GCSE Biology Higher Paper 3
Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 9 more.
Questions
Question 1 [4 marks]
Cell Structure and Transport
A student views a cheek cell using a light microscope.
The image of the cell measured on the photograph is 24 mm wide.
The actual width of the cheek cell is 60 micrometres.
Calculate the magnification of the image. Give your answer as a whole number.
Question 2 [4 marks]
Hormones and Homeostasis
A patient's blood glucose concentration was 5.5 mmol/l at 08:00 and rose to 9.5 mmol/l at 08:30 after breakfast.
Calculate the mean rate of increase in blood glucose concentration, in mmol/l per minute, over this time.
Question 3 [4 marks]
Cell Division and Stem Cells
A single cell divides by mitosis every 20 minutes.
Starting from one cell, calculate how many cells will be present after 2 hours, assuming every cell divides on schedule.
Question 4 [4 marks]
Digestion and the Circulatory System
In an investigation of lipase activity, 12 cm^3 of gas was produced in 4 minutes.
Calculate the mean rate of gas production, in cm^3 per minute, then calculate how long it would take to produce 45 cm^3 of gas at this rate.
Question 5 [5 marks]
Pathogens and Communicable Disease
At the start of an outbreak, one person is infected. Each infected person then infects 3 new people before recovering, and this repeats for further rounds of infection.
Calculate the total number of people who have been infected in total after 3 further rounds of infection beyond the original case.
Question 6 [5 marks]
Genetics and Inheritance
In a large sample of 240 pea plants grown from a cross expected to give a 1:2:1 ratio of genotypes, calculate how many plants would be expected of each genotype.
Question 7 [5 marks]
The Nervous System and Reflexes
Nerve impulses must cross a synapse (gap) to pass from one neurone to the next.
Explain how a nerve impulse crosses a synapse.
Question 8 [5 marks]
Human Defence Systems and Treating Disease
Explain why new medical drugs must be tested and trialled before they are prescribed to patients, describing the main stages involved.
Question 9 [6 marks]
Ecosystems and Material Cycles
In a food chain of producer -> primary consumer -> secondary consumer -> tertiary consumer, only 10% of energy is transferred between each trophic level. A population of tertiary consumers (eagles) contains 80 kJ of energy stored as biomass.
Calculate the energy that must have been available at the secondary consumer, primary consumer and producer levels to support this, and explain why an ecosystem can typically support far fewer tertiary consumers than producers.
Question 10 [6 marks]
Respiration and Exercise
Aerobic respiration of one mole of glucose releases about 2900 kJ of energy. Anaerobic respiration (producing lactic acid) of one mole of glucose releases only about 120 kJ of energy.
Calculate how many times more energy is released by aerobic respiration than anaerobic respiration of the same amount of glucose, and explain why muscle cells still respire anaerobically during vigorous exercise despite this being far less efficient.
Question 11 [6 marks]
Evolution and Natural Selection
Two populations of the same species become separated by a new river, and cannot interbreed for many thousands of years.
Explain how this separation could eventually lead to the formation of two separate species, and explain how scientists could use DNA evidence to confirm that speciation has occurred.
Question 12 [6 marks]
Photosynthesis
A student uses the inverse square law to investigate how distance from a lamp affects light intensity reaching pondweed.
Light intensity is proportional to 1/distance^2.
At a distance of 20 cm, the light intensity is 4 arbitrary units.
Calculate the light intensity at a distance of 40 cm from the lamp, and explain what this means for the expected rate of photosynthesis at 40 cm compared with 20 cm.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| converting the actual width to the same units as the image: 60 micrometres = 0.06 mm | M1 |
| using magnification = image size / actual size | M1 |
| substituting 24 / 0.06 | M1 |
| magnification = x400 | A1 |
| Final answer: x400 | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the change in concentration = 9.5 - 5.5 = 4 mmol/l | M1 |
| identifying the time taken = 30 minutes | M1 |
| using rate = change / time | M1 |
| a rate of 4/30 = 0.13 mmol/l per minute (2 significant figures) | A1 |
| Final answer: 0.13 mmol/l per minute | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| recognising 2 hours contains 6 divisions (120 / 20) | M1 |
| using the pattern that the cell number doubles at each division | M1 |
| calculating 2^6 | M1 |
| 64 cells | A1 |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| using rate = volume / time | M1 |
| substituting 12 / 4 | M1 |
| 3 cm^3 per minute | A1 |
| time = 45 / 3 = 15 minutes | A1independent |
| Final answer: Rate = 3 cm^3 per minute; it would take 15 minutes to produce 45 cm^3 of gas | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| round 1 = 1 x 3 = 3 newly infected | M1 |
| round 2 = 3 x 3 = 9 newly infected | M1 |
| round 3 = 9 x 3 = 27 newly infected | M1 |
| the total number infected = 1 + 3 + 9 + 27 = 40 | A1 |
| identifying this pattern as exponential growth, which is why an outbreak can spread rapidly if unchecked | B1independent |
| Final answer: 40 people in total have been infected after 3 further rounds; this is an example of exponential growth | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the ratio 1:2:1 represents 4 total parts | M1 |
