AS & A Level Biology (9700)
Control & coordination Banker
59 questions · about 6% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1Neurones & action potentials · Core · Paper 4 style · 6 marks- (a) State the function of the myelin sheath and name the cells that form it. [2]
- (b) Explain how a resting potential of about −70 mV is maintained across the membrane of a neurone. [3]
- (c) State the direction in which an impulse travels along a motor neurone. [1]
Show answer
- (a) Insulates the axon so that depolarisation occurs only at the nodes of Ranvier, speeding up conduction; formed by Schwann cells.
B1 insulation / speeds conduction (saltatory). B1 Schwann cells.
- (b) Na⁺/K⁺ pumps actively transport 3 Na⁺ out for every 2 K⁺ in, using ATP; the membrane is more permeable to K⁺ than Na⁺ (K⁺ leak channels open), so K⁺ diffuses out; large negatively charged proteins/organic anions stay inside; so the inside is negative relative to the outside.
1 mark each for any three of: Na⁺/K⁺ pump, 3 Na⁺ out : 2 K⁺ in; active transport / ATP; membrane more permeable to K⁺ / K⁺ diffuses out through open channels; few Na⁺ channels open so little Na⁺ diffuses back in; organic anions inside / inside more negative.
- (c) From the cell body (in the CNS) along the axon to the effector (muscle).
B1
- (a) Describe the changes in the membrane that cause the depolarisation and repolarisation phases of an action potential. [4]
- (b) Explain the importance of the refractory period. [2]
Show answer
- (a) A stimulus opens some voltage-gated Na⁺ channels; when the threshold (about −50 mV) is reached, many more open and Na⁺ diffuses in rapidly, making the inside positive (to about +30 mV); the Na⁺ channels then close and voltage-gated K⁺ channels open; K⁺ diffuses out, so the potential falls back (repolarisation) and overshoots (hyperpolarisation) before the resting potential is restored.
1 mark each for any four of: voltage-gated Na⁺ channels open; threshold reached → more open (positive feedback); Na⁺ diffuses in → inside becomes positive; Na⁺ channels close; voltage-gated K⁺ channels open; K⁺ diffuses out → repolarisation; hyperpolarisation, then K⁺ channels close.
- (b) During the refractory period a new action potential cannot be generated at that part of the membrane (Na⁺ channels inactive); this makes impulses travel in one direction only and keeps impulses separate (discrete), limiting the maximum frequency.
B1 impulses travel in one direction only. B1 action potentials separate / discrete / limits frequency.
Electrodes were placed 0.90 m apart along a myelinated motor neurone in a volunteer's arm. An action potential took 15 ms to travel between them.
- (a) Calculate the speed of conduction in m s⁻¹. [1]
- (b) Explain how myelination allows such a high speed of conduction. [3]
- (c) Suggest why multiple sclerosis, in which the myelin sheath is damaged, causes muscle weakness. [1]
Show answer
- (a) 60 m s⁻¹
0.90 ÷ 0.015 = 60
- (b) Myelin insulates the axon so ions cannot move across the membrane except at the nodes of Ranvier; depolarisation occurs only at the nodes; local circuits of current flow between nodes so the action potential 'jumps' from node to node (saltatory conduction), covering more distance per action potential.
1 mark each for any three of: myelin insulates / prevents ion movement; depolarisation only at nodes of Ranvier; local circuits between nodes; saltatory conduction – impulse jumps; fewer depolarisations needed per length of axon.
- (c) Impulses are conducted more slowly or fail to reach the muscle, so muscle fibres are stimulated less.
B1
Conduction speeds of axons from different animals:Unmyelinated, diameter 2 µm: 1.2 m s⁻¹Unmyelinated, diameter 20 µm: 4.0 m s⁻¹Myelinated, diameter 2 µm: 12 m s⁻¹Myelinated, diameter 20 µm: 100 m s⁻¹
- (a) Use the data to describe how myelination and axon diameter affect conduction speed. [2]
- (b) Explain why wider axons conduct impulses faster. [2]
- (c) The squid has some unmyelinated axons about 500 µm in diameter. Suggest an advantage of these giant axons to the squid, and why mammals do not need them. [2]
Show answer
- (a) Myelinated axons conduct much faster than unmyelinated axons of similar diameter (at 2 µm: 12 v 1.2 m s⁻¹; at 20 µm: 100 v 4.0 m s⁻¹); in both types, wider axons conduct faster.
