Genetics Quiz
Practice Incomplete Dominance And Codominance in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.
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A geneticist crosses two chickens with gray feathers and obtains 32 black, 61 gray, and 28 white offspring. A subsequent cross is performed between a gray chicken from this F1 generation and a black chicken. What is the expected phenotypic ratio in the offspring of this second cross?
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.
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A geneticist crosses two chickens with gray feathers and obtains 32 black, 61 gray, and 28 white offspring. A subsequent cross is performed between a gray chicken from this F1 generation and a black chicken. What is the expected phenotypic ratio in the offspring of this second cross?
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.
In radishes, root shape is controlled by a single gene with two alleles ((S^L) and (S^S)) that exhibit incomplete dominance. The phenotypes are long ((S^L S^L)), oval ((S^L S^S)), and round ((S^S S^S)). A breeder possesses a large stock of oval-rooted radishes and wishes to establish a true-breeding line of round-rooted radishes as quickly as possible. Which of the following crosses represents the most efficient first step?
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.
The human MN and ABO blood group systems assort independently. For the MN system, alleles L^M and L^N are codominant. For the ABO system, I^A and I^B are codominant and dominant to i. A man with blood type A and M marries a woman with blood type B and N. Their first child has blood type O and MN. What is the probability that their next child will have blood type AB and M?
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.
A hypothetical genetic disorder is caused by a single gene. The (A^1) allele codes for an enzyme with 100 units of activity, while the (A^2) allele codes for a related enzyme with only 20 units of activity. An individual is considered clinically affected by the disorder if their total enzyme activity is below 50 units. What is the pattern of inheritance for the clinical disorder phenotype?
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.
In four o'clock plants, flower color is determined by incomplete dominance. A cross between a true-breeding red-flowered plant ((C^R C^R)) and a true-breeding white-flowered plant ((C^W C^W)) produces all pink-flowered F1 offspring ((C^R C^W)). An F1 plant is then backcrossed to its red-flowered parent. If the progeny from this backcross are allowed to randomly pollinate each other, what will be the phenotypic ratio in the next generation?
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.
The color of a fish species is controlled by a single gene with two incompletely dominant alleles, (D^1) and (D^2). (D^1D^1) fish have a dark blue phenotype, corresponding to a pigment concentration of 50 mg/g. (D^2D^2) fish are white, with 0 mg/g of pigment. In a large, randomly mating population in Hardy-Weinberg equilibrium, 16% of the fish are white. What is the expected pigment concentration in a fish with the most common phenotype in this population?
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.
A species of clover has a gene for leaf markings with three alleles: (S^A), (S^B), and (s). Alleles (S^A) and (S^B) are codominant, producing spots and stripes, respectively. Both (S^A) and (S^B) are completely dominant over the recessive allele (s), which results in no markings. A cross is performed between a clover with genotype (S^A s) and a clover with genotype (S^B s). What is the expected proportion of offspring that will exhibit both spots and stripes?
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)).
In a certain animal species, a cross between two gray-furred individuals produces offspring in the ratio of 1 black : 2 gray : 1 white. In a different species, a cross between two gray-furred individuals produces offspring in the ratio of 3 gray : 1 white. Which statement provides the best genetic explanation for these different outcomes?
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.
In snapdragons, flower color exhibits incomplete dominance (RR = red, Rr = pink, rr = white), and leaf width exhibits complete dominance (B = broad, b = narrow). A plant that is heterozygous for both traits is self-pollinated. What proportion of the offspring is expected to have pink flowers and broad leaves?
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.
Sickle-cell anemia is a human genetic disorder where alleles for normal hemoglobin (HbA) and sickle-cell hemoglobin (HbS) are codominant. Individuals with genotype HbS HbS have severe anemia, while those with HbA HbS have the milder sickle-cell trait and are resistant to malaria. Two individuals who both have sickle-cell trait have a child. What is the probability that this child will have normal hemoglobin and be susceptible to malaria?
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.
The L^M and L^N alleles for the human MN blood group code for two distinct glycoprotein variants on the surface of red blood cells. In an individual with genotype L^M L^N, which statement best describes the composition of glycoproteins on the surface of a single red blood cell?
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.
