2023 Cambridge IGCSE Biology (0610) 模拟试题及答案详解

(a) The anthers are the pollen-producing organs of the stamen.
(b) Petals of insect-pollinated flowers are typically large and brightly colored to provide a clear visual cue and a landing platform for foraging insects.

Part (a): [1 mark]
- Anther (ignore stamen / pollen sac)

Part (b): [1.5 marks]
- Feature: Large / brightly colored (petals) [1 mark]
- Explanation: Acts as a visual attractant / landing platform [0.5 marks]

(a) The palisade mesophyll cells are vertically elongated and packed with chloroplasts to maximize light absorption near the upper surface of the leaf.
(b) Upper epidermal cells are thin and transparent (lacking chloroplasts), allowing light rays to pass freely into the palisade layer below.

Part (a): [1 mark]
- Palisade mesophyll (layer / cells)

Part (b): [1.5 marks]
- Transparent / lack of chloroplasts [1 mark]
- Thin layer / thin cells [0.5 marks]

(a) The pulmonary artery delivers deoxygenated blood from the right ventricle of the heart to the alveolar capillaries.
(b) The wall of each alveolus is composed of a single layer of flattened epithelial cells (one cell thick), minimizing the physical barrier and distance for diffusion.

Part (a): [1 mark]
- Pulmonary artery

Part (b): [1.5 marks]
- Feature: Alveolar wall is one cell thick / thin epithelium [1 mark]
- Explanation: Reduces the diffusion distance / short diffusion pathway [0.5 marks]

(a) The hypothalamus monitors the temperature of blood flowing through the brain and receives sensory input from skin thermoreceptors.
(b) Under cold conditions, smooth muscles in the arteriole walls contract, causing vasoconstriction. This diverts blood away from the surface capillaries to deeper tissues, minimizing heat loss.

Part (a): [1 mark]
- Hypothalamus

Part (b): [1.5 marks]
- Response: Arterioles constrict / vasoconstriction [1 mark]
- Explanation: Less blood flows to the skin surface capillaries / reduces heat loss by radiation [0.5 marks]
- Reject: capillaries constrict

(a) Carbon dioxide gas bubbles are released by yeast, which expand inside the dough during proving and baking.
(b) Yeast respires anaerobically (fermentation) inside the dough where oxygen is limited, producing ethanol alongside carbon dioxide.

Part (a): [1 mark]
- Carbon dioxide / \(CO_2\)

Part (b): [1.5 marks]
- Anaerobic (respiration) / fermentation [1 mark]
- Ethanol / alcohol [0.5 marks]
- Reject: lactic acid

(a) The waxy cuticle acts as a physical barrier preventing evaporation directly from the epidermal cells.
(b) Rolled leaves or sunken stomata trap a microclimate of humid air around the stomatal pores, lowering the water potential gradient between the inside and outside of the leaf, which slows transpiration.

Part (a): [1 mark]
- (Waxy) cuticle

Part (b): [1.5 marks]
- Adaptation: Sunken stomata / rolled leaves / hairs on leaf surface / reduced leaf surface area / spines [1 mark]
- Explanation: Traps water vapor / reduces water potential gradient / reduces wind effect [0.5 marks]

Actual size = Image size / Magnification. Convert the image length to micrometres: \(75\text{ mm} = 75\ 000\ \mu\text{m}\). Divide by magnification: \(75\ 000\ /\ 15\ 000 = 5.0\ \mu\text{m}\).

1. Correct conversion of mm to \(\mu\text{m}\) (75 mm = 75 000 \(\mu\text{m}\)) [1 mark]
2. Correct substitution of values into the magnification formula (\(75\ 000\ /\ 15\ 000\)) [1 mark]
3. Correct final calculation of actual length (5 or 5.0) [1 mark]

First, calculate the volume of water uptake: \(\text{Volume} = \text{distance} \times \text{cross-sectional area} = 45\text{ mm} \times 0.8\text{ mm}^2 = 36\text{ mm}^3\). Next, calculate the rate of water uptake per minute: \(36\text{ mm}^3 / 15\text{ minutes} = 2.4\text{ mm}^3/\text{minute}\).

