High School Biology — Mendelian Genetics and Heredity — Solutions

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Section A · Worked SolutionsA 部分 · 详细解答

Q1EASY易 🇺🇸 US美 AP-style MCQAP 风格选择题 §1 Genetics terminology遗传学术语 · HS-LS3-1 [3 marks][3 分]

A pea plant has genotype $Tt$. Which correctly describes this plant?一株豌豆基因型为 $Tt$，以下哪项正确描述了该植株？

Answer:答案： (B) Heterozygous and will show the dominant phenotype杂合子，表现出显性表现型

Identify genotype and phenotype判断基因型与表现型 M1·A1·A1

The genotype $Tt$ contains two different alleles (one dominant $T$, one recessive $t$), so the plant is heterozygous. Because the dominant allele $T$ masks $t$, the plant expresses the dominant phenotype. Option (B) is correct. 基因型 $Tt$ 含两个不同等位基因（一个显性 $T$，一个隐性 $t$），因此该植株为杂合子。由于显性等位基因 $T$ 掩盖了 $t$，植株表现出显性表现型。选 (B)。

Why the distractors fail.干扰项分析。

(A) Homozygous dominant would be $TT$, not $Tt$; recessive phenotype cannot show when a dominant allele is present.纯合显性应为 $TT$；有显性等位基因时不会表现出隐性表现型。

(D) Showing both phenotypes simultaneously describes codominance, which requires a different inheritance pattern (e.g., $C^R C^W$).同时表现两种表现型描述的是共显性，需要不同的遗传模式（如 $C^R C^W$）。

Heterozygous vs. homozygous is about allele sameness; dominant vs. recessive is about which allele is expressed.杂合与纯合说的是等位基因是否相同；显性与隐性说的是哪个等位基因被表达。 Heterozygous ($Tt$) means the two alleles differ. Because $T$ is dominant, only one copy is needed to produce the dominant phenotype. This is Mendel's principle of dominance: the dominant allele fully masks the recessive in the $F_1$ heterozygote. The recessive phenotype only appears in $tt$ homozygotes. Locking in the definitions of homozygous, heterozygous, dominant, and recessive early in a genetics problem prevents cascading errors downstream.杂合（$Tt$）意味着两个等位基因不同。由于 $T$ 为显性，只需一个拷贝即可产生显性表现型。这正是孟德尔显性原则：显性等位基因在 $F_1$ 杂合子中完全掩盖隐性等位基因。隐性表现型只出现在 $tt$ 纯合子中。在遗传题起步时先明确纯合、杂合、显性、隐性的定义，可避免后续错误。

Q2EASY易 🇨🇦 ON安 ON Provincial-style安大略省考风格 §2 Monohybrid cross单杂交 · SBI3U D1 [5 marks][5 分]

In pea plants, round seeds ($R$) are dominant over wrinkled ($r$). Two heterozygous round-seeded plants are crossed ($Rr \times Rr$).豌豆中圆粒（$R$）对皱粒（$r$）为显性。两株杂合圆粒植株杂交（$Rr \times Rr$）。

Answer:答案： (a) Punnett square below旁氏表见下 · (b) $1\,RR : 2\,Rr : 1\,rr$ · (c) $3\ \text{round} : 1\ \text{wrinkled}$ · (d) $25\%$

(a) Punnett square for $Rr \times Rr$$Rr \times Rr$ 的旁氏表 M1·A1

Parent gametes: $R$ and $r$ from each parent.亲本配子：每个亲本均产生 $R$ 和 $r$。

	$R$	$r$
$R$	$RR$	$Rr$
$r$	$Rr$	$rr$

(b) Genotypic ratio基因型比例 A1

From the four cells: $1\,RR : 2\,Rr : 1\,rr$.四格结果：$1\,RR : 2\,Rr : 1\,rr$。

(c) Phenotypic ratio表现型比例 A1

Both $RR$ and $Rr$ show round seeds (dominant phenotype). Only $rr$ shows wrinkled. Ratio: $3\ \text{round} : 1\ \text{wrinkled}$.$RR$ 和 $Rr$ 均表现圆粒（显性）；只有 $rr$ 表现皱粒。比例：$3\ \text{圆粒} : 1\ \text{皱粒}$。

(d) Percentage homozygous recessive纯合隐性的比例 A1

$rr$ occupies 1 out of 4 cells: $\frac{1}{4} = 25\%$.$rr$ 占四格中的一格：$\frac{1}{4} = 25\%$。

The 3:1 phenotypic ratio is the hallmark of a monohybrid cross between two heterozygotes.3:1 表现型比例是两杂合子单杂交的标志性结果。 This ratio arises directly from Mendel's Law of Segregation: each parent passes either the dominant or recessive allele with equal probability (1/2 each). The probability of a dominant phenotype in the offspring is $P(R\_) = P(RR) + P(Rr) = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$, and the probability of the recessive phenotype is $P(rr) = \frac{1}{4}$. In a large sample, observed ratios converge on 3:1. In small samples (e.g., a family of 4) deviations are expected by chance alone.该比例直接来自孟德尔分离定律：每个亲本以相等概率（各 1/2）传递显性或隐性等位基因。后代中显性表现型概率为 $P(R\_) = \frac{1}{4} + \frac{2}{4} = \frac{3}{4}$，隐性表现型概率为 $P(rr) = \frac{1}{4}$。大样本下观察比例趋近 3:1；小样本中偏差纯属机遇。

Q3MEDIUM中 🇺🇸 US美 AP-style MCQAP 风格选择题 §3 Law of Segregation分离定律 · HS-LS3-2 [3 marks][3 分]

$TT \times tt$ cross. Proportion of $F_1$ that is tall?$TT \times tt$ 杂交，$F_1$ 中高茎比例？

Answer:答案： (D) $100\%$

Apply the Law of Segregation to a homozygous dominant x homozygous recessive cross将分离定律应用于纯合显性 x 纯合隐性杂交 M1·A1·A1

$TT$ produces only $T$ gametes; $tt$ produces only $t$ gametes. All $F_1$ offspring receive one $T$ and one $t$, giving genotype $Tt$ with probability 1. Since $T$ is dominant, all $Tt$ plants show the tall phenotype: $100\%$ tall, option (D). $TT$ 只产生 $T$ 配子；$tt$ 只产生 $t$ 配子。所有 $F_1$ 后代均接受一个 $T$ 和一个 $t$，基因型为 $Tt$（概率为 1）。由于 $T$ 为显性，所有 $Tt$ 植株均表现高茎：$100\%$ 高茎，选 (D)。

