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Food Science & Technology 100A Professor S.R. Dungan Final Examination December 13, 2006
Version A
Name: KEY Last First Signature: INSTRUCTIONS Correctly completing the boxed instructions are worth 4 pts. on the exam PRINT your last and first names above. Then SIGN the exam copy. PRINT your name on the front of your Scantron #2000. WRITE the test version that appears at the top of this page on your Scantron #2000. WRITE your student identification number in the appropriate area of your Scantron # on Part 1 and follow the instructions to CODE your ID number in the bubbles. CHECK that there are 70 multiple choice questions. There are a total of 16 pages to the exam. Use a #2 pencil to mark your Scantron answers. You are responsible for the accuracy and completeness of your Scantron. Do not put any marks (other than filling in the answer bubbles) on the left side(s) of the Scantron sheet.
Please respect the UC Davis Honor Code in the taking of this exam.
There should be plenty of time to take this exam without being rushed. However, you will need to use your time efficiently. Graded exams can be picked up in 110 FSTB. Instructions / 4 Multiple Choice / Short answers / 21 Total 200
THIS BEGINS THE MULTIPLE CHOICE SECTION: Mark answers on your Scantron
- Hydroperoxides are a. formed upon hydrogenation of polyunsaturated fatty acid groups b. formed when a peroxy free radical takes off a lipid hydrogen @ c. another name for singlet oxygen d. a product of protein hydrolysis e. enzymes for starch hydrolysis
- The solubility of proteins will often decrease a. above the denaturation temperature b. when the protein denatures to expose hydrophobic residues c. when salting-in salts are added d. Answers a and b are both true @ e. Answers a, b and c are all true
- Which of the following is true of cocoa butter? a. It has a very narrow melting range b. Its fatty acid composition is quite similar to lard c. Its triglycerides are quite homogeneous in structure d. Answer a and b only e. Answers a, b and c are all true @
- Why does the solubility of sugars usually go up with temperature? a. Because entropy becomes more important with increasing temperature @ b. Because sugar and water start to form covalent bonds at high temperature c. Because sugars are mainly in the open chain form at high temperature d. Answers a and b are both true e. Answers a, b and c are all true
- Above the gelatinization temperature a. pectin becomes a gel b. starch granules lose their crystallinity @ c. collagen protein turns to gelatin d. trans fatty acid chains tend to melt e. cellulose fibers break apart
- Starch granules a. typically contain more amylopectin than amylose @ b. are completely crystalline solids c. are a rich source of gelatin d. are most soluble at cold temperatures e. can form reversibly and spontaneously in cold water
- When adding a salting-in salt to a β-lactoglobulin aqueous solution, we expect a. the protein's native conformation to be stabilized @ b. the protein's denaturation temperature to decrease c. the protein to aggregate due to hydrophobic interactions d. answer b and c would both be expected e. answer a and b would both be expected
- Which of the following molecules would help emulsions to form by lowering oil/water interfacial tension? a. lecithin @ b. a saturated triglyceride c. salt d. glucose e. beeswax
- Aging of meats causes a. coagulation of casein micelles b. chemical and enzymatic breakdown of actin and myosin @ c. starch retrogradation d. caramelization e. denaturation of gliadins
- Starch contains a. amylose b. pectin c. amylopectin d. answer a and c only @ e. answer a, b and c
- Emulsion formation a. requires high energy input @ b. occurs spontaneously under most conditions c. occurs spontaneously only if surfactants are present d. answer a and c only are true e. answer a and b only are true
- There is a greater energy penalty to create emulsions with a. smaller droplets @ b. larger droplets c. lower interfacial tensions d. answer a and c are both correct e. lower concentrations of oil droplets
- A β[1,4] glycosidic linkage is found in a. glucose b. maltose c. lactose @ d. amylose e. sucrose
- Which of the following types of food molecules produce tastes and flavors after hydrolysis? a. proteins b. triglycerides with short chain fatty acids c. oligosaccharides d. answer a and c only e. answers a, b and c all are likely to produce tastes and flavors @ The following 3 questions refer to the list below: A. Amylose C. Glucose B. Maltose D. Maltodextrins
- Which of these molecules has the largest number of reducing ends per molecule? a. Molecule A b. Molecule B c. Molecule C d. Molecule D e. The above molecules all have equal numbers of reducing ends @
- Which of the following solutions of 1 wt% carbohydrate would have the highest DE value? a. A solution containing only molecule A b. A solution containing only molecule B c. A solution containing only molecule C@ d. A solution containing only molecule D e. All of the above solutions would have equal DE values
