High School Biology

Mendelian Genetics and Heredity孟德尔遗传学与遗传

Gregor Mendel's pea-plant experiments uncovered the rules that govern how traits pass from parents to offspring. This guide builds from core vocabulary (gene, allele, genotype, phenotype, dominant, recessive) through monohybrid and dihybrid Punnett squares, the two laws of inheritance, and then extends to incomplete dominance, codominance, multiple alleles, sex-linked inheritance, and pedigree analysis. Every section pairs worked crosses with quiz questions grounded in real biological contexts.格雷戈尔·孟德尔的豌豆实验揭示了性状从亲代传递给后代的规律。本指南从核心词汇（基因、等位基因、基因型、表现型、显性、隐性）出发，经由单杂交与双杂交旁氏表、两大遗传定律，延伸至不完全显性、共显性、复等位基因、伴性遗传与系谱分析。每节均配有实际生物情境下的例题与测验。

7 sections7 节内容 US NGSS · ON · BC · ABUS NGSS · ON · BC · AB AB Biology 30 (Grade 12) content marked Honors阿尔伯塔 Biology 30（12 年级）内容标为荣誉级
Read me first先读这里

How to use this guide如何使用本指南

Mendelian genetics appears in all four curricula we map to, but at different grade levels and depths. BC introduces Punnett squares as early as Science 10 (Grade 10). Ontario SBI3U (Grade 11) expects full monohybrid, dihybrid, incomplete dominance, codominance, and sex-linked problems (D2.3). Alberta defers the full genetics treatment to Biology 30 (Grade 12, Honors Biology 30). NGSS frames heredity through variation and probability (HS-LS3-2, HS-LS3-3) rather than Mendel's named laws. The table below locates each section in your curriculum.孟德尔遗传学出现在我们对照的所有四套大纲中，但年级和深度各有不同。BC 早在 Science 10（10 年级）即引入旁氏表。安大略 SBI3U（11 年级）要求完整的单杂交、双杂交、不完全显性、共显性和伴性遗传题（D2.3）。阿尔伯塔将完整遗传学内容推迟至 Biology 30（12 年级，荣誉 Biology 30）。NGSS 通过变异与概率（HS-LS3-2、HS-LS3-3）而非孟德尔命名定律来框架遗传学。下表定位各节在你大纲中的位置。

If you are in…如果你在… Focus on these sections重点学习 Defer / lighter可推迟 / 减负 Source依据
🇺🇸 US NGSS HS Life Sciences美国 NGSS 生命科学 §1 (gene/allele/chromosome vocabulary), §2–§3 (Punnett squares and probability), §5–§6 (variation through incomplete dominance, codominance, multiple alleles) — HS-LS3-1, HS-LS3-2, HS-LS3-3§1（基因/等位基因/染色体词汇）、§2–§3（旁氏表与概率）、§5–§6（通过不完全显性、共显性、复等位基因产生的变异）——HS-LS3-1、HS-LS3-2、HS-LS3-3 NGSS HS-LS3-3 Assessment Boundary excludes Hardy-Weinberg calculations; pedigree analysis (§7) is not explicitly required but is highly useful contextNGSS HS-LS3-3 评估边界排除哈迪-温伯格计算；系谱分析（§7）并非明确要求，但具有很高的背景价值 NGSS HS Life Science — HS-LS3-1, HS-LS3-2, HS-LS3-3 PEs— HS-LS3-1、HS-LS3-2、HS-LS3-3 表现期望
🇨🇦 ON Grade 11 — SBI3U安大略 11 年级 — SBI3U All 7 sections in full. D2.3 requires Punnett square solutions for monohybrid, incomplete dominance, codominance, dihybrid, and sex-linked crosses. D3.2–D3.3 require Mendelian laws and all dominance/recessiveness concepts全部 7 节完整学习。D2.3 要求旁氏表解决单杂交、不完全显性、共显性、双杂交和伴性遗传题。D3.2–D3.3 要求孟德尔定律与所有显/隐性概念 Nothing — genetics is a core SBI3U strand (Strand D)无 — 遗传学是 SBI3U 的核心单元（D 单元） Ontario SBI3U/4U Biology — SBI3U Strand D D2.3, D3.2, D3.3— SBI3U D 单元 D2.3、D3.2、D3.3
🇨🇦 BC Science 10BC Science 10 All 7 sections — BC Science 10 explicitly names "Punnett squares, complete dominance, co-dominance, incomplete dominance, sex-linked inheritance" as core content全部 7 节——BC Science 10 明确将"旁氏表、完全显性、共显性、不完全显性、伴性遗传"列为核心内容 Formal pedigree construction (§7 going-deeper) is not explicitly named in Science 10 but follows from the inheritance patterns正式系谱构建（§7 深入内容）在 Science 10 中未明确提名，但由遗传模式推导而来 BC Life Sciences 11 / Anatomy 12 — BC Science 10 patterns of inheritance— BC Science 10 遗传模式内容
🇨🇦 AB Biology 30 Honors阿尔伯塔 Biology 30 荣誉 All 7 sections. 30–C2.1k (dominance, segregation, independent assortment), 30–C2.2k (genotype/phenotype ratios for all allele types), 30–C2.5k (sex-chromosome vs. autosome inheritance)全部 7 节。30–C2.1k（显性、分离、自由组合）、30–C2.2k（各等位基因类型的基因型/表现型比例）、30–C2.5k（性染色体与常染色体遗传对比） Hardy-Weinberg population genetics (30–D1.3k) is a separate Biology 30 Unit D topic — not covered in this guide哈迪-温伯格种群遗传学（30–D1.3k）属 Biology 30 Unit D 的独立主题——本指南不涵盖 Alberta Biology 20/30 — Biology 30 Unit C GO2— Biology 30 Unit C GO2

Once you have located your row, use the two cards below for the approach that fits your timeline.找到所在行后，用下面两张卡片选择适合你时间安排的方式。

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If you are cramming the night before如果你在临阵磨枪

Know the vocabulary cold (gene, allele, genotype, phenotype, dominant, recessive, homozygous, heterozygous). Know how to set up a monohybrid Punnett square and read off the 3:1 phenotype ratio. Know the two Mendelian laws by name. Read every cram-cheat box. Skip the going-deeper sections on dihybrid linkage and pedigree construction on a first pass.熟记词汇（基因、等位基因、基因型、表现型、显性、隐性、纯合子、杂合子）。掌握如何列单杂交旁氏表并读出 3:1 表现型比。牢记两大孟德尔定律的名称。读每个速记框，第一遍跳过双杂交连锁与系谱构建的深入内容。

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If you are going for the top mark如果你目标顶分

Be precise about when to use a 4×4 dihybrid Punnett square (two independently assorting gene pairs) versus a 2×2 (one gene pair or linked genes). Distinguish incomplete dominance (blended phenotype, three phenotypic classes from F1 × F1) from codominance (both alleles expressed, e.g. A and B blood types). For sex-linked problems, always write X-linked alleles on the X chromosome symbol. For pedigrees, use the carrier female symbol correctly and apply the rule of parsimony when choosing the simplest inheritance pattern.精准判断何时用 4×4 双杂交旁氏表（两对独立分配基因）与 2×2（一对基因或连锁基因）。区分不完全显性（混合表现型，F1×F1 产生三类表现型）与共显性（两个等位基因均表达，如 A 型和 B 型血）。处理伴性遗传题时，始终在 X 染色体符号上标注 X 连锁等位基因。分析系谱时，正确使用携带者雌性符号，并运用简约原则选择最简单的遗传模式。

Honors flag.荣誉级标记。 Alberta places the full genetics treatment in Biology 30 (Grade 12), flagged Honors Biology 30 throughout this guide. If you are in BC Science 10 or Ontario SBI3U, all sections are required at your grade level. NGSS students need the probability and variation framing (HS-LS3-2, HS-LS3-3) but not the named Mendelian laws as a memorized list.阿尔伯塔将完整遗传学内容置于 Biology 30（12 年级），本指南全程标注 荣誉 Biology 30。BC Science 10 或安大略 SBI3U 的学生在其年级均需学习全部章节。NGSS 学生需要概率与变异框架（HS-LS3-2、HS-LS3-3），但不需要将孟德尔命名定律作为记忆列表。
Section 1 · Mendel and genetics terminology第 1 节 · 孟德尔与遗传学术语