1. Correct calculation of water volume absorbed (\(45 \times 0.8 = 36\text{ mm}^3\)) [1 mark]
2. Dividing volume by time (\(36 / 15\)) [1 mark]
3. Correct final value of 2.4 (allow ecf) [1 mark]

First, determine the number of doublings in 12 hours: \(12 / 3 = 4\) doublings. Calculate the final population using the exponential growth equation: \(2.0 \times 10^5 \times 2^4 = 2.0 \times 10^5 \times 16 = 3.2 \times 10^6\) cells per \(\text{cm}^3\) (or \(3\ 200\ 000\)).

1. Calculating number of generation doublings as 4 [1 mark]
2. Showing exponential progression or doubling calculation sequence [1 mark]
3. Correct final answer of \(3.2 \times 10^6\) or \(3\ 200\ 000\) [1 mark]

First, calculate the energy transferred to the rabbits: \(1\ 200\ 000\text{ kJ} \times 0.085 = 102\ 000\text{ kJ}\). Next, calculate the energy transferred to the foxes: \(102\ 000\text{ kJ} \times 0.12 = 12\ 240\text{ kJ}\).

1. Calculating energy transferred to rabbits (\(102\ 000\text{ kJ}\)) [1 mark]
2. Method to find \(12\%\) of rabbits' energy (\(\text{rabbit energy} \times 0.12\)) [1 mark]
3. Correct final answer (\(12\ 240\text{ kJ}\) or \(1.224 \times 10^4\text{ kJ}\)) [1 mark]

The pancreas detects the increase in blood glucose levels and secretes insulin. Insulin travels in the blood to the liver and muscle cells, stimulating them to take up more glucose and convert it to glycogen, which reduces blood glucose concentration back to normal.

Award up to 4.5 marks (0.5 marks per point): 1. Rise in blood glucose detected by the pancreas; 2. Pancreas secretes the hormone insulin; 3. Insulin is released into the blood; 4. Insulin travels to the liver and/or muscles; 5. Increases permeability of target cells to glucose / increases glucose uptake; 6. Stimulates the conversion of glucose to glycogen; 7. Glycogen is stored inside the cells; 8. Increases respiration rate of glucose; 9. Blood glucose concentration decreases back to the normal set point.

The contraction of the external intercostal muscles and the diaphragm flattens the dome, increasing thoracic volume and decreasing internal pressure below atmospheric pressure, causing air to rush in.

Award up to 4.5 marks (0.5 marks per point): 1. External intercostal muscles contract; 2. Ribcage moves upwards and outwards; 3. Diaphragm contracts; 4. Diaphragm flattens / dome shape is lost; 5. Volume of the thoracic cavity increases; 6. Pressure inside the thorax/lungs decreases; 7. Pressure drops below atmospheric pressure; 8. Air moves into the lungs down a pressure gradient.

The pollen grain germinates on the stigma, growing a pollen tube down the style to the ovary. The male nuclei travel through this tube, enter the ovule via the micropyle, and fuse with the female nucleus to complete fertilization.

Award up to 4.5 marks (0.5 marks per point): 1. Sugary secretions on the stigma trigger pollen germination; 2. Pollen tube grows down the style; 3. Growth of the pollen tube is controlled by the tube nucleus / digestive enzymes; 4. Male gamete nuclei travel down the pollen tube; 5. Pollen tube enters the ovary; 6. Pollen tube enters the ovule via the micropyle; 7. Male gamete nucleus fuses with the egg cell / female nucleus; 8. Produces a diploid zygote; 9. Fertilization is complete.

Carbon dioxide diffuses through the stomata into the spongy mesophyll air spaces, dissolves in the moisture on the cell walls, and diffuses into the palisade cell cytoplasm and chloroplasts.

Award up to 4.5 marks (0.5 marks per point): 1. Carbon dioxide diffuses through the stomata down a concentration gradient; 2. Guard cells control stomatal opening; 3. Gas moves through intercellular air spaces; 4. Air spaces are in the spongy mesophyll layer; 5. Carbon dioxide dissolves in the moisture film on the cell walls; 6. Dissolved gas diffuses across the cellulose cell wall; 7. Diffuses across the partially permeable cell membrane; 8. Enters the cytoplasm; 9. Enters the chloroplasts.