B1 myelinated faster, with data. B1 larger diameter → faster, in both.
- (b) A wider axon has a lower resistance to the flow of ions inside it (larger cross-sectional area of cytoplasm); so local currents spread further and faster along the axon, depolarising the next region sooner.
B1 lower resistance / less resistance to ion flow. B1 local currents spread further / faster.
- (c) Giant axons conduct fast enough for a rapid escape response without myelin; mammals achieve high speed with narrow myelinated axons, so many more axons can fit in a small space (e.g. nerves, spinal cord).
B1 fast escape response. B1 myelination gives speed with small diameter / saves space and energy.
- (a) Explain what is meant by the all-or-nothing principle. [2]
- (b) Explain how the nervous system distinguishes between a weak stimulus and a strong stimulus. [2]
- (c) The absolute refractory period of a neurone lasts 2.0 ms. Calculate the maximum number of action potentials that could pass along it in one second. [1]
Show answer
- (a) If a stimulus causes depolarisation to reach the threshold, an action potential is produced and it is always the same size; if threshold is not reached, no action potential is produced.
B1 threshold must be reached. B1 action potentials all the same size (amplitude).
- (b) A stronger stimulus produces a higher frequency of action potentials (not bigger ones); it may also stimulate more receptors / neurones with different thresholds.
B1 higher frequency of impulses. B1 more neurones / receptors stimulated.
- (c) 500
1 ÷ 0.002 = 500
- (a) State which of the following change in length when a sarcomere contracts: A band, I band, H zone. [2]
- (b) Explain why the A band does not change in length. [1]
- (c) Name the protein that makes up the thin filaments and the protein that makes up the thick filaments. [1]
Show answer
- (a) I band and H zone become shorter; the A band stays the same length.
B1 I band and H zone shorten. B1 A band unchanged.
- (b) The A band is the length of the myosin (thick) filaments, which do not shorten – the filaments slide past each other.
B1
- (c) Thin: actin (with troponin and tropomyosin); thick: myosin.
B1 both needed.
- (a) Describe the role of calcium ions and ATP in the contraction of a sarcomere. [5]
- (b) After death, muscles become stiff (rigor mortis). Suggest why. [1]
Show answer
- (a) Ca²⁺ released from the sarcoplasmic reticulum binds to troponin; troponin changes shape and moves tropomyosin away from the myosin-binding sites on actin; myosin heads (carrying ADP and Pi) bind to actin forming cross-bridges; release of ADP and Pi causes the power stroke, pulling actin towards the centre (M line); ATP binds to the myosin head, causing it to detach from actin; hydrolysis of ATP by ATPase in the myosin head re-cocks the head ready to bind again.
1 mark each for any five of: Ca²⁺ from sarcoplasmic reticulum; Ca²⁺ binds troponin; tropomyosin moves, exposing binding sites on actin; myosin heads bind → cross-bridges; power stroke when ADP + Pi released; actin pulled towards the M line / sarcomere shortens; ATP binds → myosin head detaches; ATP hydrolysed → head returns to original position.
- (b) No ATP is made after death, so myosin heads cannot detach from actin; cross-bridges remain attached.
B1
- (a) Describe how an impulse arriving at a neuromuscular junction leads to depolarisation of the sarcolemma. [4]
- (b) Explain the role of T-tubules in muscle contraction. [2]
- (c) State the role of acetylcholinesterase at the neuromuscular junction. [1]
Show answer
- (a) The action potential opens voltage-gated Ca²⁺ channels in the presynaptic membrane and Ca²⁺ diffuse in; vesicles of acetylcholine move to and fuse with the presynaptic membrane, releasing ACh by exocytosis; ACh diffuses across the synaptic cleft and binds to receptors on the sarcolemma (motor end plate); these ligand-gated channels open and Na⁺ diffuse in, depolarising the sarcolemma.
1 mark each for any four of: Ca²⁺ channels open, Ca²⁺ enter; vesicles fuse, ACh released by exocytosis; ACh diffuses across cleft; binds to receptors on sarcolemma / motor end plate; Na⁺ channels open, Na⁺ enter; depolarisation / action potential in sarcolemma.
- (b) T-tubules carry the depolarisation (action potential) deep into the muscle fibre; this causes the sarcoplasmic reticulum to release Ca²⁺ throughout the fibre, so all myofibrils contract at the same time.
