Non-Mendelian Genetics - AP Biology

1. In humans, ABO blood type is determined by three alleles: $I^A$, $I^B$, and i. $I^A$ and $I^B$ are codominant, and both are dominant to i. A person with type AB blood ($I^AI^B$) has a child with a person with type O blood (ii). Which set lists all possible child blood types from this cross?
   Answer: Types A and B only This question examines non-Mendelian inheritance through multiple alleles and codominance in human ABO blood types. The AB parent ($I^AI^B$) can produce gametes with either $I^A$ or $I^B$, while the O parent (ii) can only produce gametes with i. Crossing I^AI^B × ii yields offspring that are 50% I^Ai (type A blood, as $I^A$ dominates i) and 50% I^Bi (type B blood, as $I^B$ dominates i). No offspring can be type AB (requires both $I^A$ and $I^B$) or type O (requires ii), making option C correct. Students often incorrectly choose option D, thinking all blood types are possible from any cross, but this ignores that the O parent cannot contribute dominant alleles. When solving ABO blood type problems, list each parent's possible gametes and remember that $I^A$ and $I^B$ are codominant to each other but both dominate i.
2. In a plant, leaf pattern is codominant. Allele $L^S$ produces solid green leaves and allele $L^M$ produces mottled leaves. Homozygotes ($L^SL^S$) are solid, ($L^ML^M$) are mottled, and heterozygotes ($L^SL^M$) show both patterns on the same leaf. Two heterozygous plants are crossed. Which phenotype ratio is expected among offspring?
   Answer: 1 solid : 2 both : 1 mottled This problem involves analyzing non-Mendelian inheritance through codominance in plant leaf patterns. When two heterozygous plants ($L^S$ $L^M$) are crossed, the offspring genotypes are: 25% $L^S$ $L^S$ (solid), 50% $L^S$ $L^M$ (both patterns), and 25% $L^M$ $L^M$ (mottled). This produces a 1 solid : 2 both : 1 mottled phenotypic ratio, making answer B correct. Students often confuse codominance with incomplete dominance and expect a blended phenotype rather than both traits expressed simultaneously (answer D). The key strategy for codominance problems is to remember that heterozygotes express both alleles fully, creating a distinct phenotype that shows both parental traits.
3. In humans, red-green color blindness is caused by a recessive allele $c$ on the X chromosome; allele $C$ results in normal color vision. A woman with normal vision whose father was color-blind has a child with a man with normal vision ($X^CY$). Which outcome best predicts the probability that a son will be color-blind?
   Answer: 50% This question examines non-Mendelian inheritance of X-linked recessive traits in humans. The woman's father was color-blind ($X^c$ Y), so she must have inherited $X^c$ from him, making her $X^C$ $X^c$ (carrier). When crossed with a normal-vision man ($X^C$ Y), sons inherit Y from father and either $X^C$ or $X^c$ from mother with equal probability. This results in 50% $X^C$ Y (normal) and 50% $X^c$ Y (color-blind) sons, making answer C correct. A common misconception is thinking carriers cannot have affected offspring (answer A), forgetting that X-linked carriers can pass the recessive allele to sons. The strategy is to deduce the mother's genotype from her father's phenotype, then track X chromosome inheritance to sons.
4. A lizard species has multiple alleles for skin pattern at one locus: S (striped) is dominant to s (spotted). A third allele $s^n$ produces no pattern and is recessive to both S and s. A striped lizard with genotype $S s^n$ is crossed with a spotted lizard with genotype $s s^n$. Which offspring phenotype ratio is expected?
   Answer: 2 striped : 1 spotted : 1 no pattern This question tests analysis of non-Mendelian inheritance involving multiple alleles with a dominance hierarchy. The S $s^n$ parent can contribute S or $s^n$ gametes, while the s $s^n$ parent can contribute s or $s^n$ gametes. The possible offspring are: Ss (striped), S $s^n$ (striped), s $s^n$ (spotted), and $s^n$ $s^n$ (no pattern), each with 25% probability. This gives 2 striped : 1 spotted : 1 no pattern ratio. Choice D incorrectly reverses the striped and spotted frequencies, likely confusing which alleles came from which parent. For multiple allele problems, carefully track the dominance hierarchy and use a Punnett square to determine all possible genotype combinations and their corresponding phenotypes.
