2022 OCR GCSE Gateway Science - Biology A - J247 模擬試題連答案詳解

Plant cells are eukaryotic cells and contain membrane-bound organelles such as mitochondria. Prokaryotic cells (such as bacteria) do not contain membrane-bound organelles like mitochondria or a true nucleus. Both plant and bacterial cells have cell walls, cytoplasm, and ribosomes.

C (Mitochondria) [1]

Photosynthesis is controlled by enzymes. At high temperatures above the optimum (such as \(45\,^\circ\text{C}\)), the active site of the enzymes changes shape (denatures), meaning substrates can no longer bind and the reaction rate drops to zero.

B (Enzymes controlling the reactions become denatured.) [1]

Active transport moves substances from a region of lower concentration to a region of higher concentration (against a concentration gradient), which requires energy from cellular respiration. Osmosis refers specifically to the movement of water, and diffusion is passive movement down a concentration gradient.

A (Active transport) [1]

In a spinal reflex arc, the receptor detects a stimulus and generates an electrical impulse that travels along the sensory neurone to the central nervous system, across a synapse to a relay neurone, and then to a motor neurone which carries the impulse to the effector.

D (Sensory neurone \(\rightarrow\) relay neurone \(\rightarrow\) motor neurone) [1]

A single organism (one producer) has a huge biomass and can supply food for a large number of primary consumers (caterpillars), resulting in the lowest bar representing a count of 1 while the next bar represents thousands of individuals.

B (One producer has a very large biomass and supports many primary consumers.) [1]

A mutation is defined as a rare, random change in the base sequence of DNA. This can alter the triplet code and potentially result in a different amino acid sequence in the protein produced.

A (A change in the base sequence of DNA) [1]

Magnification is calculated using the formula:
\[\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\]
\[\text{Magnification} = \frac{20\text{ mm}}{0.04\text{ mm}} = 500\]
Therefore, the correct magnification is \(\times 500\).

C (\(\times 500\)) [1]

Aerobic respiration breaks down glucose completely in the presence of oxygen to produce carbon dioxide and water, releasing a large amount of energy. In yeast, anaerobic respiration (fermentation) produces ethanol and carbon dioxide, releasing less energy.

C (Aerobic respiration produces water; anaerobic respiration produces ethanol.) [1]

Phloem tissue transports dissolved sugars (sucrose) and amino acids around the plant by translocation. Xylem transports water and mineral ions.

C (Phloem) [1]

When focusing on a distant object, the ciliary muscles relax, causing the suspensory ligaments to pull tight. This pulls the lens into a thinner, less curved shape, refracting light less strongly.

B (Ciliary muscles relax and suspensory ligaments tighten) [1]

Crossing two heterozygous parents (\(Pp \times Pp\)) gives genotypes \(PP\), \(Pp\), \(Pp\), and \(pp\) in a \(1:2:1\) ratio. Only homozygous recessive (\(pp\)) plants produce white flowers, which is \(1\) out of \(4\) or \(25\%\).

B (\(25\%\)) [1]

Abiotic factors are non-living physical and chemical components of the environment, such as light intensity, temperature, and pH. Predators, pathogens, and prey are biotic (living) factors.

A (Light intensity) [1]

Amylase is a carbohydrase enzyme that catalyses the breakdown of starch into simpler sugars like maltose.

D [1] - 1 mark for the correct letter.

Transpiration occurs most rapidly in warm, windy, and dry (low humidity) conditions because these factors maintain a steep water vapour concentration gradient between the inside of the leaf and the surrounding air.

A [1] - 1 mark for the correct letter.

When blood glucose levels are too low, the pancreas secretes glucagon, which stimulates the breakdown of glycogen into glucose in the liver, raising blood glucose back to normal.

B [1] - 1 mark for the correct letter.

1. Calculate the proportionality constant \(k = \text{light intensity} \times d^2 = 25 \times (0.20)^2 = 25 \times 0.04 = 1.0\).
2. Use the formula for the new distance: \(\text{New light intensity} = \frac{k}{d^2} = \frac{1.0}{(0.50)^2} = \frac{1.0}{0.25} = 4\text{ a.u.}\)

Calculation of constant \(k = 25 \times 0.20^2 = 1.0\) OR correct ratio setup \(\frac{I_2}{25} = \frac{0.20^2}{0.50^2}\) [1]
Division by \(0.50^2\) or \(0.25\) [1]
Correct final answer of 4 (a.u.) [1]

1. Total surface area of cube = \(6 \times (3\text{ cm} \times 3\text{ cm}) = 6 \times 9 = 54\text{ cm}^2\).
2. Volume of cube = \(3\text{ cm} \times 3\text{ cm} \times 3\text{ cm} = 27\text{ cm}^3\).
3. Surface area to volume ratio = \(54 : 27 = 2 : 1\).