B1 carry action potential into fibre. B1 sarcoplasmic reticulum releases Ca²⁺ / all myofibrils contract together.
- (c) Hydrolyses (breaks down) acetylcholine in the cleft so the muscle fibre is not continuously stimulated / the receptors are freed.
B1
In a relaxed skeletal muscle fibre, each sarcomere is 2.4 µm long, the A band is 1.6 µm long and the H zone is 0.6 µm long. During contraction each sarcomere shortens to 2.0 µm.
- (a) Calculate the total length of the I band within one relaxed sarcomere and within one contracted sarcomere. [2]
- (b) Calculate the percentage decrease in sarcomere length. Give your answer to 1 decimal place. [1]
- (c) Calculate the length of the H zone in the contracted sarcomere, and explain your answer. [2]
- (d) Suggest why a muscle fibre produces less force when its sarcomeres are stretched to 3.6 µm. [1]
Show answer
- (a) Relaxed: 0.8 µm; contracted: 0.4 µm
I band (both halves) = sarcomere length − A band. B1 2.4 − 1.6 = 0.8. B1 2.0 − 1.6 = 0.4.
- (b) 16.7 %
(2.4 − 2.0) ÷ 2.4 × 100 = 16.7 %
- (c) 0.2 µm; the sarcomere shortened by 0.4 µm and this is taken up by the thin filaments sliding further into the A band, so the H zone (the part of the A band with no actin) shortens by 0.4 µm.
M1 0.6 − 0.4. A1 0.2 µm with explanation that A band is constant and actin slides in.
- (d) Thin and thick filaments overlap very little or not at all, so few or no myosin heads can form cross-bridges with actin.
B1
Organophosphate pesticides are sprayed on cotton crops in southern Punjab. Farm workers who handle them without protection may be poisoned. Organophosphates inhibit the enzyme acetylcholinesterase.
- (a) Explain why organophosphate poisoning causes uncontrolled muscle twitching and spasms. [3]
- (b) Suggest why severe poisoning can lead to death by failure of breathing. [2]
- (c) Atropine, which blocks some types of acetylcholine receptor, is used as a treatment. Suggest how it helps. [1]
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- (a) Acetylcholinesterase is inhibited, so acetylcholine is not broken down in the synaptic cleft; ACh stays bound to receptors on the sarcolemma; Na⁺ channels keep opening, so the sarcolemma is repeatedly depolarised and the muscle is stimulated continuously.
B1 acetylcholinesterase inhibited / ACh not hydrolysed. B1 ACh remains bound to receptors / accumulates. B1 repeated / continuous depolarisation and contraction.
- (b) Breathing muscles (diaphragm and intercostal muscles) are skeletal muscles; continuous stimulation leads to spasm then paralysis (receptors desensitised), so ventilation stops.
B1 diaphragm / intercostal muscles affected. B1 cannot contract and relax rhythmically / paralysis → no ventilation.
- (c) It prevents excess acetylcholine from binding to (those) receptors, reducing overstimulation.
B1
- (a) Describe how the Venus fly trap detects an insect and closes. [4]
- (b) Suggest why the trap only closes after two stimulations within a short time. [1]
- (c) Suggest why Venus fly traps, which can photosynthesise, catch insects. [1]
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- (a) The insect touches sensitive trigger hairs on the inner surface of the lobes; bending a hair causes a receptor potential and an action potential; a second stimulation within about 20–30 s is needed (prevents closing for raindrops / debris); the action potentials spread across the lobes; ions (Ca²⁺) and water move quickly, so cells change turgor / outer cells expand; the lobes flip from convex to concave and the trap snaps shut in a fraction of a second.
1 mark each for any four of: trigger / sensory hairs bent; action potential generated; two stimulations within a short time needed; action potential spreads across lobe; rapid ion and water movement / change in turgor / cell walls loosened; lobes change shape (convex → concave), trap closes.
- (b) To avoid wasting energy closing on non-food objects such as raindrops or falling debris.
B1
- (c) They grow in soils poor in nitrogen (and other minerals), so insects provide nitrogen (for amino acids / proteins).
