Form 5 Biology Practice Paper: Genetics And Ecology

Form 5 Biology Practice Paper: Genetics and Ecology

2026-11-05

Form 5 Biology – 12 questions – 100 marks – 75 minutes

Form 5 Biology practice paper: 12 questions, 100 marks in 75 minutes on genetics, energy flow, population sampling, osmosis and enzymes, with answers and marking points.

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Paper details　Form 5 Biology practice paper, 12 questions and 100 marks in 75 minutes: genetics, energy flow, population sampling, osmosis and enzymes.

Section A: Genetics

Answer all questions and show the working and the unit in each calculation.

#	Question	Answer and marking notes
1	Two pea plants, both of genotype Tt, are crossed (T = tall dominant, t = dwarf recessive). (a) Use a Punnett square to list the genotypes of the offspring and their ratio. (b) Give the phenotypic ratio. (c) If there are 400 offspring, how many tall plants are expected?	(a) TT : Tt : tt = 1 : 2 : 1 (4 marks). (b) Tall : dwarf = 3 : 1 (2 marks). (c) Tall = 400 × 3/4 = 300 plants (2 marks).
2	A tall pea plant of unknown genotype is test-crossed with a dwarf plant (tt) and the offspring are close to 1 : 1 tall to dwarf. (a) Deduce the genotype of the tall parent and explain. (b) If there are 200 offspring, how many dwarf plants are expected? (c) What would the parent genotype be if all offspring were tall?	(a) The parent is Tt; a 1 : 1 ratio shows it produced both T and t gametes (4 marks). (b) Dwarf = 200 × 1/2 = 100 plants (3 marks). (c) The parent would be TT (2 marks).
3	Producers in an ecosystem fix 50 000 kJ/m²/year and only 10% of the energy passes to the next trophic level. (a) Calculate the energy of the primary consumers. (b) Calculate the energy of the secondary consumers. (c) Calculate the energy of the tertiary consumers and explain why food chains rarely have more than four or five levels.	(a) Primary consumers = 50 000 × 10% = 5 000 kJ/m²/year (3 marks). (b) Secondary consumers = 500 kJ/m²/year (3 marks). (c) Tertiary consumers = 50 kJ/m²/year; energy is lost at each level so the fifth or sixth level cannot support a population (2 marks).
4	Plants in a grassland are sampled with 0.25 m² quadrats. Twenty random quadrats contain a mean of 6 plants of the target species. (a) Calculate the plant density per square metre. (b) Estimate the total number of plants if the grassland is 2 000 m². (c) Explain why random sampling is used with quadrats.	(a) Density = 6 / 0.25 = 24 plants/m² (3 marks). (b) Total = 24 × 2 000 = 48 000 plants (3 marks). (c) Random sampling avoids bias so the sample is representative (2 marks).

Section B: Energy Flow and Populations

Answer all questions and show genetic crosses with a Punnett square or probability working.

#	Question	Answer and marking notes
5	40 fish are caught in a pond, marked and released. A week later 50 fish are caught and 8 of them are marked. (a) Use the Lincoln index to estimate the population size. (b) Write down the formula used. (c) Give two reasons why the estimate may be too high.	(a) Population ≈ 40 × 50 / 8 = 250 fish (4 marks). (b) N = (M × C) / R (3 marks). (c) Marking affects survival or behaviour, marks are lost, marked fish are recaptured more easily (any two, 2 marks).
6	An aquatic plant is placed in light and the volume of gas collected per minute is used to compare the rate of photosynthesis. Group A gives 0.6 cm³ per minute; group B, with double the light intensity, gives 0.9 cm³ per minute. (a) Calculate the total gas volume collected by A in 30 minutes. (b) Calculate the percentage increase in the rate of B over A. (c) Explain why doubling the light intensity did not double the rate.	(a) 0.6 × 30 = 18 cm³ (3 marks). (b) (0.9 − 0.6) / 0.6 × 100% = 50% (3 marks). (c) Light is no longer limiting; carbon dioxide concentration or temperature is limiting (2 marks).
7	Two plants of genotype AaBb are crossed (the two genes are on different chromosomes and show complete dominance). (a) Give the phenotypic ratio of the offspring. (b) If there are 320 offspring, how many are expected to show both dominant traits? (c) Explain why four phenotypes appear.	(a) 9 : 3 : 3 : 1 (3 marks). (b) 320 × 9/16 = 180 plants (3 marks). (c) The two gene pairs assort independently and gametes combine at random, giving four phenotypes (2 marks).
8	Four potato strips of equal mass and length are placed in distilled water, 0.2 M, 0.5 M and 1.0 M sucrose solution for 30 minutes. In distilled water a strip increases from 5.0 g to 5.6 g. (a) Calculate the percentage change in mass. (b) Explain why the mass increases in distilled water. (c) Predict and explain the change in 1.0 M sucrose solution.	(a) (5.6 − 5.0) / 5.0 × 100% = 12% (3 marks). (b) Distilled water is more dilute than the cell sap, so water enters the cells by osmosis and they become turgid (3 marks). (c) The mass falls: the external solution is much more concentrated, water leaves the cells and they lose turgor (2 marks).

Section C: Plant Physiology and Applications

Answer all questions and support your explanations with biological principles or examples.

#	Question	Answer and marking notes
9	A man of blood group A (genotype I^A i) marries a woman of blood group B (genotype I^B i). (a) List the possible blood groups and genotypes of their children. (b) Give the probability of a child being group AB. (c) Explain how the ABO blood group shows multiple alleles and codominance.	(a) AB (I^A I^B), A (I^A i), B (I^B i) and O (ii) (4 marks). (b) Probability = 1/4 (2 marks). (c) There are three alleles I^A, I^B and i; I^A and I^B are both expressed in group AB, which is codominance (2 marks).
10	A student studies the effect of temperature on amylase activity, measured as starch digested per minute in arbitrary units: 0 °C = 5, 37 °C = 30, 60 °C = 8, 100 °C = 0. (a) State the optimum temperature. (b) Explain the fall in activity at 60 °C. (c) Explain why some activity remains at 0 °C.	(a) About 37 °C (2 marks). (b) High temperature denatures the enzyme, changing the shape of the active site so the substrate cannot bind (4 marks). (c) Low temperature only slows molecular movement; the enzyme is not denatured and activity returns on warming (2 marks).
11	A food chain in a lake is: phytoplankton → small shrimp → small fish → large fish. The DDT concentration in the water is 0.00005 ppm and that in the large fish is 5 ppm. (a) Calculate the biological magnification factor. (b) Explain why the DDT concentration rises along the food chain. (c) State one effect of this on humans.	(a) 5 / 0.00005 = 100 000 times (3 marks). (b) DDT is not easily broken down and is fat-soluble, so it accumulates in body tissue while biomass falls at each level (4 marks). (c) People eating the large fish may take in a high concentration of DDT, harming health (2 marks).
12	A development plan proposes to fill a wetland for housing. Discuss: (a) three functions of a wetland in an ecosystem; (b) two effects of the reclamation on biodiversity; (c) two sustainable measures that could reduce the impact.	(a) A wetland stores flood water and regulates flow, filters pollutants, and provides habitat that maintains biodiversity (3 marks). (b) Loss of habitat reduces species numbers, breaks food webs and isolates populations (any two, 4 marks). (c) Protect a core wetland reserve, carry out an environmental impact assessment, translocate affected species, create compensatory artificial wetlands (any two, 3 marks).

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