S5 Biology Practice Paper: Genetics, Sex Linkage And Molecular Biology (100 Marks)

S5 Biology Practice Paper: Genetics, Sex Linkage and Molecular Biology (100 marks)

2026-11-12

Biology – Genetics and Molecular Biology

S5 Biology paper: 6 questions, 24 parts, 100 marks, 60 minutes. Genetics, sex linkage, DNA bases, transcription, translation, mutation and PCR. Full answers included.

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Question 1 – Monohybrid Cross (18 marks)

In garden peas the allele T for tall stem is dominant to the allele t for dwarf stem. Two heterozygous tall plants (Tt) are crossed and produce 400 offspring.

#	Question	Answer and marking notes
1	(a) State the gametes produced by the parents and give the genotypic and phenotypic ratios of the offspring.	Answer: Each parent produces two kinds of gamete, T and t, in equal numbers. Combining them gives TT 1/4, Tt 2/4 and tt 1/4, that is 1 : 2 : 1. Plants carrying T are tall, so the phenotypic ratio is tall : dwarf = 3 : 1. Marking: gametes 1, genotypic ratio 2, phenotypic ratio 2.
2	(b) Find the expected numbers of tall and dwarf offspring among the 400 plants.	Answer: expected tall = 400 x 3/4 = 300 plants; expected dwarf = 400 x 1/4 = 100 plants. The two add up to 400. Marking: tall 2, dwarf 2.
3	(c) A tall plant is crossed with a dwarf plant and half of the offspring are tall and half are dwarf. Explain how this identifies the genotype of the tall parent.	Answer: the dwarf parent must be tt and produces only t gametes. A TT tall parent would give all Tt offspring, all tall. Since half the offspring are dwarf (tt), the tall parent must pass on t gametes, so its genotype is Tt. Crossing with a homozygous recessive is called a test cross. Marking: dwarf parent tt 1, deduction 2, test cross idea 1.
4	(d) Another experiment gives 290 tall and 110 dwarf plants. Explain why the observed numbers differ from the expected numbers.	Answer: fertilisation is a random union of T-bearing and t-bearing gametes, so it is a matter of probability. Each cross produces only a limited number of offspring, and a finite sample cannot match 3 : 1 exactly, so random deviation appears. With a larger sample the observed ratio approaches the expected one. Marking: random fertilisation 2, limited sample 2, larger sample 1.

Question 2 – Sex-Linked Inheritance (17 marks)

Red-green colour blindness is caused by a recessive allele b carried on the X chromosome; the normal allele is written B. A carrier mother (XBXb) marries a man with normal colour vision (XBY), and they have sons and daughters.

#	Question	Answer and marking notes
5	(a) State the types of gamete produced by the mother and by the father.	Answer: the mother carries B on one X and b on the other, so she produces XB and Xb eggs in equal numbers. The father has one X and one Y, so he produces XB and Y sperm in equal numbers. Marking: mother 2, father 2.
6	(b) Find the probability that a son is colour blind.	Answer: a son receives his single X from his mother. There is a 1/2 chance she passes on Xb, and Xb with Y gives XbY, which is colour blind. The probability is therefore 1/2, that is 50 per cent. Marking: X from mother 1, XbY genotype 1, probability 2.
7	(c) Find the probability that a daughter has normal colour vision but is a carrier.	Answer: a daughter always receives XB from her father, so she has normal colour vision. She receives Xb from her mother with probability 1/2, giving the carrier genotype XBXb. The probability is therefore 1/2, that is 50 per cent; the other 1/2 are XBXB and not carriers. Marking: paternal XB 1, XBXb 1, probability 2.
8	(d) Explain why far more males than females are colour blind.	Answer: a male has only one X chromosome, so a single recessive b on that X is enough to cause the condition; he is hemizygous. A female has two X chromosomes and needs two b alleles (XbXb) to be affected, which is much less likely. Recessive sex-linked conditions are therefore commoner in males. Marking: single X in males 2, two alleles needed in females 2, conclusion 1.

Question 3 – Base Composition of DNA (17 marks)

A double-stranded DNA molecule contains 2,000 bases in total. Chemical analysis shows that adenine (A) makes up 30 per cent of all the bases.

#	Question	Answer and marking notes
9	(a) Find the percentage of thymine (T) and explain your answer.	Answer: T makes up 30 per cent. DNA is double-stranded and A always pairs with T by two hydrogen bonds, so every A on one strand faces a T on the other. The number of A equals the number of T, hence the same percentage. Marking: value 2, pairing reason 2.
10	(b) Calculate the percentage of guanine (G).	Answer: A and T together account for 60 per cent, leaving 40 per cent shared by G and C. Since G equals C, G is 20 per cent and C is also 20 per cent. Marking: G plus C is 40 per cent 2, value of G 2.
11	(c) Find the actual numbers of A and G bases in this DNA molecule.	Answer: number of A = 2,000 x 30 per cent = 600; number of G = 2,000 x 20 per cent = 400. There are also 600 T and 400 C, making 2,000 in total. Marking: number of A 3, number of G 2.
12	(d) Heating this DNA to 95 degrees separates the two strands. Explain this in terms of base pairing.	Answer: the two strands are held by hydrogen bonds between A and T and between G and C, and hydrogen bonds are relatively weak. Heating supplies energy that breaks them, so the strands separate, while the sugar-phosphate backbone stays intact. Because pairing is specific, the strands can re-pair on cooling. Marking: hydrogen bonds 2, heat breaks them 1, backbone intact or re-pairing 1.

