OxfordAQA International AS and A-level Biology: HIV and plant virus transmission by aphids (9610) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions practise HIV (section 3.2.4) and the spread of plant viruses by aphids (section 3.2.9) from the OxfordAQA International AS and A-level Biology (9610) specification, Version 5.1, Unit 2, plus the principle that viruses are acellular and non-living. Learn the content first from the study guide and the revision notes. More links: the course hub, the checklist and the free diagnostics.

Questions

1. State the part of an HIV particle that:

(a) originates from the host cell plasma membrane [1] (b) binds to CD4 receptors [1] (c) makes DNA using viral RNA as a template [1] (d) encloses the RNA and the viral enzymes [1]

2. (a) A drawing of an HIV particle is 48 mm across. The actual diameter is 120 nm. Calculate the magnification. [2]

(b) A rod-shaped plant virus is 650 nm long. A drawing of it is 52 mm long. Calculate the magnification. [2]

3. (a) Explain why viruses are described as acellular and non-living. [3] (b) An antibiotic blocks the synthesis of bacterial cell walls. Explain why it has no effect on HIV. [2]

4. Describe the replication cycle of HIV in a helper T-cell, from binding to release of new virus particles. [5]

5. Three drugs are tested on cells infected with HIV. Observations:

• Drug P: viral RNA is found inside the cells, but no viral DNA is detected.
• Drug Q: viral DNA is found in the cytoplasm, but none is found in the host chromosomes, and no new viral RNA is made.
• Drug R: viral RNA and viral proteins are made and particles are released. The particles contain uncut protein chains and cannot infect new cells.

For each drug, name the viral enzyme it most likely inhibits and explain your choice. [6]

6. A patient starts drug treatment. Before treatment: viral load 64 000 copies per mm³ and CD4 helper T-cell count 150 cells per mm³. After 6 months: viral load 2 000 copies per mm³ and CD4 count 360 cells per mm³.

(a) Calculate the percentage increase in CD4 count. [2] (b) Calculate how many times smaller the viral load is after 6 months. [1] (c) The fall is also given as a change in log₁₀ of viral load. Calculate this change to 2 significant figures. [1] (d) Explain the link between the fall in viral load and the rise in CD4 count. [3]

7. For this question, assume reverse transcriptase makes one error in every 5 000 bases it copies and that the HIV genome is 10 000 bases long (figures simplified).

(a) Calculate the expected number of errors in one copy of the genome. [1] (b) Suggest why drug-resistant forms of HIV can arise during treatment. [2] (c) Patients are often given three drugs that act on different viral enzymes together. Suggest why this reduces resistance. [2]

8. (a) State two symptoms of a plant virus disease. [2] (b) Explain why yellowing of leaves reduces crop yield. [2] (c) Suggest why the transmission of plant viruses usually involves a vector. [1]

9. (a) Describe how an aphid obtains food from a plant. [3] (b) An aphid excretes 9 drops of liquid, each of volume 0.4 µL, in 20 minutes. Calculate the rate of excretion in µL per hour. [2]

10. A farmer monitors 500 potato plants for a virus spread by aphids. A trap catches winged aphids each week. The cumulative number of infected plants is recorded.

Week	1	2	3	4	5
Winged aphids caught	6	20	75	40	12
Cumulative infected plants	4	10	25	61	94

(a) Calculate the percentage of plants infected at the end of week 5. [1] (b) Calculate the number of new infections in week 4. [1] (c) Describe the relationship between the trap catches and the new infections. [2] (d) The farmer concludes that trapped aphids caused the infections. Evaluate this conclusion. [2]

11. Compare how HIV and an aphid-borne plant virus reach and take over host cells. [8]

Answers

1. (a) The envelope [1]; (b) glycoproteins [1]; (c) reverse transcriptase [1]; (d) the capsid [1] [4] Examiner insight: Name the part, not just its job. “Glycoprotein” gains credit where “protein” alone may be too vague.

2. (a) 48 mm = 48 000 000 nm [1]; 48 000 000 / 120 = ×400 000 [1] (b) 52 mm = 52 000 000 nm; 52 000 000 / 650 [1] = ×80 000 [1] [4] Examiner insight: Convert units before dividing. Writing the working shows the unit step even when the final value is slightly wrong.

3. (a) Not made of cells, with no cytoplasm or ribosomes [1]; no metabolism of its own, so no respiration or growth [1]; can only replicate by using host cell machinery [1]. (b) Viruses have no cell wall to attack [1]; HIV replicates inside host cells using host synthesis, not bacterial processes [1] [5] Examiner insight: “Not living” alone is a statement. Reasons that refer to missing cell structures or metabolism are what the question is asking for.

