| Big Idea | Learning Objective | Essential Knowledge |
|---|---|---|
Big Idea 3 — Information Storage and Transmission | 5.1.A |
|
5.1.B |
|
Heredity
AP Biology · Topic 5
5.1
Meiosis
Syllabus
Source: College Board AP Course and Exam Description
Meiosis 减数分裂 makes gametes 配子 (eggs and sperm) with half the chromosome number, so fertilization restores the full set. 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 halves the chromosome number in two divisions
Divide a cell's chromosomes
Meiosis halves the chromosome number and shuffles genes, making four genetically varied gametes. Step through to see the divisions.
| English | Chinese | Pinyin |
|---|---|---|
| Meiosis | 减数分裂 | jiǎn shù fēn liè |
| gametes | 配子 | pèi zi |
| homologous chromosomes | 同源染色体 | tóng yuán rǎn sè tǐ |
5.2
Meiosis and Genetic Diversity
Syllabus
| Big Idea | Learning Objective | Essential Knowledge |
|---|---|---|
Big Idea 3 — Information Storage and Transmission | 5.2.A |
|
Source: College Board AP Course and Exam Description
Meiosis shuffles genes three ways, so offspring differ from parents and each other:
Independent assortment produces many gamete combinations
Crossing over swaps segments between homologous chromosomes
- Crossing over 交叉互换: homologous chromosomes swap segments in meiosis 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.
| English | Chinese | Pinyin |
|---|---|---|
| Crossing over | 交叉互换 | jiāo chā hù huàn |
| Independent assortment | 自由组合 | zì yóu zǔ hé |
5.3
Mendelian Genetics
Syllabus
| Big Idea | Learning Objective | Essential Knowledge |
|---|---|---|
Big Idea 3 — Information Storage and Transmission | 5.3.A |
|
Source: College Board AP Course and Exam Description
A gene's alternative versions are alleles 等位基因. An organism's genotype 基因型 (its alleles) produces its phenotype 表型 (its traits). Mendel's rules: a dominant 显性 allele masks a recessive 隐性 one; the two alleles segregate into different gametes (law of segregation); genes for different traits assort independently. A Punnett square 庞纳特方格 predicts offspring ratios (a heterozygous cross gives 3:1). Homozygous 纯合 means two identical alleles; heterozygous 杂合 means two different.
A monohybrid Punnett square giving a 3:1 ratio
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.
Cross two parents
A Punnett square combines each parent's alleles to predict the offspring ratios. Set the parent genotypes and read off the expected proportions.
| English | Chinese | Pinyin |
|---|---|---|
| alleles | 等位基因 | děng wèi jī yīn |
| genotype | 基因型 | jī yīn xíng |
| phenotype | 表型 | biǎo xíng |
| dominant | 显性 | xiǎn xìng |
| recessive | 隐性 | yǐn xìng |
| Punnett square | 庞纳特方格 | páng nà tè fāng gé |
| Homozygous | 纯合 | chún hé |
| heterozygous | 杂合 | zá hé |
5.4
Non-Mendelian Genetics
Syllabus
| Big Idea | Learning Objective | Essential Knowledge |
|---|---|---|
Big Idea 3 — Information Storage and Transmission | 5.4.A |
|
Source: College Board AP Course and Exam Description
Many traits do not follow simple dominance:
Sex linkage gives different results for sons and daughters
- 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 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.
| English | Chinese | Pinyin |
|---|---|---|
| Incomplete dominance | 不完全显性 | bù wán quán xiǎn xìng |
| Codominance | 共显性 | gòng xiǎn xìng |
| polygenic | 多基因 | duō jī yīn |
| sex-linked | 伴性 | bàn xìng |
5.5
Environmental Effects on Phenotype
Syllabus
| Big Idea | Learning Objective | Essential Knowledge |
|---|---|---|
Big Idea 4 — Systems Interactions | 5.5.A |
|
Source: College Board AP Course and Exam Description
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.
5.5
Exam tips
- 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.