Mendel and Core Genetics Vocabulary孟德尔与核心遗传学词汇

Eight terms you must know cold before doing any cross.做任何遗传交叉之前必须熟记的八个术语。
  • Gene:基因： a segment of DNA on a chromosome that codes for a specific trait. Each gene occupies a specific locus.染色体上编码特定性状的 DNA 片段。每个基因占据特定的基因座。
  • Allele:等位基因： alternative versions of a gene. Represented by letters (e.g. A vs a). Diploid organisms carry two alleles per gene locus — one from each parent.基因的不同版本。用字母表示（如 A 与 a）。二倍体生物每个基因座携带两个等位基因——各来自一个亲本。
  • Genotype:基因型： the allele combination an organism carries. E.g. AA, Aa, aa.生物体携带的等位基因组合。如 AA、Aa、aa。
  • Phenotype:表现型： the observable trait produced by the genotype. E.g. "tall" or "short" in pea plants.基因型产生的可观察性状。如豌豆中的"高茎"或"矮茎"。
  • Dominant:显性： the allele whose phenotype is expressed when paired with any other allele. Written with an uppercase letter (A).与任何其他等位基因配对时其表现型均可表达的等位基因。用大写字母（A）表示。
  • Recessive:隐性： the allele whose phenotype is only expressed in the homozygous state (aa). Written with a lowercase letter (a).仅在纯合状态（aa）下其表现型才能表达的等位基因。用小写字母（a）表示。
  • Homozygous:纯合子： both alleles are the same (AA or aa). A homozygous dominant (AA) and a homozygous recessive (aa) both breed true.两个等位基因相同（AA 或 aa）。显性纯合子（AA）与隐性纯合子（aa）均可稳定遗传。
  • Heterozygous:杂合子： the two alleles differ (Aa). A heterozygous individual shows the dominant phenotype but carries the recessive allele.两个等位基因不同（Aa）。杂合子个体表现显性表现型但携带隐性等位基因。
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Mendel's experimental approach孟德尔的实验方法

Gregor Mendel (1822–1884) cross-pollinated pea plants (Pisum sativum) over eight years, tracking seven discrete traits (seed shape, seed colour, pod shape, pod colour, flower colour, flower position, plant height). He chose pea plants because they: (1) had clearly distinct, discontinuous trait variants; (2) could be self-fertilized to produce true-breeding lines; (3) had a short generation time. His P generation were true-breeding parents; F1 was the first filial (hybrid) generation; F2 was produced by crossing F1 individuals. The 3:1 F2 phenotype ratio in monohybrid crosses was the key numerical pattern he explained with "units of heredity" (now called alleles).格雷戈尔·孟德尔（1822–1884）历时八年对豌豆（Pisum sativum）进行杂交授粉，追踪七个离散性状（种子形状、种子颜色、豆荚形状、豆荚颜色、花色、花位置、株高）。他选择豌豆是因为：(1) 性状变异明显且不连续；(2) 可自花传粉产生稳定遗传系；(3) 世代时间短。P 代为纯系亲本；F1 为第一子代（杂合代）；F2 由 F1 个体杂交产生。单杂交中 F2 表现型 3:1 比是他用"遗传单元"（现称等位基因）解释的关键数字规律。

Worked Example 1 · Identifying genotypes from descriptions例题 1 · 从描述中识别基因型

In pea plants, tall (T) is dominant over short (t). A pea plant is tall and breeds true when self-fertilized. Write its genotype, phenotype, and state whether it is homozygous or heterozygous.在豌豆中，高茎（T）对矮茎（t）显性。一株豌豆为高茎，自花传粉时稳定遗传。写出其基因型、表现型，并说明是纯合子还是杂合子。

Genotype: TT.基因型：TT。 "Breeds true" means all offspring match the parent — only possible if both alleles are the same dominant allele. A heterozygous Tt parent crossed with itself would produce 1/4 tt (short) offspring."稳定遗传"意味着所有后代与亲本相同——只有两个等位基因均为显性时才可能实现。杂合子 Tt 亲本自交会产生 1/4 tt（矮茎）后代。

Phenotype: Tall.表现型：高茎。 Homozygous dominant.显性纯合子。

A pea plant has the genotype Tt, where T (tall) is dominant over t (short). What is its phenotype?一株豌豆的基因型为 Tt，其中 T（高茎）对 t（矮茎）显性。其表现型是什么？
§1 · Q1
Short, because it carries the recessive allele矮茎，因为它携带隐性等位基因
A blend of tall and short高茎与矮茎的混合
Tall, because T is dominant and masks t高茎，因为 T 是显性且掩盖 t
Cannot be determined without knowing the parent genotypes不知道亲本基因型就无法确定
In complete dominance, one dominant allele (T) is sufficient to produce the dominant phenotype. The recessive allele (t) is masked whenever T is present. Blending only occurs in incomplete dominance (§5).在完全显性中，一个显性等位基因（T）足以产生显性表现型。隐性等位基因（t）在 T 存在时被掩盖。混合仅发生于不完全显性（§5）。
T is dominant over t, so Tt is tall. Blending would require incomplete dominance. Parental genotypes are not needed once we know the organism's own genotype.T 对 t 显性，故 Tt 为高茎。混合需要不完全显性。一旦知道生物体自身基因型，就不需要亲本基因型。
Which genotype is heterozygous?哪个基因型是杂合子？
§1 · Q2
TTTT
tttt
TTTTTTTT
TtTt
Heterozygous means the two alleles at a locus differ. Tt has one dominant (T) and one recessive (t) allele. TT and tt are both homozygous (same allele at both positions).杂合子是指一个基因座的两个等位基因不同。Tt 有一个显性（T）和一个隐性（t）等位基因。TT 和 tt 均为纯合子（两个位置等位基因相同）。
Heterozygous = two different alleles at the same locus. Only Tt satisfies this. TT and tt are homozygous. TTTT would be tetraploid, not standard diploid genetics.杂合子 = 同一基因座两个不同等位基因。只有 Tt 符合。TT 和 tt 是纯合子。TTTT 是四倍体，不是标准二倍体遗传学。
Section 2 · Monohybrid crosses and Punnett squares第 2 节 · 单杂交与旁氏表

Monohybrid Crosses and Punnett Squares单杂交与旁氏表

How to set up a Punnett square in four steps.旁氏表四步设置法。
  1. Write the gametes of Parent 1 along the top (one allele per gamete).将亲本 1 的配子写在顶部（每个配子一个等位基因）。
  2. Write the gametes of Parent 2 along the left side.将亲本 2 的配子写在左侧。
  3. Fill each cell by combining one top allele with one side allele.将顶部等位基因与侧部等位基因组合，填写每个格子。
  4. Read off the genotype ratio and infer the phenotype ratio.读取基因型比并推断表现型比。

Classic F1 × F1 monohybrid result (Aa × Aa):经典 F1×F1 单杂交结果（Aa×Aa）：

  • Genotype ratio: 1 AA : 2 Aa : 1 aa基因型比：1 AA : 2 Aa : 1 aa
  • Phenotype ratio: 3 dominant : 1 recessive (because AA and Aa both show the dominant phenotype)表现型比：3 显性 : 1 隐性（因为 AA 和 Aa 均表现显性表现型）

Test cross:测交： crossing an organism of unknown genotype with a homozygous recessive (aa). If any offspring show the recessive phenotype, the unknown parent must be heterozygous (Aa). If all offspring show the dominant phenotype, the unknown parent is homozygous dominant (AA).将基因型未知的生物与隐性纯合子（aa）杂交。若任何后代表现隐性表现型，则未知亲本必为杂合子（Aa）。若所有后代均表现显性表现型，则未知亲本为显性纯合子（AA）。

