GCSE Biology Higher Paper 2
Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 9 more.
Questions
Question 1 [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 2 [4 marks]
Human Defence Systems and Treating Disease
A new painkiller was tested in a double-blind trial. Of 200 patients given the drug, 150 reported reduced pain. Of 200 patients given a placebo, 60 reported reduced pain.
Calculate the percentage of each group that reported reduced pain, and use your answer to comment on the effectiveness of the drug.
Question 3 [4 marks]
Respiration and Exercise
Before exercise, a student's resting heart rate was 68 beats per minute.
Immediately after five minutes of running, their heart rate was 119 beats per minute.
Calculate the percentage increase in the student's heart rate.
Question 4 [4 marks]
The Nervous System and Reflexes
A student measured reaction time using a ruler drop test.
The ruler fell 18 cm before being caught.
Using the relationship distance (m) = 0.5 x 9.8 x time^2, calculate the student's reaction time in seconds, to 2 significant figures.
Question 5 [5 marks]
Evolution and Natural Selection
A population of cheetahs went through a period, thousands of years ago, when only a very small number of individuals survived (a genetic bottleneck), before the population later grew large again.
Explain why modern cheetah populations are thought to have unusually low genetic variation, and explain why this low genetic variation puts the species at greater risk today.
Question 6 [5 marks]
Cell Division and Stem Cells
Explain why the four gametes produced by meiosis from one parent cell are not usually genetically identical to each other.
Question 7 [5 marks]
Photosynthesis
Commercial plant growers sometimes provide extra carbon dioxide and artificial lighting in a greenhouse during winter days, but not at night.
Explain why providing extra carbon dioxide and light during the day can increase crop yield, and explain why it would not be worth providing these at night.
Question 8 [5 marks]
Pathogens and Communicable Disease
In a village of 2000 people, 800 mosquito nets were distributed, each protecting 2 people.
Calculate the percentage of the village population protected by the nets, and calculate how many more nets would be needed to protect the whole village, assuming each net still protects 2 people.
Question 9 [6 marks]
Ecosystems and Material Cycles
In a food chain, producers contain 12000 kJ of energy stored in biomass. The primary consumers that feed on them contain 1200 kJ of energy stored in biomass.
Calculate the percentage of energy transferred from producers to primary consumers, and explain three reasons why not all of the energy is transferred between these trophic levels.
Question 10 [6 marks]
Genetics and Inheritance
Human height is an example of continuous variation, showing a wide range of values influenced by both genes and environment. Human blood group (A, B, AB or O) is an example of discontinuous variation, controlled entirely by genes, with a small number of distinct categories.
Explain the difference between continuous and discontinuous variation, and explain why environmental factors can affect a characteristic controlled by continuous variation (such as height) but generally do not affect a characteristic controlled by discontinuous variation (such as blood group).
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]
Digestion and the Circulatory System
A patient's heart rate is 72 beats per minute and their stroke volume (the volume of blood pumped by the heart with each beat) is 70 cm^3.
Calculate the patient's cardiac output in cm^3 per minute, using cardiac output = heart rate x stroke volume. Then calculate the cardiac output in dm^3 per minute (1 dm^3 = 1000 cm^3), and explain what would happen to cardiac output during exercise and why this is useful.
Model solutions
| Question 1[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 2[4 marks] | |
|---|---|
| Answer or working | Marks |
| drug group percentage = (150/200) x 100 = 75% | M1 |
| placebo group percentage = (60/200) x 100 = 30% | M1 |
| comparing the two percentages | M1 |
| the drug is effective because a much greater percentage of patients on the drug reported reduced pain than on the placebo (75% compared with 30%) | A1 |
| Final answer: 75% (drug) versus 30% (placebo); the drug appears effective since a far higher proportion improved than with the placebo | |
| Question 3[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the increase = 119 - 68 = 51 beats per minute | M1 |
| using percentage change = (increase / original) x 100 | M1 |
| substituting (51/68) x 100 | M1 |
| 75% (to 2 significant figures) | A1 |
| Final answer: 75% increase | |
| Question 4[4 marks] | |
|---|---|
| Answer or working | Marks |
| converting 18 cm to 0.18 m | M1 |
| rearranging to time = sqrt(distance / (0.5 x 9.8)) | M1 |
| substituting 0.18 into the rearranged equation | M1 |
| 0.19 seconds | A1 |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| when the population was reduced to a very small number of surviving individuals, only the alleles carried by those few individuals remained in the population | B1 |
| many alleles that had existed in the larger, earlier population were lost, because the individuals carrying them did not survive | B1 |
| as the population grew again from this small number of survivors, all of the new individuals shared alleles with this small starting group, so overall genetic variation remained low | B1 |
