This set mixes the five lessons in human nutrition and digestion: tracing food, enzyme links, villus structure, absorption versus assimilation, and dietary data. All data are invented for practice. Questions run from easier to harder, and they do not say which lesson to use.
Write each answer fully on paper first. Then open the solution and compare your steps with it. Questions about your own diet or health belong with a doctor or registered dietitian; nothing here is personal advice.
Questions
Q1. Match each word to its meaning: digestion, absorption, assimilation, egestion. (a) Removal of undigested food as faeces. (b) Use of absorbed molecules by cells. (c) Breakdown of large insoluble molecules into small soluble ones. (d) Movement of small molecules from the gut into the blood or lymph.
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Digestion = (c). Absorption = (d). Assimilation = (b). Egestion = (a).
Q2. Name the enzyme, its substrate and its product for each of these steps: (a) in saliva; (b) in the stomach; (c) in the small intestine acting on fat.
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(a) Amylase, substrate starch, product maltose. (b) Pepsin (a protease), substrate protein, products shorter chains of amino acids, with amino acids completed in the small intestine. (c) Lipase, substrate fats, products fatty acids and glycerol.
Q3. Trace the starch in a slice of bread from the mouth to absorption. Name the organ, the enzyme and the product at each stage where chemical digestion happens.
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Mouth: salivary amylase digests starch to maltose. Stomach: no starch digestion, because the acid stops amylase working. Small intestine: pancreatic amylase digests more starch to maltose, then maltase in the gut lining digests maltose to glucose. Absorption: glucose is absorbed into blood capillaries in the villi.
Q4. A fat droplet is a cube of side 1 cm. Bile breaks it into 1000 small cubes, each of side 0.1 cm. (a) Find the total surface area before and after. (b) Explain why this helps digestion.
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(a) Before: 6 × 1 × 1 = 6 cm². One small cube: 6 × 0.1 × 0.1 = 0.06 cm². After: 1000 × 0.06 = 60 cm², which is 10 times larger. Check: volume 1000 × 0.001 = 1 cm³, unchanged. (b) Bile emulsifies fat, giving a larger surface area. Lipase can then reach more fat at once, so digestion is faster.
Q5. In an invented experiment, protein is mixed with pepsin. At pH 2 the protein is digested in 5 minutes. At pH 7 it is not digested after 30 minutes. (a) Calculate the rate at pH 2 as 1 ÷ time. (b) Explain the result at pH 7.
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(a) Rate = 1 ÷ 5 = 0.2 per minute. (b) Pepsin is most active in acid, which matches the stomach. At pH 7 its active site changes shape, so the enzyme is denatured or almost inactive, and the protein does not fit and stays undigested.
Q6. A model of gut lining has a flat 1 cm² patch with 40 square-column villi, each 0.1 cm wide and 0.3 cm tall. Find the total surface area and compare it with the flat patch.
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One villus adds 4 × (0.1 × 0.3) = 0.12 cm². For 40 villi: 40 × 0.12 = 4.8 cm². Total = 1 + 4.8 = 5.8 cm², which is 5.8 times the flat patch. Check: bases use 40 × 0.01 = 0.4 cm², which fits inside 1 cm².
Q7. Explain how two features of a villus help absorption of glucose. Each explanation must include a function.
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Any two of: thin wall, one cell thick, giving a short diffusion distance so glucose crosses quickly; many capillaries, which carry glucose away, keeping the concentration lower in the blood so absorption continues; large surface area from the folded lining, so more glucose crosses at once.
Q8. After a meal, an amino acid is absorbed and later used by a cell to make a protein. Which step is absorption and which is assimilation? Give the point where one ends and the other begins.
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Absorption is the movement into the blood capillary in the villus. Assimilation is the cell joining the amino acid into a new protein. They meet when the molecule is in the blood or lymph and taken up by body cells.
Q9. Invented data per 100 g: food S has 300 kJ and 10 g protein; food T has 1500 kJ and 6 g protein. A model meal has 250 g of S and 80 g of T. Find (a) total energy, (b) total protein, (c) the percentage of energy from T.
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Factors: S = 2.5, T = 0.8. (a) S 300 × 2.5 = 750 kJ; T 1500 × 0.8 = 1200 kJ. Total = 1950 kJ. (b) S 10 × 2.5 = 25 g; T 6 × 0.8 = 4.8 g. Total = 29.8 g. (c) 1200 ÷ 1950 × 100 = 61.5%, so about 62%.
Q10. An invented survey finds that students who ate breakfast had higher test scores. A student writes, “Eating breakfast causes higher scores.” Evaluate this claim.
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The data show an association only. Other factors, such as sleep or time to study, may differ between the groups, and a survey does not control them. A supported conclusion is “In this survey, breakfast was linked to higher scores.” The claim also says nothing about any one student.
Q11. Sort into digestion, absorption, assimilation: (a) glucose stored as glycogen in liver cells; (b) protease splitting protein in the stomach; (c) fatty acids entering a lacteal.
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(a) Assimilation. (b) Digestion. (c) Absorption.
If you got these wrong
| Error you made | Go to |
|---|---|
| Wrong organ, enzyme or product when tracing food (Q2, Q3) | Trace a food component through digestion |
| Mixed up substrate and product, or a rate calculation (Q2, Q4, Q5) | Link an enzyme with substrate and products |
| Villus area calculation or feature without function (Q6, Q7) | Explain absorption using villus structure |
| Swapped absorption, assimilation or egestion (Q1, Q8, Q11) | Distinguish assimilation from absorption |
| Scaling per 100 g, percentage share or over-claiming (Q9, Q10) | Interpret a dietary dataset |
The mistake log and retest queue helps you record each slip by type and retest it later. The probability tree and counting board and inheritance model board are separate modelling tools for other topics.
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