Food Components: Carbohydrates, Proteins, Lipids, Pigments, Flavours and Enzymes

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. In the Kjeldahl method, crude protein is commonly calculated as % nitrogen × 6.25. What assumption lies behind the factor 6.25?

   1. Food proteins contain, on average, 16 % nitrogen by mass, and 100 ÷ 16 = 6.25
   2. Part of the ammonia escapes during digestion, and the factor restores the lost nitrogen
   3. The factor converts the mass of ammonium sulphate formed into the mass of ammonia distilled
   4. Non-protein nitrogen such as urea is excluded by the factor, leaving only protein nitrogen
   Show answer

   Answer: A — Food proteins contain, on average, 16 % nitrogen by mass, and 100 ÷ 16 = 6.25

   The factor is 100/16: the method measures nitrogen, and an average protein is taken to be 16 % N, so protein = N × 6.25 (6.38 for milk and 5.70 for wheat flour, whose proteins contain different proportions of N). The tempting idea that the factor removes non-protein nitrogen is the opposite of the truth: Kjeldahl counts urea and other non-protein N as protein. Ammonia is not lost in a correctly run digestion, and the factor has nothing to do with the mass of ammonium sulphate.

2. A 0.800 g sample of pulse flour is analysed by the Kjeldahl method. The ammonia distilled into boric acid needs 14.20 mL of 0.1000 N HCl, and a reagent blank needs 0.20 mL. Taking the atomic mass of nitrogen as 14.007 and a nitrogen-to-protein factor of 6.25, the crude protein content of the flour, in % (w/w) to one decimal place, is ____.

   Numerical answer — type the value.

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   Answer: 15.3

   Net titre = 14.20 − 0.20 = 14.00 mL, i.e. 14.00 × 0.1000 = 1.400 mmol of N = 1.400 × 14.007 = 19.61 mg N. %N = 19.61 ÷ 800 × 100 = 2.451 %. Protein = 2.451 × 6.25 = 15.32 %, which is 15.3. Forgetting the blank gives 15.5 %, and using the wheat factor 5.70 gives 14.0 %.

3. Three identical loaves of white bread are stored, well wrapped, at −18 °C, 4 °C and 25 °C. After two days, which loaf has the firmest, most stale crumb, and why?

   1. The loaf at 25 °C, because staling is driven mainly by moisture loss, which is fastest at room temperature
   2. The loaf at −18 °C, because ice crystals squeeze water out of the gluten and harden the crumb
   3. The loaf at 4 °C, because amylopectin recrystallises fastest at refrigerator temperatures
   4. All three equally, because retrogradation depends only on time and not on storage temperature
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   Answer: C — The loaf at 4 °C, because amylopectin recrystallises fastest at refrigerator temperatures

   Staling is mainly retrogradation of amylopectin in the crumb, a crystallisation that is fastest a little above freezing: cold enough to favour nucleation, warm enough for chains to move. Freezing immobilises the chains and nearly stops it, and room temperature is slower than the refrigerator. Moisture loss, the tempting answer, is minor in well-wrapped bread; staling happens without it, and reheating reverses much of it, which moisture loss could not explain.

4. Which of the following statements about food carbohydrates are correct?

   1. Starch granules lose birefringence on gelatinisation because their crystalline regions melt
   2. Amylopectin retrogrades faster than amylose because its many branch points nucleate crystals
   3. Low-methoxyl pectin can form a gel with calcium ions even when little sugar is present
   4. Sucrose reduces Fehling's solution because its fructose unit keeps a free anomeric carbon
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   Answer: A — Starch granules lose birefringence on gelatinisation because their crystalline regions melt; C — Low-methoxyl pectin can form a gel with calcium ions even when little sugar is present

   Gelatinisation melts the ordered regions, so the Maltese cross disappears in polarised light. Low-methoxyl pectin has many free carboxyl groups that Ca²⁺ bridges in an egg-box gel, which is why it is used in low-sugar jams. Amylose, not amylopectin, retrogrades fast (hours); the short outer branches of amylopectin crystallise slowly over days. Sucrose is non-reducing because its glycosidic bond joins the anomeric carbons of both glucose and fructose.

5. Rats fed raw soybean flour grow poorly and develop enlarged pancreases, while rats fed the same flour after moist-heat toasting grow normally. Which explanation fits best?

