Secondary 5 Biology Practice Paper: Cell Division And Inheritance, Molecular Genetics And Gene Expression, Ecosystem Energy Flow And Agricultural Productivity, Population Dynamics And Biodiversity, Health And Circulation Data, Biotechnology And Applications (100 Marks)

Secondary 5 Biology Practice Paper: Cell Division and Inheritance, Molecular Genetics and Gene Expression, Ecosystem Energy Flow and Agricultural Productivity, Population Dynamics and Biodiversity, Health and Circulation Data, Biotechnology and Applications (100 marks)

2026-11-14

Biology – Secondary 5 – Practice Paper

Secondary 5 Biology practice paper with 24 parts and worked answers on cell doubling, monohybrid and dihybrid ratios, carrier inheritance, base composition, codons,

Paper details　Secondary 5 Biology practice paper: 24 parts with worked answers on genetics ratios, energy flow, population growth, health data and biotechnology calculations.

Section A: Cell Division and Inheritance

Show the working and give genetic ratios as fractions or percentages.

#	Question	Answer and marking notes
1	A tissue starts with 8 cells and each mitotic division takes 30 minutes. Find the number of cells after 4 hours and the increase, assuming no cell death.	Answer: divisions = 4 x 60 / 30 = 8; cells = 8 x 2^8 = 2,048; increase = 2,048 – 8 = 2,040 cells.
2	Two heterozygous tall pea plants (Tt) are crossed and 480 seeds germinate. Find the expected numbers of tall and short plants and the number with genotype TT.	Answer: Tt x Tt gives 3 tall to 1 short; tall = 480 x 3/4 = 360; short = 480 x 1/4 = 120; TT = 480 x 1/4 = 120 plants.
3	Two round yellow peas (RrYy) are crossed and 320 offspring are obtained. Find the expected numbers of the four phenotypes and state the ratio.	Answer: the ratio is 9:3:3:1; round yellow = 320 x 9/16 = 180; round green = 320 x 3/16 = 60; wrinkled yellow = 320 x 3/16 = 60; wrinkled green = 320 x 1/16 = 20, giving 320 plants in total.
4	A heterozygous carrier (Aa) has children with a partner who carries no disease allele, the normal phenotype being aa. For 500 children, find the expected numbers with the disease phenotype and with the normal phenotype but carrying the allele, and state whether a homozygous affected child is possible.	Answer: Aa x aa gives half Aa and half aa; affected children = 500 x 1/2 = 250; unaffected carriers = 250. A homozygous affected child (AA) is impossible because the partner supplies only the a allele.

Section B: Molecular Genetics and Gene Expression

Show the working and the units; base numbers are whole numbers.

#	Question	Answer and marking notes
5	A double-stranded DNA molecule is 1,200 base pairs long and adenine (A) makes up 30 per cent of all bases. Find the number of each of the four bases.	Answer: total bases = 1,200 x 2 = 2,400; A = 2,400 x 30 per cent = 720; T pairs with A, so T = 720; G + C = 2,400 – 720 – 720 = 960, hence G = C = 480.
6	A messenger RNA has 363 bases and one of its codons is a stop codon. Find the number of codons, the number of amino acids coded for and the number of peptide bonds formed.	Answer: codons = 363 / 3 = 121; amino acids = 121 – 1 = 120; peptide bonds = 120 – 1 = 119.
7	A gene contains 210 base pairs and is replicated 3 times. Find the number of DNA molecules produced and the number of free nucleotides incorporated.	Answer: molecules = 2^3 = 8; each molecule has 210 x 2 = 420 nucleotides; nucleotides added = 420 x (8 – 1) = 2,940.
8	A polypeptide contains 45 amino acids. Find the minimum number of mRNA bases needed to code for it, counting the stop codon.	Answer: each amino acid needs 3 bases and the stop codon needs another 3; minimum bases = 45 x 3 + 3 = 138.

Section C: Ecosystem Energy Flow and Agricultural Productivity

Energy is in kJ m per square metre per year; give percentages to 1 decimal place.

#	Question	Answer and marking notes
9	Gross primary production in an ecosystem is 88,000 kJ per square metre per year and producers respire 52,000 kJ per square metre per year. Find net primary production and state where the rest of the energy goes.	Answer: net primary production = 88,000 – 52,000 = 36,000 kJ per square metre per year; the respired part is lost as heat while the remainder is stored as biomass available to primary consumers.
10	Producers in a food chain fix 250,000 kJ and only 10 per cent of the energy passes to the next trophic level. Find the energy reaching the primary, secondary and tertiary consumers and the percentage of energy lost between producers and tertiary consumers.	Answer: primary = 250,000 x 10 per cent = 25,000 kJ; secondary = 2,500 kJ; tertiary = 250 kJ; loss = (250,000 – 250) / 250,000 x 100 per cent = 99.9 per cent.
11	A rice field produces 12,000 kg of total biomass per hectare, of which 5,400 kg is grain. Find the harvest index and explain what it means for food production.	Answer: harvest index = 5,400 / 12,000 x 100 per cent = 45 per cent; a higher index means more photosynthate goes to the harvestable grain, so food yield is relatively greater.
12	If a tertiary consumer is to obtain 400 kJ and each trophic level transfers 10 per cent, find the energy the producers must fix, and explain why food chains rarely exceed four or five trophic levels.	Answer: producers = 400 / 10 per cent / 10 per cent / 10 per cent = 400,000 kJ; each step keeps only one tenth, so too little energy remains to support a viable population at higher levels, which limits food-chain length.

