Year 10 Paper 6: Full Topic Review
Covers cell structure and transport, cell division and stem cells, digestion and the circulatory system, pathogens and communicable disease, the body's defence systems and how disease is treated, photosynthesis, and respiration and exercise.
Year 10 here means a typical teaching order, not a syllabus rule. No exam board defines what belongs to Year 10, and schools sequence the course differently. Check it against your own scheme of work before using it to decide what a class has covered.
Questions
Question 1 [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 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 [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 4 [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 5 [5 marks]
Cell Structure and Transport
Glucose can move from the gut into the blood by diffusion when its concentration in the gut is higher than in the blood, but sometimes glucose must instead be absorbed by active transport.
Explain why active transport is sometimes needed to absorb glucose from the gut into the blood, and explain one way active transport differs from diffusion in terms of energy.
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
Measles is a viral disease spread by inhalation of droplets from an infected person's coughs and sneezes.
Explain how vaccination against measles can protect an individual, and explain how high vaccination rates in a population can protect people who are not vaccinated.
Question 9 [6 marks]
Respiration and Exercise
An untrained person begins to respire anaerobically (and produce lactic acid) at a lower exercise intensity than a well-trained athlete.
Explain why an untrained person's muscles begin anaerobic respiration sooner during increasing exercise, referring to oxygen supply to the muscles.
Question 10 [6 marks]
Human Defence Systems and Treating Disease
In a double-blind drug trial, neither the patients nor the doctors giving the treatment know which patients are receiving the real drug and which are receiving a placebo until after the results have been collected.
Explain why using a placebo group is important in a drug trial, and explain why the trial is made double-blind rather than just giving patients a placebo or the real drug openly.
Question 11 [6 marks]
Cell Division and Stem Cells
A red blood cell is a specialised cell produced by differentiation of a stem cell in the bone marrow. As it differentiates, a red blood cell loses its nucleus and most other organelles, and produces large amounts of the protein haemoglobin.
Explain how these changes make the red blood cell well adapted to its function of transporting oxygen, and explain why, once fully differentiated, a red blood cell cannot itself divide to produce more red blood cells.
Question 12 [6 marks]
Cell Structure and Transport
The concentration of nitrate ions in root hair cells is normally much higher than in the surrounding soil water.
Soil nitrate concentration is 0.002 arbitrary units and the concentration inside a root hair cell is 0.014 arbitrary units.
Calculate how many times more concentrated the nitrate ions are inside the cell than in the soil, then explain how root hair cells absorb nitrate ions against this concentration gradient, and explain why this process requires energy from respiration.
Question 13 [6 marks]
Cell Division and Stem Cells
A laboratory needs to grow 1,000,000 skin cells from a single starting cell for a skin graft, by mitosis. The cells divide every 24 hours.
Calculate the minimum number of complete divisions needed to produce at least 1,000,000 cells, and calculate how many days this would take.
Question 14 [6 marks]
Digestion and the Circulatory System
Large food molecules such as starch, proteins and fats are insoluble and cannot pass through the wall of the digestive system into the blood.
Explain how carbohydrase, protease and lipase enzymes allow these large molecules to be absorbed into the blood, naming the products each enzyme produces.
Question 15 [6 marks]
Pathogens and Communicable Disease
Doctors are increasingly concerned about bacteria that are resistant to antibiotics.
Explain how a population of bacteria can evolve to become resistant to an antibiotic, and explain one way the emergence of antibiotic-resistant bacteria can be reduced.
Model solutions
| Question 1[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 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[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 4[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 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| diffusion is the net movement of particles from a region of higher concentration to a region of lower concentration | B1 |
| diffusion cannot move glucose from the gut into the blood once the gut concentration has fallen below that in the blood (against the concentration gradient) | B1 |
| active transport can move glucose from a lower to a higher concentration, against the concentration gradient, using carrier proteins in the cell membrane | B1 |
| diffusion does not require energy from respiration, whereas active transport does require energy from respiration | B1 |
| this energy being needed because moving particles against a concentration gradient could not otherwise happen | B1 |
| Final answer: Active transport is needed once the gut glucose concentration is no longer higher than the blood's, because diffusion cannot work against a gradient; unlike diffusion, active transport requires energy from respiration to move particles from a lower to a higher concentration | |
| 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 |
| a vaccine contains a small, safe, dead or inactive form of the pathogen, or its antigens | B1 |
| this stimulates the immune system to produce antibodies specific to the pathogen | B1 |
| memory cells remain in the body, so a future infection is destroyed quickly before symptoms develop | B1 |
| if a large proportion of the population is vaccinated, the pathogen cannot easily spread between people | B1 |
| this gives herd immunity, reducing the chance of an unvaccinated person coming into contact with the pathogen | B1 |
