GCSE Biology Higher Paper 4
Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 9 more.
Questions
Question 1 [4 marks]
Evolution and Natural Selection
In a population of 5000 ladybirds, 800 have a rare colour pattern that camouflages them well against a newly introduced plant.
After several generations, the number of ladybirds with this colour pattern rises to 3500 in a population that has grown to 8000. Calculate the percentage of the population with this colour pattern before and after this change.
Question 2 [4 marks]
Genetics and Inheritance
In snapdragon plants, flower colour shows codominance. The allele for red flowers (R) and the allele for white flowers (W) are both expressed if present together, producing pink flowers.
A red-flowered plant (RR) is crossed with a white-flowered plant (WW). Use a genetic diagram to determine the genotype and phenotype of all the offspring, and explain why this phenotype appears.
Question 3 [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 4 [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 5 [5 marks]
Photosynthesis
A grower measures that one tomato plant produces 40 g of new dry biomass per week under standard greenhouse conditions. Under improved lighting, this rises to 58 g per week.
Calculate the percentage increase in biomass production per plant, and calculate the total extra biomass produced per week across a greenhouse of 250 plants, assuming every plant shows the same increase.
Question 6 [5 marks]
Human Defence Systems and Treating Disease
In a trial, 3% of 2000 vaccinated people caught a disease, compared with 15% of 2000 unvaccinated people.
Calculate the number of people who caught the disease in each group, and calculate the vaccine's efficacy using efficacy (%) = (1 - (rate in vaccinated group / rate in unvaccinated group)) x 100.
Question 7 [5 marks]
Digestion and the Circulatory System
A student tested the activity of pepsin (a protease enzyme found in the stomach) at pH 2 and at pH 8, and found it broke down protein much faster at pH 2.
Explain, in terms of enzyme structure, why pepsin is much less active at pH 8.
Question 8 [5 marks]
Ecosystems and Material Cycles
A student used a transect line across a rocky shore and placed a 0.5 m^2 quadrat every 2 metres along the line. In one quadrat, seaweed covered an estimated 60% of the quadrat's area.
Calculate the area, in m^2, of the quadrat that was covered by seaweed. If the whole study area along the shore covers 400 m^2 and this percentage cover is representative of the whole area, estimate the total area of the shore covered by seaweed.
Question 9 [6 marks]
The Nervous System and Reflexes
A voluntary action, such as deciding to pick up a cup, involves conscious thought, while a reflex action, such as pulling a hand away from something hot, does not.
Explain the difference in the neural pathway between a voluntary action and a reflex action, and explain why this difference makes reflex actions faster.
Question 10 [6 marks]
Pathogens and Communicable Disease
Before a hand-washing campaign in a hospital, 40 out of 800 patients developed a healthcare-associated infection in one month. After the campaign, 12 out of 750 patients developed an infection in one month.
Calculate the percentage of patients infected before and after the campaign, and explain, in terms of pathogen transmission, why improved hand hygiene reduces the spread of infection.
Question 11 [6 marks]
Cell Structure and Transport
The rate of diffusion across an exchange surface is proportional to its surface area and inversely proportional to its thickness.
Membrane A has a surface area of 20 mm^2 and a thickness of 0.04 mm. Membrane B has a surface area of 15 mm^2 and a thickness of 0.05 mm.
Calculate the ratio of surface area to thickness for each membrane, then determine which membrane would allow a faster rate of diffusion, explaining your reasoning.
Question 12 [6 marks]
Respiration and Exercise
A trained athlete and an untrained person perform the same short bout of vigorous exercise. The untrained person builds up an oxygen debt of 8 litres and takes 25 minutes to recover. The trained athlete builds up an oxygen debt of only 5 litres and recovers in 10 minutes.
