S6 Biology Practice Paper: Genetics, DNA And Gene Expression (12 Questions, 100 Marks, 75 Minutes)

S6 Biology Practice Paper: Genetics, DNA and Gene Expression (12 Questions, 100 Marks, 75 Minutes)

2026-11-09

S6 Biology: Genetics, DNA and Gene Expression

S6 Biology: DNA structure, replication, transcription and translation, Mendelian crosses, linkage, sex-linked inheritance and genetic engineering. 12 questions with answers.

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Instructions: This paper has three sections, A, B and C. Answer all questions. Total marks: 100. Time allowed: 75 minutes.

Show the formula used, the values substituted and the correct units. Give numerical answers to three significant figures.

In explanations, name the principle used, for example limiting factors, cohesion and adhesion, or source and sink.

Spend about 20 minutes on Section A, 30 minutes on Section B and the last 25 minutes on Section C.

Section A: DNA Structure, Replication and Gene Expression (Questions 1-4)

Four questions, 28 marks in total. Suggested time: 20 minutes.

#	Question	Answer and marking notes
1	Describe three structural features of the DNA molecule and explain how complementary base pairing ensures accuracy during replication. (6 marks)	Two antiparallel polynucleotide strands, with the sugar-phosphate backbones on the outside and the bases on the inside: 2 marks. Bases pair by hydrogen bonds, adenine with thymine by two bonds and guanine with cytosine by three, and purine-pyrimidine pairing keeps the helix width constant: 2 marks. During replication each parent strand acts as a template so only complementary bases can pair, giving daughter strands identical in sequence to the parent strand, and DNA polymerase proofreads mismatches: 2 marks.
2	Describe the experiment of Meselson and Stahl and explain how its results support semi-conservative replication. (6 marks)	E. coli was first grown in a medium containing the heavy isotope 15N so that its DNA was dense: 1 mark. The bacteria were transferred to 14N medium and DNA was sampled and centrifuged in a density gradient; the first generation gave a single band of intermediate density: 2 marks. The second generation gave an intermediate band and a light band in an approximately 1:1 ratio: 2 marks. Each daughter DNA therefore has one parental strand and one new strand, which is semi-conservative replication and rules out conservative and dispersive models: 1 mark.
3	Describe how a eukaryotic cell produces a polypeptide from a gene, and state two post-transcriptional modifications. (8 marks)	RNA polymerase in the nucleus uses the template strand to synthesise pre-mRNA by complementary base pairing: 2 marks. Post-transcriptional modification: introns are removed and exons joined, a cap is added at the 5 end, and a poly-A tail is added at the 3 end; any two for 2 marks. Mature mRNA leaves through a nuclear pore to the cytoplasm: 1 mark. On the ribosome, tRNA anticodons pair with codons starting at the start codon AUG, peptide bonds form between amino acids, and the polypeptide is released at a stop codon: 3 marks.
4	The coding region of an mRNA has 903 nucleotides, including one stop codon, and no poly-A tail. Calculate the maximum number of amino acids it can encode, and explain the effect of inserting one base after the fourth nucleotide. (8 marks)	903 divided by 3 gives 301 codons: 2 marks. Subtracting one stop codon, at most 300 amino acids can be encoded: 2 marks. Inserting one base shifts the reading frame from that point onwards: 2 marks. All later codons are regrouped, the amino acid sequence changes greatly, and a stop codon usually appears early, giving a shortened non-functional polypeptide: 2 marks.

Section B: Mendelian Genetics, Genetic Diagrams and Sex Linkage (Questions 5-8)

Four questions, 32 marks in total. Suggested time: 30 minutes.

