Year 10 Paper 4: Cells, Disease and Bioenergetics
Covers cell structure and transport, cell division and stem cells, digestion and the circulatory system, pathogens and communicable disease, 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 [5 marks]
Cell Division and Stem Cells
In a tissue culture, one cell divides by mitosis every 15 minutes.
If a single cell starts dividing at 09:00, calculate how many complete divisions will have occurred by 10:15, and calculate the total number of cells present at that time.
Question 3 [5 marks]
Digestion and the Circulatory System
A student burned a 0.5 g sample of food beneath a boiling tube containing 50 cm^3 of water. The temperature of the water rose from 20 deg C to 44 deg C.
Using energy (J) = mass of water (g) x 4.2 x temperature rise (deg C), calculate the energy released per gram of food, in J/g.
Question 4 [5 marks]
Cell Structure and Transport
Root hair cells are specialised plant cells found on the surface of roots.
Explain how the structure of a root hair cell makes it well adapted for absorbing water and mineral ions from the soil.
Question 5 [5 marks]
Photosynthesis
In an investigation, the rate of photosynthesis increased steadily as light intensity increased from 0 to 40 units, then stayed constant as light intensity increased further to 80 units.
Explain the shape of this graph, referring to limiting factors.
Question 6 [6 marks]
Digestion and the Circulatory System
The small intestine is adapted to efficiently absorb the products of digestion.
Explain how the structure of the small intestine, including villi, is adapted for efficient absorption of digested food.
Question 7 [6 marks]
Cell Structure and Transport
An egg cell (ovum) is a large, specialised cell that fuses with a sperm cell during fertilisation, then must supply everything a developing embryo needs before it implants.
Explain how the structure of an egg cell is adapted for these functions.
Question 8 [6 marks]
Photosynthesis
At 10 deg C, increasing carbon dioxide concentration from 0.04% to 0.4% increased the rate of photosynthesis in a crop from 2 to 2.4 arbitrary units.
At 25 deg C, the same increase in carbon dioxide concentration increased the rate from 2 to 8 arbitrary units.
Calculate the percentage increase in rate at each temperature, and explain why increasing carbon dioxide had a much greater effect at 25 deg C than at 10 deg C.
Question 9 [6 marks]
Cell Structure and Transport
Exchange surfaces in living organisms, such as the alveoli in the lungs or gill filaments in fish, are adapted to maximise the rate of diffusion of substances across them.
Describe five features that a good exchange surface needs to maximise the rate of diffusion, explaining why each feature increases the rate.
Question 10 [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 11 [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.
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[5 marks] | |
|---|---|
| Answer or working | Marks |
| finding the time elapsed = 75 minutes | M1 |
| finding the number of divisions = 75 / 15 | M1 |
| 5 divisions | A1 |
| using cell number = 2^(number of divisions) | M1 |
| 32 cells | A1 |
| Final answer: 5 complete divisions occur by 10:15, giving a total of 32 cells | |
| Question 3[5 marks] | |
|---|---|
| Answer or working | Marks |
| temperature rise = 44 - 20 = 24 deg C | M1 |
| substituting into energy = mass of water x 4.2 x temperature rise | M1 |
| energy released = 50 x 4.2 x 24 = 5040 J | M1 |
| energy per gram of food = 5040 / 0.5 = 10080 J/g | A1 |
| converting to 10.08 kJ/g (or equivalent to appropriate significant figures) | B1independent |
| Final answer: 5040 J released in total, equivalent to 10080 J/g (10.08 kJ/g) of food | |
| Question 4[5 marks] | |
|---|---|
| Answer or working | Marks |
| the cell has a long, thin extension (hair) that increases the surface area | B1 |
| the increased surface area increases the rate of water and mineral ion uptake | B1 |
| the cell wall and membrane are thin, reducing the distance for diffusion and osmosis | B1 |
| the cell contains many mitochondria | B1 |
| the mitochondria release energy for active transport of mineral ions against a concentration gradient | B1 |
| Final answer: Large surface area from the hair-like extension, a thin cell wall/membrane for a short diffusion path, and many mitochondria to provide energy for active transport of mineral ions | |
| Question 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| from 0 to 40 units, light intensity is the limiting factor | B1 |
| as light intensity increases in this range, the rate of photosynthesis increases because more light energy is available for the reaction | B1 |
