Nutrition, Food Adulteration and Chemical Changes during Processing

Practice questions (18)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

1. A meal of rice and dal has a higher protein quality than either food eaten alone. Which explanation is correct?

   1. Rice protein is short of methionine, which dal supplies, while dal is short of lysine, which rice supplies
   2. Both foods are limiting in tryptophan, and cooking them together releases the tryptophan that is bound to starch granules
   3. Rice protein is short of lysine, which dal supplies, while dal is short of methionine and cysteine, which rice supplies
   4. Dal supplies phytase that removes the phytic acid of rice, so that the rice protein becomes completely digestible
   Show answer

   Answer: C — Rice protein is short of lysine, which dal supplies, while dal is short of methionine and cysteine, which rice supplies

   Cereal proteins are limited by lysine and pulse proteins by the sulphur amino acids (methionine plus cysteine); because each is relatively rich in what the other lacks, the mixture scores higher than either. The reversed pairing is the tempting distractor. Tryptophan is not bound to starch, and dal does not make rice protein completely digestible.

2. A biscuit contains, per 100 g, moisture 3.0 g, protein 7.0 g, fat 22.0 g, ash 1.5 g and crude fibre 1.5 g, with carbohydrate taken by difference. Using Atwater factors of 4 kcal g⁻¹ for protein and carbohydrate and 9 kcal g⁻¹ for fat, and counting fibre and ash as supplying no energy, the percentage of the biscuit's energy that comes from fat, to one decimal place, is ____.

   Numerical answer — type the value.

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   Answer: 40.7

   Carbohydrate = 100 − (3.0 + 7.0 + 22.0 + 1.5 + 1.5) = 65.0 g. Energy = 7.0 × 4 + 22.0 × 9 + 65.0 × 4 = 28 + 198 + 260 = 486 kcal per 100 g. Fat share = 198 ÷ 486 × 100 = 40.7 %. Quoting fat's share of mass (22 %), or forgetting to subtract fibre and ash (carbohydrate 68 g, giving 39.8 %), are the usual slips.

3. In a 28-day rat assay, a group fed a diet containing 10.0 % of a test protein ate 380 g of diet and gained 68.4 g. The casein reference group in the same assay gave a PER of 2.85. Taking the standard PER of casein as 2.5, the adjusted PER of the test protein, to two decimal places, is ____.

   Numerical answer — type the value.

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   Answer: 1.58

   Protein eaten = 10.0 % of 380 g = 38.0 g, so PER = 68.4 ÷ 38.0 = 1.80. Scaling to casein = 2.5: adjusted PER = 1.80 × 2.5 ÷ 2.85 = 1.58. Dividing the gain by the diet eaten instead of the protein eaten (giving 0.18) is the commonest error.

4. In a nitrogen-balance study a subject's daily nitrogen intake is 8.0 g, faecal N is 1.6 g and urinary N is 3.4 g. On a protein-free diet the same subject excretes 0.6 g of endogenous faecal N and 1.2 g of endogenous urinary N per day. Using true digestibility and true biological value, the net protein utilisation (NPU, %) of the dietary protein is ____.

   Numerical answer — type the value.

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   Answer: 60

   N absorbed = I − (F − Fk) = 8.0 − (1.6 − 0.6) = 7.0 g, so true digestibility = 87.5 %. N retained = 7.0 − (U − Uk) = 7.0 − (3.4 − 1.2) = 4.8 g, so BV = 4.8 ÷ 7.0 = 68.6 %. NPU = retained ÷ intake = 4.8 ÷ 8.0 = 60 %, equal to BV × D = 0.686 × 0.875. Ignoring the endogenous corrections gives (8.0 − 1.6 − 3.4) ÷ 8.0 = 37.5 %.

5. A day's diet provides 14 mg of preformed niacin and 720 mg of tryptophan. Taking 60 mg of dietary tryptophan as equivalent to 1 mg of niacin, the total niacin equivalents supplied, in mg, are ____.

   Numerical answer — type the value.

