CCEA A-Level · thinka-original Practice Paper

2024 CCEA A-Level Biology 1010 Practice Paper with Answers

Thinka Jun 2024 CCEA A Level-Style Mock — Biology 1010

260 marks345 mins2024
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 CCEA A Level Biology 1010 paper. Not affiliated with or reproduced from CCEA.

Assessment Unit A2 1 - Section A

Answer all eight questions in the spaces provided. Show working in all calculations.
32 Question · 82 marks
Question 1 · Short Answer / Structured Recall
1 marks
State precisely the role of the myelin sheath in increasing the speed of nerve impulse conduction.
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Worked solution

Myelin acts as an electrical insulator around the axon. Because voltage-gated ion channels are concentrated only at the nodes of Ranvier, the action potential appears to jump from node to node (saltatory conduction) rather than travelling continuously along the membrane, greatly increasing conduction velocity.

Marking scheme

[1] Insulates the axon, restricting depolarisation to the nodes of Ranvier / enables saltatory conduction (impulse 'jumps' node to node).
Question 2 · Short Answer / Structured Recall
2 marks
Describe how the arrangement of the loop of Henle creates the concentration gradient in the medulla that allows urine to be concentrated.
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Worked solution

The ascending limb actively transports sodium and chloride ions into the surrounding medulla tissue but is impermeable to water, so the interstitial fluid becomes increasingly salty towards the base of the loop. Because the descending limb is permeable to water and fluid flows in the opposite direction (countercurrent), water is progressively drawn out of the descending limb by osmosis, establishing a steep osmotic gradient that later allows the collecting duct to reabsorb water and concentrate urine.

Marking scheme

[1] Ascending limb actively pumps out Na+/Cl- and is impermeable to water; [1] countercurrent (opposing-direction) flow in the two limbs builds an increasing salt gradient towards the base of the medulla, from which water leaves the descending limb by osmosis.
Question 3 · Short Answer / Structured Recall
1 marks
Define the term 'antigen'.
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Worked solution

Antigens are the surface molecules by which the immune system distinguishes self from non-self, and their presence triggers B- and T-lymphocyte responses, including antibody production.

Marking scheme

[1] A (usually protein/glycoprotein) molecule on the surface of a cell/pathogen recognised as non-self / that triggers an immune response (e.g. antibody production).
Question 4 · Short Answer / Structured Recall
2 marks
Explain how unequal auxin distribution causes a shoot to bend towards a unidirectional light source.
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Worked solution

Auxin (IAA) produced at the shoot tip is transported laterally away from the light and accumulates on the shaded side. Because auxin promotes cell elongation in shoot cells, the shaded side elongates faster than the illuminated side, bending the shoot towards the light source (positive phototropism).

Marking scheme

[1] Auxin is transported laterally to, and accumulates on, the shaded side of the shoot; [1] the higher auxin concentration causes greater cell elongation on the shaded side, so the shoot curves towards the light.
Question 5 · Short Answer / Structured Recall
1 marks
Define the term 'carrying capacity' of a population.
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Worked solution

Carrying capacity (K) is set by the resources (food, space, nesting sites) and other limiting factors of a habitat; population size stabilises around K once births and immigration balance deaths and emigration.

Marking scheme

[1] The maximum population size that a given environment/habitat can support/sustain indefinitely (with the resources available).
Question 6 · Short Answer / Structured Recall
2 marks
State the relationship between gross primary productivity (GPP), respiratory loss (R) and net primary productivity (NPP), and explain what NPP represents.
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Worked solution

Plants fix energy as GPP through photosynthesis, but a proportion of this is used in their own respiration (R). What remains, \( NPP = GPP - R \), is the energy stored as new plant biomass that is available to be passed on to the next trophic level or to increase plant growth.

Marking scheme

[1] \( NPP = GPP - R \); [1] NPP is the energy/biomass remaining after respiratory loss, available to primary consumers / for plant growth.
Question 7 · Short Answer / Structured Recall
1 marks
Name the genus of bacteria found in the root nodules of leguminous plants that fixes atmospheric nitrogen.
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Worked solution

Rhizobium bacteria live symbiotically in the root nodules of legumes (e.g. clover, peas) and use the enzyme nitrogenase to convert atmospheric N2 into ammonia, which the plant uses to build amino acids.

Marking scheme

[1] Rhizobium.
Question 8 · Short Answer / Structured Recall
2 marks
State two physiological effects of adrenaline that prepare the body for a 'fight or flight' response.
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Worked solution

Adrenaline, released from the adrenal medulla, binds to receptors on target tissues and triggers rapid changes that increase the availability of oxygen and glucose to skeletal muscle, while suppressing non-essential functions such as digestion.

Marking scheme

[1] each for any two of: increased heart rate/cardiac output; increased blood glucose (glycogenolysis in the liver); vasodilation of arterioles supplying skeletal muscle / vasoconstriction of arterioles supplying the gut; increased ventilation rate; pupil dilation. Max [2].
Question 9 · Short Answer / Structured Recall
1 marks
Name the region of the nephron in which selective reabsorption of glucose by active transport mainly occurs.
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Worked solution

Glucose is reabsorbed from the filtrate into the blood by co-transport with Na+ across the epithelium of the proximal convoluted tubule, which is adapted with microvilli and numerous mitochondria.

Marking scheme

[1] Proximal convoluted tubule.
Question 10 · Short Answer / Structured Recall
2 marks
Describe the sequence of events by which a phagocyte destroys a bacterium.
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Worked solution

Phagocytosis is a non-specific immune response. The phagocyte's cell-surface membrane engulfs the bacterium, enclosing it within a vesicle called a phagosome. A lysosome then fuses with the phagosome (forming a phagolysosome) and releases hydrolytic enzymes that digest the pathogen.

Marking scheme

[1] Phagocyte engulfs the bacterium (by endocytosis), enclosing it in a phagosome/vesicle; [1] a lysosome fuses with the phagosome and releases (hydrolytic/digestive) enzymes that break down the bacterium.
Question 11 · Short Answer / Structured Recall
1 marks
State the effect of a high auxin concentration on the growth of root cells.
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Worked solution

Root cells are far more sensitive to auxin than shoot cells; concentrations that promote elongation in a shoot are inhibitory in a root, contributing to the positive gravitropism/negative phototropism of roots.

Marking scheme

[1] Inhibits (elongation/growth of) root cells.
Question 12 · Short Answer / Structured Recall
2 marks
Distinguish between a density-dependent and a density-independent factor limiting population growth, giving one example of each.
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Worked solution

Density-dependent factors intensify as more individuals compete for the same finite resources or as disease spreads more readily in crowded populations, so they regulate population size around the carrying capacity. Density-independent factors, such as a severe storm or drought, reduce population size by roughly the same proportion regardless of how dense the population was.

Marking scheme

[1] Density-dependent factor has a greater effect at higher population density, with a valid example (e.g. food shortage, disease, predation, intraspecific competition); [1] density-independent factor affects the population regardless of density, with a valid example (e.g. extreme weather, fire, natural disaster).
Question 13 · Short Answer / Structured Recall
1 marks
State the approximate percentage of energy transferred between successive trophic levels in a food chain.
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Worked solution

Only around 10% of the energy available at one trophic level is typically incorporated into the biomass of the next level; the rest is lost as heat through respiration, or in faeces/urine and uneaten parts of the organism.

Marking scheme

[1] Approximately 10% (accept a value in the range 5-20%).
Question 14 · Short Answer / Structured Recall
2 marks
Describe the roles of Nitrosomonas and Nitrobacter in the process of nitrification.
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Worked solution

Both are chemoautotrophic soil bacteria that obtain energy by oxidising nitrogen compounds. Nitrosomonas converts NH4+ to NO2-, and Nitrobacter then converts this NO2- to NO3-, the form of nitrogen most readily absorbed by plant roots.

Marking scheme

[1] Nitrosomonas oxidises ammonium (NH4+) to nitrite (NO2-); [1] Nitrobacter oxidises nitrite (NO2-) to nitrate (NO3-).
Question 15 · Data Analysis & Scientific Application
3 marks
A student measured a volunteer's heart rate and stroke volume at rest and during moderate exercise. At rest: heart rate 68 beats min⁻¹, stroke volume 70 cm³. During exercise: heart rate 142 beats min⁻¹, stroke volume 110 cm³. Calculate the increase in cardiac output, in dm³ min⁻¹, between rest and exercise. Show your working.
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Worked solution

Cardiac output \( CO = HR \times SV \). Resting: \( 68 \times 70 = 4760 \text{ cm}^3\,\text{min}^{-1} = 4.76 \text{ dm}^3\,\text{min}^{-1} \). Exercise: \( 142 \times 110 = 15620 \text{ cm}^3\,\text{min}^{-1} = 15.62 \text{ dm}^3\,\text{min}^{-1} \). Increase \( = 15.62 - 4.76 = 10.86 \text{ dm}^3\,\text{min}^{-1} \).

Marking scheme

[1] Resting CO = 68 × 70 = 4760 cm³ min⁻¹ (4.76 dm³ min⁻¹); [1] exercise CO = 142 × 110 = 15620 cm³ min⁻¹ (15.62 dm³ min⁻¹); [1] correct increase = 10.86 dm³ min⁻¹ with correct unit conversion (allow ecf from earlier values).
Question 16 · Data Analysis & Scientific Application
3 marks
The table below shows the conduction velocity of a nerve impulse along a myelinated axon at four temperatures:
Temperature (°C) Conduction velocity (m/s)
10 25
20 42
30 61
37 70
Describe and explain the trend shown by these data.
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Worked solution

As temperature rises from 10°C to 37°C, conduction velocity rises steadily from 25 to 70 m/s. Higher temperature increases the kinetic energy of ions, increasing the rate of ion diffusion through voltage-gated channels, and increases the rate of activity of membrane transport proteins (e.g. the Na+/K+ pump), so depolarisation and repolarisation occur faster at each node, increasing conduction velocity.