| low genetic variation means the population is less likely to already contain an allele giving resistance or an advantage against a new threat, such as a disease | B1 |
| this makes it more likely that a new disease or environmental change could affect a very large proportion of the population at once, increasing the risk of extinction | B1 |
| Final answer: A past bottleneck left only a few surviving individuals, losing most of the population's alleles; because today's cheetahs descend from this small group, genetic variation remains low, making it less likely the population already has an allele that could help it survive a new disease or environmental change, increasing extinction risk | |
| Question 6[5 marks] | |
|---|---|
| Answer or working | Marks |
| chromosomes are copied at the start of meiosis, then the cell divides twice to form four cells | B1 |
| during meiosis, each of the four cells gets a different combination of chromosomes | B1 |
| one chromosome from each pair originally came from the organism's mother and one from its father, and these are mixed randomly between the gametes | B1 |
| this means each gamete receives a different mixture of the organism's chromosomes | B1 |
| this random mixing (plus mutation) is a source of genetic variation between gametes, and therefore between offspring | B1 |
| Final answer: During meiosis each of the four gametes receives a different, randomly mixed combination of the parent's chromosomes (originally from its own two parents), so the gametes are genetically different from each other | |
| Question 7[5 marks] | |
|---|---|
| Answer or working | Marks |
| during a winter day, light intensity and/or carbon dioxide concentration are often limiting factors for photosynthesis | B1 |
| raising carbon dioxide concentration and light intensity above the natural winter level increases the rate of photosynthesis, provided temperature is not limiting | B1 |
| a faster rate of photosynthesis produces more glucose, which the plant can use for growth, increasing yield | B1 |
| no photosynthesis occurs at night, regardless of light or carbon dioxide levels, because photosynthesis requires light energy | B1 |
| providing extra light or carbon dioxide at night would therefore have no effect on the rate of photosynthesis, so would waste money without increasing yield | B1 |
| Final answer: Extra light and carbon dioxide remove those limiting factors during the day, increasing the rate of photosynthesis and so yield; at night no photosynthesis occurs at all (it needs light), so providing them then would have no effect and would waste money | |
| Question 8[5 marks] | |
|---|---|
| Answer or working | Marks |
| people protected = 800 x 2 = 1600 | M1 |
| percentage protected = (1600 / 2000) x 100 | M1 |
| 80% | A1 |
| remaining people = 2000 - 1600 = 400, nets needed = 400 / 2 | M1 |
| 200 more nets | A1 |
| Final answer: 80% of the village is protected; 200 more nets would be needed to protect everyone | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| using percentage transferred = (energy in consumers / energy in producers) x 100 | M1 |
| substituting (1200/12000) x 100 | M1 |
| 10% | A1 |
| a reason: not all parts of the producer are eaten by the consumer, such as roots not being consumed | B1 |
| a reason: some absorbed material is not digested and is lost as waste (egested) | B1 |
| a reason: energy is lost as heat to the surroundings, especially from respiration, or is used for movement and other life processes rather than being stored as biomass | B1 |
| Final answer: 10% of the energy is transferred; energy is lost because not all of the producer is eaten, some is egested as waste, and much is lost as heat from respiration and other life processes | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| continuous variation showing a wide range of values between two extremes, with no distinct categories | B1 |
| discontinuous variation falling into a small number of distinct categories, with no intermediate values | B1 |
| continuous variation usually being controlled by a combination of many genes (polygenic) as well as the environment | B1 |
| discontinuous variation usually being controlled by a single gene, with limited or no environmental influence | B1 |
| height being influenced by many genes plus factors such as diet and exercise, so environmental factors can shift a person's height within the range set by their genes | B1 |
| blood group being determined entirely by which alleles a person inherits, so no environmental factor can change which blood group alleles are present | B1 |
| Final answer: Continuous variation (e.g. height) shows a range of values controlled by many genes plus environment, so environmental factors can affect it; discontinuous variation (e.g. blood group) falls into distinct categories controlled by a single gene, so it is unaffected by the environment | |
| 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 |
| using cardiac output = heart rate x stroke volume | M1 |
| substituting 72 x 70 | M1 |
| cardiac output = 5040 cm^3 per minute | A1 |
| converting to 5.04 dm^3 per minute | A1 |
| during exercise, cardiac output increases because both heart rate and stroke volume increase | B1 |
| a higher cardiac output delivers oxygen and glucose to respiring muscles faster and removes carbon dioxide more quickly, meeting the increased demand | B1 |
| Final answer: Cardiac output = 5040 cm^3 per minute (5.04 dm^3 per minute); during exercise cardiac output increases, as heart rate and stroke volume both rise, delivering oxygen and glucose to muscles faster | |