Revision Library

revisionlibrary.co.uk

A-Level · Biology

AB6 Organisms Respond to Changes in their Environments

AQA 7402 · Calculator allowed · about 155 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Woodlice are commonly found in damp, dark places such as under logs, stones and loose bark. A student studied the behavioural response of woodlice to differences in humidity using a choice chamber divided into a dry side and a humid side (with light and temperature kept the same on both sides). For woodlice found in each side, the student recorded the mean rate at which individual woodlice changed direction (the turning rate).

(a)State what is meant by a kinesis.(1)

(b)The student recorded a mean turning rate of 9.6 turns per minute for woodlice in the dry side of the chamber and 1.2 turns per minute for woodlice in the humid side. Using these data, explain how this kinesis response results in woodlice accumulating in the humid side of the chamber over time.(4)

(c)Explain one advantage to the woodlouse of relying on a simple kinesis, rather than a taxis, to respond to changes in humidity.(2)

(d)State one difference between a taxis and a kinesis.(1)

(Total for Question 1 is 8 marks)

Required practical. A choice chamber was used to study how humidity affects the distribution of woodlice. The chamber had two connected sections: one containing dry silica gel and one containing damp cotton wool, with a mesh floor so woodlice could not tell the two sides apart by sight. Twenty woodlice were placed in the centre of the chamber and left for 10 minutes, after which the number on each side was counted. The procedure was repeated five times using fresh woodlice each time.

(a)Describe how the investigation should be set up to ensure that humidity is the only variable affecting the distribution of the woodlice.(4)

(b)Explain why the same number of woodlice should be used in each repeat, and why the count should be taken after a set, fixed period of time in every repeat.(2)

(c)In one repeat, 4 of the 20 woodlice were found on the dry side and the rest on the humid side after 10 minutes. Calculate the percentage of woodlice found on the humid side.(2)

(d)The student wants to test whether this distribution is significantly different from an even (50:50) split due to chance. Name an appropriate statistical test for this data, and state a suitable null hypothesis.(2)

(Total for Question 2 is 10 marks)

In a classic experiment, the tips of oat coleoptiles (seedling shoots) were removed and placed upside-down on blocks of agar jelly, allowing auxin (indoleacetic acid, IAA) produced by the tips to diffuse into the agar. Each agar block (now containing auxin) was then placed off-centre on top of a separate decapitated coleoptile stump, touching only one side of the cut surface. The stumps were left in the dark and their growth was monitored.

(a)Predict, with a reason, the direction in which a coleoptile stump would bend as it grew.(3)

(b)In an intact coleoptile exposed to unilateral (one-sided) light, explain how light brings about lateral (side-to-side) redistribution of auxin, and how this causes the shoot to curve towards the light (positive phototropism).(4)

(c)The coleoptile in the intact plant curved through 35 degrees over a period of 90 minutes. Calculate the mean rate of curvature, in degrees per hour.(2)

(d)Suggest one improvement to the experimental procedure that would increase confidence in a conclusion drawn from a single curvature measurement like this.(2)

(Total for Question 3 is 11 marks)

The Pacinian corpuscle is a pressure receptor found deep in the skin, especially in the fingers and soles of the feet.

(a)Describe the structure of a Pacinian corpuscle.(2)

(b)Explain how pressure applied to the skin above a Pacinian corpuscle leads to the generation of a generator potential in the sensory neurone.(4)

(c)Explain why increasing the pressure applied to the corpuscle increases the frequency of action potentials passing along the sensory neurone, but does not change the size (amplitude) of each individual action potential.(3)

(Total for Question 4 is 9 marks)

A resting neurone maintains a potential difference of about -70 mV across its axon membrane (inside negative relative to outside).

(a)Explain how the sodium-potassium pump and ion channels in the axon membrane maintain this resting potential.(5)

(b)State the term used to describe the state of the axon membrane when it is at its resting potential and not currently transmitting an impulse.(1)

(Total for Question 5 is 6 marks)

The table shows the membrane potential of an axon recorded every 0.5 milliseconds during a single nerve impulse.

