PART I · SHORT RESPONSE第一部分 · 短答题AP-style MCQ + ON/BC short answer · 25 marksAP 风格选择题 + 安/卑省考短答 · 共 25 分
Section A · Short ResponseA 部分 · 短答题
For MCQs, circle the letter of the best answer; for short-answer items, show your Punnett square or reasoning clearly. Use standard notation: dominant allele upper-case, recessive lower-case (e.g. $T$ / $t$). No calculator needed for this section.选择题请圈出最佳答案字母；短答题请清楚画出旁氏表或写出推理过程。采用标准符号：显性等位基因用大写，隐性用小写（如 $T$ / $t$）。本节无需计算器。
In pea plants, round seeds ($R$) are dominant over wrinkled seeds ($r$). Two heterozygous round-seeded plants are crossed.在豌豆中，圆粒种子（$R$）对皱粒种子（$r$）为显性。两株杂合圆粒植株进行杂交。
(a)Draw the Punnett square for this cross.画出该杂交的旁氏表。[2]
(b)State the genotypic ratio of the offspring.写出后代的基因型比例。[1]
(c)State the phenotypic ratio of the offspring.写出后代的表现型比例。[1]
(d)What percentage of the offspring are expected to be homozygous recessive?预期后代中纯合隐性个体的比例是多少？[1]
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]
A tall pea plant ($TT$) is crossed with a dwarf pea plant ($tt$). According to Mendel's Law of Segregation, what proportion of the $F_1$ offspring will be tall?一株高茎豌豆（$TT$）与一株矮茎豌豆（$tt$）杂交。根据孟德尔分离定律，$F_1$ 后代中高茎个体的比例是多少？
In snapdragons, flower color shows incomplete dominance. Red flowers have genotype $C^R C^R$, white flowers have genotype $C^W C^W$, and pink flowers have genotype $C^R C^W$. Two pink-flowered plants are crossed.在金鱼草中，花色表现出不完全显性。红花基因型为 $C^R C^R$，白花基因型为 $C^W C^W$，粉花基因型为 $C^R C^W$。两株粉花植株进行杂交。
(a)Draw the Punnett square for this cross.画出该杂交的旁氏表。[2]
(b)State the expected phenotypic ratio of the offspring (red : pink : white).写出后代预期的表现型比例（红色：粉色：白色）。[2]
(c)Explain why this cross does not produce a 3:1 phenotypic ratio.解释为什么该杂交不产生 3:1 的表现型比例。[2]
Color blindness in humans is caused by an X-linked recessive allele ($X^b$). Normal vision is dominant ($X^B$). A carrier mother ($X^B X^b$) has children with a color-blind father ($X^b Y$).人类色盲由 X 连锁隐性等位基因（$X^b$）引起，正常视力为显性（$X^B$）。一位携带者母亲（$X^B X^b$）与一位色盲父亲（$X^b Y$）生育后代。
(a)Draw the Punnett square for this cross, showing all possible genotypes of the offspring.画出该杂交的旁氏表，列出后代所有可能的基因型。[3]
(b)What is the probability that a daughter is a carrier?女儿为携带者的概率是多少？[2]
(c)What is the probability that a son is color blind?儿子为色盲的概率是多少？[2]
(d)State one reason why color blindness is more common in males than in females.说明一个色盲在男性中比女性更常见的原因。[1]
PART II · EXTENDED RESPONSE第二部分 · 简答题AP-feeder FRQ + honors · 30 marksAP 衔接简答题 + 荣誉级 · 共 30 分
Section B · Extended ResponseB 部分 · 简答题
Show every step of reasoning. Draw Punnett squares where applicable. State genotypic and phenotypic ratios explicitly. For pedigree questions, justify your conclusions with evidence from the pedigree. Calculator permitted on Q6-Q9.每一步推理都要写出。适用时须画旁氏表。明确写出基因型和表现型比例。系谱题须用系谱中的证据支持结论。Q6-Q9 可用计算器。
In peas, seed color (yellow $Y$ dominant over green $y$) and seed shape (round $R$ dominant over wrinkled $r$) are controlled by genes on different chromosomes. Two plants heterozygous for both traits ($YyRr \times YyRr$) are crossed.在豌豆中，种子颜色（黄色 $Y$ 对绿色 $y$ 为显性）和种子形状（圆粒 $R$ 对皱粒 $r$ 为显性）由位于不同染色体上的基因控制。两株双杂合植株（$YyRr \times YyRr$）进行杂交。
(a)List all gamete types produced by one parent and state which Mendel's law governs their formation.列出一个亲本产生的所有配子类型，并说明哪条孟德尔定律支配其形成。[2]
(b)Using a $4 \times 4$ Punnett square or the probability method, determine the phenotypic ratio of the offspring. State the four phenotypic classes and their expected ratio.用 $4 \times 4$ 旁氏表或概率法，确定后代的表现型比例。写出四种表现型及其预期比例。[3]
