📚 Sickle Cell Anaemia and Malaria: A Case Study Practical Walkthrough | 镰刀型细胞贫血与疟疾案例分析实战演练
Case studies form a vital part of Year 13 CAIE Biology, requiring you to integrate genetics, evolution, and data handling. This walkthrough uses the classic example of sickle cell anaemia and malaria to guide you through every step of decoding an exam-style scenario, from calculating allele frequencies to evaluating natural selection.
案例分析是 Year 13 CAIE 生物的重要组成部分，需要你将遗传学、进化和数据处理结合起来。本次实战演练使用镰刀型细胞贫血与疟疾这个经典例子，一步步带你解读考试情境，从计算等位基因频率到评估自然选择。
1. Case Study Scenario | 案例场景设置
A research team studied 1000 residents of a West African village where malaria is endemic. Blood samples revealed three genotypes for the beta-globin gene: HbA HbA (normal haemoglobin), HbA HbS (carrier), and HbS HbS (sickle cell disease). The observed numbers were 700 HbA HbA, 270 HbA HbS, and 30 HbS HbS individuals.
某研究团队调查了一个疟疾流行的西非村庄中的 1000 名居民。血液样本显示 β-珠蛋白基因有三种基因型：HbA HbA（正常血红蛋白）、HbA HbS（携带者）和 HbS HbS（镰刀型细胞病）。观察到的个体数量为 700 名 HbA HbA、270 名 HbA HbS 和 30 名 HbS HbS。
Your task is to determine whether this population is in Hardy–Weinberg equilibrium and to interpret the selective forces maintaining the sickle cell allele.
你的任务是判断该群体是否符合哈代-温伯格平衡，并解释维持镰刀型等位基因的选择力量。
2. Genetic Basis of Sickle Cell Anaemia | 镰刀型细胞贫血的遗传基础
The sickle cell trait is caused by a single nucleotide substitution in the beta-globin gene, changing GAG to GTG and replacing glutamic acid with valine. The allele HbS produces abnormal haemoglobin that polymerises under low oxygen, distorting red blood cells.
镰刀型细胞性状由 β-珠蛋白基因中的一个单核苷酸替换引起，将 GAG 变为 GTG，使谷氨酸被缬氨酸取代。等位基因 HbS 产生异常血红蛋白，在低氧条件下聚合，使红细胞变形。
The inheritance pattern is autosomal codominant. HbA HbA individuals have normal round red cells, HbA HbS carriers show some sickling only at very low oxygen, and HbS HbS homozygotes suffer severe haemolytic anaemia.
该遗传模式为常染色体共显性。HbA HbA 个体拥有正常的圆盘状红细胞，HbA HbS 携带者仅在极低氧时出现部分镰变，而 HbS HbS 纯合子则患有严重的溶血性贫血。
Despite the severe disease, the HbS allele is maintained at frequencies as high as 0.1–0.2 in malarial regions due to heterozygote advantage.
尽管疾病严重，由于杂合子优势，HbS 等位基因在疟疾区的频率仍可高达 0.1 至 0.2。
3. Understanding Heterozygote Advantage | 理解杂合子优势
Heterozygote advantage, also called overdominance, occurs when the heterozygous genotype has a higher fitness than either homozygote. In malaria-endemic zones, HbA HbS individuals are protected against severe falciparum malaria because the parasite cannot complete its life cycle efficiently in sickled cells or in cells undergoing accelerated sickling removal.
杂合子优势，也称超显性，指杂合基因型的适合度高于任一纯合子。在疟疾流行区，HbA HbS 个体能抵抗重症恶性疟疾，因为疟原虫无法在镰变细胞或被加速清除的细胞中有效完成生活史。
The relative fitness values can be approximated as: HbA HbA = 0.89 (susceptible to malaria), HbA HbS = 1.0 (protected), HbS HbS = 0.2 (lethal anaemia). This creates a balanced polymorphism, sustaining the harmful HbS allele.
相对适合度可近似为：HbA HbA = 0.89（易感疟疾），HbA HbS = 1.0（受保护），HbS HbS = 0.2（致死性贫血）。这形成了平衡多态，维持了有害的 HbS 等位基因。
In the case study, you must use the measured genotype counts to infer whether selection is acting, and whether the observed allelic proportions match this well-known selective balance.
