Genetics Quiz: Incomplete Dominance And Codominance

What this quiz covers

This quiz focuses on Incomplete Dominance And Codominance, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

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. 1 black : 1 gray (correct answer)
2. 1 black : 2 gray : 1 white
3. All gray
4. 3 black : 1 gray

Explanation: The initial cross (gray x gray) producing offspring in a ratio of approximately 1 black : 2 gray : 1 white (32:61:28 ≈ 1:2:1) is characteristic of incomplete dominance. This means gray is the heterozygous phenotype (let's use BW), while black (BB) and white (WW) are homozygous. The second cross is between a gray chicken (BW) and a black chicken (BB). The Punnett square for this cross (BW x BB) yields offspring with genotypes 1/2 BB and 1/2 BW. Therefore, the expected phenotypic ratio is 1 black : 1 gray.

Question 2

1. Cross two oval radishes.
2. Cross an oval radish with a long radish.
3. Cross an oval radish with a round radish. (correct answer)
4. Cross two round radishes once they are obtained.

Explanation: The goal is to produce round-rooted ($S^S S^S$) radishes to start a true-breeding line. We need to find the cross that yields the highest proportion of $S^S S^S$ offspring from the available oval ($S^L S^S$) stock. Let's analyze the options. A) Oval x Oval ($S^L S^S \times S^L S^S$) produces 1/4 round offspring. B) Oval x Long ($S^L S^S \times S^L S^L$) produces 0 round offspring. C) Oval x Round ($S^L S^S \times S^S S^S$) produces 1/2 round offspring. To do this, the breeder must first find or produce at least one round radish, but this cross gives the best yield. D) This is the definition of a true-breeding line, but it is the final goal, not the most efficient first step to produce the necessary individuals from the oval stock. Comparing the yields, cross C is the most efficient way to generate round-rooted individuals.

Question 3

1. 1/8
2. 1/4
3. 1/2
4. 0 (correct answer)

Explanation: First, deduce the parental genotypes. The child is type O (genotype ii), so both parents must carry the recessive i allele. The man is type A, so he is I^A i. The woman is type B, so she is I^B i. The child is type MN (genotype $L^M$ $L^N$), meaning the child received L^M from one parent and L^N from the other. Since the man is type M and the woman is type N, their genotypes must be L^M L^M and L^N L^N, respectively. Thus, the parents are I^A i L^M L^M (man) and I^B i L^N L^N (woman). For the next child, the probability of type AB ($I^A$ $I^B$) is 1/4 from the I^A i x I^B i cross. The probability of type M ($L^M$ $L^M$) requires inheriting an L^M allele from both parents. However, the woman's genotype is L^N L^N, so she can only pass on an L^N allele. Therefore, it is impossible for them to have a child with genotype L^M L^M, and the probability is 0.

Question 4

1. Incomplete dominance, because the heterozygote has intermediate enzyme activity.
2. Codominance, because both alleles produce a quantifiable protein product.
3. Complete dominance, with the disorder being a recessive trait. (correct answer)
4. Complete dominance, with the disorder being a dominant trait.

Explanation: We must evaluate the phenotype for each genotype based on the activity threshold. Assume activity is additive. Genotype $A^1A^1$ has 100 + 100 = 200 units of activity (unaffected). Genotype $A^1A^2$ has 100 + 20 = 120 units of activity (unaffected, as 120 > 50). Genotype $A^2A^2$ has 20 + 20 = 40 units of activity (affected, as 40 < 50). Because the heterozygote ($A^1A^2$) has the same clinical phenotype (unaffected) as the homozygous $A^1A^1$ individual, the $A^1$ allele shows complete dominance over the $A^2$ allele with respect to the disorder. The disorder phenotype only appears in the $A^2A^2$ genotype, making it a recessive trait.