#	Question	Answer and marking notes
5	In guinea pigs the allele B for black coat is dominant to b for white coat. Two black guinea pigs were crossed and produced 32 offspring: 24 black and 8 white. Deduce the genotypes of the parents and use a Punnett square to explain the ratio. (8 marks)	White offspring are bb, so one b allele came from each parent and both parents carry b: 1 mark. Both parents are black, so both are Bb: 2 marks. Punnett square gives one quarter BB, one half Bb and one quarter bb: 3 marks. BB and Bb are black, together three quarters, that is 32 times three quarters equals 24; bb is one quarter, that is 8, matching the observed numbers: 2 marks.
6	In peas, the allele R for round seeds is dominant to r for wrinkled seeds, and Y for yellow cotyledons is dominant to y for green. A test cross of RrYy with rryy produced 500 offspring: 210 round yellow, 40 round green, 35 wrinkled yellow and 215 wrinkled green. Calculate the recombination frequency and explain the result. (8 marks)	If the two genes assort independently, a test cross should give a 1:1:1:1 ratio, about 125 of each: 1 mark. The observed offspring are mostly parental types (round yellow and wrinkled green): 2 marks. Recombinants are 40 round green plus 35 wrinkled yellow, which is 75: 1 mark. Recombination frequency is 75 divided by 500 times 100%, which is 15%: 2 marks. A frequency below 50% shows the two genes are linked on the same chromosome and that crossing over occurred during meiosis: 2 marks.
7	A couple with normal colour vision has a son who is red-green colour blind. Explain this using X-linked recessive inheritance, and calculate the probability that each of their daughters is a carrier. (8 marks)	The colour blindness allele is recessive and carried on the X chromosome, and a male has only one X chromosome, so one recessive allele is enough to cause the condition: 1 mark. The son receives a Y chromosome from his father and an X chromosome from his mother: 1 mark. The allele must therefore come from the mother, so she is a carrier: 2 marks. The father has a normal X and a Y, so he cannot pass the allele to a daughter: 2 marks. A daughter receives the normal X from her father and either the normal X or the allele from her mother with equal chance, so the probability of being a carrier is one half, that is 50%: 2 marks.
8	The human ABO blood group is controlled by three alleles I^A, I^B and i. A father is blood group AB and the mother is group O. List the possible blood groups of the children and their probabilities, and explain the expression of group AB. (8 marks)	The AB father is I^A I^B and the O mother is ii: 2 marks. A child receives I^A or I^B from the father and must receive i from the mother: 2 marks. Half the children are therefore I^A i, group A, and half are I^B i, group B: 2 marks. Groups AB and O are impossible because the mother has neither I^A nor I^B and the father has no i: 1 mark. I^A and I^B are codominant, so both antigens are produced and the red cells carry A and B antigens together: 1 mark.

Section C: Mutation, Genetic Engineering, Data Analysis and Counselling (Questions 9-12)

Four questions, 40 marks in total. Suggested time: 25 minutes.

#	Question	Answer and marking notes
9	Compare silent, missense and nonsense mutations, and use sickle-cell anaemia to explain the consequence of a missense mutation. (10 marks)	A gene mutation is a permanent change in the base sequence of DNA, by substitution, insertion or deletion: 2 marks. Silent mutation: the base changes but, because the genetic code is degenerate, the amino acid is unchanged: 1 mark. Missense mutation: one codon changes so one amino acid is replaced by another: 1 mark. Nonsense mutation: a codon becomes a stop codon and the polypeptide ends early: 1 mark. In sickle-cell anaemia the sixth codon of the haemoglobin gene changes from GAG to GUG, so glutamic acid is replaced by valine: 2 marks. The haemoglobin polymerises at low oxygen tension and the red cells become sickle shaped: 2 marks. Sickle cells haemolyse easily and block capillaries, causing anaemia and pain, but give some resistance to malaria: 1 mark.
10	Describe the main steps of using genetic engineering to make Escherichia coli produce human insulin, and give one advantage and one concern. (10 marks)	A restriction enzyme cuts out the human insulin gene: 2 marks. The same enzyme cuts the plasmid vector to give complementary sticky ends: 2 marks. DNA ligase joins the gene into the plasmid to form a recombinant plasmid: 2 marks. The plasmid is introduced into E. coli and transformed colonies are selected using markers such as antibiotic resistance: 2 marks. The bacteria are cultured on a large scale and the insulin is extracted and purified: 1 mark. Advantage: large-scale production of pure insulin without dependence on donated blood; concern: antibiotic resistance genes may spread to bacteria in the environment: 1 mark.
11	DNA short tandem repeats of four people were analysed by gel electrophoresis. The band positions in bp are: alleged father A 300, 500; alleged father B 250, 300, 500; child 250, 500; mother 300, 700. State which alleged father cannot be excluded and explain why. (10 marks)	Every band of the child must come from either the father or the mother: 2 marks. The child’s 250 bp band is absent from the mother, who has 300 bp and 700 bp, so it must come from the biological father: 2 marks. Alleged father A has no 250 bp band and is excluded: 2 marks. Alleged father B has the 250 bp band and his 300 bp and 500 bp bands are consistent with the child’s 500 bp band: 2 marks. Conclusion: alleged father B cannot be excluded, but matching short tandem repeats only indicate compatibility and more loci are needed before a conclusion: 2 marks.
12	A man’s younger brother has cystic fibrosis, an autosomal recessive condition. The man is healthy and his wife has no family history. Assuming a carrier frequency of one in twenty-five in the population, calculate the probability that their child is affected and state two points for genetic counselling. (10 marks)	The affected brother is homozygous recessive, so both parents must each carry one recessive allele: 2 marks. The man is healthy, so he is either homozygous dominant or heterozygous, and among healthy children two thirds are heterozygous: 2 marks. The probability that his wife is a carrier is one in twenty-five: 1 mark. An affected child requires both parents to be carriers and to pass the recessive allele, so the probability is two thirds times one twenty-fifth times one quarter, which is one in one hundred and fifty, about 0.67%: 3 marks. Counselling points: explain that the figure is a probability, not a certainty, and offer carrier testing with support and follow-up: 2 marks.

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