A Level Biology (A2) MCQs with answers

Practise A Level Biology (A2) (9700) with 437 exam-style MCQs, each with the answer and a short explanation. Every test is marked the moment you finish and shows your score chapter by chapter, so you know what to revise next. It is free and needs no sign-up.

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A Level Biology (A2) MCQs with answers

Practise A Level Biology (A2) (9700) with 437 exam-style MCQs, each with the answer and a short explanation. Every test is marked the moment you finish and shows your score chapter by chapter, so you know what to revise next. It is free and needs no sign-up.

437 questions · 8 chapters

What each chapter covers 8 chapters

The questions follow the syllabus chapter by chapter. You can test the whole subject or one chapter at a time.

  1. Energy and respiration 56 questionsEnergy, Respiration
  2. Photosynthesis 48 questionsPhotosynthesis as an energy transfer process, Investigation of limiting factors
  3. Homeostasis 53 questionsHomeostasis in mammals, Homeostasis in plants
  4. Control and coordination 55 questionsControl and coordination in mammals, Control and coordination in plants
  5. Inheritance 61 questionsPassage of information from parents to offspring, The roles of genes in determining the phenotype, Gene control
  6. Selection and evolution 48 questionsVariation, Natural and artificial selection, Evolution
  7. Classification, biodiversity and conservation 58 questionsClassification, Biodiversity, Conservation
  8. Genetic technology 58 questionsPrinciples of genetic technology, Genetic technology applied to medicine, Genetically modified organisms in agriculture
Sample A Level Biology (A2) MCQs with answers 12 questions

12 questions from the test, one or two from each chapter. Try each one, then open the answer.

1. Which gives the usual order of energy value per gram of respiratory substrate, from highest to lowest?

  1. A
    carbohydrate, protein, lipid
  2. B
    lipid, protein, carbohydrate
  3. C
    lipid, carbohydrate, protein
  4. D
    protein, lipid, carbohydrate
Show answer

Answer: B. Typical values are about 39 kJ g−1 for lipid, 17 kJ g−1 for protein and 16 kJ g−1 for carbohydrate. Protein is only slightly higher than carbohydrate.

2. Methylene blue can be used instead of DCPIP with a chloroplast suspension. In the light, which colour change shows that the light-dependent reactions are taking place?

  1. A
    colourless to blue, as it is oxidised
  2. B
    blue to colourless, as it is reduced
  3. C
    blue to colourless, as it is oxidised
  4. D
    colourless to blue, as it is reduced
Show answer

Answer: B. Like DCPIP, methylene blue is blue when oxidised and becomes colourless when it is reduced by electrons from the light-dependent reactions.

3. Glucose appears in the urine only when blood glucose concentration rises above a certain value. Why?

  1. A
    above this value, the glomerulus stops filtering glucose
  2. B
    above this value, the cotransporter proteins in the proximal tubule are saturated
  3. C
    above this value, glucose is actively secreted into the filtrate in the collecting duct
  4. D
    above this value, insulin prevents glucose from being reabsorbed
Show answer

Answer: B. Reabsorption depends on a limited number of cotransporter proteins. When the filtrate contains more glucose than they can carry, the excess stays in the tubule and passes into the urine.

4. How does auxin cause cells in a shoot to elongate?

  1. A
    it stimulates pumping of protons into the cytoplasm, which raises the pH of the cell wall and loosens it
  2. B
    it stimulates proton pumping into the cell wall, and the low pH activates expansins that loosen it
  3. C
    it breaks down all the cellulose so that the cell wall is removed
  4. D
    it makes the cell wall more rigid by adding lignin
Show answer

Answer: B. Auxin causes protons to be pumped from the cytoplasm into the cell wall. The lower pH activates expansins, which loosen the links between cellulose microfibrils, so the wall stretches as water enters by osmosis.

5. A dwarf pea plant of genotype lele is sprayed regularly with active gibberellin and grows as tall as a plant of genotype LeLe. Which conclusion is supported?

