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ICAR AICE-JRF/SRF Food Chemistry I: Water, Carbohydrates, Proteins & Lipids: practice questions

ICAR AICE-JRF/SRF (PhD) — Food Processing Technology, Food Safety & Quality, Processing and Food Engineering · India · ICAR through the National Testing Agency

Original practice questions written to the published syllabus · not past papers
10worked here, free
69written to this exam in this subject
14syllabus units

What this subject covers

The four constituents that make up most of any food and how they behave during processing and storage: water and its mobility, sugars and starch, proteins and their functional properties, and lipids and their oxidation.

In the exams: GATE XE-G and XL-U Section 1 (carbohydrates, proteins, lipids); ICAR AIEEA-PG Code 18 Unit I; ICAR AICE 35.b Unit 5 (emulsions, foams, gels, functional properties)

What ICAR AICE-JRF/SRF asks from it

ICAR AICE-JRF/SRF (PhD) — Food Processing Technology, Food Safety & Quality, Processing and Food Engineering

35.b Unit 5 names water activity and shelf life, autoxidation of milk fat and thermal oxidation, emulsions, foams and gels, functional properties of starch, proteins and lipids, and hydrocolloid–protein interactions; 35.a Units 2–3 name modified starches and proteins, vanaspati, meat tenderisation and enzymes. Expect deeper mechanism questions: glass transition and WLF, sorption hysteresis, singlet-oxygen versus radical oxidation, polymorphic transitions, calpain–calpastatin.

Where the 69 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Food Chemistry I: Water, Carbohydrates, Proteins & Lipids. Difficulty: 3 easy, 37 medium, 29 hard.

Protein denaturation and functional properties: solubility, gelation, emulsification and foaming7
Lipid oxidation: autoxidation, photo-oxidation, rancidity and antioxidants7
Glass transition and molecular mobility6
Starch: structure, gelatinisation, pasting and retrogradation6
Cellulose, pectic substances, gums and dietary fibre6
Hydrolytic rancidity and thermal polymerisation of frying oils6
Chemical and physical changes in macronutrients during processing6
Post-mortem changes and tenderisation of muscle proteins5
Amino acids, peptides and protein structure4
Physical properties of fats: crystallisation and polymorphism4
Water in foods: structure, bound and free water, sorption isotherms and water activity3
Monosaccharides, disaccharides and oligosaccharides3
Food proteins: classification and sources3
Lipids: classification, structure and nomenclature3

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Water in foods: structure, bound and free water, sorption isotherms and water activity · hard

On the food stability map, lipid oxidation in a dried food is slowest at

  1. the very lowest water activity, as close to aw 0 as possible
  2. an intermediate water activity of about aw 0.7–0.8
  3. about the monolayer water activity, roughly aw 0.2–0.3
  4. the highest water activity, just below aw 1.0
Answer and working

C. about the monolayer water activity, roughly aw 0.2–0.3

Lipid oxidation has a minimum near the monolayer; drier foods oxidise faster because water no longer shields metals and hydroperoxides, and intermediate aw raises mobility and catalyst activity. Browning, not oxidation, peaks near 0.6–0.8.

Further reading: Damodaran and Parkin, Fennema's Food Chemistry

2 · Glass transition and molecular mobility · hard

Using the Gordon–Taylor equation with Tg(sucrose) = 62 °C, Tg(water) = −135 °C and k = 4.7, the Tg of amorphous sucrose containing 5 % water is closest to

  1. 52 °C
  2. 23 °C
  3. 60 °C
  4. 27 °C
Answer and working

B. 23 °C

Tg = (0.95 × 62 + 4.7 × 0.05 × (−135))/(0.95 + 4.7 × 0.05) = (58.9 − 31.725)/1.185 = 22.9 °C. 52 °C ignores k (a simple mass-fraction average); 60 °C applies k to the sugar instead of the water; 27 °C stops at the numerator without dividing by 1.185.

Further reading: Roos, Phase Transitions in Foods

3 · Glass transition and molecular mobility · medium

Tg′ in frozen foods refers to the glass transition of

  1. pure ice at atmospheric pressure
  2. the food before any freezing begins
  3. the packaging film around the food
  4. the maximally freeze-concentrated unfrozen phase
Answer and working

D. the maximally freeze-concentrated unfrozen phase

As ice forms, the remaining solution concentrates until it vitrifies at Tg′; storage below Tg′ greatly slows deterioration. Ice is crystalline and has no glass transition in this sense.

Further reading: Roos, Phase Transitions in Foods

4 · Starch: structure, gelatinisation, pasting and retrogradation · hard

A frozen sauce shows weeping and a spongy texture after freeze–thaw cycles. Which starch would best correct this?

