. Intentional use of various chemicals, such as pesticides, animal drugs, and fertilizers. 3. Natural sources, such as mycotoxins produced by certain types of microorganisms in foods. 4. A process, such as the chemicals that are formed during deep frying as a result of decomposition of lipids. Indirect/unintentional food additives may imply a short- or long-term risk to human health, and measures must be taken to minimize contaminants in foodstuffs. The summary of food additives is given in Table 9.1. TABLE 9.1 Summary of Food Additives

Purpose of Use

The Most Common Categories

Examples

Examples of Uses in Foods

Intentional/Direct Food Additives To maintain and ımprove nutritional quality

Nutrients, primarily vitamins, minerals, essential fatty acids

Vitamin C, vitamin D, vitamin A, omega 3, iodine, calcium

Flour, breads, cereals, rice, macaroni, margarine, milk, fruit juices

To preserve or ımprove quality and freshness

Antioxidants

Citric acid, phosphoric acid, ascorbic acid, vitamin E, BHA, BHT

Oils and margarines, cereals

Antimicrobial preservatives

Sorbates, sodium nitrite, calcium propionate, sulfites, salt, and sugar

Cured meats, sauces, baked goods, cheese, snack foods (Continued)

310  COOKING AS A CHEMICAL REACTION

TABLE 9.1 (Continued) Summary of Food Additives

Purpose of Use To help in processing or preparation

To make food more appealing

The Most Common Categories

Examples

Examples of Uses in Foods

Emulsifiers

Monoglycerides, diglyceride, lecithin, polysorbate 60

Salad dressings, mayonnaise, chocolate, margarine, butter

Leavening agents

Sodium bicarbonate, calcium carbonate

Baked goods

Color

Blue 1 and 2, citrus red 2, yellow 5 and 6, paprika oleoresin, caramel color, saffron, fruit and vegetable juices

Confection, fruit juices, cakes, frostings, jams, many processed foods

Flavoring

Quinine, spices, plant extracts

Snack foods, sauces, soda, confection, gelatin desserts, jams, marmalades

Flavor enhancers

Monosodium glutamate, hydrolyzed vegetable protein

Dry soups, cereals, many processed foods

Indirect Additives The most common indirect food additives are the antibiotics, growthpromoting hormones, pesticide residues, radioactive residues, chemicals from environmental pollution, polluted water, packaging materials, metals, and industrial wastes.

Food Additives in Culinary Transformations  311

Consumers should feel confident about the safety of the foods that they consume. The intentional food additives must be considered safe in order to be used in food products. The safety of food additives is approved only on the basis of scientific studies and is strictly regulated and controlled by governmental bodies. The risks and benefits of all food additives are scientifically assessed by considering the typical amount of consumption, chemical structure of the food additive, and short- and long-term health effects among different consumer groups, such as children, the elderly, and people with chronic diseases. All food additives are subject to ongoing safety studies. If new scientific studies suggest that a food additive is approved to be safe, but still may not be safe, the legal authorities may prohibit its use. Food legislation in many countries require clear labeling of food additives in the list of ingredients. The E-code given to each food additive indicates that it is a “European Union approved” food additive. Food additives cannot be used 1. to mask faulty processing and food spoilage; 2. if they decrease the nutritional value of the food products; or 3. if the same quality of the foods can be produced without food additives.

312  COOKING AS A CHEMICAL REACTION

Food Additives in Culinary Transformations  313

EXPERIMENT 9.1 OBJECTIVE To evaluate how color additives influence the acceptance of food products by consumers. Lemonade Recipe Ingredients and Equipment • 400 g (14.1 oz) sugar • 800 mL (27 fl oz) water • 8–10 lemons • Lemon rinds • Yellow food coloring • 2 pitchers • 200 transparent drinking cups • Saucepan • Bowl • Stove Method 1. Squeeze juice from lemons into a bowl. 2. Heat the sugar, lemon rinds, and 500 mL (16.9 fl oz) of water in a saucepan until the sugar is dissolved completely. Set aside. 3. Discard rinds. 4. Combine the lemon juice, the sugar syrup, and rest of the water, and mix well. 5. Divide the lemonade into two pitchers. 6. Add two to three drops of yellow food coloring in one of the pitchers and mix well. 7. Refrigerate both pitchers until well chilled (both should come to the same temperature). 8. Carry out sensory analysis tests. Sensory Test Method Prepare two different scorecards 9.1 and 9.2 for each of 60 consumers. Give codes to your sample, i.e., 397 and 298. Carry out the acceptance sensory analysis test (follow the procedure explained in Chapter 7).

298

397

1 Sample (Dislike Code Extremely)

2 (Dislike Very Much) 3 (Dislike Moderately)

4 (Dislike Slightly)

5 (Neither Like or Dislike)

6 7 (Like (Like Slightly) Moderately)

8 (Like Very Much)

9 (Like Extremely)

You Are Provided with Two Coded Samples. Taste the Samples and Indicate Your Degree of Liking. Check a Box, from 1 to 9, to Indicate Your Preference

SCORECARD 9.1

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Food Additives in Culinary Transformations  315

SCORECARD 9.2 Check the Corresponding Boxes to Indicate the Reasons for Liking or Disliking Each Sample

Sample Number Attributes

397

298

It looks appealing It smells good It tastes great It is a good-quality product It does not look appealing It does not smell good It does not taste nice It is not a good-quality product Analysis of the Results 1. Calculate the average scores of scorecard 1 for each sample using the equation given in Chapter 1. 2. Count the number of responses for each attribute of both samples and record in Data Table 9.1. 3. Draw a graph to compare the results for Data Table 9.1. 4. Compare and discuss the results. Tips: Let’s assume the overall responses are found as given in Sample Table 9.1. Figure 9.1 is how a chart on that dataset would look.

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DATA TABLE 9.1 Total Number of Responses for Each Attribute Attributes

Sample 397

Sample 298

It looks appealing It smells good It tastes great It is a good-quality product It does not look appealing It does not smell good It does not taste nice It is not a good-quality product

SAMPLE TABLE 9.1 Total Number of Responses for Each Sample Attributes

Sample 397

Sample 298

It looks appealing

60

39

It smells good

55

43

It tastes great

52

24

It is a good-quality product

80

19

It does not look appealing

30

51

It does not smell good

25

47

It does not taste nice

38

66

It is not a good-quality product

10

71

Food Additives in Culinary Transformations  317

90 Sample 397

80

Sample 298

Number of Responses

70 60 50 40 30 20 10 0

It looks appealing

It smells good

It tastes great

It is a goodquality product

It does not look appealing

It does not smell good

It does not It is not taste nice a goodquality product

Figure 9.1  Sample figure to compare the attributes of the samples.

HYDROCOLLOIDS IN CULINARY TRANSFORMATIONS Food hydrocolloids are biopolymers, which are made of long chains of carbohydrate or protein monomers with side groups (R) on the backbone chain (Figure 9.2). They are known with their affinity for binding water molecules. The size and type of monomers (sugar or protein) and the side groups affect the properties of hydrocolloids such as solubility, gelling property, shear stability, and susceptibility to acid and salts. Carboxyl (COO−), sulfate (OSO3−), and carboxymethyl (OCH2COO−) are the good examples for the side groups found in the ­structures of the most common hydrocolloids used in food processing (Figure 9.3). R

R R

Figure 9.2  Representation of a typical hydrocolloid.

318  COOKING AS A CHEMICAL REACTION

Figure 9.3  K appa-carrageenan (Example).

Food hydrocolloids are extracted from the natural sources such as seaweed, cereal grains, plants, animals, and fermentation byproducts. Some are slightly modified or functionalized to serve more specific purposes. Food hydrocolloids are widely used as intentional food additives in many food products to produce functional food products, to increase the shelf life of the products, to keep or improve the sensory attributes of the foods and drinks, to make the production processes easier and efficient.

Food Additives in Culinary Transformations  319

EXPERIMENT 9.2 OBJECTIVE To explain the functional properties of the most common hydrocolloids used in culinary transformations. Case 1 Direct spherification Apple juice spheres Ingredients and Equipment • 2 L (67.6 fl oz) apple juice (with no pulp) • 10 g (0.022 lb) sodium alginate • 10 g (0.022 lb) calcium chloride • Water • Syringe or pipette • Colander • Measuring cup • Immersion blender • 3 bowls • Plate • Scale Method 1. Using an immersion blender, mix 4 g (0.14 oz) sodium alginate into 150 mL (5.07 fl oz) apple juice in a bowl until it is completely dissolved. 2. Add in 350 mL (11.8 fl oz) apple juice, and continue mixing. 3. Rest the mixture for 45 min. 4. Using an immersion blender, dissolve the calcium chloride in 1.5 L (50.7 fl oz) cold water in a bowl (this mixture is usually called as a calcium bath). 5. Fill the last bowl with cold fresh water (plain water bath) and set aside. 6. Fill syringe or pipette with the apple juice and sodium alginate mixture. 7. Extract it drop by drop into the calcium bath. 8. Collect the apple juice spheres with a colander. 9. Place them carefully in a plain water bath to rinse. 10. Collect them with a clean colander and place them on a plate (EXP 9.1). 11. Evaluate the transparency, shape, color, and hardness; record your observations in Data Table 9.2. EXP 9.1

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Complete the same procedure for 3 g (0.11 oz), 2 g (0.07 oz), and 1 g (0.035 oz) sodium alginate. Keep the amount of the apple juice the same (EXP 9.2).

EXP 9.2

DATA TABLE 9.2 Amount of Sodium Alginate (g)

Transparencya

Shapeb

Colorc

4 3 2 1 Use the terms transparent, translucent, and opaque to evaluate the transparency. Use the term sphere, irregular sphere, and flat to describe the shape. c Use the terms bright golden, dark golden, and brown to describe the color. d Use the terms soft, semisoft, and firm to describe the texture. a

b

Textured

Food Additives in Culinary Transformations  321

Case 2 Reverse spherification Ingredients and Equipment • 100 g (3.53 oz) yogurt • 7.5 g (0.26 oz) sodium alginate • Water • Syringe or pipette • Colander • Measuring cup • Spoon • 2 bowls • Plate Method 1. Dissolve the sodium alginate in 1.5 L (50.7 fl oz) cold water in a bowl (this mixture is usually called as a sodium bath). 2. Leave in the fridge 12 h to eliminate excess air. 3. Fill the last bowl with cold fresh water (plain water bath) and set aside. 4. Fill syringe or pipette with the yogurt. 5. Extract it drop by drop into the sodium bath. 6. Collect the yogurt spheres with a colander. 7. Place them carefully in a plain water bath to rinse for 2 min. 8. Collect them with a clean colander and place them on a plate. Complete the same procedure but use plain carrot juice (or any other non-calcium containing food with pH > 3.6) instead of yogurt. Compare your results with your results from Case 1 in Data Table 9.3.

DATA TABLE 9.3 Food

Shapea

Yogurt Carrot juice a b

Use the terms sphere, irregular sphere, and flat to describe the shape. Use the terms liquid, soft, semisoft, and firm to describe the texture.

Textureb

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Case 3 Coating Ingredients and Equipment • 2 red apples • 12.5 g (0.44 oz) chitosan • 5 mL (0.17 fl oz) vinegar • 2 perforated trays • Water • Colander • Measuring cup • Bowl • Magnetic stirrer • Knife • Spoon Method 1. Label the two trays as “coated apples” and “un-coated apples.” 2. Mix 500 mL (16.9 fl oz) water with vinegar in a bowl. 3. Dissolve the chitosan in water–vinegar solution over a magnetic stirrer for 90 min. 4. Place the perforated trays on the same bench with their upsides down. 5. Peel the apples and slice them in equal pieces. 6. Immediately, place half of the slices into chitosan solution for 2–3 min (for coating). 7. Take them out with a colander and place them on the tray labeled “coated apples” in a single layer. 8. Place rest of the apples (un-coated) on the tray labeled “un-coated apples.” 9. Keep both trays on the bench throughout the observation periods. 10. Compare the samples and record your observations in Data Table 9.4. 11. After the first 5 h, continue to check the samples every 24 h for the first mold growth sign on the surface of one of the sample groups.

