S5 Biology Practice Paper: Genetics, Populations And Ecosystems (12 Questions, 100 Marks, 75 Minutes)

S5 Biology Practice Paper: Genetics, Populations and Ecosystems (12 questions, 100 marks, 75 minutes)

2026-11-08

S5 Biology: Genetics, Populations and Ecosystems

S5 biology paper covering monohybrid crosses and phenotypic ratios, allele frequencies, DNA base composition and codons, energy transfer along food chains, population growth and mark-recaptu

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Section A: Genetics and Molecular Basis (25 marks)

Answer all questions and show your working.

#	Question	Answer and marking notes
1	Two heterozygous black mice (Bb) are crossed; black (B) is dominant to white (b). (a) List the genotypes of the offspring and their ratio. (b) State the ratio of black to white offspring and explain why.	Bb x Bb gives BB : Bb : bb = 1 : 2 : 1. Black (BB and Bb) accounts for 3/4 and white (bb) for 1/4, i.e. 3 : 1. Marking: (a) 3; (b) 2 (ratio 1, reason 1).
2	An insect population of 500 individuals contains 320 AA, 160 Aa and 20 aa. (a) Calculate the frequencies of alleles A and a. (b) Calculate the frequency of genotype Aa.	Number of A alleles = 320×2 + 160 = 800, so frequency = 800 / 1000 = 0.80; frequency of a = 1 – 0.80 = 0.20. Frequency of Aa = 160 / 500 = 0.32. Marking: (a) 3; (b) 2.
3	A DNA molecule contains 900 base pairs and adenine (A) makes up 30% of all bases. (a) Calculate the number of guanine (G) bases. (b) If the whole molecule is coding sequence, what is the maximum number of amino acids it can encode (excluding the stop codon)?	Total bases = 900 x 2 = 1800; A = 30% of 1800 = 540 and T = 540; G + C = 1800 – 1080 = 720, so G = C = 360. Codons = 900 / 3 = 300; minus one stop codon gives 299 amino acids. Marking: (a) 3; (b) 2.
4	Producers in a grassland fix 40,000 kJ per square metre of solar energy per year. Assimilation efficiency is 10% in primary consumers and 15% in secondary consumers. (a) Calculate the energy assimilated by primary consumers. (b) Calculate the energy assimilated by secondary consumers. (c) State two other fates of the remaining energy. (d) Explain why food chains rarely exceed four to five trophic levels.	(a) 40,000 x 10% = 4000 kJ; (b) 4000 x 15% = 600 kJ; (c) lost as heat in respiration, not eaten or lost in faeces, broken down by decomposers; (d) only about 10% of energy passes to the next level, so after four or five levels too little energy remains to support another trophic level. Marking: (a) 2; (b) 2; (c) 3; (d) 3.

Section B: Populations and Genetic Ratios (30 marks)

Show the genetic cross or formula; give frequencies to 3 significant figures.

#	Question	Answer and marking notes
5	A bird population of 250 individuals grows at 12% per year with no immigration or emigration. (a) Estimate the population after 6 years (integer). (b) Calculate the doubling time in years (2 d.p.).	(a) 250 x 1.12^6 = 493.5, about 493 birds. (b) Doubling time = ln2 / ln1.12 = 6.12 years. Marking: (a) 4; (b) 3.
6	A researcher uses mark-recapture on a snail population: 120 snails are caught, marked and released; one week later 200 are caught and 45 carry marks. (a) Estimate the population size. (b) State two assumptions of this method.	(a) N = 120 x 200 / 45 = 533. (b) Marks do not affect survival or catchability; the population is well mixed with no immigration, emigration, births or deaths during the study; marks do not fall off. Marking: (a) 4; (b) 3.
7	A recessive disease affects 1 in 2500 people in a population at Hardy-Weinberg equilibrium. (a) Calculate the recessive allele frequency q. (b) Calculate the dominant allele frequency p. (c) Calculate the carrier (heterozygote) frequency to 3 significant figures.	(a) q squared = 1/2500 so q = 0.02. (b) p = 1 – q = 0.98. (c) 2pq = 2 x 0.98 x 0.02 = 0.0392, i.e. 0.039 (about 3.92%). Marking: (a) 3; (b) 2; (c) 3.
8	Human ABO blood groups are controlled by alleles IA, IB and i. (a) A father of group AB and a mother of group O have children; give the blood groups and their probabilities. (b) If the father is AO and the mother is BO, give the ratio of the four blood groups.	(a) AB x ii gives IAi (group A) and IBi (group B), 50% each. (b) IAi x IBi gives AB : A : B : O = 1 : 1 : 1 : 1. Marking: (a) 4; (b) 4.

Section C: Long Questions (45 marks)

Answer in full steps, with a table, or in paragraphs.

#	Question	Answer and marking notes
9	A test cross AaBb x aabb gives 640 seeds; the two genes are on different chromosomes. (a) Calculate the expected number of each phenotype. (b) What phenotypes would appear if the genes were completely linked with no crossing over? (c) Only two phenotypes of 320 seeds each are observed; explain the result.	(a) 1 : 1 : 1 : 1, i.e. 160 seeds each. (b) Only the two parental phenotypes appear, 50% each (320 seeds). (c) The genes are on the same chromosome and completely linked, so the test cross shows only parental combinations and no independent assortment. Marking: (a) 4; (b) 3; (c) 4.
10	Gross primary production (GPP) of a forest is 50,000 kJ per square metre per year and producer respiration uses 20,000 of it. (a) Calculate net primary production (NPP). (b) If primary consumers eat 15% of NPP and assimilate 40% of it, calculate the energy assimilated. (c) State two roles of decomposers in the carbon cycle.	(a) NPP = 50,000 – 20,000 = 30000. (b) 30000 x 15% x 40% = 1800 kJ per square metre per year. (c) Decomposers release carbon from dead organic matter as CO2 by respiration, and mineralise organic matter into inorganic nutrients returned to the soil so carbon can be fixed again by producers. Marking: (a) 3; (b) 4; (c) 4.
11	A field of 60 hectares holds voles at a density of 25 per hectare. (a) Calculate the total number of voles. (b) If the carrying capacity K is 1800, calculate the current population as a percentage of K (3 s.f.). (c) Describe how the birth rate and death rate change as the population approaches K.	(a) 60 x 25 = 1500 voles. (b) 1500 / 1800 x 100% = 83.3%. (c) Density-dependent factors (food, space, disease, predation) intensify, birth rate falls and death rate rises, net growth approaches zero and the population fluctuates around K. Marking: (a) 3; (b) 3; (c) 5.
12	A student investigates light intensity and photosynthesis rate using pondweed. Bubble counts over 5 minutes at 10, 20, 30 and 40 cm from the lamp are 42, 28, 17 and 9; a dark control gives 2 bubbles in 5 minutes. (a) Calculate the bubble rate per minute for each group. (b) Calculate how many times faster the 10 cm group is than the 40 cm group (3 s.f.). (c) State two other variables that must be controlled. (d) Explain why the same plant and three repeats are used.	(a) 8.4, 5.6, 3.4 and 1.8 bubbles per minute; dark control 0.4 per minute. (b) 8.4 / 1.8 = 4.667 times. (c) Temperature, CO2 concentration of the hydrogencarbonate solution, plant species and size, volume of water. (d) Using one plant reduces individual variation, and three repeats averaged reduce random error so the results are more reliable. Marking: (a) 3; (b) 2; (c) 3; (d) 4.

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