Pearson Edexcel International A Level Biology: Sensory Reception, Plant Hormones and the Brain (YBI11) – 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 cover statements 8.8 to 8.16 (including Core Practical 18) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology specification (YBI11), Issue 2. Topic 8 is International A2 (IA2) and there are no tiers. Learn the content in the study guide and revision notes first. The course hub and checklist show the wider course.

Questions

1. Describe how the closing of cation channels in a rod cell leads to an action potential in an optic neurone. [4]

2. The membrane potential of a rod cell was -40 mV in the dark and -65 mV in bright light. (a) Calculate the change in membrane potential. [1] (b) Name this change. [1] (c) Explain why the potential changes in the light. [2]

3. Tapping the glass of a fish tank every 10 seconds made fish dart away. The number of darts in one minute was 36 in minute 1 and 9 in minute 20. (a) Define habituation. [1] (b) Calculate the percentage decrease in darting. [2] (c) Suggest one advantage to the fish of habituation. [1] (d) Suggest how you could show that the fish were not simply tired. [1]

4. Seeds of a small plant were exposed to light. Germination was 84% after red light, 9% after red light then far-red light, and 81% after red, far-red, then red light. (a) State which form of phytochrome is present after the far-red light. [1] (b) Explain these results. [3] (c) State one other response controlled by phytochrome. [1]

5. The tip of a coleoptile was removed and an agar block containing IAA was placed on the left half of the cut surface. The coleoptile was kept in darkness. Three coleoptiles curved by 12°, 15° and 18°. (a) Calculate the mean angle. [1] (b) State the direction of curvature. [1] (c) Explain the curvature. [3]

6. Amylase activity (arbitrary units) in germinating barley grains was 2 at 0 h, 5 at 12 h, 18 at 24 h, 41 at 36 h and 66 at 48 h. (a) Describe the change in activity. [2] (b) Calculate the mean rate of increase between 24 h and 48 h. [2] (c) Explain how gibberellin causes the increase. [3]

7. In Core Practical 18, grain halves were placed on starch agar. After incubation the plates were flooded with iodine solution. Mean clear zone diameters were 15.0 mm for halves with an embryo, 0 mm for halves without an embryo, and 13.0 mm for halves without an embryo plus gibberellic acid. (a) Calculate the area of the clear zone around the halves with an embryo. [2] (b) Explain the three results. [3] (c) State why boiled grain halves were used as a control. [1] (d) Suggest why the grain surfaces were sterilised. [1]

8. (a) Compare nervous and hormonal coordination in animals. [4] (b) State what the central nervous system and the peripheral nervous system are. [2]

9. Name the region of the brain that: (a) controls ventilation rate; (b) monitors blood temperature; (c) coordinates balance; (d) is responsible for language; (e) releases antidiuretic hormone. [5]

10. Scans are used to investigate the brain. (a) Explain how fMRI shows which region is active. [2] (b) A PET tracer labelled with fluorine-18 (half-life 110 minutes) has an activity of 800 MBq when made. Calculate its activity 330 minutes later. [2] (c) Suggest why the tracer must be made close to the hospital. [1] (d) A patient may have bleeding in the brain after a head injury. Suggest which scan is used and why. [2]

11. (a) Explain how loss of dopamine-releasing neurones affects movement and why L-DOPA is given. [3] (b) Explain how a selective serotonin reuptake inhibitor (SSRI) helps in depression. [2] (c) In a trial, 54 of 120 patients on an SSRI improved and 30 of 120 on a placebo improved. Calculate the difference in percentage improved. [2] (d) Explain why the trial was double-blind. [1]

Answers

1. Sodium ions no longer enter the outer segment, so the rod is hyperpolarised [1]. The rod releases less neurotransmitter [1]. The bipolar neurone is no longer inhibited, so it is depolarised and releases neurotransmitter [1]. If the optic neurone reaches threshold, an action potential is generated [1]. [4] Examiner insight: A sequence of linked steps is rewarded, so write the answer in order; stopping at “the rod is hyperpolarised” misses the later steps.

2. (a) -65 - (-40) = -25 mV (25 mV more negative) [1]. (b) Hyperpolarisation [1]. (c) In the light, rhodopsin splits and the cation channels close, so sodium ions no longer enter [1]. Sodium ions are still pumped out, so the inside becomes more negative [1]. [4] Examiner insight: State the sign and the units. Giving “25 mV” without saying that it is a fall loses the point.

3. (a) A decrease in response to a stimulus that is repeated many times and is neither harmful nor rewarding [1]. (b) (36 - 9) ÷ 36 × 100 [1] = 75.0% [1]. (c) The fish save energy by not responding to a harmless stimulus, and stay alert to real threats [1]. (d) Give a new stimulus, such as a shadow, and show that the fish still respond fully [1]. [5] Examiner insight: On “define”, include both repetition and the lack of consequence; “gets used to it” alone is too vague.