Worked Example 2 · Monohybrid cross with test cross例题 2 · 单杂交与测交

In guinea pigs, black coat (B) is dominant over white coat (b). A black guinea pig is test-crossed with a white guinea pig. Of 20 offspring, 9 are black and 11 are white. (a) What is the genotype of the black parent? (b) What offspring ratio did Mendel predict for this type of cross?豚鼠中，黑色毛（B）对白色毛（b）显性。一只黑色豚鼠与一只白色豚鼠测交。20 只后代中，9 只为黑色，11 只为白色。(a) 黑色亲本的基因型是什么？(b) 孟德尔预测这类杂交的后代比例是什么？

(a) The white guinea pig is bb (homozygous recessive). White offspring appeared, so the black parent contributed a b allele to some offspring. Therefore, the black parent is Bb (heterozygous). A BB parent could never produce white offspring in a test cross.(a) 白色豚鼠为 bb（隐性纯合子）。出现了白色后代，说明黑色亲本向部分后代贡献了 b 等位基因。因此，黑色亲本为 Bb（杂合子）。BB 亲本在测交中绝不会产生白色后代。

(b) The predicted ratio for Bb × bb is 1 Bb : 1 bb, or 1 black : 1 white (50:50). The observed 9:11 is close to this 1:1 expectation; the deviation from exactly 10:10 is due to chance sampling variation.(b) Bb×bb 的预测比为 1 Bb : 1 bb，即 1 黑色 : 1 白色（50:50）。观察到的 9:11 接近此 1:1 期望；与恰好 10:10 的偏差是由于随机抽样变异。

Two heterozygous tall pea plants (Tt × Tt) are crossed. What fraction of the offspring will be short (tt)?两株杂合高茎豌豆（Tt×Tt）杂交。后代中矮茎（tt）占多少？
§2 · Q1
1/4 (25%)1/4（25%）
1/2 (50%)1/2（50%）
3/4 (75%)3/4（75%）
0 (0%) — short is impossible from tall parents0（0%）— 高茎亲本不可能产生矮茎后代
Punnett square for Tt × Tt gives TT : Tt : tt = 1:2:1. Only tt shows the short phenotype, so 1/4 of offspring are short. Short offspring CAN appear from two heterozygous tall parents because each carries a hidden t allele.Tt×Tt 的旁氏表给出 TT:Tt:tt = 1:2:1。只有 tt 表现矮茎表现型，故 1/4 后代为矮茎。两株杂合高茎亲本确实可以产生矮茎后代，因为每株都携带隐藏的 t 等位基因。
Tt × Tt gives 1 TT : 2 Tt : 1 tt. The tt fraction is 1/4. Short offspring can come from heterozygous tall parents because each carries the recessive t.Tt×Tt 产生 1 TT:2 Tt:1 tt。tt 比例为 1/4。矮茎后代可来自杂合高茎亲本，因为每株都携带隐性 t。
A test cross is performed between an organism of unknown genotype and a homozygous recessive (aa). All 40 offspring show the dominant phenotype. What is the most likely genotype of the unknown parent?将基因型未知的生物与隐性纯合子（aa）测交。40 只后代全部表现显性表现型。未知亲本最可能的基因型是什么？
§2 · Q2
aaaa
AaAa
AAAA
Cannot be determined无法确定
AA × aa produces all Aa offspring (dominant phenotype). Aa × aa produces 1/2 Aa : 1/2 aa, so roughly half would show the recessive phenotype. With 40 offspring all dominant, AA is the most likely genotype (though not proven with absolute certainty).AA×aa 产生全部 Aa 后代（显性表现型）。Aa×aa 产生 1/2 Aa:1/2 aa，约一半会表现隐性表现型。40 只后代全为显性，AA 是最可能的基因型（尽管不能绝对确定）。
If the unknown were Aa, roughly half of 40 offspring would be recessive (aa). Since all 40 are dominant, AA is the most likely conclusion from a test cross.若未知亲本为 Aa，约一半后代（aa）会表现隐性。由于全部 40 只为显性，测交结论最可能是 AA。
Section 3 · Laws of segregation and independent assortment第 3 节 · 分离定律与自由组合定律

Mendel's Two Laws of Inheritance孟德尔遗传的两大定律

Two laws — know the mechanism behind each.两大定律 — 掌握各自背后的机制。
  • Law of Segregation:分离定律： each organism carries two alleles for each trait. During gamete formation (meiosis), the two alleles separate so that each gamete carries only one allele. Which allele goes into which gamete is random. This is why Aa produces 1/2 A gametes and 1/2 a gametes.每个生物体的每个性状携带两个等位基因。配子形成（减数分裂）期间，两个等位基因分离，每个配子只携带一个等位基因。哪个等位基因进入哪个配子是随机的。这就是为什么 Aa 产生 1/2 A 配子和 1/2 a 配子。
  • Law of Independent Assortment:自由组合定律： alleles for different genes (on non-homologous chromosomes) segregate independently of each other during meiosis. Knowing what allele a gamete carries for Gene 1 tells you nothing about which allele it carries for Gene 2. This law applies only to genes on different chromosomes (or far apart on the same chromosome) — linked genes violate it.不同基因（位于非同源染色体上）的等位基因在减数分裂过程中独立地相互分离。知道配子携带基因 1 的哪个等位基因，并不能告诉你它携带基因 2 的哪个等位基因。该定律仅适用于不同染色体上（或同一染色体上相距较远）的基因——连锁基因违反该定律。

Meiosis connection:与减数分裂的关联： Segregation occurs at Meiosis I (homologous chromosomes separate). Independent assortment occurs because each homologous pair lines up randomly at the metaphase plate — which pole the maternal vs. paternal chromosome of one pair goes to is independent of every other pair.分离发生于减数第一次分裂（同源染色体分离）。自由组合之所以发生，是因为每对同源染色体在分裂中期板上随机排列——一对中的母方或父方染色体去哪一极，与其他每对无关。

T
Mendel's original ratios — the numbers that revealed the laws孟德尔的原始比例 — 揭示定律的数据
Cross type杂交类型Genotype ratio (F2)基因型比（F2）Phenotype ratio (F2)表现型比（F2）
Monohybrid (Aa × Aa)单杂交（Aa×Aa）1 AA : 2 Aa : 1 aa3 dominant : 1 recessive3 显性 : 1 隐性
Test cross (Aa × aa)测交（Aa×aa）1 Aa : 1 aa1 dominant : 1 recessive1 显性 : 1 隐性
Dihybrid (AaBb × AaBb)双杂交（AaBb×AaBb）9 combinations9 种组合9:3:3:1 (four phenotypic classes)9:3:3:1（四种表现型类别）
Worked Example 3 · Applying the Law of Segregation例题 3 · 应用分离定律

Mendel crossed true-breeding round seeds (RR) with true-breeding wrinkled seeds (rr). The F1 were all round (Rr). He then self-fertilized the F1 plants. Predict the F2 genotype and phenotype ratios. Explain the result using the Law of Segregation.孟德尔将纯系圆形种子（RR）与纯系皱形种子（rr）杂交。F1 全为圆形（Rr）。然后他将 F1 植株自花传粉。预测 F2 的基因型和表现型比，并用分离定律解释结果。

F1 gametes:F1 配子： Rr segregates into 1/2 R gametes and 1/2 r gametes (Law of Segregation).Rr 分离产生 1/2 R 配子和 1/2 r 配子（分离定律）。

F2 genotype ratio: 1 RR : 2 Rr : 1 rr.F2 基因型比：1 RR : 2 Rr : 1 rr。

F2 phenotype ratio: 3 round (RR + Rr) : 1 wrinkled (rr).F2 表现型比：3 圆形（RR + Rr）: 1 皱形（rr）。 The wrinkled phenotype reappears in F2 because the recessive r allele, hidden in F1 Rr plants, segregates back out. Mendel counted 5,474 round and 1,850 wrinkled seeds in this actual experiment — a ratio of 2.96:1, close to the predicted 3:1.皱形表现型在 F2 重新出现，是因为隐藏在 F1 Rr 植株中的隐性 r 等位基因重新分离出来。孟德尔在这一实际实验中计数了 5,474 粒圆形和 1,850 粒皱形种子——比例为 2.96:1，接近预测的 3:1。