Time (ms): 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5
Membrane potential (mV): -70 -70 -20 +40 +30 -60 -80 -70

(a)Using the data, state the resting potential of the axon and identify the time at which depolarisation begins.(2)

(b)Explain the ionic basis of the rising phase (depolarisation) and the falling phase (repolarisation) of the action potential shown in the table.(4)

(c)Explain why the membrane potential becomes more negative than the resting potential at t = 3.0 ms (hyperpolarisation), and state the importance of the refractory period that follows an action potential.(3)

(d)The absolute refractory period of this axon is 1 millisecond. Calculate the maximum theoretical frequency, in Hz, at which action potentials could be generated.(2)

(Total for Question 6 is 11 marks)

A student used two pairs of electrodes placed 45 mm apart along an isolated, myelinated sensory neurone to measure the time taken for a nerve impulse to travel between them. The impulse took 0.5 milliseconds to travel from the first pair of electrodes to the second.

(a)Calculate the speed of conduction of the impulse, in m/s. Give your answer to 2 significant figures. (speed = distance / time)(3)

(b)Explain why myelination increases the speed of conduction of a nerve impulse.(3)

(c)Explain how each of the following would affect the speed of conduction of a nerve impulse: (i) an increase in axon diameter; (ii) a decrease in temperature.(4)

(d)A technician suggests that raising the temperature of an isolated nerve preparation above 40 degrees C would continue to increase the speed of impulse conduction. Evaluate this suggestion.(2)

(Total for Question 7 is 12 marks)

Required practical. A simple ruler-drop test was used to measure a person's reaction time. One person held a ruler vertically with the zero mark between the thumb and forefinger of a second person, who was not looking at the ruler. The first person released the ruler without warning, and the second person caught it as quickly as possible. The distance the ruler fell before being caught was recorded.

(a)Describe how this test should be carried out to obtain a reliable estimate of a person's mean reaction time.(3)

(b)In one trial, the ruler fell 18.0 cm before being caught. Using s = 1/2 x g x t² (where g = 9.81 m/s²), calculate the person's reaction time in milliseconds.(4)

(c)Explain one variable that should be controlled when comparing reaction times between different people, and suggest one way to improve the reliability of the comparison.(2)

(d)The reaction time measured in this test includes the time taken for sensory processing, decision-making in the brain, and the motor response, as well as synaptic delays. Explain why this measured reaction time is much longer than the time taken for a single action potential to cross one synapse.(2)

(Total for Question 8 is 11 marks)

Acetylcholine is the neurotransmitter released at a cholinergic synapse. Describe the sequence of events at a cholinergic synapse that leads to the generation of an action potential in the postsynaptic neurone, starting from the arrival of an action potential at the presynaptic knob.

(Total for Question 9 is 6 marks)

Synaptic transmission at a cholinergic synapse can be modified by the number and pattern of impulses arriving at the presynaptic neurone(s).

(a)Explain why transmission across a cholinergic synapse can only occur in one direction (from the presynaptic to the postsynaptic neurone).(2)

(b)Explain the difference between spatial summation and temporal summation, and how each can help a postsynaptic neurone reach the threshold needed to generate an action potential.(3)

(Total for Question 10 is 5 marks)

Heart rate is controlled by the cardiovascular (cardiac) centre in the medulla oblongata, which receives information from receptors in the walls of the aorta and carotid arteries and sends impulses to the heart via the autonomic nervous system.

(a)Explain how baroreceptors help to reduce heart rate when blood pressure rises above normal.(4)

(b)Explain how chemoreceptors help to increase heart rate in response to a fall in blood pH (caused by a rise in dissolved carbon dioxide).(4)

(c)A person at rest has a heart rate of 68 beats per minute and a stroke volume of 72 cm^3. Using cardiac output (cm³/min) = heart rate x stroke volume, calculate their cardiac output in dm³ per minute.(3)

(Total for Question 11 is 11 marks)

A sarcomere is the repeating structural unit of a myofibril in a skeletal muscle fibre, containing overlapping thick (myosin) and thin (actin) filaments.

(a)Describe the roles of tropomyosin and troponin in a relaxed muscle, and explain how an increase in calcium ion concentration in the sarcoplasm triggers muscle contraction.(5)

(b)During contraction, a sarcomere shortens from 2.6 micrometres to 2.1 micrometres. Calculate the percentage decrease in the length of the sarcomere.(2)

(c)Explain why, during this contraction, the I band and the H zone both decrease in width, but the width of the A band stays the same.(3)

(d)Explain why the observation that the A band stays a constant width during contraction supports the sliding filament theory, rather than an alternative hypothesis in which the thick and thin filaments themselves become shorter during contraction.(2)

(Total for Question 12 is 12 marks)

Animals coordinate responses to changes in their external environment using both the nervous system and the hormonal (endocrine) system. Compare and contrast nervous and hormonal communication as means of coordinating a response to a change in the external environment.