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   Answer: 26

   Tryptophan contributes 720 ÷ 60 = 12 mg niacin equivalents, so the total is 14 + 12 = 26 mg NE. This conversion is why diets poor in both niacin and tryptophan, such as maize-based diets whose niacin is also largely bound, lead to pellagra.

6. In a community living mainly on maize, with little milk, pulses or meat, many people show dermatitis on sun-exposed skin, diarrhoea and mental confusion. Deficiency of which vitamin is most likely?

   1. Thiamine, whose deficiency causes beriberi
   2. Niacin, whose deficiency causes pellagra
   3. Riboflavin, whose deficiency causes cheilosis
   4. Ascorbic acid, whose deficiency causes scurvy
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   Answer: B — Niacin, whose deficiency causes pellagra

   Dermatitis, diarrhoea and dementia are the three Ds of pellagra. Maize niacin is largely bound and maize protein is poor in tryptophan, from which niacin can be made. Beriberi presents with neuropathy or heart failure, riboflavin deficiency with cracks at the corners of the mouth, and scurvy with bleeding gums.

7. Which of the following statements about vitamins, minerals and deficiency diseases are correct?

   1. Ascorbic acid eaten in the same meal reduces the absorption of non-haem iron
   2. Strict vegetarians are at risk of vitamin B12 deficiency because the vitamin is made only by microbes and occurs naturally in animal foods
   3. Kwashiorkor, unlike marasmus, is marked by severe wasting without oedema
   4. Riboflavin in milk is destroyed mainly by light, so milk keeps it better in opaque packs
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   Answer: B — Strict vegetarians are at risk of vitamin B12 deficiency because the vitamin is made only by microbes and occurs naturally in animal foods; D — Riboflavin in milk is destroyed mainly by light, so milk keeps it better in opaque packs

   B12 is synthesised only by microorganisms and reaches the diet through animal foods (and fortified products), so vegans need fortified foods or supplements. Riboflavin is heat-stable but photolabile. Ascorbic acid enhances non-haem iron absorption by reducing Fe³⁺ to Fe²⁺ and chelating it. Kwashiorkor is the oedematous form of protein–energy malnutrition; severe wasting without oedema describes marasmus.

8. Why must humans obtain linoleic and α-linolenic acids from the diet?

   1. Human cells cannot elongate C16 fatty acids to C18, so all C18 acids must come from food
   2. Humans lack Δ9 desaturase, so they cannot make any monounsaturated fatty acid such as oleic acid
   3. Both acids are destroyed in the liver as soon as they are synthesised from acetyl-CoA
   4. Humans lack the Δ12 and Δ15 desaturases that insert double bonds beyond carbon 9
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   Answer: D — Humans lack the Δ12 and Δ15 desaturases that insert double bonds beyond carbon 9

   Human desaturases insert double bonds at Δ9, Δ6 and Δ5 but not beyond carbon 9, so the n-6 and n-3 double bonds of linoleic (Δ9,12) and α-linolenic (Δ9,12,15) acids cannot be made. Humans do have elongases and Δ9 desaturase (oleic acid is made from stearic acid), and nothing destroys these acids on synthesis, so the other options are false.

9. Red kidney beans soaked and then cooked for several hours in a slow cooker at about 75 °C caused vomiting and diarrhoea within a few hours, whereas the same beans boiled vigorously for 10 minutes before simmering caused no illness. Which constituent is responsible?

   1. Phytic acid, which chelates calcium and zinc in the gut and so triggers acute vomiting
   2. Phytohaemagglutinin, a lectin destroyed by boiling but not by warm sub-boiling heat
   3. Oxalic acid, which crystallises in the kidneys within hours of the meal
   4. Raffinose oligosaccharides, which are fermented in the colon to toxic gases
   Show answer

   Answer: B — Phytohaemagglutinin, a lectin destroyed by boiling but not by warm sub-boiling heat

   Phytohaemagglutinin, the lectin of red kidney beans, binds the intestinal lining and causes vomiting and diarrhoea within hours. It is a protein destroyed by vigorous boiling, but heating below boiling may leave it active or even make matters worse. Phytic acid and oxalate impair mineral absorption over time rather than causing acute illness, and raffinose sugars cause flatulence, not poisoning.

