遗传

Two brown-eyed parents can have a blue-eyed child. Colour blindness is far more common in boys than girls. Two genes on the same chromosome are inherited together, except when they are not. Each of these follows from what chromosomes do in meiosis, and each can be turned into a probability you can calculate.

In animals, Meiosis 减数分裂 produces haploid cells used in forming gametes 配子 (eggs and sperm), with half the diploid chromosome number, so fertilization restores the full set. Not all four meiotic products necessarily become functional gametes. One diploid cell divides twice to give four haploid cells. Meiosis I separates homologous chromosomes 同源染色体 (reducing the number); meiosis II separates sister chromatids (like mitosis).

Meiosis shuffles genes three ways, so offspring differ from parents and each other:

• Crossing over 交叉互换: non-sister chromatids of homologous chromosomes exchange corresponding segments during prophase I.
• Independent assortment 自由组合: each homologous pair lines up and separates randomly.
• Random fertilization: any sperm can meet any egg.

Together these create enormous variation – the raw material for evolution.

A gene's alternative versions are alleles 等位基因. An organism's genotype 基因型 (its alleles), together with environmental influences, contributes to its phenotype 表型 (its traits). Under complete dominance, a dominant 显性 allele masks a recessive 隐性 one in a heterozygote. In the diploid Mendelian model, the two alleles segregate into different gametes, and unlinked genes assort independently. A Punnett square 庞纳特方格 predicts offspring probabilities; an Aa × Aa cross gives a 3:1 phenotype ratio under complete dominance. Homozygous 纯合 means two identical alleles; heterozygous 杂合 means two different.

Worked example. For a dihybrid cross of two independent genes, $AaBb\times AaBb$, use the multiplication rule instead of a $16$-box square. Each gene alone gives $\tfrac34$ dominant, so the chance an offspring shows both dominant traits is $\tfrac34\times\tfrac34=\tfrac{9}{16}$, and the chance of the fully recessive $aabb$ is $\tfrac14\times\tfrac14=\tfrac{1}{16}$. Multiplying two independent $3{:}1$ ratios is what produces the classic $9{:}3{:}3{:}1$ pattern.

Many traits do not follow simple dominance:

• Incomplete dominance 不完全显性: heterozygotes are a blend (red × white → pink).
• Codominance 共显性: both alleles show fully (AB blood type).
• Multiple alleles, polygenic 多基因 traits (many genes, like height), pleiotropy (one gene, many effects), and sex-linked 伴性 genes (on sex chromosomes, often the X chromosome) all give more complex ratios.

Worked example (chi-square test). To check whether real data fit a predicted ratio, use $\chi^2=\sum\dfrac{(o-e)^2}{e}$. A monohybrid cross predicts $3{:}1$, so of $80$ offspring you expect $60$ dominant and $20$ recessive, but you observe $55$ and $25$. Then $\chi^2=\dfrac{(55-60)^2}{60}+\dfrac{(25-20)^2}{20}=\dfrac{25}{60}+\dfrac{25}{20}=0.42+1.25=1.67$. With $1$ degree of freedom the critical value at $p=0.05$ is $3.84$; since $1.67<3.84$, we fail to reject the null hypothesis – the deviation is within chance.

Worked example (a sex-linked cross). Red–green colour blindness is caused by a recessive allele on the X chromosome. A carrier mother ($X^BX^b$) has children with a father of normal vision ($X^BY$).

• Daughters get the father's $X^B$, so none is colour-blind; half are carriers ($X^BX^b$).
• Sons get the father's $Y$ and either maternal X, so half the sons are colour-blind ($X^bY$).

Boys need only one recessive allele to show the trait, because the Y carries no copy of the gene. That is why X-linked recessive conditions are far more common in males.