Calculate each person's mean rate of recovery (oxygen consumed above resting level, in litres per minute), and explain why the trained athlete's oxygen debt is smaller for the same exercise.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| before percentage = (800 / 5000) x 100 | M1 |
| 16% | A1 |
| after percentage = (3500 / 8000) x 100 | M1 |
| 43.75% | A1 |
| Final answer: 16% of the population had the colour pattern before, rising to 43.75% afterwards | |
| Question 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| a correctly completed genetic diagram (Punnett square) crossing RR with WW | M1 |
| all offspring having the genotype RW | A1 |
| all offspring having a pink phenotype | A1 |
| explaining that because R and W are codominant, both alleles are expressed together in a heterozygous plant, producing a blended pink phenotype rather than a plant being purely red or white | B1 |
| Final answer: All offspring have the genotype RW and a pink phenotype, because in codominance both the red and white alleles are expressed together | |
| Question 3[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 4[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 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| the increase = 58 - 40 = 18 g | M1 |
| percentage increase = (18 / 40) x 100 | M1 |
| 45% | A1 |
| total extra biomass = 18 x 250 | M1 |
| 4500 g | A1 |
| Final answer: A 45% increase per plant; across 250 plants this is an extra 4500 g of biomass per week | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| vaccinated cases = 3% of 2000 = 60 | M1 |
| unvaccinated cases = 15% of 2000 = 300 | M1 |
| using efficacy = (1 - (3 / 15)) x 100 | M1 |
| substituting to get (1 - 0.2) x 100 | A1 |
| 80% efficacy | A1 |
| Final answer: 60 people caught the disease in the vaccinated group and 300 in the unvaccinated group; the vaccine's efficacy is 80% | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| enzymes have an active site with a specific shape that is complementary to their substrate | B1 |
| pepsin's active site is adapted to work best (has an optimum) at a low, acidic pH, such as pH 2, matching conditions in the stomach | B1 |
| at a pH far from the optimum, such as pH 8, bonds holding the enzyme's structure together are disrupted | B1 |
| this changes the shape of the enzyme's active site | B1 |
| the substrate (protein) can no longer fit (bind to) the active site, so the enzyme cannot catalyse the reaction and appears much less active | B1 |
| Final answer: Pepsin's active site has an optimum at a low pH; at pH 8, far from this optimum, bonds holding its shape are disrupted, changing the active site's shape so the substrate no longer fits, greatly reducing its activity | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| area covered in the quadrat = 60% of 0.5 m^2 | M1 |
| 0.3 m^2 | A1 |
| using the same percentage (60%) applied to the total area | M1 |
| substituting 60% of 400 m^2 | M1 |
| 240 m^2 | A1 |
| Final answer: 0.3 m^2 of the quadrat was covered by seaweed; an estimated 240 m^2 of the total 400 m^2 shore is covered by seaweed | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| in a voluntary action the impulse from the sensory neurone travels to the brain | B1 |
| the brain processes the information, involves conscious decision-making, and sends an impulse down a motor neurone to the effector | B1 |
| in a reflex action the impulse from the sensory neurone passes to a relay neurone in the spinal cord, not the brain | B1 |
| the relay neurone passes the impulse directly to a motor neurone, which carries it to the effector | B1 |
| the reflex pathway not needing to pass to the brain and not involving conscious decision-making, so it involves fewer synapses and a shorter distance | B1 |
| impulses taking a short but finite time to cross each synapse, so a shorter pathway with fewer synapses and no processing time in the brain results in a much faster response | B1 |
| Final answer: A voluntary action routes through the brain for conscious processing before a response; a reflex bypasses the brain via a relay neurone in the spinal cord, giving a shorter pathway with fewer synapses and no decision-making time, making it faster | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| before percentage = (40 / 800) x 100 | M1 |
| 5% | A1 |
| after percentage = (12 / 750) x 100 | M1 |
| 1.6% | A1 |
| pathogens can be transferred from a contaminated hand to another person, or to a surface then another person, causing infection | B1 |
| washing hands removes/kills pathogens, reducing the chance of pathogens being transferred between patients (or between staff and patients) | B1 |
| Final answer: Infection rate fell from 5% to 1.6%; hand hygiene reduces spread because pathogens are transferred between people (or via surfaces) on contaminated hands, and washing removes or kills them before they can be passed on | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| using ratio = surface area / thickness | M1 |
| substituting membrane A's values, 20 / 0.04 | M1 |
| membrane A ratio = 500 | A1 |
| substituting membrane B's values, 15 / 0.05 | M1 |
| membrane B ratio = 300 | A1 |
| identifying membrane A as giving the faster rate of diffusion, because it has the larger surface area to thickness ratio, and a greater ratio increases the rate of diffusion | B1 |
| Final answer: Membrane A ratio = 500, membrane B ratio = 300; membrane A would allow the faster rate of diffusion because it has the greater surface area to thickness ratio | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| untrained rate = 8 / 25 | M1 |
| 0.32 litres per minute | A1 |
| trained rate = 5 / 10 | M1 |
| 0.5 litres per minute | A1 |
| regular training increasing the number of mitochondria and capillaries in muscle cells, improving oxygen delivery and the capacity for aerobic respiration | B1 |
| this meaning the trained athlete's muscles can respire aerobically for longer during the exercise before needing to respire anaerobically, producing less lactic acid and so a smaller oxygen debt | B1 |
| Final answer: The untrained person recovers at 0.32 litres per minute and the trained athlete at 0.5 litres per minute; the athlete's smaller oxygen debt is because training has increased their muscles' mitochondria and capillaries, letting them respire aerobically for longer before needing anaerobic respiration | |