Why the distractors fail.干扰项分析。

(A) $0\%$ would require all offspring to be $tt$, impossible when one parent is $TT$.需要所有后代均为 $tt$，但亲本之一为 $TT$，不可能。

(B) $25\%$ and (C) $50\%$ apply to $F_2$ crosses ($Tt \times Tt$), not $F_1$.和 (C) $50\%$ 适用于 $F_2$（$Tt \times Tt$），不适用于 $F_1$。

A cross between two true-breeding parents is a classic Mendelian $P \times P$ cross; the $F_1$ is uniformly heterozygous and shows only the dominant phenotype.两个纯系亲本间的杂交是经典孟德尔 $P \times P$ 杂交；$F_1$ 一致为杂合子，只表现显性表现型。 This is precisely how Mendel discovered dominance: he crossed pure-breeding tall and pure-breeding dwarf plants and found all $F_1$ offspring tall. The disappearance of the dwarf trait in $F_1$ is not because it is gone; it reemerges in $F_2$ in a 3:1 ratio. The Law of Segregation explains both results: alleles segregate into gametes with equal probability, and reunion restores the diploid genotype in offspring.这正是孟德尔发现显性的方式：他杂交纯系高茎与纯系矮茎植株，发现所有 $F_1$ 均为高茎。矮茎性状在 $F_1$ 中消失并非因为消除，而是在 $F_2$ 中以 3:1 的比例重现。分离定律解释了这两个结果：等位基因以相等概率分配到配子中，再结合恢复后代的二倍体基因型。

Q4MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §5 Incomplete dominance不完全显性 · Biology 12 [6 marks][6 分]

Snapdragons: $C^R C^R$ = red, $C^W C^W$ = white, $C^R C^W$ = pink. Cross: two pink plants ($C^R C^W \times C^R C^W$).金鱼草：$C^R C^R$ = 红色，$C^W C^W$ = 白色，$C^R C^W$ = 粉色。杂交：两株粉花植株（$C^R C^W \times C^R C^W$）。

Answer:答案： (a) see Punnett square见旁氏表 · (b) $1\ \text{red} : 2\ \text{pink} : 1\ \text{white}$ · (c) neither allele is fully dominant两等位基因均非完全显性

(a) Punnett square for $C^R C^W \times C^R C^W$$C^R C^W \times C^R C^W$ 的旁氏表 M1·A1

	$C^R$	$C^W$
$C^R$	$C^R C^R$ (red)	$C^R C^W$ (pink)
$C^W$	$C^R C^W$ (pink)	$C^W C^W$ (white)

(b) Phenotypic ratio表现型比例 A1·A1

Offspring: 1 $C^R C^R$ (red) : 2 $C^R C^W$ (pink) : 1 $C^W C^W$ (white). Phenotypic ratio = $1 : 2 : 1$ (red : pink : white). 后代：1 $C^R C^R$（红色）: 2 $C^R C^W$（粉色）: 1 $C^W C^W$（白色）。表现型比例 = $1 : 2 : 1$（红色：粉色：白色）。

(c) Why no 3:1 ratio为何不出现 3:1 比例 A1·A1

In complete dominance one allele fully masks the other, collapsing $C^R C^R$ and $C^R C^W$ into the same phenotype and giving 3:1. In incomplete dominance, neither $C^R$ nor $C^W$ completely dominates. The heterozygote $C^R C^W$ produces an intermediate phenotype (pink) distinct from both homozygotes, so all three genotypic classes are phenotypically distinguishable and the ratio is 1:2:1. 完全显性时，一个等位基因完全掩盖另一个，$C^R C^R$ 和 $C^R C^W$ 表现型相同，得到 3:1。不完全显性时，$C^R$ 和 $C^W$ 均无法完全掩盖对方。杂合子 $C^R C^W$ 产生介于两者之间的表现型（粉色），三种基因型类别在表现型上均可区分，因此比例为 1:2:1。

Incomplete dominance: genotypic ratio equals phenotypic ratio because each genotype has a unique phenotype.不完全显性：基因型比例等于表现型比例，因为每种基因型对应独特的表现型。 This is the diagnostic feature of incomplete dominance. Because the heterozygote is visually distinct, you can determine genotype by phenotype alone. This makes incomplete dominance useful in breeding programs: a pink plant is always $C^R C^W$, never $C^R C^R$ or $C^W C^W$. Contrast with codominance (next question), where heterozygotes show both parental traits simultaneously rather than a blend. A common exam error is confusing "intermediate phenotype" (incomplete dominance) with "both phenotypes expressed" (codominance). 这是不完全显性的诊断特征。由于杂合子表现型独特，仅凭表现型即可确定基因型。因此不完全显性在育种中很有用：粉花植株必然为 $C^R C^W$，而非 $C^R C^R$ 或 $C^W C^W$。与共显性（下一题）对比：共显性的杂合子同时表现两个亲本特征，而非混合。考试常见错误是将"中间表现型"（不完全显性）与"两种表现型同时出现"（共显性）相混淆。

Q5MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §6 Sex-linked inheritance伴性遗传 · 30-C1.3k [8 marks][8 分]

Color blindness: X-linked recessive $X^b$. Carrier mother $X^B X^b$ x color-blind father $X^b Y$.色盲：X 连锁隐性 $X^b$。携带者母亲 $X^B X^b$ x 色盲父亲 $X^b Y$。

Answer:答案： (a) see Punnett square见旁氏表 · (b) $\frac{1}{2}$ of daughters的女儿 · (c) $\frac{1}{2}$ of sons的儿子 · (d) males have only one X chromosome男性只有一条 X 染色体

(a) Punnett square: $X^B X^b \times X^b Y$旁氏表：$X^B X^b \times X^b Y$ M1·A1·A1

	$X^b$	$Y$
$X^B$	$X^B X^b$ (carrier daughter)	$X^B Y$ (normal son)
$X^b$	$X^b X^b$ (color-blind daughter)	$X^b Y$ (color-blind son)