(Total for Question 13 is 6 marks)

Organophosphate compounds are used as pesticides because they inhibit the enzyme acetylcholinesterase in the nervous systems of insects.

(a)Predict and explain the short-term effect of an organophosphate pesticide on transmission at a cholinergic synapse in an insect's nervous system.(4)

(b)A researcher placed woodlice in a choice chamber with dry and humid sides, as in Question 2. Woodlice treated with a sub-lethal dose of an organophosphate pesticide showed a significantly lower rate of turning in the dry side of the chamber compared with untreated woodlice. Suggest an explanation for this observation, and evaluate why it might reduce the survival chances of the treated woodlice population in a pesticide-treated habitat.(5)

(Total for Question 14 is 9 marks)

Mark scheme · AB6 Organisms Respond to Changes in their Environments

Question 1

(a) B1 a non-directional response in which the rate of movement (or rate of turning) changes according to the intensity of the stimulus, not the direction of movement (oe)
(a) Answer: A kinesis is a non-directional response in which the organism's rate of movement or turning changes with the intensity of the stimulus.
(b) B1 in the dry side, the high turning rate (9.6 turns/min) means woodlice frequently change direction, i.e. they move along a very indirect/random path (oe)
(b) B1 this random, indirect movement increases the chance that a woodlouse will, by chance, leave the dry side
(b) B1 in the humid side, the low turning rate (1.2 turns/min) means woodlice move in a straighter path / change direction much less often
(b) B1 this increases the chance that a woodlouse will, by chance, remain in the humid side, so over time more woodlice accumulate there even though the response is not directed towards the humidity (oe)
(b) Answer: The much higher turning rate in dry air (9.6 vs 1.2 turns per minute) means woodlice move about randomly and are statistically more likely to leave the dry side, while the low turning rate in humid air means they are statistically more likely to stay there, so woodlice accumulate in the humid side purely by chance rather than by directed movement.
(c) B1 a kinesis does not require the woodlouse to detect the direction of the stimulus (useful because humidity does not spread as a clear directional gradient in the way light or a chemical source might) (oe)
(c) B1 so a kinesis can still keep the woodlouse in suitable conditions using a much simpler nervous response/less complex sensory processing, which is less costly for a simple invertebrate (oe)
(c) Answer: A kinesis does not need the direction of the humidity gradient to be detected, so it can be produced by a much simpler nervous system than a directional taxis would require.
(d) B1 a taxis is a directional response (movement towards or away from the stimulus), whereas a kinesis is non-directional (only the rate of movement/turning changes) (oe)
(d) Answer: A taxis involves directed movement towards or away from a stimulus; a kinesis involves a change in the rate of movement or turning with no fixed direction.

Question 2

(a) B1 use a covered/opaque chamber (or cover with foil/black paper) so that both sides receive the same, or no, light (controls light as a variable)
(a) B1 carry out the investigation in a single location/at a constant room temperature so both sides of the chamber are at the same temperature
(a) B1 use a mesh or perforated partition/floor (not solid, and not one that lets woodlice sense a difference other than humidity) so woodlice can move freely between sides but cannot see or smell a difference other than humidity
(a) B1 use the same number of similarly sized/acclimatised woodlice and the same time period in each repeat, to allow a fair comparison
(a) Answer: Cover the chamber (or keep it in the dark) and keep it at a constant temperature so that light and temperature are controlled and only humidity differs between the two sides; use a mesh floor so woodlice can move freely but cannot see which side is which; use the same number and type of woodlice and the same time period for each repeat.
(b) B1 using the same number of woodlice each time means the percentage/proportion choosing each side can be validly compared between repeats (oe)
(b) B1 using a fixed time ensures woodlice in every repeat have had the same opportunity to move/respond, so differences between repeats are not simply due to some being given longer to move (oe)
(b) Answer: Keeping the number of woodlice constant allows fair comparison of the proportions choosing each side between repeats, and using a fixed time ensures every repeat gives woodlice the same opportunity to respond, so time is not a confounding variable.
(c) M1 correct method: (20 - 4)/20 x 100, i.e. 16/20 x 100
(c) A1 80% (cao)
(c) Answer: 80%
(d) B1 chi-squared (chi²) test
(d) B1 null hypothesis: there is no significant difference between the observed distribution of woodlice and the expected (even, 50:50) distribution between the humid and dry sides; any difference is due to chance (oe)
(d) Answer: Chi-squared test; H0: there is no significant difference between the observed and expected (50:50) distribution of woodlice between the two sides of the chamber.

