S5 Biology Paper: Mendelian Genetics, Blood Groups, DNA And Genetic Engineering (100 Marks)

S5 Biology Paper: Mendelian Genetics, Blood Groups, DNA and Genetic Engineering (100 marks)

2026-11-11

S5 Biology paper: 6 questions, 100 marks, 75 minutes. Covers Mendelian crosses, ABO blood groups, DNA replication, transcription, mutation and genetic engineering.

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Paper details　

Section A: Monohybrid cross (16 marks)

Seed shape in garden peas is controlled by one pair of alleles; round R is dominant to wrinkled r. Two round-seeded plants are crossed and the offspring show both round and wrinkled seeds.

#	Question	Answer and marking notes
1	(a) If both parents are Rr, state the gametes each parent produces and their ratio, and list the possible offspring genotypes. (3 marks)	Answer: Each parent produces two kinds of gamete, R and r, in the ratio 1:1 (each 1/2). Fertilisation gives the genotypes RR, Rr and rr.
2	(b) Use a Punnett square to find the genotypic and phenotypic ratios of the offspring. (4 marks)	Answer: The 2 by 2 square gives 4 cells: 1 RR, 2 Rr and 1 rr, so the genotypic ratio is 1:2:1. Round = 1/4 + 2/4 = 3/4 and wrinkled = 1/4, so the phenotypic ratio is 3:1.
3	(c) If there are 400 offspring plants, how many round and how many wrinkled are expected? (3 marks)	Answer: Round = 400 x 3/4 = 300 plants; wrinkled = 400 x 1/4 = 100 plants.
4	(d) How can a test cross determine the genotype of a round-seeded plant? State the two possible outcomes. (6 marks)	Answer: Cross the plant with a wrinkled-seeded plant (rr). If all offspring are round, the plant is RR; if round and wrinkled appear in about a 1:1 ratio, the plant is Rr. (Marking: 2 marks for the design, 2 marks for each outcome.)

Section B: Dihybrid cross (18 marks)

In garden peas, seed shape (round R dominant to wrinkled r) and cotyledon colour (yellow Y dominant to green y) are each controlled by one pair of alleles located on different chromosomes.

#	Question	Answer and marking notes
5	(a) Which gametes does a plant of genotype RrYy produce, and in what ratio? (4 marks)	Answer: Because the two genes lie on different chromosomes and assort independently, four gametes are produced: RY, Ry, rY and ry in the ratio 1:1:1:1 (each 1/4).
6	(b) If the RrYy plant is self-crossed, state the ratio of the four phenotypes in the offspring. (5 marks)	Answer: Of the 16 combinations, round yellow is 9, round green is 3, wrinkled yellow is 3 and wrinkled green is 1, giving the ratio 9:3:3:1.
7	(c) Calculate the probability that an offspring is round and green. (4 marks)	Answer: The probability of round is 3/4 and of green is 1/4; the genes assort independently, so the probability is 3/4 x 1/4 = 3/16.
8	(d) If there are 320 offspring plants, how many round yellow and how many wrinkled green are expected? (5 marks)	Answer: Round yellow = 320 x 9/16 = 180 plants; wrinkled green = 320 x 1/16 = 20 plants.

Section C: Inheritance of ABO blood groups (16 marks)

Human ABO blood groups are controlled by three alleles, IA, IB and i. IA and IB are codominant and i is recessive. The father is group A (genotype IA i) and the mother is group B (genotype IB i).

#	Question	Answer and marking notes
9	(a) State the gametes the father and the mother can produce, with their ratios. (4 marks)	Answer: The father (IA i) produces IA and i gametes, each 1/2; the mother (IB i) produces IB and i gametes, each 1/2.
10	(b) Use a Punnett square to find the possible blood groups of the children and their ratio. (4 marks)	Answer: The four combinations each occur with probability 1/4: IA IB gives group AB, IA i gives group A, IB i gives group B and ii gives group O, so the ratio is 1:1:1:1.
11	(c) Can this couple have a group O child? Explain. (4 marks)	Answer: Yes. The i gamete of the father can combine with the i gamete of the mother to give genotype ii, which is group O, with probability 1/4.
12	(d) If the couple has 4 children, what is the probability that all four blood groups appear once each, in any order? (4 marks)	Answer: The number of orders is 4 factorial = 24, and each specified combination has probability (1/4)^4 = 1/256, so the probability is 24/256 = 3/32, about 9.4 per cent.