(b) Probability a daughter is a carrier女儿为携带者的概率 A1·A1

Daughters are $X^B X^b$ or $X^b X^b$ in equal numbers. Only $X^B X^b$ is a carrier (has normal vision but carries the allele). Among daughters: probability of being a carrier $= \frac{1}{2}$. 女儿分别为 $X^B X^b$ 或 $X^b X^b$，各占一半。只有 $X^B X^b$ 为携带者（视力正常但携带该等位基因）。女儿中携带者概率 $= \frac{1}{2}$。

(c) Probability a son is color blind儿子为色盲的概率 A1·A1

Sons are $X^B Y$ (normal) or $X^b Y$ (color blind) in equal numbers. Probability a son is color blind $= \frac{1}{2}$. 儿子分别为 $X^B Y$（正常）或 $X^b Y$（色盲），各占一半。儿子为色盲的概率 $= \frac{1}{2}$。

(d) Why color blindness is more common in males色盲在男性中更常见的原因 A1

Males have only one X chromosome ($XY$), so a single recessive allele $X^b$ is sufficient to produce color blindness. Females ($XX$) require two copies of $X^b$ (one from each parent) to be color blind; otherwise one normal $X^B$ allele masks the condition. 男性只有一条 X 染色体（$XY$），因此单个隐性等位基因 $X^b$ 即足以导致色盲。女性（$XX$）需要两个 $X^b$（分别来自双亲）才会患色盲；否则一个正常 $X^B$ 等位基因即可掩盖该病。

X-linked recessive traits skip generations and appear predominantly in males because males are hemizygous for X-linked genes.X 连锁隐性性状隔代遗传，主要出现在男性中，因为男性对 X 连锁基因为半合子。 A male with $X^b Y$ is called hemizygous: he has only one allele at this locus, so there is no second allele to mask the recessive. Females need $X^b X^b$ to be affected; $X^B X^b$ females are unaffected carriers. This hemizygosity explains why X-linked recessive disorders (color blindness, hemophilia, Duchenne muscular dystrophy) are far more common in males. The carrier frequency in females equals roughly the square of the disease frequency in males (Hardy-Weinberg), so a rare trait in males is much rarer in females. $X^b Y$ 男性称为半合子：该位点只有一个等位基因，没有第二个等位基因来掩盖隐性。女性需 $X^b X^b$ 才会患病；$X^B X^b$ 女性为无症状携带者。这种半合子性解释了为何 X 连锁隐性疾病（色盲、血友病、杜氏肌营养不良）在男性中远比女性普遍。女性中携带者频率约等于男性患病频率的平方（哈迪-温伯格定律），因此男性中罕见的性状在女性中更为罕见。

Section B · Worked SolutionsB 部分 · 详细解答

Q6MEDIUM中 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §4 Dihybrid cross双杂交 · HS-LS3-2 [8 marks][8 分]

Peas: seed color (yellow $Y$ dominant over green $y$) and seed shape (round $R$ dominant over wrinkled $r$) on different chromosomes. $YyRr \times YyRr$.豌豆：种子颜色（黄色 $Y$ 对绿色 $y$ 为显性）和形状（圆粒 $R$ 对皱粒 $r$ 为显性）位于不同染色体上。$YyRr \times YyRr$。

Answer:答案： (a) 4 gamete types; Law of Independent Assortment4 种配子；自由组合定律 · (b) $9:3:3:1$ · (c) $30$ · (d) genes must be on different chromosomes (independent assortment)基因须位于不同染色体上（独立分配）

(a) Gametes from $YyRr$ and governing law$YyRr$ 产生的配子及支配定律 M1·A1

Each locus segregates independently. Gametes: $YR$, $Yr$, $yR$, $yr$ each with probability $\frac{1}{4}$. The governing principle is Mendel's Law of Independent Assortment: alleles of different genes (on different chromosomes) sort into gametes independently of one another. 每个位点独立分离。配子：$YR$、$Yr$、$yR$、$yr$，各占 $\frac{1}{4}$。支配原则为孟德尔自由组合定律：位于不同染色体上的不同基因的等位基因彼此独立地分配到配子中。

(b) Phenotypic ratio using probability method用概率法求表现型比例 M1·A1·A1

For seed color alone: $Yy \times Yy$ gives $\frac{3}{4}$ yellow, $\frac{1}{4}$ green. 仅考虑种子颜色：$Yy \times Yy$ 给出 $\frac{3}{4}$ 黄色，$\frac{1}{4}$ 绿色。 For seed shape alone: $Rr \times Rr$ gives $\frac{3}{4}$ round, $\frac{1}{4}$ wrinkled. 仅考虑种子形状：$Rr \times Rr$ 给出 $\frac{3}{4}$ 圆粒，$\frac{1}{4}$ 皱粒。 Combined (independent):结合（独立）：

Yellow round (Y_ R_): $\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$黄色圆粒 (Y_ R_)：$\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$
Yellow wrinkled (Y_ rr): $\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$黄色皱粒 (Y_ rr)：$\frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$
Green round (yy R_): $\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$绿色圆粒 (yy R_)：$\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$
Green wrinkled (yy rr): $\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$绿色皱粒 (yy rr)：$\frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

Phenotypic ratio: $\mathbf{9:3:3:1}$ (yellow round : yellow wrinkled : green round : green wrinkled). 表现型比例：$\mathbf{9:3:3:1}$（黄色圆粒：黄色皱粒：绿色圆粒：绿色皱粒）。

(c) Expected yellow wrinkled from 160 offspring160 株后代中预期黄色皱粒数量 M1·A1

$$ \frac{3}{16} \times 160 \;=\; 30 \;\text{plants.} $$

(d) Condition for 9:3:3:1 to apply9:3:3:1 成立的条件 A1

The two genes must be on different chromosomes (or far enough apart on the same chromosome to assort independently). If the genes were linked (on the same chromosome without recombination), the ratio would deviate from 9:3:3:1. 两个基因须位于不同染色体上（或同一染色体上距离足够远可独立分配）。若基因连锁（同一染色体上且无重组），比例将偏离 9:3:3:1。