  1. A
    the spraying has changed the plant's genotype, so its offspring will also grow tall
  2. B
    the le allele codes for a faulty receptor
  3. C
    gibberellin makes the stem taller only by increasing the rate of cell division in the roots
  4. D
    the plant can respond to gibberellin but cannot make enough of it
Show answer

Answer: D. Adding the hormone restores normal growth, so the plant's response pathway works and its problem is making active gibberellin, as expected for a non-functional enzyme. A change in phenotype caused by spraying does not change the genotype, so the offspring will still be lele dwarfs.

6. An antibiotic was added to a culture of bacteria. The number of living bacteria fell sharply for several hours, then rose again, even though the antibiotic was still present. Which explanation is best?

  1. A
    the antibiotic had all been broken down
  2. B
    a few bacteria were already resistant; they survived and multiplied
  3. C
    the surviving bacteria were exposed to the antibiotic for long enough to develop resistance
  4. D
    the antibiotic made the survivors mutate
Show answer

Answer: B. With the antibiotic still present and working, only bacteria able to survive it can explain the rise. They were resistant before exposure because of a chance mutation; selection removed the rest, so the new population is mainly resistant.

7. A population of an endangered frog lives in a single protected pond and has declined to 40 individuals. Why is it still at high risk of extinction even though its habitat is now protected?

  1. A
    small populations always have higher mutation rates, producing harmful alleles
  2. B
    inbreeding in a small population increases the proportion of heterozygotes
  3. C
    low genetic diversity may leave it unable to adapt to disease or climate change
  4. D
    a small population cannot be affected by competition from other species
Show answer

Answer: C. A small population has lost much of its genetic variation, so it is less likely to include individuals with alleles that allow survival when conditions change, such as a new disease or a warmer climate. Inbreeding increases homozygosity, and chance events can also wipe out a small population.

8. The herbicide glyphosate inhibits a plant enzyme needed to make certain amino acids. GM soybean contains a bacterial gene for a form of this enzyme that is not inhibited by glyphosate. Why does the GM soybean survive being sprayed?

  1. A
    it can still make these amino acids, and so its proteins
  2. B
    it breaks down all the glyphosate before it can enter any of its cells
  3. C
    it no longer needs these amino acids, so blocking the enzyme has no effect
  4. D
    it absorbs the amino acids from soil
Show answer

Answer: A. The bacterial enzyme keeps working in the presence of glyphosate, so the GM plant continues to make the amino acids needed for protein synthesis and survives. Weeds, with only the sensitive enzyme, die.

9. Palmitic acid is respired aerobically: C16H32O2 + 23O2 → 16CO2 + 16H2O. What is the RQ?

  1. A
    1.00
  2. B
    1.44
  3. C
    0.70
  4. D
    0.35
Show answer

Answer: C. RQ = 16 ÷ 23 = 0.70, typical of lipids. 1.44 is the ratio upside down, and 1.00 compares carbon dioxide with water instead of with oxygen.

10. In a DCPIP experiment with a chloroplast suspension, which control shows that the colour change depends on light?

  1. A
    a tube with DCPIP and buffer but no chloroplasts
  2. B
    an identical tube containing boiled chloroplasts
  3. C
    a tube with chloroplasts and water instead of DCPIP
  4. D
    an identical tube wrapped in aluminium foil
Show answer

Answer: D. Only light is changed in the foil-wrapped tube, so if it stays blue, the reduction of DCPIP needs light. The boiled chloroplast tube tests whether active chloroplasts are needed.

11. Which feature distinguishes guard cells from most other cells of the lower epidermis?

  1. A
    guard cells have no cell wall at all
  2. B
    guard cells have no vacuole in their cytoplasm
  3. C
    guard cells contain chloroplasts
  4. D
    guard cells are dead, with lignified walls
Show answer

Answer: C. Guard cells have chloroplasts, whereas most other epidermal cells do not. They have living contents, a large vacuole and unevenly thickened cellulose walls.

12. Which changes are seen in a sarcomere when a muscle contracts?

  1. A
    the A band shortens, but the I band and H zone stay the same length
  2. B
    all the bands shorten by the same amount
  3. C
    the A band stays the same, but the I band and H zone shorten
  4. D
    the H zone lengthens and the I band shortens
Show answer

Answer: C. The filaments themselves do not shorten; they slide past each other. The A band (the length of the myosin filaments) is unchanged, while the regions without overlap, the I band and H zone, get shorter.

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