Hardy-Weinberg Principle & Gene Frequency Calculations | 哈迪-温伯格定律与基因频率计算

📚 Hardy-Weinberg Principle & Gene Frequency Calculations | 哈迪-温伯格定律与基因频率计算

The Hardy-Weinberg principle is a cornerstone of population genetics and a guaranteed topic in A-Level biology exams. This article breaks down the principle, its equations, and step-by-step methods for solving gene frequency problems, including the tricky X-linked cases.

哈迪-温伯格定律是群体遗传学的基石，也是A-Level生物考试的必考内容。本文将系统梳理该定律的核心概念、方程公式，以及基因频率计算的完整解题方法，包括容易出错的伴X染色体遗传案例。

1. What Is the Hardy-Weinberg Principle? | 什么是哈迪-温伯格定律？

The Hardy-Weinberg principle states that in a large, randomly mating population, allele and genotype frequencies remain constant from generation to generation, provided that no evolutionary forces act upon the population. This state is called Hardy-Weinberg equilibrium.

哈迪-温伯格定律指出：在一个大的、随机交配的群体中，等位基因频率和基因型频率逐代保持不变——前提是没有进化力量作用于该群体。这种状态称为哈迪-温伯格平衡。

Think of it as a null hypothesis for evolution: if the frequencies change over time, then evolution is occurring. If they stay the same, the population is not evolving at that locus.

可以将它理解为进化的零假设：如果频率随时间改变，说明进化正在发生；如果频率保持不变，说明该位点上群体未发生进化。

The five conditions required for Hardy-Weinberg equilibrium are:

维持哈迪-温伯格平衡需要满足以下五个条件：

No mutations at the gene locus / 基因位点上不发生突变
Random mating — all individuals have equal chance to mate / 随机交配——所有个体交配机会均等
No natural selection — all genotypes have equal survival/reproductive success / 无自然选择——所有基因型的存活和繁殖成功率相同
Very large population (no genetic drift) / 群体极大（无遗传漂变）
No migration in or out of the population / 无个体迁入或迁出

In real populations, these conditions are rarely met perfectly, which is why the principle is a model. However, it remains a powerful tool for estimating allele frequencies from phenotype data.

在真实群体中，这些条件几乎无法完美满足，因此它是一个理想化模型。但该定律仍是利用表现型数据估算等位基因频率的强大工具。

2. The Two Key Equations | 两个核心方程

Consider a gene with two alleles, A (dominant) and a (recessive). Let p represent the frequency of allele A, and q represent the frequency of allele a. The first equation is:

考虑一个具有两个等位基因的基因：A（显性）和 a（隐性）。设 p 为等位基因 A 的频率，q 为等位基因 a 的频率。第一个方程为：

p + q = 1

This simply means that the total frequency of all alleles for this gene in the population must equal 100% (or 1).

这表示该基因在群体中所有等位基因的频率之和等于100%（即1）。

The second equation describes genotype frequencies. The three genotypes are AA (homozygous dominant), Aa (heterozygous), and aa (homozygous recessive). Since mating is random, the genotype frequencies follow the binomial expansion of (p + q)²:

第二个方程描述基因型频率。三种基因型分别是：AA（显性纯合子）、Aa（杂合子）和 aa（隐性纯合子）。由于随机交配，基因型频率遵循 (p + q)² 的二项展开式：

p² + 2pq + q² = 1

Where p² = frequency of AA, 2pq = frequency of Aa, and q² = frequency of aa.

其中 p² = AA 的频率，2pq = Aa 的频率，q² = aa 的频率。

These two equations are all you need for most exam problems. Let’s see how they connect to phenotype data.

这两个方程足以应对大多数考题。接下来我们看看如何将表现型数据与它们联系起来。

3. Step-by-Step Calculation Method | 分步计算方法

The most common exam scenario gives you the frequency of a recessive phenotype and asks you to find allele frequencies or carrier (heterozygote) frequencies. Follow these steps:

最常见的考题情境是：题目给出隐性表现型的频率，要求你求等位基因频率或携带者（杂合子）频率。按以下步骤操作：

Step 1: Identify the recessive phenotype frequency. In a Mendelian trait, only the homozygous recessive genotype (aa) shows the recessive phenotype. Thus, the recessive phenotype frequency equals q².

步骤1： 确定隐性表现型频率。在孟德尔性状中，只有隐性纯合基因型（aa）表现隐性性状。因此，隐性表现型频率等于 q²。

Step 2: Take the square root to find q.

步骤2： 开平方根求出 q。

Step 3: Calculate p using p + q = 1.