4. Glycoproteins bind to CD4 receptors on the helper T-cell [1]; capsid contents (RNA and enzymes) released into the cell [1]; reverse transcriptase makes double-stranded DNA from the RNA [1]; integrase inserts the DNA into a host chromosome [1]; viral DNA transcribed, proteins made, then RNA and proteins packaged and released as new particles [1] [5] Examiner insight: Order and the enzyme names are what separate answers. Ending with the new particles being released completes the cycle.

5. Drug P: reverse transcriptase [1]; RNA is present, so entry occurred, but no DNA made from it [1]. Drug Q: integrase [1]; DNA is made but not inserted into a chromosome, so no viral RNA is made from it [1]. Drug R: protease [1]; proteins are made but not cut to working lengths, so particles are not infectious [1] [6] Examiner insight: Show the evidence that the earlier steps worked (RNA present, DNA made, proteins made) before naming the blocked step.

6. (a) (360 − 150) / 150 × 100 [1] = 140% [1] (b) 64 000 / 2 000 = 32 times smaller [1] (c) log₁₀ 64 000 − log₁₀ 2 000 = log₁₀ 32 = 1.5 [1] (d) Less virus means fewer helper T-cells are infected [1]; fewer are destroyed as particles are released [1]; so the count can recover [1] [7] Examiner insight: A percentage change divides by the starting value. Dividing by the final value gives a different, wrong percentage.

7. (a) 10 000 / 5 000 = 2 errors [1] (b) Copying makes errors, so mutations occur in the genome [1]; a mutation may alter an enzyme so the drug no longer fits, and that form survives and multiplies [1] (c) A form resistant to one drug is unlikely to also be resistant to the other two [1]; so the virus is still blocked at a different step [1] [5] Examiner insight: Use the number from part (a) in your argument. Linking a mutation to an altered enzyme shape explains resistance rather than just naming it.

8. (a) Any two from: yellowing of leaves; leaf distortion; abnormalities of other parts of the plant [2] (b) Less chlorophyll, so less light absorbed [1]; less photosynthesis, so less organic material for growth [1] (c) Plants are immobile, so the virus cannot pass between them directly [1] [5] Examiner insight: “Plants are immobile” is the idea tied to vectors. Leave out speculation about contact between plants.

9. (a) Aphid inserts its stylets into a phloem vessel [1]; phloem sap is under high pressure [1]; the sap is forced into the aphid’s gut once the vessel is punctured [1] (b) Volume = 9 × 0.4 = 3.6 µL in 20 minutes [1]; 3.6 × 3 = 10.8 µL per hour [1] [5] Examiner insight: For a rate, state the volume and the time first. Converting 20 minutes to an hour is an easy step to miss.

10. (a) 94 / 500 × 100 = 18.8% [1] (b) 61 − 25 = 36 [1] (c) New infections rise as catches rise [1]; the trap peak (week 3) comes before the infection peak (week 4), so there is a time lag [1] (d) The data show a correlation only [1]; further evidence such as testing trapped aphids for the virus, or a control field, is needed [1] [6] Examiner insight: “Correlation does not prove causation” gains most credit when you also say what extra evidence would test the claim.

11. Any eight from: HIV binds by glycoproteins to CD4 receptors on helper T-cells [1]; plant virus is carried to the plant by a vector, an aphid [1]; the aphid pushes stylets into phloem [1]; sap under pressure enters the aphid gut, so virus can be taken up and later passed on [1]; winged females spread it to new plants; HIV releases RNA and enzymes into the host cell [1]; reverse transcriptase makes DNA, and integrase inserts it into a chromosome [1]; both viruses are acellular and replicate using host machinery [1]; new particles are made and released, and HIV takes host membrane as an envelope [1] [8] Examiner insight: A comparison needs both viruses in each point. Listing HIV facts and then plant virus facts separately may not show the link.

Where marks are usually lost

• Writing that a virus divides, or treating it as a small cell.
• Writing “protein” for glycoprotein, or saying the virus makes its own envelope.
• Mixing reverse transcriptase and integrase.
• Saying the virus “enters the nucleus” without mentioning integration into a chromosome.
• Naming a drug rather than the step blocked and what is missing afterwards.
• Dividing percentage changes by the final value.
• Forgetting unit conversion in magnification and rate calculations.
• Saying aphids bite leaves or suck sap by muscle power.
• Treating a correlation as proof of causation.

Next steps

• Revise with the revision notes and re-read the study guide.
• Course hub: OxfordAQA A-level Biology and printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Biology (9610), Version 5.1, published by OxfordAQA: sections 3.2.4 and 3.2.9 of Unit 2.