S5 Biology Practice Paper (Set 2): Cell Division And Genetics, Enzymes And Metabolic Rate, Water Transport In Plants, Body Defence, Energy Flow And Population Estimation, Genetic Ratios And Data Handling

S5 Biology Practice Paper (Set 2): Cell Division and Genetics, Enzymes and Metabolic Rate, Water Transport in Plants, Body Defence, Energy Flow and Population Estimation, Genetic Ratios and Data Handling

2026-11-15

Senior Secondary Biology

S5 Biology practice paper, Set 2: 24 parts, 100 marks on cell division, enzymes, plant transport, defence, ecology and genetics, with full answers.

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Paper details　S5 Biology practice paper with answers: cell division, enzymes, plant transport, defence, ecology, genetics.

Cell Division and Genetics

A human body cell has a chromosome number of 2n = 46.

#	Question	Answer and marking notes
1	What is the chromosome number of each daughter cell after a human body cell completes mitosis?	Still 46. DNA is replicated before division and one complete set goes to each daughter cell, so the number equals that of the parent cell.
2	How many chromosomes are in a human gamete (sperm or egg), and why does this matter at fertilisation?	A gamete has 23 chromosomes; after fertilisation the number is restored to 46, keeping the chromosome number constant in every generation.
3	How many gametes arise from one primary reproductive cell by meiosis, and at which stage is the chromosome number halved?	It gives 4 gametes; the number is halved in the first meiotic division, when homologous chromosomes separate and the count falls from 46 to 23.
4	What is the main difference between mitosis and meiosis in the genetic make-up of the products?	Mitosis gives products genetically identical to the parent cell; in meiosis, independent assortment and crossing over make the four gametes genetically different from one another.

Enzymes and Metabolic Rate

The table shows the activity of an enzyme at different temperatures (units of product per minute): 10 C: 12, 20 C: 28, 30 C: 45, 40 C: 60, 50 C: 60, 60 C: 38, 70 C: 8.

#	Question	Answer and marking notes
5	From the data, state the optimum temperature range of the enzyme.	Between 40 C and 50 C, where the activity is highest at 60 units per minute.
6	Calculate the activity at 30 C as a percentage of the maximum activity.	45 / 60 x 100 = 75.0 per cent.
7	When the temperature rises from 40 C to 70 C, by what percentage of the maximum does the activity fall?	Activity falls from 60 to 8, a drop of 52 units, i.e. 52 / 60 x 100 = 86.7 per cent.
8	Why does the activity fail to recover after exposure to 70 C?	The enzyme is a protein; high temperature destroys the shape of its active site (denaturation) so the substrate can no longer bind, and denaturation is irreversible.

Water Transport in Plants

A shoot with 3 leaves is placed in a potometer; it absorbs 24.0 cm3 of water in 30 minutes.

#	Question	Answer and marking notes
9	Calculate the mean rate of water uptake of the shoot in cm3 per minute.	24.0 / 30 = 0.8 cm3 per minute.
10	Convert this rate to water uptake per hour in cm3 per hour.	0.8 x 60 = 48.0 cm3 per hour.
11	Assuming each leaf transpires at the same rate, how much water would a shoot with 5 leaves absorb per hour?	48.0 / 3 x 5 = 80.0 cm3 per hour, i.e. the 5-leaf shoot accounts for about 60.0 per cent of the uptake, in line with its share of leaves.
12	Explain how the number of leaves affects the rate of water uptake.	More leaves mean more stomata, so more water is lost by transpiration; the stronger tension (negative pressure) in the xylem columns pulls water up from the roots faster.

Body Defence

Two weeks after vaccination, a student shows a marked rise in blood antibody level and later shows no illness after contact with the same pathogen.

#	Question	Answer and marking notes
13	Which type of immunity does vaccination produce? Give one feature of it.	Active immunity (artificially acquired active immunity); the body makes its own antibodies and memory cells, so the protection lasts longer.
14	Which type of white blood cell produces antibodies?	Lymphocytes (B cells, which differentiate into plasma cells that secrete antibodies).
15	Why does a second dose of the same vaccine raise the antibody level faster and higher?	The first dose produced memory cells; on re-exposure these divide quickly into plasma cells, the secondary response, giving a faster and larger antibody rise.
16	Can a vaccine treat an infection that has already developed? Explain.	No. It takes days to weeks to build the immune response, far too slow for immediate treatment; a bacterial infection needs antibiotics and a viral one needs antiviral drugs and supportive care.

Energy Flow and Population Estimation

Snails are estimated by mark-recapture: 40 are caught, marked and released; a second catch of 50 contains 8 marked individuals. Another survey shows producers fix 50000 kJ of energy.

#	Question	Answer and marking notes
17	Estimate the population size by mark-recapture, giving the formula and the calculation.	N = (marked in first catch x size of second catch) / marked in second catch = (40 x 50) / 8 = 250 snails.
18	State one assumption of this estimation method.	For example: marking does not affect survival or the chance of being caught; the population is closed (no immigration or emigration); marks do not rub off.
19	If some marks rub off before the second catch, is the estimate too high or too low? Explain.	The estimate becomes too high: fewer marked individuals are recorded, the denominator shrinks and the calculated population is inflated.
20	Using the ten per cent rule, how much energy is available to primary and to secondary consumers?	Primary consumers: 50000 / 10 = 5000 kJ; secondary consumers: 5000 / 10 = 500 kJ, with the remaining 45000 kJ lost mainly as heat and in excretory products.

Genetic Ratios and Data Handling

In peas the tall allele T is dominant to the dwarf allele t. Two tall plants (Tt x Tt) are crossed and 400 seeds are planted.

#	Question	Answer and marking notes
21	Write the genotypes of the two parents and the gametes each produces.	Both parents are Tt (heterozygous); each produces T and t gametes in a 1 : 1 ratio.
22	Use a genetic diagram to derive the genotype and phenotype ratios of the offspring.	Genotypes TT : Tt : tt = 1 : 2 : 1; phenotypes tall : dwarf = 3 : 1.
23	Among 400 offspring, how many are expected to be tall and how many dwarf?	Tall: 400 x 3 / 4 = 300 plants; dwarf: 400 x 1 / 4 = 100 plants.
24	If the actual result is 290 tall and 110 dwarf, how do you explain the difference from the expectation?	A 3 : 1 ratio is an expected probability, not an exact count; with a finite sample random deviation always occurs, and 290 : 110 is close enough to 3 : 1 to be consistent.

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