The Law of Segregation states that alleles for a trait separate during the formation of gametes. At which stage of meiosis does this occur?分离定律指出，性状的等位基因在配子形成时分离。这发生在减数分裂的哪个阶段？
§3 · Q1
During fertilization, when egg and sperm fuse受精时，当卵细胞与精子融合
During Meiosis I, when homologous chromosomes separate减数第一次分裂时，同源染色体分离
During DNA replication in interphase间期 DNA 复制时
During Meiosis II, when sister chromatids separate减数第二次分裂时，姐妹染色单体分离
Segregation of alleles (homologous chromosome pairs) occurs at Meiosis I. Each homolog carries one allele for a gene; when they separate, each daughter cell gets only one allele. Meiosis II separates sister chromatids (which are copies of one allele, not different alleles).等位基因（同源染色体对）的分离发生在减数第一次分裂时。每条同源染色体携带一个基因的一个等位基因；当它们分离时，每个子细胞只获得一个等位基因。减数第二次分裂分离姐妹染色单体（它们是同一等位基因的拷贝，而非不同等位基因）。
Meiosis I is when homologous chromosomes (carrying different alleles) separate — this is the physical basis of the Law of Segregation.减数第一次分裂时同源染色体（携带不同等位基因）分离——这是分离定律的物理基础。
The Law of Independent Assortment applies ONLY when genes are located on:自由组合定律仅适用于基因位于以下情况：
§3 · Q2
The same chromosome, close together同一染色体上，彼此相邻
The X chromosome only仅在 X 染色体上
Autosomes only, never sex chromosomes仅在常染色体上，从不在性染色体上
Different (non-homologous) chromosomes, or far apart on the same chromosome不同（非同源）染色体上，或同一染色体上相距较远
Independent assortment requires that the two gene pairs are on different chromosomes (non-homologous) or far enough apart on the same chromosome that recombination makes them behave independently. Genes close together on the same chromosome are "linked" and do not assort independently.自由组合要求两对基因位于不同染色体（非同源）上，或在同一染色体上相距足够远使得重组让它们独立行为。同一染色体上靠近的基因是"连锁"的，不独立分配。
Independent assortment applies when genes are on different chromosomes or far apart on the same one. Closely linked genes do not assort independently and violate this law.自由组合定律适用于基因位于不同染色体或同一染色体上相距较远的情况。紧密连锁的基因不独立分配，违反该定律。
Section 4 · Dihybrid crosses第 4 节 · 双杂交

Dihybrid Crosses and the 9:3:3:1 Ratio双杂交与 9:3:3:1 比

A dihybrid cross tracks two gene pairs simultaneously.双杂交同时追踪两对基因。
  • Set-up: use a 4×4 Punnett square. Each parent (AaBb) produces four types of gametes: AB, Ab, aB, ab. Each in equal proportions (1/4 each), because A/a and B/b assort independently.设置：使用 4×4 旁氏表。每个亲本（AaBb）产生四种配子：AB、Ab、aB、ab，各占 1/4（因为 A/a 和 B/b 独立分配）。
  • F2 phenotype ratio (AaBb × AaBb, two independently assorting genes):F2 表现型比（AaBb×AaBb，两对独立分配基因）：
    • 9 A_B_ (dominant for both)9 A_B_（两者均显性）
    • 3 A_bb (dominant A, recessive B)3 A_bb（A 显性，B 隐性）
    • 3 aaB_ (recessive A, dominant B)3 aaB_（A 隐性，B 显性）
    • 1 aabb (recessive for both)1 aabb（两者均隐性）
  • Shortcut: multiply the monohybrid probabilities. P(A_) = 3/4; P(aa) = 1/4; P(B_) = 3/4; P(bb) = 1/4. So P(A_B_) = 3/4 × 3/4 = 9/16, etc.捷径：将单杂交概率相乘。P(A_) = 3/4；P(aa) = 1/4；P(B_) = 3/4；P(bb) = 1/4。故 P(A_B_) = 3/4×3/4 = 9/16，等等。
Worked Example 4 · Dihybrid cross in pea plants例题 4 · 豌豆双杂交

In peas, round seed shape (R) is dominant over wrinkled (r), and yellow seed colour (Y) is dominant over green (y). Two dihybrid plants (RrYy × RrYy) are crossed. Predict the fraction of offspring that will have round, green seeds.在豌豆中，圆形种子（R）对皱形（r）显性，黄色种子（Y）对绿色（y）显性。两株双杂合植株（RrYy×RrYy）杂交。预测后代中圆形绿色种子的比例。

Use the product rule (genes assort independently):用乘积法则（基因独立分配）：

P(round, R_) = 3/4 (from Rr × Rr monohybrid)P(圆形, R_) = 3/4（来自 Rr×Rr 单杂交）

P(green, yy) = 1/4 (from Yy × Yy monohybrid)P(绿色, yy) = 1/4（来自 Yy×Yy 单杂交）

P(round AND green) = 3/4 × 1/4 = 3/16.P(圆形且绿色) = 3/4×1/4 = 3/16。 This is the "3 A_bb" class from the 9:3:3:1 ratio. Mendel confirmed this ratio experimentally (315:108:101:32 ≈ 9:3:3:1 from 556 seeds).这是 9:3:3:1 比中的"3 A_bb"类。孟德尔通过实验证实了这一比例（556 粒种子中 315:108:101:32 ≈ 9:3:3:1）。

In a dihybrid cross AaBb × AaBb (A and B independently assorting), what fraction of offspring will be homozygous recessive for BOTH genes (aabb)?在双杂交 AaBb×AaBb（A 和 B 独立分配）中，两个基因均为隐性纯合子（aabb）的后代比例是多少？
§4 · Q1
3/163/16
1/161/16
9/169/16
1/41/4
P(aa) = 1/4 from Aa × Aa. P(bb) = 1/4 from Bb × Bb. P(aabb) = 1/4 × 1/4 = 1/16. This is the "1 aabb" class in the 9:3:3:1 ratio, representing the doubly recessive phenotypic class.Aa×Aa 中 P(aa) = 1/4。Bb×Bb 中 P(bb) = 1/4。P(aabb) = 1/4×1/4 = 1/16。这是 9:3:3:1 比中的"1 aabb"类，代表双隐性表现型类别。
Multiply the two independent probabilities: P(aa) = 1/4 and P(bb) = 1/4. Product = 1/16. The 9:3:3:1 ratio has 1/16 in the doubly recessive class.将两个独立概率相乘：P(aa) = 1/4，P(bb) = 1/4。乘积 = 1/16。9:3:3:1 比中双隐性类别占 1/16。
The 9:3:3:1 phenotype ratio from a dihybrid cross assumes which condition?双杂交产生的 9:3:3:1 表现型比假设满足哪个条件？
§4 · Q2
The two genes assort independently (are on different chromosomes or far apart)两对基因独立分配（位于不同染色体或相距较远）
Both genes show incomplete dominance两个基因均表现不完全显性
The two genes are tightly linked on the same chromosome两个基因紧密连锁在同一染色体上
One gene epistatic to the other一个基因对另一个具有上位性
The 9:3:3:1 ratio depends on the Law of Independent Assortment — the two gene pairs must be on different chromosomes (or far apart on the same chromosome). Tightly linked genes produce ratios closer to 3:1. Epistasis and incomplete dominance alter phenotypic ratios from 9:3:3:1 in different ways.9:3:3:1 比依赖于自由组合定律——两对基因必须位于不同染色体上（或同一染色体上相距较远）。紧密连锁基因产生的比例更接近 3:1。上位性和不完全显性以不同方式改变 9:3:3:1 的表现型比。
The 9:3:3:1 ratio requires independent assortment (different chromosomes). Linkage, incomplete dominance, and epistasis all deviate from this ratio.9:3:3:1 比需要独立分配（不同染色体）。连锁、不完全显性和上位性均偏离此比例。
Section 5 · Incomplete dominance and codominance第 5 节 · 不完全显性与共显性