5. In snapdragons, flower color shows incomplete dominance: allele R produces red pigment and allele W produces no pigment. RR plants have red flowers, WW plants have white flowers, and RW plants have pink flowers. A student crosses two pink-flowered plants (RW × RW) and grows 160 offspring under the same conditions. Assuming fertilization is random and the trait is controlled by this single gene, which outcome best predicts the numbers of red, pink, and white flowers among the offspring?
   Answer: 40 red, 80 pink, 40 white This question tests your ability to analyze non-Mendelian inheritance patterns, specifically incomplete dominance in snapdragons. In incomplete dominance, heterozygotes (RW) show a blended phenotype (pink) rather than expressing one allele over another. When crossing two pink-flowered plants (RW × RW), the Punnett square yields: 1 RR (red) : 2 RW (pink) : 1 WW (white), which is a 1:2:1 ratio. With 160 offspring, this translates to 40 red, 80 pink, and 40 white flowers, making answer A correct. Students often mistakenly apply complete dominance rules and expect a 3:1 ratio (answer C), forgetting that incomplete dominance produces three distinct phenotypes. The key strategy is to recognize incomplete dominance from the problem description (heterozygotes have an intermediate phenotype) and apply the 1:2:1 ratio for F2 crosses.
6. In snapdragons, flower color shows incomplete dominance: allele R produces red pigment, allele W produces no pigment. RR plants have red flowers, WW plants have white flowers, and RW plants have pink flowers. A gardener crosses two pink (RW) snapdragon plants and grows 200 offspring under the same conditions. Which outcome best predicts the distribution of flower colors among the offspring?
   Answer: About 25% red, 50% pink, 25% white, because RW × RW yields a 1:2:1 ratio. This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically incomplete dominance in snapdragon flower color. The cross involves two heterozygous RW plants, where R produces red pigment and W produces no pigment, resulting in pink flowers for RW due to incomplete dominance. A Punnett square for RW × RW yields 25% RR (red), 50% RW (pink), and 25% WW (white) offspring, matching the 1:2:1 phenotypic ratio observed in incomplete dominance. This prediction holds because the gardener grows 200 offspring under identical conditions, ensuring the ratio approximates the expected probabilities. Choice E is tempting but incorrect as it assumes complete dominance where R fully masks W, a common misconception confusing incomplete dominance with Mendelian dominance. To analyze similar inheritance patterns, always construct a Punnett square to visualize allele combinations and resulting phenotypes.
7. In a plant, leaf shape is determined by a gene with multiple alleles. Allele $L^1$ and $L^2$ are codominant, producing a speckled leaf when together ($L^1L^2$). Genotypes L^1L^1 and L^2L^2 produce solid leaves of two different colors. A speckled plant ($L^1L^2$) is crossed with a solid $L^1$ plant ($L^1L^1$). Which outcome best predicts the leaf phenotypes of the offspring?
   Answer: About 50% solid $L^1$ and 50% speckled, because offspring are L^1L^1 or L^1L^2. This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically codominance with multiple alleles in plant leaf shapes. The speckled plant ($L^1L^2$) contributes $L^1$ or $L^2$, while the solid $L^1$ plant ($L^1L^1$) contributes only $L^1$, yielding 50% L^1L^1 (solid $L^1$) and 50% L^1L^2 (speckled). Codominance ensures both alleles express in heterozygotes, producing the speckled phenotype without dominance. No solid $L^2$ offspring occur because the second parent lacks $L^2$. Choice C is tempting but wrong as it assumes a 1:2:1 ratio like a heterozygous cross, a misconception ignoring the homozygous parent's contribution. In codominance problems, identify gametes from each parent to accurately predict phenotypic ratios.
8. In a fish, body pattern shows incomplete dominance: allele S produces stripes and allele s produces no stripes. SS fish are striped, ss fish are unstriped, and Ss fish are lightly striped. A lightly striped fish is crossed with an unstriped fish (Ss × ss). Which outcome best illustrates the expected offspring phenotypes?
