S5 Biology Practice Paper: Genetics, DNA And Cell Division (12 Questions, 100 Marks)

S5 Biology Practice Paper: Genetics, DNA and Cell Division (12 questions, 100 marks)

2026-11-08

S5 Biology: Genetics, DNA and Cell Division

S5 Biology: crosses and ratios, DNA structure and replication, meiosis and sex-linked inheritance. 12 questions, 100 marks, 75 min, answers included.

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Section A: Mendelian Genetics and Cross Ratios (4 questions, 25 marks)

Cross ratios: write parental genotypes and gametes first, then count with a Punnett square; reduce ratios to lowest whole numbers and name the phenotypes.

#	Question	Answer and marking notes
1	Define the terms gene, allele, genotype and phenotype, and explain how genotype and phenotype are related.	A gene is a length of DNA on a chromosome that controls a characteristic; alleles are different versions of the same gene (T and t); genotype is the combination of alleles present (TT, Tt or tt); phenotype is the observable characteristic (tall or dwarf). Genotype and environment together determine phenotype, and a dominant allele is expressed in a heterozygote, so Tt and TT look the same. Marking: one mark per definition, two marks for the relationship including dominance and environment.
2	In peas, tall stem (T) is dominant to dwarf stem (t). Two tall plants, both of genotype Tt, are crossed: (a) state the gametes each parent produces; (b) use a Punnett square to give the genotypic and phenotypic ratios of the offspring; (c) what fraction of the tall offspring are homozygous tall (TT)?	(a) Each parent makes two kinds of gamete, T and t, in a 1:1 ratio. (b) Punnett square gives TT : Tt : tt = 1 : 2 : 1, so tall : dwarf = 3 : 1. (c) Among tall offspring TT : Tt = 1 : 2, so the fraction that is TT is 1/3. Marking: gametes 2, genotypic ratio 1.5, phenotypic ratio 1.5, 1/3 one mark.
3	Explain what a test cross is and why it can reveal the genotype of an individual showing a dominant phenotype. A tall pea plant is test crossed with a dwarf plant, and 60 of the 120 offspring are dwarf. Deduce the genotype of the tall parent and the expected ratio.	A test cross is a cross between the individual showing the dominant phenotype and a homozygous recessive (tt) individual. The recessive parent contributes only t gametes, so the offspring phenotypes show directly which gametes the unknown parent made. A Tt parent gives a 1:1 ratio (60 tall : 60 dwarf), while a TT parent gives all tall offspring. Since 60 of 120 are dwarf, the ratio is 1:1 and the parent is Tt. Marking: definition 2, explanation 2, genotype 1, ratio 1.
4	In a Mendelian dihybrid cross the parents are AaBb x AaBb, with the two genes on different chromosomes: (a) how many kinds of gamete does each parent make? List them; (b) what is the phenotypic ratio of the offspring? (c) If 320 offspring are produced, how many fall into each of the four phenotypic classes?	(a) Four kinds of gamete: AB, Ab, aB, ab, in equal numbers. (b) The phenotypic ratio is 9 : 3 : 3 : 1. (c) With 320 offspring the classes are 180 : 60 : 60 : 20, that is 180 dominant-dominant, 60 dominant-recessive, 60 recessive-dominant and 20 recessive-recessive. Marking: gametes 2, 9:3:3:1 two marks, class numbers 3 (each must match the ratio).

Section B: DNA Structure, Replication, Transcription and Translation (4 questions, 33 marks)

Molecular genetics: A pairs with T (2 hydrogen bonds), G with C (3); multiply base pairs by 2 first; one codon per amino acid plus one stop codon.