步骤3： 利用 p + q = 1 计算 p。

Step 4: Substitute p and q into 2pq to find the carrier frequency, or into p² for the homozygous dominant frequency.

步骤4： 将 p 和 q 代入 2pq 求携带者频率，或代入 p² 求显性纯合子频率。

A crucial point: the dominant phenotype includes both AA and Aa genotypes. You cannot take the square root of the dominant phenotype frequency to get p. Only the recessive phenotype gives you q² directly.

关键点：显性表现型包含 AA 和 Aa 两种基因型。你不能对显性表现型频率开平方来得到 p。只有隐性表现型才能直接给出 q²。

4. Worked Example: Autosomal Recessive Disease | 例题解析：常染色体隐性遗传病

Let’s apply this to a classic exam question about cystic fibrosis, an autosomal recessive disorder.

我们以囊性纤维化为例——这是一种常染色体隐性遗传病，也是经典考题。

Problem: In a population of 10,000 people, 4 individuals have cystic fibrosis. Calculate: (a) the frequency of the recessive allele, (b) the frequency of the dominant allele, (c) the expected number of carriers.

题目： 在一个10000人的群体中，有4人患有囊性纤维化。求：(a) 隐性等位基因频率；(b) 显性等位基因频率；(c) 预期携带者人数。

Solution:

解答：

Step	Calculation	Result
(a) q²	4 ÷ 10000	q² = 0.0004
q	√0.0004	q = 0.02
(b) p	p = 1 − q = 1 − 0.02	p = 0.98
(c) carrier frequency	2pq = 2 × 0.98 × 0.02	2pq = 0.0392
carrier number	0.0392 × 10000	≈ 392 people

Notice that even though only 0.04% of individuals are affected, about 3.92% of the population are carriers. This demonstrates how a harmful recessive allele can persist at relatively high frequency in heterozygotes.

注意：尽管只有0.04%的个体患病，却有约3.92%的人口是携带者。这说明有害的隐性等位基因能以杂合子的形式在群体中保持较高的频率。

5. Worked Example: Codominance | 例题解析：共显性

For codominant traits, each genotype produces a distinct phenotype, so you can read genotype frequencies directly from the phenotype counts. The classic example is the MN blood group system in humans.

对于共显性性状，每种基因型产生独特的表现型，因此可以直接从表现型计数中读出基因型频率。经典例子是人类的 MN 血型系统。

Problem: In a sample of 200 people, 80 are MM, 100 are MN, and 20 are NN. Calculate the allele frequencies of M and N.

题目： 在一个200人的样本中，80人为 MM，100人为 MN，20人为 NN。计算 M 和 N 的等位基因频率。

Solution: Each person has 2 alleles, so the total number of alleles is 2 × 200 = 400.

解答： 每人有2个等位基因，所以等位基因总数为 2 × 200 = 400。

Number of M alleles = (2 × 80) + (1 × 100) = 160 + 100 = 260

M 等位基因数 = (2 × 80) + (1 × 100) = 160 + 100 = 260

Number of N alleles = (2 × 20) + (1 × 100) = 40 + 100 = 140

N 等位基因数 = (2 × 20) + (1 × 100) = 40 + 100 = 140

Frequency of M: p = 260 ÷ 400 = 0.65. Frequency of N: q = 140 ÷ 400 = 0.35. Check: p + q = 0.65 + 0.35 = 1 ✓

M 的频率：p = 260 ÷ 400 = 0.65。N 的频率：q = 140 ÷ 400 = 0.35。检验：p + q = 0.65 + 0.35 = 1 ✓

This method — the allele counting method — always works for codominance, and it also works as an alternative approach for any genetic system when you have genotype data.

这种等位基因计数法对共显性永远适用，同时它也是任何遗传系统下、当你拥有基因型数据时的通用替代方法。

6. X-Linked Genes: Special Considerations | 伴X染色体基因：特殊情况

Sex-linked genes, especially X-linked ones in mammals, require extra care. Males have only one X chromosome (XY), so their genotype frequency equals the allele frequency.

性连锁基因——尤其是哺乳动物的伴X染色体基因——需要格外小心。雄性只有一条X染色体（XY），因此其基因型频率等于等位基因频率。

For an X-linked gene with two alleles (dominant Xᴬ and recessive Xᵃ):

对于一个具有两个等位基因（显性 Xᴬ 和隐性 Xᵃ）的伴X基因：

In males: The frequency of affected males = q (not q²!). The frequency of unaffected males = p.