   Answer: 1 lightly striped : 1 unstriped This question involves analyzing non-Mendelian inheritance through incomplete dominance in fish body patterns. The cross between a lightly striped fish (Ss) and an unstriped fish (ss) produces offspring genotypes: 50% Ss (lightly striped) and 50% ss (unstriped). This yields a 1 lightly striped : 1 unstriped ratio, making answer D correct. Students often apply complete dominance thinking and expect all offspring to show stripes (answer A) or expect a 3:1 ratio (answer E). The key strategy for incomplete dominance is to recognize that heterozygotes have an intermediate phenotype and to set up a simple Punnett square for the testcross.
9. In snapdragons, flower color shows incomplete dominance: allele R produces red, allele W produces white, and heterozygotes (RW) are pink. A gardener crosses two pink plants (RW × RW). Which offspring outcome is most consistent with this inheritance pattern?
   Answer: Offspring occur in a 1 red : 2 pink : 1 white phenotypic ratio This question tests your ability to analyze non-Mendelian inheritance patterns, specifically incomplete dominance in snapdragons. In incomplete dominance, heterozygotes show a blended phenotype (pink) that is intermediate between the two homozygous phenotypes (red and white). When crossing two heterozygotes (RW × RW), the Punnett square yields: 1 RR (red) : 2 RW (pink) : 1 WW (white), giving a 1:2:1 phenotypic ratio. Choice A incorrectly assumes heterozygotes breed true, which contradicts Mendel's law of segregation where heterozygotes produce gametes with different alleles. To solve incomplete dominance problems, remember that heterozygotes have a distinct intermediate phenotype, and crosses between heterozygotes always produce a 1:2:1 ratio.
10. In humans, red-green color blindness is an X-linked recessive trait. A woman with normal vision whose father was color-blind has children with a man who has normal vision. Which outcome best predicts the sons' phenotypes?
    Answer: Half of the sons are expected to be color-blind This question tests understanding of non-Mendelian inheritance through X-linked recessive traits. The woman's father was color-blind ($X^c$ Y), so she must have inherited his $X^c$ chromosome, making her a carrier ($X^C$ $X^c$). When crossed with a normal-vision male ($X^C$ Y), half her sons will inherit her $X^c$ and be color-blind ($X^c$ Y), while half will inherit her $X^C$ and have normal vision ($X^C$ Y). Choice A incorrectly assumes all sons are affected because the maternal grandfather was, ignoring that the mother has two different X chromosomes to pass on. For X-linked problems, track X chromosomes carefully and remember that males express whatever allele is on their single X chromosome.
11. In humans, red-green color blindness is caused by an X-linked recessive allele c; the normal-vision allele C is dominant. A woman with normal vision whose father was color-blind mates with a man with normal vision ($X^CY$). The woman's genotype is X^CX^c. Which outcome best predicts the probability that a son will be color-blind?
    Answer: 50%, because each son has a one-half chance of inheriting $X^c$ from the mother. This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically sex-linked recessive inheritance in human color blindness. The carrier mother ($X^CX^c$) contributes $X^C$ or $X^c$ with equal probability, while the normal father ($X^CY$) contributes Y to sons, so each son has a 50% chance of being X^cY (color-blind). This probability arises because sons inherit their single X from the mother, and c is recessive but expresses in hemizygous males. Daughters would be carriers or normal, but the question focuses on sons. Choice A is tempting but wrong as it assumes sons inherit X from the father, a misconception about sex chromosome inheritance. For X-linked recessive traits, calculate probabilities separately for sons and daughters based on maternal allele transmission.
12. In a rabbit population, coat color is controlled by multiple alleles at one locus with a dominance hierarchy: C (full color) > c^ch (chinchilla) > $c^h$ (Himalayan) > c (albino). A chinchilla rabbit with genotype c^ch c is crossed with a Himalayan rabbit with genotype $c^h$ c. Assuming simple dominance by the hierarchy, which offspring phenotype is expected to be most common?