Food Additives in Culinary Transformations  323

DATA TABLE 9.4 Time (min)

Coated apples

Un-coated apples

b

Color: Texture: cMicrobial growth:

Color: Texture: Microbial growth:

60

Color: Texture: Microbial growth:

Color: Texture: Microbial growth:

120

Color: Texture: Microbial growth:

Color: Texture: Microbial growth:

180

Color: Texture: Microbial growth:

Color: Texture: Microbial growth:

240

Color: Texture: Microbial growth:

Color: Texture: Microbial growth:

300

Color: Texture: Microbial growth:

Color: Texture: Microbial growth:

0

a

Time (day) 2

3

Coated Apples

Un-coated Apples

b

Color: Texture: cMicrobial growth:

Color: Texture: Microbial growth:

Color: Texture: cMicrobial growth:

Color: Texture: Microbial growth:

a

a

b

…  se the terms no color change, no significant color change, and color change to evaluate the U samples. b Use the terms firm, soft, and wilted to evaluate the samples. c Use the terms no growth, not significant sign of mold growth, and significant mold growth to evaluate the samples. a

324  COOKING AS A CHEMICAL REACTION

Food Additives in Culinary Transformations  325

EXPERIMENT 9.3 OBJECTIVES • To explain the functional properties of the most common hydrocolloids used in culinary transformations. • To explain the effects of concentration of the hydrocolloids, processing temperature, and the other ingredients in the recipe on thickening and gelling properties of hydrocolloids. Case 1 Concentration effect Ingredients and Equipment • 990 g (2.18 lb) gum Arabic • 5 L (169.1 fl oz) water at 25°C (77°F) • Five 2 L (68 fl oz) beakers or cups • Measuring cups • Spoons • Thermometer Method 1. Label the beakers “40 g (1.41 oz) gum Arabic,” “100 g (3.53 oz) gum Arabic,” “150 g (5.29 oz) gum Arabic,” “300 g (10.58 oz) gum Arabic,” and “400 g (14.11 oz) gum Arabic.” 2. Place 40 g (1.41 oz) gum Arabic and 1 L (33.81 fl oz) water into “40 g (1.41 oz) gum Arabic” beaker. Do not forget to check the temperature of water to make sure it is at 25°C (77°F). 3. Stir with a spoon thoroughly. 4. Record your observations in Data Table 9.5. 5. Repeat Steps 2–4 for 100 g (3.53 oz), 150 g (5.29 oz), 300 g (10.58 oz), and 400 g (14.11 oz) gum Arabic.

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DATA TABLE 9.5 Amount of Gum Arabic g (oz)

Viscosity of the Solutiona

40 g (1.41 oz) 100 g (3.53 oz) 150 g (5.29 oz) 300 g (10.58 oz) 400 g (14.11 oz) a

Use the terms thin, thick, viscous, and less viscous to evaluate the viscosity.

Case 2 Temperature effect Ingredients and Equipment • 20 g (0.71 oz) agar • 1 L (33.81 fl oz) water at 25°C (77°F) • Pan • Thermometer • Spoon • Stove Method 1. In a saucepan, combine the agar and cold water. 2. Stir gently with a spoon, measure the initial temperature of the mixture, and record your observations in Data Table 9.6. 3. Place the thermometer into the solution. Thermometer should not touch the sides or bottom of the saucepan. 4. Move the pan to the stove over low heat. Stir the mixture gently throughout the experiment. 5. Record your observations in Data Table 9.6 when the solution reaches 30°C (86°F). 6. Continue heating, and record your observation at temperatures indicated in Data Table 9.6. 7. Remove the saucepan from the stove when the solution comes to 100°C (212°F). 8. Record your observations during the cooling period at temperatures indicated in Data Table 9.6.

Food Additives in Culinary Transformations  327

DATA TABLE 9.6 Temperature (ºC/ºF)

Solubility

a

Viscosity of the solutionb

Gel formationc

Initial: 30/86 40/104 50/122 60/140 70/158 80/176 90/194 100/212 90/194 80/176 70/158 60/140 50/122 40/104 30/86 a

Use the terms soluble, insoluble, and partially soluble to evaluate the solubility.

b

Use the terms thin, thick, more viscous, and less viscous to evaluate the viscosity. Use the terms gel formation and no gel formation to evaluate the gel formation.

c

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Case 3 Temperature effect Ingredients and Equipment • 3 g (0.11 oz) guar gum • 1 L (33.81 fl oz) water at 25°C (77°F) • Pan • Thermometer • Stove Method 1. In a saucepan, combine the guar gum and water. 2. Stir with a spoon, measure the initial temperature of the mixture, and record your observations in Data Table 9.7. 3. Place the thermometer into the solution. Thermometer should not touch the sides or bottom of the saucepan. 4. Move the pan to the stove over low heat. Stir the mixture gently throughout the experiment. 5. Record your observations in Data Table 9.7 when the solution reaches 30°C (86°F). 6. Continue heating, and record your observation at temperatures indicated in Data Table 9.7. 7. Remove the saucepan from the stove when the solution comes to 100°C (212°F). 8. Place the pan on a bench, and leave it to cool down to 25°C (77°F) and record your observations in Data Table 9.7.

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DATA TABLE 9.7 Temperature (ºC/ºF)

Solubilitya

Viscosity of the Solutionb

Gel Formationc

Initial: 30/86 40/104 50/122 60/140 70/158 80/176 90/194 100/212 25/77 Use the terms soluble, insoluble, and partially soluble to evaluate the solubility. Use the terms thin, thick, more viscous, and less viscous to evaluate the viscosity. c Use the terms gel formation and no gel formation to evaluate the gel formation. a

b

Case 4 Ingredient effect For the effects of other ingredients in the recipe on thickening and gelling properties of hydrocolloids, go over Experiment 4.9 and the results obtained from the experiment.

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Food Additives in Culinary Transformations  331

The Science Behind the Results Based on their molecular weight and confirmation, surface charge density, and polymer chain interactions, hydrocolloids play specific roles such as binding, stabilizing, thickening, gelling, film forming, and coating in food processing (see Table 9.2). Nowadays, these functional properties make their application obvious in the professional kitchens as in molecular gastronomy. Understanding the basic structure of hydrocolloids is crucial for chefs because the functional properties of hydrocolloids are primarily related to their structures. By definition, hydrocolloids, to various degrees, are water-soluble. A high number of polar groups in chemical structures of hydrocolloids markedly increases their affinity for binding to water molecules. They also directly interact and cross-link with each other to form a three-dimensional network, which holds a large amount of water. These properties make them to form a dispersion that exhibits the properties of colloid (that is, because they are termed as “hydrocolloids”); they add viscosity to a solution and/or form gels. Hydrocolloids that have been commonly used as thickening agents in culinary transformations include starch, xanthan gum, guar gum, gum Arabic, carboxymethyl cellulose (CMC), and locust bean gum (LGB). Hydrocolloids are either linear (mostly) or branched. In the solution, hydrocolloids come in contact with each other to form a network. The complex structure of branched hydrocolloids restricts their movement, and hence their interaction with the other hydrocolloids in the solution. Therefore, branched hydrocolloids typically show lower viscosities than linear hydrocolloids of the same size. While all hydrocolloids modify the viscosity of solutions, a few of them also have another major property of being able to form gels (see Table 9.2). Gel formation involves three-dimensional hydrated network formation as a result of association or cross-linking of the polymer chains. Pectin, agar, alginate, carrageenan, and gelatin are the major hydrocolloids that have been used as gelling agents in culinary transformations. Sodium alginate is a salt of alginic acid, which can be found in cell walls of brown algae. In direct spherification, when liquid is dropped into a calcium bath, calcium ions displace the sodium ion and cross-link the chains of alginate, since it has positive two charges (Ca2+). The chemical reaction can be shown as:

Na + − alginate  +  CaCl 2 → Ca − alginate + NaCl

The calcium ions act like a bridge between alginate polymers, improving their interactions. This is commonly termed “egg-box” model (Figure 9.4).

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-

COO Na

-

COO Na

+

+

-

COO Na

-

COO Na

+

-

COO Na

-

+

Ca +2

-

COO Na

-

COO Na

+

+

-

COO Na

-

COO Na

Sodium Alginate

+

-

COO Na

+ COO- Na +

Ca+2 -

-

COO

COO

Ca+2

Ca+2

-

COO

+

COO

-

COO

COO

-

COO

-

-

+ COO- Na +

-

COO

-

COO

-

COO

COO

-

COO

Calcium ion salt bridge

Figure 9.4  Representation of spherification.

The droplet solidifies from the outside inward; the result is a liquid encapsulated in a gel film. When the solidified droplet is removed from the calcium bath and rinsed in fresh water, gel formation reaction continues because some calcium ions have diffused into the droplet. In reverse spherification, sodium ions in a sodium alginate bath are switched with calcium ions on the surface of the liquid droplet, forming a sphere gel film outward from the liquid–liquid interface. When the solidified droplet is removed from the alginate bath and rinsed in fresh water, gel formation stops since sodium alginate cannot diffuse into the droplet. Therefore, reverse spherification can be manipulated easily, and gives better results when the thickness of the gel is a critical factor in the process. Several hydrocolloids such as chitosan, and cellulose and cellulose derivatives are used as coating materials or in production of edible films in culinary transformations due to their gel formation properties. In general, film forming hydrocolloids form networks with hydrogen bonds, and/or electrostatic, and hydrophobic interactions. These films play a role in controlling biological and chemical reactions such as lipid oxidation and enzymatic browning reactions, and moisture loss in foods because they act as moisture and gas barriers. Coating foods with edible films also reduces lipid uptake in food frying processes. In certain cases, macroand/or micronutrients or flavor additives are incorporated in edible films to improve the nutritional values of food products. Overall, functional properties of hydrocolloids mainly depend on the type of hydrocolloid, chemical structures of hydrocolloids, chemical properties of foods, concentration of hydrocolloids, pH of the medium, and temperature (Table 9.2).

Red algae extract

Brown algae

Alginate

Main Source

Agar

Major Hydrocolloids Used in Culinary Transformations

A gelling agent. Soluble in cold water. Forms thermoirreversible gel in the presence of calcium ions at pH values above 3.6. Gel does not melt on heating; it must be mixed in cold liquid. In culinary transformations, required concentration is mostly in between 0.3% and 1.5%.

A gelling agent. It is a heterogeneous polysaccharide. Forms gels at 35°C–45°C (95ºF–113ºF), upon cooling solutions at 80°C–90°C (176°F–194°F). Insoluble in cold liquids, becomes soluble above 80°C (176°F). Stable in a wide pH range. In culinary transformations, required concentration is mostly in between 0.1%–2%.

Major Properties

Properties of Major Hydrocolloids and Their Most Common Applications

TABLE 9.2

(Continued)

• encapsulation • film formation • coating • quick set dessert mix gel • dressings • ice-cream • bakery products

• vegetarian gelatin substitute • jellies • water- and milkbased desserts • confectionary • bakery products

Examples For Uses in the Food Industry

Food Additives in Culinary Transformations  333

Plant extract/ modified

Red seaweeds

Carrageenan (kappa, iota, lamda)

Main Source

Carboxymethyl cellulose (CMC)

Major Hydrocolloids Used in Culinary Transformations

A gelling agent. Forms thermo-reversible transparent gels on cooling. All forms are soluble in water at 80°C (176°F). At room temperature, viscosity is stable over a wide pH range. Kappa carrageenan forms firm, brittle gel with the addition of potassium. Iota type forms soft, elastic gel in presence of calcium. Iota gives higher viscosity compared to kappa. Lamda carrageenan is non-gelling, and mostly used as a thickener in cold liquid. In culinary transformations, required concentration is mostly in between 0.02% and 1.5%.

