Higher Tier - Grades 4-9

GCSE Biology Higher Paper 3

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]

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

Mark scheme for Question 1 [4 marks]
Question 1[4 marks]
Answer or workingMarks
converting the actual width to the same units as the image: 60 micrometres = 0.06 mmM1
using magnification = image size / actual sizeM1
substituting 24 / 0.06M1
magnification = x400A1
Final answer: x400
Mark scheme for Question 2 [4 marks]
Question 2[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 3 [4 marks]
Question 3[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 4 [4 marks]
Question 4[4 marks]
Answer or workingMarks
using rate = volume / timeM1
substituting 12 / 4M1
3 cm^3 per minuteA1
time = 45 / 3 = 15 minutesA1independent
Final answer: Rate = 3 cm^3 per minute; it would take 15 minutes to produce 45 cm^3 of gas
Mark scheme for Question 5 [5 marks]
Question 5[5 marks]
Answer or workingMarks
round 1 = 1 x 3 = 3 newly infectedM1
round 2 = 3 x 3 = 9 newly infectedM1
round 3 = 9 x 3 = 27 newly infectedM1
the total number infected = 1 + 3 + 9 + 27 = 40A1
identifying this pattern as exponential growth, which is why an outbreak can spread rapidly if uncheckedB1independent
Final answer: 40 people in total have been infected after 3 further rounds; this is an example of exponential growth
Mark scheme for Question 6 [5 marks]
Question 6[5 marks]
Answer or workingMarks
recognising the ratio 1:2:1 represents 4 total partsM1
finding the value of one part = 240 / 4 = 60M1
60 plants of each of the two homozygous genotypesA1
120 plants of the heterozygous genotypeA1
checking the total, 60 + 120 + 60 = 240, matches the sample sizeB1independent
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)
Mark scheme for Question 7 [5 marks]
Question 7[5 marks]
Answer or workingMarks
an electrical impulse arrives at the end of the first (presynaptic) neuroneB1
this triggers the release of a chemical called a neurotransmitterB1
the neurotransmitter diffuses across the synapse between the two neuronesB1
the neurotransmitter binds to specific receptor molecules on the membrane of the next (postsynaptic) neuroneB1
this triggers a new electrical impulse in the second neuroneB1
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
Mark scheme for Question 8 [5 marks]
Question 8[5 marks]
Answer or workingMarks
preclinical testing is carried out in a laboratory, using cells, tissues or live animalsB1
this tests for toxicity, efficacy and the correct dosageB1
clinical trials then test the drug on healthy volunteers, at a very low dose, to check for harmful side effectsB1
if safe, the drug is trialled on a small number of patients to find the optimum doseB1
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 effectivenessB1
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
Mark scheme for Question 9 [6 marks]
Question 9[6 marks]
Answer or workingMarks
using previous level's energy = current level's energy / 10%M1
secondary consumer = 80 / 0.1M1
800 kJA1
primary consumer = 800 / 0.1 = 8000 kJM1
producer = 8000 / 0.1 = 80,000 kJA1
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 consumersB1
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
Mark scheme for Question 10 [6 marks]
Question 10[6 marks]
Answer or workingMarks
using ratio = 2900 / 120M1
approximately 24 times more energyA1
during vigorous exercise, the heart and lungs cannot supply oxygen to the muscles fast enough to meet the high demand for aerobic respirationB1
anaerobic respiration not requiring oxygen, so it can continue to release some energy even when oxygen supply is insufficientB1
this allowing muscle contraction (and exercise) to continue, even though anaerobic respiration releases far less energy per glucose moleculeB1
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
Mark scheme for Question 11 [6 marks]
Question 11[6 marks]
Answer or workingMarks
the two populations are geographically isolated and cannot interbreedB1
each population experiences different environmental conditions or independently accumulates different mutationsB1
natural selection acts differently on each population, favouring different characteristics in eachB1
over many generations the two populations become increasingly genetically differentB1
eventually, the two populations become so different that they can no longer interbreed to produce fertile offspring, meaning they have become separate speciesB1
scientists can compare DNA sequences between the two populations, and a large enough difference confirms they can no longer be classified as the same speciesB1
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
Mark scheme for Question 12 [6 marks]
Question 12[6 marks]
Answer or workingMarks
using intensity is proportional to 1/distance^2M1
finding the constant of proportionality, k = intensity x distance^2 = 4 x 20^2 = 1600M1
substituting distance = 40 into intensity = k/distance^2M1
intensity = 1600/1600 = 1 arbitrary unitA1
recognising light intensity at 40 cm is a quarter of that at 20 cmB1
explaining the rate of photosynthesis would be expected to be lower at 40 cm because light intensity is a limiting factor and is much reducedB1
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