GCSE Biology Foundation Paper 4
Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 9 more.
Questions
Question 1 [2 marks]
Digestion and the Circulatory System
State the function of the stomach in digestion, and state two things the stomach does to food.
Question 2 [2 marks]
Genetics and Inheritance
State what is meant by a gene, and state what a chromosome is made of.
Question 3 [2 marks]
Respiration and Exercise
State the word equation for aerobic respiration in humans, and state one place in a cell where it occurs.
Question 4 [3 marks]
Photosynthesis
Describe the role of stomata and guard cells in gas exchange in a leaf.
Question 5 [3 marks]
Ecosystems and Material Cycles
State the name of the process by which carbon dioxide is removed from the atmosphere by plants, and state one process that returns carbon dioxide to the atmosphere.
Question 6 [3 marks]
Cell Division and Stem Cells
Stem cells can be found in embryos and in some adult tissues.
Describe one difference between embryonic stem cells and adult stem cells.
Question 7 [3 marks]
The Nervous System and Reflexes
State what makes up the central nervous system (CNS), and describe the general role of the CNS in coordinating a response to a stimulus.
Question 8 [3 marks]
Evolution and Natural Selection
Describe how the Linnaean classification system organises living organisms into groups.
Question 9 [3 marks]
Human Defence Systems and Treating Disease
When culturing bacteria on agar plates in a school laboratory, describe two precautions that should be taken to reduce the risk of contamination or infection.
Question 10 [4 marks]
Cell Structure and Transport
A student views a sample of onion cells under a microscope.
The diameter of the field of view is 0.6 mm, and exactly 15 onion cells fit side by side across this diameter.
Calculate the mean width of one onion cell, in micrometres.
Question 11 [4 marks]
Hormones and Homeostasis
A doctor prescribes insulin at a dose of 0.4 units per kg of body mass.
Calculate the dose, in units, needed for a patient with a body mass of 65 kg, and calculate the dose needed for a patient with a body mass of 80 kg.
Question 12 [4 marks]
Pathogens and Communicable Disease
A school recorded the number of pupils absent with a flu-like illness.
On Monday, 18 out of 600 pupils were absent with symptoms. By Friday, 54 out of 600 pupils were absent with symptoms.
Calculate the percentage of pupils absent on each day, then calculate the percentage change in absence over the week.
Question 13 [4 marks]
The Nervous System and Reflexes
A nerve impulse travels 1.2 m along a neurone in 0.01 seconds.
Calculate the speed of the nerve impulse in metres per second, and calculate how long it would take an impulse to travel 3 m along a similar neurone at this speed.
Question 14 [5 marks]
Evolution and Natural Selection
Scientists compared a particular gene in four species and found the following percentage similarity in its DNA base sequence to species A: species B 98%, species C 85%, species D 60%.
Calculate the percentage difference between species A and each of the other species, and state which species is most closely related to species A and which is most distantly related.
Question 15 [5 marks]
Ecosystems and Material Cycles
A student used a transect line across a rocky shore and placed a 0.5 m^2 quadrat every 2 metres along the line. In one quadrat, seaweed covered an estimated 60% of the quadrat's area.
Calculate the area, in m^2, of the quadrat that was covered by seaweed. If the whole study area along the shore covers 400 m^2 and this percentage cover is representative of the whole area, estimate the total area of the shore covered by seaweed.
Question 16 [5 marks]
Photosynthesis
A student wants to test whether light is needed for a plant to produce starch. They cover part of one leaf with aluminium foil to exclude light, and leave the rest of the leaf exposed to light for a day, then test the whole leaf with iodine solution.
Describe the result you would expect, and describe one step the student should take before this test to make sure any starch found is produced only during the experiment.
Question 17 [5 marks]
Human Defence Systems and Treating Disease
Home pregnancy tests use monoclonal antibodies to detect a hormone called hCG in urine, which is only present if a woman is pregnant.