Correct calculation of surface area (\(54\text{ cm}^2\)) and volume (\(27\text{ cm}^3\)) [1]
Correct simplified ratio given as \(2 : 1\) (or \(n = 2\)) [1]

When blood glucose drops, the pancreas secretes glucagon into the bloodstream. Glucagon targets liver (and muscle) cells, stimulating the conversion of stored glycogen into glucose. This glucose enters the blood, raising blood glucose levels back to normal.

(Pancreas releases) glucagon which travels to the liver / target cells [1]
Causes (stored) glycogen to be converted/broken down into glucose (and released into the blood) [1]

Gametes for parent 1: F, f
Gametes for parent 2: F, f
Offspring genotypes: FF, Ff, Ff, ff
Phenotypes: 3 unaffected (FF, Ff, Ff) : 1 affected with cystic fibrosis (ff)
Probability of cystic fibrosis (ff) = 1/4 = 25% = 0.25.

Correct gametes shown for both parents (F and f) [1]
Correct genotypes of offspring shown in grid/diagram (FF, Ff, Ff, ff) [1]
Correct probability of having cystic fibrosis stated as 1/4 / 25% / 0.25 / 1 in 4 [1]

An electrical impulse triggers the release of neurotransmitter chemical molecules from the presynaptic neurone. The neurotransmitters diffuse across the synapse (gap) down a concentration gradient. They bind to complementary receptor proteins on the membrane of the postsynaptic neurone, initiating a new electrical impulse.

Chemical/neurotransmitter released (from first neurone/presynaptic knob) [1]
(Neurotransmitter) diffuses across the synapse / gap [1]
Binds to receptors on the second neurone / triggers a new electrical impulse in the next neurone [1]

Estimated population = \(\frac{\text{number in 1st sample} \times \text{total in 2nd sample}}{\text{number marked in 2nd sample}} = \frac{40 \times 30}{8} = \frac{1200}{8} = 150\).

Correct substitution into formula: \(\frac{40 \times 30}{8}\) [1]
Correct final answer = 150 [1]

1. Convert the measured image width into micrometres or find the actual size in mm first:
\[\text{Actual size (mm)} = \frac{\text{Image size}}{\text{Magnification}} = \frac{18\text{ mm}}{600} = 0.03\text{ mm}\]
2. Convert to micrometres:
\[0.03\text{ mm} \times 1000 = 30\ \mu\text{m}\]
(Alternatively: \(18\text{ mm} = 18000\ \mu\text{m}\); \(\frac{18000}{600} = 30\ \mu\text{m}\)).

• Method mark: \(18 \div 600\) OR \(18000 \div 600\) [1]
• Accuracy mark: \(30\) (\(\mu\text{m}\)) [1]

ALLOW award 2 marks for correct answer on answer line without working.
ALLOW 1 mark for \(0.03\) (if unit conversion not done).

(a) \(\text{RQ} = \frac{36}{45} = 0.8\)
(b) An RQ of 0.8 is between 0.7 (lipid) and 1.0 (carbohydrate), which indicates that the seeds are not respiring pure carbohydrate alone, but a mixture of substrates (or proteins).

(a) • \(0.8\) [1]

(b) Any two from:
• Value is less than 1.0 / greater than 0.7 [1]
• Indicates the seeds are not using only glucose / carbohydrate [1]
• Suggests respiration of a mixture of lipids and carbohydrates / respiration of protein [1]

Reflex arcs do not involve conscious thought from the brain. Because electrical impulses travel across fewer synapses through the spinal cord/relay neurones directly to effectors, the response is automatic and rapid. This rapid removal of the foot prevents or minimises tissue injury.

• The response is faster / automatic / does not require conscious thought / has fewer synapses [1]
• Minimises damage / protects the body / prevents further injury [1]

DO NOT ALLOW 'avoids pain' alone without reference to speed/preventing damage.

When blood water concentration decreases, osmoreceptors detect this change and stimulate the pituitary gland to release more ADH into the bloodstream. ADH increases the permeability of the kidney collecting ducts, so more water is reabsorbed back into the blood. This results in a smaller volume of more concentrated urine.