B1
- (a) Explain how auxin causes elongation of cells in a shoot. [4]
- (b) A student puts thin slices of shoot tissue in a buffer solution at pH 4.5 without auxin. Predict and explain what happens. [2]
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- (a) Auxin binds to a receptor in/on the cell; this stimulates proton pumps in the cell surface membrane to pump H⁺ into the cell wall (ATP used); the pH of the cell wall falls; low pH activates expansins, which loosen the bonds between cellulose microfibrils; the wall becomes more extensible, so water enters by osmosis and the cell elongates (acid growth hypothesis).
1 mark each for any four of: auxin binds receptor; proton pumps activated / H⁺ pumped into cell wall; cell wall pH lowered; expansins activated; bonds between cellulose microfibrils loosened / wall extensible; water enters by osmosis → cell elongates.
- (b) The cells elongate; the low pH activates expansins in the cell wall directly, mimicking the effect of auxin.
B1 cells elongate / grow. B1 acidic conditions activate expansins without needing auxin.
When a wheat grain absorbs water, the embryo begins to produce gibberellin.
- (a) Explain how gibberellin causes the starch stored in the endosperm to be broken down. [4]
- (b) Explain how the breakdown of starch helps the seedling to grow. [2]
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- (a) The embryo releases gibberellin, which diffuses to the aleurone layer; gibberellin binds to a receptor and causes DELLA proteins to be broken down; DELLA proteins normally inhibit a transcription factor (PIF); the transcription factor can now bind to the promoter of the amylase gene, so amylase is synthesised; amylase is secreted into the endosperm and hydrolyses starch to maltose.
1 mark each for any four of: gibberellin from embryo to aleurone layer; binds receptor; DELLA proteins broken down; transcription factor no longer inhibited / binds promoter; transcription of amylase gene; amylase hydrolyses starch to maltose (glucose).
- (b) Maltose/glucose is used by the embryo in respiration to provide ATP for cell division and growth (and to make cellulose / new cells) until the seedling can photosynthesise.
B1 sugars used in respiration / ATP. B1 for growth / until photosynthesis begins.
In pea plants, the gene Le codes for an enzyme that converts an inactive gibberellin (GA20) into active gibberellin (GA1). The allele le is recessive. Plants with genotype le le are dwarf.
- (a) Explain why plants with genotype le le are dwarf. [3]
- (b) Predict the height of le le plants sprayed with GA1 and explain your prediction. [2]
- (c) State the phenotype of plants with genotype Le le and explain why. [1]
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- (a) The recessive allele le codes for a non-functional (less active) enzyme; so inactive GA20 is not converted into active GA1; without GA1, DELLA proteins are not broken down and continue to inhibit transcription factors needed for stem elongation (cell elongation and division in internodes).
B1 le gives non-functional enzyme. B1 no / little active GA1 produced. B1 DELLA proteins remain → growth genes not transcribed / short internodes.
- (b) They would grow tall (similar to Le plants), because GA1 is the active form, so the missing enzyme step is bypassed and DELLA proteins are broken down.
B1 tall. B1 active GA supplied directly / missing enzyme not needed.
- (c) Tall; Le is dominant – one functional allele produces enough enzyme to make active GA1.
B1
- (a) Give three differences between nervous and endocrine communication. [3]
- (b) Explain why a hormone travelling in the blood affects only certain cells. [1]
- (c) Name the gland that secretes ADH into the blood. [1]
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- (a) Nervous: electrical impulses along neurones (and chemical at synapses); endocrine: hormones (chemicals) carried in the blood. Nervous: very fast; endocrine: slower. Nervous: short-lived responses; endocrine: longer-lasting. Nervous: response localised to specific effectors; endocrine: can affect target cells widely throughout the body.
1 mark for each correct comparative difference, max 3.
- (b) Only target cells have specific receptors complementary to the hormone.
B1
- (c) Posterior pituitary gland.
B1
- (a) Define the term endocrine gland. [1]
- (b) Adrenaline and insulin are both hormones. Adrenaline is released within seconds of a fright; the effects of insulin are needed over hours after a meal. Explain why the endocrine system, rather than the nervous system, is used for these responses. [3]
- (c) Suggest why the reflex withdrawal of a bare foot from a sharp thorn uses the nervous system. [1]
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- (a) A gland that secretes hormones directly into the blood (it has no duct).
B1
- (b) Both hormones need to reach many different target cells throughout the body (e.g. liver, muscle, fat cells); blood carries hormones to all tissues; hormone effects (e.g. on enzymes or membrane transporters) last as long as the hormone remains, giving a longer, sustained response.