Marking scheme

[1] Conduction velocity increases as temperature increases (positive correlation); [1] higher temperature increases the kinetic energy of ions / rate of diffusion of ions through channels; [1] higher temperature increases the rate of activity of transport proteins (e.g. Na+/K+ pump, ion channels), speeding depolarisation/repolarisation at each node.
Question 17 · Data Analysis & Scientific Application
3 marks
After a meal, a person's blood glucose concentration rose from 4.5 mmol dm⁻³ to 7.2 mmol dm⁻³ within 30 minutes, then fell back to 4.8 mmol dm⁻³ by 90 minutes, while plasma insulin concentration rose sharply between 30 and 60 minutes before falling. Explain these changes in terms of negative feedback.
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Worked solution

The initial rise in blood glucose after the meal is detected by beta cells in the pancreatic islets of Langerhans, which respond by secreting insulin. Insulin binds to receptors on liver and muscle cells, stimulating increased uptake of glucose and its conversion to glycogen (glycogenesis), so blood glucose concentration falls back towards the normal set point. As glucose concentration returns to normal, less insulin is secreted, illustrating negative feedback.

Marking scheme

[1] Rise in blood glucose detected by beta cells of the islets of Langerhans, which secrete insulin; [1] insulin stimulates glucose uptake by cells and conversion of glucose to glycogen (glycogenesis), lowering blood glucose; [1] as glucose falls back to the set point, insulin secretion decreases (negative feedback).
Question 18 · Data Analysis & Scientific Application
3 marks
A graph of antibody concentration in the blood following exposure to antigen X shows: after a first exposure at day 0, antibody concentration rises slowly to a low peak around day 12, then declines by day 25. After a second exposure to the same antigen at day 28, antibody concentration rises rapidly to a much higher peak by day 35. Explain the difference between these primary and secondary immune responses.
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Worked solution

During the primary response, only a small number of B cells specific to antigen X exist, so it takes time for clonal selection and expansion to produce enough plasma cells, giving a slow, low rise in antibody concentration. The first exposure also generates memory B cells (and memory T cells) that persist in the body. On the second exposure, these memory cells recognise the antigen immediately and divide rapidly into plasma cells, producing a faster and much larger secondary antibody response.

Marking scheme

[1] Primary response is slow/low because few B cells are initially specific to the antigen and must undergo clonal selection/expansion; [1] first exposure produces memory B cells that persist; [1] on second exposure, memory cells respond immediately, dividing rapidly into plasma cells, giving a faster and greater secondary response.
Question 19 · Data Analysis & Scientific Application
3 marks
The basic reproduction number (\( R_0 \)) for a pathogen in an unvaccinated population is 6. Using the formula \( H = 1 - \frac{1}{R_0} \), calculate the minimum percentage of the population that must be immune to achieve herd immunity, showing your working.
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Worked solution

\( H = 1 - \frac{1}{R_0} = 1 - \frac{1}{6} = 1 - 0.167 = 0.833 \). As a percentage this is 83.3%, so at least 83.3% of the population must be immune to prevent sustained transmission.

Marking scheme

[1] Substitution: \( H = 1 - \frac{1}{6} \); [1] \( H = 0.833 \); [1] correctly expressed as a percentage, 83.3% (accept 83%).
Question 20 · Data Analysis & Scientific Application
3 marks
A dehydrated volunteer's plasma ADH concentration and urine concentration were recorded: as plasma ADH rose from 2 to 14 pg cm⁻³ over several hours, urine concentration rose from 300 to 1150 mosmol kg⁻¹, while urine volume fell from 1.8 to 0.4 dm³ per day. Describe and explain the relationship shown by these data.
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Worked solution

The data show a positive correlation between plasma ADH concentration and urine concentration, and a corresponding fall in urine volume. Higher plasma ADH binds to receptors on collecting duct cells, causing more aquaporin channels to be inserted into the cell-surface membrane; this increases the permeability of the collecting duct to water, so more water is reabsorbed by osmosis into the blood, producing a smaller volume of more concentrated urine.

Marking scheme

[1] As ADH concentration increases, urine concentration increases and urine volume decreases (positive/negative correlations identified); [1] ADH increases the number of aquaporins / permeability of the collecting duct to water; [1] more water is reabsorbed by osmosis from the collecting duct into the blood, concentrating the urine.
Question 21 · Data Analysis & Scientific Application
3 marks
The table below shows the size of a population of bacteria introduced into a fresh culture medium:
Time (hours) Population size
0 50
4 400
8 3200
12 9800
16 10200
20 10250
Describe the shape of the growth curve shown by these data and explain the biological cause of each phase.
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Worked solution

Between 0 and 8 hours the population grows exponentially (doubling repeatedly) because nutrients and space are abundant and there is little competition or waste accumulation. Between 8 and 12 hours growth slows as nutrients become depleted and metabolic waste accumulates, increasing the death rate and slowing the birth rate. From 12 to 20 hours the population size levels off at around 10 200-10 250, the carrying capacity of the culture, as the birth rate falls to equal the death rate.

Marking scheme

[1] Sigmoid/S-shaped growth curve identified with an exponential (log) phase followed by a stationary phase; [1] exponential phase explained by abundant resources/space and low competition allowing unrestricted division; [1] stationary phase explained by resources becoming limiting so birth rate falls to equal death rate at the carrying capacity.
Question 22 · Data Analysis & Scientific Application
3 marks
In a grassland ecosystem, the gross primary productivity of producers was measured as 20 000 kJ m⁻² yr⁻¹, of which 12 000 kJ m⁻² yr⁻¹ was lost through plant respiration. Primary consumers were found to have a net productivity of 640 kJ m⁻² yr⁻¹. Calculate the percentage efficiency of energy transfer from net primary productivity to primary consumer net productivity, showing your working.
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Worked solution

\( NPP = GPP - R = 20000 - 12000 = 8000 \text{ kJ m}^{-2}\text{yr}^{-1} \). Efficiency of transfer \( = \frac{640}{8000} \times 100 = 8\% \).

Marking scheme

[1] NPP of producers = 20000 - 12000 = 8000 kJ m⁻² yr⁻¹; [1] correct efficiency calculation set up, \( \frac{640}{8000} \times 100 \); [1] correct answer, 8%.
Question 23 · Data Analysis & Scientific Application
3 marks
A soil sample treated with a bactericide that kills nitrifying bacteria showed nitrate concentration falling from 45 to 8 mg kg⁻¹ over four weeks, while ammonium concentration rose from 12 to 51 mg kg⁻¹ over the same period. Explain these changes.
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Worked solution

Nitrifying bacteria (Nitrosomonas, which oxidises ammonium to nitrite, and Nitrobacter, which oxidises nitrite to nitrate) are killed by the bactericide, so nitrification stops. Ammonium continues to be released into the soil by decomposers breaking down organic matter (ammonification/deamination) but is no longer converted to nitrate, so ammonium accumulates while nitrate concentration falls as the existing nitrate is taken up by plants, leached, or denitrified without replacement.

Marking scheme

[1] Bactericide kills nitrifying bacteria (Nitrosomonas/Nitrobacter), stopping nitrification; [1] ammonium continues to be released by decomposers (ammonification) but is no longer oxidised, so it accumulates; [1] nitrate concentration falls because it is no longer produced but continues to be removed (uptake/leaching/denitrification).
Question 24 · Data Analysis & Scientific Application
3 marks
Shoot cuttings were treated with increasing concentrations of synthetic auxin (IAA) and the degree of curvature towards a unidirectional light source was measured after 2 hours:
IAA concentration (mg dm⁻³) Curvature (°)
0 2
1 14
5 29
10 41
20 38
Describe and explain the trend shown by these data.
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Worked solution

As IAA concentration rises from 0 to 10 mg dm⁻³, curvature increases steadily, since auxin promotes cell elongation, so a greater difference in auxin concentration between the shaded and illuminated sides produces greater differential growth and curvature. Above the optimum concentration (here between 10 and 20 mg dm⁻³), curvature decreases because the auxin concentration becomes supra-optimal and starts to inhibit, rather than promote, cell elongation.

Marking scheme

[1] Curvature increases with IAA concentration up to 10 mg dm⁻³; [1] curvature then decreases at the highest concentration tested (20 mg dm⁻³); [1] explanation that low/moderate auxin concentrations promote cell elongation (more differential growth) but supra-optimal concentrations inhibit elongation.
Question 25 · Mechanistic Explanations
4 marks
Describe and explain the events occurring at a cholinergic synapse that result in transmission of a nerve impulse from the pre-synaptic to the post-synaptic neurone.
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Worked solution

An action potential arriving at the presynaptic knob depolarises the membrane, opening voltage-gated calcium channels; Ca2+ ions diffuse into the presynaptic knob. This influx of Ca2+ causes synaptic vesicles containing acetylcholine (ACh) to move to and fuse with the presynaptic membrane, releasing ACh into the synaptic cleft by exocytosis. ACh diffuses across the cleft and binds to specific receptors on the postsynaptic membrane, opening ligand-gated Na+ channels; Na+ influx depolarises the postsynaptic membrane, and if threshold is reached, a new action potential is generated.

Marking scheme

[1] Depolarisation of the presynaptic membrane opens voltage-gated Ca2+ channels, and Ca2+ diffuses into the presynaptic knob; [1] Ca2+ causes synaptic vesicles to fuse with the presynaptic membrane; [1] acetylcholine is released into the synaptic cleft by exocytosis and diffuses across; [1] ACh binds to receptors on the postsynaptic membrane, opening (ligand-gated) Na+ channels and depolarising it.
Question 26 · Mechanistic Explanations
4 marks
Describe and explain the role of the sliding filament mechanism in skeletal muscle contraction.
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Worked solution

An action potential travelling along the sarcolemma and down the T-tubules triggers release of Ca2+ from the sarcoplasmic reticulum. Ca2+ binds to troponin, changing its shape and pulling the attached tropomyosin away from the myosin-binding sites on the actin filament. Myosin heads, energised by ATP hydrolysis, bind to actin forming cross-bridges; the heads then flex, pulling the actin filament towards the centre of the sarcomere (the power stroke), before detaching (using a further ATP molecule) and re-attaching further along. This repeated cycle slides the actin filaments past the myosin filaments, shortening the sarcomere and the muscle fibre.