    Answer: Chinchilla, because c^ch is dominant to $c^h$ and c This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically multiple alleles with a dominance hierarchy in rabbit coat color. The chinchilla rabbit is c^ch c, producing c^ch or c gametes, and the Himalayan is $c^h$ c, producing $c^h$ or c gametes. Offspring genotypes are c^ch $c^h$ (chinchilla, since $c^ch$ > $c^h$), c^ch c (chinchilla, $c^ch$ > c), $c^h$ c (Himalayan, $c^h$ > c), and c c (albino), with chinchilla in 50% of cases. This matches choice B, as c^ch dominates over both $c^h$ and c in the hierarchy. A tempting distractor is choice D, predicting mostly albino by ignoring the hierarchy, a misconception treating all alleles as equally recessive. For a transferable strategy, always reference the dominance order when predicting phenotypes in multiple-allele systems and count frequencies accordingly.
13. Human ABO blood type is controlled by three alleles: $I^A$, $I^B$, and $i$. $I^A$ and $I^B$ are codominant, and both are dominant to $i$. A person with type AB ($I^A I^B$) has a child with a person with type O ($ii$). Which blood type(s) can their child have?
    Answer: Only type A or type B This question tests analysis of non-Mendelian inheritance involving multiple alleles and codominance in human ABO blood types. The AB parent ($I^A$ $I^B$) can contribute either $I^A$ or $I^B$, while the O parent (ii) can only contribute i. The possible offspring genotypes are $I^A$ i (type A) and $I^B$ i (type B), with equal probability. Choice D incorrectly suggests all blood types are possible, but this cross cannot produce type AB (requires both $I^A$ and $I^B$) or type O (requires ii). When solving ABO blood type problems, systematically determine which alleles each parent can contribute and remember that i is recessive to both $I^A$ and $I^B$.
14. In a cat species, fur pattern is X-linked. Allele $X^O$ produces orange fur and allele $X^o$ produces black fur. Heterozygous females ($X^O$ $X^o$) show patches of orange and black. A patchy female is crossed with a black male ($X^o$ Y). Assume no other genes affect color. Which outcome best predicts the phenotypes of male offspring?
    Answer: Half the male offspring are orange and half are black This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically X-linked traits in cats. The patchy female is $X^O$ $X^o$, contributing $X^O$ or $X^o$ to sons, while the black male is $X^o$ Y, contributing Y to sons. Male offspring (XY) are thus 50% $X^O$ Y (orange) and 50% $X^o$ Y (black), with no patchy pattern possible in males due to single X. This matches choice B, reflecting the equal inheritance from the heterozygous mother. A tempting distractor is choice C, predicting all black males by assuming the father's allele dominates, a misconception ignoring that sons inherit their X from the mother only. For transferable strategy, focus on X inheritance from the mother for male phenotypes in X-linked traits and calculate ratios based on her genotype.
15. In a plant, leaf shape shows codominance: allele L produces lobed leaves, allele R produces round leaves, and heterozygotes (LR) have both lobed and round sections on the same leaf. A lobed plant (LL) is crossed with a plant showing both sections (LR). Which offspring phenotypes are expected?
    Answer: Half lobed and half show both sections This question examines non-Mendelian inheritance through codominance in plant leaf shape. The lobed parent (LL) can only contribute L alleles, while the codominant parent (LR) can contribute either L or R. The possible offspring genotypes are LL (lobed) and LR (showing both sections) in equal proportions. Choice C incorrectly suggests round offspring are possible, but this would require RR genotype, which cannot occur when one parent contributes only L alleles. When solving codominance problems, remember that heterozygotes express both traits simultaneously and determine which allele combinations are actually possible from the given parents.
16. A human blood type gene has three alleles: $I^A$, $I^B$, and $i$. Alleles $I^A$ and $I^B$ are codominant, and both are dominant to $i$. A person with type AB blood ($I^AI^B$) has a child with a person with type O blood ($ii$). Which set of possible child blood types is expected?
    Answer: Only A and B This problem involves analyzing non-Mendelian inheritance with codominant alleles in human blood types. The AB parent ($I^A$ $I^B$) can contribute either $I^A$ or $I^B$, while the O parent (ii) can only contribute i. The possible offspring genotypes are $I^A$ i (type A blood) and $I^B$ i (type B blood), each with 50% probability. This makes answer B correct, as children can only have type A or B blood. Students often mistakenly include AB as a possibility (answer C), forgetting that the O parent cannot contribute $I^A$ or $I^B$ alleles. The key strategy is to recognize that each parent contributes one allele and that the O parent (ii) can only pass on the recessive i allele.