The 9:3:3:1 ratio is the product of two independent 3:1 ratios and is the hallmark prediction of the Law of Independent Assortment.9:3:3:1 是两个独立 3:1 比例的乘积，是自由组合定律的标志性预测。 The probability method is faster and less error-prone than a full 4x4 Punnett square (16 cells). Multiply the single-locus probabilities: the product law holds because the two loci are independent. Memorize the four classes and their fractions: 9/16, 3/16, 3/16, 1/16. These fractions let you quickly compute expected counts for any sample size. The 9:3:3:1 ratio breaks down when genes are linked, when epistasis occurs, or when sample sizes are very small. 概率法比完整的 4x4 旁氏表（16 格）更快且不易出错。将单位点概率相乘：乘积法则成立，因为两个位点独立。记住四种类别及其分数：9/16、3/16、3/16、1/16。这些分数让你能快速计算任意样本量下的预期数量。当基因连锁、发生上位性或样本量极小时，9:3:3:1 比例会失效。

Q7MEDIUM中 🇨🇦 ON安 ON Provincial-style安大略省考风格 §7 Pedigree analysis系谱分析 · SBI3U D2 [7 marks][7 分]

Gen I: unaffected parents. Gen II: two unaffected daughters, one affected son. Gen III: one Gen II daughter (unaffected) x unaffected male, one affected daughter.第 I 代：未患病双亲。第 II 代：两名未患病女儿，一名患病儿子。第 III 代：第 II 代一名未患病女儿与未患病男性婚配，生一名患病女儿。

Answer:答案： (a) Autosomal常染色体 · (b) Recessive隐性 · (c) $Aa$ · (d) $\frac{1}{2}$ ($50\%$)

(a) Autosomal or X-linked?常染色体还是 X 连锁？ M1·A1

The condition is autosomal. Evidence: a daughter in Generation III is affected. If the trait were X-linked recessive, an affected female would need two copies of $X^a$, requiring her father to be affected. However, the Gen III father is described as unaffected. Therefore, the affected Gen III daughter cannot have inherited the condition as an X-linked recessive trait from an unaffected father. The trait must be autosomal. 该病为常染色体遗传。证据：第 III 代有一名患病女儿。若该病为 X 连锁隐性，患病女性需要两个 $X^a$，要求其父亲也患病。但第 III 代父亲未患病。因此患病的第 III 代女儿不可能从未患病父亲处遗传 X 连锁隐性性状。该性状必定为常染色体遗传。

(b) Dominant or recessive?显性还是隐性？ M1·A1

The condition is autosomal recessive. Evidence: two unaffected Generation I parents produce an affected Generation II son. Unaffected parents producing an affected child is the hallmark of a recessive trait: both parents must be carriers ($Aa$) who pass the recessive $a$ allele to the child ($aa$). 该病为常染色体隐性遗传。证据：两名未患病的第 I 代双亲生出一名患病的第 II 代儿子。未患病双亲生出患病子女是隐性性状的标志：双亲均为携带者（$Aa$），将隐性 $a$ 等位基因传给子女（$aa$）。

(c) Genotype of Gen I female第 I 代女性的基因型 M1·A1

Gen I female is unaffected, so she has at least one $A$ allele. Her Gen II son is affected ($aa$), so she must have contributed an $a$ allele. Therefore, she is heterozygous: genotype $\mathbf{Aa}$. 第 I 代女性未患病，至少有一个 $A$ 等位基因。她的第 II 代儿子患病（$aa$），说明她必须提供了一个 $a$ 等位基因。因此她为杂合子：基因型 $\mathbf{Aa}$。

(d) Probability the affected Gen II son has an affected child with a heterozygous female患病的第 II 代儿子与杂合子女性婚配后子女患病的概率 A1

The Gen II affected son has genotype $aa$. A heterozygous female has genotype $Aa$. Cross: $aa \times Aa$ gives $\frac{1}{2}\,Aa$ (carriers, unaffected) and $\frac{1}{2}\,aa$ (affected). Probability of an affected child $= \mathbf{\frac{1}{2}}$ ($50\%$). 第 II 代患病儿子基因型为 $aa$。杂合子女性基因型为 $Aa$。杂交：$aa \times Aa$ 给出 $\frac{1}{2}\,Aa$（携带者，未患病）和 $\frac{1}{2}\,aa$（患病）。子女患病概率 $= \mathbf{\frac{1}{2}}$（$50\%$）。

Pedigree analysis: use two key tests to determine inheritance pattern before assigning genotypes.系谱分析：先用两个关键检验确定遗传方式，再分配基因型。 Test 1 (autosomal vs. X-linked): an affected female rules out X-linked recessive if her father is unaffected, because X-linked recessive requires an affected father for an affected daughter. Test 2 (dominant vs. recessive): unaffected parents with an affected child points to recessive. These two tests together narrow the mode of inheritance before any genotype is written. Apply them in this order every time. Then trace obligate carrier genotypes (Gen I and Gen II unaffected carriers must be $Aa$) to find the requested probability. 检验 1（常染色体 vs. X 连锁）：若父亲未患病而女儿患病，可排除 X 连锁隐性，因为 X 连锁隐性要求患病女儿的父亲也患病。检验 2（显性 vs. 隐性）：未患病双亲生出患病子女指向隐性遗传。这两项检验合起来可在写下任何基因型前确定遗传方式。每次都按此顺序操作。然后追踪必然携带者基因型（第 I 代和第 II 代未患病携带者必为 $Aa$），找出所求概率。

Q8MEDIUM中 🇨🇦 BC卑 BC Provincial-style卑诗省考风格 §6 Multiple alleles (ABO blood type)复等位基因（ABO 血型） · Biology 12 [7 marks][7 分]

Father blood type AB, mother blood type O. ABO locus: $I^A$, $I^B$ codominant; $i$ recessive.父亲 AB 血型，母亲 O 血型。ABO 位点：$I^A$、$I^B$ 共显性；$i$ 为隐性。

Answer:答案： (a) Father $I^A I^B$; Mother $ii$父亲 $I^A I^B$；母亲 $ii$ · (b) see Punnett square见旁氏表 · (c) $\frac{1}{2}$ type A, A 型，$\frac{1}{2}$ type BB 型 · (d) No不能