Incomplete Dominance and Codominance不完全显性与共显性

When neither allele completely dominates the other.当两个等位基因均不完全支配另一个时。
  • Incomplete dominance:不完全显性： the heterozygote shows an intermediate (blended) phenotype. Neither allele is fully dominant. Example: red snapdragon (CRCR) × white snapdragon (CWCW) → F1 all pink (CRCW). F1 × F1 gives 1 red : 2 pink : 1 white (three phenotypic classes).杂合子表现出中间（混合）表现型。两个等位基因均非完全显性。例：红花金鱼草（CRCR）×白花金鱼草（CWCW）→ F1 全为粉花（CRCW）。F1×F1 产生 1 红:2 粉:1 白（三种表现型类别）。
  • Codominance:共显性： both alleles are fully expressed simultaneously in the heterozygote. There is no blending — both phenotypes are visible side by side. Example: roan cattle (CRCW = red and white hairs mixed in the same coat). ABO blood group M/N antigens are also codominant.杂合子中两个等位基因同时完全表达。没有混合——两种表现型并排可见。例：沙毛牛（CRCW = 红色与白色毛混于同一皮毛中）。ABO 血型 M/N 抗原也是共显性的。
Feature特征Complete dominance完全显性Incomplete dominance不完全显性Codominance共显性
Heterozygote phenotype杂合子表现型Identical to dominant homozygote与显性纯合子相同Intermediate / blend中间 / 混合Both alleles visible simultaneously两个等位基因同时可见
F1×F1 phenotype ratioF1×F1 表现型比3:11:2:11:2:1
Example例子Pea tall/short豌豆高茎/矮茎Snapdragon red/pink/white金鱼草红/粉/白ABO blood types (IAIB = type AB)ABO 血型（IAIB = AB 型）
Worked Example 5 · Incomplete dominance cross例题 5 · 不完全显性杂交

In four o'clock flowers, red (CRCR) crossed with white (CWCW) gives all pink F1 (CRCW). Two pink F1 plants are crossed. Predict the F2 genotype ratio, F2 phenotype ratio, and explain why the F2 ratio differs from a complete dominance monohybrid cross.茉莉花中，红花（CRCR）×白花（CWCW）产生全为粉花的 F1（CRCW）。两株粉花 F1 植株杂交。预测 F2 的基因型比与表现型比，并解释为何 F2 比例与完全显性单杂交不同。

F2 genotype ratio: 1 CRCR : 2 CRCW : 1 CWCWF2 基因型比：1 CRCR : 2 CRCW : 1 CWCW (same as any monohybrid cross).（与任何单杂交相同）。

F2 phenotype ratio: 1 red : 2 pink : 1 white.F2 表现型比：1 红 : 2 粉 : 1 白。

In complete dominance, CRCR and CRCW are indistinguishable in phenotype (both appear dominant), giving a 3:1 ratio. In incomplete dominance, each genotype class has its own distinct phenotype, so the 1:2:1 genotype ratio is also the phenotype ratio. This demonstrates that the alleles do not fully mask each other.在完全显性中，CRCR 和 CRCW 在表现型上无法区分（均表现显性），产生 3:1 比。在不完全显性中，每种基因型类别有其自己的独特表现型，因此 1:2:1 的基因型比也是表现型比。这表明等位基因不能完全掩盖彼此。

A person with blood type AB has genotype IAIB. Their red blood cells display both A and B antigens on the surface. This is an example of:AB 型血的人基因型为 IAIB。其红细胞表面同时显示 A 和 B 抗原。这是一个什么例子？
§5 · Q1
Complete dominance完全显性
Incomplete dominance不完全显性
A mutation in the ABO geneABO 基因的突变
Codominance共显性
Both IA and IB alleles are fully expressed simultaneously (both A and B antigens are present). This is codominance: no blending, both phenotypes visible. Incomplete dominance would give a single intermediate phenotype (e.g. a blend), not two distinct antigens side by side.IA 和 IB 等位基因同时完全表达（A 和 B 抗原均存在）。这是共显性：无混合，两种表现型均可见。不完全显性会产生单一中间表现型（如混合），而非两种不同抗原并列。
Both alleles fully expressed = codominance. Incomplete dominance would produce a single blend. Complete dominance would hide one allele.两个等位基因完全表达 = 共显性。不完全显性会产生单一混合表现型。完全显性会隐藏一个等位基因。
Two pink snapdragons (CRCW × CRCW, incomplete dominance) are crossed. What proportion of offspring will be pink?两株粉花金鱼草（CRCW×CRCW，不完全显性）杂交。后代中粉花占多少比例？
§5 · Q2
3/43/4
1/41/4
1/2 (2/4)1/2（2/4）
All offspring (4/4)全部后代（4/4）
CRCW × CRCW gives 1 CRCR (red) : 2 CRCW (pink) : 1 CWCW (white). Pink = CRCW = 2/4 = 1/2 of offspring.CRCW×CRCW 产生 1 CRCR（红）:2 CRCW（粉）:1 CWCW（白）。粉花 = CRCW = 2/4 = 1/2 后代。
Punnett: CRCW × CRCW gives ratio 1:2:1 (red:pink:white). Pink fraction = 2/4 = 1/2.旁氏表：CRCW×CRCW 产生比例 1:2:1（红:粉:白）。粉花比例 = 2/4 = 1/2。
Section 6 · Multiple alleles and sex-linked inheritance第 6 节 · 复等位基因与伴性遗传

Multiple Alleles and Sex-Linked Inheritance复等位基因与伴性遗传

Beyond two alleles — and genes on sex chromosomes.超越两个等位基因——以及性染色体上的基因。
  • Multiple alleles:复等位基因： a gene can have more than two alleles in the population, even though any individual diploid organism carries only two. ABO blood group is the classic example: three alleles (IA, IB, i) produce four phenotypes (A, B, AB, O). IA and IB are codominant; both are dominant over i (recessive). Result: six possible genotypes, four phenotypes.一个基因在种群中可有两个以上的等位基因，尽管任何二倍体个体只携带两个。ABO 血型是经典例子：三个等位基因（IA、IB、i）产生四种表现型（A、B、AB、O）。IA 和 IB 共显性；两者均对 i（隐性）显性。结果：六种可能的基因型，四种表现型。
  • Sex determination:性别决定： human females are XX; human males are XY. The Y chromosome is smaller and carries fewer genes. Eggs always carry X; sperm carry either X (produces daughter) or Y (produces son).人类女性为 XX；人类男性为 XY。Y 染色体较小，携带基因较少。卵细胞总携带 X；精子携带 X（产生女儿）或 Y（产生儿子）。
  • X-linked inheritance:X 连锁遗传（伴性遗传）： genes on the X chromosome with no counterpart on Y. Males (XY) are hemizygous — they have only one copy of each X-linked allele, so even recessive alleles are expressed. Classic X-linked recessive traits: colour blindness (XcXc or XcY), haemophilia A (XhY). A carrier female (XCXc) shows the dominant phenotype but can pass the recessive allele to sons. Affected fathers cannot pass X-linked alleles to sons (sons get Y from father).X 染色体上在 Y 染色体上没有对应位点的基因。男性（XY）是半合子——每个 X 连锁等位基因只有一份拷贝，因此即使是隐性等位基因也会表达。经典 X 连锁隐性性状：色盲（XcXc 或 XcY）、血友病 A（XhY）。携带者女性（XCXc）表现显性表现型，但可将隐性等位基因传给儿子。患病父亲不能将 X 连锁等位基因传给儿子（儿子从父亲获得 Y）。
AB
ABO blood group genotypes and phenotypesABO 血型基因型与表现型
Genotype基因型Blood type (phenotype)血型（表现型）Antigens on RBC红细胞上的抗原
IAIA or IAiAA 型A antigenA 抗原
IBIB or IBiBB 型B antigenB 抗原
IAIBAB (codominant)AB 型（共显性）Both A and B antigensA 和 B 两种抗原
iiOO 型Neither均无
Worked Example 6 · X-linked colour blindness cross例题 6 · X 连锁色盲杂交

A carrier woman (XCXc) and a man with normal vision (XCY) have children. What is the probability that a son is colour blind? What is the probability that a daughter is a carrier?一位携带者女性（XCXc）与一位视觉正常的男性（XCY）育有孩子。儿子色盲的概率是多少？女儿是携带者的概率是多少？

