HKDSE Biology Paper 1B Practice Paper: Enzymes, Genetics, Ecology and Osmosis

2026-11-10

HKDSE Biology · Paper 1B

HKDSE Biology Paper 1B Practice Paper: Enzymes, Genetics, Ecology and Osmosis: 6 questions, 23 parts, 100 marks, 60 minutes.

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Question 1 Enzyme properties and activity

Equal volumes of an enzyme solution and a substrate solution were mixed. At 25 degrees Celsius, 60.0 micromol of product formed in 5 minutes. At 40 degrees Celsius the mean rate of product formation was 30.0 micromol per minute. At 60 degrees Celsius the rate fell to almost zero.

#	Question	Answer and marking notes
1	(a) Calculate the mean rate of product formation at 25 degrees Celsius.	60.0 / 5.0 = 12.0 micromol per minute.
2	(b) Calculate the ratio of the mean rate at 40 degrees Celsius to that at 25 degrees Celsius.	30.0 / 12.0 = 2.5, so the rate at 40 degrees Celsius is 2.5 times the rate at 25 degrees Celsius.
3	(c) Explain why the rate at 40 degrees Celsius is higher than at 25 degrees Celsius, yet almost zero at 60 degrees Celsius.	A higher temperature gives molecules more kinetic energy, so successful collisions with the active site become more frequent and the rate rises. Above the optimum the three-dimensional protein structure of the enzyme, including the active site, is disrupted, so the enzyme is denatured and almost no catalysis occurs at 60 degrees Celsius.
4	(d) Design a test to show that the enzyme is chemically a protein.	Add a protease to the enzyme solution and measure the activity again; a large fall in activity supports the conclusion that the enzyme is a protein. A biuret test may also be used, since a purple colour indicates the presence of protein.

Question 2 Monohybrid cross and chi-squared test

In a plant species a single pair of alleles controls flower colour, and the red allele (R) is completely dominant to the white allele (r). Two heterozygous red-flowered plants were crossed and produced 320 offspring, of which 68 had white flowers.

#	Question	Answer and marking notes
5	(a) State the genotypes of the two parents and the expected phenotypic ratio of the offspring.	Both parents are Rr, so the cross is Rr x Rr and the expected ratio is 3 red : 1 white.
6	(b) Calculate the expected numbers of white-flowered and red-flowered offspring.	Expected white = 320 / 4 = 80 plants; expected red = 320 – 80 = 240 plants.
7	(c) Use a chi-squared test to compare the observed and expected numbers and decide whether the 3 : 1 hypothesis is acceptable (1 degree of freedom, critical value 3.84).	X2 = (68 – 80)^2 / 80 + (252 – 240)^2 / 240 = 144 / 80 + 144 / 240 = 1.8 + 0.6 = 2.4. The calculated value 2.4 is smaller than the critical value 3.84, so the 3 : 1 hypothesis is accepted and the deviation is not significant.
8	(d) Give two reasons why genetic ratios often deviate from theoretical values.	The number of offspring is limited and fertilisation is random, which produces sampling error; in addition, some genotypes may affect survival (for example a homozygous dominant lethal allele) or mutation may occur, so the observed ratio departs from the expected ratio.

Question 3 Gas exchange in photosynthesis and respiration

A water plant released 4.8 cm3 of oxygen in 10 minutes in the light. When moved to the dark it absorbed 1.2 cm3 of oxygen in 10 minutes. The temperature was the same in both conditions.

#	Question	Answer and marking notes
9	(a) Calculate the net rate of oxygen release in the light.	4.8 / 10 = 0.48 cm3 per minute.
10	(b) Calculate the rate of oxygen uptake in the dark.	1.2 / 10 = 0.12 cm3 per minute.
11	(c) Calculate the true rate of photosynthesis.	True rate of photosynthesis = net release + respiration = 0.48 + 0.12 = 0.60 cm3 per minute.
12	(d) Explain why the direction of gas exchange in the dark is the opposite of that in the light.	In the dark photosynthesis stops and only respiration continues, so the cells take up oxygen and release carbon dioxide; oxygen therefore enters the plant and carbon dioxide leaves it, which is the reverse of the direction seen in the light.