在男性中： 患病男性的频率 = q（不是 q²！）。正常男性的频率 = p。

In females: The frequencies follow the usual equations: p² (XᴬXᴬ), 2pq (XᴬXᵃ), q² (XᵃXᵃ).

在女性中： 频率遵循常规方程：p²（XᴬXᴬ）、2pq（XᴬXᵃ）、q²（XᵃXᵃ）。

Affected males are usually far more common than affected females for rare recessive X-linked disorders. For example, if q = 0.01, then affected male frequency = 0.01 (1%) but affected female frequency = q² = 0.0001 (0.01%). This is why conditions like red-green colour blindness and haemophilia are much more common in males.

对于罕见的X连锁隐性遗传病，患病男性通常远多于患病女性。例如，若 q = 0.01，则男性患病频率 = 0.01（1%），而女性患病频率 = q² = 0.0001（0.01%）。这就是红绿色盲和血友病在男性中更常见的原因。

Exam tip: When the question involves X-linked traits, first count the allele frequency from the male phenotypes, then use that frequency to work out the female genotype frequencies.

考试提示： 当题目涉及伴X遗传时，先用男性表现型推算出等位基因频率，再用该频率计算女性的基因型频率。

7. Common Exam Traps | 常见考试陷阱

Students often lose marks on Hardy-Weinberg calculations due to avoidable mistakes. Here are the most common traps:

学生在哈迪-温伯格计算中常因可避免的错误丢分。以下是最常见的陷阱：

Trap	Correct Approach
Taking √ of dominant phenotype frequency	Only √ recessive phenotype frequency (q²)
Forgetting that q² means affected individuals, not disease allele frequency	Always separate q (allele) from q² (genotype)
Applying q² to males for X-linked traits	Males are hemizygous: male frequency = q
Confusing carrier frequency with affected frequency	Carriers = 2pq; affected recessive = q²; dominant phenotype = p² + 2pq
Not converting percentages to decimals	Always work in decimal form: 4% → 0.04

Another common error involves wording. When a question says “the frequency of the recessive allele is 0.1”, that is q directly. But when it says “1 in 100 individuals shows the recessive phenotype”, that is q². Read carefully.

另一个常见错误涉及题干措辞。当题目说”隐性等位基因频率为0.1″时，这直接就是 q。但当题目说”每100人中1人表现出隐性表现型”时，这是 q²。请仔细审题。

8. Worked Example: X-Linked Haemophilia | 例题解析：伴X血友病

Let’s put the X-linked rules into practice with a full example.

我们用一道完整例题来实践伴X遗传规律。

Problem: In a population of 1,200 males and 1,200 females, 6 males have haemophilia (XᵃY). Assuming Hardy-Weinberg equilibrium: (a) Find q and p. (b) Calculate the expected number of affected females (XᵃXᵃ). (c) Calculate the expected number of carrier females (XᴬXᵃ).

题目： 在一个由1200名男性和1200名女性组成的群体中，有6名男性患血友病（XᵃY）。假设处于哈迪-温伯格平衡：(a) 求 q 和 p；(b) 计算患病女性（XᵃXᵃ）的预期人数；(c) 计算携带者女性（XᴬXᵃ）的预期人数。

Solution:

解答：

(a) In males, the frequency of affected individuals equals q. q = 6 ÷ 1200 = 0.005. Therefore p = 1 − 0.005 = 0.995.

(a) 在男性中，患病个体频率等于 q。q = 6 ÷ 1200 = 0.005。因此 p = 1 − 0.005 = 0.995。

(b) Affected female frequency = q² = 0.005² = 0.000025. Expected number = 0.000025 × 1200 = 0.03. This means fewer than 1 affected female is expected in this population.

(b) 患病女性频率 = q² = 0.005² = 0.000025。预期人数 = 0.000025 × 1200 = 0.03。这意味着该群体中预期患病女性不到1人。

This example highlights a striking ratio: 6 affected males versus 12 carriers and 0 affected females out of the same population size — a perfect illustration of why X-linked recessive disorders are male-dominated.

这个例子展示了一个鲜明对比：在相同人口规模下，6名患病男性对应12名女性携带者和0名患病女性——完美诠释了为什么X连锁隐性遗传病以男性患者为主。

9. Practice Problems | 练习题目

Test yourself with these problems before looking at the answers below.

先自行完成以下练习，再对照下方的答案。

Problem 1: In a population of 2,000 plants, 500 have white flowers (recessive). What is the frequency of the heterozygous genotype? What is the number of heterozygous plants?

练习1： 在一个2000株植物的群体中，500株开白花（隐性性状）。杂合基因型的频率是多少？杂合植株的数量是多少？

Problem 2: 1 in 400 babies is affected by an autosomal recessive disorder. What proportion of the population is carriers?