• (Pituitary gland) releases more ADH / increases ADH secretion [1]
• Kidney tubules / collecting ducts become more permeable / reabsorb more water into the blood [1]
• Urine produced has a smaller volume AND is more concentrated / darker [1]

1. The dark beetles have a selective advantage because they are better camouflaged against the dark lava rocks from predators.
2. Consequently, they are more likely to survive and reproduce (survival of the fittest).
3. They pass on the beneficial allele/gene for dark colour to their offspring, increasing the frequency of the dark allele over generations.

• Dark beetles are better camouflaged / less visible to predators (on dark rock) [1]
• They are more likely to survive and reproduce / breed [1]
• They pass on the beneficial allele / gene for dark colour to their offspring [1]

IGNORE 'pass on the characteristic' unqualified (must mention gene/allele).

1. Use the percentage efficiency formula:
\[\text{Percentage efficiency} = \left(\frac{\text{Energy stored in new biomass}}{\text{Total energy consumed}}\right) \times 100\]
2. Substitute values:
\[\text{Percentage efficiency} = \left(\frac{9.6 \times 10^4}{8.0 \times 10^5}\right) \times 100 = \left(\frac{96000}{800000}\right) \times 100 = 0.12 \times 100 = 12\%\]

• Method mark: \(\frac{9.6 \times 10^4}{8.0 \times 10^5} \times 100\) OR \(\frac{96000}{800000} \times 100\) [1]
• Accuracy mark: \(12\) (%) [1]

ALLOW 2 marks for correct answer \(12\) without working shown.

First convert image size to micrometres: \(18\text{ mm} = 18 \times 1000 = 18\,000\ \mu\text{m}\).

Next, use the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{18\,000}{3} = 6000\).

1 mark for unit conversion: \(18\text{ mm} = 18\,000\ \mu\text{m}\) OR \(3\ \mu\text{m} = 0.003\text{ mm}\).
1 mark for correct final answer: \(\times 6000\) (ALLOW \(6000\)).

Substitute \(d = 0.2\) into the inverse square law equation:
\(d^2 = 0.2^2 = 0.04\)
\(\text{Light intensity} = \frac{1}{0.04} = 25\text{ (arbitrary units)}\).

1 mark for correct calculation of \(d^2 = 0.04\) or showing \(\frac{1}{0.2^2}\).
1 mark for correct calculation of \(25\).

Under high pressure pumped by the ventricles, elastic fibres stretch to accommodate blood surges without rupturing and recoil between heartbeats to maintain blood pressure and keep blood flowing smoothly. Muscular tissue provides wall strength and contracts/relaxes to alter lumen diameter.

1 mark for: Elastic fibres stretch (under high pressure/during systole) AND recoil (to push blood/maintain pressure).
1 mark for: Thick muscular wall prevents bursting / withstands high pressure / contracts to control lumen size.

The pancreas detects high blood glucose concentrations in the blood and responds by secreting the hormone insulin, which causes liver and muscle cells to take up glucose and convert it into glycogen.

1 mark for naming the pancreas (ALLOW beta cells / islets of Langerhans).
1 mark for naming insulin.

Crossing two heterozygous individuals (\(Cc \times Cc\)) produces genotypes:
- \(1/4\ CC\) (unaffected homozygous dominant)
- \(2/4\ Cc\) (unaffected carrier)
- \(1/4\ cc\) (affected homozygous recessive)

The probability of an affected child is \(1/4 = 0.25 = 25\%\).

1 mark for identifying the affected genotype as \(cc\) or working showing \(1\text{ in }4\) / \(\frac{1}{4}\) / \(0.25\).
1 mark for \(25\%\) (ALLOW \(0.25\) or \(1/4\) if units omitted).

Efficiency is calculated using the formula:
\[\text{Efficiency} = \left(\frac{\text{Energy transferred to primary consumer}}{\text{Total energy in producer}}\right) \times 100\]
\[\text{Efficiency} = \left(\frac{5760}{48000}\right) \times 100 = 0.12 \times 100 = 12\%\]

1 mark for setting up the ratio: \(\frac{5760}{48000}\).
1 mark for multiplying by 100: \(0.12 \times 100\).
1 mark for correct final answer: \(12\%\) (award 3 marks for correct answer without working).

1. Calculate total magnification: \(\text{Total magnification} = 10 \times 40 = 400\).
2. Use the formula: \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} = \frac{36\text{ mm}}{400} = 0.09\text{ mm}\).
3. Convert mm to \(\mu\text{m}\): \(0.09 \times 1000 = 90\ \mu\text{m}\).