1 mark each for any three of: many / widespread target cells; hormone reaches all tissues in blood; long-lasting effect; one hormone can produce several coordinated effects; nervous responses would be too short / localised.
- (c) It needs a very fast, localised response to prevent damage.
B1
Hormone concentrations in a woman's blood (arbitrary units):Day of cycle: 2, 7, 12, 14, 17, 21, 26FSH: 8, 6, 9, 15, 4, 3, 5LH: 6, 5, 18, 50, 7, 4, 4Oestrogen: 2, 6, 14, 10, 6, 9, 3Progesterone: 1, 1, 1, 3, 12, 20, 4
- (a) Explain the role of FSH and oestrogen in the first half of the menstrual cycle. [3]
- (b) Using the data, identify the day on which ovulation is most likely and explain your answer. [2]
- (c) Explain how a contraceptive pill containing synthetic oestrogen and progesterone prevents pregnancy. [2]
Show answer
- (a) FSH from the anterior pituitary stimulates the development of a follicle in the ovary; the follicle secretes oestrogen; oestrogen stimulates repair/thickening of the endometrium; rising oestrogen initially inhibits FSH (negative feedback), then high oestrogen triggers the LH surge.
1 mark each for any three of: FSH from anterior pituitary; stimulates follicle development; follicle secretes oestrogen; oestrogen thickens endometrium; oestrogen inhibits FSH; high oestrogen stimulates LH surge.
- (b) Day 14; LH concentration peaks on day 14 (LH surge, 50 units), which triggers ovulation.
B1 day 14 (accept 14–15). B1 LH peak / surge causes ovulation.
- (c) The hormones inhibit secretion of FSH and LH by the anterior pituitary (negative feedback); so follicles do not develop and no LH surge occurs, so ovulation does not happen.
B1 FSH and LH secretion inhibited. B1 no follicle development / no ovulation. (Also accept: cervical mucus thickened.)
- (a) Describe the sequence of events at a cholinergic synapse, from the arrival of an action potential at the presynaptic knob to the production of an action potential in the postsynaptic neurone. [5]
- (b) Explain why presynaptic knobs contain many mitochondria. [1]
Show answer
- (a) Depolarisation opens voltage-gated Ca²⁺ channels and Ca²⁺ diffuse into the presynaptic knob; vesicles containing acetylcholine move to and fuse with the presynaptic membrane; ACh is released by exocytosis; ACh diffuses across the synaptic cleft and binds to receptor proteins on the postsynaptic membrane; Na⁺ channels (ligand-gated) open and Na⁺ diffuse in, depolarising the postsynaptic membrane; if threshold is reached, an action potential is produced; acetylcholinesterase hydrolyses ACh.
1 mark each for any five of: Ca²⁺ channels open / Ca²⁺ enter; vesicles fuse with presynaptic membrane; exocytosis of ACh; diffusion across cleft; binds to receptors on postsynaptic membrane; Na⁺ channels open, Na⁺ enter; depolarisation – threshold → action potential; ACh broken down by acetylcholinesterase; products reabsorbed and ACh resynthesised using ATP.
- (b) To provide ATP for resynthesis of acetylcholine / packaging into vesicles / pumping Ca²⁺ out.
B1
- (a) Explain why impulses can only travel in one direction across a synapse. [2]
- (b) Describe how synapses allow one neurone to influence many others, and many neurones to influence one, and state one benefit of each. [3]
- (c) Suggest why repeated stimulation of the same synapse at a very high frequency can stop transmission. [1]
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- (a) Vesicles of neurotransmitter are only in the presynaptic knob; receptors are only on the postsynaptic membrane.
B1 neurotransmitter released only from presynaptic side. B1 receptors only on postsynaptic membrane.
- (b) Divergence: one presynaptic neurone synapses with many postsynaptic neurones, so one signal can spread to several parts of the nervous system / several effectors; convergence: many presynaptic neurones synapse with one postsynaptic neurone, allowing information from several sources to be integrated (summed) and weak stimuli to be amplified.
B1 divergence described. B1 convergence described. B1 a benefit of each (spread of signal / integration of information).
- (c) Acetylcholine is used up faster than it can be resynthesised (vesicles depleted), so the synapse fatigues.