The human insulin gene is isolated with restriction enzymes and inserted into a cut bacterial plasmid using DNA ligase. The recombinant plasmid is put into bacteria, which are cultured in a fermenter to express and produce insulin.

Award up to 4.5 marks (0.5 marks per point): 1. Human insulin gene is isolated / cut using a restriction enzyme; 2. This leaves single-stranded sticky ends; 3. Bacterial plasmid is cut using the same restriction enzyme; 4. This produces complementary sticky ends; 5. DNA ligase is used to join the gene and plasmid together; 6. Recombinant plasmid is inserted back into a bacterium; 7. Genetically modified bacteria are grown in a fermenter; 8. Under optimum conditions (pH, temperature, nutrients); 9. Bacteria express the gene and produce insulin, which is extracted and purified.

Evaporation from mesophyll cells draws water from the xylem, creating a tension that pulls the continuous column of water up due to cohesion. In dry conditions, guard cells lose turgor and close the stomata, reducing water loss.

Award up to 4.5 marks (0.5 marks per point): 1. Water evaporates from mesophyll cell walls; 2. Water vapour diffuses out of the leaf through stomata; 3. Creates a water potential gradient; 4. Tension / transpiration pull is created; 5. Cohesion between water molecules keeps the column continuous; 6. Adhesion between water molecules and xylem walls supports the column; 7. In dry conditions, guard cells lose turgor / become flaccid; 8. Stomata close; 9. This minimizes transpiration / conserves water.

Water enters the red blood cells by osmosis down a water potential gradient across the partially permeable membrane. The cell swells and, because it lacks a cell wall to resist the pressure, it bursts (lysis).

Award up to 4.5 marks (0.5 marks per point): 1. Solution has a higher water potential than the cytoplasm / is hypotonic; 2. Water moves into the red blood cells; 3. Movement occurs by osmosis; 4. Across a partially permeable cell membrane; 5. Down a water potential gradient; 6. Red blood cells swell / increase in volume; 7. Internal pressure increases; 8. Animal cells do not have a cell wall; 9. Cells burst / undergo haemolysis / lysis.

Anaerobic respiration occurs when oxygen supply is insufficient, producing lactic acid from glucose. Post-exercise, elevated breathing and heart rates supply the oxygen needed by the liver to break down lactic acid into carbon dioxide and water, or convert it to glucose.

Award up to 4.5 marks (0.5 marks per point): 1. Vigorous exercise leads to insufficient oxygen supply in muscles; 2. Muscle cells respire anaerobically; 3. Glucose is incompletely broken down; 4. Lactic acid is produced; 5. Lactic acid builds up in muscles / blood; 6. After exercise, high heart rate and deep breathing are maintained; 7. Lactic acid is transported in the blood to the liver; 8. Lactic acid is broken down aerobically using oxygen; 9. Converted to carbon dioxide and water, or glucose / glycogen.

As the concentration of carbon dioxide in the inspired air increases from 0% to 1.0%, the breathing rate remains constant at approximately 12 breaths per minute. Above 1.0% carbon dioxide, the breathing rate increases rapidly and linearly up to 34 breaths per minute at 6.0% carbon dioxide.

This rapid increase occurs because an elevated concentration of carbon dioxide in the inspired air prevents efficient diffusion of carbon dioxide out of the blood in the alveoli. This leads to an accumulation of carbon dioxide in the blood, which dissolves to form carbonic acid, lowering the blood pH (making it more acidic). Specialized receptors in the body detect this change in pH and send nerve impulses to the brain. The brain then transmits more frequent electrical impulses via motor neurones to the diaphragm and the external intercostal muscles, stimulating them to contract more rapidly, thus increasing the breathing rate to expel the excess carbon dioxide.