17. In a bird species, feather color shows incomplete dominance: allele B yields black, allele W yields white, and BW heterozygotes are gray. A gray bird (BW) is crossed with a white bird (WW). Which outcome best predicts the offspring phenotypes?
    Answer: Half the offspring are gray and half are white This question examines non-Mendelian inheritance through incomplete dominance in bird feather color. The gray bird (BW) can contribute either B or W alleles, while the white bird (WW) can only contribute W. The possible offspring genotypes are BW (gray) and WW (white) in equal proportions, resulting in half gray and half white offspring. Choice C incorrectly applies simple dominance thinking with a 3:1 ratio, but incomplete dominance produces distinct phenotypes for each genotype. To solve incomplete dominance crosses, identify the possible gametes from each parent and remember that heterozygotes have an intermediate phenotype distinct from both homozygotes.
18. In a cat breed, tail length shows incomplete dominance: allele T produces long tails, allele t produces short tails, and heterozygotes (Tt) have medium tails. Two medium-tailed cats are crossed (Tt × Tt). Which prediction best matches the offspring?
    Answer: 1 long : 2 medium : 1 short This question tests understanding of non-Mendelian inheritance through incomplete dominance in cat tail length. When crossing two heterozygotes (Tt × Tt) showing incomplete dominance, the Punnett square yields: 1 TT (long) : 2 Tt (medium) : 1 tt (short), producing a 1:2:1 phenotypic ratio. Choice B incorrectly applies simple dominance thinking with a 3:1 ratio, failing to recognize that incomplete dominance produces three distinct phenotypes. The key to solving incomplete dominance problems is recognizing that heterozygotes have an intermediate phenotype and that crosses between heterozygotes always yield a 1:2:1 ratio, not the 3:1 ratio seen in simple dominance.
19. In a bird species, feather pattern is sex-linked on the Z chromosome. Allele $Z^B$ produces barred feathers and is dominant to $Z^b$, which produces nonbarred feathers. Males are ZZ and females are ZW. A barred female ($Z^BW$) is crossed with a nonbarred male ($Z^bZ^b$). Which prediction best describes the offspring phenotypes?
    Answer: All sons are barred and all daughters are nonbarred This question requires analyzing non-Mendelian inheritance in a ZW sex determination system where males are ZZ and females are ZW. The barred female ($Z^BW$) can pass either $Z^B$ or W to offspring, while the nonbarred male ($Z^bZ^b$) can only pass $Z^b$. All sons receive one Z from each parent ($Z^BZ^b$), making them barred due to $Z^B$ dominance. All daughters receive $Z^b$ from father and W from mother ($Z^bW$), making them nonbarred since they express the single Z chromosome's allele. This produces sex-specific phenotypes: all sons barred, all daughters nonbarred. Students often incorrectly choose option B, reversing the inheritance pattern by forgetting that in ZW systems, females are the heterogametic sex. When solving ZW sex-linked problems, track which parent contributes which chromosome to each sex.
20. A human blood type gene has three alleles: $I^A$, $I^B$, and i. $I^A$ and $I^B$ are codominant to each other, and both are dominant over i. A parent with blood type AB has genotype I^AI^B. The other parent has blood type O with genotype ii. Which outcome best predicts the possible blood types of their children?
    Answer: Types A or B only, because children can be I^Ai or I^Bi. This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically multiple alleles and codominance in human blood types. The AB parent ($I^AI^B$) contributes $I^A$ or $I^B$, while the O parent (ii) contributes only i, leading to offspring genotypes I^Ai (type A) or I^Bi (type B). This results in only types A or B because i is recessive, and $I^A$ and $I^B$ are codominant but each pairs separately with i. No AB or O offspring occur since children lack both $I^A$ and $I^B$ together or ii. Choice E is tempting but incorrect as it assumes all allele combinations are possible, a misconception overlooking parental genotypes limiting inheritance. To solve multiple-allele problems, list possible gametes from each parent and determine resulting genotypes and phenotypes.

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