Cross: XCXc × XCY. Gametes: mother produces XC or Xc eggs; father produces XC or Y sperm.杂交：XCXc×XCY。配子：母亲产生 XC 或 Xc 卵细胞；父亲产生 XC 或 Y 精子。

Offspring: XCXC (normal female, 1/4), XcXC (carrier female, 1/4), XCY (normal male, 1/4), XcY (colour-blind male, 1/4).后代：XCXC（正常女性，1/4）、XcXC（携带者女性，1/4）、XCY（正常男性，1/4）、XcY（色盲男性，1/4）。

P(son colour blind) = 1/2 of all sons = 50%.P(儿子色盲) = 所有儿子中的 1/2 = 50%。 P(daughter is carrier) = 1/2 of all daughters = 50%.P(女儿为携带者) = 所有女儿中的 1/2 = 50%。

Key rule: sons receive their X only from their mother. So any son who inherits the Xc allele from his carrier mother will be colour blind (no second X to mask it).关键规则：儿子的 X 只来自母亲。因此，任何从携带者母亲遗传到 Xc 等位基因的儿子都会色盲（没有第二条 X 来掩盖它）。

A colour-blind man (XcY) and a woman with normal vision who is NOT a carrier (XCXC) have children. Which statement about their offspring is correct?一位色盲男性（XcY）与一位视觉正常且非携带者的女性（XCXC）育有孩子。关于他们的后代，下列哪项正确？
§6 · Q1
All daughters will be carriers; no sons will be colour blind所有女儿都将是携带者；没有儿子会色盲
Half the sons will be colour blind; half the daughters will be carriers一半儿子将色盲；一半女儿将是携带者
All sons will be colour blind所有儿子都将色盲
All daughters will be colour blind所有女儿都将色盲
Cross: XCXC × XcY. All daughters receive XC from dad and XC from mum — wait, that is wrong: daughters get Xc from dad and XC from mum, so all daughters are XCXc (carriers, normal vision). Sons get Y from dad and XC from mum, so all sons are XCY (normal vision).杂交：XCXC×XcY。所有女儿从父亲获得 Xc，从母亲获得 XC，因此所有女儿均为 XCXc（携带者，视觉正常）。儿子从父亲获得 Y，从母亲获得 XC，因此所有儿子均为 XCY（视觉正常）。
XCXC × XcY: daughters = XCXc (all carriers, not colour blind); sons = XCY (all normal). No offspring are colour blind in this generation.XCXC×XcY：女儿 = XCXc（全为携带者，不色盲）；儿子 = XCY（全部正常）。这一代无色盲后代。
Two parents have blood types A (genotype IAi) and B (genotype IBi). What blood types are possible among their offspring?两位亲本的血型分别为 A 型（基因型 IAi）和 B 型（基因型 IBi）。其后代可能出现哪些血型？
§6 · Q2
A and B only仅 A 型和 B 型
A, B, AB, and OA、B、AB 和 O 型
AB only仅 AB 型
A, B, and O only仅 A、B 和 O 型
IAi × IBi gives offspring: IAIB (AB), IAi (A), IBi (B), ii (O) — all four blood types in a 1:1:1:1 ratio. This is why two parents with A and B blood types can have a child with type O (both are carriers of the recessive i allele).IAi×IBi 产生后代：IAIB（AB）、IAi（A）、IBi（B）、ii（O）——四种血型按 1:1:1:1 比出现。这就是为什么 A 型和 B 型血的父母可以生出 O 型血孩子（两者均是隐性 i 等位基因的携带者）。
IAi × IBi produces IAIB (AB), IAi (A), IBi (B), and ii (O). All four blood types are possible in equal proportions.IAi×IBi 产生 IAIB（AB）、IAi（A）、IBi（B）和 ii（O）。四种血型均等可能出现。
Section 7 · Pedigree analysis第 7 节 · 系谱分析

Pedigree Analysis系谱分析

Reading a pedigree — symbols and rules.读懂系谱——符号与规则。
  • Symbols:符号： squares = males; circles = females; filled = affected (shows the trait); half-filled = carrier (e.g. X-linked carrier female); horizontal line between a square and circle = mating; vertical line = offspring; horizontal line connecting siblings.方框 = 男性；圆圈 = 女性；实心 = 患者（表现该性状）；半实心 = 携带者（如 X 连锁携带者女性）；方框与圆圈之间的水平线 = 婚配；垂直线 = 后代；连接兄弟姐妹的水平线。
  • Autosomal recessive:常染色体隐性： affected individuals can have unaffected parents (both carriers); trait skips generations; affects both sexes equally; two affected parents always produce affected offspring.患者可有未患病的亲本（均为携带者）；性状可跳代；男女均等受影响；两个患病亲本的后代总是患病。
  • Autosomal dominant:常染色体显性： at least one parent of an affected individual is usually affected; trait appears in every generation; unaffected individuals do not carry the allele (in simple dominant inheritance).患者的至少一个亲本通常也患病；性状出现在每一代；未患病个体不携带该等位基因（在简单显性遗传中）。
  • X-linked recessive:X 连锁隐性： more males affected than females; affected males often have unaffected carrier mothers; affected males cannot transmit the allele to sons; daughters of affected fathers are obligate carriers.男性受影响多于女性；患病男性通常有未患病的携带者母亲；患病男性不能将等位基因传给儿子；患病父亲的女儿必然是携带者。
Strategy for determining inheritance pattern from a pedigree.从系谱确定遗传模式的策略。
  1. Count affected individuals by sex. If many more males than females are affected, suspect X-linked.按性别计数受影响个体。若男性受影响远多于女性，怀疑 X 连锁。
  2. Check if affected individuals appear in every generation. If yes, likely dominant. If the trait skips a generation, likely recessive.检查受影响个体是否出现在每一代。若是，可能为显性。若性状跳代，可能为隐性。
  3. Check unaffected parent combinations. Two unaffected parents with affected offspring = autosomal recessive (both parents are Aa carriers).检查未患病亲本组合。两个未患病亲本生出患病后代 = 常染色体隐性（两个亲本均为 Aa 携带者）。
  4. Apply the rule of parsimony: choose the simplest inheritance pattern that explains all affected and unaffected individuals.应用简约原则：选择能解释所有患病和未患病个体的最简单遗传模式。
Worked Example 7 · Pedigree interpretation例题 7 · 系谱解读

A pedigree shows: Generation I has an unaffected father and unaffected mother. Generation II has four children: two sons (one affected, one unaffected) and two daughters (both unaffected). Generation III shows that the affected son's children are: one affected son and two unaffected daughters. The unaffected son's children are all unaffected. Determine the most likely inheritance pattern and the genotypes of the Generation I parents.一张系谱显示：第一代有一位未患病父亲和一位未患病母亲。第二代有四个孩子：两个儿子（一个患病，一个未患病）和两个女儿（均未患病）。第三代显示：患病儿子的孩子中有一个患病儿子和两个未患病女儿。未患病儿子的孩子均未患病。确定最可能的遗传模式及第一代亲本的基因型。

Pattern clues:模式线索： Two unaffected Gen I parents have an affected son → trait is recessive. Only males are affected → suspect X-linked recessive. Affected son passes trait to a son (in Gen III) → this is NOT possible for X-linked (a son receives Y from his father), so this must be autosomal recessive, not X-linked.两个未患病的第一代亲本有一个患病儿子 → 性状为隐性。只有男性患病 → 怀疑 X 连锁隐性。患病儿子将性状传给了一个儿子（第三代）→ 这对于 X 连锁不可能（儿子从父亲获得 Y），所以这必然是常染色体隐性，而非 X 连锁。

Gen I genotypes: both parents are Aa (carriers). Affected son is aa.第一代基因型：两个亲本均为 Aa（携带者）。患病儿子为 aa。 The affected son's affected son in Gen III received one a from his father (aa) and one a from his mother (who must be Aa). The equal sex distribution of unaffected individuals across all generations is consistent with autosomal inheritance.第三代中患病儿子的患病儿子从父亲（aa）获得一个 a，从母亲（必须为 Aa）获得一个 a。所有代中未患病个体的性别分布均等，与常染色体遗传一致。

