2025 DSE 生物 答案詳解與評分準則

(a) Polysaccharides alone cannot be presented to T cells, while the carrier protein is processed and presented to activate helper T cells. (b) Memory B and T cells quickly recognize the pathogen, proliferating and differentiating into plasma cells to release a massive amount of antibodies rapidly.

(a) Glucose: Active transport; Fructose: Facilitated diffusion. (b) Less active transport reduces ATP consumption, increasing the ATP/ADP ratio, which feedback-inhibits cellular respiration and lowers oxygen consumption.

(a) Excess sucrose ensures rapid fungal growth and maintains a high rate of glycolysis to produce precursors for citric acid. (b) Without iron, aconitase is inactive, blocking the conversion of citric acid to isocitric acid, causing citric acid to accumulate.

(a) Large volume of dilute urine. Lack of ADH reduces collecting duct permeability to water, decreasing water reabsorption. (b) Desmopressin (an ADH analogue) binds to collecting duct receptors, increasing water reabsorption.

(a) ABA triggers potassium efflux from guard cells, raising their water potential so water leaves by osmosis, causing them to become flaccid and close. (b) Closed stomata limit carbon dioxide entry, reducing photosynthesis and lowering organic matter accumulation below respiratory loss.

(a) Independent variable: Surface area-to-volume ratio (or size of the agar cube). Dependent variable: The rate of diffusion. (b) Hydrochloric acid diffuses into the agar cube. As HCl is acidic, it neutralizes the alkaline sodium hydroxide in the agar, causing the pH indicator phenolphthalein to turn from pink to colorless. (c) (i) The rate of diffusion depends on intrinsic factors like temperature and concentration gradient of HCl, which are identical for all cubes. Thus, the rate of penetration is constant regardless of cube size. (ii) Side length of 1 cm cube = 10 mm. Original volume = \(10 \times 10 \times 10 = 1000\text{ mm}^3\). Depth of penetration = 2 mm on each side. Side length of the remaining pink region = \(10 - (2 \times 2) = 6\text{ mm}\). Volume of pink region = \(6 \times 6 \times 6 = 216\text{ mm}^3\). Percentage remaining pink = \(\frac{216}{1000} \times 100\% = 21.6\%\). (iii) As size increases, the surface area-to-volume ratio decreases. Simple diffusion is too slow to transport substances to and from the cells deep inside a large multicellular organism.

(a) Precaution 1: Cut the stem of the leafy shoot under water. Explanation: This prevents air bubbles from entering the xylem vessels, which would block water transport. Precaution 2: Seal all joints of the apparatus with petroleum jelly. Explanation: This ensures the apparatus is completely airtight so that water uptake is solely due to transpiration from the leafy shoot. (b) (i) In still air, water vapor accumulates around the stomata, forming a humid boundary layer, which decreases the water vapor concentration gradient between the air inside the leaf and the surrounding air, resulting in a slower transpiration rate. Under windy conditions, the wind blows away the accumulated water vapor around the leaf surface, maintaining a steep water vapor concentration gradient, which increases the transpiration rate and water uptake. (ii) A small portion of the water taken up by the leafy shoot is used in photosynthesis or to maintain cell turgidity within the plant cells, rather than being transpired into the atmosphere.

(a) Lay a line transect from the low tide mark to the high tide mark, perpendicular to the shoreline. Place quadrats at regular intervals along the transect line. Record the abundance of Species A and Species B within each quadrat. (b) (i) Abiotic factor: Desiccation/exposure to air. Near the high tide mark, organisms are exposed to air for longer periods. Since Species A is highly susceptible to desiccation, it cannot survive high water loss and is restricted to the lower shore. (ii) Although Species A is a stronger competitor, it cannot tolerate the dry conditions of the upper shore. Since Species B is tolerant to desiccation, it can colonize the upper shore where competition from Species A is absent/minimal. (c) Limitation: Mobile animals may move out of the quadrat before or during counting, leading to inaccurate counting. Alternative method: Mark-release-recapture method.

(a) To exclude oxygen from entering the yeast suspension, ensuring anaerobic conditions for fermentation. (b) (i) Glucose is the substrate for respiration. A higher concentration provides more respiratory substrate, increasing the rate of glycolysis and fermentation, leading to more carbon dioxide gas production. (ii) At 25%, the high solute concentration creates a hypertonic environment. Water leaves the yeast cells by osmosis, causing dehydration which inhibits metabolic enzymes and slows respiration. (c) Set up an identical apparatus but use boiled and cooled yeast suspension (where yeast cells are killed) with the same glucose concentrations.

(a) After the first injection, the antibody level increases slowly and reaches a low peak. After the second injection, the antibody level increases rapidly, reaches a much higher peak, and remains high for longer. This is because the first injection triggers a primary response, producing memory B cells. Upon the second injection, the memory B cells quickly recognize the antigen, rapidly proliferate and differentiate into plasma cells, producing a large amount of antibodies quickly. (b) On Day 35, the antibody concentration against Antigen Y in Group 2 is much lower than that against Antigen X in Group 1. This is because Group 2 mice are exposed to Antigen Y for the first time on Day 28, triggering a primary immune response, which has a longer lag phase. Group 1 mice are exposed to Antigen X for the second time, triggering a strong secondary response. (c) These factors act as control variables. Age, gender, and genetic background can affect the baseline immune capability, ensuring that differences in antibody level are solely due to the antigens injected.

Detailed comparative essay comparing natural selection and artificial selection across four main aspects: driving forces, criteria for selection, genetic diversity, and long-term survival.