Question 3

(a) B1 the stump bends away from the side where the agar block is placed (i.e. curves towards the side with no auxin block)
(a) B1 the side of the stump in contact with the agar block receives a higher concentration of auxin
(a) B1 auxin stimulates cell elongation, so this side elongates more than the opposite side, causing unequal growth and curvature away from the block (oe)
(a) Answer: The stump curves away from the side with the agar block, because that side receives more auxin and its cells elongate more than those on the opposite side, so unequal growth causes the stem to bend towards the side without the block.
(b) B1 auxin (IAA) is produced in the tip of the shoot
(b) B1 when the shoot is illuminated from one side, auxin is transported laterally towards the shaded side of the tip
(b) B1 this results in a higher concentration of auxin on the shaded side than on the illuminated side, and auxin then moves down the shoot from the tip
(b) B1 auxin stimulates cell elongation in the region just behind the tip; because the shaded side has more auxin it elongates more than the illuminated side, causing the shoot to bend/curve towards the light
(b) Answer: Auxin made in the tip is moved to the shaded side when light comes from one direction, giving a higher auxin concentration there; since auxin promotes cell elongation, the shaded side grows faster than the illuminated side, bending the shoot towards the light.
(c) M1 correct method: 35/90 x 60 (or equivalent unit conversion)
(c) A1 23.3 degrees per hour (awrt 23)
(c) Answer: 23.3 degrees per hour
(d) B1 repeat the experiment using several different coleoptiles (of the same age/species) and calculate a mean curvature, to reduce the effect of natural variation between individual plants (oe)
(d) B1 include a control coleoptile with a plain agar block (containing no auxin) placed off-centre, to confirm that curvature is caused specifically by auxin and not by the presence of the agar block, gravity or handling of the stump (oe)
(d) Answer: Repeat with several coleoptiles and calculate a mean, and include a control using a plain (auxin-free) agar block to confirm the curvature is due to auxin rather than the block itself or another factor.

Question 4

(a) B1 a sensory nerve ending wrapped in layers (lamellae) of connective tissue, separated by a gel/viscous fluid
(a) B1 the lamellae/layers surround the end of a sensory neurone, which has stretch-mediated/mechanically-gated sodium ion channels in its membrane
(a) Answer: It consists of the end of a sensory neurone wrapped in concentric layers (lamellae) of connective tissue separated by gel, with stretch-mediated sodium channels in the neurone's membrane.
(b) B1 pressure deforms/squashes the lamellae, which in turn deforms/stretches the membrane of the sensory neurone ending
(b) B1 this stretching widens/opens the stretch-mediated sodium ion channels in the neurone membrane
(b) B1 sodium ions diffuse into the neurone down their electrochemical gradient
(b) B1 the influx of positive charge depolarises the membrane, producing a generator potential; if this reaches the threshold, it triggers an action potential in the sensory neurone
(b) Answer: Pressure deforms the lamellae and stretches the neurone membrane, opening stretch-mediated sodium channels; sodium ions diffuse in, depolarising the membrane and producing a generator potential which, if large enough, triggers an action potential.
(c) B1 a greater pressure causes greater deformation of the lamellae/membrane, opening more stretch-mediated sodium channels and producing a larger generator potential
(c) B1 a larger generator potential exceeds the threshold more often per unit time, generating more frequent action potentials (this is how stimulus intensity is coded)
(c) B1 action potentials obey the all-or-nothing law: once the threshold is reached, an action potential of fixed size is always produced, so amplitude cannot vary with stimulus strength
(c) Answer: A greater pressure produces a bigger generator potential, which reaches the threshold more frequently, generating more action potentials per second; but because action potentials are all-or-nothing, each one is always the same size regardless of stimulus strength - stimulus intensity is coded by frequency, not amplitude.