(a) Genotypes of father and mother父亲和母亲的基因型 A1·A1

Blood type AB is only possible with genotype $I^A I^B$ (the only genotype that expresses both A and B antigens). Blood type O requires genotype $ii$ (no A or B antigen). Father: $\mathbf{I^A I^B}$; Mother: $\mathbf{ii}$. AB 血型只能由基因型 $I^A I^B$ 产生（唯一能同时表达 A 和 B 抗原的基因型）。O 血型需要基因型 $ii$（无 A 或 B 抗原）。父亲：$\mathbf{I^A I^B}$；母亲：$\mathbf{ii}$。

(b) Punnett square: $I^A I^B \times ii$旁氏表：$I^A I^B \times ii$ M1·A1

	$i$	$i$
$I^A$	$I^A i$ (type A)	$I^A i$ (type A)
$I^B$	$I^B i$ (type B)	$I^B i$ (type B)

(c) Possible blood types and probabilities可能的血型及其概率 A1·A1

Two cells $I^A i$ (type A) and two cells $I^B i$ (type B). Probability of type A $= \frac{2}{4} = \frac{1}{2}$; probability of type B $= \frac{2}{4} = \frac{1}{2}$. No other blood types are possible. 两格 $I^A i$（A 型），两格 $I^B i$（B 型）。A 型概率 $= \frac{1}{2}$；B 型概率 $= \frac{1}{2}$。无其他血型。

(d) Can they have an AB child?能否生育 AB 血型子女？ A1

No. An AB child requires genotype $I^A I^B$. The mother can only contribute $i$ alleles. Every child receives one $i$ from the mother, so no child can be $I^A I^B$; all children have genotype $I^A i$ or $I^B i$. 不能。AB 血型子女需要基因型 $I^A I^B$。母亲只能提供 $i$ 等位基因。每个子女都从母亲获得一个 $i$，因此没有子女能成为 $I^A I^B$；所有子女基因型为 $I^A i$ 或 $I^B i$。

ABO blood typing illustrates multiple alleles and codominance in the same system: three alleles produce four phenotypes.ABO 血型在同一系统中体现了复等位基因与共显性：三个等位基因产生四种表现型。 The ABO locus has three alleles ($I^A$, $I^B$, $i$) but each individual carries only two. $I^A$ and $I^B$ are codominant with each other (both expressed in $I^A I^B$ = type AB) and both dominant over $i$ (recessive). This produces four blood type phenotypes from six possible genotypes: $I^A I^A$ and $I^A i$ are both type A, $I^B I^B$ and $I^B i$ are both type B, $I^A I^B$ is type AB, and $ii$ is type O. Because the mother is $ii$, she can only donate $i$ to each offspring, limiting offspring to types A and B only. Blood type is often used in forensic and paternity contexts for this reason. ABO 位点有三个等位基因（$I^A$、$I^B$、$i$），但每个个体只携带两个。$I^A$ 和 $I^B$ 相互共显性（在 $I^A I^B$ = AB 型中同时表达），均对 $i$（隐性）为显性。六种可能基因型产生四种血型表现型：$I^A I^A$ 和 $I^A i$ 均为 A 型，$I^B I^B$ 和 $I^B i$ 均为 B 型，$I^A I^B$ 为 AB 型，$ii$ 为 O 型。由于母亲为 $ii$，她只能向每个后代提供 $i$，将后代血型限制为 A 型和 B 型。血型常因此用于法医和亲子鉴定。

Q9HARD难Honors荣誉级 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §5 Codominance + test cross design共显性 + 测交设计 · HS-LS3-2 [8 marks][8 分]

Cattle: $C^R C^R$ = red, $C^W C^W$ = white, $C^R C^W$ = roan. Roan bull ($C^R C^W$) crossed with a red cow of unknown genotype ($C^R C^R$ or $C^R C^W$).牛：$C^R C^R$ = 红色，$C^W C^W$ = 白色，$C^R C^W$ = 沙毛。沙毛公牛（$C^R C^W$）与基因型未知的红色母牛（$C^R C^R$ 或 $C^R C^W$）杂交。

Answer:答案： (a) $\frac{1}{2}$ red : 红：$\frac{1}{2}$ roan沙毛 · (b) $\frac{1}{4}$ red : 红：$\frac{2}{4}$ roan : 沙毛：$\frac{1}{4}$ white白色 · (c) any white offspring proves $C^R C^W$任何白色后代证明 $C^R C^W$

(a) If cow is $C^R C^R$: cross $C^R C^W \times C^R C^R$若母牛为 $C^R C^R$：杂交 $C^R C^W \times C^R C^R$ M1·A1·A1

	$C^R$	$C^R$
$C^R$	$C^R C^R$ (red)	$C^R C^R$ (red)
$C^W$	$C^R C^W$ (roan)	$C^R C^W$ (roan)

Offspring: $\frac{1}{2}\,C^R C^R$ (red) : $\frac{1}{2}\,C^R C^W$ (roan). No white offspring can appear. 后代：$\frac{1}{2}\,C^R C^R$（红色）：$\frac{1}{2}\,C^R C^W$（沙毛）。不会出现白色后代。

(b) If cow is $C^R C^W$: cross $C^R C^W \times C^R C^W$若母牛为 $C^R C^W$：杂交 $C^R C^W \times C^R C^W$ M1·A1·A1

	$C^R$	$C^W$
$C^R$	$C^R C^R$ (red)	$C^R C^W$ (roan)
$C^W$	$C^R C^W$ (roan)	$C^W C^W$ (white)

Offspring: $\frac{1}{4}\,C^R C^R$ (red) : $\frac{2}{4}\,C^R C^W$ (roan) : $\frac{1}{4}\,C^W C^W$ (white). Ratio $1:2:1$. 后代：$\frac{1}{4}\,C^R C^R$（红色）：$\frac{2}{4}\,C^R C^W$（沙毛）：$\frac{1}{4}\,C^W C^W$（白色）。比例 $1:2:1$。

