GCSE Biology Foundation Paper 2
Covers Cell Structure and Transport, Cell Division and Stem Cells, Digestion and the Circulatory System and 9 more.
Questions
Question 1 [2 marks]
Ecosystems and Material Cycles
State what is meant by a producer and a decomposer in an ecosystem.
Question 2 [2 marks]
Hormones and Homeostasis
State what a hormone is, and state which system of the body produces hormones.
Question 3 [2 marks]
Photosynthesis
State where in a plant cell photosynthesis takes place, and state the raw materials needed for photosynthesis.
Question 4 [2 marks]
Pathogens and Communicable Disease
State two differences between a virus and a bacterium.
Question 5 [3 marks]
Respiration and Exercise
During vigorous exercise, muscle cells may not receive enough oxygen.
Describe what happens to glucose in muscle cells under these conditions, and describe one effect this has on the muscles.
Question 6 [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 7 [3 marks]
Digestion and the Circulatory System
State three differences in structure between an artery and a vein.
Question 8 [3 marks]
The Nervous System and Reflexes
Describe why reflex actions are important, and describe one feature of the reflex arc that makes reflex actions fast.
Question 9 [3 marks]
Genetics and Inheritance
Describe how DNA can be extracted from fruit cells in a simple school practical, including the general purpose of adding detergent to the mixture.
Question 10 [4 marks]
Evolution and Natural Selection
In a polluted area, a population of moths was studied over 10 years. The percentage of dark-coloured moths increased from 20% to 80% of the population.
Calculate the percentage increase in the proportion of dark-coloured moths over the 10 years.
Question 11 [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 12 [4 marks]
Cell Division and Stem Cells
A single cell divides by mitosis every 20 minutes.
Starting from one cell, calculate how many cells will be present after 2 hours, assuming every cell divides on schedule.
Question 13 [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 14 [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 15 [4 marks]
Ecosystems and Material Cycles
A student placed ten 1 m^2 quadrats randomly in a field measuring 500 m^2 and counted a total of 60 daisies across all ten quadrats.
Calculate an estimate for the total number of daisies in the field.
Question 16 [4 marks]
Digestion and the Circulatory System
A student investigated the effect of temperature on the rate of starch digestion by amylase.
The time taken for starch to disappear was 240 seconds at 20 deg C and 60 seconds at 35 deg C.
Calculate the rate of reaction at each temperature in arbitrary units of 1/time (s^-1), then state which temperature gave the faster rate.
Question 17 [4 marks]
Ecosystems and Material Cycles
A student estimates the population of snails in a garden using the mark-release-recapture method. On the first day, 40 snails are caught, marked, and released. The next day, 50 snails are caught, of which 8 are found to be marked.
Calculate an estimate of the total population of snails in the garden, using population estimate = (number marked in first sample x total number in second sample) / number of marked individuals recaptured.
Question 18 [5 marks]
Cell Structure and Transport
A student sets the eyepiece lens of a microscope to a magnification of x10 and the objective lens to a magnification of x40.
Calculate the total magnification produced by the microscope. The student then measures a cell using this microscope and finds its magnified image is 8000 micrometres wide. Calculate the actual width of the cell in micrometres.