Question 5

(a) B1 the sodium-potassium pump actively transports (using ATP) 3 sodium ions out of the axon for every 2 potassium ions moved into the axon
(a) B1 this creates a concentration gradient with a higher concentration of sodium ions outside the axon and a higher concentration of potassium ions inside the axon
(a) B1 the membrane has (some) potassium ion channels open at rest, making it more permeable to potassium ions than to sodium ions
(a) B1 potassium ions diffuse back out of the axon down their concentration gradient through these channels (facilitated diffusion), while few sodium ions can re-enter as most sodium channels are closed
(a) B1 the resulting net loss of positive charge from inside the axon makes the inside of the membrane negatively charged relative to the outside, maintaining the resting potential of about -70 mV
(a) Answer: The sodium-potassium pump actively moves 3 Na+ out for every 2 K+ in, using ATP, creating concentration gradients; because the membrane is more permeable to K+ (via open potassium channels) than to Na+, K+ diffuses back out, leaving the inside relatively negative and maintaining the resting potential of about -70 mV.
(b) B1 polarised (accept 'resting state')
(b) Answer: Polarised

Question 6

(a) B1 resting potential = -70 mV
(a) B1 depolarisation begins at t = 0.5 ms (accept 'between 0.5 and 1.0 ms')
(a) Answer: Resting potential = -70 mV; depolarisation begins at t = 0.5 ms.
(b) B1 a stimulus causes some sodium channels to open; if depolarisation reaches the threshold, voltage-gated sodium channels open
(b) B1 sodium ions diffuse rapidly into the axon down their electrochemical gradient, causing rapid depolarisation and reversing the potential (inside becomes positive, e.g. reaching +40 mV)
(b) B1 at the peak, sodium channels close (become inactivated) and voltage-gated potassium channels open
(b) B1 potassium ions diffuse out of the axon down their concentration gradient, removing positive charge from inside and restoring a negative internal potential (repolarisation)
(b) Answer: Depolarisation: voltage-gated sodium channels open once threshold is reached, sodium ions flood in, reversing the membrane potential. Repolarisation: sodium channels close, potassium channels open, potassium ions diffuse out, restoring the negative interior.
(c) B1 potassium channels are slower to close than they were to open, so potassium ions continue to diffuse out after the membrane has returned to -70 mV, causing a brief overshoot to a more negative potential (e.g. -80 mV)
(c) B1 during the refractory period, sodium channels are inactivated/cannot reopen immediately, so a new action potential cannot be generated
(c) B1 this ensures action potentials only travel in one direction (away from the previous site of depolarisation) and are discrete, separate impulses, which also limits the maximum frequency at which impulses can be generated (oe)
(c) Answer: Potassium channels stay open slightly too long, so more K+ leaves than needed, causing a brief hyperpolarisation below resting potential. The refractory period that follows (during which sodium channels cannot reopen) ensures impulses travel in one direction only and are transmitted as discrete impulses, limiting the maximum impulse frequency.
(d) M1 correct method: f = 1/T, using T = 1 ms = 0.001 s
(d) A1 1000 Hz (cao)
(d) Answer: 1000 Hz

Question 7

(a) M1 converts distance and time to SI units: 45 mm = 0.045 m and 0.5 ms = 0.0005 s
(a) dM1 correctly substitutes into speed = distance/time = 0.045/0.0005
(a) A1 90 m/s (cao, 2 sf)
(a) Answer: 90 m/s
(b) B1 the myelin sheath is an electrical insulator, so ion movement (and depolarisation) can only occur at the gaps between myelin cells, the nodes of Ranvier
(b) B1 depolarisation jumps from node to node rather than travelling continuously along the whole membrane (saltatory conduction)
(b) B1 this means fewer sections of membrane need to be depolarised over a given distance, so the impulse travels faster than in a non-myelinated (continuously conducting) neurone
(b) Answer: Myelin insulates the axon so depolarisation can only occur at the nodes of Ranvier; the impulse jumps from node to node (saltatory conduction) rather than depolarising the whole membrane, which greatly increases conduction speed.
(c) B1 a larger diameter axon has less resistance to the flow of ions (less charge leakage relative to the volume of cytoplasm)
(c) B1 so local currents spread further/faster along the axon before dissipating, increasing conduction speed
(c) B1 a decrease in temperature reduces the kinetic energy of ions and slows the rate of diffusion of ions through channels (and reduces the rate of respiration providing ATP for the sodium-potassium pump)
(c) B1 this slows the rate of depolarisation/repolarisation at each point along the axon, decreasing conduction speed
(c) Answer: A larger diameter reduces relative ion leakage so local currents spread faster, increasing speed. Lower temperature reduces ion diffusion rate and ATP-dependent pump activity, slowing depolarisation and repolarisation, decreasing speed.
(d) B1 the suggestion is not correct/only partly correct; although conduction speed does increase with temperature up to a point, above about 40 degrees C the proteins involved (ion channels and the sodium-potassium pump) begin to denature
(d) B1 this would disrupt normal ion movement and could slow, or even stop, conduction rather than continuing to speed it up (oe)
(d) Answer: The suggestion is incorrect above a certain point: while conduction speed rises with temperature initially, temperatures above roughly 40 degrees C denature the channel and pump proteins, which would slow or stop conduction rather than keep increasing its speed.