Section D: DNA structure and replication (18 marks)

A double-stranded DNA molecule contains 1000 base pairs, and adenine (A) makes up 30 per cent of all the bases.

#	Question	Answer and marking notes
13	(a) Find the percentage of thymine (T), guanine (G) and cytosine (C) in all the bases. (5 marks)	Answer: A pairs with T, so T = A = 30 per cent. G and C together make up 100 − 30 − 30 = 40 per cent, and G = C, so G = C = 20 per cent.
14	(b) Find the total number of hydrogen bonds in the molecule. (5 marks)	Answer: The molecule has 1000 x 2 = 2000 bases. A = 2000 x 30 per cent = 600 bases, giving 600 A-T pairs with 2 hydrogen bonds each, which is 1200. G = 2000 x 20 per cent = 400 bases, giving 400 G-C pairs with 3 each, which is 1200. The total is 2400 hydrogen bonds.
15	(c) Describe the process of semi-conservative replication of DNA. (4 marks)	Answer: The double helix unwinds and the two old strands separate, each acting as a template. New strands are built by complementary base pairing (A with T, G with C). Each daughter molecule keeps one old strand and one new strand, hence semi-conservative replication.
16	(d) If one DNA molecule replicates three times, how many molecules are produced, and what fraction of them contain an original strand? (4 marks)	Answer: The number of molecules is 2 to the power 3 = 8. Two molecules contain an original strand, one for each old strand, so the fraction is 2/8 = 1/4.

Section E: Transcription and translation (16 marks)

The mRNA transcribed from a gene contains 303 nucleotides. Its last codon is a stop codon and every other codon codes for one amino acid.

#	Question	Answer and marking notes
17	(a) How many codons does the mRNA contain? (4 marks)	Answer: Every three nucleotides form one codon, so 303 / 3 = 101 codons.
18	(b) How many amino acids are joined, and how many peptide bonds are formed? (4 marks)	Answer: The stop codon codes for no amino acid, so the number of amino acids is 101 – 1 = 100. A chain of 100 amino acids has 100 – 1 = 99 peptide bonds.
19	(c) What is the minimum number of tRNA molecules needed for translation? Explain. (4 marks)	Answer: Each amino acid is carried to the ribosome by one tRNA, so at least 100 tRNA molecules are used. The stop codon is not matched by any tRNA.
20	(d) State where transcription and translation take place in a eukaryotic cell. (4 marks)	Answer: Transcription occurs in the nucleus, where mRNA is made from the DNA template strand. The mRNA passes through a nuclear pore into the cytoplasm, where translation occurs on ribosomes.

Section F: Gene mutation and genetic engineering (16 marks)

Gene mutation can alter the structure and function of a protein, while genetic engineering can insert a foreign gene into another organism, for example using bacteria to make human insulin.

#	Question	Answer and marking notes
21	(a) Distinguish between a point mutation (base substitution) and a frameshift mutation. (4 marks)	Answer: A substitution involves one base pair and usually changes only one codon, leaving the rest unchanged. An insertion or deletion of one base shifts the reading frame from the mutation point onwards, so all later codons are changed, which is usually more damaging.
22	(b) A coding region has 300 bases, that is 100 codons. If one base is deleted just before codon 20, how many codons are affected? (4 marks)	Answer: The reading frame shifts from codon 20 onwards, so the number of affected codons is 100 – 19 = 81.
23	(c) Outline the three main steps in making human insulin by genetic engineering. (4 marks)	Answer: First, cut out the human insulin gene with a restriction enzyme. Second, join it to a plasmid vector with DNA ligase to form recombinant DNA. Third, introduce the recombinant DNA into bacteria, select the transformed cells, culture them on a large scale and extract and purify the insulin. (1 mark per step, 1 mark for a correct sequence.)
24	(d) Discuss two benefits and one risk of genetically modified food. (4 marks)	Answer: Benefits (1 mark each, any two): higher yields to meet food demand; built-in pest and disease resistance that reduces pesticide use; improved nutritional content or longer shelf life. Risk (2 marks): the inserted gene may spread to wild relatives through pollen and disturb the ecosystem, or the food may cause allergic reactions in some people, or pests may develop resistance.

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