A thickener. Forms thermo-reversible soft gel at 50°C–60°C (122°F–140°F). Soluble in cold water. Viscosity reversibly decreases with increasing temperature. Tolerates a wide range of pH values. Shear thinning but re-thickening a while after the applied force is removed. In culinary transformations, required concentration is mostly in between 0.05% and 0.8%.

Major Properties

Properties of Major Hydrocolloids and Their Most Common Applications

TABLE 9.2 (Continued)

(Continued)

• desserts • puddings • milk shakes • ice-cream • soups • sauces • water-based jellies • cheese spreads • salad dressings • meat texturizer

• salad dressings • deserts • cake batters • beverages • whipped toppings • stabilizer • fat replacer

Examples For Uses in the Food Industry

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Main Source

Plant extract

Exudate from two species of the Acacia tree

Major Hydrocolloids Used in Culinary Transformations

Guar gum

Gum Arabic

A thickener. Soluble in cold water. Highly branched chemical structure. Viscosity is very low. Stable over a wide range of pH values (4–9). Shear thinning at low shear rates. It does not form gels. 40%–50% of gum Arabic must be used to achieve high viscosities.

A thickener. Soluble in cold water. Viscosity is high in cold solutions, low in hot solutions. Shear thinning. Viscosity decreases only at extreme pH values and at high temperatures (above 90°C (194°F)). It does not form a gel. In culinary transformations, required concentration is mostly in between 0.05% and 0.5%.

Major Properties

Properties of Major Hydrocolloids and Their Most Common Applications

TABLE 9.2 (Continued)

(Continued)

• soft drinks • fruit juices • chewing gums • confections • brewing

• pastry fillings • dairy products • cake batter • ice-cream • fruit juices • bakery products • soups • sauces

Examples For Uses in the Food Industry

Food Additives in Culinary Transformations  335

Main Source

Plant

Plant extract

Major Hydrocolloids Used in Culinary Transformations

Locust bean gum

Pectin (highmethoxyl, HM and low-methoxyl, LM)

A gelling agent. It has a high film formation ability. Viscosity of solutions are low. HMP is soluble in cold water. Viscosity increases with concentration. LMP is only soluble as Na or K salt. Shear stability is high. LMP forms thermo-reversible gels on cooling with Ca2+. HMP forms thermo-irreversible gels on cooling at low pH values (below 3.5) in the presence of sugar. Both LMP and HMP form transparent gels. In culinary transformations, required concentration is mostly in between 0.5% and 6.0%.

A thickener. Solubility in water is limited at low temperatures. Forms a viscous opaque solution above 80°C (176°F) upon cooling. Stable over a wide range of pH values. Sheer thinning. Forms gels only with xanthan or kappa-carrageenan. In culinary transformations, required concentration is mostly in between 0.2% and 2.0%.

Major Properties

Properties of Major Hydrocolloids and Their Most Common Applications

TABLE 9.2 (Continued)

(Continued)

• jam • jelly • marmalade • glazes • dairy • dairy-based desserts • bake stable fillings

• diabetic products • ice-cream • chilled dairy dessert • meat products

Examples For Uses in the Food Industry

336  COOKING AS A CHEMICAL REACTION

Main Source

Microbial

Major Hydrocolloids Used in Culinary Transformations

Xanthan gum

A thickener. Soluble either in cold and hot water. Viscosity is high even at very low concentrations. Viscosity of solution is high independent of temperature and is not affected by salts, and pH changes. Forms gels only in combination with locust bean gum. Gel is thermo-reversible, and gel transparency is low. In culinary transformations, required concentration is mostly in between 0.1% and 1.5%.

Major Properties

Properties of Major Hydrocolloids and Their Most Common Applications

TABLE 9.2 (Continued)

• salad dressings • sauces • dairy desserts • gluten-free products • soups • gravies • ketchup • instant dry mix products • bakery fillings

Examples For Uses in the Food Industry

Food Additives in Culinary Transformations  337

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Food Additives in Culinary Transformations  339

EXPERIMENT 9.4 OBJECTIVES • To explain the functional properties of the most common hydrocolloids used in culinary transformations. • To explain the synergistic effects of hydrocolloids Case 1 Gluten free cupcake Ingredients and Equipment • 1.2 g (0.042 oz) xanthan • 1.2 g (0.042 oz) guar gum • 75 g (2.65 oz) soy flour • 75 g (2.65 oz) rice, potato, or corn flour • 50 g (1.76 oz) cacao powder • 200 g (7.05 oz) sugar • 1 teaspoon baking powder • ¼ teaspoon baking soda • 115 g (4.06 oz) butter • 2 large eggs • 125 mL (4.23 fl oz) milk • Electric mixer • Cupcake tray • Cupcake cases • Spoon • Knife • Oven • Two bowls • Cooling racks Method 1. Preheat oven to 177°C (350°F). Fill cupcake tray with cupcake cases. 2. In a large bowl add the dry ingredients and mix to combine. 3. In a separate bowl combine the softened butter and sugar, and beat with the electric mixer until fluffy. 4. Whisk in eggs, one at a time. 5. Set the electric mixer at low speed, add the wet ingredients to the dry ingredients and continue mixing for about 2–3 min until smooth. 6. Spoon the mixture into each cupcake case about 2/3 full. 7. Bake for 18–20 min until golden brown. 8. Place the pan on a cooling rack and rest for 2–3 min.

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9. Gently remove the cupcakes from the pan and cool completely. 10. Remove one cupcake from the case and cut it in half. 11. Evaluate the chewiness, color, taste, and texture; record your observations in Data Table 9.8. Case 2 Gluten free cupcake without xanthan gum Ingredients and Equipment • 1.2 g (0.042 oz) guar gum • 75 g (2.65 oz) soy flour • 75 g (2.65 oz) rice, potato, or corn flour • 50 g (1.76 oz) cacao powder • 200 g (7.05 oz) sugar • 1 teaspoon baking powder • ¼ teaspoon baking soda • 115 g (4.05 oz) butter • 2 large eggs • 125 mL (4.23 fl oz) milk Method Complete the same method from Case 1, and record your observations in Data Table 9.8. Case 3 Gluten free cupcake without guar gum Ingredients and Equipment • 1.2 g (0.042 oz) xanthan gum • 75 g (2.65 oz) soy flour • 75 g (2.65 oz) rice, potato, or corn flour • 50 g (1.76 oz) cacao powder • 200 g (7.05 oz) sugar • 1 teaspoon baking powder • ¼ teaspoon baking soda • 115 g (4.05 oz) butter • 2 large eggs • 125 mL (4.23 fl oz) milk Method Complete the same method from Case 1, and record your observations in Data Table 9.8.

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DATA TABLE 9.8 Muffin with Xanthan Gum and Guar Gum

Muffin with Xanthan Gum

Muffin with Guar Gum

Chewiness

a

b

Color

Taste

c

Structure

d

Use the terms soft, chewy, and hard to evaluate the chewiness of the samples. Use your own terms to evaluate the taste and color of the samples. d Use the terms elastic, firm, with holes inside, and without holes inside to evaluate the structure of the samples. a

b,c

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Food Additives in Culinary Transformations  343

The Science Behind the Results In food processing, single or combination of hydrocolloids are used to attain the desired functional properties. A synergistic interaction occurs between some hydrocolloids; when combined they create something different from what they form when using them alone. For example, a blend of guar gum and xanthan gum adds viscosity and provides the texture that gluten-free breads and bakeries often lack. Locust bean gum and xanthan gum mixture increases the viscosity and enhances the gel formation in solutions. Locust bean gum and agar (or alginates) improve the gel strength and elasticity when they are used together in certain ratios. The differences in the degree of synergism and the nature of interactions between hydrocolloids can be attributed to the chemical structure of the hydrocolloids. The ratios of the hydrocolloids and their concentrations in the solution are important to attain the desired functional properties.

Study Question Explain the differences between direct and reverse spherifications.

More Ideas to Try Place 100 mL (3.38 fl oz) vegetable oil in a refrigerator a few hours before the experiment. Add 1 g (0.035 oz) agar into 100 mL (3.38 fl oz) apple juice, and mix gently. Bring it to boil over low heat. With a syringe or a pipette, add the liquid drop by drop into a very cold oil. Strain and rinse. Observe the fast gelling capability of agar. Repeat Experiment 9.1, but color the lemonade with blue food coloring instead of yellow. Analyze the consumer’s attitudes toward the samples that have unusual colors. Compare your results with your results from Experiment 9.1.

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POINTS TO REMEMBER Food additives are any substances added to foods during production, preparation, processing, packaging, and transportation. Food additives are divided into two groups: intentional/direct and unintentional/indirect. Intentional/direct food additives are often added during processing to perform a specific purpose: a. To maintain and improve nutritional quality b. To preserve or improve quality and freshness c. To help in processing or preparation d. To make food more appealing Intentional/direct food additives may be natural, synthetic, or nature identical. Food additives are approved as safe only on the basis of scientific studies and are strictly regulated and controlled by governmental bodies. Food additives cannot be used: a. To mask faulty processing and food spoilage b. If they decrease the nutritional value of the food products c. If the same quality of the foods can be produced without food additives The E-code given to the food additive indicates that it is a “European Union approved” food additive. Unintentional/indirect additives enter foods in trace quantities during their production, processing, storage, or packaging. Antibiotics, growth-promoting hormones, pesticide residues, radioactive residues and chemicals from the packaging materials, metals, and industrial wastes are the most common unintentional/indirect additives in foods. Unintentional/indirect additives may show short- or long-term adverse health effects. Food hydrocolloids are biopolymers, which are made of high molecular weight polysaccharides or proteins with side groups. They are widely used as intentional food additives in many food products. Hydrocolloids are known with their affinity for water. The molecular weight and confirmation, surface charge density, and polymer chain interactions of hydrocolloids determine their functional properties in culinary transformations. Chemical properties of foods, concentration of hydrocolloids, pH of the medium, and temperature also affect the efficiency of hydrocolloids.

Food Additives in Culinary Transformations  345

Hydrocolloids play specific roles such as binding, stabilizing, thickening, gelling, film forming, and coating in food processing. In food processing, single or a combination of hydrocolloids are used to attain the desired functional properties.

SELECTED REFERENCES Banerjee, S., and S. Bhattacharya. 2012. Food gels: Gelling process and new applications. Critical Reviews in Food Science and Nutrition 52(4):334–346. Brown, A. C. 2007. Understanding food: Principles and preparation, 3rd ed. Belmont, CA: Wadsworth Publishing. Dipjoti, S., and S. Bhattacharya. 2010. Hydrocolloids a thickening and gelling agents in food: A critical review. Journal of Food Science and Technology 47(6):587–597. Food texture and nutrition: The changing roles of hydrocolloids and food fibers. Online at: www.aocs.org/stay-informed/read-inform/featured-articles/food-texture-andnutrition-the-changing-roles-of-hydrocolloids-and-food-fibers-march-2015. Gómez-Díaz, D., Navaza, J. M., and L. C. Quintáns-Riveiro. 2008. Intrinsic viscosity and flow behaviour of arabic gum aqueous solutions. International Journal of Food Properties 11(4):773–780. Hoefler, A. C. Introduction to food gums chemistry, functionality, and applications. Online at: http://web.utk.edu/~jmount/Classes/515/gums.pdf. International Food Information Council (IFIC) and U.S. Food and Drug Administration. 2004, revised 2010. Food ingredients and colors. Online at: www.fda.gov/­downloads/ Food/FoodIngredientsPackaging/ucm094249.pdf. Karaman, S., Kesler, K., Goksel, M., Dogan, M., and A. Kayacier. 2014. Rheological and some physicochemical properties of selected hydrocolloids and their interactions with guar gum: Characterization using principal component analysis and viscous synergism index. International Journal of Food Properties 17(8):1655–1667. Lersch, M. 2014. Texture: A hydrocolloid recipe collection. Online at: https://blog.­khymos. org/wp-content/2009/02/hydrocolloid-recipe-collection-v3.0.pdf. McGee, H. 2004. On food and cooking. The science and lore of the kitchen, 1st Scribner rev. ed. New York: Scribner. McGee, H. 2010. Modern gastronomy A to Z. Boca Raton, FL: CRC Press/Taylor & Francis Group. Nussinovitch, A., and M. Hirashima. 2013. Cooking innovations: Using hydrocolloids for thickening, gelling, and emulsification. Boca Raton, FL: CRC Press. Ozilgen, S., and S. Malta. 2018. Functional properties of biopolymers in food manufacturing in “Handbook of Food Bioengineering, Volume 20”. Elsevier Publishing Company. Purdue University. 2002. Food additives. Online at: www.four_h.purdue.edu/foods/ Food%20additives.htm.