In a pedigree, two unaffected parents have an affected daughter. What inheritance pattern does this rule out?在一张系谱中，两位未患病的亲本有一个患病的女儿。这排除了哪种遗传模式？
§7 · Q1
Autosomal recessive常染色体隐性
X-linked recessiveX 连锁隐性
Autosomal dominant常染色体显性
X-linked recessive in a female女性的 X 连锁隐性
Two unaffected parents having an affected child rules out autosomal dominant (if the trait were dominant, at least one parent would show it). The affected daughter could result from autosomal recessive (parents both Aa) or X-linked recessive if her father carries XcY and her mother is XCXc.两个未患病亲本生出患病孩子，排除常染色体显性（若为显性，至少一个亲本会表现该性状）。患病女儿可能来自常染色体隐性（亲本均为 Aa）或 X 连锁隐性（若父亲为 XcY，母亲为 XCXc）。
Autosomal dominant requires at least one affected parent. If neither parent is affected, the trait cannot be autosomal dominant.常染色体显性要求至少一个亲本患病。若两个亲本均未患病，则性状不可能是常染色体显性。
Going deeper — distinguishing autosomal recessive from X-linked recessive when only males are affected深入 — 当仅男性患病时区分常染色体隐性与 X 连锁隐性

The key test: can an affected male pass the trait to his son? For X-linked recessive, the answer is no (sons receive Y from their father, not X). For autosomal recessive, the answer is yes, if the mother is a carrier (she passes an a allele). In a pedigree, if an affected male's sons are also affected, the trait cannot be X-linked — it must be autosomal. Conversely, if no sons of affected males are ever affected across many generations, X-linked is more parsimonious. Also note: an X-linked recessive condition appears in females only if they are homozygous (XcXc), requiring an affected father and a carrier or affected mother; this is rarer than the male condition.关键测试：患病男性能否将性状传给儿子？对于 X 连锁隐性，答案是不能（儿子从父亲获得 Y，而非 X）。对于常染色体隐性，答案是可以，如果母亲是携带者（她传递一个 a 等位基因）。在系谱中，如果患病男性的儿子也患病，则性状不可能是 X 连锁——必然是常染色体。相反，如果在多代中患病男性的儿子从未患病，X 连锁更为简约。还要注意：X 连锁隐性在女性中仅在纯合子（XcXc）时出现，需要患病父亲和携带者或患病母亲；这比男性病例更为罕见。

Exam Tactics考试策略

Exam Strategy and Common Pitfalls考试策略与常见陷阱

Punnett square questions旁氏表题
  • Always write gametes, not genotypes, along the margins.始终沿边界写配子，而非基因型。 Each gamete carries one allele per gene. A Aa parent produces A and a gametes (not "Aa" gametes).每个配子每个基因携带一个等位基因。Aa 亲本产生 A 和 a 配子（而非"Aa"配子）。
  • Distinguish genotype ratio from phenotype ratio.区分基因型比与表现型比。 Aa × Aa gives a 1:2:1 genotype ratio but a 3:1 phenotype ratio (under complete dominance). In incomplete dominance, genotype and phenotype ratios are both 1:2:1.Aa×Aa 给出 1:2:1 的基因型比，但在完全显性下表现型比为 3:1。在不完全显性下，基因型和表现型比均为 1:2:1。
Dominance questions显性问题
  • Incomplete dominance vs codominance: the crucial distinction.不完全显性与共显性：关键区别。 Incomplete dominance = one blended intermediate phenotype (e.g. pink, not red or white). Codominance = both original phenotypes expressed side by side (e.g. both A and B antigens present in AB blood).不完全显性 = 一种混合的中间表现型（如粉色，既非红色也非白色）。共显性 = 两种原始表现型同时并排表达（如 AB 血中同时存在 A 和 B 两种抗原）。
  • Heterozygous does not always mean "shows dominant phenotype."杂合子并不总意味着"表现显性表现型"。 In incomplete dominance and codominance, the heterozygote has its own distinctive phenotype. Reserve "dominant" and "recessive" for complete dominance.在不完全显性和共显性中，杂合子有其自己独特的表现型。将"显性"和"隐性"保留用于完全显性。
Sex-linkage questions伴性遗传题
  • Always write alleles on the X chromosome symbol.始终将等位基因写在 X 染色体符号上。 Write XCXc for a carrier female, XcY for an affected male. Never write X and Y without the allele superscript for sex-linked problems.将携带者女性写为 XCXc，患病男性写为 XcY。对于伴性遗传题，绝不要在不加等位基因上标的情况下写 X 和 Y。
  • Sons get X from mother, Y from father. Daughters get one X from each parent.儿子从母亲获得 X，从父亲获得 Y。女儿从每个亲本各获得一条 X。 This means a colour-blind father cannot have colour-blind sons (unless the mother is also a carrier or affected).这意味着色盲父亲不可能有色盲儿子（除非母亲也是携带者或患者）。
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Gene vs allele — difference?基因与等位基因 — 区别？
Gene: a DNA segment coding for a trait. Allele: a specific version of that gene (e.g. A or a). One gene can have multiple alleles in a population.基因：编码某一性状的 DNA 片段。等位基因：该基因的特定版本（如 A 或 a）。一个基因在种群中可有多个等位基因。
Genotype vs phenotype?基因型与表现型？
Genotype: the allele combination (e.g. Tt). Phenotype: the observable trait produced (e.g. tall). Same phenotype can arise from different genotypes (TT and Tt both = tall under complete dominance).基因型：等位基因组合（如 Tt）。表现型：产生的可观察性状（如高茎）。不同基因型可产生相同表现型（TT 和 Tt 在完全显性下均 = 高茎）。
Monohybrid F2 ratios (Aa × Aa)?单杂交 F2 比（Aa×Aa）？
Genotype: 1 AA : 2 Aa : 1 aa. Phenotype (complete dominance): 3 dominant : 1 recessive.基因型：1 AA : 2 Aa : 1 aa。表现型（完全显性）：3 显性 : 1 隐性。
What is a test cross and what does it reveal?什么是测交，它揭示什么？
Cross of unknown genotype × homozygous recessive (aa). If any recessive offspring appear → unknown parent is heterozygous (Aa). All dominant offspring → unknown parent is likely AA.未知基因型×隐性纯合子（aa）的杂交。若出现隐性后代 → 未知亲本为杂合子（Aa）。全为显性后代 → 未知亲本可能为 AA。
Law of Segregation?分离定律？
Each organism has two alleles per gene. They separate during meiosis so each gamete gets only one. Which goes where is random.每个生物体每个基因有两个等位基因。它们在减数分裂时分离，每个配子只获得一个。哪个去哪里是随机的。
Law of Independent Assortment?自由组合定律？
Alleles for genes on different chromosomes segregate independently. Knowing allele for Gene 1 tells you nothing about Gene 2. Applies only to non-linked genes.不同染色体上基因的等位基因独立分离。知道基因 1 的等位基因不能告诉你基因 2 的情况。仅适用于非连锁基因。
Dihybrid F2 phenotype ratio (AaBb × AaBb, independent assortment)?双杂交 F2 表现型比（AaBb×AaBb，独立分配）？
9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb. Shortcut: multiply monohybrid probabilities (3/4 × 3/4 = 9/16, etc.).9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb。捷径：将单杂交概率相乘（3/4×3/4 = 9/16，等等）。
Incomplete dominance — key feature?不完全显性 — 关键特征？
Heterozygote shows intermediate (blended) phenotype. F1×F1 gives 1:2:1 phenotype ratio (e.g. red : pink : white snapdragons).杂合子表现中间（混合）表现型。F1×F1 产生 1:2:1 表现型比（如红:粉:白金鱼草）。
Codominance — key feature?共显性 — 关键特征？
Both alleles fully expressed simultaneously. No blending. Example: IAIB blood type AB shows both A and B antigens.两个等位基因同时完全表达。无混合。例：IAIB 血型 AB 同时显示 A 和 B 抗原。
ABO blood group — alleles and dominance?ABO 血型 — 等位基因与显性关系？
Three alleles: IA, IB, i. IA and IB codominant; both dominant over i. Genotypes: IAIA or IAi = A; IBIB or IBi = B; IAIB = AB; ii = O.三个等位基因：IA、IB、i。IA 和 IB 共显性；两者均对 i 显性。基因型：IAIA 或 IAi = A；IBIB 或 IBi = B；IAIB = AB；ii = O。
X-linked recessive — why more males affected?X 连锁隐性（伴性遗传）— 为何男性受影响更多？
Males are hemizygous (XcY): one X means one copy of the allele, and even a recessive allele is expressed with no second X to mask it. Females need XcXc to be affected.男性为半合子（XcY）：一条 X 意味着一份等位基因拷贝，无第二条 X 来掩盖，即使隐性等位基因也会表达。女性需 XcXc 才会患病。
Carrier female in X-linked inheritance?X 连锁遗传中的携带者女性？
Genotype XCXc: normal phenotype but carries recessive allele. Passes Xc to 1/2 of sons (who become affected) and 1/2 of daughters (who become carriers).基因型 XCXc：表现正常但携带隐性等位基因。将 Xc 传给 1/2 的儿子（成为患者）和 1/2 的女儿（成为携带者）。
Autosomal recessive pedigree clues?常染色体隐性系谱的线索？
Affected child from two unaffected parents; trait skips generations; both sexes equally affected; two affected parents always produce affected offspring.两个未患病亲本生出患病孩子；性状跳代；男女均等受影响；两个患病亲本总是产生患病后代。
Key rule: can a colour-blind father (XcY) have colour-blind sons?关键规则：色盲父亲（XcY）能有色盲儿子吗？
Only if the mother is also a carrier (XCXc) or affected (XcXc). Sons receive Y from father, X from mother — so a son's X-linked trait always comes from his mother.仅当母亲也是携带者（XCXc）或患者（XcXc）时。儿子从父亲获得 Y，从母亲获得 X——因此儿子的 X 连锁性状总是来自母亲。
Mixed Practice综合练习