练习2： 每400名新生儿中有1人患常染色体隐性遗传病。携带者在人群中的比例是多少？

Problem 3: In a population, the frequency of a dominant allele is 0.7. The population size is 500. How many individuals are expected to be homozygous dominant?

练习3： 在一个群体中，显性等位基因频率为0.7，群体规模为500人。预期有多少个体是显性纯合子？

Answers:

答案：

Problem 1: q² = 500/2000 = 0.25, so q = 0.5, p = 0.5. Heterozygote frequency 2pq = 2 × 0.5 × 0.5 = 0.5. Number = 0.5 × 2000 = 1000 plants.

练习1：q² = 500/2000 = 0.25，故 q = 0.5，p = 0.5。杂合子频率 2pq = 2 × 0.5 × 0.5 = 0.5。数量 = 0.5 × 2000 = 1000 株。

Problem 2: q² = 1/400 = 0.0025, so q = 0.05, p = 0.95. Carrier frequency = 2pq = 2 × 0.95 × 0.05 = 0.095, or about 9.5%.

练习2：q² = 1/400 = 0.0025，故 q = 0.05，p = 0.95。携带者频率 = 2pq = 2 × 0.95 × 0.05 = 0.095，即约9.5%。

Problem 3: p = 0.7, so p² = 0.49. Number = 0.49 × 500 = 245 individuals.

练习3：p = 0.7，所以 p² = 0.49。数量 = 0.49 × 500 = 245 人。

10. Applications Beyond Exams | 考试之外的应用

The Hardy-Weinberg principle is not just an exam tool — it has real-world applications in medicine, conservation biology, and evolutionary studies.

哈迪-温伯格定律不仅是考试工具——它在医学、保护生物学和进化研究中有实际应用。

In medicine, it helps estimate carrier frequencies for genetic disorders, which is essential for genetic counselling. For example, if the incidence of a disease is known, doctors can calculate the probability that an individual is a carrier and advise on family planning.

在医学领域，它用于估算遗传病的携带者频率，这对遗传咨询至关重要。例如，已知某种疾病的发病率后，医生可据此计算个体为携带者的概率，并给出生育建议。

In conservation biology, the principle helps assess whether a small, endangered population is losing genetic diversity. If observed heterozygosity is significantly lower than the Hardy-Weinberg expected 2pq, it may indicate inbreeding — a warning sign for conservationists.

在保护生物学中，该定律用于评估小型濒危种群是否正在丧失遗传多样性。如果观察到的杂合度显著低于哈迪-温伯格预期的 2pq，可能提示近亲繁殖——这对保育工作者是一个警示信号。

In evolutionary biology, deviations from Hardy-Weinberg equilibrium can reveal the action of natural selection, non-random mating, or population structure, giving researchers clues about the evolutionary forces shaping a population.

在进化生物学中，偏离哈迪-温伯格平衡可以揭示自然选择、非随机交配或群体结构的效应，为研究者提供关于塑造种群进化力量的线索。

11. Summary Checklist | 要点总结清单

By now you should be confident with the core concepts. Here is a final revision checklist — tick off each item before your exam.

至此，你应该已对核心概念充满信心。以下是最终复习清单——考试前请逐项确认。

State the five conditions for Hardy-Weinberg equilibrium / 能陈述哈迪-温伯格平衡的五个条件
Know when to use p + q = 1 versus p² + 2pq + q² = 1 / 知道何时使用 p + q = 1 与 p² + 2pq + q² = 1
Only square-root the recessive phenotype frequency, never the dominant / 只对隐性表现型频率开平方，绝不对显性表现型开平方
Remember that dominant phenotype = p² + 2pq, not p² alone / 记住显性表现型 = p² + 2pq，而不只是 p²
For X-linked genes, male frequency equals q, not q² / 对于伴X基因，男性频率等于 q 而非 q²
Always check that your answers sum to 1 (p + q = 1) / 始终检查答案之和为1（p + q = 1）
Use decimal form, not percentages, in equations / 在方程中使用小数而非百分数

Mastering the Hardy-Weinberg principle is a matter of practice. Work through as many past paper questions as you can — the method is always the same, and once you have internalised the steps, these questions become easy marks.

掌握哈迪-温伯格定律的关键在于练习。尽可能多做真题——解题方法始终如一，一旦内化这些步骤，这类题目就会变成送分题。

Published by TutorHao | Biology Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Hardy-Weinberg Principle & Gene Frequency Calculations | 哈迪-温伯格定律与基因频率计算