(c) Diagnostic phenotype proving cow is $C^R C^W$证明母牛为 $C^R C^W$ 的诊断性表现型 M1·A1

A white calf ($C^W C^W$) proves the cow is $C^R C^W$. A white calf must inherit one $C^W$ from the bull and one $C^W$ from the cow. The bull is $C^R C^W$ and can donate $C^W$. If the cow were $C^R C^R$, she could only donate $C^R$; no white calf would be possible. Observing even one white calf is conclusive evidence that the cow carries $C^W$, i.e., her genotype is $C^R C^W$. 白色小牛（$C^W C^W$）证明母牛为 $C^R C^W$。白色小牛必须从公牛处获得一个 $C^W$，从母牛处获得一个 $C^W$。公牛为 $C^R C^W$，可提供 $C^W$。若母牛为 $C^R C^R$，她只能提供 $C^R$，不可能出现白色小牛。观察到哪怕一头白色小牛，即为母牛携带 $C^W$（即基因型为 $C^R C^W$）的确凿证据。

This is a test cross adapted for codominance: the "test" parent is the roan bull, and the diagnostic offspring is the white calf that only appears when the unknown parent contributes $C^W$.这是适用于共显性的测交："测验"亲本为沙毛公牛，诊断性后代是只有在未知亲本提供 $C^W$ 时才出现的白色小牛。 In classical test crosses (for complete dominance), you use a homozygous recessive individual to reveal a hidden recessive allele. With codominance the logic is identical but the "hidden" allele ($C^W$) is revealed by the unique white phenotype of the $C^W C^W$ offspring. No white calf means the cow might be $C^R C^R$ or might just be an unlucky $C^R C^W$ (with many trials, the white class at 25% would eventually appear). Absence of evidence is not evidence of absence in genetics; more crosses increase confidence. 在经典测交（完全显性）中，使用纯合隐性个体来揭示隐藏的隐性等位基因。共显性的逻辑相同，但"隐藏"的等位基因（$C^W$）通过 $C^W C^W$ 后代独特的白色表现型而显现。没有白色小牛意味着母牛可能为 $C^R C^R$，或可能只是运气不好的 $C^R C^W$（多次杂交后，白色类别在 25% 的概率下终将出现）。遗传学中无阳性证据并非阴性证据；增加杂交次数可提高确信度。

Section C · Worked SolutionsC 部分 · 详细解答

Q10MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §2-3 Probability and test cross概率与测交 · 30-C1.1k [8 marks][8 分]

Rabbits: black fur $B$ dominant over brown $b$. Black rabbit of unknown genotype test-crossed with brown rabbit. 40 offspring: 21 black, 19 brown.兔子：黑色毛 $B$ 对棕色毛 $b$ 为显性。基因型未知的黑色兔子与棕色兔子测交。40 只后代：21 只黑色，19 只棕色。

Answer:答案： (a) $bb$ · (b) $Bb$ · (c) all 40 black, 0 brown40 只黑色，0 只棕色 · (d) reveals presence of hidden recessive alleles揭示隐藏的隐性等位基因

(a) Genotype of the brown rabbit棕色兔子的基因型 A1

Brown fur is the recessive phenotype; the rabbit must be homozygous recessive: genotype $\mathbf{bb}$. 棕色毛为隐性表现型；该兔子必须为纯合隐性：基因型 $\mathbf{bb}$。

(b) Most likely genotype of the black rabbit, with Punnett square黑色兔子最可能的基因型及旁氏表 M1·A1·A1

The offspring ratio is approximately $21:19 \approx 1:1$. A 1:1 ratio from a test cross occurs only when the tested individual is heterozygous ($Bb$). Cross: $Bb \times bb$: 后代比例约为 $21:19 \approx 1:1$。测交中出现 1:1 比例，仅当被测个体为杂合子（$Bb$）时。杂交：$Bb \times bb$：

	$b$	$b$
$B$	$Bb$ (black)	$Bb$ (black)
$b$	$bb$ (brown)	$bb$ (brown)

Expected ratio 1 black : 1 brown, consistent with observed 21:19. Most likely genotype of the black rabbit: $\mathbf{Bb}$. 预期比例 1 黑：1 棕，与观察到的 21:19 一致。黑色兔子最可能的基因型：$\mathbf{Bb}$。

(c) If black rabbit were $BB$: expected offspring from 40若黑色兔子为 $BB$：40 只后代中各类型预期数量 M1·A1

Cross $BB \times bb$: all offspring are $Bb$ (black). Expected: 40 black, 0 brown. The actual result of 19 brown rabbits is very unlikely if the genotype were $BB$; this strongly supports $Bb$. 杂交 $BB \times bb$：所有后代均为 $Bb$（黑色）。预期：40 只黑色，0 只棕色。实际出现 19 只棕色兔子，若基因型为 $BB$ 则概率极低；这强烈支持 $Bb$。

(d) Why the test cross is useful测交为何有用 A1·A1

A test cross reveals whether a dominant-phenotype individual carries a hidden recessive allele ($Bb$) or is homozygous dominant ($BB$) because the homozygous recessive test parent contributes only recessive alleles, making any recessive allele in the test individual visible in the offspring phenotype. 测交揭示了表现显性表现型的个体是否携带隐性等位基因（$Bb$）或为纯合显性（$BB$），因为纯合隐性测验亲本只提供隐性等位基因，使被测个体中的任何隐性等位基因都能在后代表现型中显现。

The test cross is the classic tool for uncovering hidden recessive alleles: a 1:1 offspring ratio reveals heterozygosity; an all-dominant offspring ratio reveals homozygosity.测交是揭示隐藏隐性等位基因的经典工具：后代 1:1 比例揭示杂合性；后代全为显性揭示纯合性。 Mendel invented the test cross precisely because dominant-phenotype individuals could not be visually distinguished as $BB$ vs. $Bb$. By crossing with a known $bb$ tester, the hidden genotype is "read out" in the offspring. Deviations from expected ratios in small samples (like 21:19 instead of 20:20) are due to chance and are expected; statistical tests such as chi-square assess whether the deviation is within normal sampling variation. The 19 brown offspring here are fully consistent with a $Bb$ parent and simple binomial sampling. 孟德尔发明测交正是因为表现显性表现型的个体无法从外表区分 $BB$ 与 $Bb$。通过与已知 $bb$ 测验亲本杂交，隐藏的基因型在后代中被"读出"。小样本（如 21:19 而非 20:20）与预期比例的偏差是机遇所致，属正常现象；卡方检验等统计方法评估偏差是否在正常抽样变异范围内。此题 19 只棕色后代与 $Bb$ 亲本及简单二项分布抽样完全一致。