在本案例中，你必须利用测得的基因型数量推断选择是否在起作用，以及观察到的等位基因比例是否与这一已知的选择平衡相符。
4. The Hardy–Weinberg Equation Explained | 哈代-温伯格方程解读
For a gene with two alleles A and S, with frequencies p and q respectively, the expected genotype frequencies under random mating and no evolutionary forces are given by:
对于具有等位基因 A 和 S（频率分别为 p 和 q）的基因，在随机交配且无进化力量作用下，期望基因型频率由下式给出：
p + q = 1
(p + q)² = p² + 2pq + q² = 1
Here p² represents the expected frequency of HbA HbA, 2pq the frequency of HbA HbS, and q² the frequency of HbS HbS. These ratios only hold if the population is in equilibrium.
在此，p² 表示 HbA HbA 的期望频率，2pq 表示 HbA HbS 的频率，q² 表示 HbS HbS 的频率。这些比例仅当群体处于平衡时才成立。
We can estimate p and q from the observed genotype counts and then calculate the expected numbers to compare with the observed data, often using a chi-squared (χ²) test.
我们可以从观察到的基因型计数中估计 p 和 q，然后计算期望数量并与观察数据比较，通常使用卡方（χ²）检验。
5. Step-by-Step Calculation of Allele Frequencies | 逐步计算等位基因频率
From the data, first determine the total number of HbA and HbS alleles in the population of 1000 individuals (2000 alleles). HbA HbA individuals carry two HbA alleles, HbA HbS carry one HbA and one HbS, and HbS HbS carry two HbS.
根据数据，首先确定 1000 人群体中 HbA 和 HbS 等位基因的总数（共 2000 个等位基因）。HbA HbA 个体携带两个 HbA 等位基因，HbA HbS 携带一个 HbA 和一个 HbS，HbS HbS 携带两个 HbS。
Number of HbA alleles = 2 × 700 + 1 × 270 = 1400 + 270 = 1670. So p (frequency of HbA) = 1670 / 2000 = 0.835.
Number of HbS alleles = 2 × 30 + 1 × 270 = 60 + 270 = 330. So q (frequency of HbS) = 330 / 2000 = 0.165.
HbA 等位基因数 = 2 × 700 + 1 × 270 = 1400 + 270 = 1670。故 p（HbA 频率）= 1670 / 2000 = 0.835。
HbS 等位基因数 = 2 × 30 + 1 × 270 = 60 + 270 = 330。故 q（HbS 频率）= 330 / 2000 = 0.165。
Always check that p + q = 1: 0.835 + 0.165 = 1.00, which confirms the arithmetic.
务必检查 p + q = 1：0.835 + 0.165 = 1.00，确认计算无误。
6. Predicting Genotype Frequencies Under Equilibrium | 预测平衡下的基因型频率
Using p and q, compute the expected Hardy–Weinberg proportions:
使用 p 和 q 计算期望哈代-温伯格比例：
p² = (0.835)² = 0.6972
2pq = 2 × 0.835 × 0.165 = 0.2756
q² = (0.165)² = 0.0272
Check total: 0.6972 + 0.2756 + 0.0272 = 1.0000.
计算总和：0.6972 + 0.2756 + 0.0272 = 1.0000。
Now convert these to expected numbers in a sample of 1000: HbA HbA expected = 697.2, HbA HbS expected = 275.6, HbS HbS expected = 27.2.
现在将这些转换为 1000 人的期望数量：期望 HbA HbA = 697.2，期望 HbA HbS = 275.6，期望 HbS HbS = 27.2。
Compare with observed: 700, 270, 30. The slight excess of HbS HbS individuals already suggests a deviation, but a statistical test is needed.
与观察值 700、270、30 进行比较。HbS HbS 个体数略微过剩已提示存在偏差，但需要统计学检验。
7. Performing a Chi-Squared Test | 执行卡方检验
The chi-squared test assesses the goodness of fit between observed (O) and expected (E) numbers. The formula is:
卡方检验评估观察值（O）与期望值（E）之间的拟合优度。公式为：
χ² = Σ (O – E)² / E
Calculate for each genotype:
对每种基因型进行计算：
| Genotype | Observed (O) | Expected (E) | (O–E)²/E |
|---|---|---|---|
| HbA HbA | 700 | 697.2 | (2.8)²/697.2 ≈ 0.011 |
| HbA HbS | 270 | 275.6 | (-5.6)²/275.6 ≈ 0.114 |
| HbS HbS | 30 | 27.2 | (2.8)²/27.2 ≈ 0.288 |
Sum of χ² = 0.011 + 0.114 + 0.288 = 0.413.