B1
Scientists studied a synapse between two neurones in a laboratory preparation.Without drug X: presynaptic action potential normal; postsynaptic depolarisation 25 mV, action potential produced.With drug X: presynaptic action potential normal; postsynaptic depolarisation 4 mV, no action potential.With drug X, acetylcholine applied directly to the postsynaptic membrane: postsynaptic depolarisation 25 mV, action potential produced.
- (a) Describe the effect of drug X on the postsynaptic response. [1]
- (b) Use the data to suggest how drug X acts. Explain your reasoning. [3]
- (c) Nicotine has a similar shape to acetylcholine. Suggest how nicotine affects transmission at synapses with nicotinic receptors. [2]
Show answer
- (a) Drug X reduces the depolarisation of the postsynaptic membrane from 25 mV to 4 mV, so no action potential is produced.
B1 with data.
- (b) Drug X probably blocks the voltage-gated Ca²⁺ channels in the presynaptic membrane (or prevents vesicle fusion); when acetylcholine was applied directly to the postsynaptic membrane the response was normal, so the receptors and Na⁺ channels still work; the presynaptic action potential is normal, so the drug acts between the arrival of the impulse and release of ACh – less ACh is released.
B1 acts on presynaptic side / reduces ACh release (e.g. blocks Ca²⁺ channels). B1 postsynaptic response to applied ACh normal → receptors unaffected. B1 presynaptic action potential normal → conduction unaffected.
- (c) Nicotine binds to the acetylcholine receptors on the postsynaptic membrane, opening Na⁺ channels and causing depolarisation; it is not broken down by acetylcholinesterase, so it causes prolonged stimulation.
B1 binds to ACh receptors / mimics ACh → depolarisation. B1 not hydrolysed by acetylcholinesterase → prolonged effect.
A boy in Faisalabad accidentally touches a hot tawa (griddle) and pulls his hand away immediately, before he feels the pain.
- (a) Name the parts of the reflex arc in order, starting with the receptor. [3]
- (b) Explain why reflex actions are fast. [2]
- (c) State why a reflex is described as involuntary. [1]
Show answer
- (a) Receptor (thermoreceptor/pain receptor in skin) → sensory neurone → relay (intermediate) neurone in the spinal cord → motor neurone → effector (biceps muscle).
B1 receptor and sensory neurone. B1 relay neurone in spinal cord. B1 motor neurone and muscle as effector.
- (b) Few synapses are involved (only two or three), and the impulse does not need to pass to the brain for a decision; myelinated neurones conduct quickly.
B1 few synapses / short pathway. B1 brain not involved in initiating the response / no conscious decision (myelinated neurones).
- (c) It is automatic – it happens without conscious thought.
B1
- (a) Describe how a chemoreceptor cell in a taste bud responds to sodium ions in food and how this results in impulses in a sensory neurone. [4]
- (b) Explain why a sensory receptor is described as a transducer. [1]
- (c) Suggest why very salty food produces a stronger sensation of saltiness than slightly salty food. [1]
Show answer
- (a) Na⁺ ions from salty food diffuse into the receptor cell through Na⁺ channels in its cell surface membrane; this depolarises the receptor cell (a receptor / generator potential); the bigger the stimulus, the bigger the receptor potential; if threshold is reached, the receptor cell releases neurotransmitter onto the sensory neurone; this depolarises the sensory neurone, producing action potentials that travel to the brain.
1 mark each for any four of: Na⁺ diffuse in through channels; depolarisation / receptor potential; size depends on stimulus strength; threshold; neurotransmitter released to sensory neurone; action potentials in sensory neurone to the brain.
- (b) It converts energy of one form (the stimulus, e.g. chemical) into electrical energy (a nerve impulse).
B1
- (c) More Na⁺ enters, giving a larger receptor potential and a higher frequency of action potentials in the sensory neurone (and more receptors stimulated).
B1
Percentage change in growth of roots and shoots of maize seedlings treated with auxin, compared with untreated seedlings:Auxin concentration / mol dm⁻³: 10⁻¹², 10⁻¹⁰, 10⁻⁸, 10⁻⁶, 10⁻⁴, 10⁻³Roots: +5, +20, −10, −40, −60, −70Shoots: 0, +5, +30, +80, +40, −15
- (a) Describe the effect of auxin concentration on the growth of roots and shoots. [3]
- (b) A student concludes: 'Auxin always increases plant growth.' Use the data to explain why this conclusion is not valid. [2]
Show answer
- (a) Root growth is stimulated by very low concentrations (maximum +20 % at 10⁻¹⁰ mol dm⁻³) and inhibited at 10⁻⁸ mol dm⁻³ and above (−60 % at 10⁻⁴); shoot growth is unaffected or slightly stimulated at low concentrations, increases to a maximum (+80 %) at 10⁻⁶ mol dm⁻³ and is inhibited at 10⁻³ mol dm⁻³ and above; roots are far more sensitive to auxin than shoots.