   1. Toasting hydrolyses soy protein to free amino acids, which are absorbed faster than intact protein
   2. Toasting creates essential amino acids through Maillard reactions between sugars and lysine
   3. Toasting volatilises the phytic acid of soy, releasing the bound zinc and iron for absorption
   4. Toasting denatures the trypsin inhibitors and lectins and unfolds the proteins for proteases
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   Answer: D — Toasting denatures the trypsin inhibitors and lectins and unfolds the proteins for proteases

   Kunitz and Bowman–Birk trypsin inhibitors block intestinal trypsin, so protein is poorly digested and the pancreas enlarges as it over-secretes; lectins damage the gut wall. Both are proteins and are inactivated by moist-heat denaturation, which also opens other proteins to digestion. Toasting does not hydrolyse peptide bonds to free amino acids, Maillard reactions destroy lysine rather than create amino acids, and phytic acid is not volatile.

6. A fat sample has a saponification value of 255 mg KOH g⁻¹ and an iodine value of 9 g I₂ per 100 g. Which fat is it most likely to be?

   1. Groundnut oil, rich in oleic and linoleic acids esterified in long-chain triacylglycerols
   2. Coconut oil, dominated by saturated lauric and myristic acids of medium chain length
   3. Linseed oil, a drying oil in which α-linolenic acid makes up the largest share
   4. Cocoa butter, built mainly from palmitic, stearic and oleic acids in symmetrical triacylglycerols
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   Answer: B — Coconut oil, dominated by saturated lauric and myristic acids of medium chain length

   A high SV means a low mean molar mass, i.e. short or medium chains, and a very low IV means almost no double bonds: coconut oil (lauric C12:0 and myristic C14:0) fits both. Groundnut oil and cocoa butter have C16–C18 chains (SVs near 190) and IVs well above 9 (groundnut near 90, cocoa butter about 35); linseed has an IV above 170. Cocoa butter tempts because it is solid and rich in saturated acids, but its long chains keep its SV low.

7. Calculate the iodine value of pure trilinolein, the triacylglycerol of linoleic acid (C18:2), with molecular formula C₅₇H₉₈O₆ and M = 879.4 g mol⁻¹. Take M(I₂) = 253.8 g mol⁻¹ and assume that each C=C double bond adds one molecule of I₂. Give the answer to the nearest whole number.

   Numerical answer — type the value.

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   Answer: 173

   Each linoleoyl chain has 2 C=C, so one molecule has 6 and absorbs 6 I₂ = 6 × 253.8 = 1522.8 g per 879.4 g of fat. IV = 1522.8 ÷ 879.4 × 100 = 173.2, i.e. 173. Counting one double bond per chain gives 86.6, close to triolein's 86.0, which is the commonest slip.

8. A 2.000 g sample of an oil is refluxed with 25.0 mL of about 0.5 N alcoholic KOH until saponification is complete. The excess alkali is back-titrated with 0.500 N HCl, needing 12.0 mL; a blank run without oil needs 25.4 mL. Taking M(KOH) = 56.1 g mol⁻¹, the saponification value of the oil (mg KOH g⁻¹), to the nearest whole number, is ____.

   Numerical answer — type the value.

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   Answer: 188

   KOH consumed by the oil = (blank − sample) titre = 25.4 − 12.0 = 13.4 mL of 0.500 N = 6.70 meq, i.e. 6.70 × 56.1 = 375.9 mg KOH for 2.000 g of oil. SV = 375.9 ÷ 2.000 = 187.9, i.e. 188, typical of a C18-rich seed oil. Using the sample titre alone instead of the difference gives 168, a common error.

9. To find the peroxide value of a stored groundnut oil, 5.00 g of the oil is treated with acetic acid–chloroform and saturated KI, and the liberated iodine is titrated with 0.0100 N sodium thiosulphate, needing 3.10 mL. A blank needs 0.10 mL. The peroxide value, in milliequivalents of active oxygen per kg of oil, is ____.

   Numerical answer — type the value.

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   Answer: 6

   PV = (S − B) × N × 1000 ÷ W = (3.10 − 0.10) × 0.0100 × 1000 ÷ 5.00 = 6 meq kg⁻¹. Each mL of 0.01 N thiosulphate is 0.01 meq of iodine freed by the hydroperoxides, and the factor 1000 converts per gram to per kilogram. Because PV falls again as hydroperoxides decompose, one reading is judged together with the oil's history.