Q11MEDIUM中 🇨🇦 AB阿 AB Diploma-style阿尔伯塔毕业考风格 §3-4 Independent assortment applied自由组合定律应用 · 30-C1.2k [8 marks][8 分]

Labrador coat: $E/e$ gene (pigment deposition; $ee$ = yellow regardless), $B/b$ gene ($BB$ or $Bb$ = black, $bb$ = chocolate). Cross: $BbEe \times BbEe$.拉布拉多毛色：$E/e$ 基因（色素沉积；$ee$ 无论其他基因均为黄色），$B/b$ 基因（$BB$ 或 $Bb$ = 黑色，$bb$ = 巧克力色）。杂交：$BbEe \times BbEe$。

Answer:答案： (a) $BE,\,Be,\,bE,\,be$ · (b) $9\ \text{black} : 3\ \text{chocolate} : 4\ \text{yellow}$ · (c) $16$ yellow黄色 · (d) Law of Independent Assortment自由组合定律

(a) Gamete types from $BbEe$$BbEe$ 产生的配子类型 A1

Independent assortment of two loci gives $2^2 = 4$ gamete types: $BE$, $Be$, $bE$, $be$, each with probability $\frac{1}{4}$. 两个位点独立分配给出 $2^2 = 4$ 种配子：$BE$、$Be$、$bE$、$be$，各占 $\frac{1}{4}$。

(b) Expected ratio using probability method用概率法求预期比例 M1·A1·A1·A1

Yellow ($ee$ epistasis): $P(ee) = \frac{1}{4}$ regardless of $B/b$ locus. All $ee$ dogs are yellow. 黄色（$ee$ 上位性）：$P(ee) = \frac{1}{4}$，与 $B/b$ 位点无关。所有 $ee$ 犬均为黄色。 $$ P(\text{yellow}) = P(ee) = \frac{1}{4} = \frac{4}{16}. $$ For dogs with $E\_$ ($P = \frac{3}{4}$), pigment is deposited and $B/b$ determines color: 对于 $E\_$（$P = \frac{3}{4}$）的犬，色素沉积，$B/b$ 决定颜色：

Black ($B\_E\_$): $P(B\_) \times P(E\_) = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$黑色（$B\_E\_$）：$\frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$
Chocolate ($bbE\_$): $P(bb) \times P(E\_) = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$巧克力色（$bbE\_$）：$\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$
Yellow ($\_\_ee$): $P(ee) = \frac{1}{4} = \frac{4}{16}$ (includes both $B\_ee$ and $bbee$)黄色（$\_\_ee$）：$P(ee) = \frac{1}{4} = \frac{4}{16}$（包含 $B\_ee$ 和 $bbee$）

Ratio: $\mathbf{9\ \text{black} : 3\ \text{chocolate} : 4\ \text{yellow}}$ (note: this is a modified 9:3:4 due to epistasis; the yellow class absorbs $3/16 + 1/16 = 4/16$). 比例：$\mathbf{9\ \text{黑色} : 3\ \text{巧克力色} : 4\ \text{黄色}}$（注意：由于上位性，这是修正的 9:3:4；黄色类别吸收了 $3/16 + 1/16 = 4/16$）。

(c) Expected yellow offspring from 6464 只后代中预期黄色数量 M1·A1

$$ \frac{4}{16} \times 64 \;=\; \frac{1}{4} \times 64 \;=\; \mathbf{16} \;\text{yellow dogs.} $$

(d) Mendel's law explaining discrete coat color classes解释离散毛色类别的孟德尔定律 A1

The Law of Independent Assortment (Mendel's second law) explains why discrete phenotypic classes appear: alleles of the $B/b$ gene and the $E/e$ gene segregate independently into gametes, producing the four genotypic combinations that then sort into three coat color phenotypes. 自由组合定律（孟德尔第二定律）解释了为何出现离散表现型类别：$B/b$ 基因和 $E/e$ 基因的等位基因独立地分配到配子中，产生四种基因型组合，进而分类为三种毛色表现型。

Epistasis modifies the standard 9:3:3:1 dihybrid ratio. When one gene masks another, classes collapse and new ratios emerge (9:3:4 here).上位性修改了标准的 9:3:3:1 双杂交比例。当一个基因掩盖另一个时，类别合并产生新比例（此处为 9:3:4）。 Epistasis is gene-to-gene interaction where one gene's alleles mask the expression of another gene. Here the $e$ allele is recessive epistatic: homozygous $ee$ blocks pigment deposition entirely, making coat color independent of the $B/b$ locus. The 4/16 yellow class combines the $3/16$ that would have been chocolate ($bbee$) with the $1/16$ that would have been yellow ($B\_ee$) from a standard 9:3:3:1. Recognizing epistasis requires asking: "Does one gene's phenotypic effect depend on the genotype of another gene?" If yes, the 9:3:3:1 will be modified. 上位性是基因间相互作用，其中一个基因的等位基因掩盖另一个基因的表达。此处 $e$ 等位基因为隐性上位性：纯合 $ee$ 完全阻断色素沉积，使毛色与 $B/b$ 位点无关。4/16 的黄色类别将标准 9:3:3:1 中的 3/16 巧克力色（$bbee$）与 1/16 黄色（$B\_ee$）合并。识别上位性需问："一个基因的表现型效应是否取决于另一个基因的基因型？"若是，9:3:3:1 将被修改。

Q12HARD难 🇺🇸 US美 AP-feeder FRQAP 衔接简答题 §1-7 Integrated genetics analysis遗传学综合分析 · HS-LS3-1 [10 marks][10 分]

Familial hypercholesterolemia (FH): autosomal dominant allele $H$. $Hh$ = mild form; $HH$ = severe, often fatal. Both parents have mild FH; four children: two affected (mild), two unaffected.家族性高胆固醇血症（FH）：常染色体显性等位基因 $H$。$Hh$ = 轻度；$HH$ = 重度，常致命。双亲均患轻度 FH；四个子女：两名患轻度，两名未患病。

Answer:答案： (a) Both parents $Hh$双亲均为 $Hh$ · (b) $\frac{1}{4}HH : \frac{2}{4}Hh : \frac{1}{4}hh$ · (c) No; small sample size不；样本量小 · (d) $P(HH) = 0$ · (e) Heterozygotes are affected (dominant); recessive conditions require two copies杂合子受影响（显性）；隐性需要两个拷贝

