Genetics Quiz: Multiple Alleles And Blood Types

What this quiz covers

This quiz focuses on Multiple Alleles And Blood Types, 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 Type A : 1 Type B
2. 1 Type A : 1 Type B : 1 Type AB
3. 1 Type A : 1 Type B : 1 Type AB : 1 Type O (correct answer)
4. 3 Type A : 1 Type B

Explanation: A cross between $I^A i$ and $I^B i$ can be visualized with a Punnett square. The possible genotypes of the offspring are $I^A I^B$, $I^A i$, $I^B i$, and $ii$, each with a probability of 1/4. The corresponding phenotypes are Type AB, Type A, Type B, and Type O. Therefore, the expected phenotypic ratio is 1:1:1:1.

Question 2

1. Incomplete dominance
2. Complete dominance (correct answer)
3. Epistasis
4. Pleiotropy

Explanation: In complete dominance, the heterozygote's phenotype is indistinguishable from that of the homozygous dominant individual. A person with genotype $I^A i$ has blood type A, the same phenotype as a person with genotype $I^A I^A$. Therefore, the $I^A$ allele shows complete dominance over the $i$ allele. Incomplete dominance results in a blended intermediate phenotype. Epistasis is when one gene masks the effect of another. Pleiotropy is when one gene affects multiple traits.

Question 3

1. $ii hh$
2. $I^B i hh$ (correct answer)
3. $I^A I^B Hh$
4. $ii Hh$

Explanation: The woman has a Type O phenotype despite having Type AB parents, indicating she has the Bombay phenotype ($hh$) which prevents expression of ABO antigens. Her child with a Type O man ($ii$) has Type B blood, meaning the child's genotype is $I^B i$. Since the father contributed $i$, the mother must have contributed $I^B$. Therefore, her ABO genotype contains $I^B$. Combined with her $hh$ H-locus genotype that masks ABO expression, her complete genotype is $I^B i hh$.

Question 4

1. Type O-negative, because this is the universal donor.
2. Type A-positive, because it matches one of the recipient's antigens.
3. Type AB-positive, because their plasma lacks anti-A and anti-B antibodies. (correct answer)
4. Any Rh-negative donor to prevent Rh sensitization.

Explanation: Plasma transfusion rules are the reverse of red blood cell transfusion rules. The goal is to avoid giving the recipient antibodies that will attack their own red blood cells. The Type AB recipient has both A and B antigens on their cells but lacks anti-A and anti-B antibodies in their plasma. Therefore, they need plasma that is also free of these antibodies. Type AB plasma has no anti-A or anti-B antibodies, making it safe for any recipient (universal plasma donor). Plasma from a Type O donor contains both anti-A and anti-B antibodies and would be dangerous.

Question 5

1. 9 Type A : 3 Type O : 4 Type Bombay
2. 3 Type A : 1 Type O
3. 9 Type A : 7 Type O (correct answer)
4. 12 Type A : 4 Type O

Explanation: This is a dihybrid cross with epistasis. First, analyze the ABO cross: $I^A i \times I^A i$ yields genotypes in a 1 $I^A I^A$ : 2 $I^A i$ : 1 $ii$ ratio. This corresponds to a 3 Type A : 1 Type O phenotypic ratio. Next, analyze the H-locus cross: $Hh \times Hh$ yields genotypes in a 1 $HH$ : 2 $Hh$ : 1 $hh$ ratio. This corresponds to a 3 H_ (normal expression) : 1 $hh$ (Bombay phenotype) ratio. Now, combine the two. The individuals with H_ genotype (3/4 of offspring) will show their normal ABO phenotype. The individuals with $hh$ genotype (1/4 of offspring) will be phenotypically Type O.

• P(Type A) = P(I^A_) * P(H_) = 3/4 * 3/4 = 9/16.
• P(Genetically Type O) = P($ii$) * P(H_) = 1/4 * 3/4 = 3/16.
• P(Bombay Phenotype O) = P(any ABO) * P($hh$) = 1 * 1/4 = 4/16. Total phenotypic ratio: 9/16 are Type A. The remaining (3/16 + 4/16) = 7/16 are phenotypically Type O. The ratio is 9 Type A : 7 Type O.

Question 6

1. The infant must belong to the Smiths, as the Joneses cannot have a Type B child. (correct answer)
2. The infant must belong to the Joneses, as they are both Type A.
3. The infant could belong to either couple, so blood typing is not useful here.
4. The infant cannot belong to either couple, indicating a mix-up with a third family.