10. Which finding is an example of nutrigenomics, the effect of a dietary component on gene expression?

    1. Polyunsaturated fatty acids bind PPARα and switch on liver genes for fatty-acid oxidation
    2. Heating soybean flour raises its protein efficiency ratio by destroying trypsin inhibitors
    3. Iodising salt lowers the prevalence of goitre across a whole population
    4. Boiling vegetables in excess water leaches their vitamin C into the cooking water
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    Answer: A — Polyunsaturated fatty acids bind PPARα and switch on liver genes for fatty-acid oxidation

    PUFA acting as ligands of the nuclear receptor PPARα to change transcription is a nutrient–gene interaction, the core of nutrigenomics. Heat inactivation of an anti-nutrient, population iodisation and vitamin leaching are nutrition and processing effects with no gene-level mechanism.

11. Which reagent is used in the simple test for urea added to milk, and what does a positive result look like?

    1. Iodine solution, which turns the milk a deep blue colour within a minute
    2. Rosolic acid in alcohol, which turns the milk a distinct rose-red colour
    3. p-Dimethylaminobenzaldehyde, which gives a distinct yellow colour
    4. Furfural with hydrochloric acid, which gives a crimson colour
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    Answer: C — p-Dimethylaminobenzaldehyde, which gives a distinct yellow colour

    Urea reacts with p-dimethylaminobenzaldehyde (DMAB) to give a yellow colour clearly deeper than that of normal milk, which has only traces of natural urea. Iodine detects starch, rosolic acid detects neutralisers such as sodium carbonate, and furfural with HCl is the Baudouin test for sesame oil.

12. Genuine milk from a region has a freezing point of −0.520 °C; a market sample freezes at −0.416 °C. Using % added water = (T − T′) × 100 ÷ T, where T and T′ are the freezing-point depressions of the genuine milk and of the sample, the percentage of added water in the sample is ____.

    Numerical answer — type the value.

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    Answer: 20

    T = 0.520 and T′ = 0.416, so % added water = (0.520 − 0.416) × 100 ÷ 0.520 = 20. The freezing point depends on dissolved lactose and salts, which the cow keeps nearly constant, so added water moves it towards 0 °C in proportion. It is more reliable than the lactometer, whose reading can be restored by skimming a watered milk.

13. Which of the following statements about the detection of adulterants are correct?

    1. Argemone oil in mustard oil gives a reddish-brown colour with concentrated nitric acid
    2. A crimson colour in the Baudouin test proves that a ghee sample is pure milk fat
    3. Mineral oil in an edible oil shows up as turbidity after saponification because it does not form soap
    4. Starch added to paneer is revealed by a rose-red colour with rosolic acid
    Show answer

    Answer: A — Argemone oil in mustard oil gives a reddish-brown colour with concentrated nitric acid; C — Mineral oil in an edible oil shows up as turbidity after saponification because it does not form soap

    The nitric acid test colours the argemone alkaloids reddish-brown, and mineral oil, being hydrocarbon, stays unsaponified and clouds the soap solution. A crimson Baudouin test shows sesame oil, so in ghee it points to vegetable oil, not purity. Starch is detected with iodine (blue); rosolic acid detects neutralisers in milk.

14. A honey sample from bees foraging on C3 plants is suspected of being blended with cane-sugar or maize syrup that has been inverted to match honey's glucose and fructose levels. Which measurement can detect the blend?

    1. Total reducing sugars by Lane–Eynon titration, which fall below the honey standard
    2. The ¹³C/¹²C ratio, since C4 plants such as cane and maize are richer in ¹³C than C3 nectar
    3. Refractometric total soluble solids, which are lower in syrups than in genuine honey
    4. The iodine test, which turns blue because inverted maize syrup still carries much native starch
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    Answer: B — The ¹³C/¹²C ratio, since C4 plants such as cane and maize are richer in ¹³C than C3 nectar

    C4 photosynthesis (PEP carboxylase) discriminates less against ¹³C than C3 photosynthesis (Rubisco), so cane and maize sugars have a less negative δ¹³C than the C3 nectar sugars of honey; stable-isotope analysis detects the blend however well the sugar profile is matched. An inverted syrup gives normal reducing sugars, its soluble solids can be set to any value, and a hydrolysed syrup has no starch left to react with iodine.

