S6 Biology Paper: Mendelian Genetics, DNA Structure And Replication, Transcription And Translation, Mutation And Genetic Engineering

S6 Biology Paper: Mendelian Genetics, DNA Structure and Replication, Transcription and Translation, Mutation and Genetic Engineering

2026-11-13

S6 Biology – genetics and genetic engineering

S6 Biology paper: 6 questions, 24 parts, 100 marks, 60 minutes on genetics, DNA replication, transcription, translation, mutation and genetic engineering, with answers.

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Paper details　S6 Biology practice paper with 6 questions and 24 parts on Mendelian genetics, DNA replication, transcription, translation, mutation and genetic engineering, with marked answers.

Section A: Mendelian Genetics and Probability (16 marks)

Answer the following genetics questions and show all working and ratios.

#	Question	Answer and marking notes
1	Two pea plants, both of genotype Aa, are cross-pollinated and give 800 offspring. Assuming complete dominance and equal survival, find the expected number of plants of each phenotype.	Answer: genotype ratio AA:Aa:aa = 1:2:1; dominant phenotype 3/4 x 800 = 600 plants and recessive 1/4 x 800 = 200 plants. (Marking: ratio 2, numbers 2)
2	A test cross is carried out between an Aa plant and an aa plant, giving 240 offspring. Find the expected number of each genotype and state the purpose of a test cross.	Answer: Aa x aa gives a 1:1 ratio, so 120 Aa and 120 aa; a test cross shows whether a dominant-looking individual is heterozygous. (Marking: ratio and numbers 2, purpose 2)
3	Two independently assorting gene pairs are crossed, AaBb x AaBb, giving 960 offspring. Using the 9:3:3:1 ratio, find the expected number of each of the four phenotypes.	Answer: total parts 9 + 3 + 3 + 1 = 16, each part 960 / 16 = 60 plants, so the four phenotypes number 540, 180, 180 and 60. (Marking: ratio 2, four numbers 2)
4	A father is blood group A (genotype I^A i) and the mother group B (I^B i). Give the probability of each of the four ABO groups among their children, and the probability that all 4 children are group O.	Answer: AB : A : B : O = 1:1:1:1, each 1/4; all four children group O = (1/4)^4 = 1/256, about 0.39%. (Marking: four groups 2, 1/256 2)

Section B: DNA Structure and Semi-conservative Replication (16 marks)

Answer the following questions on DNA structure and replication, showing your calculations.

#	Question	Answer and marking notes
5	A double-stranded DNA molecule has 2000 bases in total, of which adenine (A) accounts for 30%. Find the number of A, T, G and C bases.	Answer: A = T = 2000 x 30% = 600; G = C = (2000 – 1200) / 2 = 400. (Marking: A and T 2, G and C 2)
6	A DNA double helix contains 1200 base pairs, of which 400 pairs are A-T and the rest are G-C. Find the total number of hydrogen bonds in this segment.	Answer: G-C pairs = 1200 – 400 = 800; hydrogen bonds = 400 x 2 + 800 x 3 = 800 + 2400 = 3200. (Marking: G-C pairs 2, total 2)
7	One DNA molecule is replicated 3 times in a cell cycle. Find the number of DNA molecules formed and how many of them contain an original (old) strand, and state the replication model supported by this result.	Answer: 2^3 = 8 molecules; 2 of them each keep one old strand, which supports semi-conservative replication. (Marking: 8 molecules 2, 2 molecules and model 2)
8	A DNA segment has 1200 base pairs and each pair rises 0.34 nm along the helix. Find the length of the segment and explain why DNA can store a huge amount of genetic information.	Answer: length = 1200 x 0.34 = 408 nm; information lies in the base sequence and the possible combinations of four bases are enormous. (Marking: 408 nm 2, explanation 2)

Section C: Transcription, Splicing and Translation (18 marks)

Answer the questions on gene expression; write codons and anticodons in the 5 to 3 direction unless stated otherwise.

#	Question	Answer and marking notes
9	The template strand of a gene reads 3-TAC CGG AAT-5. Write the mRNA sequence (5 to 3) and the amino acid sequence translated from it (AUG = methionine, GCC = alanine, UUA = leucine).	Answer: mRNA is 5-AUG GCC UUA-3, giving methionine – alanine – leucine. (Marking: mRNA 3, amino acids 2)
10	A polypeptide consists of 150 amino acids. Find the minimum number of bases in the coding region of its mRNA, and the number of bases when the stop codon is included.	Answer: each amino acid is coded by 3 bases, so 150 x 3 = 450 bases; adding one stop codon of 3 bases gives 453 bases. (Marking: 450 3, 453 2)
11	A gene is 900 base pairs long, of which the exons total 660 base pairs and the rest are introns. Find the number of amino acids in the polypeptide after splicing.	Answer: 660 / 3 = 220 codons; minus one stop codon gives 219 amino acids. (Marking: 220 codons 2, 219 amino acids 2)
12	A codon on mRNA reads 5-GCA-3. Write the tRNA anticodon that pairs with it (in the 3 to 5 direction) and state the direction of pairing.	Answer: the anticodon is 3-CGU-5 (written 5-UGC-3); pairing is antiparallel, the 5 end of mRNA facing the 3 end of tRNA. (Marking: sequence 2, direction 2)

Section D: Mutation and Genetic Diseases (16 marks)

Answer the following questions on mutation and genetic disease, explaining the mechanisms.