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Shepherd, E. Hydrocolloids for cooks: A simple introduction to hydrocolloids & other modern ingredients for chefs & home cooks. Online at: www.veggiechef.co.uk/Blog/ files/hydrocolloids-for-cooks.pdf. Tanja Wüstenberg, T. 2014. General overview of food hydrocolloids in cellulose and cellulose derivatives in the food industry. Wiley online library. doi:10.1002/9783527682935. ch01. This, H. 2007. Kitchen mysteries: Revealing the science of cooking. New York: Columbia University Press. Vaclavic, V. A., and E. W. Christian. 2008. Essentials in food science, 3rd ed. Berlin: Springer. Van der Linden, E., McClements, D., and Ubbink, J. 2008. Molecular gastronomy: A food fad or an interface for science-based cooking? Food Biophysics 3(2): 246–254. Wang, K., Lu, F., Li, Z., Zhao, L., and C. Han. 2017. Recent developments in gluten-free bread baking approaches: A review Food Science and Technology 37: 1–9.

Chapter 10 Food Safety and Hygiene in Culinary Transformations

CONSUMERS HAVE A RIGHT TO EXPECT THAT THE FOODS THEY CONSUME WILL BE SAFE AND OF HIGH QUALITY Consumption of contaminated foods or drinks is the most common cause of ­foodborne illnesses. Food allergies (such as proteins allergies) and sensitivities (such as gluten and lactose intolerance) may also be considered as foodborne illnesses because some foods or food ingredients that most consumers can tolerate without any problem may have adverse health effects on certain people. For the food business, safe food production is critical to protect the health of the consumers, to minimize the risk of foodborne illnesses, to build and maintain the loyal consumer volume, to avoid financial loss due to loss of business, to satisfy government laws and regulations, and to avoid unexpected expenses on disposal, recalls, and fines.

347

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FOOD SAFETY HAS TO DO WITH CONTROLLING POTENTIAL FOODBORNE HAZARDS A foodborne hazard refers to any agent with the potential to cause illness or injury when consumed by the people along with the food. Foodborne hazards may be: 1. Biological The most common are: • • • •

bacteria molds viruses parasites

Bacteria are the major concern since they are the major causes of 90% of biological contaminations. They can be found everywhere, and most of them grow fast under proper conditions. Molds rarely cause illnesses but they spoil the foods. Viruses need a host cell to grow. They cannot reproduce in the foods; they may cause foodborne illness once consumed. Parasites need a host cell to grow. Proper heat treatment can kill the parasites. The primary sources of microorganisms into foods are the soil, water, environment, tools and utensils, plants and animals, insects and pets, food handlers, and so on. 2. Chemical Such as detergents, sanitizers, antibiotics, pesticides, herbicides, natural toxins, heavy metals, and unapproved uses of food additives. 3. Physical Such as hair, fingernail, stones, pips, bone pieces, glass, staples, and toothpicks. Taking safe steps at every stage of food production is crucial for chefs to prevent foodborne hazards and related foodborne illness. This is commonly known as “food safety from farm to table or food safety from farm to fork.”

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EXPERIMENT 10.1 OBJECTIVES • To understand the types of bacteria in foods. • To explain the beneficial bacteria in foods. Case 1 Sour Cream Production Ingredients and Equipment • 242 g (0.53 lb) pasteurized cream (20%–36% butterfat) • 0.4 g (0.014 oz) sour cream starter culture (bacterial culture) • 1 tall large saucepan (to make a double boiler) • 1 medium saucepan (must fit on the large pan) • 1 spoon • 1 food thermometer • 1 medium-sized jar or a container with a lid • Water • Paper towel • Elastic band Method 1. Fill the large saucepan with about 4 cm (1.57 in.) of water. 2. Place the saucepan on the stove and heat until boiling. 3. Reduce heat to the temperature at which the water reaches a steady simmer. 4. Analyze the cream and record your observations in Data Table 10.1. 5. Pour the cream into the medium saucepan. 6. Place the medium saucepan over the simmering pot. 7. Place the thermometer in the saucepan. Be sure the thermometer does not touch the bottom of the saucepan. 8. Heat until the cream reaches 65°C (149°F). 9. Remove the medium saucepan from heat and hold it at this temperature for 30 min. 10. Check the temperature of the cream frequently and place the saucepan over the large saucepan if temperature drops during this period. 11. Cool the cream to 25°C (77°F). 12. Add the starter culture and mix it into the cream thoroughly. 13. Pour the mixture into the jar and cover with double folded paper towel and secure the towel with an elastic band. 14. Place the jar in an environment at around 25°C (77°F) for 7–8 h. 15. Remove the paper towel and the elastic band.

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DATA TABLE 10.1 Cream

Sour Cream

Texturea Smellb Tastec Colord Use the terms smooth, soft, runny, thick, and thin to evaluate the texture. Use the terms fresh, buttery, and acidic to evaluate the smell. c Use the terms fresh, milky, buttery, and acidic to evaluate the taste. d Use the terms white and yellowish white to evaluate the color. a

b

6. Record your observations in Data Table 10.1. 1 17. Place the lid on the jar and refrigerate. Case 2 • To understand types of bacteria in foods. • To explain the spoilage bacteria in foods. Ingredients and equipment • 100 mL (3.38 fl oz) whole milk • 1 medium-sized jar or a water glass Methods 1. Pour the milk in the jar. 2. Set the jar in open air at room temperature. 3. Record your observation in Data Table 10.2 daily. 4. Stop the experiment when you notice an unpleasant smell and curdling in the milk. 5. Pour out the milk. Note: Never taste the samples

FOOD SAFETY AND HYGIENE  351

DATA TABLE 10.2 Smell Day 1 Day 2 Day 3 …

Use your own words to describe the attributes.

Color

Appearance

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FOOD SAFETY AND HYGIENE  353

The Science Behind the Results Are all microorganisms harmful? The answer is NO! Foodborne microorganisms can be classified in three groups: 1. Beneficial/Fermentative (The Good) 2. Pathogenic (The Bad) 3. Spoilage (The Ugly) The beneficial microorganisms are intentionally added into foods to preserve and/or to produce food products. For example, lactic acid bacteria are used in yogurt, sour cream, cheese, and kefir production. Special types of yeasts are used to make bread and beer, and some types of molds are used to make certain cheeses such as blue cheese. The pathogenic microorganisms are the major causes of foodborne illness. They cause health problems in two primary ways: • Infection: caused by ingestion of live bacteria in or on food. Bacteria can grow in the human digestive system. • Intoxication: usually caused by ingestion of toxins produced by bacteria in food. The live microorganism does not have to be consumed for intoxication. Pathogenic microorganisms usually do not change the sensory characteristics of foods. No one can see, smell, or taste the changes in foods caused by pathogenic bacteria, and this makes them difficult to detect. Tasting, smelling, and/or looking does not predict the safety of the foods. Therefore, do not count on your five senses to judge the safety of the foods. The spoilage microorganisms change the sensory properties of foods that make foods undesirable (spoiled). The food contaminated with spoilage microorganisms usually look and/or smell unappealing, and this makes them easy to detect. They usually do not cause short-term illness. Discolored, mushy, or fuzzy fruits and vegetables, moldy bread, sour milk, and slimy meat are the examples for spoiled foods.

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FOOD SAFETY AND HYGIENE  355

EXPERIMENT 10.2 OBJECTIVE To explain microbial growth requirements Case 1 Temperature and time effects Ingredients and Equipment • 300 mL (10.14 fl oz) pasteurized whole milk • 3 drinking glasses • Oven • Plastic wrap • Graduated cup Methods 1. Label the drinking glasses “room temperature,” “refrigeration temperature,” and “oven temperature.” 2. Set the oven at 40°C (104°F). 3. Pour 100 mL (3.38 fl oz) (milk in each glass and cover them tightly with a plastic wrap. 4. Place the glass labeled “room temperature” in a safe place in the room with no direct sunlight, the jar labeled “refrigeration temperature” in the refrigerator, and the jar “labeled oven temperature” in the oven. 5. After 24 h, record your observations in Data Table 10.3. 6. Continue observation until clear spoilage in all the samples are detected. 7. Discard the samples and clean drinking glasses thoroughly. 8. Compare the results. Note: Never taste the samples. Case 2 Ingredients and Equipment Initial microbial load effect • • • • •

400 mL (13.5 fl oz) UHT whole milk 3 drinking glasses Oven Plastic wrap Graduated cup

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Method 1. Complete the same stages of Case 1 with UHT milk, and record the results in Data Table 10.3. 2. Compare the results. Note: Never taste the samples. DATA TABLE 10.3 Day 1

Day 2

Day 3

Day 4

…

Changes in pasteurized milk sample stored at room temperature (Case 1) Changes in UHT milk sample stored at room temperature (Case 2) Changes in pasteurized milk sample stored at refrigeration temperature (Case 1) Changes in UHT milk sample stored at refrigeration temperature (Case 2) Changes in pasteurized milk sample stored at oven temperature (Case 1) Changes in UHT milk sample stored at oven temperature (Case 2)  Use the terms no changes in consistency, no changes in color, yellowish in color, smells milk, slightly smells sour, smells sour, thin white layer is formed around the surface, thick white layer is formed around the surface, some curdling, and more curdling to evaluate the samples.

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Case 3 Food moisture content and time effects Ingredients and Equipment • 5 equal slices of bread (from the same loaf or the package) • Bread toaster or a toaster oven • 5 Ziploc© bags • Kitchen scale • Paper towel Methods 1. Label the Ziploc© bags “control,” “1 min heated,” “2 min heated,” “3 min heated,” and “4 min heated.” 2. Place a slice of bread into “control” Ziploc© bag. 3. Weigh another slice and record the weight in Data Table 10.4. 4. Place it in the toaster and heat for 1 min. 5. Cool it down on the paper towel for 2 min. 6. Weigh and record the result in Data Table 10.4. 7. Place the slice into “1 min heated” Ziploc© bag and close tight. 8. Calculate the moisture loss, and record the result in Data Table 10.4.

DATA TABLE 10.4 İnitial Weight Control 1 min heated bread sample 2 min heated bread sample 3 min heated bread sample 4 min heated bread sample

Final Weight -

% Moisture Loss after Heating

Time for the First Sign of Mold Growth (days)

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9. Repeat Steps 3–8 with “2 min heated,” “3 min heated,” and “4 min heated” bread samples. 10. Store all the samples at room temperature under the same conditions. 11. Observe the samples every day for the first sign of mold growth and record your observations. 12. Discard all the samples after clear sign of mold growth on every sample. 13. Compare the results. Note: Never taste the samples. Method to calculate the moisture loss: Moisture loss (g) = Initial weight of the slice (g) − Final weight of the slice (g)

% Moisture loss =

Moisture loss ( g ) × 100 Initial weight (Initial weight of the slice ( g ))



Case 4 Acidity effect Cucumber Pickle Ingredients and Equipment • 2 kg (4.41 lb) cucumber (gherkin or ridget) • 3 L (0.8 gal) water • 450 mL (15.2 fl oz) vinegar • 20 garlic cloves • 4 tablespoons salt • 4 large jars with lids • Saucepan • Bowl • Kitchen scale • Graduated cup • Thermometer • Stove Methods 1. Label the jars “control,” “250 mL vinegar,” “150 mL vinegar,” and “50 mL vinegar.” 2. Place 500 g (1.10 lb.) cucumber and 4 garlic cloves into each jar. 3. Place 750 mL (25.4 fl oz) water in the bowl. 4. Add 1 tablespoon salt to the water. Mix until all salt crystals dissolve.