10. Which of the following statements about the deterioration and crystallisation of fats are correct?

    1. Blanching soybeans before grinding prevents the beany flavour made by lipoxygenase
    2. Hydrolytic rancidity of butter is followed by a rise in its iodine value
    3. β′ crystals are preferred in margarine because they are small and give a smooth texture
    4. Photo-oxidation sensitised by chlorophyll is fully stopped by adding BHT to the oil
    Show answer

    Answer: A — Blanching soybeans before grinding prevents the beany flavour made by lipoxygenase; C — β′ crystals are preferred in margarine because they are small and give a smooth texture

    Lipoxygenase is heat-labile, so blanching before the cells are broken prevents the hydroperoxides that give soy its beany note. β′ crystals are small and give plasticity and smoothness, whereas β crystals grow large and make margarine grainy. Hydrolysis frees fatty acids without touching the double bonds, so it raises the acid value, not the iodine value. Chlorophyll-sensitised photo-oxidation runs through singlet oxygen, which adds directly to double bonds; chain-breaking antioxidants such as BHT do not stop it, while carotenoids and opaque packaging do.

11. A properly tempered chocolate bar is left in a warm car, partly melts and then sets again. It now has a dull grey surface film that smears and melts away under a warm finger and does not dissolve in water. What has most probably happened?

    1. Moisture condensed on the bar, dissolved surface sugar, and the sugar recrystallised as a white crust
    2. Lipid oxidation formed hydroperoxides that coat the surface as a grey, waxy layer of polymer
    3. The cocoa butter recrystallised in unstable forms that changed towards form VI, giving fat bloom
    4. Maillard reactions between milk proteins and lactose formed melanoidins at the surface on reheating
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    Answer: C — The cocoa butter recrystallised in unstable forms that changed towards form VI, giving fat bloom

    A film that smears and melts under a finger is fat; one that dissolves in water and feels gritty would be sugar bloom, the tempting alternative. Melting destroyed the form V crystals set by tempering; on uncontrolled resetting the cocoa butter crystallised in unstable forms that move towards the stable form VI with large, light-scattering crystals, and liquid fat drawn to the surface recrystallised there. Oxidation gives off-flavour, not a crystalline film, and a few hours in a car do not cause visible Maillard browning.

12. Green beans simmered for 20 minutes in a covered pan turn olive-brown, while the same beans cooked for 5 minutes in an open pan of boiling water stay bright green. What change in the chlorophyll explains the olive colour?

    1. Chlorophyllase removed the phytol tail, forming water-soluble chlorophyllide that leached into the water
    2. Copper from the pan replaced the magnesium ion, forming a stable olive copper–chlorophyll complex
    3. Heat oxidised the carotenoids that mask chlorophyll, and their breakdown products turned brown
    4. Plant acids retained by the lid replaced Mg²⁺ with two H⁺, converting chlorophyll to pheophytin
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    Answer: D — Plant acids retained by the lid replaced Mg²⁺ with two H⁺, converting chlorophyll to pheophytin

    Heating releases plant acids; a lid keeps the volatile ones in, and long cooking lets H⁺ displace the central Mg²⁺, giving olive-brown pheophytin. Chlorophyllide is still green, so its leaching would pale the beans, not turn them olive. Copper–chlorophyll complexes are bright green, which is why copper or zinc is used for regreening, and carotenoids do not mask chlorophyll; chlorophyll masks them.

13. Which of the following statements about food pigments are correct?

    1. Vacuum-packed fresh beef looks purple because its myoglobin is in the deoxygenated Fe²⁺ form
    2. Anthocyanins in a fruit drink are most intensely red at neutral pH and fade as acid is added
    3. Lycopene of tomato has no provitamin A activity because it lacks a β-ionone ring
    4. Beetroot owes its red colour to anthocyanins that occur together with betalains in its cells
    Show answer

    Answer: A — Vacuum-packed fresh beef looks purple because its myoglobin is in the deoxygenated Fe²⁺ form; C — Lycopene of tomato has no provitamin A activity because it lacks a β-ionone ring

    Without oxygen myoglobin stays as purple deoxymyoglobin (Fe²⁺); when the pack is opened O₂ binds and the meat blooms red. Lycopene is an acyclic carotene, and vitamin A activity needs at least one unsubstituted β-ionone ring. Anthocyanins are red as the flavylium cation at low pH and turn colourless and then blue as pH rises, the reverse of the claim. Beetroot is coloured by betalains (betanin), and betalains and anthocyanins never occur in the same plant.