Marking scheme

[1] Ca2+ (released from the sarcoplasmic reticulum) binds to troponin, moving tropomyosin to expose the myosin-binding sites on actin; [1] myosin heads bind to actin, forming cross-bridges; [1] ATP hydrolysis powers the myosin head 'power stroke', pulling the actin filament inward; [1] repeated cross-bridge cycling slides the actin and myosin filaments past each other, shortening the sarcomere.
Question 27 · Mechanistic Explanations
4 marks
Explain how negative feedback maintains core body temperature within narrow limits, using a rise in core temperature above the set point as your example.
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Worked solution

Thermoreceptors in the hypothalamus (and skin) detect a rise in blood/core temperature above the set point. The hypothalamus acts as a co-ordination centre and sends nervous impulses to effectors: arterioles supplying the skin capillaries dilate (vasodilation), increasing blood flow near the skin surface and heat loss by radiation, and sweat glands increase sweat secretion, increasing heat loss by evaporation. As core temperature falls back towards the set point, the thermoreceptors detect the change and the effector responses are reduced/switched off, completing the negative feedback loop.

Marking scheme

[1] Thermoreceptors (in the hypothalamus) detect the rise in blood/core temperature; [1] hypothalamus co-ordinates a response via effectors; [1] vasodilation of skin arterioles increases heat loss by radiation, and/or increased sweating increases heat loss by evaporation; [1] as temperature returns to the set point, the effector response is reduced, illustrating negative feedback.
Question 28 · Mechanistic Explanations
4 marks
Explain the roles of T-helper cells and B-lymphocytes in the humoral immune response to a bacterial infection.
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Worked solution

A macrophage phagocytoses the bacterium and displays fragments of its antigens on its cell-surface membrane (antigen presentation). A T-helper cell bearing a complementary receptor binds to the presented antigen and becomes activated, releasing cytokines. These cytokines stimulate a B-lymphocyte with a complementary antibody to undergo clonal selection and mitotic clonal expansion; the resulting clone of B cells differentiates into plasma cells, which secrete large quantities of specific antibody, and memory B cells, which persist to provide future immunity.

Marking scheme

[1] Macrophage engulfs the bacterium and presents its antigens on its surface; [1] T-helper cell with complementary receptor binds and is activated, releasing cytokines; [1] cytokines stimulate clonal selection/expansion of the complementary B-lymphocyte; [1] B cells differentiate into antibody-secreting plasma cells and memory cells.
Question 29 · Mechanistic Explanations
4 marks
Explain how memory cells produced during a primary immune response provide long-term immunity to a pathogen.
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Worked solution

During a primary immune response, some of the activated B and T lymphocytes differentiate into long-lived memory cells rather than short-lived effector cells; these persist in the blood and lymphoid tissue for years. If the same pathogen is encountered again, the memory cells recognise its antigens immediately, without the delay required to activate naive lymphocytes, and rapidly proliferate into large numbers of plasma cells. These plasma cells secrete antibody more quickly and in far greater quantity than in the primary response, so the pathogen is destroyed before it can cause symptomatic disease.

Marking scheme

[1] Memory B and T cells persist in the body long after the primary response; [1] on re-exposure, memory cells recognise the antigen immediately (no delay for clonal selection of naive cells); [1] memory cells divide rapidly into plasma cells; [1] antibody is produced faster and in greater quantity than in the primary response, destroying the pathogen before symptoms develop.
Question 30 · Mechanistic Explanations
4 marks
Explain how a fall in blood water potential is detected and corrected by the action of ADH, restoring water balance.
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Worked solution

Osmoreceptor cells in the hypothalamus lose water by osmosis when blood water potential falls (i.e. solute concentration rises), causing them to shrink and fire more frequently. This stimulates the posterior pituitary gland to release more antidiuretic hormone (ADH) into the blood. ADH travels to the kidney and binds to receptors on the cells of the collecting duct, causing more aquaporin channels to be inserted into their cell-surface membranes. This increases the permeability of the collecting duct to water, so more water is reabsorbed by osmosis into the surrounding blood, producing a smaller volume of more concentrated urine and restoring blood water potential (negative feedback).

Marking scheme

[1] Osmoreceptors in the hypothalamus detect the fall in blood water potential (shrink by osmosis); [1] hypothalamus stimulates the posterior pituitary to release more ADH; [1] ADH increases the number of aquaporins/permeability of the collecting duct to water; [1] more water is reabsorbed by osmosis, restoring blood water potential (negative feedback).
Question 31 · Mechanistic Explanations
4 marks
Explain the mechanism by which abscisic acid (ABA) causes stomatal closure during water stress.
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Worked solution

When a plant experiences water stress, abscisic acid is synthesised (mainly in the roots, but also in stressed leaves) and transported to the guard cells surrounding the stomata. ABA binds to specific receptors on the guard cell plasma membrane, triggering signalling that opens ion channels and causes efflux of K+ (and associated anions such as malate) out of the guard cells. This raises the guard cells' water potential, so water leaves by osmosis; the guard cells lose turgor and become flaccid, and because their cell walls are unevenly thickened, this causes the stoma to close, reducing water loss by transpiration.

Marking scheme

[1] ABA is synthesised (in roots/leaves) in response to water stress and transported to guard cells; [1] ABA binds to guard cell receptors, triggering efflux of K+ (and other solutes) from the guard cells; [1] water potential of guard cells rises, so water leaves by osmosis; [1] guard cells lose turgor/become flaccid, closing the stoma and reducing transpiration.
Question 32 · Mechanistic Explanations
3 marks
Explain how earthworms and saprotrophic fungi contribute to decomposition and nutrient cycling in soil.
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Worked solution

Earthworms are detritivores: they physically break dead leaves and other organic matter into smaller fragments as they feed and burrow, greatly increasing the surface area available for microbial and fungal attack, and mixing organic matter through the soil. Saprotrophic fungi cannot ingest food, so they secrete extracellular (saprotrophic) enzymes onto dead organic matter, digesting complex molecules externally into simpler soluble products, which the fungal hyphae then absorb by diffusion/active transport. This process releases inorganic ions such as nitrate, phosphate and ammonium back into the soil (mineralisation), making them available for uptake by plant roots.

Marking scheme

[1] Earthworms physically fragment/break down organic matter, increasing surface area for microbial/fungal decomposition; [1] saprotrophic fungi secrete extracellular enzymes that digest organic matter externally; [1] fungi absorb the soluble products, releasing (mineralising) inorganic ions such as nitrate/phosphate/ammonium back into the soil for plant uptake.

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Practice This Topic

Assessment Unit A2 1 - Section B

Answer the extended prose question in continuous prose. Quality of written communication will be assessed.
2 Question · 18 marks
Question 1 · Extended Essay (QWC Level of Response)
9 marks
Discuss the roles of the nervous system and the endocrine system in maintaining homeostasis in mammals, using named examples to illustrate your answer.
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Worked solution

A strong answer identifies homeostasis as the maintenance of a constant internal environment via negative feedback, and contrasts the nervous system (electrical impulses along neurones, fast-acting, short-lived, e.g. thermoregulation via hypothalamic control of vasodilation and sweating) with the endocrine system (hormones travelling in the blood, slower-acting, longer-lasting, e.g. insulin/glucagon controlling blood glucose, or ADH controlling osmoregulation). Both systems are co-ordinated by the hypothalamus/pituitary axis, and the best answers link named receptors, co-ordinators and effectors to specific negative feedback loops.

Marking scheme

Level of response marking. Band 3 (7-9 marks): wide-ranging, accurate coverage of both nervous and hormonal homeostatic mechanisms with at least two well-explained named examples (e.g. thermoregulation and blood glucose regulation), correct use of specialist terminology (negative feedback, receptor, co-ordinator, effector), clear and coherent written communication. Band 2 (4-6 marks): reasonable coverage of both systems with at least one clear named example, mostly accurate terminology, generally clear communication. Band 1 (1-3 marks): basic, largely descriptive points with limited named examples or unclear links to homeostasis, weak use of terminology. Band 0: no creditworthy material.
Question 2 · Extended Essay (QWC Level of Response)
9 marks
Discuss how intraspecific competition, interspecific competition and predator-prey interactions regulate the size of natural populations.
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Worked solution

A strong answer explains intraspecific competition as competition between members of the same species for limited resources (food, space, mates), which intensifies as population density rises and acts as a density-dependent factor regulating population size around the carrying capacity. It explains interspecific competition between different species for the same resources, potentially leading to competitive exclusion of the less well-adapted species or resource partitioning/niche differentiation allowing coexistence. It explains predator-prey interactions using a named example (e.g. lynx and snowshoe hare), describing how an increase in prey numbers provides more food for predators, whose numbers subsequently rise and reduce prey numbers, causing predator numbers to fall again, producing cyclical oscillations that regulate both populations.

Marking scheme

Level of response marking. Band 3 (7-9 marks): accurate, detailed coverage of intraspecific competition, interspecific competition and predator-prey cycles, with a named example (e.g. lynx/hare) and clear explanation of density-dependent regulation, coherent and well-organised prose with accurate terminology. Band 2 (4-6 marks): coverage of at least two of the three mechanisms with reasonable explanation, some named examples, generally clear communication. Band 1 (1-3 marks): basic descriptive points on one or two mechanisms, limited exemplification, weak terminology. Band 0: no creditworthy material.

