AP Biology Quiz: Non Mendelian Genetics

What this quiz covers

This quiz focuses on Non Mendelian Genetics, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Biology.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. About 25% green, 50% green-and-silver, 25% silver, because genotypes follow a 1:2:1 ratio. (correct answer)
2. About 75% green and 25% silver, because green is completely dominant to silver.
3. All offspring green-and-silver, because codominant alleles always remain paired.
4. About 50% green and 50% silver, because heterozygotes split into parental phenotypes.
5. All offspring silver, because S alleles are more common in heterozygous parents.

Explanation: This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically codominance in fish scale color. Both parents are heterozygous GS, contributing G or S alleles, leading to a Punnett square with 25% GG (green), 50% GS (green-and-silver), and 25% SS (silver). Codominance causes both scale colors to appear in heterozygotes, producing the mixed phenotype. The 1:2:1 ratio reflects independent segregation of alleles in this cross. Choice D is tempting but wrong as it assumes heterozygotes split into pure phenotypes, a misconception confusing codominance with blending or dominance. To handle codominance, use Punnett squares and remember that both alleles express equally in heterozygotes.

Question 2

1. Only AB and O
2. Only A and B (correct answer)
3. Only A, B, and AB
4. Only A, B, and O
5. All four types: A, B, AB, and O

Explanation: 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.

Question 3

1. Full color, because C will appear in half the offspring
2. Chinchilla, because c^ch is dominant to c^h and c (correct answer)
3. Himalayan, because c^h is dominant to c^ch and c
4. Albino, because recessive alleles combine in most offspring
5. All phenotypes occur equally because there are four alleles

Explanation: 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.

Question 4

1. All offspring are gray
2. Half the offspring are gray and half are white (correct answer)
3. Three-fourths gray and one-fourth white
4. Half the offspring are black and half are white
5. All offspring are black

Explanation: 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.

Question 5

1. Gold only
2. Silver only
3. Gold-and-silver (both colors visible)
4. No reflective pigment (correct answer)
5. All offspring show gold-and-silver

Explanation: This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically multiple alleles with codominance in fish scale color. The Gg fish produces G or g gametes, and the Sg fish produces S or g gametes, yielding GS, Gg, Sg, and gg offspring each at 25%. The gg genotype lacks reflective pigment since g is recessive to both G and S, matching the no pigment phenotype. This matches choice D, as one-quarter of offspring are gg with this phenotype. A tempting distractor is choice C, assuming gold-and-silver for 25% by misidentifying gg as codominant, a misconception confusing recessivity with codominance. A transferable strategy is to list all genotype combinations and apply the dominance rules to predict phenotype frequencies accurately.

Question 6

1. 3 long : 1 round
2. 1 long : 2 intermediate : 1 round (correct answer)
3. 1 intermediate : 1 round
4. All intermediate
5. 2 long : 1 intermediate : 1 round

Explanation: This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically incomplete dominance in plant leaf shape. Both parents are heterozygous Ll, showing intermediate leaves, and each contributes L or l equally in gametes. The cross yields 25% LL (long), 50% Ll (intermediate), and 25% ll (round), producing a 1:2:1 phenotypic ratio. This matches choice B, as incomplete dominance results in a distinct heterozygous phenotype without blending into dominance. A tempting distractor is choice A, predicting a 3:1 ratio by assuming complete dominance, a misconception that applies Mendelian rules to non-Mendelian traits. For similar questions, use Punnett squares to visualize ratios and remember that incomplete dominance often yields three phenotypes in heterozygous crosses.

Question 7

1. All offspring are roan because both alleles are expressed
2. Half roan and half white offspring (correct answer)
3. Three-fourths roan and one-fourth white offspring
4. Half red and half white offspring
5. All offspring are white because C^W is dominant

Explanation: This question requires analyzing non-Mendelian inheritance through codominance in cattle coat color. In codominance, both alleles in a heterozygote are fully expressed simultaneously, resulting in roan cattle ($C^RC^W$) that have both red and white hairs. When crossing a roan bull ($C^RC^W$) with a white cow ($C^WC^W$), we can predict offspring using a Punnett square: the roan parent can contribute either C^R or C^W, while the white parent can only contribute C^W. This produces offspring that are 50% C^RC^W (roan) and 50% C^WC^W (white), giving a 1:1 ratio. Students often incorrectly choose option A, thinking codominance means all offspring express both traits, but this ignores the white parent's homozygous genotype. When solving codominance problems, remember that both alleles are expressed in heterozygotes, but offspring ratios still depend on parental genotypes.