#	Question	Answer and marking notes
13	One base pair is inserted in the middle of a coding region. Name this type of mutation and explain why its effect on the protein is usually more serious than that of a base substitution.	Answer: a frameshift mutation; one inserted base shifts the reading frame, so every codon after the insertion changes and the protein usually loses function, while a substitution affects only one codon. (Marking: name 1, mechanism 2, comparison 1)
14	In sickle cell anaemia a haemoglobin codon changes from GAG to GUG. State the amino acid change and explain how one base change affects the shape of red blood cells.	Answer: GAG codes glutamate and GUG codes valine, so glutamate is replaced by valine; haemoglobin structure and solubility change, it polymerises at low oxygen and red cells become sickle-shaped and fragile. (Marking: amino acid change 2, mechanism 2)
15	A gene mutates at a rate of about 1 in 100000 per replication. If the gene is replicated 1000000 times in an experiment, how many mutations are expected, and why are mutations still common although the rate is low?	Answer: 1000000 / 100000 = 10 mutations; genomes are large and cell divisions are numerous, so mutations remain common in individuals and populations. (Marking: 10 mutations 2, explanation 2)
16	In a population, 1 in 25 people is a carrier (heterozygote) of a recessive disease. Find (a) the chance that a child of two carriers is affected, and (b) the chance that a child of two randomly chosen people is affected.	Answer: (a) heterozygote x heterozygote gives 1/4; (b) both carriers = (1/25)^2 = 1/625, times 1/4 gives 1/2500. (Marking: (a) 2, (b) 2)

Section E: PCR and Genetic Engineering (18 marks)

Answer the following questions on molecular biotechnology, showing your calculations.

#	Question	Answer and marking notes
17	A 5000 bp plasmid has two EcoRI sites, located after base 1200 and after base 3600. Find the lengths of the three fragments produced by EcoRI digestion and state the use of restriction enzymes in genetic engineering.	Answer: the fragments are 1200 bp, 3600 – 1200 = 2400 bp and 5000 – 3600 = 1400 bp, totalling 5000 bp; restriction enzymes cut DNA at specific sequences and are used to isolate the target gene and open the vector. (Marking: three lengths 3, use 2)
18	PCR amplifies a DNA segment through 20 cycles. Assuming 100% efficiency per cycle, find the amplification factor (as a power of 2 and as a number) and state the role of the high-temperature step.	Answer: 2^20 = 1048576 fold; the high temperature (about 95 C) denatures the double helix so the strands separate, then cooling lets primers anneal and polymerase extend them. (Marking: 2^20 and value 3, role 2)
19	Gel electrophoresis of digested DNA gives three fragments of 4000 bp, 1500 bp and 500 bp. State which fragment moves fastest and which slowest, and explain why.	Answer: 500 bp moves fastest and 4000 bp slowest; the gel mesh lets shorter fragments pass through more easily, so they travel further. (Marking: order 2, explanation 2)
20	To produce insulin in bacteria, why is cDNA made from mRNA used instead of genomic DNA that contains introns?	Answer: bacteria lack splicing machinery and cannot remove introns; cDNA contains only exons, so it is translated correctly into a functional polypeptide in bacteria. (Marking: no splicing 2, exons only 2)

Section F: Genetic Data Analysis and Biotechnology Applications (16 marks)

Answer the following questions and show your calculations of genetic ratios.

#	Question	Answer and marking notes
21	Two long-winged fruit flies are crossed and give 960 offspring, of which 720 are long-winged and 240 are short-winged. Find the phenotypic ratio and deduce the parental genotypes and which allele is dominant.	Answer: 720 : 240 = 3 : 1; both parents are heterozygous (Aa x Aa), long wing is dominant and short wing recessive, with genotypes 1 AA : 2 Aa : 1 aa. (Marking: ratio 1, genotypes 2, dominance 1)
22	In a population at Hardy-Weinberg equilibrium the frequency of albinism (recessive) is 1 in 10000. Find the recessive allele frequency q, the dominant allele frequency p and the proportion of carriers.	Answer: q^2 = 1/10000 so q = 0.01; p = 1 – 0.01 = 0.99; carriers 2pq = 2 x 0.99 x 0.01 = 0.0198, about 1.98% (1 in 50). (Marking: q and p 2, carriers 2)
23	The human haploid genome contains about 3 x 10^9 base pairs, of which only about 1.5% codes for protein. Find the number of coding base pairs and state one function of the non-coding sequences.	Answer: 3 x 10^9 x 1.5% = 4.5 x 10^7 base pairs; non-coding DNA can regulate gene expression (promoters, enhancers) or be transcribed into rRNA and tRNA. (Marking: 4.5 x 10^7 2, function 2)
24	Gene editing such as CRISPR can modify a genome precisely. Give one application and state one risk or ethical limitation of using this technology.	Answer: example: correcting a mutation that causes a genetic disease, or improving pest resistance in crops; risk or limit: off-target edits may alter other genes, and editing germline genes passes changes to later generations, raising safety and ethical concerns. (Marking: example 2, risk or limit 2)

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