Question 4 Energy flow and population estimation

In a grassland ecosystem the producers fix 50 000 kJ m-2 yr-1, primary consumers assimilate 5 000 kJ m-2 yr-1 and secondary consumers assimilate 400 kJ m-2 yr-1. The snail population was also estimated by mark-recapture: 80 snails were marked and released, and a second sample of 100 snails contained 16 marked individuals.

#	Question	Answer and marking notes
13	(a) Calculate the efficiency of energy transfer from producers to primary consumers.	5 000 / 50 000 x 100 = 10 per cent.
14	(b) Calculate the assimilation efficiency of secondary consumers relative to primary consumers and compare it with the previous level.	400 / 5 000 x 100 = 8 per cent, which is lower than the 10 per cent from producers to primary consumers.
15	(c) Estimate the size of the snail population.	Population size = 80 x 100 / 16 = 500 snails.
16	(d) State three assumptions of the mark-recapture method and explain the direction of bias if one assumption fails.	The assumptions are that marked and unmarked animals mix randomly, that there is no immigration or emigration, no birth or death during the study, and that marks are not lost and do not affect the chance of capture. If marked animals are caught more easily, the proportion of marked individuals in the second sample is too high, so the estimated population size is too low.

Question 5 Osmosis and water potential

Five potato strips of equal length were placed in 0.0, 0.2, 0.4, 0.6 and 0.8 mol dm-3 sucrose solutions for 30 minutes. The strip in 0.6 mol dm-3 sucrose changed in mass from 2.50 g to 2.425 g. The strip in 0.4 mol dm-3 sucrose changed in mass by -1.2 per cent.

#	Question	Answer and marking notes
17	(a) Calculate the percentage change in mass of the strip in 0.6 mol dm-3 sucrose.	(2.425 – 2.50) / 2.50 x 100 = -3.0 per cent.
18	(b) Use the results for the 0.4 and 0.6 mol dm-3 solutions to estimate the concentration at which the mass of the strip would not change.	The mass falls by 1.2 per cent at 0.4 mol dm-3 and by 3.0 per cent at 0.6 mol dm-3, so no change in mass would occur between about 0.45 and 0.50 mol dm-3, which is close to the concentration of the cell sap.
19	(c) Use water potential to explain the change in mass of the potato strips.	The more concentrated the solution, the lower its water potential. When the external solution has a lower water potential than the cell sap, water leaves the cells by osmosis, so the cells lose water and the mass of the strip falls.
20	(d) Predict the result for the strip in 0.0 mol dm-3 sucrose and explain it.	Pure water has the highest water potential, so water enters the cells by osmosis and the cells become turgid; the mass of the strip increases the most, and the cell walls prevent the cells from bursting.

Question 6 Bacterial growth and genetic engineering

The human insulin gene was inserted into a plasmid and introduced into bacteria. The number of bacteria was 100 at the start of culture and rose to 6 400 after some time, with each generation taking 20 minutes.

#	Question	Answer and marking notes
21	(a) Calculate the number of generations that the bacteria went through.	6 400 / 100 = 64 = 2 to the power 6, so 6 generations took place.
22	(b) Calculate the time taken for this culture period.	6 x 20 = 120 minutes, which is 2 hours.
23	(c) Explain why adding an antibiotic to the medium selects bacteria that have taken up the plasmid.	The recombinant plasmid carries an antibiotic resistance gene, so bacteria that have taken up the plasmid survive and multiply on medium containing the antibiotic. Bacteria without the plasmid have no resistance and are killed, so only the surviving colonies are candidates carrying the target gene.

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