χ² 总和 = 0.011 + 0.114 + 0.288 = 0.413。
Degrees of freedom (df) = number of classes – number of parameters estimated – 1 = 3 – 1 (for p) – 1 = 1. At df=1, the critical value at p=0.05 is 3.841. Since 0.413 < 3.841, we fail to reject the null hypothesis: the population does not significantly deviate from Hardy–Weinberg expectations.
自由度 (df) = 类别数 – 估算参数个数 – 1 = 3 – 1（估算 p）– 1 = 1。自由度为 1 时，ρ=0.05 的临界值为 3.841。由于 0.413 < 3.841，不能拒绝原假设：群体未显著偏离哈代-温伯格期望。
This result suggests that, at this moment, the genotypic frequencies are still broadly consistent with random mating; however, the slight overrepresentation of HbS HbS could be a sampling artefact or reflect an early signal of selection.
这一结果表明，目前基因型频率仍大体符合随机交配预期；然而 HbS HbS 轻微偏多可能是抽样误差，或反映了选择的早期信号。
8. Interpreting Selection and Heterozygote Advantage Over Time | 解读选择与杂合子优势随时间的变化
Even though a chi-squared test might not show deviation today, long-term exposure to malaria will shift allele frequencies if heterozygote advantage persists. The equilibrium frequency of the sickle cell allele (q_eq) under balancing selection can be predicted from fitness values using the formula:
尽管卡方检验今天可能未显示偏差，但如果杂合子优势持续存在，长期暴露于疟疾将改变等位基因频率。在平衡选择下，镰刀型等位基因的平衡频率 (q_eq) 可从适合度中用以下公式预测：
q_eq = s / (s + t)
where s is the selection coefficient against the normal homozygote (fitness HbA HbA = 1 – s) and t is the selection coefficient against the sickle cell homozygote (fitness HbS HbS = 1 – t).
其中 s 是针对正常纯合子的选择系数（适合度 HbA HbA = 1 – s），t 是针对镰刀型纯合子的选择系数（适合度 HbS HbS = 1 – t）。
Given HbA HbA fitness = 0.89, s = 1 – 0.89 = 0.11. HbS HbS fitness = 0.2, so t = 1 – 0.2 = 0.8. Then q_eq = 0.11 / (0.11 + 0.8) ≈ 0.121. This is lower than the observed q of 0.165, hinting that the population may not yet have reached equilibrium or that migration and genetic drift are also acting.
已知 HbA HbA 适合度 = 0.89，s = 1 – 0.89 = 0.11。HbS HbS 适合度 = 0.2，所以 t = 1 – 0.2 = 0.8。则 q_eq = 0.11 / (0.11 + 0.8) ≈ 0.121。这低于观察到的 q 值 0.165，暗示该群体可能尚未到达平衡，或者迁移与遗传漂变也在起作用。
In an exam, you could be asked to calculate the expected equilibrium frequency and compare it with the measured q, then discuss reasons for any discrepancy, such as gene flow from neighbouring populations with higher HbS incidence.
在考试中，你可能被要求计算预期的平衡频率并与实测 q 值比较，然后讨论差异的原因，例如来自邻近高 HbS 发病率的群体的基因流动。
9. Graphical Representation of Allele Frequency Change | 等位基因频率变化的图形表示
Plotting q against time can reveal how selection alters allele trajectories. Starting with an initial q of 0, a single mutation introducing HbS would initially rise slowly due to drift, then accelerate under heterozygote advantage until approaching q_eq. You may be given a graph showing a sigmoidal curve approaching 0.12 and asked to label the lag phase, exponential phase, and plateau phase.
绘制 q 随时间的变化可以揭示选择如何改变等位基因轨迹。若 q 从 0 开始，单个突变引入 HbS 起初会因漂变而缓慢上升，随后在杂合子优势作用下加速，直至接近 q_eq。你可能会遇到一张显示 S 形曲线接近 0.12 的图，并被要求标注滞后期、指数期和平稳期。
Understanding that the plateau represents the stable equilibrium frequency is crucial. If malaria were eradicated, the fitness advantage for heterozygotes would disappear, and q would decline due to selection against HbS HbS. An exam question could ask you to redraw the curve after eradication, showing a rapid drop in q towards zero over generations.