(c)Out of 160 offspring, how many are expected to be yellow and wrinkled?在 160 株后代中，预期黄色皱粒有多少株？[2]
(d)State one condition required for the 9:3:3:1 ratio to apply.写出 9:3:3:1 比例成立所需的一个条件。[1]
A family pedigree for an inherited condition is described below. In Generation I, an unaffected male and an unaffected female have three children in Generation II: two unaffected females and one affected male. In Generation III, one of the unaffected Generation II females marries an unaffected male and has one affected daughter.以下描述一个遗传病的家族系谱。第 I 代：一名未患病男性与一名未患病女性。第 II 代：他们有三个子女，两名未患病女儿和一名患病儿子。第 III 代：其中一名未患病的第 II 代女儿与一名未患病男性结婚，生有一名患病女儿。
(a)Is the condition autosomal or X-linked? Justify your answer using evidence from the pedigree.该病是常染色体遗传还是 X 连锁遗传？用系谱中的证据支持你的判断。[2]
(b)Is the condition dominant or recessive? Justify your answer.该病是显性遗传还是隐性遗传？请给出理由。[2]
(c)Determine the genotype of the Generation I female, using $A$ for the dominant allele and $a$ for the recessive allele.确定第 I 代女性的基因型，以 $A$ 表示显性等位基因，$a$ 表示隐性等位基因。[2]
(d)What is the probability that the affected son in Generation II will have an affected child, if he mates with a heterozygous female?第 II 代患病儿子若与一名杂合子女性婚配，其子女中患病的概率是多少？[1]
ABO blood type is controlled by three alleles: $I^A$ (codominant with $I^B$), $I^B$ (codominant with $I^A$), and $i$ (recessive to both). A father with blood type AB and a mother with blood type O have children.ABO 血型由三个等位基因控制：$I^A$（与 $I^B$ 共显性）、$I^B$（与 $I^A$ 共显性）和 $i$（对两者均为隐性）。一位 AB 血型父亲与一位 O 血型母亲生育后代。
(a)State the genotypes of the father and mother.写出父亲和母亲的基因型。[2]
(b)Draw the Punnett square for this cross.画出该杂交的旁氏表。[2]
(c)List all possible blood types of their children and the probability of each.列出其子女所有可能的血型及各自的概率。[2]
(d)Could this couple have a child with blood type AB? Explain.这对夫妇能生育 AB 血型的子女吗？请解释。[1]
In cattle, coat color shows codominance: $C^R C^R$ = red, $C^W C^W$ = white, $C^R C^W$ = roan (mixed red and white hairs). A rancher observes a roan bull and wishes to determine whether a particular red cow is homozygous ($C^R C^R$) or heterozygous ($C^R C^W$) roan-producing.在牛中，毛色表现共显性：$C^R C^R$ = 红色，$C^W C^W$ = 白色，$C^R C^W$ = 沙毛（红白混杂）。一位牧场主观察到一头沙毛公牛，希望确定一头特定红色母牛是纯合（$C^R C^R$）还是杂合（$C^R C^W$）。
(a)The rancher crosses the roan bull with the red cow. If the cow is $C^R C^R$, draw the Punnett square and state all expected offspring phenotypes with their ratios.牧场主将沙毛公牛与红色母牛杂交。若母牛为 $C^R C^R$，画出旁氏表，并写出所有预期后代表现型及其比例。[3]
(b)If the cow is $C^R C^W$, draw the Punnett square and state all expected offspring phenotypes with their ratios.若母牛为 $C^R C^W$，画出旁氏表，并写出所有预期后代表现型及其比例。[3]
(c)What single offspring phenotype would allow the rancher to conclude with certainty that the cow is $C^R C^W$? Explain.出现哪种后代表现型可让牧场主确定母牛为 $C^R C^W$？请解释。[2]
PART III · MODELING / APPLIED第三部分 · 建模与应用AB Diploma + Universal · 26 marks阿省毕业考 + 通用题型 · 共 26 分
Section C · Modeling and ApplicationsC 部分 · 建模与应用
Define all allele symbols at the start of each question. Draw Punnett squares where applicable and show probability calculations explicitly. Conclude each question with a one-sentence answer in context. Calculator permitted throughout Part III.每题开始时定义所有等位基因符号。适用时须画旁氏表，并明确写出概率计算过程。每题以一句结合情境的完整句子作答。第三部分全程可用计算器。
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]
In rabbits, black fur ($B$) is dominant over brown fur ($b$). A black rabbit of unknown genotype is test-crossed with a brown rabbit. Out of 40 offspring, 21 are black and 19 are brown.在兔子中，黑色毛（$B$）对棕色毛（$b$）为显性。一只基因型未知的黑色兔子与棕色兔子进行测交。在 40 只后代中，21 只为黑色，19 只为棕色。
(a)State the genotype of the brown rabbit used in the test cross.写出测交中棕色兔子的基因型。[1]
