S6 Biology Paper: Mendelian Genetics, Crosses And Inheritance Probability (100 Marks)

S6 Biology Paper: Mendelian Genetics, Crosses and Inheritance Probability (100 marks)

2026-11-11

S6 Biology – genetics and probability

S6 Biology paper: six questions on monohybrid and dihybrid crosses, test cross, ABO codominance, sex-linked inheritance and allele frequencies, with answers and marking notes.

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Paper details　S6 Biology practice paper with six questions on Mendelian monohybrid and dihybrid crosses, F2 ratios, test cross and parental genotype, ABO codominance, sex-linked inheritance and Hardy-Weinberg allele frequencies. Full answers with marking notes.

Section A: Monohybrid Cross and F2 Ratio (16 marks)

In peas the tall allele T is completely dominant over the dwarf allele t. A pure-breeding tall plant (TT) is crossed with a pure-breeding dwarf plant (tt), and the F1 plants are then self-crossed to give the F2 generation.

#	Question	Answer and marking notes
1	State the genotype and phenotype of the F1 plants and explain why.	Answer: every F1 plant is Tt and all are tall, because each parent contributes one allele and T is completely dominant, so Tt shows the tall phenotype. (Marking: 2 marks for the genotype, 1 mark for the phenotype, 1 mark for the dominance explanation.)
2	State the genotypic and phenotypic ratios of the F2 generation.	Answer: the F2 genotypic ratio is 1 TT to 2 Tt to 1 tt, that is 25 per cent, 50 per cent and 25 per cent, and the phenotypic ratio is 3 tall to 1 dwarf. (Marking: 2 marks for the genotypic ratio, 2 marks for the phenotypic ratio.)
3	If there are 800 F2 plants, how many dwarf plants are expected?	Answer: dwarf plants make up one quarter of the total, so 800 times one quarter gives 200 plants. (Marking: 2 marks for the ratio, 2 marks for the calculation and answer.)
4	Among the tall F2 plants, what percentage are heterozygotes?	Answer: there are 600 tall plants, of which 400 are heterozygotes, so the percentage is 400 divided by 600, that is 66.7 per cent. (Marking: 1 mark for the number of tall plants, 1 mark for the heterozygotes, 2 marks for the percentage.)

Section B: Test Cross and Deducing Parental Genotype (16 marks)

A tall pea plant is test-crossed with a dwarf plant (tt). The offspring total 230 plants, of which 118 are tall and 112 are dwarf.

#	Question	Answer and marking notes
5	Calculate the ratio of tall to dwarf offspring, to two decimal places.	Answer: 118 divided by 112 is 1.05, so the ratio is about 1.05 to 1, close to 1 to 1. (Marking: 2 marks for the calculation, 2 marks for stating that it is close to 1 to 1.)
6	Deduce the genotype of the tall parent and explain your reasoning.	Answer: the genotype is Tt, because dwarf offspring appear in the test cross, so each offspring must receive a t from this parent; the parent is tall, so its other allele must be T. (Marking: 2 marks for the genotype, 2 marks for the reasoning from the dwarf offspring.)
7	If the parent had been TT, what phenotypic ratio would be expected?	Answer: all offspring would be Tt and therefore tall, so there would be no dwarf plants at all. (Marking: 3 marks for all tall, 1 mark for the explanation.)
8	Why do the actual numbers not match the expected numbers exactly?	Answer: fertilisation is random, so with a limited sample the observed ratio departs from the expected value by chance, and larger samples give ratios closer to the theoretical value. (Marking: 2 marks for random fertilisation, 2 marks for sample size.)

Section C: Dihybrid Cross and Independent Assortment (16 marks)

In peas the round allele R is dominant over wrinkled r, and the yellow allele Y is dominant over green y, with the two gene pairs on different chromosomes. Two plants both of genotype RrYy are crossed, giving 1600 seeds.

#	Question	Answer and marking notes
9	State the phenotypic ratio of the offspring.	Answer: 9 round yellow to 3 round green to 3 wrinkled yellow to 1 wrinkled green. (Marking: 2 marks for all four classes, 2 marks for the correct numbers.)
10	How many wrinkled yellow seeds are expected?	Answer: wrinkled yellow seeds make up three sixteenths, so 1600 times three sixteenths gives 300 seeds. (Marking: 2 marks for the ratio, 2 marks for the calculation and answer.)
11	How many round green seeds are expected?	Answer: round green seeds also make up three sixteenths, so the expected number is 300. (Marking: 2 marks for the ratio, 2 marks for the calculation and answer.)
12	How many seeds of genotype RRYY are expected?	Answer: RRYY makes up one sixteenth, so 1600 times one sixteenth gives 100 seeds. (Marking: 2 marks for deriving one sixteenth, 2 marks for the calculation and answer.)

