Worked Examples for AQA A-Level Biology | AQA 生物典型例题详解

📚 Worked Examples for AQA A-Level Biology | AQA 生物典型例题详解

This article unpacks a series of model questions that mirror the style and demand of AQA A-Level Biology papers. Each worked example breaks down the required knowledge, common pitfalls, and step‑by‑step logic needed to secure full marks. Topics span biological molecules, cells, exchange and transport, genetics, energy transfers, and ecology.

本文详解一系列贴近 AQA A-Level 生物考试的典型例题。每道例题不仅梳理核心考点与常见失分点，更通过逐步推演展示满分答题思路。选题覆盖生物分子、细胞、交换与运输、遗传、能量转换及生态学等关键领域。

1. Benedict’s Test & Colorimetry for Reducing Sugars | 本尼迪克特测试与比色法定量还原糖

Example: A student prepares a series of glucose standards and treats each with Benedict’s reagent. After heating and filtering to remove the red precipitate, the absorbance of the filtrate is measured at 590 nm.

一学生配制一系列葡萄糖标准溶液，均用本尼迪克特试剂处理。加热、过滤除去红色沉淀后，在 590 nm 波长下测量滤液吸光度。

Glucose concentration / mmol dm⁻³	0	0.2	0.4	0.6	0.8	1.0	Fruit juice (unknown)
Absorbance	0.00	0.12	0.23	0.35	0.46	0.58	0.41

Step 1: Plot a calibration curve of absorbance against glucose concentration and draw a line of best fit. The curve should be a straight line passing through the origin because absorbance is directly proportional to the concentration of remaining Cu²⁺ ions, which decreases as more reducing sugar is present.

绘制吸光度对葡萄糖浓度的标准曲线，并作出最佳拟合线。曲线应为通过原点的直线，因为吸光度与剩余 Cu²⁺ 离子浓度成正比，而还原糖越多，Cu²⁺ 离子消耗越多，吸光度越低——本实验通过滤液吸光度间接反映糖浓度，实际标准曲线常呈负相关，但教师可依据试剂方案修正；这里我们采用正向校准。

Step 2: Locate the absorbance of the unknown juice (0.41) on the y‑axis and read across to the curve, then drop vertically to the x‑axis. The reading gives approximately 0.70 mmol dm⁻³. If the juice was diluted, multiply by the dilution factor.

在 y 轴上找到未知果汁吸光度 0.41，横向交于标准曲线，再垂直下引至 x 轴，读出约 0.70 mmol dm⁻³。若样品经过稀释，需乘以稀释倍数还原原始浓度。

Step 3: Always convert to the units requested and comment on reliability: “The result relies on the calibration curve being linear; an outlier standard point should be ignored if it does not lie on the line.”

按要求换算单位，并评价可靠性：“结果依赖于标准曲线的线性；若某标准点明显偏离最佳拟合线，应视为异常值剔除。”

2. Enzyme Kinetics – Initial Rate from a Tangent | 酶动力学 – 由切线求初始反应速率

Example: Catalase activity is followed by measuring the volume of O₂ produced over time. The data are:

过氧化氢酶活性通过记录 O₂ 产生体积随时变化来跟踪。数据如下：

Time / s	0	10	20	30	40	50	60
O₂ volume / cm³	0.0	2.8	5.0	6.6	7.8	8.4	8.6

Step 1: Plot the graph of O₂ volume (y‑axis) against time (x‑axis). To find the initial rate, draw a tangent to the curve at time = 0. The tangent must touch only that point and lean along the steepest part of the early curve.

绘制 O₂ 体积（y 轴）对时间（x 轴）的曲线。为求初始速率，在时间 = 0 处作曲线的切线。切线仅在该点相切，并贴合曲线早期最陡部分。

Step 2: Select two well‑separated points on the tangent, e.g. (0 s, 0 cm³) and (20 s, 5.6 cm³). Calculate the slope:

在切线上取两个相距较远的点，如 (0 s, 0 cm³) 和 (20 s, 5.6 cm³)，计算斜率：

Initial rate = (5.6 cm³ − 0 cm³) ÷ (20 s − 0 s) = 0.28 cm³ s⁻¹

初始速率 = (5.6 cm³ − 0 cm³) ÷ (20 s − 0 s) = 0.28 cm³ s⁻¹

Step 3: State the units and explain why the rate declines later – substrate concentration falls, product may inhibit, or enzyme denaturation if temperature is uncontrolled. AQA often asks for such explanations after rate calculation.

写明单位并解释为何后期速率下降 —— 底物浓度降低、产物反馈抑制，或温度失控导致酶变性。AQA 常在速率计算后要求此类解释。

3. Osmosis and Water Potential of Potato Tissue | 渗透与马铃薯组织水势

Example: Cylinders of potato are placed in sucrose solutions of different concentrations. Their masses are recorded before and after 24 hours.

将马铃薯圆柱体放入不同浓度的蔗糖溶液中，24 小时后记录质量变化。

Sucrose / mol dm⁻³	0.0	0.2	0.4	0.6	0.8	1.0
% change in mass	+14.2	+5.8	−0.1	−6.3	−11.7	−16.0

Step 1: Plot percentage change in mass against sucrose concentration. Draw a line of best fit. The point where the line crosses the x‑axis (zero mass change) indicates the solution that is isotonic to the potato tissue.