Describe how a monoclonal antibody pregnancy test works to give a positive result.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| the stomach digests protein (using the enzyme protease/pepsin) | B1 |
| the stomach churns/mixes food with digestive juices, and produces hydrochloric acid, which kills bacteria in the food and gives the right pH for protease to work | B1 |
| Final answer: The stomach digests protein (using protease) and churns food with acid, which kills bacteria and gives the right pH for the enzyme to work | |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| a gene is a small section of DNA (on a chromosome) that codes for a particular protein/characteristic | B1 |
| a chromosome is a long molecule of DNA, tightly coiled | B1 |
| Final answer: A gene is a section of DNA that codes for a characteristic; a chromosome is a long, coiled molecule of DNA | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| glucose + oxygen -> carbon dioxide + water | B1 |
| the mitochondria | B1 |
| Final answer: Glucose + oxygen -> carbon dioxide + water; occurs in the mitochondria | |
| Question 4[3 marks] | |
|---|---|
| Answer or working | Marks |
| stomata are small pores/openings, mostly on the underside of a leaf, that allow gases to diffuse in and out | B1 |
| carbon dioxide diffuses in through the stomata for photosynthesis, and oxygen (and excess water vapour) diffuses out | B1 |
| guard cells surround each stoma and can change shape to open or close it, controlling gas exchange (and water loss) | B1 |
| Final answer: Stomata are pores that let carbon dioxide diffuse in and oxygen/water vapour diffuse out; guard cells around each stoma open or close it to control this exchange | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| photosynthesis | B1 |
| a valid return process: respiration (by plants, animals or microorganisms) | B1 |
| another valid return process, such as combustion (burning fossil fuels or wood), or the decomposition of dead organisms | B1 |
| Final answer: Photosynthesis removes carbon dioxide; respiration, combustion and decomposition all return carbon dioxide to the atmosphere | |
| Question 6[3 marks] | |
|---|---|
| Answer or working | Marks |
| embryonic stem cells can differentiate into any type of cell (they are pluripotent) | B1 |
| adult stem cells can only differentiate into a limited range of cell types | B1 |
| a named source, such as adult stem cells being found in bone marrow while embryonic stem cells are found in early embryos | B1 |
| Final answer: Embryonic stem cells can become any cell type; adult stem cells (e.g. in bone marrow) can only become a limited range of cell types | |
| Question 7[3 marks] | |
|---|---|
| Answer or working | Marks |
| the central nervous system is made up of the brain and spinal cord | B1 |
| information from receptors is sent to the CNS as electrical impulses along neurones | B1 |
| the CNS coordinates the response by sending electrical impulses to effectors (muscles or glands) along neurones | B1 |
| Final answer: The CNS is made up of the brain and spinal cord; it receives impulses from receptors and coordinates a response by sending impulses to effectors | |
| Question 8[3 marks] | |
|---|---|
| Answer or working | Marks |
| organisms are classified into a hierarchy of groups based on their similarities and differences | B1 |
| the largest group is the kingdom, which is progressively subdivided into smaller and smaller groups | B1 |
| the smallest group is the species, and organisms are given a scientific name based on their genus and species (binomial naming) | B1 |
| Final answer: Organisms are classified into a hierarchy of groups from kingdom down to species, based on their similarities and differences, with each organism given a two-part (binomial) scientific name from its genus and species | |
| Question 9[3 marks] | |
|---|---|
| Answer or working | Marks |
| sterilising the inoculating loop (e.g. in a flame) before and after use, to kill any unwanted microorganisms | B1 |
| taping the lid of the Petri dish rather than sealing it completely, allowing gas exchange while restricting contaminating microorganisms from entering | B1 |
| incubating the plate at a maximum of 25 deg C in a school laboratory, to reduce the growth of pathogens that are harmful to humans | B1 |