Practice Quiz综合测验

In peas, yellow seed (Y) is dominant over green (y). A cross produces offspring in a ratio of 1 yellow : 1 green. What were the parental genotypes?在豌豆中，黄色种子（Y）对绿色（y）显性。一次杂交产生比例为 1 黄色 : 1 绿色的后代。亲本基因型是什么？
Q1
YY × YYYY×YY
YY × yyYY×yy
Yy × YyYy×Yy
Yy × yyYy×yy
A 1:1 ratio of dominant : recessive offspring is the result of a test cross (Yy × yy). Yy gives Y and y gametes; yy gives only y gametes; offspring are 1/2 Yy (yellow) and 1/2 yy (green).显性:隐性后代 1:1 的比例是测交的结果（Yy×yy）。Yy 产生 Y 和 y 配子；yy 只产生 y 配子；后代为 1/2 Yy（黄色）和 1/2 yy（绿色）。
A 1:1 phenotype ratio (1 dominant : 1 recessive) comes from a test cross: Yy × yy. Yy × Yy gives 3:1; YY × yy gives all yellow; YY × YY gives all yellow.1:1 表现型比（1 显性:1 隐性）来自测交：Yy×yy。Yy×Yy 给出 3:1；YY×yy 全为黄色；YY×YY 全为黄色。
In Mendel's dihybrid cross (RrYy × RrYy), what fraction of offspring are predicted to be round AND yellow (R_Y_)?在孟德尔的双杂交（RrYy×RrYy）中，预计圆形且黄色（R_Y_）后代的比例是多少？
Q2
1/161/16
3/163/16
9/169/16
6/166/16
P(R_) = 3/4; P(Y_) = 3/4. Since R and Y assort independently, P(R_Y_) = 3/4 × 3/4 = 9/16. This is the largest class in the 9:3:3:1 ratio, as both dominant phenotypes are expressed together.P(R_) = 3/4；P(Y_) = 3/4。由于 R 和 Y 独立分配，P(R_Y_) = 3/4×3/4 = 9/16。这是 9:3:3:1 比中最大的类别，因为两种显性表现型同时表达。
Multiply independent probabilities: P(R_) = 3/4 and P(Y_) = 3/4 from each monohybrid cross. 3/4 × 3/4 = 9/16.将独立概率相乘：每个单杂交中 P(R_) = 3/4，P(Y_) = 3/4。3/4×3/4 = 9/16。
Flower colour in snapdragons shows incomplete dominance. A red plant (CRCR) is crossed with a pink plant (CRCW). What fraction of offspring will be pink?金鱼草花色表现不完全显性。一株红花植株（CRCR）与一株粉花植株（CRCW）杂交。后代中粉花比例是多少？
Q3
1/21/2
All offspring全部后代
None无
1/41/4
CRCR × CRCW gives 1/2 CRCR (red) and 1/2 CRCW (pink). No white (CWCW) offspring appear because the CRCR parent provides only CR gametes.CRCR×CRCW 产生 1/2 CRCR（红）和 1/2 CRCW（粉）。不出现白色（CWCW）后代，因为 CRCR 亲本只提供 CR 配子。
Gametes from CRCR are all CR; from CRCW are CR or CW. Offspring: 1/2 CRCR (red) + 1/2 CRCW (pink). Pink = 1/2.CRCR 的配子全为 CR；CRCW 的配子为 CR 或 CW。后代：1/2 CRCR（红）+ 1/2 CRCW（粉）。粉花 = 1/2。
Haemophilia A is X-linked recessive. A carrier mother (XHXh) and an unaffected father (XHY) have a son. What is the probability that the son has haemophilia?血友病 A 为 X 连锁隐性。一位携带者母亲（XHXh）和一位未患病父亲（XHY）有一个儿子。儿子患血友病的概率是多少？
Q4
0% — the father is unaffected so sons cannot be affected0%——父亲未患病，所以儿子不可能患病
50% — sons receive their X from the carrier mother50%——儿子从携带者母亲处获得 X
25% — half the probability applies to both sexes25%——一半概率适用于两种性别
100% — all sons of a carrier mother are affected100%——携带者母亲的儿子全部患病
Sons get their X only from their mother. The mother is XHXh, so she passes XH or Xh with equal probability. A son who receives Xh (probability 1/2) will have haemophilia (XhY). The father's genotype is irrelevant to sons.儿子只从母亲获得 X。母亲为 XHXh，所以她等概率传递 XH 或 Xh。获得 Xh（概率 1/2）的儿子将患血友病（XhY）。父亲的基因型与儿子无关。
Sons inherit their X from their mother, not their father. The carrier mother (XHXh) gives 1/2 her sons Xh, making 50% of sons affected.儿子从母亲而非父亲遗传 X。携带者母亲（XHXh）将 Xh 传给 1/2 的儿子，使 50% 的儿子患病。
A pedigree shows that two unaffected parents (generation I) have both affected and unaffected children, and the affected children include both sons and daughters in roughly equal numbers. What inheritance pattern best fits?一张系谱显示，两位未患病的亲本（第一代）有患病和未患病的孩子，且患病孩子中儿子和女儿数量大致相等。哪种遗传模式最符合？
Q5
Autosomal dominant常染色体显性
X-linked recessiveX 连锁隐性
Autosomal recessive常染色体隐性
X-linked dominantX 连锁显性
Two unaffected parents with affected children rules out autosomal dominant (would require at least one affected parent). Both sexes equally affected rules out X-linked recessive (males more often affected). Autosomal recessive (both parents Aa) produces affected children of both sexes in 1/4 of offspring.两位未患病亲本生出患病孩子，排除常染色体显性（需要至少一位患病亲本）。两性均等受影响，排除 X 连锁隐性（男性受影响更多）。常染色体隐性（两个亲本均为 Aa）产生两性患病孩子，占后代 1/4。
Autosomal dominant requires an affected parent. X-linked recessive affects more males. Autosomal recessive fits: unaffected carrier parents (Aa × Aa) give 1/4 affected offspring of both sexes.常染色体显性需要患病亲本。X 连锁隐性男性受影响更多。常染色体隐性符合：未患病的携带者亲本（Aa×Aa）产生 1/4 的两性患病后代。
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High School Biology · Mendelian Genetics and Heredity · v1高中生物 · 孟德尔遗传学与遗传 · v1