14. Limonene, the main component of orange-peel oil, is best classified as a

    1. monoterpene hydrocarbon built from two isoprene units
    2. sesquiterpene ketone built from three isoprene units
    3. fruity ester formed from a C10 alcohol and acetic acid
    4. phenolic aldehyde formed by Strecker degradation
    Show answer

    Answer: A — monoterpene hydrocarbon built from two isoprene units

    Limonene, C₁₀H₁₆, is a cyclic monoterpene: two C5 isoprene units and no oxygen. Sesquiterpenes have 15 carbons; nootkatone of grapefruit is a sesquiterpene ketone, the tempting confusion for a citrus note. Esters such as isoamyl acetate carry fruity notes, and Strecker aldehydes come from amino acids during heating.

15. Which enzyme is most commonly used as the indicator that vegetables have been adequately blanched before freezing?

    1. Alkaline phosphatase, because it is the most heat-stable enzyme of plant tissue
    2. Peroxidase, because it is among the most heat-stable enzymes in vegetables
    3. Polyphenol oxidase, because it is the only enzyme active in frozen vegetables
    4. Invertase, because it is the main cause of flavour loss in frozen vegetables
    Show answer

    Answer: B — Peroxidase, because it is among the most heat-stable enzymes in vegetables

    Peroxidase is one of the most heat-resistant enzymes in vegetables and is easy to test (a colour with guaiacol and H₂O₂), so its loss shows that less stable enzymes are also destroyed; lipoxygenase, the cause of off-flavour, is sometimes used for peas and beans. Alkaline phosphatase is the indicator for milk pasteurisation, which makes it the tempting distractor. Polyphenol oxidase is not the only active enzyme in frozen vegetables, and invertase is not a blanching indicator.

16. An enzyme follows Michaelis–Menten kinetics with Km = 2.0 mM and Vmax = 120 µmol min⁻¹. It is assayed at [S] = 3.0 mM in the presence of a competitive inhibitor at [I] = 4.0 µM, whose Ki = 2.0 µM. Compared with the uninhibited rate at the same [S], the percentage inhibition of the rate, to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 44.4

    Uninhibited: v₀ = 120 × 3.0 ÷ (2.0 + 3.0) = 72.0. Competitive inhibition multiplies Km by (1 + [I]/Ki) = 1 + 4.0/2.0 = 3, so apparent Km = 6.0 mM and vᵢ = 120 × 3.0 ÷ (6.0 + 3.0) = 40.0. Inhibition = (72.0 − 40.0) ÷ 72.0 × 100 = 44.4 %. Treating it as non-competitive (Vmax ÷ 3) would give 66.7 %.

17. Spray-dried egg white turns brown and loses solubility during storage, but egg white treated with glucose oxidase and catalase before drying does not. What type of browning was taking place?

    1. Enzymatic browning, in which polyphenol oxidase oxidises tyrosine residues of the egg proteins
    2. Maillard browning between the small amount of free glucose and the amino groups of the proteins
    3. Caramelisation of the free glucose, which glucose oxidase prevents by lowering the drying temperature
    4. Ascorbic acid browning, because glucose oxidase removes the oxygen that degrades vitamin C
    Show answer

    Answer: B — Maillard browning between the small amount of free glucose and the amino groups of the proteins

    Egg white contains a little free glucose, a reducing sugar, which reacts with protein amino groups at the low water activity of the powder: Maillard browning, with loss of solubility. Glucose oxidase converts the glucose to gluconic acid (catalase removes the H₂O₂), so the reaction has no sugar to use. Egg white has no phenolic substrate pool for polyphenol oxidase, caramelisation needs far higher temperatures than storage, and egg white contains no vitamin C.

18. Freshly cut apple slices are to be kept from browning for a few hours. Which of the following measures would slow the browning?

    1. Dipping the slices in ascorbic acid solution, which reduces the o-quinones back to diphenols
    2. Lowering the pH of the dip to about 3 with citric acid
    3. Keeping the slices immersed in sugar syrup to limit their contact with oxygen
    4. Adding a little copper sulphate to the dip to saturate the enzyme with its metal
    Show answer

    Answer: A — Dipping the slices in ascorbic acid solution, which reduces the o-quinones back to diphenols; B — Lowering the pH of the dip to about 3 with citric acid; C — Keeping the slices immersed in sugar syrup to limit their contact with oxygen

    Polyphenol oxidase needs O₂, a phenolic substrate and its copper centre, and works best near pH 5 to 7. Ascorbic acid reduces quinones before they polymerise (until it is consumed), citric acid takes the pH below the optimum and chelates copper, and syrup limits oxygen. Adding Cu²⁺ does not inactivate the enzyme; copper ions, if anything, promote oxidation and browning.