- Which of the following would have the highest viscosity? a. Pure water b. An aqueous solution containing 1 wt% of molecule B c. An aqueous solution containing 1 wt% of molecule C d. An aqueous solution containing 1 wt% of molecule D @ e. All of the above would have equal viscosity
- α-D-Mannopyranose and α-D-glucopyranose are a. epimers b. diastereomers c. stereoisomers d. answers a and c are both correct e. answers a, b and c are all correct @
The following 3 questions refer to the molecule below right:
- During isomerization of this molecule, which carbon first loses a proton? a. Carbon 1 b. Carbon 2 @ c. Carbon 3 d. Carbon 5 e. Carbon 6
- This molecule is a. D-glucose @ b. D-mannose c. D-mannitol d. L-gulose e. D-galactose
- During isomerization reactions, this molecule could become a. D-glucitol b. D-maltose c. D-galactose d. sucrose e. D-mannose @
- Which of the following would be most likely to participate in Maillard browning reactions? a. brown sugar b. corn syrup @ c. molasses d. cane sugar e. beet sugar
- Which of the following would NOT make caramelization more likely? a. The presence of non-reducing sugars b. High temperatures c. High concentrations of sugar d. The presence of reducing sugars e. All of the above are conditions favoring caramelization @
- A Newtonian fluid a. is a fluid with very low viscosity b. has a constant viscosity for different shear rates (stir speeds) @ c. is a fluid that becomes more viscous at higher temperatures d. is another word for a gel e. None of the above describe a Newtonian fluid C C OH HO C H H C OH H C OH CH 2 OH H H O 1 2 3 4 5 6
- Compared to stearic acid, the rate of oxidation of a diunsaturated group (2 double bonds) with methylene interruption would likely be a. Half as fast b. Twice as fast c. Ten times faster d. More than 1000 times faster @ e. almost the same
- Olestra is produced a. by hydrogenation of peanut oil b. by esterification of saturated fatty acid groups to sucrose @ c. through a cellulose modification reaction d. through the formation of peptide bonds e. through a saponification reaction
- Why is the percentage of trans fat now required to be labelled on US foods? a. Because trans fats are particularly unstable to oxidation b. Because trans fats are examples of saturated fats c. Because trans fats have health benefits like other unsaturated fats d. Because trans fats are another name for cholesterol e. None of the above are reasons @
- Why do conditions of very low pH slow down Maillard browning? a. Because low pH inactivates browning enzymes b. Because the amine becomes protonated and less effective as a nucleophile @ c. Because sugar hydrolyzes at low pH d. Because sugars isomerize at low pH e. Answers a and b are both reasons
- Soybean oil is rich in linolenic acid fatty acid groups. This creates problems in the use of soybean oil because a. linolenic acid groups are especially susceptible to oxidation @ b. linolenic acid groups contain mostly trans double bonds c. linolenic acid tastes bad d. linolenic acid is always solid e. All of the above are reasons
- Which of the following is NOT reduced at low water activity? a. microbial growth b. water vapor pressure c. the boiling point (temperature) of water @ d. rates of hydrolysis reactions e. none of the above are reduced at low water activity
- Which of the following is NOT true of lipid hydrolysis? a. It is important in the digestion of triglycerides b. It most rapidly cleaves fatty acids from sn-1 and sn-3 positions c. It is important in the production of trans fats @ d. It is important in the production of flavors from milkfat e. It produces sn-2 monoglycerides
- A portion of milkfat remains liquid even at cool temperatures (below room temperature). For milkfat, this liquid most likely is enriched in a. triglycerides with fatty acids of 4-6 carbons each @ b. triglycerides with palmitic and stearic acid groups c. triglycerides with polyunsaturated triglycerides of 22-26 carbons d. limonene e. oxidized triglycerides
- Which of the following is true of an enediol? a. It contains two hydroxyl groups on adjacent carbons b. It is produced by adding a proton to the enolate ion c. It is an intermediate in the isomerization of glucose to fructose d. It contains a carbon-carbon double bond e. All of the above are true of an enediol @
- Which of the following describes dried pasta at room temperature? a. It is below its glass transition temperature @ b. It contains ice crystals c. It has a very high water activity d. It does not contain any hydrophilic solutes e. None of the above are true of dried pasta at room temperature
- A food lipid is rapidly cooled to form a crystal whose tail orientations are random, with a relatively low melting point (compared to the slowly cooled crystal). This crystal is most likely to be a. in the β form b. in the β′ form c. in the α form @ d. answers a and b are equally likely e. answers a, b and c are equally likely
- Water activity can be decreased a. by adding fructose b. by dehydration c. by lowering temperature d. by freezing ice e. All of the above will decrease water activity @