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5. Pour the salt solution into the “control” jar and cover the jar tightly with lids. 6. Place 750 mL (25.4 fl oz) of water in a saucepan. 7. Add 50 mL (1.69 fl oz) of vinegar and 1 tablespoon of salt. Mix until all salt crystals dissolve. 8. Boil the solution for a minute or two. 9. Cool the solution down to 25°C (77°F). 10. Pour the solution into the “50 mL (1.69 fl oz) vinegar” jar and cover the jar tightly with lids. 11. Repeat Steps 6–10 with 150 mL (5.07 fl oz) vinegar and 250 mL (8.45 fl oz) vinegar, respectively (always start with lower concentration if you are going to use the same saucepan for all concentrations). 12. Store all the samples at room temperature protected from direct light under the same conditions. 13. Observe the surface of the liquids (do not remove the lids) every day at the same time for signs of microbial growth. 14. Record your observations in Data Table 10.5. Note: Never taste the samples.

DATA TABLE 10.5

Time (day)

Microbial Growth on the Surface of Control Sample

Microbial Growth on the Surface of 50 mL (1.69 fl oz) Vinegar Sample

Microbial Growth on the Surface of 150 mL (5.07 fl oz) Vinegar Sample

Microbial Growth on the Surface of 250 mL (8.45 fl oz) Vinegar Sample

1 2 3 4 …  Use the terms no growth, no significant sign of microbial growth, and significant microbial growth to evaluate the samples.

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Case 5 Environmental moisture content effect Ingredients and Equipment • 1 cube feta cheese slice (approximately 10 cm × 10 cm × 10 cm/3.9 in. × 3.9 in. × 3.9 in.) feta cheese slices • 2 equal-sized (approximately 10 cm × 10 cm × 10 cm/3.9 in. × 3.9 in. × 3.9 in.) hard cheese slices • 2 medium-sized glass storage containers with lids Method 1. Label the containers “control” and “higher environmental humidity.” 2. Place a slice of hard cheese into the containers marked “control.” 3. Place a slice of hard cheese and a slice of feta cheese into the same ­containers marked “higher environmental humidity.” The slices should not touch each other. Note: During storage, due to the difference in free water contents of cheeses, the relative humidity of the environment will increase. 4. Cover the containers tight with lids. 5. Store both containers at room temperature in a safe place. 6. Check the surface of the samples every day at the same time for signs of microbial growth. Do not open the lids. 7. Record your observations in Data Table 10.6. Note: Never taste the samples. DATA TABLE 10.6

Time (day)

Microbial Growth on the Surface of Control Sample

Microbial Growth on the Surface of the Samples Stored at Higher Environmental Humidity

1 2 3 4 …  Use the terms no growth, no significant sign of microbial growth, and significant microbial growth to evaluate the samples.

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The Science Behind the Results Bacteria are living organisms. Under optimum conditions, they multiply through binary fission process, in which a single bacterium grows and divides to form two new bacteria. This means production of billions of bacteria from one cell in a few hours (Figure 10.1). Microorganisms have six major requirements to grow: • • • • • •

Food Acidity Temperature Time Oxygen Moisture

Proper control of these requirements is necessary to control microbial growth. Foods are used as an energy source by microorganisms to grow and multiply. Foods that promote the rapid growth of microorganisms are called Potentially Hazardous Foods (PHF). These foods are rich in especially proteins and/or

Figure 10.1  Binary fission process.

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carbohydrates. Some examples are meat, poultry, milk, some dairy products, fish, eggs, cooked potatoes, cooked dry beans, cooked rice, some bakery products, melons, and tomatoes. The amount and type of nutrients required for growth depend on the microorganisms. Acidity is measured with a scale called pH (potential of hydrogen) that ranges from 0 to 14. Foods with pH values 7 are considered neutral, lower than 7 are acidic, and above 7 are basic. Low-acid foods have pH values higher than 4.6, while high-acid foods have pH values 4.6 or lower. Most foodborne bacteria prefer neutral to mildly acidic foods. That’s why most of them do not grow on/in acidic foods like pickles, citrus fruits, yogurt, and vinegar; however, they may survive. Molds predominate in low-pH foods because they can grow in a wide range changing from pH 2 to 9. Approximate pH values of selected foods are given in Table 10.1. Temperature is one of the key requirements for microbial growth. While every bacterial species has specific optimum growth temperature, almost all can grow in a temperature range between 5°C (41°F) and 60°C (140°F), called Temperature Danger Zone (TDZ). Most pathogenic bacteria grow best between 20°C (68°F) and 45°C (113°F) having optimum growth temperature 37°C (98.6°F). These are called Mesophiles (middle-loving). Mesophilic bacteria are the most common causes of foodborne poisonings and food spoilage. Psychrophiles (cold-loving) are responsible for most refrigerated food spoilage. They grow best between −5°C (23°F) and 20°C (68°F) having optimum growth temperature 10°C (50°F). Thermophiles (hot-loving) have high optimum growth temperatures (55°C/131°F). They grow best between 45°C (113°F) and 80°C (176°F). Always remember!!! Bacteria do not die when stored at or below 5°C/41°F; only their rate of growth decreases. That means most bacteria grow during cold storage, but the generation time is high. Recommended cold storage temperature and time combination of food depends on types of food (Table 10.3). Time required for a single bacterium to grow and divide to form two new bacteria is called generation time. When bacteria are first introduced to a food, they need time to get used to the medium and new conditions. During this period (known as lag phase), generation time is long, and rate of growth is low. If the conditions are optimum (growth requirements are satisfied), they start to grow rapidly after the lag phase (known as log phase). Generation time in the log phase varies considerably from one bacterium to another; under the optimum conditions, it is 15–20 min for most foodborne bacteria. Log phase should be delayed/avoided to keep the food safe. On the contrary, lag phase should be kept short, and growth in the log phase must be controlled in food production (such as yogurt).

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TABLE 10.1 Approximate pH Values of Selected Foods

Name of the Food Fresh egg white Fish (most of them) Milk, cow

pH Value 7.96 6.6–6.8 6.40–6.80

Chicken

6.2–6.4

Pork

5.9–6.1

Fresh egg yolk

6.10

Cooked white rice

6.00–6.70

Spaghetti

5.97–6.40

Beef

5.1–6.2

Salmon, cooked (boiled or broiled)

5.36–6.50

Watermelon

5.18–5.60

Cucumbers

5.12–5.78

Tomatoes

4.30–4.90

Pickled cucumbers

4.20–4.60

Yogurt

3.8–4.2

Orange

3.60–4.34

Apple eating

3.30–4.00

Vinegar

2.40–3.40

Lemon juice

2.00–2.60

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Microorganisms have different needs for oxygen. Some need oxygen to grow (aerobes). Others can only grow without oxygen (anaerobes). Most can grow with or without oxygen (facultative anaerobes). Moisture is another major requirement for bacterial growth. Microorganisms require certain amount of free water to survive and grow. The shelf life of foods increases as their free water content decreases. That is why foods like dried legumes, dried meat and fish, and dried fruits last longer compared to their fresh forms. Curing, drying, freezing, and binding free water with hydrophilic substances such as salt and sugar are the most common methods for decreasing free water content in foods. Rapid growth is usually wanted with beneficial microorganisms in controlled food processes, such as in sour cream production. However, it is undesirable when it occurs with spoilage and pathogenic microorganisms. It is required to focus on these requirements to control the rate of microbial growth based on the purpose. Combined effects of growth factors, such as heat and moisture, time and temperature, oxygen and heat, and heat, oxygen, and cold storage can be used to produce safe and high-quality food (hurdle concept). Time and temperature combination is the most commonly used food safety control in the kitchen. Example includes: • Not leaving foods in the TDZ for more than 2 h. Proper hot and cold holding. • Proper cooking temperature-time combination (Table 10.2). • Estimation of shelf life of foods under different storage temperatures (Table 10.3).

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TABLE 10.2 Safe Cooking Time-Temperature Combinations of Selected Foods

Internal (Center Temperature) (°C/°F)

Holding Time (s)

• Poultry (including patties) • Duck and goose • Casseroles • Stuffing with poultry, meat, fish, and/or duck and goose • Stuffed foods

74/165

15

• Eggs and egg dishes • Ground meat

71/160

15

Steaks and chops from: • Pork • Veal • Lamb • Beef

63/145

180

Fruits and vegetables

57/135

Hold hot

Food

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TABLE 10.3 Safe Refrigerator and Freezer Storage and Time Combinations of Selected Foods

Time at Refrigeration Temperature

Time at Freezer Storage

Eggs (fresh, in shell)

3–5 weeks

Don’t freeze

Bacon

7 days

1 month

Meat Steaks Chops Roasts

3–5 days 3–5 days 3–5 days

6–12 months 4–6 months 4–12 months

Raw, whole chicken, or turkey Raw, whole chicken, or turkey parts

1–2 days 1–2 days

1 year 9 months

Fried or boiled chicken

3–4 days

4 months

Ground meat and hamburger

1–2 days

3–4 months

Lean fish

1–2 days

6–8 months

Offal

1–2 days

3–4 months

Whipped cream

2–3 h

1 month

Food

FOOD SAFETY AND HYGIENE  367

EXPERIMENT 10.3 OBJECTIVE To understand safe practices to prevent potential hazards in foods. Case Analysis Last Saturday, 783 of 1,220 guests got food poisoning with the symptoms of bad gastroenteritis, which caused nausea, vomiting, cramps, and diarrhea after the wedding dinner. The local authority inspected the establishment, and also talked to the head chef, Sibel, about the production process in the kitchen. She says, she woke up a bit late that day. When she got out of bed, her stomach had a few cramps and she did not feel her best, but blamed that on the drinks she consumed from the previous night. Sibel did not have time to see the doctor since she was already late. On the way to the restaurant, she stopped by the butcher, bought 60 kg (132.3 lb) chicken breast, and placed them in the car. When she arrived at the restaurant the manager was very angry and put her to work right away, she says. First, she grabbed the sanitizer stored in the dry food storage room, and wiped the bench with a sponge soaked in sanitizer. Afterwards, she prepared the soup and put it on the stove for cooking. Meanwhile, she grabbed one of the cutting boards and chopped the vegetables, and set them aside. After finishing her first preparation, Sibel went to the car to get the chicken breast that she bought on her way to the restaurant. Then wiped the cutting board off and sliced the chicken, placed them in a big shallow tray, added some oil and seasonings, covered it tight, and kept it refrigerated until roasting. There was a little chicken on the cutting board, so she put some detergent on the board and gave it a quick wipe. Then she cut the bread on the board, and prepared the garlic bread. The soup must be ready by now. She dipped a spoon into the soup for a taste, seasoned, and tasted again with the same spoon. Half an hour before the wedding, Sibel quickly roasted the surface of chicken slices and placed them back to its tray until service. The food was ready on time. 1. Analyze the case and record your findings in Data Table 10.7. 2. Go over your findings and correct/complete Data Table 10.7 after in class discussion hour.

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DATA TABLE 10.7 Mistakes Observed

…

Suggested Solutions

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The Science Behind the Results Safe Practices are Important to Avoid Foodborne Illness Sick food handlers should not work with food since they can cause foodborne disease outbreaks. They must be excluded from food handling and food handling area especially if they have diarrhea, vomiting, fever, continual sneezing, coughing, any discharge from the nose or eyes, infected wounds, and any diagnosed infection that can be spread through foods. The length of exclusion is usually 48 h after the symptoms stop. Medical report is usually required to ensure the person has recovered and can return to work. People can also be carriers of disease without showing any symptoms, so good managerial control and appropriate actions are very important to ensure the health conditions of the staff. Poor personal hygiene of food handlers is one of the major factors in food contamination. To avoid physical, chemical, and biological hazards that may arise from poor personal hygiene, food handlers should: • • • •

• • • • • • • • •

bathe daily wear clean and proper uniforms change into uniforms in the designated dressing rooms wash hands with warm water and soap for at least 20 s when they first arrive at work, before touching foods, after handling raw foods such as raw meat, between different processes, after handling dirty equipment, tools, dishes or utensils, after using bathroom, after touching nose, ears or any part of the body, after coughing and sneezing, after taking out the trash restrain their hair have short and clean nails have no nail polish, and no fake nails have no jewelries (only a plain wedding ring is allowed) have no make-up have no mustache or beard (or restrain them properly) not smoke in the food preparation area not eat or drink in the food preparation area take the uniforms off before going to bathroom, and outside of the establishment.