Assessment Unit A2 2 - Section A

Answer all seven questions in the spaces provided. Statistical tables are provided.
28 Question · 82 marks
Question 1 · Structured Diagnostic / Classification
2 marks
State two features that distinguish angiosperms (flowering plants) from gymnosperms (conifers).
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Worked solution

Angiosperms are characterised by flowers as their reproductive structures and by seeds enclosed within an ovary that develops into a fruit, whereas gymnosperms lack flowers, reproduce using cones, and their seeds are 'naked', not enclosed in an ovary.

Marking scheme

[1] Angiosperms produce flowers / seeds enclosed in an ovary (fruit); [1] gymnosperms produce cones / naked seeds not enclosed in an ovary.
Question 2 · Structured Diagnostic / Classification
2 marks
State whether mosses (bryophytes) possess true vascular tissue, and state one adaptation, or lack of adaptation, that restricts them to damp habitats.
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Worked solution

Bryophytes such as mosses do not possess true vascular (xylem and phloem) tissue, so they cannot transport water efficiently over large distances or support tall growth. They also lack a well-developed waterproof cuticle and true roots, and their motile male gametes must swim through a film of water to reach the female gametangium, so they are restricted to damp, low-growing habitats.

Marking scheme

[1] Mosses lack true vascular tissue (no xylem/phloem); [1] valid reason restricting them to damp habitats, e.g. male gametes must swim through water to fertilise the egg / lack a waterproof cuticle / lack true roots.
Question 3 · Structured Diagnostic / Classification
2 marks
State the net number of ATP molecules produced by substrate-level phosphorylation during glycolysis (per glucose molecule), and name the 3-carbon end product.
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Worked solution

Glycolysis uses 2 ATP to phosphorylate glucose and produces 4 ATP by substrate-level phosphorylation later in the pathway, giving a net gain of 2 ATP per glucose molecule, along with 2 molecules of the 3-carbon compound pyruvate and reduced NAD.

Marking scheme

[1] Net 2 ATP; [1] pyruvate (pyruvic acid).
Question 4 · Structured Diagnostic / Classification
2 marks
State the location within the mitochondrion of (i) the Krebs cycle and (ii) oxidative phosphorylation.
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Worked solution

The Krebs cycle enzymes are located in solution in the mitochondrial matrix, whereas the electron transport chain and ATP synthase enzymes responsible for oxidative phosphorylation are embedded in the inner mitochondrial membrane, which is folded into cristae to increase surface area.

Marking scheme

[1] Krebs cycle: mitochondrial matrix; [1] oxidative phosphorylation: inner mitochondrial membrane / cristae.
Question 5 · Structured Diagnostic / Classification
2 marks
State the base-pairing rule in DNA and name the type of bond that holds complementary base pairs together.
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Worked solution

In the DNA double helix, adenine always pairs with thymine (via 2 hydrogen bonds) and cytosine always pairs with guanine (via 3 hydrogen bonds); these complementary base pairs are held together by hydrogen bonds, which are individually weak but numerous, giving stability while still allowing the strands to separate for replication and transcription.

Marking scheme

[1] A pairs with T, C pairs with G (complementary base pairing); [1] hydrogen bonds.
Question 6 · Structured Diagnostic / Classification
2 marks
Name the enzyme that unwinds the DNA double helix during replication, and the enzyme that joins Okazaki fragments on the lagging strand.
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Worked solution

DNA helicase breaks the hydrogen bonds between complementary base pairs, unwinding the double helix and separating the two strands to expose them for replication. On the lagging strand, DNA polymerase synthesises short Okazaki fragments discontinuously; these are subsequently joined together by DNA ligase, which catalyses formation of phosphodiester bonds between adjacent fragments.

Marking scheme

[1] DNA helicase; [1] DNA ligase.
Question 7 · Structured Diagnostic / Classification
2 marks
State two ways in which mRNA differs structurally from DNA.
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Worked solution

mRNA differs from DNA in that it is a single polynucleotide strand rather than a double helix, it contains the sugar ribose rather than deoxyribose, and it contains the base uracil in place of thymine; mRNA molecules are also generally much shorter than DNA.

Marking scheme

[1] mRNA is single-stranded, DNA is double-stranded; [1] mRNA contains ribose (not deoxyribose) and uracil (not thymine) (accept either sugar or base difference for the second mark).
Question 8 · Structured Diagnostic / Classification
2 marks
Name the type of enzyme used to cut DNA at specific recognition sequences during genetic engineering, and state what is meant by a 'sticky end'.
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Worked solution

Restriction endonucleases cut DNA at specific recognition sequences. Many make a staggered cut across the two strands, leaving short single-stranded overhangs called sticky ends; because these overhangs are complementary to the sticky ends produced by the same enzyme cutting elsewhere, they can base-pair with any other DNA fragment cut with the same enzyme, which is essential for inserting a gene into a plasmid vector.

Marking scheme

[1] Restriction enzyme / restriction endonuclease; [1] a short single-stranded overhang of unpaired/exposed bases (produced by a staggered cut) that can base-pair with a complementary sticky end.
Question 9 · Structured Diagnostic / Classification
2 marks
Name the enzyme used to join a gene of interest into a plasmid vector, and state the type of bond it forms.
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Worked solution

DNA ligase catalyses the formation of phosphodiester bonds between the sugar-phosphate backbones of the gene of interest and the cut plasmid, sealing the two pieces of DNA together to form a single recombinant DNA molecule.

Marking scheme

[1] DNA ligase; [1] phosphodiester bond(s).
Question 10 · Structured Diagnostic / Classification
2 marks
State two features of a plasmid that make it suitable for use as a vector in genetic engineering.
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Worked solution

Plasmids are small, circular molecules of DNA found naturally in bacteria, separate from the main bacterial chromosome. Their small size and circular structure make them easy for bacterial cells to take up (e.g. during transformation) and structurally stable, while their own origin of replication allows them to be copied independently of the host chromosome; they can also be engineered to carry marker genes that allow transformed cells to be identified.

Marking scheme

[1] each for any two of: small (readily taken up by bacteria); circular (structurally stable); contains an origin of replication (replicates independently); can carry a marker/selectable gene. Max [2].
Question 11 · Structured Diagnostic / Classification
2 marks
State the role of a marker gene, such as an antibiotic-resistance gene, in identifying genetically transformed bacteria.
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Worked solution

After the transformation procedure, only a small proportion of bacteria will have taken up the recombinant plasmid. By growing all the bacteria on agar containing the relevant antibiotic, only those cells carrying the plasmid (and therefore the antibiotic-resistance marker gene) will survive and form colonies, while untransformed bacteria are killed; this allows the transformed colonies to be selected and identified.

Marking scheme

[1] Only bacteria that have taken up the plasmid carry the marker/resistance gene and survive on antibiotic-containing agar; [1] non-transformed bacteria (lacking the gene) are killed, allowing transformed colonies to be identified/selected.
Question 12 · Structured Diagnostic / Classification
2 marks
Name the technique used to amplify a specific DNA sequence in vitro, and name the heat-stable enzyme it requires.
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Worked solution

The polymerase chain reaction (PCR) is used to amplify a specific target DNA sequence in vitro. Because the reaction is repeatedly heated to high temperatures to separate the DNA strands, it requires a heat-stable DNA polymerase, typically Taq polymerase, originally isolated from the thermophilic bacterium Thermus aquaticus, which is not denatured by the high temperatures used.

Marking scheme

[1] Polymerase chain reaction (PCR); [1] Taq polymerase (heat-stable/thermostable DNA polymerase).
Question 13 · Data Interpretation & Epigenetic Analysis
3 marks
The table below shows four plant classification characteristics recorded for specimens P, Q, R and S:
Specimen Vascular tissue Seeds Flowers
P Yes Yes Yes
Q Yes Yes No
R Yes No No
S No No No
Using the data, identify which specimen is most likely a fern, a conifer, a moss and a flowering plant, explaining your reasoning.
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Worked solution

Specimen S has no vascular tissue, so it must be a moss (bryophyte), the only group in the table lacking xylem/phloem. Specimen R has vascular tissue but no seeds, matching a fern (pteridophyte), which reproduces by spores rather than seeds. Specimen Q has vascular tissue and seeds but no flowers, matching a conifer (gymnosperm), which produces naked seeds in cones. Specimen P has vascular tissue, seeds and flowers, matching a flowering plant (angiosperm), whose seeds are enclosed within an ovary/fruit.

Marking scheme

[1] S identified as moss, reasoning: no vascular tissue; [1] R identified as fern, reasoning: vascular tissue present but no seeds; [1] Q and P correctly identified as conifer (seeds, no flowers) and flowering plant (seeds and flowers) respectively.
Question 14 · Data Interpretation & Epigenetic Analysis
3 marks
A germinating pea seed in a respirometer absorbed 2.4 cm³ of oxygen and released 2.4 cm³ of carbon dioxide over a one-hour period. Calculate the respiratory quotient (RQ) and use your answer to identify the main respiratory substrate being used.
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Worked solution

\( RQ = \frac{\text{volume of } CO_2 \text{ produced}}{\text{volume of } O_2 \text{ consumed}} = \frac{2.4}{2.4} = 1.0 \). An RQ of 1.0 indicates that carbohydrate is the main substrate being fully aerobically respired, since equal volumes of O2 are consumed and CO2 produced when respiring carbohydrate.

