S5 Biology Paper: Human Digestion, Nutrition And Enzyme Kinetics (100 Marks)

S5 Biology Paper: Human Digestion, Nutrition and Enzyme Kinetics (100 marks)

2026-11-11

S5 Biology – digestion and nutrition

S5 Biology paper: six questions on digestion, enzymes, bile emulsification, villi absorption, dietary energy and water balance, with answers.

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Paper details　S5 Biology practice paper with six questions on the alimentary canal, peristalsis, enzyme optimum temperature and pH, bile emulsification, villi absorption area, dietary energy calculations and water balance, with full answers and marking notes.

Part 1: Structure of the alimentary canal and mechanical digestion (18 marks)

Answer all parts and show your working.

#	Question	Answer and marking notes
1	The large intestine is 1.5 m long and 1.2 m of it absorbs water. Name the main region that absorbs water and find its percentage of the total length.	Answer: The colon. 1.2 ÷ 1.5 = 0.8, about 80%. (Marking: 2 marks for the name, 2 marks for the percentage.)
2	Peristalsis in the oesophagus occurs 3 times per minute. Find the number of contractions in 15 minutes and name the tissue in the gut wall that carries them out.	Answer: 3 × 15 = 45 contractions. They are produced by the smooth muscle of the gut wall, namely the circular and longitudinal muscle, under involuntary control. (Marking: 3 marks for the number, 2 marks for the tissue.)
3	Teeth grind food from particles of 8 mm to particles of 2 mm. Assuming the total volume is unchanged, find the factor by which the number of particles increases.	Answer: (8 ÷ 2)³ = 4³ = 64 times. (Marking: 2 marks for the diameter ratio, 2 marks for cubing it.)
4	Bile emulsifies fat globules of diameter 2.0 mm into droplets of diameter 0.10 mm. Find the factor by which the number of droplets increases and the factor by which the total surface area increases.	Answer: The diameter falls by a factor of 20. The number of droplets rises by 20³ = 8000 times. Each droplet has 1/400 of the original surface area, so the total surface area rises by 8000 ÷ 400 = 20 times. (Marking: 2 marks for the number, 1 mark for the single droplet area, 2 marks for the total area.)

Part 2: Properties of enzymes and enzyme kinetics (16 marks)

The questions below concern salivary amylase, pepsin and trypsin.

#	Question	Answer and marking notes
5	Salivary amylase completely hydrolyses 5.0 g of starch in 12 minutes. Find the mean rate of reaction in g per minute to three significant figures.	Answer: 5.0 ÷ 12 = 0.417 g per minute. (Marking: 2 marks for the working, 2 marks for the value with unit.)
6	Enzyme activity falls sharply above 60 °C. Explain this and state the optimum temperature of human enzymes.	Answer: High temperature breaks the hydrogen and ionic bonds of the protein, so the three dimensional shape changes, the active site is distorted and the substrate can no longer bind; the enzyme is denatured and this is irreversible. The optimum temperature for human enzymes is about 37 °C. (Marking: 2 marks for the shape change, 1 mark for the active site, 1 mark for the optimum temperature.)
7	Pepsin works best at pH 2.0 and trypsin at pH 8.0. Find the factor by which their hydrogen ion concentrations differ.	Answer: They differ by 6 pH units, so the hydrogen ion concentrations differ by a factor of 10⁶ = 1 000 000, with the higher concentration in the stomach where pepsin acts. (Marking: 2 marks for 6 units, 1 mark for 10⁶, 1 mark for the direction.)
8	The same reaction finishes in 2 minutes with an enzyme and needs 40 minutes without it. Find the increase in rate and state whether the enzyme changes the equilibrium position.	Answer: 40 ÷ 2 = 20 times. The enzyme lowers the activation energy and so reaches equilibrium sooner, but it does not change the equilibrium position or the final proportions of reactants. (Marking: 2 marks for the factor, 1 mark for activation energy, 1 mark for the equilibrium position.)

Part 3: Chemical digestion, bile and emulsification (18 marks)

The questions below concern the hydrolysis of the three main food molecules.

#	Question	Answer and marking notes
9	The molar mass of maltose is 342 g/mol and of glucose 180 g/mol. 342 g of maltose is completely hydrolysed. Find the amount in moles and the mass of glucose produced.	Answer: 342 ÷ 342 = 1.0 mol of maltose, which gives 2.0 mol of glucose; 2.0 × 180 = 360 g. (Marking: 2 marks for the moles, 1 mark for the ratio, 2 marks for the mass.)
10	Name the enzymes and organs involved in the chemical digestion of protein in the stomach and the small intestine, and state the main product of each stage.	Answer: Stomach: pepsin breaks protein into polypeptides. Small intestine: trypsin and peptidases break polypeptides into short peptides and amino acids. (Marking: 2 marks for the stomach stage, 2 marks for the small intestine stage.)
11	The molar mass of a triglyceride is 885 g/mol. 15 g of the triglyceride is completely hydrolysed by lipase. Find the amount in moles of fatty acids formed, to four decimal places.	Answer: 15 ÷ 885 = 0.0169 mol of triglyceride; each molecule gives 3 fatty acid molecules, so 0.0169 × 3 = 0.0508 mol. (Marking: 2 marks for the moles, 2 marks for the three to one ratio, 1 mark for the value.)
12	People with lactose intolerance become bloated after eating dairy food. Explain this using enzyme specificity.	Answer: These people lack lactase, so lactose cannot be hydrolysed into glucose and galactose. The undigested lactose reaches the large intestine and is fermented by bacteria, producing gas and causing bloating. Enzymes are specific, so no other enzyme can replace lactase. (Marking: 1 mark for the missing enzyme, 1 mark for no digestion, 1 mark for bacterial fermentation, 1 mark for specificity.)

