S6 Biology Practice Paper: Genetics, Gene Expression, Photosynthesis And Respiration, Ecosystems, Immunity And Evolution

S6 Biology Practice Paper: Genetics, Gene Expression, Photosynthesis and Respiration, Ecosystems, Immunity and Evolution

2026-11-12

S6 Biology

S6 biology practice paper with worked answers on dihybrid crosses, gene expression, photosynthesis, respiration, energy flow, immunity and Hardy-Weinberg equilibrium.

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Paper details　S6 biology practice paper: dihybrid cross and test cross counts, gene expression and frameshift mutation, photosynthesis and ATP yield, energy pyramid and mark-recapture, immunity and Hardy-Weinberg.

Part 1: Dihybrid cross and test cross (20 marks)

In peas, seed shape is controlled by one pair of alleles (R round dominant, r wrinkled recessive) and cotyledon colour by another pair (Y yellow dominant, y green recessive); the two pairs assort independently. Two plants of genotype RrYy are crossed and 3,200 seeds are collected. The same RrYy plant is also crossed with a plant of genotype rryy in a test cross, giving 800 seeds.

#	Question	Answer and marking notes
1	State the types of gametes produced by an RrYy plant and their ratio.	Answer: Independent assortment during meiosis gives four gamete types, RY, Ry, rY and ry, in the ratio 1 : 1 : 1 : 1, that is 25% each. Marking: four gamete types (3 marks), correct ratio (2 marks).
2	Work out the number of each of the four F2 phenotypes (round yellow, round green, wrinkled yellow, wrinkled green).	Answer: the F2 phenotypic ratio is 9 : 3 : 3 : 1, so round yellow = 3,200 x 9 / 16 = 1800, round green = 600, wrinkled yellow = 600, wrinkled green = 200 seeds; total 3200.
3	How many seeds of each phenotype are expected in the test cross progeny?	Answer: RrYy x rryy gives a 1 : 1 : 1 : 1 ratio, so each class = 800 / 4 = 200 seeds. A test cross reveals the gamete ratio of the dominant parent and is a direct test of independent assortment.
4	Among the round yellow F2 seeds, about how many are of genotype RRYY?	Answer: the round yellow class contains RRYY : RRYy : RrYY : RrYy in the ratio 1 : 2 : 2 : 4, nine parts in all, so RRYY = 1800 / 9 = 200 seeds. Marking: 1/9 proportion (3 marks), correct division (2 marks).

Part 2: Gene expression and mutation (18 marks)

A structural gene is 900 base pairs long, of which 300 base pairs are introns and the rest exons. After transcription the mRNA is spliced to remove intron sequences before translation. In another sample one base pair is inserted into the same gene and the protein loses its function.

#	Question	Answer and marking notes
5	Calculate the number of nucleotides in the mature mRNA, the number of codons and the number of amino acids.	Answer: exons = 900 – 300 = 600 base pairs, so the mRNA has 600 nucleotides; codons = 600 / 3 = 200; excluding one stop codon the polypeptide has 200 – 1 = 199 amino acids.
6	Estimate the molecular mass of the protein if the mean amino acid mass is 110.	Answer: mass = 199 x 110 = 21890. Marking: uses the amino acid count (3 marks), correct value and unit (2 marks). The true mass is slightly lower because water is lost in peptide bond formation.
7	If the change were a single base substitution, the protein would not necessarily lose its function. Explain why.	Answer: the genetic code is degenerate, so several codons may specify the same amino acid; a change at the third base is often silent. Even when an amino acid is replaced (missense mutation), function may be unaffected if the site is not part of the active site. Marking: degeneracy (2 marks), effect of silent or missense change (2 marks).
8	Why is the insertion of one base usually more damaging than a substitution?	Answer: an insertion shifts the reading frame from that point on, so every later codon is regrouped and the amino acid sequence changes wholesale; a premature stop codon usually follows, giving a truncated, non-functional protein. Marking: reading frame shifted (2 marks), consequence explained (2 marks).

Part 3: Photosynthesis and cellular respiration (16 marks)

A plant in bright light synthesises 5.4 g of glucose (molar mass 180). In a separate experiment 0.05 mol of glucose is respired aerobically, yielding 32 mol ATP per mol glucose; hydrolysis of one mole of ATP releases about 30.5 kJ.

#	Question	Answer and marking notes
9	Write the overall equation of photosynthesis and state which substance is oxidised and which is reduced.	Answer: 6CO2 + 6H2O -> C6H12O6 + 6O2, driven by light energy absorbed by chlorophyll. Water is oxidised to oxygen (loses electrons) and carbon dioxide is reduced to glucose (gains electrons). Marking: equation (2 marks), oxidation and reduction (1 mark each).
10	Calculate the mass of carbon dioxide used and the mass of oxygen released.	Answer: moles of glucose = 5.4 / 180 = 0.03 mol, so 0.18 mol of carbon dioxide and oxygen are involved. CO2 mass = 0.18 x 44 = 7.92 g; O2 mass = 0.18 x 32 = 5.76 g.
11	Calculate the moles of ATP produced and the energy made available when 0.05 mol of glucose is respired aerobically.	Answer: ATP = 0.05 x 32 = 1.6 mol; energy = 1.6 x 30.5 = 48.8 kJ. The remaining energy is lost as heat.
12	If the same 0.05 mol of glucose underwent lactate fermentation, how much ATP would be produced? Compare the efficiency of the two routes.	Answer: lactate fermentation yields only 2 mol ATP per mol glucose, so ATP = 0.05 x 2 = 0.1 mol, just 1/16 of the aerobic yield. Aerobic respiration oxidises glucose completely to carbon dioxide and water and extracts far more energy, whereas fermentation leaves most of the energy in the lactate molecule. Marking: value (2 marks), comparison with reason (2 marks).

