S5 Biology Paper: Cell Cycle And Mitosis, Mendelian Genetics, Energy Flow, Photosynthesis, Fermentation And Population Estimation

S5 Biology Paper: Cell Cycle and Mitosis, Mendelian Genetics, Energy Flow, Photosynthesis, Fermentation and Population Estimation

2026-11-13

S5 Biology – cells, genetics and ecology

S5 Biology paper: 6 questions, 100 marks, 60 minutes on the cell cycle, monohybrid inheritance, energy flow, photosynthesis, fermentation and population estimation.

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Paper details　S5 Biology practice paper: 6 questions on the cell cycle, monohybrid inheritance, energy flow, photosynthesis, fermentation and population estimation, with answers.

Section A: The Cell Cycle and Mitosis (16 marks)

A cell in an onion root tip has a cycle made of G1 = 9 hours, S = 6 hours, G2 = 3 hours and M = 2 hours.

#	Question	Answer and marking notes
1	Find the total length of the cycle and the hours spent in interphase (G1, S and G2 together).	Answer: total = 9+6+3+2 = 20 hours; interphase = 9+6+3 = 18 hours. (Marking: total 2, interphase 2)
2	What percentage of the whole cycle is spent in S phase?	Answer: 6 / 20 x 100% = 30.0%. (Marking: correct ratio 2, percentage 2)
3	A cell completes one mitosis every 20 hours. Starting from one cell, how many cells are present after 60 hours?	Answer: 60 / 20 = 3 divisions; cells = 2^3 = 8. (Marking: number of divisions 2, power of two 1, answer 1)
4	Explain why S phase is essential for mitosis.	Answer: DNA is replicated in S phase so each chromosome forms two identical sister chromatids; without this step the daughter cells would receive half the genetic material and the chromosome number could not be maintained. (Marking: DNA replication 2, sister chromatids 1, maintaining chromosome number 1)

Section B: Mendelian Monohybrid Inheritance (16 marks)

In peas the tall allele D is dominant to the dwarf allele d. Two heterozygous tall plants (Dd x Dd) produce 1200 offspring, of which 888 are tall and 312 are dwarf.

#	Question	Answer and marking notes
5	State the genotypic ratio and the phenotypic ratio of this cross.	Answer: genotypic ratio = 1 DD : 2 Dd : 1 dd; phenotypic ratio = 3 tall : 1 dwarf. (Marking: genotypic ratio 2, phenotypic ratio 2)
6	Using the expected 3 : 1 ratio, calculate the expected numbers of tall and dwarf offspring among 1200 plants.	Answer: tall = 1200 x 3/4 = 900; dwarf = 1200 x 1/4 = 300. (Marking: tall 2, dwarf 2)
7	Calculate the observed ratio of tall to dwarf plants (3 decimal places) and state how it differs from the expected value of 3.	Answer: 888 / 312 = 2.846; slightly below the expected 3 by about 0.154, which is random sampling variation and still close to 3 : 1. (Marking: ratio 2, difference 1, random variation 1)
8	A tall offspring is chosen at random. What is the probability that it is heterozygous? Explain.	Answer: probability = 2/3 = 0.667; among tall offspring DD : Dd = 1 : 2, so two thirds are heterozygous. (Marking: fraction 2, reason 2)

Section C: Energy Flow along a Food Chain (17 marks)

In a grassland, producers fix 60 000 kJ per square metre each year. Primary consumers obtain 12% of it, and each later trophic level transfers 10% of the energy it receives.

#	Question	Answer and marking notes
9	How much energy per square metre per year is available to primary consumers?	Answer: 60 000 x 12% = 7,200 kJ. (Marking: percentage 2, unit and answer 2)
10	Calculate the energy actually lost between the producer and primary consumer levels and its percentage of the producer energy.	Answer: loss = 60 000 – 7,200 = 52,800 kJ; percentage = 52,800 / 60 000 x 100% = 88.0%. (Marking: loss 2, percentage 2)
11	How much energy per square metre per year reaches the tertiary consumers?	Answer: 7,200 x 10% = 720 kJ for secondary consumers, then x 10% = 72 kJ for tertiary consumers. (Marking: secondary 2, tertiary 2)
12	Explain why food chains rarely have more than four or five trophic levels.	Answer: only about one tenth of the energy passes to the next level, the rest being lost as heat from respiration, in excreta and in uneaten parts; by the fifth level too little energy remains to support a viable population, so food chains are short. (Marking: energy decrease 2, loss pathways 2, insufficient for a population 1)

Section D: Limiting Factors in Photosynthesis (17 marks)

Pondweed is lit by a lamp: at a distance of 10 cm it releases 32 bubbles per minute. With temperature and carbon dioxide concentration unchanged, the rate is inversely proportional to the square of the lamp distance.