Question 8

1. 1 red : 1 roan (correct answer)
2. 3 red : 1 roan
3. 1 red : 1 white
4. All roan
5. 3 roan : 1 red

Explanation: This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically codominance in cattle coat color. The roan bull is C^R C^W, expressing both red and white hairs due to codominance, and the red cow is C^R C^R, producing only red hairs. Crossing these yields offspring where half inherit C^R from both parents (red) and half inherit C^R from the bull and C^W from the cow (roan), resulting in a 1:1 ratio. This matches choice A, as codominance allows both alleles to be visible in heterozygotes without blending. A tempting distractor is choice D, assuming all offspring are roan due to misunderstanding codominance as incomplete dominance, a misconception that ignores the equal expression of both alleles. For transferable strategy, identify codominance by checking if heterozygotes show both parental traits distinctly, then apply standard Punnett square analysis.

Question 9

1. All male offspring are patchy
2. Half the male offspring are orange and half are black (correct answer)
3. All male offspring are black
4. Three-fourths of male offspring are orange and one-fourth are black
5. Male offspring phenotypes depend only on the father's X allele

Explanation: 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.

Question 10

1. Only type O
2. Only type AB
3. Only type A or type B (correct answer)
4. Type A, type B, type AB, or type O
5. Only type A

Explanation: 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.

Question 11

1. All offspring have red eyes because W is dominant
2. All daughters are white-eyed and all sons are red-eyed
3. Half the sons are white-eyed, and all daughters are red-eyed (correct answer)
4. Half the daughters are white-eyed, and half the sons are red-eyed
5. One-fourth of all offspring are white-eyed, regardless of sex

Explanation: This question analyzes non-Mendelian inheritance patterns through X-linked traits in fruit flies. The heterozygous female ($X^W$ $X^w$) can pass either X^W or X^w, while the male ($X^W$ Y) passes X^W to daughters and Y to sons. All daughters receive X^W from father and are red-eyed (either $X^W$ $X^W$ or $X^W$ $X^w$), while sons receive Y from father and either X^W (red-eyed) or X^w (white-eyed) from mother with equal probability. Choice D incorrectly suggests daughters can be white-eyed, but they always receive X^W from their father, preventing the white-eye phenotype. When solving X-linked problems, remember that fathers pass their X chromosome only to daughters and their Y chromosome only to sons.

Question 12

1. All sons will be color-blind because their maternal grandfather was affected
2. No sons will be color-blind because the father provides the X chromosome to sons
3. Half of the sons are expected to be color-blind (correct answer)
4. Three-fourths of the sons are expected to be color-blind
5. All sons will have normal vision because the trait is autosomal recessive

Explanation: 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.

Question 13

1. All offspring are lightly striped
2. All offspring are unstriped
3. 1 striped : 1 lightly striped
4. 1 lightly striped : 1 unstriped (correct answer)
5. 3 lightly striped : 1 unstriped

Explanation: 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.

Question 14

1. 1 red : 2 roan : 1 white (correct answer)
2. All roan because both alleles must be present in every offspring
3. 3 roan : 1 white because roan is dominant to white
4. 1 red : 1 white with no roan offspring
5. 3 red : 1 white because red is dominant to white

Explanation: This question requires analyzing non-Mendelian inheritance through codominance in cattle coat color. In codominance, both alleles are fully expressed in heterozygotes, resulting in roan cattle that have both red and white hairs (not a blend). When crossing two heterozygotes (RW × RW), the Punnett square produces: 1 RR (red) : 2 RW (roan) : 1 WW (white), yielding a 1:2:1 phenotypic ratio. Choice C incorrectly applies simple dominance thinking, assuming a 3:1 ratio, but codominance produces three distinct phenotypes, not two. The key strategy for codominance problems is recognizing that heterozygotes express both traits simultaneously, leading to a 1:2:1 ratio in heterozygote crosses.