Marking scheme

[1] Correct RQ formula/substitution \( \frac{2.4}{2.4} \); [1] RQ = 1.0; [1] correctly identifies carbohydrate as the main respiratory substrate (RQ = 1.0 is characteristic of carbohydrate respiration).
Question 15 · Data Interpretation & Epigenetic Analysis
3 marks
A student compared the number of ATP molecules produced per glucose molecule in yeast under aerobic conditions (38 ATP) and anaerobic conditions (2 ATP). Using this data, explain why anaerobic respiration is far less efficient, and suggest why yeast might still use it when oxygen is limited.
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Worked solution

Aerobic respiration breaks glucose down completely to carbon dioxide and water via glycolysis, the link reaction, the Krebs cycle and oxidative phosphorylation, releasing far more chemical energy (38 ATP) than glycolysis alone. Anaerobic respiration (fermentation) in yeast stops after glycolysis (net 2 ATP), converting pyruvate to ethanol and CO2 to regenerate NAD+, so most of the chemical energy remaining in the substrate is not released. However, when oxygen is limited or absent, aerobic respiration cannot proceed (no oxygen to act as the final electron acceptor), so yeast relies on anaerobic respiration to still generate some ATP and allow glycolysis to continue by regenerating NAD+.

Marking scheme

[1] Anaerobic respiration only completes glycolysis (net 2 ATP), while aerobic respiration continues through the link reaction/Krebs cycle/oxidative phosphorylation (38 ATP); [1] most chemical energy remains in the ethanol/incompletely oxidised end product under anaerobic conditions; [1] without oxygen (as final electron acceptor) oxidative phosphorylation cannot occur, so anaerobic respiration allows continued ATP production/NAD+ regeneration despite low efficiency.
Question 16 · Data Interpretation & Epigenetic Analysis
3 marks
A graph of the rate of photosynthesis against light intensity was plotted at two CO2 concentrations (0.04% and 0.4%), at constant temperature. At low light intensity, both curves rise together and are identical. Above a light intensity of 20 arbitrary units, the 0.04% CO2 curve levels off, while the 0.4% CO2 curve continues to rise before levelling off at a higher rate. Explain these observations in terms of limiting factors.
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Worked solution

At low light intensity, both curves follow the same path because light intensity is the factor limiting the rate of photosynthesis at both CO2 concentrations, so increasing CO2 has no effect. Above 20 units of light intensity, the 0.04% CO2 curve plateaus because CO2 concentration has become the limiting factor and further increases in light intensity cannot increase the rate. At the higher CO2 concentration (0.4%), CO2 is less likely to be limiting, so photosynthesis can continue to increase with light intensity to a higher maximum rate, until light intensity itself (or another factor) again becomes limiting.

Marking scheme

[1] At low light intensity, light is the limiting factor for both curves, so they are identical; [1] above 20 units, the 0.04% CO2 curve plateaus because CO2 becomes the limiting factor; [1] at 0.4% CO2, CO2 is no longer limiting so the rate continues to rise (to a higher plateau) until another factor becomes limiting.
Question 17 · Data Interpretation & Epigenetic Analysis
3 marks
Gel electrophoresis was performed on restriction-digested DNA fragments from three suspects and a crime-scene sample. Explain how the pattern of bands produced could be used to identify the suspect responsible, referring to fragment size and migration distance.
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Worked solution

The digested DNA fragments are loaded into wells in an agarose gel and an electric current applied across it. Because DNA is negatively charged (due to its phosphate groups), the fragments migrate towards the positive electrode (anode); smaller fragments experience less resistance from the gel matrix and so travel further in a given time than larger fragments, separating fragments by size into distinct bands. Because the number and size of restriction fragments produced depends on an individual's specific base sequence, each person produces a characteristic band pattern (DNA profile). By comparing the band pattern of the crime-scene sample with those of the three suspects, the suspect whose pattern matches (same number of bands at the same migration distances) can be identified as the likely source.

Marking scheme

[1] DNA fragments are negatively charged and migrate towards the anode under the electric current; [1] smaller fragments migrate further/faster than larger fragments, separating them into a band pattern (DNA profile); [1] the suspect's DNA profile that matches the crime-scene sample (same band pattern/positions) is identified as the source.
Question 18 · Data Interpretation & Epigenetic Analysis
3 marks
State the three temperature stages of one cycle of the polymerase chain reaction (PCR) and explain the purpose of each.
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Worked solution

Each PCR cycle consists of three stages at different temperatures. During denaturation, the reaction mixture is heated to around 95°C, breaking the hydrogen bonds between the two DNA strands and separating them. During annealing, the temperature is lowered to around 50-65°C, allowing short single-stranded DNA primers to bind (anneal) to their complementary sequences on each template strand, flanking the target region. During extension, the temperature is raised to around 72°C, the optimum for Taq polymerase, which adds free nucleotides complementary to the template strand, synthesising new DNA from each primer and doubling the amount of target DNA present.

Marking scheme

[1] Denaturation (~95°C): breaks hydrogen bonds, separating the double strand into single strands; [1] annealing (~50-65°C): primers bind to complementary sequences on the single strands; [1] extension (~72°C): Taq polymerase synthesises new DNA from free nucleotides, extending from the primers.
Question 19 · Data Interpretation & Epigenetic Analysis
3 marks
Explain, with reference to the use of a marker gene, how scientists select for bacteria that have successfully taken up a recombinant plasmid carrying a gene of interest and an ampicillin-resistance marker gene.
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Worked solution

Following the transformation procedure, only a proportion of bacteria will have taken up the recombinant plasmid, and this cannot be seen directly. By spreading all the treated bacteria onto agar plates containing the antibiotic ampicillin, only those bacterial cells that have taken up the plasmid (and therefore carry the ampicillin-resistance marker gene) will be able to survive and grow into visible colonies; bacteria that failed to take up the plasmid lack the resistance gene and are killed by the antibiotic. This allows the successfully transformed colonies to be identified and selected for further use, such as culturing to produce a useful gene product.

Marking scheme

[1] Bacteria are plated on agar containing ampicillin; [1] only bacteria carrying the plasmid (with the resistance gene) survive and form colonies; [1] untransformed bacteria (without the resistance gene) are killed, allowing transformed colonies to be selected/identified.
Question 20 · Data Interpretation & Epigenetic Analysis
3 marks
State one ethical concern associated with genetically engineering crop plants to be resistant to a broad-spectrum herbicide, and explain its implication for the environment or human health.
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Worked solution

One ethical/environmental concern is that the herbicide-resistance gene could spread from the GM crop to closely related wild plant species through cross-pollination (horizontal/gene flow). If wild relatives acquire herbicide resistance, they could become difficult-to-control 'superweeds' that out-compete native, non-resistant plant species for resources such as light, water and nutrients, reducing local biodiversity and potentially requiring the use of even more toxic herbicides to control them.

Marking scheme

[1] Valid concern stated, e.g. gene flow/cross-pollination to wild relatives, or reduced biodiversity, or reliance on/overuse of herbicide; [1] explanation of the mechanism (e.g. pollen transfer of the resistance gene to a wild relative); [1] explanation of the consequence (e.g. resistant 'superweeds' out-compete native species / reduced biodiversity / increased herbicide use).
Question 21 · Genetic Cross & Chi-Squared Calculation
4 marks
In a species of flower, allele R codes for red pigment and allele W codes for white pigment; the two alleles are codominant, so heterozygotes are pink. A pink-flowered plant (RW) was crossed with a white-flowered plant (WW). Using a genetic diagram, determine the expected phenotype ratio of the offspring.
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Worked solution

Parental genotypes: RW (pink) × WW (white). Gametes from RW: R or W. Gametes from WW: W only. Combining gametes in a Punnett square gives offspring genotypes RW and WW in a 1:1 ratio. Because R and W are codominant, RW offspring are pink and WW offspring are white, giving an expected phenotype ratio of 1 pink : 1 white.

Marking scheme

[1] Correct parental gametes identified (R, W from RW parent; W, W from WW parent); [1] correctly completed genetic diagram/Punnett square showing offspring genotypes RW and WW; [1] correct genotype ratio 1 RW : 1 WW; [1] correct phenotype ratio stated, 1 pink : 1 white, with reference to codominance.
Question 22 · Genetic Cross & Chi-Squared Calculation
4 marks
In pea plants, allele T (tall) is dominant to t (dwarf), and allele G (green pods) is dominant to g (yellow pods); the two genes assort independently. A plant heterozygous for both traits (TtGg) was crossed with a plant homozygous recessive for both traits (ttgg). Using a genetic diagram, determine the expected phenotypic ratio of the offspring.
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Worked solution

Parent TtGg produces four equally frequent gamete types (independent assortment): TG, Tg, tG, tg. Parent ttgg produces only tg gametes. Combining these in a genetic diagram gives offspring genotypes TtGg, Ttgg, ttGg and ttgg in equal (1:1:1:1) proportions, corresponding to phenotypes tall green, tall yellow, dwarf green and dwarf yellow respectively, in a 1:1:1:1 ratio.

Marking scheme

[1] Correct gametes from TtGg identified (TG, Tg, tG, tg) by independent assortment; [1] correct single gamete type from ttgg (tg); [1] correctly completed genetic diagram showing all four offspring genotypes (TtGg, Ttgg, ttGg, ttgg); [1] correct phenotype ratio, 1 tall green : 1 tall yellow : 1 dwarf green : 1 dwarf yellow.
Question 23 · Genetic Cross & Chi-Squared Calculation
4 marks
In fruit flies, the allele for red eyes (R) is dominant to white eyes (r) and is carried on the X chromosome. A red-eyed female fly, heterozygous for the gene, was crossed with a white-eyed male. Using a genetic diagram, determine the expected genotype and phenotype ratios of the offspring, with reference to sex.
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Worked solution

The heterozygous female parent is X^R X^r; the white-eyed male parent is X^r Y. Female gametes: X^R or X^r. Male gametes: X^r or Y. Combining these gives offspring X^R X^r (red female), X^r X^r (white female), X^R Y (red male) and X^r Y (white male), each with a probability of 1/4. All female offspring receive an X^R from their mother's X^R gamete or an X^r, so half the females are red-eyed (X^R X^r) and half white-eyed (X^r X^r); all male offspring receive their single X chromosome from the mother, so half are red-eyed (X^R Y) and half white-eyed (X^r Y).