Question 8

(a) B1 repeat the test several times (e.g. at least 5-10 times) for each person
(a) B1 discard anticipatory catches (e.g. where the person appears to guess/close their hand before release) and use random, unpredictable time intervals between repeats
(a) B1 calculate the mean distance fallen (or mean reaction time) from the repeats, to reduce the effect of anomalous results/random variation
(a) Answer: Repeat the drop several times per person using random time intervals so the catch cannot be anticipated, discard any obvious anticipatory catches, and calculate a mean value from the repeats.
(b) M1 rearranges correctly to t = √2s/g
(b) M1 correctly substitutes s = 0.180 m: t = √(2 x 0.180)/9.81
(b) A1 t = 0.192 s (awrt 0.19 s)
(b) A1 converts to 192 ms (awrt 190-193 ms)
(b) Answer: 192 ms (0.192 s)
(c) B1 control a variable such as the hand/finger position used to catch, whether the dominant hand is used, time of day, or level of alertness/caffeine intake, with a reason why it affects reaction time (any valid confounding variable with reason)
(c) B1 suggest a way to improve reliability, e.g. increasing the number of repeats per person and/or testing multiple people to calculate a mean, to reduce the effect of random variation between individual trials
(c) Answer: For example, the hand used to catch should be the same for every person tested (dominant hand affects reaction time); reliability can be improved by increasing the number of repeats per person and averaging the results.
(d) B1 the measured reaction time involves impulses passing along several neurones (sensory, relay/interneurones in the brain, and motor) and crossing more than one synapse, each of which adds a synaptic delay
(d) B1 it also includes the time needed for processing/decision-making in the brain and for the muscle to respond and physically move the hand, not just the transmission across a single synapse
(d) Answer: The overall reaction time includes conduction along several neurones, several synaptic delays (not just one), processing time in the brain, and the time for the muscle to contract and move the hand, so it is much longer than a single synaptic crossing time.

Question 9

Level 3 (5-6): A logical, coherent and detailed account of the sequence of events, using correct scientific terminology throughout, covering most of the indicative content points in the correct order.
Level 2 (3-4): A generally correct account covering several stages of the sequence, with mostly appropriate terminology, but the account may be incomplete, contain minor inaccuracies, or lack full coherence/order.
Level 1 (1-2): Some relevant, isolated points are made, but with little logical sequence or correct use of scientific terminology; the answer may be a simple list.
Level 0 (0): No relevant content, or the response does not address the question.
Indicative content:
- An action potential arrives at the presynaptic knob and depolarises its membrane
- Depolarisation causes voltage-gated calcium ion channels in the presynaptic membrane to open
- Calcium ions diffuse into the presynaptic knob down their concentration gradient
- The influx of calcium ions causes synaptic vesicles (containing acetylcholine) to move to and fuse with the presynaptic membrane
- Acetylcholine is released into the synaptic cleft by exocytosis
- Acetylcholine diffuses across the synaptic cleft
- Acetylcholine binds to specific (cholinergic) receptors on ligand-gated sodium ion channels in the postsynaptic membrane
- Binding causes the sodium ion channels to open, and sodium ions diffuse into the postsynaptic neurone
- This depolarises the postsynaptic membrane, producing an excitatory postsynaptic potential (EPSP)
- If the depolarisation reaches the threshold, an action potential is generated in the postsynaptic neurone
- Acetylcholine is broken down by acetylcholinesterase into choline and ethanoic acid, which are reabsorbed by the presynaptic neurone, preventing continuous stimulation