In cats, the gene for coat color is X-linked. One allele produces black fur ((X^B)) and another produces orange fur ((X^O)). Heterozygous females ((X^B X^O)) have a tortoiseshell coat with patches of black and orange fur due to X-inactivation. A black male is crossed with an orange female. What are the expected phenotypes of their offspring?
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.
A breeder crosses a true-breeding blue-flowered plant with a true-breeding white-flowered plant, and the F1 generation is all light blue. The F1 plants are then self-crossed, producing 605 F2 plants. Assuming this trait is controlled by a single gene with incomplete dominance, what is the expected number of F2 plants with a genotype identical to their F1 parents?
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.
In carnations, flower color exhibits incomplete dominance: (R_1R_1) is red, (R_2R_2) is white, and (R_1R_2) is pink. A cross is made between two pink carnations. If the breeder discards all the white-flowered offspring, what is the probability that a randomly selected plant from the remaining offspring will be red?
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 (R1R2×R1R2). Using a Punnett square, the offspring ratios are: 1 R1R1 (red) : 2 R1R2 (pink) : 1 R2R2 (white). This gives us 4 total offspring in a 1:2:1 ratio. Since the breeder discards all white flowers, we remove the 1 R2R2 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 31. Looking at the wrong answers: B) 41 represents the original probability of getting red offspring before any were discarded—this ignores the conditional aspect. C) 21 incorrectly assumes equal numbers of red and pink offspring remain, forgetting that incomplete dominance produces twice as many heterozygotes. D) 32 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.
In snapdragons, flower color exhibits incomplete dominance (RR = red, Rr = pink, rr = white), and leaf width exhibits complete dominance (B = broad, b = narrow). A plant that is heterozygous for both traits is self-pollinated. What proportion of the offspring is expected to have pink flowers and broad leaves?
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.
A species of clover has a gene for leaf markings with three alleles: (S^A), (S^B), and (s). Alleles (S^A) and (S^B) are codominant, producing spots and stripes, respectively. Both (S^A) and (S^B) are completely dominant over the recessive allele (s), which results in no markings. A cross is performed between a clover with genotype (S^A s) and a clover with genotype (S^B s). What is the expected proportion of offspring that will exhibit both spots and stripes?
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)).
In cats, the gene for coat color is X-linked. One allele produces black fur ((X^B)) and another produces orange fur ((X^O)). Heterozygous females ((X^B X^O)) have a tortoiseshell coat with patches of black and orange fur due to X-inactivation. A black male is crossed with an orange female. What are the expected phenotypes of their offspring?
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.
In radishes, root shape is controlled by a single gene with two alleles ((S^L) and (S^S)) that exhibit incomplete dominance. The phenotypes are long ((S^L S^L)), oval ((S^L S^S)), and round ((S^S S^S)). A breeder possesses a large stock of oval-rooted radishes and wishes to establish a true-breeding line of round-rooted radishes as quickly as possible. Which of the following crosses represents the most efficient first step?
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.
Sickle-cell anemia is a human genetic disorder where alleles for normal hemoglobin (HbA) and sickle-cell hemoglobin (HbS) are codominant. Individuals with genotype HbS HbS have severe anemia, while those with HbA HbS have the milder sickle-cell trait and are resistant to malaria. Two individuals who both have sickle-cell trait have a child. What is the probability that this child will have normal hemoglobin and be susceptible to malaria?
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.
In carnations, flower color exhibits incomplete dominance: (R_1R_1) is red, (R_2R_2) is white, and (R_1R_2) is pink. A cross is made between two pink carnations. If the breeder discards all the white-flowered offspring, what is the probability that a randomly selected plant from the remaining offspring will be red?
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 (R1R2×R1R2). Using a Punnett square, the offspring ratios are: 1 R1R1 (red) : 2 R1R2 (pink) : 1 R2R2 (white). This gives us 4 total offspring in a 1:2:1 ratio. Since the breeder discards all white flowers, we remove the 1 R2R2 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 31. Looking at the wrong answers: B) 41 represents the original probability of getting red offspring before any were discarded—this ignores the conditional aspect. C) 21 incorrectly assumes equal numbers of red and pink offspring remain, forgetting that incomplete dominance produces twice as many heterozygotes. D) 32 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.