| finding the value of one part = 240 / 4 = 60 | M1 |
| 60 plants of each of the two homozygous genotypes | A1 |
| 120 plants of the heterozygous genotype | A1 |
| checking the total, 60 + 120 + 60 = 240, matches the sample size | B1independent |
| Final answer: 60 plants of each of the two homozygous genotypes and 120 plants of the heterozygous genotype would be expected (60 + 120 + 60 = 240) | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| an electrical impulse arrives at the end of the first (presynaptic) neurone | B1 |
| this triggers the release of a chemical called a neurotransmitter | B1 |
| the neurotransmitter diffuses across the synapse between the two neurones | B1 |
| the neurotransmitter binds to specific receptor molecules on the membrane of the next (postsynaptic) neurone | B1 |
| this triggers a new electrical impulse in the second neurone | B1 |
| Final answer: An electrical impulse causes neurotransmitter release, which diffuses across the synapse and binds to receptors on the next neurone, triggering a new electrical impulse | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| preclinical testing is carried out in a laboratory, using cells, tissues or live animals | B1 |
| this tests for toxicity, efficacy and the correct dosage | B1 |
| clinical trials then test the drug on healthy volunteers, at a very low dose, to check for harmful side effects | B1 |
| if safe, the drug is trialled on a small number of patients to find the optimum dose | B1 |
| large-scale trials on patients then compare the new drug (or a placebo) against an existing treatment, often using a double-blind design, to test effectiveness | B1 |
| Final answer: Drugs are tested in the lab (preclinical), then on healthy volunteers and small patient groups to check safety and dose, then in large double-blind clinical trials to confirm effectiveness before being prescribed | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| using previous level's energy = current level's energy / 10% | M1 |
| secondary consumer = 80 / 0.1 | M1 |
| 800 kJ | A1 |
| primary consumer = 800 / 0.1 = 8000 kJ | M1 |
| producer = 8000 / 0.1 = 80,000 kJ | A1 |
| because only about 10% of energy is transferred at each level, an enormous amount of energy is needed at the producer level to support even a small amount of biomass at the top of the food chain, so an ecosystem can support far more producers than tertiary consumers | B1 |
| Final answer: 800 kJ would be needed at the secondary consumer level, 8000 kJ at the primary consumer level, and 80,000 kJ at the producer level; because only about 10% of energy passes on at each level, a huge amount of producer biomass is needed to support even a little biomass of top consumers | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| using ratio = 2900 / 120 | M1 |
| approximately 24 times more energy | A1 |
| during vigorous exercise, the heart and lungs cannot supply oxygen to the muscles fast enough to meet the high demand for aerobic respiration | B1 |
| anaerobic respiration not requiring oxygen, so it can continue to release some energy even when oxygen supply is insufficient | B1 |
| this allowing muscle contraction (and exercise) to continue, even though anaerobic respiration releases far less energy per glucose molecule | B1 |
| the resulting build-up of lactic acid needing to be broken down using oxygen later (an oxygen debt repaid during recovery) | B1 |
| Final answer: Aerobic respiration releases about 24 times more energy than anaerobic respiration from the same glucose; muscles still respire anaerobically during vigorous exercise because oxygen cannot be supplied fast enough, and this lets contraction continue despite being far less efficient, though it creates an oxygen debt that must later be repaid | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| the two populations are geographically isolated and cannot interbreed | B1 |
| each population experiences different environmental conditions or independently accumulates different mutations | B1 |
| natural selection acts differently on each population, favouring different characteristics in each | B1 |
| over many generations the two populations become increasingly genetically different | B1 |
| eventually, the two populations become so different that they can no longer interbreed to produce fertile offspring, meaning they have become separate species | B1 |
| scientists can compare DNA sequences between the two populations, and a large enough difference confirms they can no longer be classified as the same species | B1 |
| Final answer: Geographic isolation stops interbreeding, so each population evolves separately by natural selection until they can no longer produce fertile offspring together, forming two species; comparing their DNA sequences can confirm how different they have become | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| using intensity is proportional to 1/distance^2 | M1 |
| finding the constant of proportionality, k = intensity x distance^2 = 4 x 20^2 = 1600 | M1 |
| substituting distance = 40 into intensity = k/distance^2 | M1 |
| intensity = 1600/1600 = 1 arbitrary unit | A1 |
| recognising light intensity at 40 cm is a quarter of that at 20 cm | B1 |
| explaining the rate of photosynthesis would be expected to be lower at 40 cm because light intensity is a limiting factor and is much reduced | B1 |
| Final answer: Light intensity at 40 cm = 1 arbitrary unit (a quarter of that at 20 cm), so the rate of photosynthesis would be expected to be lower at 40 cm | |