1 mark each for any three of: root optimum at 10⁻¹⁰; root inhibition above about 10⁻⁸; shoot optimum at 10⁻⁶; shoot inhibition at high concentration; roots more sensitive than shoots (comparison with data).
- (b) At high concentrations auxin inhibits growth (negative values: roots above 10⁻⁸, shoots at 10⁻³ and above); the effect depends on concentration and on the organ.
B1 inhibition at high concentration with data. B1 effect depends on organ / concentration.
Twelve students did a ruler-drop reaction test before and 20 minutes after drinking a cup of strong tea (contains caffeine).Before: mean reaction time 0.203 s, 95 % confidence interval 0.190–0.216 sAfter: mean reaction time 0.186 s, 95 % confidence interval 0.181–0.191 s
- (a) Discuss the extent to which these results support the claim that drinking a cup of strong tea improves reaction time. [3]
- (b) Suggest how the investigation could be improved to test the claim fairly. [2]
Show answer
- (a) The mean reaction time after tea (0.186 s) was shorter than before (0.203 s), which supports the claim; but the 95 % confidence intervals overlap slightly (both include 0.190–0.191 s), so the difference may not be significant; the test was not blind and students may have improved with practice (order effect) – no control group drinking water or decaffeinated tea.
B1 mean shorter after tea – supports. B1 confidence intervals overlap → difference may not be significant. B1 a design limitation (no control / practice effect / not blind / small sample).
- (b) Use a control group (or a separate session) drinking decaffeinated tea/water that looks and tastes the same, in random order, with students not knowing which drink they had; allow practice trials first; larger sample; use a t-test on the results.
1 mark each for any two.
Reaction times (ms) of ten volunteers of different ages in a computer reaction test:Age / years: 20, 25, 30, 35, 40, 45, 50, 55, 60, 65Reaction time / ms: 212, 205, 230, 226, 241, 238, 255, 262, 249, 270rs = 1 − 6Σd² ÷ (n(n² − 1))
- (a) Calculate Spearman's rank correlation coefficient, rs. Show your working. [3]
- (b) The critical value of rs for 10 pairs at p = 0.05 is 0.648. State the null hypothesis and what you can conclude. [2]
- (c) Explain why this result does not prove that ageing causes slower reactions. [1]
Show answer
- (a) Ranks of reaction time: 2, 1, 4, 3, 6, 5, 8, 9, 7, 10; Σd² = 12; rs = 0.93
M1 ranks correct for both variables (age ranks 1–10). M1 Σd² = 12. A1 rs = 1 − (6 × 12) ÷ (10 × 99) = 0.93.
- (b) Null hypothesis: there is no correlation between age and reaction time. 0.93 > 0.648, so reject the null hypothesis: there is a significant positive correlation – reaction time increases with age.
B1 null hypothesis (no correlation). B1 rs > critical value → significant positive correlation.
- (c) Correlation does not show cause; other factors (e.g. eyesight, practice, fitness) may vary with age; only one person of each age was tested.
B1
Barley grains are used to make malt in breweries and bakeries. A student wants to find out how the concentration of gibberellin affects the production of amylase by the aleurone layer of barley grains.
- (a) Describe a method to investigate the effect of gibberellin concentration on amylase production by barley grains, using starch agar plates. [6]
- (b) Explain why the embryo must be removed from each grain. [2]
Show answer
- (a) Cut grains in half and use the endosperm halves without the embryo (so they make no gibberellin of their own); sterilise the surface (e.g. dilute bleach, then rinse); soak sets of half-grains in gibberellin solutions of e.g. 0, 0.01, 0.1, 1, 10 mg dm⁻³ for a fixed time; place the same number of half-grains, cut surface down, on starch agar plates; incubate at the same temperature (e.g. 25 °C) for a fixed time (e.g. 48 h); flood with iodine and measure the diameter (or area) of the clear zone around each grain; at least 5 grains per concentration and calculate the mean.