Marking scheme

[1] Correct parental genotypes and gametes identified (X^R X^r female; X^r Y male); [1] correctly completed genetic diagram showing all four offspring genotypes; [1] correct genotype ratio, 1 X^R X^r : 1 X^r X^r : 1 X^R Y : 1 X^r Y; [1] correct phenotype ratio with reference to sex, e.g. half of female offspring red-eyed and half white-eyed, half of male offspring red-eyed and half white-eyed.
Question 24 · Genetic Cross & Chi-Squared Calculation
4 marks
A student carried out a dihybrid cross in maize and expected a 9:3:3:1 ratio of phenotypes. State an appropriate null hypothesis for a chi-squared test on these data, and calculate the expected number of each phenotype class if 320 offspring were produced in total.
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Worked solution

The null hypothesis (H0) states that there is no significant difference between the observed ratio of phenotypes and the expected 9:3:3:1 ratio; any deviation between observed and expected values is due to chance alone. Expected numbers are calculated by multiplying the total offspring by each fraction of the ratio: \( \frac{9}{16} \times 320 = 180 \), \( \frac{3}{16} \times 320 = 60 \), \( \frac{3}{16} \times 320 = 60 \), \( \frac{1}{16} \times 320 = 20 \).

Marking scheme

[1] Valid null hypothesis: no significant difference between observed and expected (9:3:3:1) ratios / any difference is due to chance; [1] correct method, \( \frac{9}{16}, \frac{3}{16}, \frac{3}{16}, \frac{1}{16} \times 320 \); [1] correct expected values 180 and 60; [1] correct expected values 60 and 20 (all four values correct for full marks).
Question 25 · Genetic Cross & Chi-Squared Calculation
3 marks
For one phenotype class in a chi-squared test, the observed number was 68 and the expected number was 60. Calculate the contribution of this class to the chi-squared value, showing your working.
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Worked solution

The chi-squared contribution of each class is \( \frac{(O-E)^2}{E} \). Here, \( \frac{(68-60)^2}{60} = \frac{8^2}{60} = \frac{64}{60} = 1.067 \) (to 3 s.f.).

Marking scheme

[1] Correct substitution, \( \frac{(68-60)^2}{60} \); [1] \( (68-60)^2 = 64 \); [1] correct final answer, 1.07 (accept 1.06-1.07).
Question 26 · Genetic Cross & Chi-Squared Calculation
3 marks
In a genetic cross investigation, the calculated chi-squared value was 2.71 with 3 degrees of freedom. The critical value at the p = 0.05 significance level is 7.82. State the decision regarding the null hypothesis and explain what this means for the observed data.
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Worked solution

Because the calculated chi-squared value of 2.71 is smaller than the critical value of 7.82 at the p = 0.05 significance level with 3 degrees of freedom, the null hypothesis is accepted (not rejected). This means that any difference between the observed and expected phenotype ratios is not statistically significant, and is likely to be due to chance rather than to a real biological cause such as linkage or gene interaction.

Marking scheme

[1] Calculated value (2.71) is less than the critical value (7.82); [1] null hypothesis is accepted (not rejected); [1] the difference between observed and expected data is not statistically significant / is due to chance.
Question 27 · Graph Sketching & Spectrum Analysis
6 marks
Describe and explain the shape of the absorption spectra of chlorophyll a and chlorophyll b across the visible light spectrum (400-700 nm), and relate these to the action spectrum of photosynthesis.
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Worked solution

Chlorophyll a and chlorophyll b each absorb strongly in two regions of the visible spectrum: the blue-violet region (approximately 430-450 nm) and the red region (approximately 640-680 nm), producing two absorption peaks. Between these peaks, in the green region of the spectrum (approximately 500-600 nm), both pigments absorb very little light, which is instead reflected or transmitted, explaining why leaves appear green. The action spectrum, which plots the rate of photosynthesis against wavelength, closely mirrors this combined absorption spectrum of the photosynthetic pigments (chlorophylls and carotenoids): the rate of photosynthesis is highest at the blue-violet and red wavelengths that are absorbed most strongly, and lowest in the green region, providing evidence that light absorbed by these pigments is used to drive photosynthesis.

Marking scheme

[1] Two absorption peaks identified, in the blue-violet region (~430-450 nm) and the red region (~640-680 nm); [1] a trough of low absorption in the green region (~500-600 nm); [1] green light is largely reflected/transmitted, explaining the green colour of leaves; [1] the action spectrum closely matches/mirrors the (combined) absorption spectrum of the pigments; [1] rate of photosynthesis is highest at wavelengths most strongly absorbed (blue-violet and red); [1] rate of photosynthesis is lowest in the green region, providing evidence that absorbed light drives photosynthesis.
Question 28 · Graph Sketching & Spectrum Analysis
6 marks
Describe and explain the shape of a graph showing the rate of photosynthesis against light intensity at a constant, non-limiting CO2 concentration and constant temperature, identifying the limiting factor(s) operating in each region of the curve.
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Worked solution

At low light intensity, the rate of photosynthesis increases linearly (in direct proportion) with increasing light intensity, because light is the factor limiting the rate of the light-dependent reactions; each additional unit of light increases the rate of photolysis and ATP/NADPH production, allowing a proportionally faster rate of carbon fixation. As light intensity continues to increase, the rate of increase slows and the curve begins to level off, and eventually plateaus at a maximum rate; beyond this point, increasing light intensity no longer increases the rate of photosynthesis because another factor, typically CO2 concentration (availability of substrate for the Calvin cycle) or temperature (affecting enzyme activity), has become limiting instead.

Marking scheme

[1] Rate rises steeply/linearly from the origin at low light intensity; [1] light intensity is the limiting factor in this initial linear region; [1] the rate of increase slows as light intensity rises further; [1] the curve levels off/plateaus at a maximum rate at high light intensity; [1] a different factor becomes limiting at the plateau, e.g. CO2 concentration (availability for the Calvin cycle) or temperature (enzyme activity); [1] the plateau would rise to a new, higher maximum if CO2 concentration or temperature were increased (i.e. these factors, not light, now limit the rate).

Assessment Unit A2 2 - Section B

Answer the extended prose question in continuous prose. Quality of written communication will be assessed.
1 Question · 18 marks
Question 1 · Extended Essay (QWC Level of Response)
18 marks
Discuss the semi-conservative replication of DNA, and explain, using named examples, how errors in this process can lead to gene mutations and their potential effects on the phenotype of an organism.
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Worked solution

A strong answer describes semi-conservative replication in detail: DNA helicase unwinds the double helix by breaking hydrogen bonds between base pairs, exposing two template strands; DNA polymerase synthesises a new complementary strand on each template by adding free nucleotides according to the base-pairing rule, continuously on the leading strand and discontinuously (as Okazaki fragments, joined by DNA ligase) on the lagging strand, producing two daughter molecules each containing one original and one new strand. It then explains that errors occurring during this process (e.g. incorrect base insertion not corrected by proofreading) produce gene mutations, and distinguishes types of mutation: base substitution (which may be silent, missense or nonsense, depending on whether the resulting codon still codes for the same amino acid, a different amino acid, or a stop codon), and insertion or deletion of bases (which causes a frameshift, altering every subsequent codon and usually the whole amino acid sequence downstream). It links this to a named example, such as sickle-cell anaemia, caused by a single base substitution in the gene coding for the beta-globin chain of haemoglobin, changing one amino acid (glutamic acid to valine) and altering the protein's structure and function, causing red blood cells to become sickle-shaped.

Marking scheme

Level of response marking (18 marks). Band 3 (13-18 marks): comprehensive, accurate account of semi-conservative replication (helicase, DNA polymerase, leading and lagging strands, ligase) and of how mutations arise and their effects (substitution vs insertion/deletion, frameshift, silent/missense/nonsense), with at least one accurately explained named example (e.g. sickle-cell anaemia); coherent, well-structured prose with accurate specialist terminology throughout. Band 2 (7-12 marks): reasonable coverage of both replication and mutation with some named exemplification, mostly accurate terminology, generally clear and organised communication, though some detail may be missing or an area under-developed. Band 1 (1-6 marks): basic, largely descriptive points on replication or mutation but not both in depth, limited or absent named examples, weak use of terminology, communication may lack clarity. Band 0: no creditworthy material.

Section Assessment Unit A2 3 - Practical Skills

Answer all eight questions based on practical procedures, experimental design, and data analysis.
27 Question · 60 marks
Question 1 · Apparatus & Practical Technique Recall
1 marks
Name the piece of apparatus used to measure the rate of water uptake by a leafy shoot, as an indirect estimate of transpiration rate.
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Worked solution

A potometer measures the rate at which a cut, leafy shoot takes up water through its xylem, which closely approximates the rate of water loss by transpiration from the leaves, since almost all water taken up is eventually lost by evaporation from the leaf surfaces.

Marking scheme

[1] Potometer.
Question 2 · Apparatus & Practical Technique Recall
1 marks
Name the stain used to make starch grains visible under a light microscope.
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Worked solution

Iodine solution reacts with starch to produce a blue-black colouration, allowing starch grains within cells to be identified and located under the light microscope.

Marking scheme

[1] Iodine (in potassium iodide) solution.
Question 3 · Apparatus & Practical Technique Recall
1 marks
Name the piece of apparatus used to count the number of cells, such as yeast cells, in a known volume of a suspension.
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Worked solution

A haemocytometer is a specialised microscope slide with an etched grid of known area and depth, allowing the number of cells within a known volume of suspension to be counted directly and the concentration of cells calculated.

Marking scheme

[1] Haemocytometer / counting chamber.
Question 4 · Apparatus & Practical Technique Recall
1 marks
State the function of soda lime in a simple respirometer investigation.
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Worked solution

Soda lime is a strong absorber of carbon dioxide. By including it in the respirometer chamber, the CO2 produced by the respiring organism is removed as it is released, so the decrease in gas volume (and pressure) recorded by the manometer reflects only oxygen consumption, allowing the rate of oxygen uptake to be measured directly.