(a) Genotypes of both parents双亲的基因型 M1·A1

Both parents have the mild form of FH. The mild form corresponds to genotype $Hh$ (one copy of the dominant allele $H$). Neither parent can be $HH$ because $HH$ causes the severe, often fatal form; individuals with $HH$ rarely survive to reproduce. Therefore both parents are $\mathbf{Hh}$. 双亲均患轻度 FH。轻度对应基因型 $Hh$（一个显性等位基因 $H$ 的拷贝）。双亲均不可能为 $HH$，因为 $HH$ 导致重度、常致命的形式；$HH$ 个体罕有存活至生育的。因此双亲均为 $\mathbf{Hh}$。

(b) Punnett square for $Hh \times Hh$ and genotypic classes$Hh \times Hh$ 的旁氏表及基因型类别 M1·A1·A1

	$H$	$h$
$H$	$HH$ (severe)	$Hh$ (mild)
$h$	$Hh$ (mild)	$hh$ (unaffected)

Genotypic classes and frequencies: $\frac{1}{4}\,HH$ (severe) : $\frac{2}{4}\,Hh$ (mild) : $\frac{1}{4}\,hh$ (unaffected). Note: phenotypically, $HH + Hh$ are "affected" ($\frac{3}{4}$) and $hh$ is "unaffected" ($\frac{1}{4}$), but the $HH$ class is clinically distinguishable as severe. 基因型类别及频率：$\frac{1}{4}\,HH$（重度）：$\frac{2}{4}\,Hh$（轻度）：$\frac{1}{4}\,hh$（未患病）。注意：表现型上 $HH + Hh$ 均"受影响"（$\frac{3}{4}$），$hh$ 为"未患病"（$\frac{1}{4}$），但 $HH$ 类别临床上可区分为重度。

(c) Does 2 affected : 2 unaffected disprove Mendel's laws?2 患病：2 未患病是否推翻孟德尔定律？ M1·A1

No. Mendel's laws are probabilistic. The expected ratio from $Hh \times Hh$ is $3$ affected $: 1$ unaffected ($\frac{3}{4}$ vs. $\frac{1}{4}$). In a family of only 4 children, random sampling variation easily produces outcomes like 2:2, 3:1, or even 4:0. With such a small sample, deviations from the expected ratio are common by chance and do not contradict the underlying law. A chi-square test on 4 children would give very low statistical power. Mendel's laws are confirmed by large-scale crosses (hundreds of offspring), not by individual family data. 不。孟德尔定律是概率性的。$Hh \times Hh$ 预期比例为 3 患病：1 未患病（$\frac{3}{4}$ vs. $\frac{1}{4}$）。在只有 4 个子女的家庭中，随机抽样变异很容易产生 2:2、3:1 甚至 4:0 的结果。样本量如此小，与预期比例的偏差纯属机遇，并不反驳基础定律。对 4 个子女进行卡方检验统计效力极低。孟德尔定律由大规模杂交（数百后代）而非个别家庭数据来验证。

(d) Probability first child of $hh \times Hh$ has severe FH ($HH$)$hh \times Hh$ 第一个孩子患重度 FH（$HH$）的概率 M1·A1

Cross $hh \times Hh$: gametes from $hh$ are all $h$; gametes from $Hh$ are $\frac{1}{2}\,H$ and $\frac{1}{2}\,h$. Offspring: $\frac{1}{2}\,Hh$ (mild) and $\frac{1}{2}\,hh$ (unaffected). $P(HH) = 0$. An $hh$ individual cannot produce an $H$ gamete, so $HH$ offspring are impossible. 杂交 $hh \times Hh$：$hh$ 的配子全为 $h$；$Hh$ 的配子为 $\frac{1}{2}\,H$ 和 $\frac{1}{2}\,h$。后代：$\frac{1}{2}\,Hh$（轻度）和 $\frac{1}{2}\,hh$（未患病）。$P(HH) = 0$。$hh$ 个体不能产生 $H$ 配子，因此 $HH$ 后代不可能出现。

(e) How FH differs from a typical autosomal recessive conditionFH 与典型常染色体隐性遗传病的不同 A1

In FH (autosomal dominant), heterozygotes ($Hh$) are affected with the mild form; only one copy of the $H$ allele is sufficient to cause disease. In a typical autosomal recessive condition, heterozygotes are carriers but are unaffected; two copies of the recessive allele are required to produce the disease phenotype. FH（常染色体显性）中，杂合子（$Hh$）受影响，表现为轻度；只需一个 $H$ 等位基因拷贝即可致病。而典型常染色体隐性遗传病中，杂合子为携带者但不受影响；需要两个隐性等位基因拷贝才能产生疾病表现型。

Autosomal dominant conditions often show a dosage effect: one copy causes mild disease, two copies cause severe disease. This is called incomplete dominance at the clinical level.常染色体显性遗传病常表现剂量效应：一个拷贝导致轻度疾病，两个拷贝导致重度疾病。这在临床层面称为不完全显性。 FH is caused by loss-of-function mutations in the LDL receptor gene ($LDLR$). Heterozygotes have roughly half the normal LDL receptor activity, leading to elevated LDL cholesterol and moderate disease risk. Homozygotes have near-zero receptor activity, leading to severe hypercholesterolemia and early cardiovascular disease (often in childhood). This dosage dependence ($HH$ more severe than $Hh$) is the hallmark of haploinsufficiency and is clinically important: knowing whether a patient is $Hh$ or $HH$ changes treatment urgency dramatically. Distinguish this from pure dominant conditions where $Hh$ and $HH$ are clinically identical. FH 由 LDL 受体基因（$LDLR$）功能丧失突变引起。杂合子约有正常 LDL 受体活性的一半，导致 LDL 胆固醇升高和中度疾病风险。纯合子受体活性接近零，导致重度高胆固醇血症和早发心血管病（常在儿童期）。这种剂量依赖性（$HH$ 比 $Hh$ 更严重）是单倍不足的标志，临床上非常重要：知道患者是 $Hh$ 还是 $HH$ 会大幅改变治疗紧迫性。将此与纯显性遗传病区分开来，后者中 $Hh$ 和 $HH$ 临床表现相同。