理解平稳期代表稳定的平衡频率至关重要。如果疟疾被消灭，杂合子的适合度优势消失，q 将因针对 HbS HbS 的选择而下降。考试题可能要求你重绘消灭疟疾后的曲线，显示 q 在数代内迅速降至零。
10. Common Mistakes and Key Tips | 常见错误与关键提示
A frequent error is dividing allele counts by the number of individuals instead of the total number of alleles. Always remember that each diploid individual contributes two alleles. Another pitfall is computing p as the frequency of the dominant phenotype; in codominance, the heterozygote is distinguishable, so p and q should be derived directly from genotypic counts.
常见错误之一是用个体数而非总等位基因数来除等位基因计数。请始终记住每个二倍体个体贡献两个等位基因。另一个陷阱是将 p 计算为显性表型的频率；在共显性情况下，杂合子可区分，因此 p 和 q 应直接从基因型计数中算得。
When performing χ² tests, ensure expected values above 5 for validity; if not, pool classes where biologically justified. Also, clearly state null and alternative hypotheses: H₀ is “the population is in Hardy–Weinberg equilibrium”, H₁ is “the population is not in equilibrium due to selection or other factors”.
进行 χ² 检验时，务必确保期望值大于 5 以确保有效性；若非如此，应在生物学合理的前提下合并类别。还要清晰陈述原假设和备择假设：H₀ 为“群体处于哈代-温伯格平衡”，H₁ 为“群体因选择或其他因素而不处于平衡”。
Lastly, when discussing selection, link your calculations to the biological context – mention the mechanism of heterozygote protection, the impact of malaria prevalence, and any assumptions you made about random mating or migration.
最后，在讨论选择时，将你的计算与生物学背景联系起来——提及杂合子保护机制、疟疾流行的影响，以及你对随机交配或迁移所做的假设。
11. Extending the Case Study: Migration and Genetic Drift | 拓展案例：迁移与遗传漂变
Real populations are not closed systems. Suppose 50 individuals from a neighbouring village with a q of 0.3 migrate into our studied population. The new allele frequency can be calculated using the weighted average formula: q_new = (m × q_migrants) + ((1 – m) × q_residents), where m is the proportion of migrants in the new population.
真实群体不是封闭系统。假设来自邻村的 50 人迁入我们的研究群体，邻村 q 为 0.3。新的等位基因频率可用加权平均公式计算：q_new = (m × q_migrants) + ((1 – m) × q_residents)，其中 m 为迁移者在新群体中的比例。
If total population becomes 1050, m = 50/1050 ≈ 0.0476. Then q_new = (0.0476 × 0.3) + (0.9524 × 0.165) ≈ 0.0143 + 0.1571 = 0.1714. This shows that even a small influx from a high-HbS area can raise q, complicating the signature of selection.
如果总人口变为 1050，m = 50/1050 ≈ 0.0476。则 q_new = (0.0476 × 0.3) + (0.9524 × 0.165) ≈ 0.0143 + 0.1571 = 0.1714。这表明即使从高 HbS 区小规模迁入也能提高 q，使选择的印记复杂化。
CAIE exams often combine migration with selection in a single question, requiring you to distinguish the effects of each evolutionary force. Always isolate the factor being asked about.
CAIE 考试常将迁移与选择合并在同一题目中，要求你区分每种进化力量的影响。务必单独分析题目所问的因素。
12. Exam Strategy and Summary | 考试策略与总结
Approach any case study by first tabulating the data, then extracting allele and genotype frequencies. Use the Hardy–Weinberg model as a null hypothesis and apply a χ² test to detect deviation. Next, relate any significant deviation to plausible evolutionary mechanisms – selection, gene flow, non-random mating or drift. Finally, support your reasoning with quantitative predictions where possible, such as equilibrium frequencies or expected numbers after migration.
处理任何案例分析时，先列表整理数据，再提取等位基因和基因型频率。将哈代-温伯格模型作为原假设，用 χ² 检验检测偏差。接着，将任何显著偏差与可能的进化机制联系起来 —— 选择、基因流动、非随机交配或漂变。最后，尽可能用定量预测（如平衡频率或迁移后的期望数值）来支持你的推理。
The sickle cell–malaria case encapsulates key Year 13 concepts: codominance, heterozygote advantage, Hardy–Weinberg analysis, and the interplay of evolutionary forces. Mastery of this example will equip you to tackle any population genetics scenario with confidence.
镰刀型细胞–疟疾案例浓缩了 Year 13 的核心概念：共显性、杂合子优势、哈代-温伯格分析和进化力量的相互作用。掌握此例将使你自信应对任何群体遗传学情境。
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