Question 5

1. 1 red : 1 pink
2. 1 red : 2 pink : 1 white
3. 3 red : 1 pink
4. 9 red : 6 pink : 1 white (correct answer)

Explanation: This is a multi-step problem. Step 1: Perform the backcross: $C^R C^W$ (F1) × $C^R C^R$ (red parent). The progeny are 1/2 $C^R C^R$ (red) and 1/2 $C^R C^W$ (pink). Step 2: Determine the allele frequencies in the gene pool of these progeny. The frequency of the $C^R$ allele (p) is (1/2) * 1 + (1/2) * (1/2) = 0.5 + 0.25 = 0.75. The frequency of the $C^W$ allele (q) is (1/2) * (1/2) = 0.25. Step 3: Use the Hardy-Weinberg principle for the next generation after random pollination. The phenotypic ratio will correspond to the genotypic ratio $p^2 : 2pq : q^2$. Red ($C^R C^R$) = $p^2 = (0.75)^2 = 0.5625 = 9/16$. Pink ($C^R C^W$) = $2pq = 2(0.75)(0.25) = 0.375 = 6/16$. White ($C^W C^W$) = $q^2 = (0.25)^2 = 0.0625 = 1/16$. The resulting ratio is 9 red : 6 pink : 1 white.

Question 6

1. 50 mg/g
2. 30 mg/g
3. 25 mg/g (correct answer)
4. 24 mg/g

Explanation: First, determine allele frequencies from the population data. White fish have genotype $D^2D^2$, so their frequency is $q^2 = 0.16$. The frequency of the $D^2$ allele is $q = \sqrt{0.16} = 0.4$. The frequency of the $D^1$ allele is $p = 1 - q = 1 - 0.4 = 0.6$. Next, calculate the genotype frequencies: $p^2 (D^1D^1) = (0.6)^2 = 0.36$, $2pq (D^1D^2) = 2(0.6)(0.4) = 0.48$, and $q^2 (D^2D^2) = 0.16$. The most common genotype is the heterozygote $D^1D^2$ with a frequency of 48%. Due to incomplete dominance, the heterozygous phenotype is intermediate. Its pigment concentration will be the average of the two homozygotes: (50 mg/g + 0 mg/g) / 2 = 25 mg/g.

Question 7

1. 1/2
2. 1/4 (correct answer)
3. 0
4. 3/4

Explanation: The cross is $S^A s \times S^B s$. The possible offspring genotypes are $S^A S^B$, $S^A s$, $S^B s$, and $ss$, each with a probability of 1/4. The phenotype of both spots and stripes occurs only in the $S^A S^B$ genotype, because $S^A$ and $S^B$ are codominant. The probability of this genotype is 1/4. The other genotypes produce spots only ($S^A s$), stripes only ($S^B s$), or no markings ($ss$).

Question 8

1. Fur color is incompletely dominant in the first species and shows complete dominance in the second. (correct answer)
2. Fur color is codominant in the first species and incompletely dominant in the second.
3. A lethal allele is involved in the first species, while the second shows complete dominance.
4. Both species show incomplete dominance, but the second cross had a significant sampling error.

Explanation: The 1:2:1 phenotypic ratio in the first species is the classic result of a monohybrid cross where the alleles exhibit incomplete dominance. The gray phenotype is heterozygous, while black and white are the two homozygous phenotypes. The 3:1 phenotypic ratio in the second species is the classic result of a monohybrid cross involving complete dominance, where the gray allele is dominant over the white allele. In this case, both homozygous dominant and heterozygous individuals would have gray fur.

Question 9

1. 9/16
2. 3/16
3. 3/8 (correct answer)
4. 1/4

Explanation: This is a dihybrid cross problem. First, consider each trait independently. For flower color (Rr x Rr), the genotypic ratio is 1 RR : 2 Rr : 1 rr, and the phenotypic ratio is 1 red : 2 pink : 1 white. The probability of pink flowers (Rr) is 1/2. For leaf width (Bb x Bb), the phenotypic ratio is 3 broad (B_) : 1 narrow (bb). The probability of broad leaves is 3/4. To find the proportion of offspring with both pink flowers and broad leaves, multiply their independent probabilities: P(pink) × P(broad) = (1/2) × (3/4) = 3/8.

Question 10

1. 0
2. 1/4 (correct answer)
3. 1/2
4. 3/4

Explanation: Both parents have sickle-cell trait, meaning their genotype is HbA HbS. The cross is HbA HbS × HbA HbS. The offspring genotypes will be in the ratio 1 HbA HbA : 2 HbA HbS : 1 HbS HbS. A child with normal hemoglobin and susceptibility to malaria has the genotype HbA HbA (homozygous normal). The probability of this genotype from the cross is 1/4.

