GCSE Biology Higher Short Paper A
Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 3 more.
Questions
Question 1 [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 2 [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 3 [6 marks]
Human Defence Systems and Treating Disease
Monoclonal antibodies can be attached to a fluorescent dye and used in a laboratory to identify a specific protein within a sample of cells under a microscope.
Explain how attaching a fluorescent dye to a monoclonal antibody that binds specifically to a target protein allows scientists to determine both whether a cell contains that protein and roughly how much of it is present.
Question 4 [6 marks]
Pathogens and Communicable Disease
The number of confirmed MRSA (antibiotic-resistant bacteria) infections in a region was 45 in 2020 and 117 in 2024.
Calculate the percentage increase in confirmed infections over this four-year period, and calculate the mean increase in the number of infections per year, assuming a steady rate of increase. Then suggest one reason, other than natural selection acting on random mutations, why resistant infections might be under-reported in official figures.
Question 5 [6 marks]
Photosynthesis
A commercial grower increases the carbon dioxide concentration in a greenhouse from 0.04% to 0.1% while keeping light intensity and temperature high and non-limiting.
Explain why increasing carbon dioxide concentration in the greenhouse is likely to increase the rate of photosynthesis of the plants, and explain why increasing carbon dioxide concentration further, to 0.5%, might not increase the rate any further.
Question 6 [6 marks]
Human Defence Systems and Treating Disease
A student investigates two antibiotics by placing paper discs soaked in each onto an agar plate spread with a bacterium, then measuring the diameter of the clear zone (no bacterial growth) around each disc after incubation.
Antibiotic A produced a clear zone of diameter 8 mm. Antibiotic B produced a clear zone of diameter 20 mm.
Calculate the area of the clear zone produced by each antibiotic, using area = pi x radius^2 and pi = 3.14, and explain what the results suggest about the relative effectiveness of the two antibiotics.
Question 7 [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.
Model solutions
| Question 1[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 2[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 3[6 marks] | |
|---|---|
| Answer or working | Marks |
| monoclonal antibodies are identical antibodies produced from a single clone of cells, and are specific to one particular protein (antigen) | B1 |
| the antibody is attached to a fluorescent dye/marker before being added to the sample of cells | B1 |
| the antibody will only bind to cells that contain the target protein, because of its specific shape | B1 |
| unbound antibody is washed away, leaving fluorescent labelling only where the target protein is present | B1 |
| the cells are viewed under a microscope, and if the target protein is present that part of the cell will fluoresce (glow) under the appropriate light | B1 |
| the brightness or amount of fluorescence gives an indication of roughly how much of the target protein is present in the cell | B1 |
| Final answer: The fluorescent-labelled antibody binds only to cells containing the target protein; after washing away unbound antibody, fluorescence under the microscope shows both the presence and, from its brightness, roughly the amount of the target protein | |
| Question 4[6 marks] | |
|---|---|
| Answer or working | Marks |
| the increase = 117 - 45 = 72 | M1 |
| percentage increase = (72 / 45) x 100 | M1 |
| 160% | A1 |
| mean increase per year = 72 / 4 | M1 |
| 18 infections per year | A1 |
| a valid reason, e.g. not everyone with an infection is tested/confirmed, so mild or undiagnosed cases would not appear in the figures | B1 |
| Final answer: A 160% increase, a mean rise of 18 infections per year; official figures may under-report resistant infections because not every case is tested and confirmed, so true numbers could be higher | |
| Question 5[6 marks] | |
|---|---|
| Answer or working | Marks |
| carbon dioxide is a raw material (reactant) needed for photosynthesis | B1 |
| at low concentrations carbon dioxide is often the limiting factor, so increasing it increases the rate provided light and temperature are not limiting | B1 |
| more carbon dioxide molecules are available to react in the light-independent reactions, increasing the rate at which glucose is produced | B1 |
| beyond a certain concentration, carbon dioxide is no longer the limiting factor | B1 |
| another factor, such as light intensity or temperature, becomes limiting instead | B1 |
| increasing carbon dioxide further having no effect on the rate because that other factor now limits how fast photosynthesis can proceed | B1 |
| Final answer: Increasing carbon dioxide increases the rate while it is the limiting factor, providing more raw material for the reaction; beyond a point, another factor such as light intensity or temperature becomes limiting, so further carbon dioxide has no effect | |
| Question 6[6 marks] | |
|---|---|
| Answer or working | Marks |
| radius A = 8 / 2 = 4 mm | M1 |
| area A = 3.14 x 4^2 | M1 |
| area A = 50.24 mm^2 | A1 |
| radius B = 10 mm, area B = 3.14 x 10^2 = 314 mm^2 | M1 |
| a larger clear zone (larger area) meaning the bacterium is more strongly inhibited by that antibiotic | B1 |
| concluding antibiotic B is more effective against this bacterium than antibiotic A, since it produced a much larger area with no bacterial growth | B1 |
| Final answer: Antibiotic A's clear zone has an area of about 50.24 mm^2; antibiotic B's is 314 mm^2; the much larger area for antibiotic B shows it more strongly inhibits this bacterium, so it is more effective | |
| Question 7[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 | |