Question 10

(a) B1 only the presynaptic neurone contains synaptic vesicles of neurotransmitter (acetylcholine) that can be released
(a) B1 only the postsynaptic membrane has the specific receptors for the neurotransmitter, to which acetylcholine can bind and open ion channels
(a) Answer: Only the presynaptic neurone has vesicles able to release acetylcholine, and only the postsynaptic membrane has receptors for it, so signals can only pass from the presynaptic to the postsynaptic neurone.
(b) B1 spatial summation: several different presynaptic neurones each release a small amount of neurotransmitter onto the same postsynaptic neurone at the same time, so their combined (summed) effect on depolarisation may reach threshold
(b) B1 temporal summation: a single presynaptic neurone releases neurotransmitter repeatedly, in quick succession, so the depolarisations combine (summate) before they fade, reaching threshold
(b) B1 in both cases, no single release of neurotransmitter is enough on its own to reach threshold, but the combined/summed depolarisations can reach threshold and trigger an action potential (oe)
(b) Answer: Spatial summation combines small depolarisations from several different presynaptic neurones acting at the same time; temporal summation combines repeated depolarisations from the same presynaptic neurone in quick succession. In both cases, individually sub-threshold signals combine to reach the threshold needed for an action potential.

Question 11

(a) B1 baroreceptors detect the increased stretching of the artery wall caused by the rise in blood pressure
(a) B1 they send an increased frequency of impulses along sensory neurones to the cardiovascular centre in the medulla oblongata
(a) B1 the medulla sends an increased frequency of impulses along parasympathetic (vagus) neurones to the sinoatrial node (SAN), and a decreased frequency of impulses along sympathetic neurones
(a) B1 acetylcholine released at the SAN reduces its rate of firing, decreasing heart rate back towards normal
(a) Answer: Baroreceptors detect the extra stretch from raised blood pressure and send more impulses to the medulla, which increases parasympathetic (vagus) and decreases sympathetic output to the SAN, releasing more acetylcholine and slowing heart rate back to normal.
(b) B1 chemoreceptors in the aorta, carotid arteries and medulla detect the fall in blood pH (rise in CO2 concentration)
(b) B1 they send an increased frequency of impulses along sensory neurones to the cardiovascular centre in the medulla oblongata
(b) B1 the medulla sends an increased frequency of impulses along sympathetic neurones to the SAN, and a decreased frequency along parasympathetic (vagus) neurones
(b) B1 noradrenaline released at the SAN increases its rate of firing, increasing heart rate, which increases blood flow to the lungs and removal of carbon dioxide (oe)
(b) Answer: Chemoreceptors detect the fall in pH/rise in CO2 and send more impulses to the medulla, which increases sympathetic and decreases parasympathetic output to the SAN, releasing noradrenaline and increasing heart rate to remove CO2 faster.
(c) M1 correct substitution: 68 x 72
(c) A1 4896 cm³/min
(c) A1 4.9 dm³/min (awrt 4.9, correct conversion of cm³ to dm³ by dividing by 1000)
(c) Answer: 4.9 dm³ per minute (4896 cm³/min)

Question 12

(a) B1 in a relaxed muscle, tropomyosin blocks the myosin-binding sites on the actin filament, and troponin is attached to both tropomyosin and actin, holding tropomyosin in this blocking position
(a) B1 an action potential arriving at the neuromuscular junction/travelling down the T-tubules causes calcium ions to be released from the sarcoplasmic reticulum
(a) B1 calcium ions bind to troponin, causing it to change shape
(a) B1 this pulls the attached tropomyosin away from the myosin-binding sites on actin, exposing them
(a) B1 myosin heads can now bind to actin, forming actomyosin cross-bridges, allowing the sliding filament mechanism (powered by ATP hydrolysis) to begin
(a) Answer: In a relaxed muscle, tropomyosin (held by troponin) blocks the myosin-binding sites on actin. A rise in calcium ion concentration causes calcium to bind troponin, changing its shape and moving tropomyosin to expose the binding sites, allowing myosin heads to attach to actin and begin the sliding filament mechanism.
(b) M1 correct method: (2.6 - 2.1)/2.6 x 100
(b) A1 19.2% (awrt 19%)
(b) Answer: 19.2%
(c) B1 the A band represents the length of the thick (myosin) filaments, which do not change length during contraction, so the A band stays the same width
(c) B1 the I band (where only thin/actin filaments are present, with no overlap) and the H zone (where only thick/myosin filaments are present, with no overlap) both narrow because the thin filaments are pulled further over/past the thick filaments
(c) B1 this increases the region of overlap between the two sets of filaments, without either type of filament itself getting shorter (oe)
(c) Answer: The A band (length of the myosin filaments) does not change because the filaments themselves do not shorten. The I band and H zone narrow because the actin filaments slide further over the myosin filaments, increasing the region of overlap, without the filaments themselves shortening.
(d) B1 the A band represents the length of the thick (myosin) filaments; if the filaments themselves were shortening, the A band would be expected to narrow along with the I band and H zone
(d) B1 because the A band stays constant while the whole sarcomere shortens, the filaments must instead be sliding past/over each other (increasing overlap) rather than shortening, supporting the sliding filament theory
(d) Answer: If filaments themselves shortened, the A band would narrow too; since it stays constant while the sarcomere shortens, the filaments must be sliding over each other rather than getting shorter, which supports the sliding filament theory over the filament-shortening hypothesis.