Marking scheme

[1] Absorbs the CO2 produced by respiration, so the (pressure/volume) change recorded is due only to O2 uptake.
Question 5 · Apparatus & Practical Technique Recall
2 marks
Describe how a colorimeter is used to measure the rate of a colour-change reaction, such as the decolourisation of DCPIP.
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Worked solution

A filter that absorbs the colour complementary to the solution's colour is fitted, maximising sensitivity to the colour change. The colorimeter is calibrated to zero absorbance using a blank (e.g. water or the uncoloured reagent) before the reaction is started. The reaction mixture is then placed in a cuvette in the colorimeter, and absorbance (or percentage transmission) is recorded at regular time intervals as the reaction proceeds; as DCPIP is reduced and decolourised, absorbance falls, and the rate of decrease can be used to calculate the rate of reaction.

Marking scheme

[1] An appropriate colour filter is selected and the colorimeter is zeroed/calibrated against a blank; [1] absorbance (or % transmission) of the reaction mixture is measured at regular time intervals, and the rate of change in absorbance used to determine the reaction rate.
Question 6 · Apparatus & Practical Technique Recall
2 marks
Describe how you would prepare a serial dilution of a bacterial culture in order to obtain a countable number of colonies on an agar plate.
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Worked solution

A measured volume of the original culture (e.g. 1 cm³) is transferred aseptically into a test tube containing a known, larger volume of sterile diluent (e.g. 9 cm³), giving a ten-fold dilution; the tube is mixed thoroughly. A sample of this diluted suspension is then transferred into the next tube of fresh sterile diluent, repeating the process through a series of tubes to achieve successively greater dilutions (e.g. 10⁻¹, 10⁻², 10⁻³...). A known volume from one or more of the more dilute tubes is then spread onto agar plates, so that individual colonies can be counted rather than an uncountable confluent lawn.

Marking scheme

[1] A fixed/known volume of culture is transferred aseptically into a known larger volume of sterile diluent and mixed, giving a known dilution factor, repeated through a series of tubes; [1] a known volume from an appropriately diluted tube is spread onto agar to give a countable number of colonies.
Question 7 · Apparatus & Practical Technique Recall
2 marks
Describe how aseptic technique is used when inoculating an agar plate with a bacterial culture, to prevent contamination.
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Worked solution

Aseptic technique minimises contamination by unwanted microorganisms. Work should be carried out close to a lit Bunsen burner, whose heat creates a rising convection current that carries airborne microbes upward and away from the work area. The inoculating loop used to transfer bacteria is sterilised by passing it through the Bunsen flame until red hot, both before and after use, and allowed to cool before touching the culture. The lid of the Petri dish is lifted only slightly and briefly (at an angle, not laid flat) while inoculating, to minimise the time it is exposed to the air, and the neck of any culture bottle is passed briefly through the flame before and after the cap is removed.

Marking scheme

[1] any two of: work near a Bunsen flame (updraft carries airborne contaminants away); loop is flamed/sterilised before and after use; culture vessel neck is flamed before/after opening; [1] lid of the agar plate is opened only briefly/at an angle to minimise exposure to air, then taped (not sealed) after inoculation.
Question 8 · Apparatus & Practical Technique Recall
2 marks
Describe how gel electrophoresis separates DNA fragments of different lengths.
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Worked solution

Digested DNA samples are loaded into wells at one end of an agarose gel, which is submerged in a conducting buffer solution, and an electric current is passed through the gel. Because DNA carries an overall negative charge due to its phosphate groups, the fragments migrate through the pores of the gel towards the positive electrode (anode). Smaller fragments encounter less resistance from the gel matrix and so move faster and travel further in a given time than larger fragments, so after a set time the fragments have separated into distinct bands ordered by size.

Marking scheme

[1] DNA fragments are negatively charged and migrate through the gel towards the anode under an electric current; [1] smaller fragments move further/faster than larger fragments, separating fragments by size into bands.
Question 9 · Apparatus & Practical Technique Recall
2 marks
Describe how paper chromatography is used to separate the photosynthetic pigments extracted from a leaf.
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Worked solution

A small, concentrated spot of the leaf pigment extract is applied to a pencil line drawn near the base of a strip of chromatography paper; the spot is allowed to dry and may be reapplied several times to concentrate it. The paper is then placed upright in a shallow layer of solvent, with the pigment spot kept above the solvent surface (so pigment does not dissolve directly into the solvent reservoir). As the solvent rises up the paper by capillary action, it carries the different pigments with it; because each pigment has a different solubility in the solvent and different degree of attraction to the paper, they travel at different rates and separate into distinct coloured bands.

Marking scheme

[1] A small, concentrated spot of pigment extract is applied near the base of the paper, above the solvent level; [1] as solvent rises up the paper by capillary action, pigments of differing solubility travel at different rates, separating into distinct bands.
Question 10 · Apparatus & Practical Technique Recall
2 marks
Explain why a control set-up, such as one containing boiled/denatured enzyme, is included in an investigation of enzyme activity.
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Worked solution

Including a control in which the enzyme has been denatured (e.g. by boiling) allows the experimenter to check that any change observed in the experimental tubes (such as a colour change or gas production) is genuinely caused by the catalytic activity of the active enzyme, rather than by some other factor in the reaction mixture (e.g. a non-enzymic chemical reaction, or a measurement artefact). Since the denatured enzyme should produce no reaction, it provides a valid baseline against which the results of the experimental (active-enzyme) tubes can be compared, increasing the validity of the conclusions drawn.

Marking scheme

[1] Shows that any change/result observed is due to the (catalytic) action of the active enzyme; [1] provides a baseline/comparison to confirm the result is not caused by another (non-enzymic) factor, increasing the validity of the investigation.
Question 11 · Microscopy & Image Analysis
2 marks
Under the ×40 objective lens, a cell measured 25 eyepiece graticule units. A stage micrometer showed that, at this magnification, 100 eyepiece units corresponded to 250 μm. Calculate the actual length of the cell, showing your working.
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Worked solution

\( 100 \text{ eyepiece units} = 250\,\mu m \), so \( 1 \text{ eyepiece unit} = \frac{250}{100} = 2.5\,\mu m \). Actual length \( = 25 \times 2.5 = 62.5\,\mu m \).

Marking scheme

[1] Correct calibration, 1 eyepiece unit = 250 ÷ 100 = 2.5 μm; [1] correct final answer, 62.5 μm (25 × 2.5), with working shown.
Question 12 · Microscopy & Image Analysis
2 marks
An electron micrograph of a chloroplast had a measured length of 40 mm, and the micrograph was stated to have a magnification of ×8000. Calculate the actual length of the chloroplast, giving your answer in μm.
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Worked solution

\( \text{actual size} = \frac{\text{image size}}{\text{magnification}} = \frac{40\,\text{mm}}{8000} = 0.005\,\text{mm} = 5\,\mu m \).

Marking scheme

[1] Correct method, \( \frac{40\,\text{mm}}{8000} \); [1] correct final answer, 5 μm, with correct unit conversion.
Question 13 · Microscopy & Image Analysis
2 marks
State two structural features of a mitochondrion visible on an electron micrograph that cannot be resolved using a light microscope.
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Worked solution

A light microscope has insufficient resolution to distinguish the fine internal structure of a mitochondrion, appearing only as a small unstructured organelle. An electron micrograph, with its much higher resolution, reveals the double membrane (a smooth outer membrane and a highly folded inner membrane), the cristae (folds of the inner membrane that increase surface area for oxidative phosphorylation), and the granular matrix, sometimes showing ribosomes or DNA loops.

Marking scheme

[1] each for any two of: cristae (folds of the inner membrane); double membrane (distinct inner and outer membranes); matrix granules/ribosomes/DNA loops within the matrix. Max [2].
Question 14 · Microscopy & Image Analysis
2 marks
Explain why an electron microscope has a much higher resolution than a light microscope.
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Worked solution

The resolution of a microscope, its ability to distinguish two points as separate, is fundamentally limited by the wavelength of the radiation used to form the image: the shorter the wavelength, the higher the resolution that can be achieved. Electrons have a much shorter wavelength than visible light, so an electron microscope, which uses a focused beam of electrons rather than light, can resolve structures that are far closer together (and so has a much higher resolution and can achieve much greater useful magnification) than a light microscope.

Marking scheme

[1] Electrons have a much shorter wavelength than (visible) light; [1] resolution is limited by/proportional to wavelength, so the shorter wavelength of electrons gives higher resolution (allows closer structures to be distinguished).
Question 15 · Quantitative Practical Calculations (Rf, RQ, Haemocytometer)
3 marks
In a chromatogram of leaf pigments, chlorophyll a travelled 5.4 cm from the origin, and the solvent front travelled 9.0 cm. Calculate the Rf value of chlorophyll a, showing your working.
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Worked solution

\( R_f = \frac{\text{distance moved by pigment}}{\text{distance moved by solvent front}} = \frac{5.4}{9.0} = 0.6 \).

Marking scheme

[1] Correct formula stated/used, \( R_f = \frac{\text{distance moved by pigment}}{\text{distance moved by solvent}} \); [1] correct substitution, \( \frac{5.4}{9.0} \); [1] correct final answer, 0.6 (no units).
Question 16 · Quantitative Practical Calculations (Rf, RQ, Haemocytometer)
3 marks
A respirometer investigation using germinating seeds recorded an oxygen uptake of 3.6 cm³ and a carbon dioxide output of 2.7 cm³ over the same time period. Calculate the respiratory quotient, and use your answer to suggest the main respiratory substrate.
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Worked solution

\( RQ = \frac{CO_2 \text{ produced}}{O_2 \text{ consumed}} = \frac{2.7}{3.6} = 0.75 \). An RQ of approximately 0.7 is characteristic of lipid respiration (lipids are relatively oxygen-poor, so more O2 is required per CO2 produced than for carbohydrate).