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Cross contamination is a transfer of microorganisms from one place to another. In the kitchen, this transfer commonly occurs from raw foods and c­ ontaminated food surfaces to ready-to-eat foods via different means. Major precautions to prevent cross contamination in safe food production can be listed as follows: • Wash hands frequently. • Always use different cutting boards to cut ready-to-eat and raw foods (especially raw meat, poultry, eggs, and fish). • Always use different tools and utensils to handle ready-to-eat and raw foods (especially raw meat, poultry, eggs, and fish). • Always clean and sanitize the food contact surfaces, utensils, equipment, and work areas. • Always separate ready-to-eat and raw foods-during receiving, storing, preparing. • Always ensure personal hygiene. • Always store ready-to-eat foods over raw foods. Store raw meat, poultry, eggs, and fish in the bottom shelves of the refrigerator. • Always store the foods at least 15 cm (6 in.) above the floor. • Always use separate spoons and forks for multiple tasting. Chemicals such as soap, detergent, grease-dissolving agent, sanitizers, and bleach are used in the kitchen to decrease the number of unwanted microorganisms (sanitation). On the other hand, misuse of these chemicals and/or using nonfood grade chemicals can contaminate foods and cause foodborne illness. Staff must be trained about types of kitchen chemicals, and proper cleaning, sanitizing, and rinsing procedure to produce safe foods. Remember! Bacterial growth requirements (Food, Acidity, Time, Temperature, Oxygen, and Moisture) must be controlled throughout the process for safe food production…!!! Systematic and well-documented analysis of potential physical, chemical, and biological hazards in product, place, and people at each stage of the process is the key for high-quality and safe food production. Food producers also must know about and comply with national standards and codes addressing design and layout of the food establishments, sanitary condition of the facility, equipment, sanitation management, food safety control procedures from farm to table, storage and transportation, garbage management, air circulation and ventilation, pest control, and staff training.

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EXPERIMENT 10.4 OBJECTIVES • To understand some people are more susceptible to foodborne illness • To understand importance of dose effect in food poisoning. Case Study Last Saturday, 783 of 1,220 guests got sick with the symptoms food poisoning after the wedding dinner. The illness had varied between people from a couple of sick-days at home to staying at the hospital. Sadly, the contaminated food was responsible for the deaths of 14 children and the serious injury of 312 other victims, including permanent stroke, kidney and brain damages. Analyze the case.

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FOOD SAFETY AND HYGIENE  373

The Science Behind the Results Does everybody on the table necessary get sick after sharing the same contaminated food? The answer is No… The immune system is one of the major factors. Some people are highly susceptible to foodborne illness and may experience severe health consequences and even death due to their weak immune systems. This group of people includes infants, young people, elderly people, pregnant women, and people with chronic illnesses, such as cancer patients, HIV positive, and diabetics. Another factor is the amount of exposure to the disease-causing agents (microorganisms or their toxins). Ingestion of a certain amount of organisms/toxins is required to cause serious illness. This amount depends on the types of bacteria. Usually ingestion of large amounts of pathogenic bacteria are required to get food poisoning, however a much smaller dose of certain types of bacteria may be fatal. People that consumed the contaminated part of the shared food may get sick while the others may not if they did not consume the contaminated part or consume enough number of pathogens to get sick.

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POINTS TO REMEMBER Consumers have a right to expect that the foods they consume will be safe and of high quality. Consumption of contaminated foods or drinks is the most common cause of foodborne illnesses. A foodborne hazard refers to any agent with the potential to cause illness or injury when consumed by the people along with the food. Foodborne hazards may be: 1. Biological 2. Chemical 3. Physical Foodborne microorganisms can be classified into three groups: 1. Beneficial/Fermentative 2. Pathogenic 3. Spoilage Under optimum conditions, bacteria multiply through binary fission process. Requirements for bacterial growth are: 1. Food 2. Acidity 3. Time 4. Temperature 5. Oxygen 6. Moisture Safe practices are important to avoid foodborne illness. Cross contamination is a transfer of microorganisms from one place to another. The immune system of the person, and the amount of exposure to the ­disease-causing agents are the most important factors on the risk of developing food poisoning. Food producers must know about and comply with national standards and codes addressing food hygiene and safety in product, place, and people.

FOOD SAFETY AND HYGIENE  375

More Ideas to Try Repeat Experiment 10.2 Case 3 but place the slices in vacuum food containers or vacuum pack them at stage 2. Compare your results with Experiment 10.2 Case 3.

Study Questions 1. Explain the safe methods for thawing frozen foods. 2. UHT milks are stored at room temperature in the markets. The label says “Refrigerate after opening.” Explain the reason for it.

SELECTED REFERENCES Factors that Influence Microbial Growth Chapter 3. 2001. Online at: http://msue.anr.msu. edu/uploads/234/48511/Safe_Practices_for_Food_Processes_Chpt._3_Factors_ that_Influence_Microbial_Growth.pdf. FDA. Approximate pH of Foods and Food Products. Online at: http://ucfoodsafety.­ ucdavis.edu/files/266402.pdf. FDA. Refrigerator and freezer storage chart. Online at: www.fda.gov/downloads/food/ resourcesforyou/healtheducators/ucm109315.pdf. Food Contamination and Risk Factors. Online at: wwwapp1.bumc.bu.edu/lphi/­ publichealthtraining/onlinecourses/foodprotection/FoodProtection5.html. Food poisoning (foodborne illness). Online at: https://food.unl.edu/food-poisoning-foodborne-illness. Hassan, A., and I. Amjad. 2010. Nutritional evaluation of yoghurt prepared by different starter cultures and their physiochemical analysis during storage. African Journal of Biotechnology 9(20):2913–2917. Introduction to food microbiology. Online at: https://aggie-horticulture.tamu.edu/ food-technology/food-processing-entrepreneurs/microbiology-of-food/. Keep food safe with time and temperature control. Online at: https://extension.umn.edu/ food-service-industry/keep-food-safe-time-and-temperature-control. King, H. 2013. Food safety management: Implementing a food safety program in a food retail business (food microbiology and food safety). New York: Springer. McSwane, D., N. R. Rue, R. Linton, A. G. Williams. 2003. Food safety fundamentals: Essentials of food safety and sanitation. Prentice Hall: New Jersey. Ozilgen, S. 2010 Application of failure mode and effect analysis model to foodservice systems operated by chefs in practice and by chefs from a culinary school in Turkey. Journal of Consumer Protection and Food Safety 5:333–343. Ozilgen, S. 2012. Failure mode and effect analysis (FMEA) for confectionery manufacturing in developing countries: Turkish delight production as a case study. Ciênc. Tecnol. Aliment 32(3):505–514.

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Potentially hazardous food. Online at: https://web.uri.edu/foodsafety/potentiallyhazardous-food/. Potentially Hazardous Foods: Safe Food is Good Business. Online at: www.health.state. mn.us/divs/eh/food/fs/hazard.html. Suleimenova, S. 2016. Biochemical and sensory profile of meat from dairy and beef cattle. Online at: http://epublications.uef.fi/pub/urn_nbn_fi_uef-20161261/urn_nbn_fi_ uef-20161261.pdf. Temperature classification. Online at: http://iws2.collin.edu/dcain/CCCCD%20Micro/ temperatureclassification.htm. World Health Organisation. 2006. Five keys to safer food manual. Online at: http://apps. who.int/iris/bitstream/handle/10665/43546/9789241594639_eng.pdf;jsessionid=3 6F5C36897B3622C17615262493D1955?sequence=1.

Extended Glossary This glossary gives brief definitions of the entire key terms used in the book, and brief explanations of the most common culinary terms used in the kitchen.

Aging of meat: A process that gives meat a flavor and also increases tenderness. After an animal is slaughtered, blood circulation and oxygen transportation to the muscles stop. Tenderness of meat decreases immediately because the meat muscles remain locked in a permanent contraction as a result of lack of oxygen supply for muscle relaxation in the energy mechanism. This process is called rigor mortis or postmortem rigidity and usually takes 6–18 h to complete. After the rigor process, muscle undergoes enzyme-caused progressive natural changes that result in a gradual increase in tenderness. Aging is simply holding a carcass or the cuts until this natural process is complete. Time for aging process varies from several hours to several days mainly depending on type, age, breed, diet, and activity level of the animals. Meat aging has been widely applied by the food industry to improve palatability attributes, but could adversely affect some quality attributes of the meat such as color and shelf life, if the temperature, relative humidity, air circulation, and sanitary quality of the aging rooms are not controlled properly. Amino acid: Monomers of proteins. They contain an amino (NH2) group and a carboxyl (COOH) group. Amino acids are joined together by peptide bonds to form polypeptide chains. Amorphous: Has no crystalline structure, such as lollipops, marshmallows, and cotton candies. Amylase: An enzyme (protein) that breaks down starch. Amylopectin: A fraction of starch. It is a highly branched polymer of glucose. Amylose: A fraction of starch. It is a linear chain polymer of glucose. Antioxidant: A substance that prevents or slows down oxidation reaction by interfering with the chain reaction. Atom: The smallest particle of a substance having the chemical properties of the substance. An atom consists of a central nucleus surrounded by negatively

377

378  EXTENDED GLOSSARY

charged electrons. The nucleus is positively charged, and contains positively charged protons and uncharged neutrons. The number of electrons, protons, and neutrons of each atom determines the chemical reactions that the atom may have with other atoms. Baking: A cooking process in which foods are surrounded by hot and dry air, usually between 163°C and 218°C (325°F and 425°F). The heat is transferred by convection in a closed environment, such as in an oven. Dry heat causes the food’s surface to have a crispy and crusty texture, brown surface, and tender and moist interior. Basting: A process of spooning or pouring liquid such as melted butter or other fat, meat drippings, and stock over food while it cooks. This process prevents foods from drying out, and helps brown the surface of the foods. Basting does not produce moisture or improve the interior flavor of the foods since it cannot penetrate the food. Blanching: A cooking technique in which food is briefly immersed in boiling water for a few seconds and then placed in cold water for sharp cooling. The primary purpose of blanching is to deactivate the enzymes in foods. If the process is not done properly, the enzymes continue to be active causing off-colors, off-flavors, and toughening. Blanching is usually applied to prepare fruits and vegetables for long-term freezer storage. Boiling point: The temperature at which a compound changes from a liquid to a gas. Bound water: Type of water, which cannot be easily removed from the food. It is not available for chemical and biological reactions. Braising: A cooking method that combines dry and moist cooking methods. It is a two-step process. First, the food is browned at high temperature. This step is key to building flavor, color, and texture in final dish. After the food has been browned, the liquid (stocks, wine, water, etc.), vegetables, and flavoring are added for moist cooking. The food cooks slowly in that liquid either on the stovetop or in the oven until very tender. This cooking method is excellent for large and tough cuts of meats such as shoulder and chuck, poultry, and some vegetables like cabbage. Broiling: A dry cooking method in which direct and intense radiant dry heat is transferred from a heat source located above the food. It accelerates the chemical reactions on the surface giving it a nutty flavor, brownish color, crispy and crusty texture on the surface while keeping the food moist on the inside. Caramelization: A nonenzymatic browning reaction. It is a complex set of sugar degradation reactions, which happens only at high temperatures. Casein: A milk protein.