Section D: ABO Blood Groups and Codominance (18 marks)

A father has blood group A with genotype I^A i, and a mother has blood group B with genotype I^B i. Both I^A and I^B are completely dominant over i, while I^A and I^B are codominant with each other.

#	Question	Answer and marking notes
13	List the possible blood groups of the children and their ratio.	Answer: the four genotypes I^A I^B, I^A i, I^B i and ii each occur with probability one quarter, giving AB, A, B and O in a 1 to 1 to 1 to 1 ratio, that is 25 per cent each. (Marking: 3 marks for the four genotypes, 2 marks for matching blood groups, 1 mark for the ratio.)
14	If the couple has four children, what is the probability that all four are group O? Give the answer as a percentage to two decimal places.	Answer: the probability for one child is one quarter, so for four children it is one quarter raised to the fourth power, that is one over 256, about 0.39 per cent. (Marking: 1 mark for the single probability, 2 marks for the fourth power, 1 mark for the decimal and percentage.)
15	If the group O child later has children with a group AB partner, what blood groups are expected, and why can no group O child appear?	Answer: group O is ii and group AB is I^A I^B, so the children can only be I^A i (group A) or I^B i (group B) in a 1 to 1 ratio; the AB partner has no i allele to pass on, so ii cannot occur and no child is group O. (Marking: 2 marks for the children and ratio, 2 marks for the missing i allele explanation.)
16	Using this example, explain the difference between codominance and complete dominance.	Answer: in codominance both I^A and I^B are fully expressed when present together, giving group AB; in complete dominance the dominant allele alone determines the phenotype and i is masked, so I^A i is group A and I^B i is group B. (Marking: 2 marks for codominance, 2 marks for complete dominance.)

Section E: Sex-linked Inheritance and Pedigree Reasoning (16 marks)

Red-green colour blindness is caused by a recessive allele b on the X chromosome, with the normal allele B dominant. A father with normal colour vision (X^B Y) marries a mother with normal colour vision who is a carrier (X^B X^b).

#	Question	Answer and marking notes
17	List the four possible genotypes of the children and their ratio.	Answer: X^B X^B, X^B X^b, X^B Y and X^b Y, each with probability one quarter. (Marking: 3 marks for the four genotypes, 1 mark for the ratio.)
18	What proportion of the sons are colour blind?	Answer: the sons are either X^B Y or X^b Y in equal numbers, so one half of the sons are colour blind. (Marking: 2 marks for noting that sons take their X from the mother, 2 marks for the proportion.)
19	What proportion of the daughters are colour blind, and why?	Answer: none of them, because every daughter receives the normal X^B from her father, so she is either X^B X^B or X^B X^b and is unaffected, at most a carrier. (Marking: 1 mark for none, 3 marks for explaining the father passes X^B.)
20	If one of the daughters marries a man with normal colour vision, what is the probability that a son is colour blind?	Answer: the daughter is a carrier with probability one half, and if she is a carrier each son has a one half chance of being X^b Y, so the probability is one half times one half, that is one quarter, or 25 per cent. (Marking: 2 marks for the carrier probability, 2 marks for multiplying the conditional probability.)

Section F: Allele Frequencies and Carrier Estimates (18 marks)

A recessive condition affects one person in 2500 in a stable population of 50000. The trait is controlled by one pair of alleles of a single gene, and the population is in Hardy-Weinberg equilibrium.

#	Question	Answer and marking notes
21	Find the frequency q of the recessive allele.	Answer: affected people are aa, so q squared is one over 2500, that is 0.0004, and q is therefore 0.02. (Marking: 2 marks for the q squared relation, 2 marks for the square root and answer.)
22	Find the frequency p of the dominant allele.	Answer: p equals one minus q, that is 1 minus 0.02, so p is 0.98. (Marking: 2 marks for the relation, 2 marks for the answer.)
23	Find the carrier frequency and express it as a percentage to two decimal places.	Answer: carriers are Aa, so the frequency is 2 p q, that is 2 times 0.98 times 0.02, giving 0.0392, or 3.92 per cent. (Marking: 2 marks for the formula, 1 mark for substitution, 2 marks for the value and percentage.)
24	Estimate the number of carriers in the population.	Answer: 50000 times 0.0392 gives about 1960 carriers. (Marking: 2 marks for the multiplication, 2 marks for the answer, 1 mark for noting it is an estimate.)

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