Question 4 – Transcription and Translation (16 marks)

A mature mRNA molecule contains 900 bases, and its sequence includes a start codon and one stop codon.

#	Question	Answer and marking notes
13	(a) Find the number of codons in this mRNA.	Answer: every three bases form one codon, so the number of codons is 900 / 3 = 300. Marking: relation to three bases 2, value 2.
14	(b) Find the number of amino acids in the polypeptide translated from this mRNA.	Answer: the stop codon codes for no amino acid and only ends translation, so the number of amino acids is 300 – 1 = 299. Marking: excluding the stop codon 2, value 2.
15	(c) Describe the function of the start codon and of the stop codon.	Answer: the start codon (AUG) marks where translation begins, the ribosome binds there and starts reading, and it also codes for methionine. A stop codon (UAA, UAG or UGA) codes for no amino acid, matches no tRNA, and makes the ribosome release the finished polypeptide. Marking: start 2, stop 2.
16	(d) If one base is replaced by another, the protein produced does not always change. Explain why.	Answer: the genetic code is degenerate, so several different codons can code for the same amino acid, for example more than one codon codes for leucine. If the new codon still specifies the same amino acid, the polypeptide sequence is unchanged and so is the protein function. A change in a non-coding region also leaves the protein unchanged. Marking: degeneracy 2, same amino acid 1, non-coding region 1.

Question 5 – Dihybrid Cross (16 marks)

In garden 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 cotyledons. The two gene pairs are on different chromosomes. Two plants of genotype RrYy are crossed and produce 1,280 offspring.

#	Question	Answer and marking notes
17	(a) State the kinds of gamete produced by each parent.	Answer: because the two gene pairs lie on different chromosomes they assort independently in meiosis, so each parent produces four kinds of gamete: RY, Ry, rY and ry, in equal numbers, each 1/4. Marking: four gametes 2, equal numbers or independent assortment 2.
18	(b) Give the phenotypic ratio of the offspring.	Answer: combining the gametes in a 4 by 4 grid gives round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1. Round is 3/4 and yellow is 3/4, and the traits are independent, so round yellow is 3/4 x 3/4 = 9/16. Marking: ratio 3, explanation 1.
19	(c) Find the expected number of round yellow offspring among the 1,280 plants.	Answer: round yellow plants make up 9/16, so the expected number is 1,280 x 9/16 = 720. Marking: using 9/16 2, result 2.
20	(d) If the two gene pairs were actually on the same chromosome, how would the offspring ratio change? Explain.	Answer: the ratio would depart from 9 : 3 : 3 : 1. Genes on the same chromosome tend to be inherited together, which is linkage, so the parental gametes RY and ry become commoner than the recombinant gametes Ry and rY. Recombinants appear only when crossing over occurs in meiosis, so the parental phenotypes exceed 9/16 and 1/16. Marking: departure 1, linkage 2, crossing over 1.

Question 6 – PCR and Gel Electrophoresis (16 marks)

A DNA sample contains 100 copies of a target sequence. A technician carries out 30 cycles of the polymerase chain reaction (PCR), and every cycle doubles the target sequences.

#	Question	Answer and marking notes
21	(a) Describe the three main steps of each PCR cycle.	Answer: first, denaturation, heating to about 95 degrees breaks the hydrogen bonds so the double helix separates into single strands. Second, annealing, cooling to about 55 degrees lets the primers bind to the complementary sequences at the ends of each template. Third, extension, raising the temperature to about 72 degrees lets a heat-stable DNA polymerase build new complementary strands from free nucleotides. Marking: one mark per step, one more for correct temperature or role.
22	(b) Find the amplification factor after 30 cycles.	Answer: the number of copies doubles each cycle, so after 30 cycles the factor is 2 to the power 30, which is 1,073,741,824. Marking: 2 raised to the number of cycles 2, value 2.
23	(c) Find the total number of target copies after 30 cycles.	Answer: total copies = 100 x (2 to the power 30) = 100 x 1,073,741,824, which is about 1.07 x 10 to the power 11 copies. Marking: starting number 2, scientific notation 2.
24	(d) PCR products are often analysed by gel electrophoresis. Explain how it separates DNA by fragment size.	Answer: the phosphate groups of DNA carry negative charges, so DNA moves towards the positive electrode. The gel forms a mesh, and longer fragments meet more resistance and travel more slowly, while shorter ones move faster, so fragments of different size separate into distinct bands. Comparing with a ladder of known sizes allows the product size to be estimated. Marking: negative charge towards positive electrode 1, gel mesh 2, shorter moves faster 1.

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