Part 4: Energy flow and population estimation (16 marks)

In a grassland, producers fix 250,000 kJ per square metre per year and about 10% of the energy passes to the next trophic level. A separate study estimates a snail population by mark-recapture: 120 snails are marked and released, then 150 are caught of which 18 are marked.

#	Question	Answer and marking notes
13	Calculate the energy available to the primary, secondary and tertiary consumers.	Answer: primary consumers = 250,000 x 10% = 25000 kJ; secondary = 2500 kJ; tertiary = 250 kJ. About nine tenths of the energy is lost at each transfer.
14	Explain why energy decreases along a food chain and why chains rarely have more than four or five levels.	Answer: most energy is lost as heat in respiration, some is not eaten and some passes out undigested in faeces. As only about one tenth survives each transfer, by the fourth or fifth level too little remains to support another viable population. Marking: routes of loss (3 marks), inference for chain length (1 mark).
15	Use the mark-recapture formula to estimate the size of the snail population.	Answer: N = (M x n) / m = (120 x 150) / 18 = 1000 snails. The formula assumes a closed population, no tag loss, no effect of tagging on survival and thorough mixing.
16	If some marks were lost before the second sample, would the estimate be too high or too low? Explain.	Answer: the estimate would be too high. Tag loss reduces m, and since N is inversely proportional to m, a smaller denominator gives a larger N. Likewise, if tags make snails more likely to be eaten, m falls and the estimate is again inflated. Marking: correct direction (2 marks), explanation via the formula (2 marks).

Part 5: Immunity and vaccination (15 marks)

After vaccination the concentration of a specific antibody in a person rises from 2.0 mg per litre to 64.0 mg per litre. Antibodies are secreted by plasma cells derived from B cells that divide after antigen stimulation. Six months later the same antigen is encountered again and the antibody concentration rises rapidly.

#	Question	Answer and marking notes
17	Describe the basic structure of an antibody.	Answer: an antibody is a protein (immunoglobulin) built from four polypeptide chains, two heavy and two light, arranged in a Y shape. Each chain has a constant and a variable region; the variable region differs between antibodies and forms the antigen-binding site complementary to the antigen. Marking: four-chain structure (2 marks), variable region and specificity (2 marks).
18	The antibody concentration rises 32-fold. If each cell secretes the same amount, what is the minimum number of divisions the B cells must undergo?	Answer: fold change = 64.0 / 2.0 = 32.0; divisions = log2 32.0 = 5.0, that is from 1 cell to 32 cells. Marking: fold change (2 marks), using powers of two for the division count (2 marks).
19	Explain how a vaccine produces immunological memory.	Answer: a vaccine contains weakened, inactivated or fragmentary antigen which is processed by macrophages and presented to helper T cells, which then activate specific B cells. These divide into antibody-secreting plasma cells and long-lived memory B cells; the memory cells persist in lymphoid tissue and respond rapidly on later exposure, giving a stronger secondary response.
20	Why do some vaccines require booster doses?	Answer: antibody levels from the primary response fall with time and the number of memory B cells may decline. A booster supplies the antigen again, so memory cells proliferate, antibody levels return to a protective titre and memory is prolonged. Marking: antibody decay (2 marks), effect of restimulation (1 mark).

Part 6: Evolution and allele frequencies (15 marks)

In a population of 5,000 people, 450 show a recessive trait (genotype aa) and the rest show the dominant phenotype. Assume the population is in Hardy-Weinberg equilibrium, with no mutation, migration or selection.

#	Question	Answer and marking notes
21	Calculate the frequencies of the recessive allele a and the dominant allele A.	Answer: q squared = 450 / 5,000 = 0.09, so q = 0.3 and p = 1 – q = 0.7. Marking: q squared (2 marks), square root and p plus q equals one (2 marks).
22	Estimate the number of heterozygotes (Aa) in the population.	Answer: heterozygote frequency = 2pq = 2 x 0.7 x 0.3 = 0.42; number = 5,000 x 0.42 = 2100 people. Most recessive alleles are therefore hidden in heterozygotes and are not easily removed by selection.
23	If the dominant phenotype confers a survival advantage in this environment, how will the allele frequencies change?	Answer: aa individuals survive and reproduce less, so allele a is passed on less often; q falls and p rises from generation to generation, and the proportion showing the dominant phenotype increases. Because a survives in heterozygotes, selection cannot act on it directly, so a declines slowly rather than disappearing at once.
24	Using antibiotic resistance in bacteria as an example, describe the process of natural selection.	Answer: a bacterial population already contains a few individuals carrying resistance genes, from mutation or gene transfer. Antibiotic treatment kills the sensitive cells while resistant ones survive and reproduce, passing the gene to their descendants; the antibiotic is the selection pressure and the frequency of resistance rises rapidly. Marking: variation pre-exists (1 mark), selection and reproduction (1 mark), change in frequency (1 mark).

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