| Final answer: Sterilise the inoculating loop before and after use, tape (rather than seal) the Petri dish lid, and incubate at a maximum of 25 deg C to reduce the growth of pathogens harmful to humans | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| converting the field of view diameter to micrometres: 0.6 mm = 600 micrometres | M1 |
| using mean cell width = field of view diameter / number of cells | M1 |
| substituting 600 / 15 | M1 |
| 40 micrometres | A1 |
| Question 11[4 marks] | |
|---|---|
| Answer or working | Marks |
| using dose = rate per kg x body mass | M1 |
| the 65 kg patient, 0.4 x 65 | M1 |
| 26 units | A1 |
| the 80 kg patient, 0.4 x 80 = 32 units | A1independent |
| Final answer: 26 units for the 65 kg patient and 32 units for the 80 kg patient | |
| Question 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| Monday percentage = (18/600) x 100 = 3% | M1 |
| Friday percentage = (54/600) x 100 = 9% | M1 |
| percentage change = ((9 - 3)/3) x 100 | M1 |
| a 200% increase | A1 |
| Final answer: Monday 3%, Friday 9%, a 200% increase over the week | |
| Question 13[4 marks] | |
|---|---|
| Answer or working | Marks |
| using speed = distance / time | M1 |
| substituting 1.2 / 0.01 | M1 |
| 120 metres per second | A1 |
| time = 3 / 120 = 0.025 seconds | A1independent |
| Final answer: Speed = 120 m/s; it would take 0.025 seconds to travel 3 m | |
| Question 14[5 marks] | |
|---|---|
| Answer or working | Marks |
| using percentage difference = 100 - percentage similarity | M1 |
| species B, 100 - 98 | M1 |
| 2% different | A1 |
| species C 15% different and species D 40% different | A1independent |
| species B being most closely related to species A (smallest difference) and species D being most distantly related (largest difference) | B1independent |
| Final answer: Species B differs by 2%, species C by 15%, and species D by 40%; species B is most closely related to species A, and species D is most distantly related | |
| Question 15[5 marks] | |
|---|---|
| Answer or working | Marks |
| area covered in the quadrat = 60% of 0.5 m^2 | M1 |
| 0.3 m^2 | A1 |
| using the same percentage (60%) applied to the total area | M1 |
| substituting 60% of 400 m^2 | M1 |
| 240 m^2 | A1 |
| Final answer: 0.3 m^2 of the quadrat was covered by seaweed; an estimated 240 m^2 of the total 400 m^2 shore is covered by seaweed | |
| Question 16[5 marks] | |
|---|---|
| Answer or working | Marks |
| the exposed (uncovered) part of the leaf turns blue-black with iodine, showing starch is present | B1 |
| the covered part of the leaf, which was kept in the dark, stays orange/brown with iodine, showing no starch is present | B1 |
| this showing light is needed for a plant to produce starch (via photosynthesis) | B1 |
| before the test, the plant should be left in the dark (destarched) for at least 24 hours, to remove any starch already stored in the leaves | B1 |
| this ensuring any starch found afterwards must have been made during the test itself, rather than being present beforehand | B1 |
| Final answer: The uncovered part turns blue-black (starch present) but the foil-covered part stays orange/brown (no starch), showing light is needed for starch production; the plant should first be destarched by leaving it in the dark for 24 hours, so any starch found is from the test itself | |
| Question 17[5 marks] | |
|---|---|
| Answer or working | Marks |
| the test strip contains monoclonal antibodies that are specific to (bind only to) hCG | B1 |
| the monoclonal antibodies are bound to a coloured/dye particle | B1 |
| if hCG is present in the urine sample, it binds to the monoclonal antibody, and this complex moves along the strip | B1 |
| the complex binds to a second set of antibodies fixed in a line on the strip, concentrating the coloured particles there | B1 |
| this produces a visible coloured line, giving a positive result; a negative result shows no line because there is no hCG to bind and be concentrated at that line | B1 |
| Final answer: hCG in the urine binds to dye-labelled monoclonal antibodies specific to it, and this complex is trapped at a fixed line on the strip, producing a visible coloured line for a positive result; with no hCG, no line forms | |