  1. An acid-thinned starch
  2. A hydroxypropylated or acetylated starch
  3. A native potato starch
  4. A pregelatinised normal maize starch
Answer and working

B. A hydroxypropylated or acetylated starch

Substituted starches resist retrogradation and are freeze–thaw stable. Acid-thinned starches form strong, retrograding gels; native potato and normal maize starches retrograde and weep.

Further reading: Belitz, Grosch and Schieberle, Food Chemistry

5 · Cellulose, pectic substances, gums and dietary fibre · hard

Oat and barley β-glucans are associated with lowering blood cholesterol mainly because they

  1. are digested to glucose that inhibits cholesterol synthesis in the liver
  2. form viscous solutions in the gut that slow absorption and bind bile acids
  3. contain plant sterols that replace cholesterol in cell membranes
  4. are insoluble fibres that only increase faecal bulk
Answer and working

B. form viscous solutions in the gut that slow absorption and bind bile acids

Mixed-linkage β-glucans are soluble and viscous, reducing bile-acid reabsorption and cholesterol uptake. They are not digested to glucose and are not insoluble bulking fibres.

Further reading: Belitz, Grosch and Schieberle, Food Chemistry

6 · Amino acids, peptides and protein structure · hard

Melamine was added to some protein ingredients as an adulterant because

  1. Kjeldahl and Dumas methods measure total nitrogen, so nitrogen-rich melamine inflates apparent protein
  2. melamine binds to proteins and makes them more soluble in standard extraction tests
  3. melamine is an essential amino acid that raises the true protein quality
  4. melamine reacts with protein to form extra peptide bonds detected by titration
Answer and working

A. Kjeldahl and Dumas methods measure total nitrogen, so nitrogen-rich melamine inflates apparent protein

Nitrogen-based methods cannot tell protein nitrogen from other nitrogen; melamine is about two-thirds nitrogen by mass. It is not an amino acid and does not form peptide bonds.

Further reading: Belitz, Grosch and Schieberle, Food Chemistry

7 · Protein denaturation and functional properties: solubility, gelation, emulsification and foaming · medium

Egg white fails to whip into a stable foam in a bowl containing traces of fat because

  1. lipids compete for the air–water interface and disrupt the protein film
  2. fat raises the pH so that the proteins become fully denatured at once
  3. fat binds the water and leaves no liquid to form the foam lamellae
  4. fat makes the egg-white proteins far too stable to unfold at the bubble surface
Answer and working

A. lipids compete for the air–water interface and disrupt the protein film

Lipids and egg yolk adsorb at the interface and break the cohesive protein film, so bubbles coalesce. The other explanations are incorrect.

Further reading: Damodaran and Parkin, Fennema's Food Chemistry

8 · Post-mortem changes and tenderisation of muscle proteins · easy

Papain, bromelain and ficin, used as meat tenderisers, are obtained respectively from

  1. papaya, pineapple and fig
  2. pineapple, fig and papaya
  3. fig, papaya and pineapple
  4. papaya, fig and pineapple
Answer and working

A. papaya, pineapple and fig

Papain comes from papaya, bromelain from pineapple and ficin from fig.

Further reading: Lawrie and Ledward, Lawrie's Meat Science

9 · Lipid oxidation: autoxidation, photo-oxidation, rancidity and antioxidants · easy

An oil has a peroxide value of 4 meq/kg and a p-anisidine value of 6. Its totox value is

  1. 10
  2. 16
  3. 14
  4. 24
Answer and working

C. 14

Totox = 2PV + AV = 2 × 4 + 6 = 14. 10 is a simple sum; 16 doubles the anisidine value instead; 24 multiplies the two.

Further reading: Frankel, Lipid Oxidation

10 · Chemical and physical changes in macronutrients during processing · medium

Caramelisation differs from the Maillard reaction in that caramelisation

  1. involves sugars alone heated to high temperature, without amino compounds
  2. requires amino acids and reducing sugars at moderate temperature
  3. occurs only in proteins and never in carbohydrates or sugar syrups heated alone
  4. takes place at refrigeration temperatures over long storage
Answer and working

A. involves sugars alone heated to high temperature, without amino compounds

Caramelisation is the thermal degradation of sugars by themselves; the Maillard reaction needs amino groups and can proceed at lower temperatures.

Further reading: Damodaran and Parkin, Fennema's Food Chemistry

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Food Chemistry I: Water, Carbohydrates, Proteins & LipidsThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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Food Chemistry I: Water, Carbohydrates, Proteins & Lipids for other exams

GATE · IndiaGATE · IndiaICAR AIEEA-PG / PGS · India

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