Extended Glossary  379

Chemical change: An irreversible change. A chemical reaction occurs and a new substance is formed. Cooking eggs, for instance, is an example of a ­chemical change; the egg white and egg yolk change from liquid to solid. Clarification: A process of removing the suspended substances that can lead to ­turbidity, taste and odor problems mainly in mainly liquid foods. Filtration, centrifugal force, use of enzymes, and use of charged s­ ubstances are the main methods applied in the kitchen for clarification. Clarification of stocks and soups with egg white is one of the common practices in the kitchen. In this process, as egg white coagulates, the bridge is formed between the coagulant and the solid particles with the action of heat that traps the solid particles into large clumps. Once solid particles are formed into larger particles, they can easily be removed from the liquid by ­filtration or straining. Coagulation: Transformation of native protein structure into a soft semisolid or solid mass. Coagulation of protein occurs when the denaturation process reaches an advanced state. Once proteins are coagulated, they cannot be returned to their native state. Compound: Made up of atoms of two or more different elements bound together, such as carbohydrates and fats. Components may be separated by c­ hemical means. Atoms of elements combine in fixed ratios to form compounds. Conduction: A mode of heat transfer. Heat always flows from a hot surface to a cold surface. In conduction, direct contact is required for heat flow. When the substances gets hot, the kinetic energy of their molecules increases. Therefore, they start to move faster, and collide against each other. As they collide, slow molecules gain energy from the fast molecules, start to move faster, and collide with other slow molecules. Heat is generated and transferred as a result of the collisions. This continues until the two surfaces come to equilibrium. Examples for conduction heating may include such cases as follows: the stove conducts heat to the pan, and the pan conducts heat to the contact surfaces of the foods being cooked in it; or a container gets hot from the hot food microwaved inside. Heat transfer from surfaces of solid and semisolid foods, and stagnant liquid to their centers also mainly occurs by conduction during cooking. Convection: A mode of heat transfer. Heat always flows from a hot to a cold surface. Convection occurs when heat is transferred by means of moving hot liquid or air. There are two types of convection: natural convection and forced convection. Density gradients are generated when liquids are heated, which cause hot liquids to rise up. This movement is the main reason for natural convection in the fluid. For example, during cooking (without mixing) the bottom of the liquid food gets hot first. The molecules gain energy, start to move faster, and spread apart.

380  EXTENDED GLOSSARY

So the heated liquid becomes less dense and rises up to the surface. As heating continues, this natural movement (natural convection) continues until it comes to equilibrium. Forced convection occurs through the forced motion of the fluid or gas. Mixing liquid during heating forces the transfer of heat (forced convection), and speeds up the heating process. Ovens with fans use the principle of forced convection heat transfer. In general, heat is transferred in fluids by combinations of conduction and convection. Covalent bond: A strong chemical bond that joins two atoms together by the sharing of electrons in the outer atomic orbitals. Crystallization: The formation of solid crystals from a solution in a precise orderly structure. Deglazing: A process of dissolving the flavorful glaze of juices and cooked food particles from the surface of a pan in which food has been cooked. In this process, the liquid such as wine, water, and stocks are added to the pan in which food has been cooked, and heated while stirring and scrubbing, and simmering to loosen the remainings. Deglazing is usually used to add flavor to sauces. Denaturation: Change of protein structure from its natural state. Denaturation involves the disruption of the secondary, tertiary, and quaternary ­structures of proteins. Dextrinization: The breakdown of starch polymers into smaller molecules when heated to high temperatures without liquid. Disaccharide: A carbohydrate that is composed of two molecules of simple sugars (monosaccharide) linked to each other upon condensation. Element: One of the basic substances that are made of atoms of only one kind and that cannot be separated by ordinary chemical means into simpler substances. Emulsion: A mixture of two immiscible liquids in which minute droplets of one liquid are dispersed in another. Enzymatic browning: A chemical change. It occurs when the phenolic substances in foods react with oxygen, in the presence of specific types of enzymes. Browning of foods occurs as a result of enzymatic browning reaction, e.g., the browning of peeled apples. Enzyme: A chemical substance produced by living organisms, such as animals, plants, and microorganisms. Enzymes are capable of initiating certain chemical changes and/or increasing the rate of chemical reactions. Fats: Substances that are composed of glycerol and fatty acids. They are solid at room temperature. Animals are the primary sources of fats. Fatty acids: Building blocks of fats and oils. They are carboxylic acids with long hydrocarbon chains. Fatty acids are either saturated or unsaturated.

Extended Glossary  381

Fermentation: Transformation of sugar primarily into acids, gases, and/or alcohol by the action of microorganisms, e.g., lactic acid bacteria ferment lactose to lactic acid in the yogurt manufacturing process. Fibrous proteins: Proteins that are composed of polypeptide chains assembled along the straight axis, e.g., collagen and elastin. They are shaped like rods or wires and are usually insoluble. They are primarily found in meat, chicken, fish, and wheat. Folding: A process of combining ingredients together using a gentle over-andunder motion without agitating the mixture. The motion is mainly top to bottom by gently lifting ingredients from the bottom of the container over the ingredients on the top rather than mixing round and round. It is a technique most commonly used to combine a lighter, airy mixture with a heavier mixture. Folding is usually executed with a rubber spatula, a wire whisk, or hand mixing without knocking air out of the mixture. Food additives: Substances that are directly or indirectly added to foods during food processing. Free water: Available water for chemical and biological reactions. Not chemically or physically bound. Free water can be easily removed from food by cutting, squeezing, pressing, or drying. Freezing point: The temperature at which a liquid changes to a solid. Fructose: A simple monosaccharide that is known as fruit sugar. It occurs naturally in many fruits. Globular proteins: Proteins composed of polypeptide chains twisted into a rounded, compact shape as globs or spheres. They are usually soluble and are primarily found in milk and eggs. Glucose: A simple monosaccharide (sugar). Gluten: A stretchy cereal protein that gives structure to baked goods. Glycerol: An alcohol. Grilling: A dry cooking method in which direct and intense dry heat is transferred from a heat source located below the food. It accelerates the chemical reactions on the surface giving it a nutty flavor, brownish color, crispy and crusty texture on the surface while keeping the food moist on the inside. Hydrocarbon: Substance made of hydrogen and carbon, and is hydrophobic. Hydrogen bonding: The attractive interaction of a hydrogen atom bearing a partial positive charge, which is covalently bonded to an electronegative atom, with another electronegative atom, such as fluorine, nitrogen, or oxygen that comes from another molecule. It is weaker than an ionic bond or a covalent bond. Hydrogenation: A chemical process that adds hydrogen atoms to an unsaturated oil to saturate it. Liquid oils become solid fats. Examples of hydrogenated fats include margarine and vegetable shortening.

382  EXTENDED GLOSSARY

Hydrolysis: A chemical reaction that breaks down the bonds in molecules using water. Hydrophilic: Substances that have an affinity for water, and are water-soluble. Hydrophobic: Substances that do not dissolve easily in water. Insoluble: A substance that is not capable of being dissolved; for example, fats are insoluble in water. Ionic bonding: The attractive interaction between two oppositely charged (+ and −) ions, e.g., common table salt (sodium chloride). Ionic bonding involves the complete transfer of electron(s) between atoms that generates two oppositely charged ions. Oppositely charged ions are attracted to each other by electrostatic forces, thus an ionic bond is formed. Kneading: A process of mixing dough by pressing in a sort of forward rolling motion, pushing, stretching, and folding, then rotating it and repeating until it stays together and is elastic. Kneading the dough made with wheat flour physically pushes protein molecules together and stretches the bonds formed. It is this squeezing and stretching that develops a gluten network that allows it to trap gas efficiently. Lactose: A disaccharide, which is made from galactose and glucose units. It is known as a milk sugar. Lard: A rendered fat obtained from swine. Leavening: A process of incorporation of gases in baked food products such as bread, muffins, pancakes, cookies, and meringue to increase volume. Leavening agents are used to produce gas. There are three major types of leavening agents: biological, chemical, and mechanical. Beneficial microorganisms, primarily yeasts, are used as biological leavening agents. During baking, yeast consumes sugar and/or sugar in a recipe and produces carbon dioxide gas (through fermentation), which is trapped by the dough (gluten) structure. Baking powder, and baking soda are the two chemical leavening agents. Baking soda is an alkaline substance that produces carbon dioxide gas when mixed with acidic ingredients such as cream of tartar, lemon juice, and yogurt. Baking powder already contains the acidic ingredients. Once combined with liquid, gas production begins. Carbon dioxide gas, which is produced by either chemical leavening agent, is trapped by the dough (gluten) structure during baking. Air and steam are two types of mechanical leavening agents. Physical forces such as mixing, and whisking can trap air in small protein matrix. During baking, gas (produced by biological, chemical, and/or mechanical means) entrapped in the food structure expands with increased heat. Therefore, the volume of the food increases. At the end of the baking process, the gas bubbles are transformed into semi-rigid structure, and small holes are formed. All leavening actions are mainly responsible for

Extended Glossary  383

a good volume, a light and fluffy texture and a uniform hole structure of the final food products. Lecithin: An emulsifier found in eggs and soybean oil. Lipids: Organic substances that are insoluble in water and soluble in nonpolar solvents. Fats and oils are known as lipids. Maillard browning reaction: A nonenzymatic browning reaction caused by the reaction of reducing sugars with proteins and amino acids in foods. The Maillard reaction is responsible for the brown color and nutty flavor of cooked foods, e.g., roasted meat, toast, and coffee. Melting point: The temperature at which solids become liquids. Mincing: A food preparation technique in which food ingredients are divided into tiny uniform pieces, smaller than diced or chopped foods. Minced foods are usually prepared with a chef’s knife, food processor, or in the case of meat, by a meat grinder. Mixture: Made of two or more different substances that are physically mixed with each other. Mixtures can be separated into their components by physical means. Molecule: Two or more atoms join together chemically to form molecules. Monomer: A molecule that can be bonded to other identical molecules to form a polymer. Monosaccharide: A simple sugar that constitutes the building blocks of disaccharides, oligosaccharides, and polysaccharides. Examples are glucose, fructose, and galactose. Monounsaturated fatty acid: Fatty acids that contain one C=C double bond in the hydrocarbon chain. Oil: Substances that are composed of glycerol and fatty acids. They are liquid at room temperature. Plants are the primary sources of oils. Osmosis: The movement of water through a semipermeable membrane from a region of high solvent concentration to a region of low solvent concentration. Pectin: A large polysaccharide that is present in plant cells. It occurs naturally in most fruits, and is concentrated in the skins and pips of different types of fruits. Peptide bonds: Bonds that connect the amino group of one amino acid and the carboxylic acid group of another amino acid. Peptides: Short chains of amino acids. Physical change: A reversible change that involves a change from one state (solid or liquid or gas) to another without a change in chemical composition, e.g., ice melting. No new substances are formed. Poaching: A process of cooking foods in small amount of gently simmering water or flavorful liquid such as salted water, broth, stock, etc. Poaching is usually used for very delicate food items such as lean poultry, eggs, fruits, or

384  EXTENDED GLOSSARY

fish. The basic steps include heating the liquid to simmer, gently submerging the food into the hot liquid, and removing food from the liquid when the cooking process is completed. The temperature of the poaching liquid is one of the most critical factors for the success of the process. It must be maintained around 71°C–82°C (160°F–180°F) throughout the process since higher heat may cause the food to fall apart or toughen. Polymers: Macromolecules that contain 10 or more monomer units. Protein is a polymer of the monomer amino acids. Polypeptides: Long chains of amino acids. Polysaccharide: A carbohydrate that is made up of a number of simple sugar units (monosaccharide) bonded together. Polyunsaturated fatty acids: Fatty acids that have multiple C=C double bonds. Primary structure of proteins: The sequential order of amino acids in a protein. Proteins: Complex polymers composed of amino acid monomers. Quaternary structure of proteins: Two or more polypeptide chains joined together to form the quaternary structure of proteins. Radiation: A mode of heat transfer. Physical contact between the energy source and food is not required in radiation. Heat is transmitted through empty space to the foods by electromagnetic waves. When the waves strike the food, they are absorbed. Kinetic energy of the molecules in the food increases so they start to move faster; increasing the temperature of the food. Microwave (light waves) and infrared energy (heat waves; such as from the walls of the hot oven, and broilers) are the two important types of radiation in the kitchen. Rancidity: The chemical deterioration of lipids. Undesirable odors and flavor may develop as a result of chain reactions. Rancidity can be either enzymatic or oxidative. Rennet: A combination of different proteolytic enzymes. It is usually only extracted from the stomachs of young animals. Rennin: A proteolytic enzyme found in rennet. It curdles milk. Retrogradation: Starch molecules, particularly the amylose fractions, reassociate in an ordered structure in gelatinized starch; eventually a crystalline order appears and water is squeezed out, e.g., staling of bread. Saturated fat: A fat that is normally solid at room temperature, e.g., butter. Fatty acids in their structures are saturated; they contain only carbon–carbon (C–C) single bonds in the hydrocarbon chain. Saturated solution: A solution in which the dissolved solute is in equilibrium with the undissolved solute. Secondary structure of proteins: The three-dimensional organization of the polypeptide chain formed by intramolecular and intermolecular hydrogen bonding.