Question 15

1. All male offspring have red eyes
2. All male offspring have white eyes
3. Half the male offspring have red eyes and half have white eyes (correct answer)
4. Three-fourths of male offspring have red eyes
5. Male eye color depends only on the father's X chromosome

Explanation: This question tests analysis of non-Mendelian inheritance through X-linked traits in fruit flies. The heterozygous red-eyed female ($X^WX^w$) can pass either X^W or X^w to offspring, while the red-eyed male ($X^WY$) passes X^W to daughters and Y to sons. Male offspring receive their single X chromosome from the mother and Y from the father, resulting in 50% X^WY (red-eyed) and 50% X^wY (white-eyed) males. This creates a 1:1 ratio of red to white eyes in male offspring only. Students often incorrectly choose option A, thinking the father's dominant phenotype ensures all offspring have red eyes, but males inherit their X chromosome solely from the mother. For X-linked problems, remember that males express whatever allele is on their single X chromosome, making them more likely to show recessive traits.

Question 16

1. All calves will be roan, because codominant alleles always appear together in offspring.
2. About 50% roan and 50% white, because C^RC^W × C^WC^W produces two genotypes. (correct answer)
3. About 75% roan and 25% red, because roan is dominant to red.
4. All calves will be white, because the white allele is recessive and both parents carry it.
5. About 50% red and 50% white, because heterozygotes separate into parental phenotypes.

Explanation: This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically codominance in cattle coat color. The roan bull has genotype C^RC^W, contributing either C^R or C^W alleles, while the white cow is C^WC^W, contributing only C^W. A Punnett square shows 50% of offspring as C^RC^W (roan) and 50% as C^WC^W (white), reflecting codominance where both alleles express fully in heterozygotes. This distribution is expected because codominance produces distinct phenotypes without blending, unlike incomplete dominance. Choice A is tempting but wrong as it assumes all offspring are roan due to the misconception that codominant alleles are always inherited together, ignoring segregation. For predicting outcomes in codominance, use Punnett squares to track allele segregation and confirm phenotypic expressions.

Question 17

1. All male kittens will be calico, because males inherit both X alleles from the mother.
2. All male kittens will be orange, because the O allele is dominant in all offspring.
3. About 50% orange and 50% black males, because sons inherit either X^O or X^o from the mother. (correct answer)
4. About 75% black and 25% orange males, because a 3:1 ratio applies to sex-linked traits.
5. All male kittens will be black, because the father provides the X chromosome to sons.

Explanation: This question assesses the skill of analyzing non-Mendelian inheritance patterns, specifically sex-linked inheritance in cat fur color. The calico female ($X^OX^o$) contributes X^O or X^o to offspring, while the black male ($X^oY$) contributes Y to males, so male kittens are X^OY (orange) or X^oY (black) at 50% each. This occurs because males express the single X allele they inherit from the mother, with no contribution from the father's X. The patchy calico pattern in females results from X-inactivation, but males show uniform color. Choice A is tempting but incorrect as it assumes males can be calico, a misconception overlooking that males have only one X chromosome. For sex-linked traits in mammals, focus on maternal X inheritance for male phenotypes to predict outcomes.

Question 18

1. All offspring speckled, because codominant alleles must always be inherited together.
2. About 50% solid L^1 and 50% speckled, because offspring are L^1L^1 or L^1L^2. (correct answer)
3. About 25% solid L^1, 50% speckled, 25% solid L^2, because a 1:2:1 ratio applies.
4. All offspring solid L^1, because L^1 is dominant over L^2 in heterozygotes.
5. About 50% solid L^2 and 50% speckled, because L^2 alleles are inherited from the solid parent.

Explanation: 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.

Question 19

1. All offspring are pink because heterozygotes always breed true
2. Half the offspring are red and half are white, with no pink offspring
3. Offspring occur in a 1 red : 2 pink : 1 white phenotypic ratio (correct answer)
4. Three-fourths of offspring are pink and one-fourth are white
5. All offspring are red because R is completely dominant to W

Explanation: 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.

Question 20

1. All offspring will have pink flowers because the heterozygous phenotype is dominant.
2. About 50% pink, 50% white, because only W alleles are passed from one parent.
3. About 75% red, 25% white, because red is completely dominant to white.
4. About 25% red, 50% pink, 25% white, because RW × RW yields a 1:2:1 ratio. (correct answer)
5. About 100% red, because the R allele masks the W allele in heterozygotes.

Explanation: 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.