Higher Tier - Grades 4-9

GCSE Biology Higher Paper 4

Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 9 more.

12 questions - 60 marks - calculator allowed

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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

Mark scheme for Question 1 [4 marks]
Question 1[4 marks]
Answer or workingMarks
before percentage = (800 / 5000) x 100M1
16%A1
after percentage = (3500 / 8000) x 100M1
43.75%A1
Final answer: 16% of the population had the colour pattern before, rising to 43.75% afterwards
Mark scheme for Question 2 [4 marks]
Question 2[4 marks]
Answer or workingMarks
a correctly completed genetic diagram (Punnett square) crossing RR with WWM1
all offspring having the genotype RWA1
all offspring having a pink phenotypeA1
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 whiteB1
Final answer: All offspring have the genotype RW and a pink phenotype, because in codominance both the red and white alleles are expressed together
Mark scheme for Question 3 [4 marks]
Question 3[4 marks]
Answer or workingMarks
finding the change in concentration = 9.5 - 5.5 = 4 mmol/lM1
identifying the time taken = 30 minutesM1
using rate = change / timeM1
a rate of 4/30 = 0.13 mmol/l per minute (2 significant figures)A1
Final answer: 0.13 mmol/l per minute
Mark scheme for Question 4 [4 marks]
Question 4[4 marks]
Answer or workingMarks
recognising 2 hours contains 6 divisions (120 / 20)M1
using the pattern that the cell number doubles at each divisionM1
calculating 2^6M1
64 cellsA1
Mark scheme for Question 5 [5 marks]
Question 5[5 marks]
Answer or workingMarks
the increase = 58 - 40 = 18 gM1
percentage increase = (18 / 40) x 100M1
45%A1
total extra biomass = 18 x 250M1
4500 gA1
Final answer: A 45% increase per plant; across 250 plants this is an extra 4500 g of biomass per week
Mark scheme for Question 6 [5 marks]
Question 6[5 marks]
Answer or workingMarks
vaccinated cases = 3% of 2000 = 60M1
unvaccinated cases = 15% of 2000 = 300M1
using efficacy = (1 - (3 / 15)) x 100M1
substituting to get (1 - 0.2) x 100A1
80% efficacyA1
Final answer: 60 people caught the disease in the vaccinated group and 300 in the unvaccinated group; the vaccine's efficacy is 80%
Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or workingMarks
enzymes have an active site with a specific shape that is complementary to their substrateB1
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 stomachB1
at a pH far from the optimum, such as pH 8, bonds holding the enzyme's structure together are disruptedB1
this changes the shape of the enzyme's active siteB1
the substrate (protein) can no longer fit (bind to) the active site, so the enzyme cannot catalyse the reaction and appears much less activeB1
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
Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or workingMarks
area covered in the quadrat = 60% of 0.5 m^2M1
0.3 m^2A1
using the same percentage (60%) applied to the total areaM1
substituting 60% of 400 m^2M1
240 m^2A1
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
Mark scheme for Question 9 [6 marks]
Question 9[6 marks]
Answer or workingMarks
in a voluntary action the impulse from the sensory neurone travels to the brainB1
the brain processes the information, involves conscious decision-making, and sends an impulse down a motor neurone to the effectorB1
in a reflex action the impulse from the sensory neurone passes to a relay neurone in the spinal cord, not the brainB1
the relay neurone passes the impulse directly to a motor neurone, which carries it to the effectorB1
the reflex pathway not needing to pass to the brain and not involving conscious decision-making, so it involves fewer synapses and a shorter distanceB1
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 responseB1
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
Mark scheme for Question 10 [6 marks]
Question 10[6 marks]
Answer or workingMarks
before percentage = (40 / 800) x 100M1
5%A1
after percentage = (12 / 750) x 100M1
1.6%A1
pathogens can be transferred from a contaminated hand to another person, or to a surface then another person, causing infectionB1
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
Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or workingMarks
using ratio = surface area / thicknessM1
substituting membrane A's values, 20 / 0.04M1
membrane A ratio = 500A1
substituting membrane B's values, 15 / 0.05M1
membrane B ratio = 300A1
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 diffusionB1
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
Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or workingMarks
untrained rate = 8 / 25M1
0.32 litres per minuteA1
trained rate = 5 / 10M1
0.5 litres per minuteA1
regular training increasing the number of mitochondria and capillaries in muscle cells, improving oxygen delivery and the capacity for aerobic respirationB1
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 debtB1
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