#	Question	Answer and marking notes
5	DNA is made of four bases. (a) State the base-pairing rule and the number of hydrogen bonds in each pair. (b) A DNA double helix has 1,000 base pairs, and adenine (A) makes up 30% of all bases. Calculate the number of cytosine (C) bases.	(a) A pairs with T by 2 hydrogen bonds and G pairs with C by 3, so A = T and G = C. (b) 1,000 base pairs means 2,000 bases. A is 30%, that is 600 bases, so T is also 600. The remaining G + C is 800, so C = 400. Marking: pairing rule with hydrogen bonds 3, total bases 2, C = 400 three marks.
6	DNA replication is semi-conservative. (a) Explain what semi-conservative means. (b) After three rounds of replication, how many DNA molecules are produced from one molecule, and how many of them contain an original parental strand?	(a) In semi-conservative replication the double helix unwinds and each parental strand acts as a template, so every daughter molecule has one old strand and one new strand. (b) Three rounds give 2^3 = 8 DNA molecules, of which 2 contain an original parental strand, one old strand in each. Marking: explanation 4, 8 molecules 2, 2 with old strand 2.
7	A protein is made of 120 amino acids. (a) What is the minimum number of bases in the corresponding mRNA? (b) What is the minimum number of tRNA molecules involved in translation? (c) Why is the number of codons one more than the number of amino acids?	(a) 120 amino acids need 120 codons, plus one stop codon, giving 121 codons x 3 = 363 bases. (b) One tRNA carries each amino acid, so at least 120 tRNA molecules are needed. (c) The stop codon does not code for an amino acid; it only terminates translation, so there is one more codon than amino acids. Marking: 363 three marks, 120 two marks, stop codon explanation 3.
8	A human body cell has 46 chromosomes. (a) How many chromosomes does a gamete have after meiosis? (b) Compare the chromosome number and DNA content in meiosis I and meiosis II. (c) Explain the significance of crossing over for genetic variation.	(a) 46 / 2 = 23 chromosomes. (b) In meiosis I homologous chromosomes separate, so the chromosome number falls from 46 to 23 per cell and DNA content falls from 4C to 2C. In meiosis II sister chromatids separate, the chromosome number stays at 23 and DNA content falls from 2C to 1C. (c) Crossing over recombines alleles between homologous chromosomes, producing new allele combinations and increasing variation for natural selection. Marking: 23 two marks, each stage 2, significance 3.

Section C: Meiosis, Sex-linked Inheritance, Mutation and Breeding (4 questions, 42 marks)

Division and disease: meiosis I separates homologous chromosomes, meiosis II separates sister chromatids; in X-linked recessive disease a son depends on the mother X; multiply independent probabilities.

#	Question	Answer and marking notes
9	Red-green colour blindness is an X-linked recessive condition. A woman who carries the allele but has normal vision (X^A X^a) marries a man with normal vision (X^A Y): (a) use a Punnett square to list the possible genotypes of the children; (b) what is the probability that a son is colour blind? (c) Why are none of the daughters affected, yet some may be carriers?	(a) The mother makes X^A and X^a gametes; the father makes X^A and Y. Children can be X^A X^A (normal daughter), X^A X^a (carrier daughter), X^A Y (normal son) or X^a Y (colour-blind son). (b) Half of the sons are affected, so the probability is 1/2. (c) Every daughter receives X^A from her father, so she has at least one normal allele and is not affected; but the mother passes X^a with probability 1/2, so half the daughters are carriers. Marking: Punnett square 4, 1/2 three marks, carrier explanation 3.
10	A recessive condition is caused by allele t. A couple, both Tt, already have one affected child. (a) What is the probability that the next child is affected? (b) If they plan three children, what is the probability that all three are affected? (c) Explain why the actual proportion in a family may differ from the theoretical probability.	(a) For Tt x Tt the probability of a tt child is 1/4. (b) Each pregnancy is independent, so three affected children have probability (1/4)^3 = 1/64. (c) A probability describes the long-run average over many families; one family has few children, so random variation is large, and other genetic factors may also act. Marking: 1/4 three marks, 1/64 four marks, explanation 3.
11	Gene mutations include substitution, insertion and deletion. (a) Give one example of how a substitution changes a protein. (b) Explain why an insertion or deletion usually has a greater effect than a substitution. (c) Distinguish between somatic and germ-line mutations in terms of inheritance.	(a) For example GAG becomes GTG, the mRNA codon changes from GAG to GUG, glutamic acid is replaced by valine, which can cause sickle-cell anaemia. (b) An insertion or deletion shifts the reading frame from the mutation point onwards, so the whole amino acid sequence after it changes and the protein usually loses function; a substitution changes only one amino acid. (c) A somatic mutation affects only that individual and its cell descendants and is not passed to children, whereas a germ-line mutation can be inherited through gametes. Marking: example 3, frameshift explanation 4, the two types 3.
12	A gardener wants a true-breeding red, large-flowered variety (R dominant to r, B dominant to b, genes assorting independently). He has one red, large-flowered plant of unknown genotype. (a) Suggest an experiment to find its genotype. (b) If the test cross gives red-large, red-small, white-large and white-small offspring, state the genotype of the plant. (c) Explain why several generations of selfing are needed to confirm a pure line. (d) Give two practical limitations in plant breeding.	(a) Cross it with a homozygous recessive rrbb and record the phenotypes and their ratio. (b) Four phenotypic classes among the offspring mean the plant made four kinds of gamete, so its genotype is RrBb. (c) A heterozygote segregates recessive offspring on selfing, so only after several generations without segregation can the line be accepted as pure (RRBB). (d) Practical limitations include linkage making the ratio depart from 9:3:3:1, the time and cost of many generations, environmental effects on phenotype, and distorted ratios in small populations. Marking: test cross design 3, RrBb three marks, selfing reason 3, each limitation 1.5.

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