Extended Glossary  385

Sifting: Passing dry ingredients such as flour, cocoa powder, baking powder, salt, or sugar through a mesh bottom sieve to fluff them up, to separate them based on their particle sizes, and to get rid of impurities and clumps, which may be in the ingredients. Dry ingredients get lighter and fluffier since this process combines air with the ingredient being sifted. Therefore, the sifted and unsifted ingredients (from the same batch) that take up the same volume do not weigh the same. For example, a spoon of sifted flour weighs less than a spoon of unsifted flour, and can change the texture and/or consistency of final products. Sifting dry ingredients together helps to combine them evenly within each other. Simmering: A moist cooking method. Food is cooked slowly in liquid at a temperature of about 82°C–96°C (180°F–205°F). A full boil is not reached; the bubbles form barely and gently rise to the surface of the liquid. Simmering is mainly used to cook soups, broths, and stocks. Smoke point: The temperature at which heated fats and oils begin to smoke and emit unpleasant odors. Soluble: A substance that is capable of being dissolved, e.g., salt is soluble in water. Solute: Substances in solutions. Solution: A mixture of one or more solutes and a solvent. Solvent: The substance in which a solute dissolves. Starch: A polymer of glucose units. It has two fractions: amylose and amylopectin. Sucrose: A disaccharide made from glucose and fructose units. Tertiary structure of proteins: The three-dimensional organization of the ­polypeptide chain, which is maintained by weak, noncovalent i­nteractions, such as hydrophobic interactions and salt bridges. Unsaturated fat: A lipid that is normally liquid at room temperature, e.g., olive oil. Fatty acids in their structures are unsaturated; they contain C=C ­double bonds in the hydrocarbon chain. Those lipids are normally liquid at room temperature, e.g., olive oil. Viscosity: The resistance to flow in a liquid. Water activity: Refers to free water in foods. Water holding capacity: Ability of a food structure to entrap water. Whey: One of the two major proteins in milk. It is the liquid left behind after milk has been curdled and strained as in cheese making. Whipping: A process of beating some foods rapidly with a utensil to incorporate air to produce volume and to provide body. During whipping, fat and protein molecules in the liquid surround the tiny air pockets and stabilize them. Formed fat bubbles clump together producing the ­characteristic stiff structures of whipped foods. Creams with at least 30% fat contents are the most common ingredients used in whipping processes in the kitchen. Whipping is usually carried out with a whisk, with rotary ­beaters, or with electric beaters.

Index A Actin, 149 Agar, 333 Aging gel, 129 Albumen, 155 Alginate, 320, 333 Amino acid, 146 Amylopectin, 115–117, 129 Amylose, 115–117, 129 Antimicrobial preservatives, 309 Antioxidant, 247, 309 Atom, 23–24

B Bacteria, 361 aerobes, 364 anaerobes, 364 beneficial/fermentative, 353 facultative anaerobes, 364 mesophiles, 362 pathogenic, 353 psychrophiles, 362 spoilage, 353 thermophiles, 362 Bacterial growth generation time, 362 hurdle concept, 364 lag phase, 362 log phase, 362 requirements, 361–366 Baking powder, 33, 382 Baking soda, 33, 382 Binary fission, 361 Biological hazards, 348 Boiling point, 61

Boiling point elevation, 61 Bound water, 49

C Caramelization, 99–109 Carbohydrates chemical structure, 90–92 functional properties, 89 Carboxymethyl cellulose, 334 Carrageenan iota carrageenan, 334 kappa carrageenan, 334 lamda carrageenan, 334 Cellulose, 137, see also Polysaccharide Chemical bonds, 24 Chemical change, 23 Chemical hazards, 348 Chemoreceptors, 275 Coagulation, 201 Collagen, 149, 165 Compound, 24 Consumer acceptance/preference tests, see Sensory evaluation tests Continuous phase, 241 Conversion between units of measures, 5–8 Cost, 13–16 Covalent bond, 24 Cross contamination, 370 Cynarin, 289

D Denaturation, see Protein denaturation Descriptive ranking tests, see Decriptive tests Descriptive rating test, see Descriptive tests Descriptive tests, see Sensory evaluation tests

387

388 Index

retrogradation, 125 safe practices, 367 saturated and unsaturated solutions, 93 shortening power of lipids, 231 smoke point, 217 starch types, 111 starch types and starch gels, 111 statistics, 17 synergistic effects of multiple factors and protein denaturation, 197 taste and flavor, 273 temperature and flavor perception, 277 temperature and sugar syrup, 99 temperature fluctuation and ice crystal size, 75 tempering, 249 types of bacteria, 349 types of water, 45 water content and water activity, 51

Dextrinization, 133 Disaccharides, 91, 97 Discrimination/difference tests, see Sensory evaluation tests Dispersed phase, 241 Duo-trio test, see Discrimination/difference tests

E Element, 24 Emulsifier, 241 Emulsion, 241–242 Enzymatic rancidity, 245, see also Lipid oxidation Experiments acidity and protein denaturation, 181, 187 basic chemistry, 27, 29 boiling point and boiling point elevation, 57 dextrinization, 131 emulsion, 237 enzyme and protein denaturation, 191 flavor pairing, 283, 287 food additives, 313 food poisoning, 371 freezing point and freezing point depression, 63 freezing rate and ice crystal size, 69 frying, 225 gel aging, 125 gelatin, 191 gluten, 203 heat and egg protein denaturation, 151, 157 heat and meat protein denaturation, 163, 167, 171 hydrocolloids, 319, 325, 339 ingredients and starch gel properties, 119 lipid oxidation, 243 lipid types, 223 mechanical force and protein denaturation, 175 microbial growth, 355 mixing and ice crystal size, 75 mixing and sugar syrup, 99 natural sugar contents, 107 osmosis, 81 pectin, 135, 139

F Fat bloom, 253 Fats and Oils, see Lipids Fermentation, 185 Fibrous proteins, 149 Flakiness, 235 Flavor chemical structure, 275 temperature effect, 281 Flavor compounds, 290–302 Flavor enhancers, 310 Flavoring, 310 Foaming, 179 Food additives classification, 307 intentional/direct, 307–310 unintentional/indirect, 308 Foodborne hazards, 348 Food pairing, 289 Fractions, 9 Free radicals, 245–247 Free water, 49 Freezing, 63–80 effect of agitation, 79 effect of storage time, 79 effect of temperature fluctuation, 79 rate of, 73 Freezing point, 67

Index 389

Freezing point depression, 67 Fructose, 90 Frying, 229 ingredient effect, 229 processing time effect, 229 temperature effect, 229 type of lipid effect, 229

G Galactose, 90 Gelatin, 195 Gliadin, 209 Globular proteins, 149 Glucose, 90 Gluten, 209 Gluten content of flours, 209 Glutenin, 209 Guar gum, 335, 339–341 Gum Arabic, 331, 335

H Hedonic rating scale, see Consumer acceptance/preference tests Hydrocolloids chemical structure, 317–318 functional properties, 331–338, 343 sources, 318 Hydrogen bond, 24

I Ice crystal, 79 Idea development, 259 Immiscible liquids, 241 Infection, 353 Intoxication, 353 Ionic bond, 24

L Lactose, 91 Leavening agents, 310, 382 Line extension, 258 Lipid oxidation, 245–247 Lipids chemical structure, 214–216 functional properties, 213

Lipoxygenase, 245 Locust bean gum, 336, 343

M Maltose, 91 Mean, 17, see also Statistics Melting point of lipids, 219 Miraculin, 289 Mode, 17, see also Statistics Monosaccharides, 90, 97 Monosodium glutamate, 310 Monounsaturated fatty acids, 214–216 Muscle fibers, 165 Myoglobin, 173 Myosin, 149, 165

O Octet Rule, 24 Odor, 275 Odor molecules, 275 Oil-in-water emulsion, 241 Osmosis, 85 Oxidative rancidity, 245–247, see also Lipid oxidation causes, 245 precautions, 245

P Paired comparison, see Discrimination/ difference tests Pectin, 137, see also Polysaccharide high-methoxyl, 336 low-methoxyl, 336 Pectin gel acid effect, 141 heat effect, 141 sugar effect, 141 Percentages, 9 Physical hazards, 348 Polysaccharides, 92, 115, 137 Polyunsaturated fatty acids, 214–216 Potentially hazardous foods, 361 Preference tests, see Consumer acceptance/ preference tests Primary structure, 146, see also Protin Product development, 259

390 Index

Protein chemical structure, 146–149 functional properties, 145 Protein denaturation, 149–201 charge effect, 149, 179 enzyme activity effect, 149, 195 fat effect, 179 heat treatment effect, 149, 155, 161, 165, 173 mechanical treatment effect, 149, 179 pH effect, 149, 185, 189 sugar effect, 149, 179 Proteolytic enzymes, see Protein denaturation actinidin, 195 bromelain, 195 ficin, 195 papain, 195

Q Qualitative observations, 1 Quantitative measurements, 2 Quaternary structure, 147, see also Protein

R Rancidity, 245–247, see also Lipid oxidation Range, 18, see also Statistics Recipe yield conversion, 11 Rennet, 201, see also Protein denaturation Retrogradation, 129

S Safe practices, 369–370 Secondary structure, 146, see also Protein Sensory attributes appearance, 260 hearing, 260 odor, 260 taste, 260 touch, 260 Sensory evaluation tests, 260–268 consumer acceptance/preference tests, 267 descriptive tests, 263–266 discrimination/difference test, 262–263 requirements, 261 terminology, 261 Shortening, 235

Smoke point, 220–221 Sodium alginate, 332 Solutions, 93–98 recrystalilization, 97 saturated, 93, 97 saturation point, 97 supersaturated, 97 Spherification direct, 331–332 reverse, 331–332 Spider graph, 266 Stages of sugar syrup cooking caramelization, 106 firm ball, 106 hard ball, 106 hard crack, 106 soft ball, 106 soft crack, 106 Starch, 111–130 functional properties, 115 gelatinization, 115 gelatinization temperature, 111, 115, 123 Starch gel acidity effect, 123 agitation effect, 123 enzyme effect, 123 lipid effect, 123 protein effect, 123 rate of cooling effect, 123 sugar effect, 123 Statistics, 16–19 Sucrose, 91 Sugar bloom, 254 Syneresis., 129

T Taste, 275 Temperature danger zone, 362 Tempering, 253–254 Tertiary structure, 146, see also Protein Triangle test, see Discrimination/difference tests

U Units of measurements SI system of units, 5 U.S. customary system, 5

Index 391

liquid phase, 43 solid phase, 43

V Vapor pressure, 61

W Water activity, 55 Water-in-oil emulsion, 241 Water molecule, 24, 41 Water phases gas phase, 42

X Xanthan gum, 337, 339–341

Y Yield percent, 11

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