Model solutions
| Question 1[2 marks] | |
|---|---|
| Answer or working | Marks |
| a producer is an organism, such as a green plant or alga, that makes its own food by photosynthesis | B1 |
| a decomposer is a microorganism that breaks down dead organic matter or waste, releasing nutrients back into the ecosystem | B1 |
| Final answer: A producer makes its own food by photosynthesis; a decomposer breaks down dead matter, releasing nutrients | |
| Question 2[2 marks] | |
|---|---|
| Answer or working | Marks |
| a hormone is a chemical messenger, produced by a gland and carried in the blood to a target organ, where it produces an effect | B1 |
| hormones are produced by glands, which make up the endocrine system | B1 |
| Final answer: A hormone is a chemical messenger made by a gland and carried in the blood to a target organ; hormones are produced by glands of the endocrine system | |
| Question 3[2 marks] | |
|---|---|
| Answer or working | Marks |
| the chloroplasts | B1 |
| carbon dioxide and water | B1 |
| Final answer: Photosynthesis takes place in the chloroplasts, using carbon dioxide and water as raw materials | |
| Question 4[2 marks] | |
|---|---|
| Answer or working | Marks |
| viruses are much smaller than bacteria | B1 |
| viruses can only reproduce inside a host cell, whereas bacteria can reproduce independently (by simple cell division) | B1 |
| Final answer: Viruses are much smaller than bacteria, and can only reproduce inside a host cell, while bacteria can reproduce independently | |
| Question 5[3 marks] | |
|---|---|
| Answer or working | Marks |
| glucose is broken down without using oxygen (anaerobic respiration) | B1 |
| the products are lactic acid and a small amount of energy | B1 |
| the build-up of lactic acid causes muscle fatigue and the muscle stops contracting efficiently | B1 |
| Final answer: Glucose is broken down anaerobically to lactic acid, releasing a small amount of energy; the build-up of lactic acid causes muscle fatigue | |
| Question 6[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 7[3 marks] | |
|---|---|
| Answer or working | Marks |
| arteries have thicker, more muscular walls than veins | B1 |
| arteries have a narrower lumen than veins | B1 |
| veins have valves to prevent the backflow of blood, but arteries do not | B1 |
| Final answer: Arteries have thicker, more muscular walls, a narrower lumen, and (unlike veins) no valves | |
| Question 8[3 marks] | |
|---|---|
| Answer or working | Marks |
| reflex actions protect the body from harm, as they are automatic, rapid responses to a stimulus | B1 |
| they do not require conscious thought or decision by the brain | B1 |
| the pathway (reflex arc) is short, with a synapse at the spinal cord rather than passing through the brain | B1 |
| Final answer: Reflexes protect the body with automatic, rapid responses that bypass conscious decision-making because the reflex arc's pathway is short and does not pass through the brain | |
| Question 9[3 marks] | |
|---|---|
| Answer or working | Marks |
| the fruit is broken down/mashed, then mixed with a salt solution and a detergent (e.g. washing-up liquid) | B1 |
| the detergent breaks down (disrupts) the cell membranes, releasing the DNA from inside the cells | B1 |
| cold ethanol (alcohol) is then carefully added/layered on top, causing the DNA to become insoluble and visible, so it can be collected | B1 |
| Final answer: Fruit is mashed with salt solution and detergent, which breaks down cell membranes to release DNA; cold ethanol is then added, making the DNA visible so it can be collected | |
| Question 10[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the increase = 80 - 20 = 60 percentage points | M1 |
| using percentage change = (increase / original) x 100 | M1 |
| substituting (60/20) x 100 | M1 |
| a 300% increase | A1 |
| Final answer: 300% increase | |
| Question 11[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 12[4 marks] | |
|---|---|
| Answer or working | Marks |
| recognising 2 hours contains 6 divisions (120 / 20) | M1 |
| using the pattern that the cell number doubles at each division | M1 |
| calculating 2^6 | M1 |
| 64 cells | A1 |
| Question 13[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 14[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 15[4 marks] | |
|---|---|
| Answer or working | Marks |
| finding the mean number of daisies per quadrat = 60 / 10 = 6 | M1 |
| recognising this represents the mean number per 1 m^2 | M1 |
| multiplying the mean by the total field area, 6 x 500 | M1 |
| 3000 daisies | A1 |
| Question 16[4 marks] | |
|---|---|
| Answer or working | Marks |
| using rate = 1 / time | M1 |
| rate at 20 deg C = 1/240 = 0.00417 s^-1 (accept 4.17 x 10^-3) | M1 |
| rate at 35 deg C = 1/60 = 0.0167 s^-1 (accept 1.67 x 10^-2) | M1 |
| 35 deg C gave the faster rate, because its rate value is larger | A1 |
| Final answer: Rate at 20 deg C = 0.00417 s^-1, rate at 35 deg C = 0.0167 s^-1; 35 deg C is faster | |
| Question 17[4 marks] | |
|---|---|
| Answer or working | Marks |
| using the given formula, population estimate = (number marked in first sample x total in second sample) / number recaptured marked | M1 |
| substituting (40 x 50) / 8 | M1 |
| calculating 40 x 50 = 2000 | M1 |
| 250 snails | A1 |
| Final answer: An estimated population of 250 snails | |
| Question 18[5 marks] | |
|---|---|
| Answer or working | Marks |
| total magnification = eyepiece magnification x objective magnification | M1 |
| total magnification = 10 x 40 = x400 | A1 |
| using actual size = image size / magnification | M1 |
| substituting 8000 / 400 | M1 |
| actual width = 20 micrometres | A1 |
| Final answer: Total magnification = x400; actual width of the cell = 20 micrometres | |