Question 11

1. A mosaic of glycoproteins, with some cells expressing only the M-type and others expressing only the N-type.
2. Only the M-type glycoprotein, as the L^M allele is dominant.
3. A novel, hybrid glycoprotein formed by the combined action of both alleles.
4. Both the M-type and the N-type glycoproteins are present and expressed simultaneously. (correct answer)

Explanation: Codominance is a mode of inheritance where two different alleles for a gene are both fully expressed in the heterozygote's phenotype. In the case of the MN blood group, an individual with genotype L^M L^N produces both the M-type glycoprotein and the N-type glycoprotein. Both variants are present on the surface of each red blood cell. Option A describes mosaicism (like X-inactivation), not codominance. Option C describes a blended or intermediate product, which is more characteristic of incomplete dominance. Option B incorrectly assumes a dominant/recessive relationship.

Question 12

1. Tortoiseshell females and black males
2. Tortoiseshell females and orange males (correct answer)
3. Black females and orange males
4. Orange females and black males

Explanation: This cross demonstrates both codominance at the cellular level (tortoiseshell coat) and X-linked inheritance. The black male's genotype is $X^B Y$. The orange female's genotype is $X^O X^O$. All female offspring will inherit an $X^B$ from the father and an $X^O$ from the mother, giving them the genotype $X^B X^O$, which results in a tortoiseshell phenotype. All male offspring will inherit a Y chromosome from the father and an $X^O$ from the mother, giving them the genotype $X^O Y$, which results in an orange phenotype.

Question 13

1. Approximately 151
2. Approximately 303 (correct answer)
3. Approximately 454
4. Approximately 605

Explanation: The F1 generation results from a cross between two true-breeding parents with different phenotypes, and it displays an intermediate phenotype (light blue). This indicates incomplete dominance. The F1 plants are heterozygous (let's use genotype $B_1B_2$). When these F1 plants are self-crossed ($B_1B_2$ x $B_1B_2$), the expected genotypic ratio in the F2 generation is 1 B_1B_1 : 2 B_1B_2 : 1 B_2B_2. The question asks for the number of F2 plants with a genotype identical to the F1 parent, which is the heterozygous genotype ($B_1B_2$). The expected proportion of heterozygotes is 2/4, or 1/2. Therefore, the expected number is (1/2) * 605 = 302.5, which is approximately 303 plants.

Question 14

1. 1/3 (correct answer)
2. 1/4
3. 1/2
4. 2/3

Explanation: When you encounter incomplete dominance problems, remember that heterozygotes show a blended phenotype, and you're often dealing with conditional probability when offspring are selectively removed. Let's work through this pink × pink cross ($R_1R_2 \times R_1R_2$). Using a Punnett square, the offspring ratios are: 1 $R_1R_1$ (red) : 2 $R_1R_2$ (pink) : 1 $R_2R_2$ (white). This gives us 4 total offspring in a 1:2:1 ratio. Since the breeder discards all white flowers, we remove the 1 $R_2R_2$ offspring from consideration. This leaves us with only 3 remaining plants: 1 red and 2 pink. The probability that a randomly selected plant from these remaining offspring will be red is therefore $\frac{1}{3}$. Looking at the wrong answers: B) $\frac{1}{4}$ represents the original probability of getting red offspring before any were discarded—this ignores the conditional aspect. C) $\frac{1}{2}$ incorrectly assumes equal numbers of red and pink offspring remain, forgetting that incomplete dominance produces twice as many heterozygotes. D) $\frac{2}{3}$ gives the probability of selecting a pink flower from the remaining offspring, which is the complement of what we want. The key insight is recognizing this as conditional probability: you're not asking about the original cross outcomes, but about a subset after certain individuals are removed. Always recalculate your denominator when offspring are selectively discarded—this changes the sample space entirely.

Question 15

1. 9/16
2. 3/16
3. 3/8 (correct answer)
4. 1/4

Explanation: This is a dihybrid cross problem. First, consider each trait independently. For flower color (Rr x Rr), the genotypic ratio is 1 RR : 2 Rr : 1 rr, and the phenotypic ratio is 1 red : 2 pink : 1 white. The probability of pink flowers (Rr) is 1/2. For leaf width (Bb x Bb), the phenotypic ratio is 3 broad (B_) : 1 narrow (bb). The probability of broad leaves is 3/4. To find the proportion of offspring with both pink flowers and broad leaves, multiply their independent probabilities: P(pink) × P(broad) = (1/2) × (3/4) = 3/8.