4. (a) Pr [1]. (b) Red light converts Pr to Pfr, and Pfr promotes germination, giving 84% [1]. Far-red light converts Pfr back to Pr, so germination falls to 9% [1]. The final red light makes Pfr again, so germination recovers to 81%, and the last light given controls the result [1]. (c) Flowering, or stem elongation / etiolation [1]. [5] Examiner insight: Use the percentages and say which light came last; naming the forms without the data does not answer “explain these results”.

5. (a) (12 + 15 + 18) ÷ 3 = 15° [1]. (b) To the right, away from the block [1]. (c) IAA diffuses from the block into the cells on the left [1]. Cells on the left elongate more, for example by wall loosening and water uptake [1]. The left side becomes longer than the right, so the coleoptile bends to the right [1]. [5] Examiner insight: “IAA makes the cells grow” loses a point. Say cell elongation and compare the two sides.

6. (a) Activity stays low up to 12 h, then increases faster, reaching 66 units at 48 h [1]. The increase is greatest between 24 and 48 h, with figures quoted [1]. (b) (66 - 18) ÷ (48 - 24) [1] = 2.0 units per hour [1]. (c) The embryo releases gibberellin, which diffuses to the aleurone layer [1]. A repressor is removed, so transcription of the amylase gene is switched on [1]. Amylase hydrolyses starch to maltose, which is converted to glucose for the embryo [1]. [7] Examiner insight: Show the subtraction and the time interval, so that your method is clear even if a digit slips.

7. (a) Radius = 7.5 mm [1]. Area = π × 7.5² = 177 mm² [1]. (b) Halves without an embryo have no gibberellin source, so the amylase gene is not switched on and no starch is digested [1]. Added gibberellic acid switches on transcription of the amylase gene in the aleurone layer [1]. Amylase digests starch, so no blue-black colour with iodine is seen and a clear zone is formed [1]. (c) Boiled grain cannot make active enzyme, so it shows that clear zones are caused by amylase made by living grain [1]. (d) To remove microorganisms that could secrete amylase [1]. [7] Examiner insight: Say that the clear zone is where starch has been digested; saying it is where starch is “present” reverses the meaning.

8. (a) Any four paired comparisons, for example: nervous signals are electrical impulses and neurotransmitter, hormonal signals are chemicals in the blood [1]. Nervous responses are faster and hormonal responses slower [1]. Nervous effects are short-lived and localised, hormonal effects are longer-lasting [1] and may act on many target cells with receptors [1]. (b) The CNS consists of the brain together with the spinal cord [1]. The PNS consists of all neurones lying outside those two organs [1]. [6] Examiner insight: A comparison needs both systems in the same sentence; writing about one system and then the other without a link may not score.

9. (a) Medulla oblongata [1]. (b) Hypothalamus [1]. (c) Cerebellum [1]. (d) Cerebral hemispheres [1]. (e) Pituitary gland [1]. [5] Examiner insight: Write the full name of the region. “Cerebrum” or “the brain” does not identify the structure being asked for.

10. (a) Active neurones use more oxygen, so blood oxygen levels change in that region [1]. fMRI detects the change in blood oxygen and maps it onto an image [1]. (b) 330 ÷ 110 = 3 half-lives [1]. 800 × (1/2)³ = 100 MBq [1]. (c) The half-life is short, so the activity falls quickly, and delay would leave too little tracer [1]. (d) CT [1], because it is quick and shows structure such as bleeding [1]. [7] Examiner insight: Link the answer to the half-life or the number of half-lives; “it decays” alone does not explain why the tracer must be near.

11. (a) Loss of the neurones means less dopamine at synapses, so control of movement is impaired [1]. L-DOPA crosses into the brain [1] and is converted to dopamine [1]. (b) An SSRI blocks reuptake of serotonin by the presynaptic neurone [1], so serotonin stays in the synapse longer and stimulates the postsynaptic neurone more [1]. (c) SSRI: 54 ÷ 120 × 100 = 45.0%; placebo: 30 ÷ 120 × 100 = 25.0% [1]. Difference = 20.0 percentage points [1]. (d) Neither patients nor researchers know who receives the drug, so expectation and bias are removed [1]. [8] Examiner insight: Subtract the percentages, not the counts, and give the unit as percentage points or per cent.

Where marks are usually lost

• Writing that light opens the cation channels in rods.
• Describing hyperpolarisation without saying sodium ions stop entering.
• Calling habituation tiredness or a permanent loss of response.
• Forgetting the time interval when calculating a rate.
• Giving the effect of IAA as “growth” without naming cell elongation.
• Explaining the amylase result without the gene being switched on.
• Reading the clear zone as the region still containing starch.
• Naming “the brain” instead of the correct region.
• Mixing up which scans show structure and which show activity.

Next steps

• Revision notes
• Study guide
• Edexcel A Level Biology hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level Biology (YBI11) specification, Issue 2, published by Pearson Education Limited.