Genes on the same chromosome are linked 连锁. Nearby genes tend to be inherited together; distant genes can behave as if unlinked because multiple crossovers can hide exchanges. In a testcross, recombination frequency 重组频率 measures recombinant offspring. It estimates short gene separations but approaches a maximum expected value of 50%; it does not keep increasing with map distance:

Worked example. A test cross of a dihybrid ($AaBb\times aabb$) gives 445 $AaBb$, 435 $aabb$, 62 $Aabb$ and 58 $aaBb$ offspring. The two large classes match the parents' combinations; the two small ones are recombinants. Recombination frequency $=\dfrac{62+58}{1000}=12\%$, so the genes are about 12 map units apart. Unlinked genes would give four roughly equal classes, near $25\%$ each.

A pedigree 系谱图 shows a trait across generations. Two quick tests decide most AP questions:

• Two unaffected parents with an affected child: the trait is recessive. Each parent must carry a hidden allele.
• An affected mother with an unaffected son: the trait cannot be X-linked recessive, since every son receives his only X from his mother.

Phenotype is not set by genes alone – the environment also matters. Temperature, nutrition, and other factors can change how genes are expressed (a Himalayan rabbit's dark fur where it is cold, a plant's height with more sunlight). So identical genotypes can give different phenotypes in different conditions.

Common mistakes.

• Thinking earlier children change the odds for the next one. Each fertilization is independent: if two carriers already have three unaffected children, the chance the fourth is affected is still $\tfrac14$.
• Adding probabilities that should be multiplied. Use the product rule for "this and that" (independent events) and the sum rule for "this or that" (different ways to get the same result).
• Saying a son inherits an X-linked allele from his father. A son gets his only X from his mother; a father passes his X to every daughter and his Y to every son.
• Treating a recombination frequency of about 50% as tight linkage. Values near 50% mean the genes assort independently (far apart or on different chromosomes); small values mean they are close together.
• Writing that a chi-square result "proves" the hypothesis. If $\chi^2$ is below the critical value you fail to reject the null hypothesis; the data are simply consistent with it.
• Using the wrong degrees of freedom. df = number of phenotype classes − 1 (for example 3 for a 9:3:3:1 test).

Answer each before reading the answers below.

1. Give the expected genotype ratio from $Aa\times aa$.
2. In $AaBbCc\times AaBbCc$, with independent genes, what is the probability of an $aabbcc$ offspring?
3. A man with haemophilia (X-linked recessive, $X^hY$) has children with a woman with two normal alleles ($X^HX^H$). Predict the sons and the daughters.
4. A cross expected to give a $1:1$ ratio produces 88 and 72 of 160 offspring. Find $\chi^2$ and decide at $p=0.05$ (critical value $3.84$ for $df=1$).
5. A student says: "Genes that are far apart on the same chromosome are inherited together most strongly." Correct the student.

1. $1\ Aa:1\ aa$.
2. $\left(\tfrac14\right)^3=\tfrac{1}{64}$.
3. No son is affected: sons receive the father's $Y$. Every daughter is a carrier, $X^HX^h$: she receives the father's only X, which carries $h$.
4. Expected $80$ and $80$: $\chi^2=\dfrac{8^2}{80}+\dfrac{8^2}{80}=1.6$. Since $1.6<3.84$, the difference is not significant; the data are consistent with $1:1$.
5. The reverse is true. Crossing over between distant genes is frequent, so they recombine often and can appear almost independent (recombination frequency near $50\%$). Genes close together rarely recombine and are the ones inherited together most strongly.

• Contrast mitosis (2 identical, full chromosome number) with meiosis (4 non-identical gametes, half the number).
• Explain variation from crossing over, independent assortment, and random fertilisation.
• Use the multiplication rule for dihybrid crosses (each gene's $3{:}1$ multiplied), and a chi-square test ($\chi^2=\sum\frac{(o-e)^2}{e}$) to judge observed vs expected ratios.
• Keep genotype (the alleles) separate from phenotype (what you see) — $AA$ and $Aa$ can look the same.
• Recognise non-Mendelian patterns: incomplete dominance, codominance, and sex linkage.