• Total magnification = 400 [1]
• \(\frac{36}{400}\) or 0.09 (mm) OR converting 36 mm to 36000 \(\mu\text{m}\) [1]
• 90 (\(\mu\text{m}\)) [1]

ALLOW 90 with no working shown for [3] marks.

1. Calculate cardiac output in \(\text{cm}^3\text{/min}\): \(145 \times 80 = 11\,600\text{ cm}^3\text{/min}\).
2. Convert \(\text{cm}^3\) to \(\text{dm}^3\): \(\frac{11\,600}{1000} = 11.6\text{ dm}^3\text{/min}\).

• \(145 \times 80\) OR 11 600 [1]
• 11.6 (\(\text{dm}^3\text{/min}\)) [1]

ALLOW 11.6 with no working shown for [2] marks.

1. Calculate \(d^2\): \(0.50^2 = 0.25\).
2. Calculate light intensity: \(\frac{1}{0.25} = 4\).
3. Give the answer to 2 significant figures: \(4.0\).

• \(0.50^2 = 0.25\) OR \(\frac{1}{0.25}\) seen in working [1]
• 4.0 [1]

ALLOW 4 with no working shown for [1] mark (fails significant figures).
ALLOW 4.0 with no working shown for [2] marks.

1. Calculate total surface area: \(2 \times (3 \times 2 + 3 \times 2 + 2 \times 2) = 2 \times (6 + 6 + 4) = 2 \times 16 = 32\text{ cm}^2\).
2. Calculate volume: \(3 \times 2 \times 2 = 12\text{ cm}^3\).
3. Calculate the ratio: \(\frac{32}{12} \approx 2.67\).
4. Written as \(X : 1\): \(2.67 : 1\) (or \(2.7 : 1\) / \(8/3 : 1\)).

• Surface area = 32 (\(\text{cm}^2\)) [1]
• Volume = 12 (\(\text{cm}^3\)) [1]
• 2.67 : 1 (ALLOW 2.7 : 1 or 8/3 : 1) [1]

ALLOW ECF from incorrect surface area or volume for final ratio mark.

1. Formula: \(\text{Efficiency} = \frac{\text{Energy in blue tits}}{\text{Energy in caterpillars}} \times 100\).
2. Calculation: \(\frac{648}{5400} \times 100 = 0.12 \times 100 = 12\%\).

• \(\frac{648}{5400} \times 100\) OR \(0.12\) seen in working [1]
• 12 (%) [1]

ALLOW 12 without working for [2] marks.

1. Substitute the values into the formula: \(\text{Population} = \frac{48 \times 60}{16}\).
2. Calculate numerator: \(48 \times 60 = 2880\).
3. Divide by 16: \(\frac{2880}{16} = 180\).

• Correct substitution: \(\frac{48 \times 60}{16}\) [1]
• 2880 seen in calculation [1]
• 180 [1]

ALLOW 180 with no working shown for [3] marks.

1. Reason for LED lamp: Traditional filament bulbs emit significant amounts of heat, which would warm up the water and introduce an uncontrolled variable (temperature). LED lamps produce minimal heat, ensuring temperature remains constant so only light intensity affects the photosynthetic rate.
2. Controlled variables (any two):
- Concentration of sodium hydrogen carbonate / dissolved carbon dioxide
- Mass / piece / length of pondweed
- Volume of water in the beaker
- Colour / wavelength of the light emitted

1 mark:
- LED lamps release less heat / do not alter the temperature of the water (so temperature does not act as an uncontrolled/confounding variable) ✓
(ALLOW filament bulbs heat up the water / affect enzyme activity)
(IGNORE just 'LED is brighter / saves energy')

2 marks (1 mark for each valid control variable, max 2):
- Concentration of sodium hydrogen carbonate / dissolved $\text{CO}_2$ ✓
- Length / mass / species / number of leaves of pondweed ✓
- Volume of water / solution in the container ✓
- Colour / wavelength of the light source ✓
(DO NOT ALLOW temperature as a controlled variable if already used to explain the LED bulb)
(IGNORE time / distance of lamp)

1. Preparation before reweighing: The student should blot/dab the cylinders dry with a paper towel to remove excess liquid/solution clinging to the surface. This ensures only internal mass changes due to osmosis are recorded.
2. Reason for percentage change:
- The initial/starting masses of the potato cylinders are not identical.
- Calculating percentage change normalises the data, allowing a fair/proportional comparison across different initial masses.