| Final answer: Vaccination stimulates antibody and memory cell production, giving immunity; when most of the population is vaccinated, the pathogen cannot spread easily, protecting unvaccinated people (herd immunity) | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| as exercise intensity increases, the muscles' demand for oxygen (for aerobic respiration) increases | B1 |
| an untrained person has fewer capillaries supplying their muscles and a heart with a smaller stroke volume than a trained athlete | B1 |
| this limits how quickly oxygen (and glucose) can be delivered to the muscles as demand rises | B1 |
| once oxygen cannot be delivered fast enough to meet the muscles' demand, the muscles must start to respire anaerobically as well | B1 |
| because an untrained person's oxygen delivery reaches its limit at a lower exercise intensity, they begin anaerobic respiration sooner than a trained athlete | B1 |
| a trained athlete's greater capillary density and stronger heart letting them deliver enough oxygen to respire aerobically up to a higher exercise intensity | B1 |
| Final answer: An untrained person has fewer muscle capillaries and a weaker heart, so oxygen delivery reaches its limit at a lower exercise intensity, forcing muscles into anaerobic respiration sooner; a trained athlete's better oxygen delivery lets them respire aerobically up to a higher intensity | |
| Question 10[6 marks] | |
|---|---|
| Answer or working | Marks |
| a placebo looks identical to the real drug but contains no active ingredient | B1 |
| comparing the drug group to the placebo group shows how much of any improvement is due to the drug itself, rather than the patient expecting to feel better (the placebo effect) | B1 |
| without a placebo group it would be difficult to know if the drug, or just receiving treatment, caused the improvement | B1 |
| if patients know they are getting the real drug, they may report feeling better due to bias/expectation rather than a real effect | B1 |
| if doctors know which patients have the real drug, this could unconsciously bias how they assess or record the patients' symptoms | B1 |
| making the trial double-blind removing bias from both the patient and the doctor, making the results more reliable | B1 |
| Final answer: A placebo group shows how much improvement is due to the drug rather than expectation; making the trial double-blind prevents bias from both patients and doctors, making the results more reliable | |
| Question 11[6 marks] | |
|---|---|
| Answer or working | Marks |
| losing the nucleus and other organelles increases the internal space available to be filled with haemoglobin | B1 |
| more haemoglobin means more oxygen can be carried by each cell | B1 |
| the resulting biconcave disc shape increases the surface area to volume ratio for the diffusion of oxygen in and out | B1 |
| differentiation being permanent in a mature specialised cell, so the genes for other cell types are switched off | B1 |
| mitosis requiring a nucleus containing the DNA/chromosomes to be copied and separated | B1 |
| the red blood cell having no nucleus (no DNA), so it cannot undergo mitosis and so cannot divide | B1 |
| Final answer: Losing the nucleus and organelles leaves more room for haemoglobin, which carries oxygen, and gives a biconcave shape with a large surface area for diffusion; without a nucleus (no DNA), the cell cannot undergo mitosis so cannot divide | |
| Question 12[6 marks] | |
|---|---|
| Answer or working | Marks |
| using ratio = concentration inside / concentration outside | M1 |
| substituting 0.014 / 0.002 | M1 |
| 7 times more concentrated | A1 |
| nitrate ions are absorbed by active transport, moving from a lower to a higher concentration (against the concentration gradient) | B1 |
| active transport requires carrier proteins in the cell membrane to move the ions | B1 |
| this process requires energy released by respiration in the mitochondria, because moving substances against a concentration gradient cannot happen by diffusion alone | B1 |
| Final answer: 7 times more concentrated; nitrate ions are absorbed against the gradient by active transport using carrier proteins, which requires energy from respiration because this cannot happen by diffusion alone | |
| Question 13[6 marks] | |
|---|---|
| Answer or working | Marks |
| recognising cell number after n divisions = 2^n | M1 |
| testing 2^19 = 524,288 (below 1,000,000) | M1 |
| testing 2^20 = 1,048,576 (exceeds 1,000,000) | M1 |
| 20 divisions | A1 |
| time = 20 x 24 = 480 hours | B1independent |
| converting to 20 days | B1independent |
| Final answer: 20 divisions are needed (2^20 = 1,048,576), taking 480 hours, which is 20 days | |
| Question 14[6 marks] | |
|---|---|
| Answer or working | Marks |
| carbohydrase enzymes (such as amylase) break down starch into simple sugars, such as glucose | B1 |
| protease enzymes break down proteins into amino acids | B1 |
| lipase enzymes break down fats (lipids) into fatty acids and glycerol | B1 |
| these smaller products (glucose, amino acids, fatty acids and glycerol) are soluble | B1 |
| being soluble and small allows them to diffuse across the wall of the small intestine into the blood | B1 |
| the products then being used by cells around the body, for example for respiration or to build new proteins | B1 |
| Final answer: Carbohydrase breaks starch into glucose, protease breaks proteins into amino acids, and lipase breaks fats into fatty acids and glycerol; these smaller, soluble molecules can diffuse into the blood and be used by cells | |
| Question 15[6 marks] | |
|---|---|
| Answer or working | Marks |
| within a large population of bacteria there is genetic variation, and some bacteria carry a mutation giving resistance | B1 |
| when the antibiotic is used, non-resistant bacteria are killed | B1 |
| resistant bacteria survive and reproduce | B1 |
| the resistant bacteria pass the resistance allele to their offspring | B1 |
| over many generations, the proportion of resistant bacteria in the population increases | B1 |
| a valid reduction method, such as doctors not prescribing antibiotics for non-serious or viral infections, or patients completing the full course prescribed | B1 |
| Final answer: Resistant bacteria survive antibiotic treatment and reproduce, passing on the resistance gene, so the resistant population increases over generations; reducing unnecessary antibiotic use slows this | |