15. Potato chips fried at 180 °C contain acrylamide, but potatoes boiled in water contain essentially none. Why?

    1. Acrylamide is produced by oxidation of the frying oil, which boiled potatoes never come into contact with
    2. Acrylamide forms when starch gelatinises, which the water of boiling prevents from happening
    3. Acrylamide forms from the glycoalkaloid solanine, which leaches into the boiling water
    4. Acrylamide forms from asparagine and reducing sugars via Maillard chemistry above about 120 °C
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    Answer: D — Acrylamide forms from asparagine and reducing sugars via Maillard chemistry above about 120 °C

    Acrylamide arises when free asparagine reacts with reducing sugars through Maillard intermediates at surface temperatures above about 120 °C and low moisture. Boiling holds the food at 100 °C in water, so it does not form. It is not an oil-oxidation product (oven-baked chips contain it too), boiling causes rather than prevents gelatinisation, and solanine is unrelated.

16. A sunflower oil is used for deep-fat frying at 180 °C for several days without replacement. Which set of changes in its properties is expected?

    1. Acid value rises, iodine value falls, viscosity rises and smoke point falls
    2. Acid value falls, iodine value rises, viscosity falls and smoke point rises
    3. Acid value rises, iodine value rises, viscosity falls and smoke point falls
    4. Acid value is unchanged, iodine value falls, viscosity rises and smoke point rises
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    Answer: A — Acid value rises, iodine value falls, viscosity rises and smoke point falls

    Steam from the food hydrolyses the oil, releasing free fatty acids (acid value up, smoke point down); oxidation and polymerisation consume double bonds (iodine value down) and build dimers and polymers (viscosity up, with foaming and darkening). Each distractor gets at least one direction wrong; in particular nothing in frying adds double bonds, so the iodine value cannot rise.

17. Which of the following statements about changes in food biopolymers during processing are correct?

    1. Blanching vegetables at about 60 °C can firm them, because pectin methylesterase demethylates pectin, which then cross-links with Ca²⁺
    2. High-pressure processing at room temperature breaks covalent bonds, so it destroys vitamins as much as retorting does
    3. Parboiling paddy before milling helps retain B vitamins, because they migrate into the endosperm
    4. Extrusion cooking raises the trypsin-inhibitor activity of soy-based snacks above that of raw soy flour
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    Answer: A — Blanching vegetables at about 60 °C can firm them, because pectin methylesterase demethylates pectin, which then cross-links with Ca²⁺; C — Parboiling paddy before milling helps retain B vitamins, because they migrate into the endosperm

    Low-temperature blanching activates pectin methylesterase, and the free carboxyl groups it creates are bridged by Ca²⁺, firming canned or frozen vegetables. During parboiling the water-soluble B vitamins of bran and germ diffuse into the endosperm and survive polishing. High pressure acts on non-covalent bonds, so small molecules such as vitamins and flavour compounds are largely spared, and the heat and shear of extrusion denature trypsin inhibitors, lowering their activity.

18. Thiamine pyrophosphate, the coenzyme form of vitamin B1, is required by which enzyme?

    1. Alanine aminotransferase, which transfers amino groups in amino-acid metabolism
    2. Acetyl-CoA carboxylase, which adds CO₂ in the first step of fatty-acid synthesis
    3. Lactate dehydrogenase, which reduces pyruvate to lactate using NADH
    4. Pyruvate dehydrogenase, which oxidatively decarboxylates pyruvate to acetyl-CoA
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    Answer: D — Pyruvate dehydrogenase, which oxidatively decarboxylates pyruvate to acetyl-CoA

    TPP is the coenzyme of oxidative decarboxylations (pyruvate and α-ketoglutarate dehydrogenases) and of transketolase, which is why thiamine deficiency impairs carbohydrate metabolism and nerve function. Aminotransferases use pyridoxal phosphate, carboxylases use biotin, and lactate dehydrogenase uses NADH derived from niacin.