Mark 1:
- Gently blot / dry the surface of the potato cylinders with a paper towel (to remove excess surface liquid/solution) ✓
(DO NOT ALLOW squeeze the potato cylinder)
(ALLOW remove excess water on the outside)

Mark 2:
- Initial / starting masses of the cylinders vary / are not identical ✓

Mark 3:
- Percentage change allows a proportional / fair comparison between cylinders of different initial masses ✓
(ALLOW compares relative gain/loss of mass)

1. Why a transect is needed: The distribution of dandelions is changing systematically along an environmental gradient (light intensity/distance from trees). A transect allows students to observe continuous/gradual changes across distance, whereas random sampling only shows overall averages and misses the spatial gradient.
2. How to set up the belt transect:
- Lay down a tape measure/transect line along the environmental gradient (from under the trees out into the open field).
- Place quadrats at regular/fixed intervals (e.g., every 1 m or 2 m) along the line and count the number/abundance of dandelions within each quadrat (or measure light intensity at each point).

Mark 1 (Why transect is suitable):
- (Belt) transect allows sampling across an environmental gradient / systematic change (e.g. from shade to light / distance from trees) ✓
(ALLOW random sampling would not show how distribution changes across a distance / along a gradient)

Mark 2 (Setup - line):
- Lay a tape measure / transect line from under the canopy / shade into the open sunny field ✓

Mark 3 (Setup - quadrats):
- Place quadrats at regular / standard intervals along the tape line (and count the dandelions / measure the abiotic factor at each point) ✓
(ALLOW placing quadrats continuously along the line)
(IGNORE throwing quadrats randomly)

Detection:
- Temperature receptors in the thermoregulatory centre (in the hypothalamus of the brain) monitor the temperature of the blood flowing through it.
- Temperature receptors in the skin detect external temperature changes and send electrical impulses along sensory neurones to the brain.

Physiological responses to reduce temperature:
1. Vasodilation: Arterioles supplying capillaries near the surface of the skin dilate (widen). This increases blood flow through surface capillaries, allowing more thermal energy / heat to radiate away from the body into the surroundings.
2. Sweating: Sweat glands secrete sweat onto the surface of the skin. As the water in sweat evaporates, it absorbs latent heat energy from the body, producing a cooling effect.
3. Hairs flatten: Erector muscles in the skin relax, causing hairs to lie flat against the skin surface so that no insulating layer of still air is trapped.

Negative Feedback Mechanism:
- The deviation of body temperature above the set point (\(37\,^\circ\text{C}\)) triggers these cooling mechanisms.
- As heat is lost and body temperature returns toward the optimum set point, the thermoregulatory centre detects this reduction and decreases the cooling responses, maintaining a constant internal body temperature.

Level 3 (5–6 marks):
- Describes accurately how the temperature increase is detected (hypothalamus / blood temperature AND skin receptors)
- AND provides a detailed physiological explanation of both vasodilation and sweating (and/or flattening of hairs)
- AND clearly explains how these mechanisms operate within a negative feedback loop to restore normal body temperature (\(37\,^\circ\text{C}\)).
There is a well-developed line of reasoning which is clear, logically structured and scientifically accurate.

Level 2 (3–4 marks):
- Describes how the temperature is detected (hypothalamus or skin receptors)
- AND gives a clear description of at least two cooling mechanisms (e.g. vasodilation and sweating) OR explains one mechanism in detail and links it to negative feedback / set point.
There is a line of reasoning presented with some structure and supporting scientific evidence.

Level 1 (1–2 marks):
- Identifies at least one mechanism used to cool the body down (e.g. sweating / blood vessels widen) OR mentions temperature detection by the brain/skin.
Information is fragmented but contains relevant basic points.

0 marks:
- No response or no response worthy of credit.

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Indicative Scientific Content:
AO1 Knowledge & Understanding (Detection & Negative Feedback):
- Thermoregulatory centre / hypothalamus in the brain detects blood temperature.
- Receptors in skin detect ambient / surface temperature and send nerve impulses to the brain.
- Negative feedback works to counteract an increase away from the set point (\(37\,^\circ\text{C}\)) and restore balance.

AO2 Application to Thermoregulation mechanisms:
- Vasodilation: Arterioles near skin surface dilate / widen; blood flow to capillaries increases; more heat lost via radiation/convection.
- Sweating: Sweat glands produce sweat / water; evaporation of water removes thermal energy / cools the skin.
- Erector muscles: Relax so hairs lie flat; reduces trapped insulating air layer.