(b)Based on the offspring results, what is the most likely genotype of the black rabbit? Draw the Punnett square to support your answer.根据后代结果，黑色兔子最可能的基因型是什么？画出旁氏表支持你的答案。[3]
(c)If the black rabbit had been homozygous dominant ($BB$), what offspring ratio would have been expected? Calculate the number of each type expected from 40 offspring.若黑色兔子为纯合显性（$BB$），预期后代比例是多少？计算 40 只后代中各类型的预期数量。[2]
(d)In one sentence, explain why the test cross is a useful tool in genetics.用一句话解释为何测交是遗传学中的有效工具。[2]
In Labrador retrievers, coat color is determined by two independently assorting genes: the $E/e$ gene (where $E$ allows pigment deposition, $ee$ gives yellow regardless of the other gene) and the $B/b$ gene (where $BB$ or $Bb$ gives black, $bb$ gives chocolate). Two dogs, both with genotype $BbEe$, are crossed.在拉布拉多寻回犬中，毛色由两个独立分配的基因决定：$E/e$ 基因（$E$ 允许色素沉积，$ee$ 无论其他基因如何均为黄色）和 $B/b$ 基因（$BB$ 或 $Bb$ 为黑色，$bb$ 为巧克力色）。两只基因型均为 $BbEe$ 的犬进行杂交。
(a)List the four gamete types produced by each parent.列出每个亲本产生的四种配子类型。[1]
(b)Using a $4 \times 4$ Punnett square or the probability method, determine the expected ratio of black : chocolate : yellow offspring. Show your working.用 $4 \times 4$ 旁氏表或概率法，确定后代中黑色：巧克力色：黄色的预期比例。写出计算过程。[4]
(c)Out of 64 offspring from many such crosses, how many would be expected to be yellow?在许多此类杂交产生的 64 只后代中，预期黄色后代有多少只？[2]
(d)Identify which of Mendel's laws explains why the black, chocolate, and yellow classes appear in discrete categories.说明孟德尔哪条定律解释了黑色、巧克力色和黄色出现为离散类别的原因。[1]
A genetic disorder called familial hypercholesterolemia (FH) is caused by a dominant allele ($H$) on an autosome. Individuals with one copy ($Hh$) have high cholesterol; individuals with two copies ($HH$) have a severe, often fatal, form. A clinician analyzes the following family: both parents are affected with the mild form of FH; they have four children: two affected (mild) and two unaffected.家族性高胆固醇血症（FH）由常染色体上的显性等位基因（$H$）引起。携带一个拷贝（$Hh$）者胆固醇偏高；携带两个拷贝（$HH$）者病情严重，常致命。一位临床医生分析以下家庭：双亲均患轻度 FH；他们有四个子女：两名患轻度 FH，两名未患病。
(a)State the genotypes of both parents. Justify your answer by explaining why neither parent can be $HH$.写出双亲的基因型。通过解释为何双亲均不可能是 $HH$ 来支持你的答案。[2]
(b)Draw the Punnett square for this cross. List all genotypic classes and their frequencies.画出该杂交的旁氏表。列出所有基因型类别及其频率。[3]
(c)The expected ratio from (b) predicts a 1 : 2 : 1 genotypic ratio ($HH : Hh : hh$). However, in the actual family only 2 affected and 2 unaffected children are observed. Does this disprove Mendel's laws? Explain, referring to sample size.（b）中预期的 1:2:1 基因型比例（$HH:Hh:hh$），但实际家庭中仅观察到 2 名患者和 2 名未患病者。这是否推翻了孟德尔定律？结合样本量给出解释。[2]
(d)One unaffected child ($hh$) marries an individual heterozygous for FH ($Hh$). What is the probability that their first child will have the severe form ($HH$)?一名未患病子女（$hh$）与一名 FH 杂合子（$Hh$）婚配。他们第一个孩子患重度 FH（$HH$）的概率是多少？[2]
(e)State one way in which this disorder differs from a typical autosomal recessive condition, in terms of who is affected.就受影响人群而言，说明该病与典型常染色体隐性遗传病的一点不同。[1]
🇺🇸 US NGSS美国 NGSSHS-LS3-1HS-LS3-2
🇨🇦 Ontario安大略SBI3U Strand D · D1 · D2
🇨🇦 British Columbia不列颠哥伦比亚Biology 12: heredity (Mendel, Punnett squares, pedigrees)生物 12：遗传（孟德尔、旁氏表、系谱）
🇨🇦 Alberta阿尔伯塔Biology 30 Unit C · 30-C1.1k · 30-C1.2k · 30-C1.3k
Full Syllabus Map lives in ../Study Guides/Unit_5_Mendelian_Genetics_and_Heredity.html. Bio 30 diploma exam uses AB-style free-response; ON SBI3U uses short-answer and extended-response formats.完整大纲对照表见 ../Study Guides/Unit_5_Mendelian_Genetics_and_Heredity.html。阿省 Bio 30 毕业考采用简答风格；安大略 SBI3U 采用短答与扩展作答格式。