The following 2 questions refer to the structure below:
- Which labeled hydrogen in this structure is most likely to be abstracted during lipid oxidation? a. A b. B @ c. C d. D e. They are all equally likely to be abstracted
- What is best explanation of why this hydrogen is abstracted to form a free radical? a. Because C-H bonds are always very weak b. Because it is pulled off by a strong acid c. Because the result is a free radical structure stabilized by resonance @ d. Because the result is a structure with conjugated double bonds e. Because hydrocarbons are very strong acids
- Protein is made more stable to denaturation a. by extremes in pH b. by high temperatures c. by the presence of many disulfide bonds @ d. by the addition of a salting-out salt e. All of the above make the protein more stable to denaturation
- Which of the following reactions involves attack by a nucleophile? a. Maillard browning b. Hydrolysis of triglycerides c. Isomerization of monosaccharides d. Answers a and b occur by nucleophilic attack @ e. Answers a, b and c all occur by nucleophilic attack
- Which of the following is true of most plant oils? a. They do not contain any saturated fatty acid groups b. They contain many short chain fatty acid groups c. Their sn-2 position is primarily saturated d. They contain only one type of fatty acid group e. None of the above characterize plant oils @
C C
H H
H
H
H H
C C
A
B
C
D
CH 2
CH 2
CH 2
C
CH 2
H H
- Which of the following are NOT true of polysaccharides? a. They are macromolecules b. They contain glycose subunits c. They are polymers d. They are always charged @ e. All of the above are true of polysaccharides
- Honey a. has a water activity close to 1 b. has a high concentration of sucrose c. is produced by bees using hydrolysis and isomerization reactions @ d. has about 20% mono and disaccharides e. is pure fructose
- Why does oil separate so quickly in natural peanut butter (ground peanuts + salt)? a. Because extensive oxidation occurs in this food b. Because the water activity is so high c. Because peanut lipids are primarily unsaturated @ d. Because peanut lipids have a high melting point e. Because peanut lipids have high levels of trans fatty acid groups
- Which of the following is a similarity between cellulose and amylose? a. They are both branched polysaccharides b. They both contain galacturonic acid c. They are both easily digestible by humans d. They both adopt a helical conformation in solution e. None of the above are similarities between cellulose and amylose @
- Lipid oxidation can be initiated by a. Ground state oxygen b. Metal ions c. lipoxygenase d. Answer b and c only @ e. All of answers a, b and c help initiate oxidation
- For a hydrophobic amino acid such as phenylalanine, the free energy to transfer it from ethanol to water (ΔGeth→w) should be a. approximately zero b. positive and large (~10 kcal/mol) @ c. very negative d. between 1 and 2 kcal/mol e. The free energy ΔGeth→w cannot tell us anything about hydrophobicity
This marks the beginning of the SHORT ANSWER SECTION: Answer the following questions briefly and specifically. (6 pts.) The following 5 questions refer to the reaction below : O CH 2 OH OH OH OH HO CH 2 OH HOH 2 C OH OH OH O CH 2 OH C O HO C H H C OH H C OH CH 2 OH
A
B
C
(1 pt.) Identify the sugar represented by Molecule A (be specific as to isomeric form) β-D-fructopyranose (1 pt.) Identify the sugar represented by Molecule C (be specific as to isomeric form) β-D-fructofuranose (1 pt.) What do we call this reaction? mutarotation (isomerization also ok) ( 2 pts.) Circle the anomeric carbon on Molecule A , and show what bond breaks in going from Molecule A to Molecule B. See circle and jagged line above. ( 1 pt.) In going from Molecule B to Molecule C , show on the diagram above what specific group acts as a nucleophile. The C5 – OH group on Molecule B acts as a nucleophile.
Melting point: Molecule B (highest) > Molecule C (middle) > Molecule A (lowest) Oxidative Stability: Molecule B (most stable) > Molecule A and Molecule C (accept any order between A and C) Effect on HDL:LDL: Molecule A (highest HDL:LDL) > Molecule B (middle) > Molecule C (lowest HDL:LDL) (6 pts.) Carboxymethylcellulose is a derivative of cellulose, where a small percentage of glycose subunits are modified to attach a – CH 2 COO − group to the C6 position. Explain how and why this modified cellulose has different solubility properties than the original cellulose. Which polysaccharide would be better at increasing aqueous viscosity? Carboxymethylcellulose is more water-soluble than cellulose. Cellulose is water insoluble because it very efficiently hydrogen bonds to other cellulose molecules to form strong, aggregated fibers. The charge on carboxymethylcellulose introduces electrostatic repulsion between the polysaccharide chains, preventing aggregation and making it water soluble. As a water soluble polymer, carboxymethylcellulose would be more effective at increasing aqueous viscosity.