Question 13

Level 3 (5-6): The answer gives a well-balanced, detailed comparison covering both similarities and differences between nervous and hormonal communication, using correct scientific terminology and, where appropriate, an example to illustrate the points made.
Level 2 (3-4): The answer identifies several valid differences and/or similarities between nervous and hormonal communication, with reasonable use of terminology, but the comparison may be less detailed, one-sided, or lack an illustrative example.
Level 1 (1-2): The answer makes one or two simple, isolated correct points about nervous or hormonal communication, with little genuine comparison between the two.
Level 0 (0): No relevant content, or the response does not address the question.
Indicative content:
- Nervous communication uses electrical impulses (action potentials) travelling along neurones; hormonal communication uses chemical messengers (hormones) carried in the blood
- Nervous transmission is much faster than hormonal transmission, which relies on the relatively slow circulation of blood
- Nervous responses tend to be short-lived/rapidly reversed once the stimulus stops; hormonal effects tend to be longer-lasting
- Nervous signals act on a specific, localised target (the effector connected by the neurone via a synapse); hormones are carried throughout the body in the blood and can affect many different target cells/organs that have the appropriate receptors
- Nervous communication produces a precise, localised response; hormonal communication can produce a widespread, generalised response affecting many organs simultaneously
- Both systems ultimately rely on a receptor/target cell having specific receptor molecules to detect the signal (electrical or chemical), and are used together in many coordinated responses
- A valid example may be used, such as detecting a sudden loud noise (nervous, rapid, localised reflex) compared with the release of adrenaline in response to a threat (hormonal, slower onset but widespread and longer-lasting effects)

Question 14

(a) B1 acetylcholinesterase normally breaks down acetylcholine in the synaptic cleft after it has triggered a response; if it is inhibited, acetylcholine is not broken down (or is broken down much more slowly)
(a) B1 acetylcholine therefore remains in the synaptic cleft and continues to bind to receptors on the postsynaptic membrane
(a) B1 this causes continuous/repeated depolarisation of the postsynaptic membrane, generating action potentials continuously in the postsynaptic neurone (or continuous stimulation of a muscle fibre at a neuromuscular junction)
(a) B1 this leads to continuous, uncontrolled stimulation, which in an insect causes convulsions, paralysis and eventually death (oe)
(a) Answer: Because acetylcholinesterase cannot break down acetylcholine, it remains bound to postsynaptic receptors, causing continuous depolarisation and repeated action potentials/continuous muscle stimulation, leading to convulsions, paralysis and death of the insect.
(b) B1 the pesticide disrupts normal synaptic/neuromuscular transmission (as described in part (a)), which is likely to interfere with the nervous control of the kinesis response, reducing the woodlouse's normal rate of turning even in dry conditions
(b) B1 a reduced turning rate in dry conditions means treated woodlice are less likely to leave the dry area by chance, compared with untreated woodlice, which turn more and are more likely to move away
(b) B1 this means treated woodlice are more likely to remain in unfavourable, dry conditions for longer
(b) B1 remaining in dry conditions for longer increases the risk of desiccation (water loss), which can be lethal to woodlice, as they require moisture to breathe and to prevent water loss through their permeable cuticle
(b) B1 impaired nervous/muscular function from the pesticide may also reduce the woodlice's ability to escape predators or find food, further reducing survival and reproductive success, potentially reducing the population size (oe)
(b) Answer: The pesticide disrupts the normal nervous control of the kinesis response, lowering the turning rate even in dry conditions, so treated woodlice are less likely to leave unfavourable dry areas by chance; this increases their risk of desiccation and reduces their ability to escape predators, lowering their survival chances and potentially reducing the size of the woodlouse population in treated habitats.

Organisms Respond to Changes in their Environments - Worksheets, Questions and Revision