Marking scheme

[1] Correct substitution, \( \frac{2.7}{3.6} \); [1] correct final answer, RQ = 0.75; [1] correctly identifies lipid as the substrate, with reference to RQ ≈ 0.7 being characteristic of lipid respiration.
Question 17 · Quantitative Practical Calculations (Rf, RQ, Haemocytometer)
3 marks
A student counted 84 yeast cells within the central 1 mm × 1 mm × 0.1 mm grid of a haemocytometer, from a sample that had been diluted 10-fold before loading. Calculate the concentration of yeast cells, in cells per cm³, in the original undiluted culture. Show your working.
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Worked solution

Volume of the counting grid \( = 1 \times 1 \times 0.1 = 0.1\,\text{mm}^3 = 1 \times 10^{-4}\,\text{cm}^3 \). Concentration in the diluted sample \( = \frac{84}{1 \times 10^{-4}} = 8.4 \times 10^5 \text{ cells cm}^{-3} \). Since the sample was diluted 10-fold, the concentration in the original culture \( = 8.4 \times 10^5 \times 10 = 8.4 \times 10^6 \text{ cells cm}^{-3} \).

Marking scheme

[1] Grid volume correctly calculated as 1 × 10⁻⁴ cm³ (0.1 mm³); [1] concentration in the diluted sample correctly calculated as 8.4 × 10⁵ cells cm⁻³ (84 ÷ 1×10⁻⁴); [1] correctly multiplied by the dilution factor (×10) to give 8.4 × 10⁶ cells cm⁻³ in the original culture.
Question 18 · Quantitative Practical Calculations (Rf, RQ, Haemocytometer)
2 marks
A drawing of a plant cell had a measured width of 60 mm. The actual width of the cell is 15 μm. Calculate the magnification of the drawing, showing your working.
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Worked solution

Converting to the same units: \( 60\,\text{mm} = 60000\,\mu m \). \( \text{magnification} = \frac{\text{image size}}{\text{actual size}} = \frac{60000}{15} = 4000 \).

Marking scheme

[1] Both measurements correctly converted to the same units (60 mm = 60000 μm); [1] correct final answer, magnification = ×4000.
Question 19 · Quantitative Practical Calculations (Rf, RQ, Haemocytometer)
2 marks
Calculate the number of degrees of freedom for a chi-squared test comparing observed and expected values across five phenotype classes.
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Worked solution

Degrees of freedom \( = n - 1 \), where \( n \) is the number of classes/categories. Here \( n = 5 \), so degrees of freedom \( = 5 - 1 = 4 \).

Marking scheme

[1] Correct formula, degrees of freedom = n − 1; [1] correct final answer, 4.
Question 20 · Quantitative Practical Calculations (Rf, RQ, Haemocytometer)
2 marks
In a water potential investigation, a potato chip had a mass of 2.15 g before immersion in a sucrose solution and a mass of 1.98 g after 24 hours. Calculate the percentage change in mass, showing your working.
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Worked solution

\( \% \text{ change} = \frac{\text{final mass} - \text{initial mass}}{\text{initial mass}} \times 100 = \frac{1.98 - 2.15}{2.15} \times 100 = \frac{-0.17}{2.15} \times 100 = -7.9\% \) (to 2 s.f.).

Marking scheme

[1] Correct substitution, \( \frac{1.98 - 2.15}{2.15} \times 100 \); [1] correct final answer, -7.9% (accept -7.9% to -8.0%), with a negative sign or the word 'decrease' shown.
Question 21 · Experimental Evaluation & Control Variables
3 marks
In an investigation into the effect of temperature on the rate of an enzyme-catalysed reaction, identify two variables, other than temperature, that should be controlled, and explain why for one of them.
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Worked solution

Variables that should be standardised include enzyme concentration, substrate concentration, pH (e.g. using a buffer) and reaction volume. If, for example, substrate concentration were not kept constant, then any difference observed in reaction rate between temperatures could be caused by the difference in substrate concentration rather than by temperature itself, making it impossible to draw a valid conclusion about the effect of temperature alone; controlling this variable ensures that temperature is the only factor being tested (a fair test).

Marking scheme

[1] each for two valid control variables identified (e.g. substrate concentration, enzyme concentration, pH/buffer used, reaction volume); [1] valid explanation for one variable, that failing to control it would independently affect reaction rate, confounding the effect of temperature and preventing a valid conclusion.
Question 22 · Experimental Evaluation & Control Variables
3 marks
Explain why it is important to use a large sample size, such as several replicate readings at each light intensity, when investigating the effect of light intensity on the rate of photosynthesis.
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Worked solution

Biological measurements are subject to natural variation and random experimental error, so a single reading at each light intensity might not be representative of the true rate. By taking several replicate readings at each light intensity, any anomalous results can be identified (and, if appropriate, excluded and repeated), and a mean value can be calculated for each light intensity. This mean is less affected by random error or unusual individual readings than a single measurement, increasing the reliability of the data and the validity of any conclusions or trend drawn from it.

Marking scheme

[1] Reduces the effect of random error/natural variation on the results; [1] allows anomalous results to be identified (and excluded/repeated); [1] allows a mean to be calculated, giving more reliable/representative data, increasing confidence in the trend/conclusion.
Question 23 · Experimental Evaluation & Control Variables
3 marks
A student investigating osmosis in potato tissue obtained an anomalous result at one sucrose concentration. Suggest one possible source of experimental error that could explain this anomaly, and describe how the investigation could be modified to reduce this error.
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Worked solution

One likely source of error is inconsistent blotting of surface moisture from the potato chip before it is weighed; if excess surface water is left on one chip but not others, its recorded mass will be anomalously high, distorting the calculated percentage change in mass. This could be reduced by standardising the blotting procedure, for example always using the same type of absorbent paper and pressing each chip a fixed number of times in the same way, for every chip and at every measurement, ensuring surface water is removed consistently across all repeats.

Marking scheme

[1] Valid source of error suggested (e.g. inconsistent blotting of surface water, chip not cut to identical dimensions, inaccurate timing, evaporation from the solution); [1] a specific modification described that would reduce this error; [1] explanation of how the modification standardises/controls the source of error across all readings.
Question 24 · Experimental Evaluation & Control Variables
3 marks
Explain why a water bath, rather than heating directly with a Bunsen burner, should be used to control temperature in an enzyme investigation.
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Worked solution

A water bath surrounds the reaction vessel with water at a controlled, even temperature, allowing heat to be transferred gradually and uniformly to the reaction mixture, so the desired temperature can be maintained accurately and consistently (checked with a thermometer) for the duration of the investigation. Heating directly with a Bunsen burner would raise the temperature of the reaction mixture unevenly and very rapidly, making it difficult to control or maintain a precise, constant temperature; this could cause local overheating that denatures the enzyme, introducing an uncontrolled source of error and making it impossible to relate the results reliably to a single, known temperature.

Marking scheme

[1] A water bath heats evenly and allows temperature to be controlled/maintained accurately at a set value; [1] direct Bunsen heating is uneven/rapid and difficult to control precisely; [1] uneven/uncontrolled heating could locally denature the enzyme or introduce an uncontrolled variable, reducing validity.
Question 25 · Experimental Evaluation & Control Variables
3 marks
In an investigation of population density using quadrats, explain why random placement of the quadrats is important for obtaining valid data.
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Worked solution

If a student chooses where to place quadrats themselves, they may unconsciously select areas that appear to have particularly high or low numbers of the organism being studied, introducing bias into the sample. Placing quadrats randomly, for example by generating random coordinates and using a measuring tape/grid to locate each point, ensures that every part of the habitat has an equal chance of being sampled, so the resulting sample is representative of the population density across the whole area, allowing a valid estimate (e.g. of mean population density) to be made for the entire habitat.

Marking scheme

[1] Random placement avoids experimenter bias (e.g. selecting unusually dense/sparse areas); [1] ensures the sample is representative of the whole habitat/area; [1] allows a valid estimate of population density to be made for the whole area (not just the sampled points).
Question 26 · Experimental Evaluation & Control Variables
3 marks
Explain the difference between the terms 'precision' and 'accuracy' as applied to quantitative biological data, using an example.
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Worked solution

Precision describes the closeness of agreement between repeated measurements of the same quantity, taken under the same conditions; a precise set of readings shows little variation/spread between repeats, regardless of whether they are close to the true value. Accuracy describes how close a measured value is to the true or accepted value of the quantity being measured. It is possible for data to be precise but not accurate (e.g. repeated readings that cluster tightly together but around the wrong value, perhaps due to a systematic error such as an uncalibrated instrument), or accurate but not precise (readings that average close to the true value but vary widely between repeats).

Marking scheme

[1] Precision: how close repeated measurements are to each other (small spread/range between repeats); [1] accuracy: how close a measurement/mean is to the true value; [1] valid example or explanation showing data can be precise without being accurate (or vice versa), e.g. due to a systematic error/uncalibrated instrument.
Question 27 · Experimental Evaluation & Control Variables
3 marks
A student wants to test whether a new fungicide affects the germination rate of seeds. State one independent variable, one dependent variable and one variable that should be standardised in this investigation.
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Worked solution

The independent variable, the factor deliberately changed by the experimenter, is the concentration (or presence/absence) of the fungicide applied to the seeds. The dependent variable, the factor measured as the outcome, is the germination rate, typically recorded as the percentage of seeds that germinate within a set time. A variable that should be standardised (kept the same for every treatment) to ensure a fair test could be the species and batch/source of seed used, the incubation temperature, the volume of water or fungicide solution applied, or the light conditions, since any of these could independently affect germination if not controlled.

Marking scheme

[1] Correct independent variable (concentration of fungicide); [1] correct dependent variable (percentage/number of seeds germinated, or germination rate); [1] one valid standardised/control variable identified (e.g. seed species/batch, temperature, volume of solution, light).

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