绘制质量变化百分比对蔗糖浓度的关系图，作最佳拟合线。线与 x 轴交点（质量无变化）对应的浓度即与马铃薯组织等渗的蔗糖浓度。

Step 2: Read the isotonic concentration – approximately 0.35 mol dm⁻³. At this concentration, the water potential of the sucrose solution equals the water potential of the potato cells. Water potential (Ψ) can be obtained from a reference table: for 0.35 mol dm⁻³ sucrose at 20 °C, Ψ ≈ −860 kPa.

读得等渗浓度约为 0.35 mol dm⁻³。该浓度下蔗糖溶液的水势与马铃薯细胞水势相等。水势 (Ψ) 可查表：20 °C 时 0.35 mol dm⁻³ 蔗糖的 Ψ 约为 −860 kPa。

Step 3: Explain the pattern: in low sucrose concentrations, cells gain water by osmosis and swell, increasing mass. In high concentrations, they lose water, become flaccid, and mass decreases.

解释趋势：低蔗糖浓度时，细胞通过渗透吸水膨胀，质量增加；高浓度时细胞失水质壁分离，质量下降。

4. DNA Sequence, Transcription, Translation and Mutation | DNA 序列、转录、翻译与突变

Example: A portion of the DNA template strand reads: TAC CGA CAT ATC.

一段 DNA 模板链序列为：TAC CGA CAT ATC。

Step 1: Transcribe the template into mRNA. Replace thymine with uracil and write the complementary bases: AUG GCU GUA UAG. (T→A, A→U, C→G, G→C).

将模板链转录为 mRNA。胸腺嘧啶 (T) 替换为尿嘧啶 (U)，写出互补碱基：AUG GCU GUA UAG。

Step 2: Translate the mRNA using a codon chart: AUG = methionine (start), GCU = alanine, GUA = valine, UAG = stop. The polypeptide is Met‑Ala‑Val.

用密码子表翻译 mRNA：AUG = 甲硫氨酸（起始），GCU = 丙氨酸，GUA = 缬氨酸，UAG = 终止密码子。多肽序列为甲硫氨酸‑丙氨酸‑缬氨酸。

Step 3: When the fifth base (C of the second codon) is deleted, the template becomes TAC CAA TAT C… Transcribe: AUG GUU AUA G… Translation: Met‑Val‑Ile… The deletion causes a frameshift, completely altering the amino acid sequence from the mutation point onward and likely producing a non‑functional protein.

当第 5 个碱基（第二个密码子中的 C）缺失时，模板变为 TAC CAA TAT C…。转录后为 AUG GUU AUA G…。翻译为甲硫氨酸‑缬氨酸‑异亮氨酸…。缺失引起移码，从突变点起氨基酸序列全部改变，很可能生成无功能蛋白。

5. Sex‑Linked Inheritance: Haemophilia | 伴性遗传：血友病

Example: Haemophilia is an X‑linked recessive disorder. A woman who is a carrier (XᴴXʰ) and a normal man (XᴴY) have children. Determine the probability of a son being haemophiliac and a daughter being a carrier.

血友病为 X 染色体隐性遗传。一位携带者女性 (XᴴXʰ) 与正常男性 (XᴴY) 生育子女。求儿子患血友病的概率及女儿为携带者的概率。

Step 1: Set up a Punnett square with female gametes Xᴴ and Xʰ, male gametes Xᴴ and Y.

建立庞纳特方格：女性配子 Xᴴ 和 Xʰ，男性配子 Xᴴ 和 Y。

	Xᴴ	Y
Xᴴ	XᴴXᴴ (normal ♀)	XᴴY (normal ♂)
Xʰ	XᴴXʰ (carrier ♀)	XʰY (haemophiliac ♂)

Step 2: Probability for a son: sons are XY; among male offspring, ½ are normal (XᴴY) and ½ are haemophiliac (XʰY). Therefore, the chance a son has haemophilia is 50%.

儿子的概率：儿子性染色体为 XY；在男性后代中，½ 正常 (XᴴY)，½ 患血友病 (XʰY)。因此儿子患病概率为 50%。

Step 3: Daughters: ½ are normal (XᴴXᴴ) and ½ are carriers (XᴴXʰ); no daughter is haemophiliac because they must receive a normal Xᴴ from the father. The probability a daughter is a carrier is 50%.

女儿情况：½ 完全正常 (XᴴXᴴ)，½ 为携带者 (XᴴXʰ)；无女儿患病，因为父亲提供了正常 Xᴴ。女儿为携带者的概率是 50%。

6. Cardiac Cycle – Pressure Changes and Heart Rate | 心动周期 – 压力变化与心率

Example: The graph shows pressure changes in the left atrium, left ventricle and aorta during one cardiac cycle. The cycle lasts 0.80 seconds.

图表显示单次心动周期中左心房、左心室和主动脉的压力变化，周期持续 0.80 秒。

Step 1: Heart rate is calculated as 60 s divided by cycle length. Published by TutorHao | A-Level Biology Revision Series | aleveler.com

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Worked Examples for AQA A-Level Biology | AQA 生物典型例题详解