Question 16

1. 1/2
2. 1/4 (correct answer)
3. 0
4. 3/4

Explanation: The cross is $S^A s \times S^B s$. The possible offspring genotypes are $S^A S^B$, $S^A s$, $S^B s$, and $ss$, each with a probability of 1/4. The phenotype of both spots and stripes occurs only in the $S^A S^B$ genotype, because $S^A$ and $S^B$ are codominant. The probability of this genotype is 1/4. The other genotypes produce spots only ($S^A s$), stripes only ($S^B s$), or no markings ($ss$).

Question 17

1. Tortoiseshell females and black males
2. Tortoiseshell females and orange males (correct answer)
3. Black females and orange males
4. Orange females and black males

Explanation: This cross demonstrates both codominance at the cellular level (tortoiseshell coat) and X-linked inheritance. The black male's genotype is $X^B Y$. The orange female's genotype is $X^O X^O$. All female offspring will inherit an $X^B$ from the father and an $X^O$ from the mother, giving them the genotype $X^B X^O$, which results in a tortoiseshell phenotype. All male offspring will inherit a Y chromosome from the father and an $X^O$ from the mother, giving them the genotype $X^O Y$, which results in an orange phenotype.

Question 18

1. Cross two oval radishes.
2. Cross an oval radish with a long radish.
3. Cross an oval radish with a round radish. (correct answer)
4. Cross two round radishes once they are obtained.

Explanation: The goal is to produce round-rooted ($S^S S^S$) radishes to start a true-breeding line. We need to find the cross that yields the highest proportion of $S^S S^S$ offspring from the available oval ($S^L S^S$) stock. Let's analyze the options. A) Oval x Oval ($S^L S^S \times S^L S^S$) produces 1/4 round offspring. B) Oval x Long ($S^L S^S \times S^L S^L$) produces 0 round offspring. C) Oval x Round ($S^L S^S \times S^S S^S$) produces 1/2 round offspring. To do this, the breeder must first find or produce at least one round radish, but this cross gives the best yield. D) This is the definition of a true-breeding line, but it is the final goal, not the most efficient first step to produce the necessary individuals from the oval stock. Comparing the yields, cross C is the most efficient way to generate round-rooted individuals.

Question 19

1. 0
2. 1/4 (correct answer)
3. 1/2
4. 3/4

Explanation: Both parents have sickle-cell trait, meaning their genotype is HbA HbS. The cross is HbA HbS × HbA HbS. The offspring genotypes will be in the ratio 1 HbA HbA : 2 HbA HbS : 1 HbS HbS. A child with normal hemoglobin and susceptibility to malaria has the genotype HbA HbA (homozygous normal). The probability of this genotype from the cross is 1/4.

Question 20

1. 1/3 (correct answer)
2. 1/4
3. 1/2
4. 2/3

Explanation: When you encounter incomplete dominance problems, remember that heterozygotes show a blended phenotype, and you're often dealing with conditional probability when offspring are selectively removed. Let's work through this pink × pink cross ($R_1R_2 \times R_1R_2$). Using a Punnett square, the offspring ratios are: 1 $R_1R_1$ (red) : 2 $R_1R_2$ (pink) : 1 $R_2R_2$ (white). This gives us 4 total offspring in a 1:2:1 ratio. Since the breeder discards all white flowers, we remove the 1 $R_2R_2$ offspring from consideration. This leaves us with only 3 remaining plants: 1 red and 2 pink. The probability that a randomly selected plant from these remaining offspring will be red is therefore $\frac{1}{3}$. Looking at the wrong answers: B) $\frac{1}{4}$ represents the original probability of getting red offspring before any were discarded—this ignores the conditional aspect. C) $\frac{1}{2}$ incorrectly assumes equal numbers of red and pink offspring remain, forgetting that incomplete dominance produces twice as many heterozygotes. D) $\frac{2}{3}$ gives the probability of selecting a pink flower from the remaining offspring, which is the complement of what we want. The key insight is recognizing this as conditional probability: you're not asking about the original cross outcomes, but about a subset after certain individuals are removed. Always recalculate your denominator when offspring are selectively discarded—this changes the sample space entirely.