Year 10 Paper 5: Disease, Defence and Energy
Covers cell division and stem cells, 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]
Cell Division and Stem Cells
A human body cell contains 46 chromosomes.
Calculate the number of chromosomes in a human gamete (sex cell) produced by meiosis, and calculate the number of chromosomes that would be present in a zygote formed when two gametes fuse at fertilisation.
Question 2 [5 marks]
Pathogens and Communicable Disease
Explain why a new flu vaccine has to be developed and given to at-risk groups each year, rather than one flu vaccine giving lifelong protection.
Question 3 [5 marks]
Cell Division and Stem Cells
A single cell divides by mitosis, and the resulting population doubles every 30 minutes.
Calculate how many complete divisions are needed for the population to first exceed 500 cells, starting from 1 cell, and calculate the total time in minutes this would take.
Question 4 [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 5 [5 marks]
Respiration and Exercise
Explain, in terms of the breakdown of glucose, why aerobic respiration releases much more energy from each glucose molecule than anaerobic respiration.
Question 6 [6 marks]
Human Defence Systems and Treating Disease
Monoclonal antibodies are identical antibodies produced from a single clone of cells.
Describe how monoclonal antibodies are produced, and explain one way they can be used, such as in pregnancy testing or in treating disease.
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.
Question 8 [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 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
Before the results of a new drug trial are accepted by the scientific and medical community, they are usually published in a peer-reviewed journal.
Explain what is meant by peer review, and explain why peer review is important before a new medical treatment is more widely accepted or used.
Question 11 [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.
Model solutions
| Question 1[4 marks] | |
|---|---|
| Answer or working | Marks |
| recognising gametes contain half the number of chromosomes of a body cell (meiosis halves the chromosome number) | M1 |
| 23 chromosomes in a gamete | A1 |
| recognising fertilisation combines one gamete from each parent | M1 |
| 46 chromosomes in the zygote | A1 |
| Final answer: A gamete contains 23 chromosomes; the zygote formed at fertilisation contains 46 chromosomes | |
| Question 2[5 marks] | |
|---|---|
| Answer or working | Marks |
| the flu virus mutates (changes) frequently, causing its surface antigens to change (antigenic variation) | B1 |
| memory cells and antibodies from a previous infection or vaccination are specific to the old antigens | B1 |
| these memory cells/antibodies no longer recognise the new, changed antigens on the mutated virus | B1 |
| so a person is not immune to the new strain and can become infected/ill again | B1 |
| a new vaccine, matched to the currently circulating strain's antigens, must therefore be developed and given each year | B1 |
| Final answer: The flu virus mutates often, changing its surface antigens; old antibodies and memory cells no longer recognise the new antigens, so people are not immune to new strains, meaning a new vaccine matching the current strain is needed each year | |
| Question 3[5 marks] | |
|---|---|
| Answer or working | Marks |
| recognising the population after n divisions = 2^n | M1 |
| testing 2^8 = 256 (below 500) | M1 |
| testing 2^9 = 512 (exceeds 500) | M1 |
| 9 divisions | A1 |
| total time = 9 x 30 = 270 minutes | B1independent |
| Final answer: 9 divisions are needed (2^9 = 512), taking a total of 270 minutes | |
| Question 4[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 5[5 marks] | |
|---|---|
| Answer or working | Marks |
| aerobic respiration uses oxygen to completely break down glucose into carbon dioxide and water | B1 |
| this complete breakdown releases all of the energy stored in the glucose molecule | B1 |
| anaerobic respiration does not use oxygen, so it can only partially break down glucose | B1 |
| glucose is only broken down as far as lactic acid, a molecule that still contains a large amount of stored (chemical) energy | B1 |
| because most of the energy originally in the glucose remains locked in the lactic acid rather than being released, anaerobic respiration releases far less usable energy per glucose molecule | B1 |
| Final answer: Aerobic respiration fully breaks glucose down into carbon dioxide and water using oxygen, releasing all the stored energy; anaerobic respiration only partially breaks glucose down to lactic acid, which still holds most of the energy, so far less is released | |
| Question 6[6 marks] | |
|---|---|
| Answer or working | Marks |
| a mouse is stimulated with a chosen antigen to produce lymphocytes that make the specific antibody | B1 |
| these lymphocytes are fused with a tumour cell to produce a hybridoma cell | B1 |
| the hybridoma cell can divide repeatedly and produces the antibody | B1 |
| many hybridoma cells are cloned to produce a large number of identical cells producing the same antibody | B1 |
| identifying a specific use, such as monoclonal antibodies binding to the hormone found in urine during pregnancy, or binding only to antigens found on cancer cells so a drug can be delivered to them | B1 |
| explaining the outcome of that use, such as this binding causing a colour change on a pregnancy test stick, or delivering the drug specifically to the cancer cells while reducing damage to other, healthy cells | B1 |
| Final answer: Monoclonal antibodies are made by fusing an antibody-producing lymphocyte with a tumour cell to form a hybridoma that is cloned; they are used, for example, in pregnancy tests or to target drugs at cancer cells | |
| 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 | |
| Question 8[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 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 |
| peer review is the process by which other scientists who are independent of the original research, and who are experts in the same field, check a scientific report before it is published | B1 |
| the reviewers check the methods used are valid/appropriate and check the conclusions are supported by the data/results presented | B1 |
| without peer review, flawed, biased or fraudulent research could be published and accepted as reliable | B1 |
| this could lead to a new treatment being used that is actually unsafe or ineffective | B1 |
| peer review helps identify errors, or gaps that need further investigation, before other scientists, doctors or the public rely on the findings | B1 |
| it therefore increases confidence that published, accepted research (and its conclusions about a treatment) can be trusted | B1 |
| Final answer: Peer review is independent experts checking a study's methods and conclusions before publication; it helps catch errors, bias or unsupported claims, so the medical community and public can trust that an accepted new treatment has been properly scrutinised | |
| Question 11[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 | |