#	Question	Answer and marking notes
13	How many bubbles per minute are produced when the lamp is 20 cm away?	Answer: doubling the distance quarters the rate: 32 / 4 = 8 bubbles per minute. (Marking: inverse square relation 2, answer 2)
14	How many bubbles per minute are produced when the lamp is 40 cm away?	Answer: the distance is four times larger, so the rate falls to 1/16: 32 / 16 = 2 bubbles per minute. (Marking: ratio 2, answer 2)
15	What is the percentage decrease in the rate when the lamp moves from 10 cm to 20 cm?	Answer: (32 – 8) / 32 x 100% = 75.0%. (Marking: difference 2, percentage 2)
16	Explain why the bubble count stops rising steeply when the lamp is very close to the pondweed.	Answer: light intensity is no longer limiting; carbon dioxide concentration or temperature becomes limiting. Extra light cannot be used, so the rate levels off, and an excessive rise in temperature may denature enzymes and lower the rate. (Marking: light saturation 2, new limiting factor 2, plateau or fall 1)

Section E: Mole Calculations for Yeast Fermentation (17 marks)

In the absence of oxygen yeast ferments glucose: C6H12O6 -> 2 C2H5OH + 2 CO2. 360 g of glucose is fermented completely (C = 12, H = 1, O = 16; molar volume of gas at room conditions = 24 dm3).

#	Question	Answer and marking notes
17	Calculate the number of moles in 360 g of glucose.	Answer: molar mass = 180 g/mol; moles = 360 / 180 = 2 mol. (Marking: molar mass 2, moles 2)
18	Calculate the mass of ethanol produced by complete fermentation.	Answer: moles of ethanol = 2 x 2 = 4 mol; mass = 4 x 46 = 184 g. (Marking: stoichiometric ratio 2, moles 1, mass 1)
19	Calculate the volume of carbon dioxide released at room conditions in dm3.	Answer: moles of carbon dioxide = 4; volume = 4 x 24 = 96 dm3. (Marking: moles 2, volume 2)
20	Explain why the same mass of glucose releases far more energy in aerobic respiration than in fermentation.	Answer: in fermentation glucose is only partly broken down to ethanol, so carbon and chemical energy remain in the ethanol. Aerobic respiration oxidises glucose completely to carbon dioxide and water, and the electron transport chain yields far more ATP. (Marking: partial breakdown 2, carbon retained 1, complete oxidation and electron transport chain 2)

Section F: Population Estimation (17 marks)

Twenty quadrats of 0.5 m2 each are used in a 5000 m2 meadow and give a mean of 3.5 plants per quadrat. Insects are estimated by mark-recapture: 200 are caught and marked first, then 250 are caught again, of which 40 carry a mark.

#	Question	Answer and marking notes
21	Estimate the total number of plants in the meadow by quadrat sampling.	Answer: density = 3.5 / 0.5 = 7 plants per m2; total = 7 x 5000 = 35,000 plants. (Marking: density 2, total 2)
22	Estimate the size of the insect population by the mark-recapture method.	Answer: (200 x 250) / 40 = 1,250 insects. (Marking: formula 2, substitution and answer 2)
23	Find the density of the insect population in the meadow in insects per m2.	Answer: 1,250 / 5000 = 0.25 insects per m2. (Marking: division 2, unit and answer 2)
24	State two assumptions required by the mark-recapture method.	Answer: (1) marked and unmarked animals have an equal chance of being caught the second time, so marking does not affect survival or movement; (2) the population is closed, with no major births, deaths, immigration or emigration between the two samples, and marks are not lost. (Marking: 2 marks per assumption, 1 mark for mixing or mark retention)

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