1 mark each for any six of: remove embryo (half-grains without embryo); at least five concentrations including 0; made by dilution / serial dilution; same soaking time and volume; surface sterilised / aseptic technique; same number of grains per plate; same temperature and time of incubation; iodine added; measure clear zone diameter / area; replicates and mean.
- (b) The embryo produces its own gibberellin; this would stimulate amylase production and mask the effect of the gibberellin supplied.
B1 embryo makes gibberellin. B1 would affect results / not controlled.
In a ruler-drop test, a partner drops a metre rule and the student catches it as quickly as possible. The distance the ruler falls is d. Reaction time t = √(2d ÷ g), with d in metres and g = 9.81 m s⁻².Distances before an energy drink / cm: 21, 19, 18, 23, 20
- (a) Calculate the mean distance before the energy drink, in cm. [1]
- (b) Use the mean distance and the formula to calculate the mean reaction time before the drink. Give your answer to 3 decimal places. [2]
- (c) After the drink the mean distance was 17.0 cm. Calculate the reaction time and the percentage decrease in reaction time. Give the percentage to 1 decimal place. [2]
Show answer
- (a) 20.2 cm
(21 + 19 + 18 + 23 + 20) ÷ 5 = 20.2
- (b) 0.203 s
M1 d = 0.202 m. A1 t = √(2 × 0.202 ÷ 9.81) = 0.203 s.
- (c) 0.186 s; 8.3 % decrease
B1 √(2 × 0.170 ÷ 9.81) = 0.186 s. B1 (0.203 − 0.186) ÷ 0.203 × 100 = 8.3 % (using unrounded times; 8.4 % from the rounded values is also accepted).
A student compares the reaction times of his dominant and non-dominant hands with a ruler-drop test.
- (a) State the independent and dependent variables. [2]
- (b) State three variables that should be standardised and explain how one of them could be controlled. [3]
- (c) Suggest why the dropper should vary the time before releasing the ruler. [1]
Show answer
- (a) Independent: whether the student uses the dominant or non-dominant hand. Dependent: distance the ruler falls (reaction time).
B1 each.
- (b) Starting position of the ruler (zero mark level with the top of the thumb); gap between thumb and finger (e.g. 2 cm); same person dropping the ruler; no warning before the drop / random delay; same time of day, posture, distractions.
B1 B1 two valid variables. B1 method of control for one, e.g. use a fixed 2 cm gap marked on a card. (Max 3.)
- (c) So the student cannot predict the drop and respond in anticipation; the test measures reaction to the stimulus.
B1
A student plans to compare reaction times when the stimulus is a sound (a beep) and when it is a light (a screen flashing), using a computer program and 20 volunteers.
- (a) Write a hypothesis for this investigation and give a biological reason for it. [2]
- (b) State a null hypothesis suitable for a t-test. [1]
- (c) Explain why a correlation test would not be appropriate for this investigation. [1]
Show answer
- (a) The mean reaction time will be shorter when a sound is used as the stimulus than when a light is used; because converting light into impulses in the retina involves a slower series of chemical reactions, whereas sound receptors are stimulated mechanically and respond faster.
B1 directional hypothesis. B1 reasonable biological reason.
- (b) There is no significant difference between the mean reaction time to a sound stimulus and to a light stimulus.
B1
- (c) The investigation compares two groups / two conditions (categoric independent variable), not a relationship between two continuous variables.
B1
Students use a ruler-drop test to measure reaction time, reading the distance the ruler falls at the top of the thumb.
- (a) Identify three sources of error in the ruler-drop method. [3]
- (b) Suggest improvements to reduce two of these errors. [2]
- (c) Explain why the student does five practice trials before recording any results. [1]
Show answer
- (a) Ruler not always released from the same height / zero not level with the thumb; distance read from where the thumb is, which is imprecise; the catcher may anticipate the drop; the ruler may not fall vertically; gap between finger and thumb varies; tiredness or learning over repeated trials.
1 mark each for any three.
- (b) Use a clamp or marked starting line so the ruler starts at the same height; read the scale at the top of the thumb each time / use a ruler with larger divisions; vary the delay before dropping; use a computer reaction timer or light gate; give practice trials and rests between trials.
1 mark each for any two improvements matched to errors.
- (c) To reduce the learning (practice) effect, so improvement during the test does not distort the comparison.
B1
