Pre-U WJEC Biology: Case Study Practice Walkthrough | Pre-U WJEC 生物：案例分析实战演练

📚 Pre-U WJEC Biology: Case Study Practice Walkthrough | Pre-U WJEC 生物：案例分析实战演练

Case studies are a key component of the Pre-U WJEC Biology specification, challenging you to apply knowledge to unfamiliar scenarios. This article provides a step-by-step walkthrough of twelve case studies spanning molecular biology, genetics, physiology and ecology. Each worked example models the analytical thinking and data-handling skills needed for high marks.

案例分析是 Pre-U WJEC 生物考试的核心部分，要求将知识应用到不熟悉的真实情境中。本文精选十二个跨越分子生物学、遗传学、生理学和生态学的案例，逐步演练分析思路与数据处理技巧，帮助你掌握高分答题方法。

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1. Enzyme Kinetics Data Analysis | 酶动力学数据分析

An enzyme was assayed at a fixed concentration, and the initial rate (v) was measured at varying substrate concentrations [S] (μM). The results are shown in the table below.

在固定酶浓度下，测定了不同底物浓度 [S]（μM）时的初始反应速率 v，数据见下表。

[S] / μM	1	2	4	8	16
v / μmol min⁻¹	0.80	1.30	2.20	3.10	4.00

To estimate the kinetic constants V_max and K_m, we construct a Lineweaver–Burk plot. Calculate 1/[S] and 1/v, then fit a straight line. The equation for the double-reciprocal plot is:

为求动力学常数 V_max 和 K_m，构建双倒数图。分别计算 1/[S] 和 1/v，拟合直线。双倒数方程如下：

1/v = (K_m / V_max)(1/[S]) + 1/V_max

Using the transformed data, the y-intercept ≈ 0.25 min μmol⁻¹, giving V_max = 4.0 μmol min⁻¹. The slope ≈ 1.0 min μM μmol⁻¹, so K_m = slope × V_max = 4.0 μM.

转换数据后，y轴截距约为 0.25 min μmol⁻¹，故 V_max = 4.0 μmol min⁻¹。斜率约为 1.0 min μM μmol⁻¹，因此 K_m = 斜率 × V_max = 4.0 μM。

A low K_m indicates high affinity; this enzyme shows moderate affinity. V_max reflects the maximum rate when enzyme is saturated.

K_m 值低表示亲和力高；该酶亲和力中等。V_max 反映酶被饱和时的最大速率。

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2. Pedigree Analysis for a Genetic Disorder | 遗传病系谱分析

A couple (I-1 and I-2) are both unaffected. They have three children: two unaffected daughters and one affected son (II-3). II-3 marries an unaffected woman (II-4), and they have a normal daughter (III-1) and an affected son (III-2). Determine the most likely mode of inheritance and the probability that III-1 is a carrier.

一对夫妇 (I-1 和 I-2) 均正常，育有三个孩子：两个正常女儿和一个患病儿子 (II-3)。II-3 与一名正常女性 (II-4) 结婚，生下正常女儿 (III-1) 和患病儿子 (III-2)。请判断最可能的遗传方式，并计算 III-1 为携带者的概率。

The pedigree suggests autosomal recessive inheritance: unaffected parents can have an affected child, and males and females are equally affected. Let allele ‘A’ be dominant, ‘a’ be recessive. I-1 and I-2 must both be heterozygous (Aa) because they have an affected son (aa).

该系谱符合常染色体隐性遗传：正常父母生出患病子女，且男女均可受累。设显性基因为 A，隐性基因为 a。因有患病儿子 (aa)，I-1 和 I-2 必然均为杂合子 (Aa)。

II-3 is affected (aa). His wife II-4 is unaffected but must be a carrier (Aa) because they have an affected child (III-2) who inherited ‘a’ from both parents. Their daughter III-1 is unaffected. Since her father is aa, he always passes ‘a’; her mother (Aa) passes either A or a. To be unaffected, III-1 must have received A from her mother. Therefore, III-1’s genotype is Aa – she is an obligate carrier. The probability is 100%.

II-3 为患者 (aa)。其妻子 II-4 正常，但由于他们生出了患病儿子 (III-2)，该儿子同时从父母得到 a，因此 II-4 必然是携带者 (Aa)。他们正常的女儿 III-1 从父亲必得 a，从母亲可能得 A 或 a；其表现为正常，则必然从母亲获得 A，故基因型为 Aa，是必然携带者，概率为 100%。

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3. Population Genetics – Hardy-Weinberg Equilibrium | 群体遗传学 – 哈迪-温伯格平衡

In a large, randomly mating population, the frequency of individuals showing a recessive phenotype (aa) is 0.16. Calculate the allele frequencies and the expected frequency of heterozygous carriers. Assume Hardy–Weinberg equilibrium.

在一个随机交配的大群体中，隐性表型 (aa) 的频率为 0.16。计算等位基因频率和预期杂合子频率。假设群体处于哈迪-温伯格平衡。

Let p = frequency of dominant allele A, q = frequency of recessive allele a. Since aa = q² = 0.16, we find q = √0.16 = 0.4. Hence p = 1 – q = 0.6.

设 p 为显性等位基因 A 的频率，q 为隐性等位基因 a 的频率。由于 aa 频率 q² = 0.16，故 q = 0.4，p = 1 – 0.4 = 0.6。

The expected frequency of heterozygotes (Aa) is 2pq = 2 × 0.6 × 0.4 = 0.48. Therefore, 48% of the population are carriers.

预期杂合子 (Aa) 频率为 2pq = 2 × 0.6 × 0.4 = 0.48，即 48% 的人群是携带者。

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4. Water Potential in Plant Tissues | 植物组织的水势

Cylinders of potato tissue were immersed in sucrose solutions of increasing concentration. The percentage change in mass was recorded:

将马铃薯圆柱体浸入不同浓度的蔗糖溶液中，记录质量百分比变化：

Sucrose / mol dm⁻³	0.0	0.2	0.4	0.6	0.8
Mass change / %	+12	+4	–2	–9	–16

The isotonic concentration (where no net mass change occurs) is found by interpolation: between 0.2 M and 0.4 M. A linear fit gives approximately 0.33 M.

等渗浓度（净质量变化为零）通过插值估算：介于 0.2 M 和 0.4 M 之间，线性拟合得约 0.33 M。

The water potential of the isotonic solution equals the tissue water potential. Use ψ = –iCRT, with i = 1 (sucrose does not ionise), C = 0.33 mol dm⁻³ = 330 mol m⁻³, R = 8.31 J mol⁻¹ K⁻¹, T = 293 K. Calculation: ψ = –1 × 330 × 8.31 × 293 ≈ –803 000 Pa = –803 kPa (approx. –0.80 MPa).

等渗溶液的水势等于组织水势。使用 ψ = –iCRT，i = 1（蔗糖不解离），C = 0.33 mol dm⁻³ = 330 mol m⁻³，R = 8.31 J mol⁻¹ K⁻¹，T = 293 K。计算得 ψ ≈ –803 kPa（约 –0.80 MPa）。

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5. Respiratory Quotient (RQ) and Metabolic Substrates | 呼吸商与代谢底物

A respirometer containing germinating seeds recorded oxygen consumption of 0.60 cm³ and carbon dioxide production of 0.72 cm³ over the same time interval. Calculate the RQ and suggest the likely metabolic substrate.

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