Part 4: Absorption by villi and the area of the small intestine (16 marks)

The questions below concern intestinal villi and the rate of absorption.

#	Question	Answer and marking notes
13	Each square centimetre of the inner wall of the small intestine bears 2000 villi of surface area 0.5 mm² each, while 1 cm² of flat wall has an area of 100 mm². Find the factor by which the villi amplify the absorbing area.	Answer: 2000 × 0.5 = 1000 mm²; 1000 ÷ 100 = 10 times. (Marking: 2 marks for the total area, 2 marks for the factor.)
14	40 g of glucose is absorbed by the small intestine in 20 minutes. Find the mean rate of absorption in g per minute and in g per hour.	Answer: 40 ÷ 20 = 2.0 g per minute; 2.0 × 60 = 120 g per hour. (Marking: 2 marks for each value.)
15	Villi contain both blood capillaries and lacteals. State the absorption route of fatty acids and of glucose inside a villus.	Answer: Fatty acids and glycerol diffuse into the epithelium, are recombined into fat and enter the lacteal, travelling through the lymphatic system into the blood. Glucose and amino acids diffuse directly into the blood capillaries and are carried by the hepatic portal vein to the liver. (Marking: 2 marks for the fatty acid route, 2 marks for the glucose route.)
16	The small intestine is 6.0 m long and food moves along it at a mean speed of 1.2 m per hour. Find the mean time spent in the small intestine and explain its importance for absorption.	Answer: 6.0 ÷ 1.2 = 5.0 hours. The long stay gives the food ample contact time with the villi, so digestion and absorption can be completed. (Marking: 2 marks for the time, 2 marks for its importance.)

Part 5: Dietary energy and nutrient requirements (16 marks)

Use energy values of 17 kJ/g for carbohydrate, 17 kJ/g for protein and 37 kJ/g for fat.

#	Question	Answer and marking notes
17	A meal contains 60 g of carbohydrate, 25 g of protein and 15 g of fat. Find the total energy it provides.	Answer: 60 × 17 = 1020 kJ; 25 × 17 = 425 kJ; 15 × 37 = 555 kJ; total 1020 + 425 + 555 = 2000 kJ. (Marking: 3 marks for the three terms, 1 mark for the total.)
18	Find the percentage of the meal energy that comes from fat, to one decimal place.	Answer: 555 ÷ 2000 × 100% = 27.8%, with an exact value of 27.75%. Fat has the highest energy density at 37 kJ per gram. (Marking: 3 marks for the percentage, 1 mark for the comment.)
19	A teenager needs about 6300 kJ per day for basal metabolism. Find this meal as a percentage of that daily requirement, to one decimal place.	Answer: 2000 ÷ 6300 × 100% = 31.7%. The remaining demand is met by activity and growth. (Marking: 3 marks for the percentage, 1 mark for the comment.)
20	Protein contains 16% nitrogen on average. Find the mass of nitrogen in the 25 g of protein in this meal.	Answer: 25 × 0.16 = 4.0 g of nitrogen. Nitrogen balance can be used to judge whether protein intake is adequate. (Marking: 3 marks for the value, 1 mark for the comment.)

Part 6: Water balance and the function of the large intestine (16 marks)

The questions below concern water balance and absorption in the large intestine.

#	Question	Answer and marking notes
21	Someone takes in 2.5 L of water a day and loses 1.5 L in urine, 0.5 L in sweat, 0.3 L in faeces and 0.2 L in breath. Find the total loss and decide whether water is balanced.	Answer: 1.5 + 0.5 + 0.3 + 0.2 = 2.5 L, equal to the intake, so the difference is zero and water is balanced. (Marking: 2 marks for the total, 2 marks for the conclusion.)
22	1.5 L of water enters the large intestine each day and 1.35 L of it is absorbed. Find the percentage of water absorbed by the large intestine.	Answer: 1.35 ÷ 1.5 × 100% = 90%. (Marking: 2 marks for the working, 2 marks for the value.)
23	During diarrhoea the large intestine absorbs only 60% of the water entering it. Find the volume of water that is not absorbed and is lost.	Answer: The fraction not absorbed is 1 − 0.60 = 0.40, so 1.5 × 0.40 = 0.6 L is lost. Excessive loss causes dehydration, so water and electrolytes must be replaced. (Marking: 2 marks for the fraction, 2 marks for the volume.)
24	Dietary fibre cannot be hydrolysed by human enzymes. State two benefits of dietary fibre in the large intestine.	Answer: First, it adds bulk to the faeces and stimulates peristalsis, shortening the time that residues remain and reducing constipation. Second, it is fermented by gut bacteria and helps maintain a healthy gut flora. (Marking: 2 marks for each benefit.)

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