Maximum 4.25 marks:
- Description: breathing rate is constant (at 12 breaths per minute) between 0% and 1.0% carbon dioxide [1]
- Description: breathing rate increases rapidly/linearly above 1.0% (reaching 34 breaths per minute at 6.0%) [1]
- Explanation: higher carbon dioxide in inspired air leads to increased carbon dioxide in the blood / lower blood pH / more acidic blood [1]
- Explanation: change in blood pH / carbon dioxide is detected by the brain (or receptors) [1]
- Explanation: brain sends more frequent nerve impulses to the intercostal muscles / diaphragm [1]

During the first 60 hours, as the nutrient concentration decreases from 100% to approximately 15%, the biomass of the fungus increases from 0 to its maximum of 80 g/dm³. This is because the fungus actively absorbs and assimilates the nutrients for primary growth, cell division, and cellular respiration.

After 60 hours, the rate of penicillin production increases rapidly. This is because penicillin is a secondary metabolite. It is not required for the normal growth of the fungus but is produced under stress conditions, such as when nutrients become severely depleted or limiting. The high biomass of the fungus achieved by 60 hours ensures there is a large population of cells capable of synthesizing and secreting penicillin into the fermenter medium once triggered by the low nutrient levels.

Maximum 4.25 marks:
- Description: as nutrient concentration decreases, biomass increases [1]
- Explanation: nutrients are used by the fungus for growth / cell division / respiration [1]
- Description: penicillin production rate increases rapidly after 60 hours (when nutrients are low / depleted) [1]
- Explanation: penicillin is a secondary metabolite / produced under stress or nutrient depletion [1]
- Explanation: a high biomass of fungal cells is present after 60 hours to produce the penicillin [1]

At low light intensities (from 0 to 150 arbitrary units), temperature has very little or no effect on the rate of photosynthesis, and the curves for both \(15^\circ\text{C}\) and \(25^\circ\text{C}\) overlap. This is because light intensity is the limiting factor in this range; there is not enough light energy to drive the light-dependent reactions of photosynthesis.

At high light intensities (above 200 arbitrary units), the rate of photosynthesis at \(15^\circ\text{C}\) plateaus at a much lower level than at \(25^\circ\text{C}\). Here, light is no longer the limiting factor, and temperature becomes the limiting factor. Photosynthesis is an enzyme-controlled process. Raising the temperature to \(25^\circ\text{C}\) provides more kinetic energy to the enzymes (such as Rubisco) and substrate molecules, increasing the frequency of successful collisions and the rate of reaction.

Maximum 4.25 marks:
- Description: at low light intensities, temperature has no/little effect on the rate of photosynthesis [1]
- Explanation: light intensity is the limiting factor at low light levels [1]
- Description: at high light intensities, the rate of photosynthesis is higher at \(25^\circ\text{C}\) than at \(15^\circ\text{C}\) [1]
- Explanation: temperature is the limiting factor at high light levels [1]
- Explanation: higher temperature increases kinetic energy of molecules / frequency of successful enzyme-substrate collisions [1]

As ambient temperature increases from \(5^\circ\text{C}\) to \(20^\circ\text{C}\), the sweat production rate remains constant and very low (near 0). Above \(20^\circ\text{C}\), the rate of sweat production increases rapidly and linearly up to the maximum measured temperature.

When the internal body temperature or external temperature rises, thermoreceptors detect this change and send electrical signals to the hypothalamus in the brain. The brain coordinates a response, stimulating sweat glands in the skin to secrete sweat onto the surface of the epidermis. Sweat consists mostly of water. As this water evaporates, it requires a significant amount of heat energy (latent heat of vaporisation), which is absorbed from the skin and the blood flowing through capillaries close to the skin surface. This transfer of heat energy away from the body cools the blood, helping to lower the internal temperature back to the normal set point of \(37^\circ\text{C}\).

Maximum 4.25 marks:
- Description: sweat production is constant/low up to \(20^\circ\text{C}\) [1]
- Description: sweat production increases rapidly/linearly above \(20^\circ\text{C}\) [1]
- Explanation: sweat is secreted by sweat glands in the skin [1]
- Explanation: water in sweat evaporates from the skin surface [1]
- Explanation: evaporation removes latent heat / heat energy from the body (cooling the skin/blood) [1]