| above 40 units, the rate levels off because light is no longer limiting | B1 |
| another factor, such as carbon dioxide concentration or temperature, has become the limiting factor | B1 |
| increasing light further beyond this point has no effect because that other factor now limits the rate | B1 |
| Final answer: Below 40 units light is limiting so rate rises with light intensity; above 40 units another factor (e.g. carbon dioxide or temperature) becomes limiting, so the rate levels off despite more light | |
| Question 6[6 marks] | |
|---|---|
| Answer or working | Marks |
| the wall of the small intestine is covered in villi, greatly increasing the surface area for absorption | B1 |
| each villus has a single layer of surface cells, giving a short diffusion path | B1 |
| villi have a good blood supply, maintaining a steep concentration gradient for diffusion | B1 |
| the blood supply carries absorbed products away quickly, maintaining the gradient | B1 |
| some villi surface cells have microvilli, which further increase the surface area | B1 |
| a large surface area increases the rate of diffusion and absorption of nutrients into the blood | B1 |
| Final answer: Villi increase surface area, provide a short diffusion path (thin walls), and have a good blood supply that maintains a steep concentration gradient, all increasing the rate of absorption | |
| Question 7[6 marks] | |
|---|---|
| Answer or working | Marks |
| the cell contains a haploid nucleus, containing half the normal number of chromosomes, so that after fertilisation with a sperm cell the resulting zygote has the full (diploid) number | B1 |
| the cell contains large stores of nutrients (cytoplasm rich in food reserves), to nourish the early embryo before it can obtain nutrients from the mother | B1 |
| immediately after fertilisation, the membrane of the egg cell changes structure, becoming impenetrable to other sperm cells | B1 |
| this change to the membrane prevents more than one sperm fertilising the egg (polyspermy), ensuring the zygote has the correct number of chromosomes | B1 |
| the cell being one of the largest cells in the human body, giving it a large volume to store the nutrients needed | B1 |
| the cell being able to divide rapidly by mitosis after fertilisation to begin forming the embryo | B1 |
| Final answer: The egg cell has a haploid nucleus so fertilisation restores the diploid number, large nutrient stores to nourish the early embryo, and a membrane that changes after fertilisation to prevent more than one sperm entering | |
| Question 8[6 marks] | |
|---|---|
| Answer or working | Marks |
| the increase at 10 deg C = 2.4 - 2 = 0.4, then percentage = (0.4 / 2) x 100 | M1 |
| a 20% increase at 10 deg C | A1 |
| the increase at 25 deg C = 8 - 2 = 6, then percentage = (6 / 2) x 100 | M1 |
| a 300% increase at 25 deg C | A1 |
| recognising that at 10 deg C, temperature (not carbon dioxide) is strongly limiting the rate, because enzyme activity is low at this temperature, so raising carbon dioxide has little effect | B1 |
| recognising that at 25 deg C, temperature is closer to the enzymes' optimum and is less limiting, so carbon dioxide becomes the (more) limiting factor and increasing it has a much greater effect | B1 |
| Final answer: The rate increased by 20% at 10 deg C but by 300% at 25 deg C; at the lower temperature, temperature itself is strongly limiting so more carbon dioxide barely helps, but at 25 deg C temperature is less limiting, so carbon dioxide becomes the limiting factor and increasing it has a much bigger effect | |
| Question 9[6 marks] | |
|---|---|
| Answer or working | Marks |
| a large surface area, which provides more space for molecules to diffuse across at any one time | B1 |
| a thin membrane (short diffusion distance), which reduces the distance molecules must travel, increasing the rate | B1 |
| a good blood supply or ventilation, which maintains a steep concentration gradient by removing or supplying substances | B1 |
| maintaining a steep concentration gradient increases the rate of diffusion | B1 |
| a moist surface (in gas exchange organs), which allows gases to dissolve so they can diffuse across the membrane | B1 |
| a permeable surface, allowing the specific molecules being exchanged to pass through | B1 |
| Final answer: Good exchange surfaces have a large surface area, a short diffusion distance (thin walls), a good blood supply or ventilation to maintain a steep concentration gradient, and (in gas exchange) a moist, permeable surface | |
| Question 10[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 11[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 | |