Section D: Population Dynamics and Biodiversity

Round to whole numbers or the decimal places asked; give the index to 2 decimal places.

#	Question	Answer and marking notes
13	A bird population now numbers 200 and grows at 12 per cent a year. Find the population after 10 years, to the nearest whole number.	Answer: after 10 years = 200 x 1.12^10 = 200 x 3.1058, which is about 621 birds.
14	Three species in a wood have 40, 30 and 30 individuals. Using the Simpson index D = 1 – sum of (n/N) squared, find D to 2 decimal places and explain its meaning.	Answer: N = 100; the sum of squared proportions is 0.16 + 0.09 + 0.09 = 0.34; D = 1 – 0.34 = 0.66. A value nearer 1 means individuals are more evenly spread and diversity is higher.
15	An insect lays 1,000 eggs, only 200 hatch and only 40 reach adulthood. Find the hatching rate, the survival of adults and the overall survival from egg to adult.	Answer: hatching = 200 / 1,000 x 100 per cent = 20 per cent; survival of the hatched to adult = 40 / 200 x 100 per cent = 20 per cent; overall = 40 / 1,000 x 100 per cent = 4 per cent.
16	A population has a carrying capacity of 5,000, stands at 800 and has an intrinsic growth rate r of 0.35. Use dN/dt = rN(1 – N/K) to find the current growth rate and explain how growth changes as the population nears capacity.	Answer: 1 – N/K = 1 – 800/5,000 = 0.84; dN/dt = 0.35 x 800 x 0.84 = 235.2 individuals a year. As N approaches K the bracket shrinks, growth slows and becomes zero at K.

Section E: Health, Breathing and Circulation Data

Write the formula, substitute the values and give the units.

#	Question	Answer and marking notes
17	A person is 1.60 m tall and weighs 68 kg. Find the body mass index to 1 decimal place and state the category under the adult cut-off of 25 or above for overweight.	Answer: body mass index = 68 / 1.60 squared = 68 / 2.56 = 26.6, which is 25 or above and therefore in the overweight category.
18	Each ventricular contraction ejects 70 mL of blood and the heart rate is 72 beats a minute. Find the cardiac output in mL per minute and in litres per minute.	Answer: cardiac output = 70 x 72 = 5,040 mL per minute = 5.04 litres per minute.
19	A person has a tidal volume of 500 mL, breathes 15 times a minute and has an anatomical dead space of 150 mL. Find the minute ventilation and the alveolar ventilation.	Answer: minute ventilation = 500 x 15 = 7,500 mL per minute; alveolar ventilation = (500 – 150) x 15 = 5,250 mL per minute, the difference of 2,250 mL staying in the dead space.
20	A vaccine is 90 per cent effective and 85 per cent of a community is vaccinated. Find the percentage of the population effectively protected; if the basic reproduction number is 3.0, find the herd-immunity threshold 1 – 1/R0 and judge whether the community has reached it.	Answer: effective protection = 90 per cent x 85 per cent = 76.5 per cent; threshold = 1 – 1/3.0 = 66.7 per cent. Since 76.5 per cent exceeds 66.7 per cent the threshold is met, though protection wanes over time and must be monitored.

Section F: Biotechnology and Applications

Give the units asked and use pi = 3.14 for areas.

#	Question	Answer and marking notes
21	In gene cloning bacteria divide every 20 minutes. Starting with 500 cells, how many minutes and hours are needed to reach 32,000?	Answer: the factor is 32,000 / 500 = 64 = 2^6, that is 6 divisions; time = 6 x 20 = 120 minutes, which is 2 hours.
22	A circular plasmid is 4,800 base pairs long; one restriction enzyme cuts at a single site and a second enzyme cuts 1,200 base pairs away. Find the number of fragments and the length of each.	Answer: two cuts in a circular molecule give 2 fragments; one is 1,200 base pairs and the other is 4,800 – 1,200 = 3,600 base pairs.
23	Yeast ferments 100 g of glucose to ethanol and carbon dioxide (C6H12O6 to 2 C2H5OH + 2 CO2, with molar masses 180, 46 and 44). Find the theoretical yields of ethanol and carbon dioxide.	Answer: 180 g of glucose gives 92 g of ethanol and 88 g of carbon dioxide; 100 g therefore gives ethanol = 92 x 100 / 180 = 51.11 g and carbon dioxide = 88 x 100 / 180 = 48.89 g, a total of 100 g.
24	In a sensitivity test an antibiotic gives an inhibition zone 24 mm across while the control disc shows 16 mm. Find both areas and the extra area of the zone, using pi = 3.14 to 2 decimal places, and explain why measuring diameter rather than radius affects the comparison.	Answer: the radii are 12 and 8 mm; areas are 3.14 x 144 = 452.16 square mm and 3.14 x 64 = 200.96 square mm; the extra area is 251.20 square mm. Diameter relates to area as a square, so comparing diameters alone understates the real difference between the zones.

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