Explanation: Let's analyze the possibilities. The Joneses are both Type A. Their genotypes could be $I^A I^A$ or $I^A i$. To have a Type B child (genotype $I^B I^B$ or $I^B i$), a parent must contribute an $I^B$ allele. Since neither Mr. nor Mrs. Jones has an $I^B$ allele, they cannot have a Type B child. The Smiths are Type A and Type B. If their genotypes are $I^A i$ and $I^B i$, they can have children with Type A, B, AB, or O blood. Therefore, the infant with Type B blood could belong to the Smiths but could not belong to the Joneses. This allows for a definitive conclusion based on exclusion.

Question 7

1. The man is definitively the father because he could have contributed the necessary allele.
2. The man is definitively excluded as the father. (correct answer)
3. The results are inconclusive; more genetic testing is required to determine paternity.
4. The man has a 50% chance of being the father, assuming no other men are involved.

Explanation: The child has blood type O, which corresponds to the genotype $ii$. This means the child inherited one $i$ allele from the mother and one $i$ allele from the father. The mother is Type A and has an O child, so her genotype must be $I^A i$. The alleged father has blood type AB, which corresponds to the genotype $I^A I^B$. A man with this genotype can only pass on an $I^A$ or an $I^B$ allele to his offspring. He cannot pass on the $i$ allele required for the child to be Type O. Therefore, he is definitively excluded as the biological father.

Question 8

1. 1/4
2. 3/16
3. 4/16
4. 7/16 (correct answer)

Explanation: The Type O phenotype can result from two genetic conditions: 1) the genotype $ii$ with at least one dominant $H$ allele (iiH_), or 2) any ABO genotype with the homozygous recessive $hh$ genotype (the Bombay phenotype). From the cross $I^A i \times I^B i$, P($ii$) = 1/4. From the cross $Hh \times Hh$, P(H_) = 3/4. So, P(iiH_) = 1/4 * 3/4 = 3/16. From the cross $Hh \times Hh$, P($hh$) = 1/4. This $hh$ genotype will mask any ABO genotype, resulting in a Type O phenotype. The probability of this is 1/4. Since these are mutually exclusive events that both result in the O phenotype, we add their probabilities: P(Type O) = P(iiH_) + P($hh$) = 3/16 + 1/4 = 3/16 + 4/16 = 7/16.

Question 9

1. The patient has Type AB blood and can receive red blood cells from any donor.
2. The patient has Type O blood and can receive red blood cells from a Type O donor only. (correct answer)
3. The patient has Type A blood and their serum contains anti-B antibodies.
4. The patient has Type B blood and their serum contains anti-A antibodies.

Explanation: Agglutination occurs when antibodies in the recipient's serum bind to antigens on the donor's red blood cells. The patient's serum agglutinates both Type A and Type B cells. This means the patient's serum contains both anti-A and anti-B antibodies. The only blood type with both of these antibodies is Type O. Individuals with Type O blood can only receive packed red blood cells from other Type O donors to avoid an immune reaction.

Question 10

1. $ii hh$
2. $I^B i hh$ (correct answer)
3. $I^A I^B Hh$
4. $ii Hh$

Explanation: The woman has a Type O phenotype despite having Type AB parents, indicating she has the Bombay phenotype ($hh$) which prevents expression of ABO antigens. Her child with a Type O man ($ii$) has Type B blood, meaning the child's genotype is $I^B i$. Since the father contributed $i$, the mother must have contributed $I^B$. Therefore, her ABO genotype contains $I^B$. Combined with her $hh$ H-locus genotype that masks ABO expression, her complete genotype is $I^B i hh$.

Question 11

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

Explanation: This is a two-gene inheritance problem. First, deduce the parents' genotypes. The man is A-positive. His father was O-negative ($ii rr$), so the man must have inherited an $i$ and an $r$ allele. His genotype is $I^A i Rr$. The woman is B-positive. Her mother was O-negative ($ii rr$), so the woman must have inherited an $i$ and an $r$ allele. Her genotype is $I^B i Rr$. Now, find the probability of an AB-negative child ($I^A I^B rr$) from the cross $I^A i Rr \times I^B i Rr$. The probability of an $I^A I^B$ child is $P(I^A \text{ from father}) \times P(I^B \text{ from mother}) = 1/2 \times 1/2 = 1/4$. The probability of an $rr$ child from an $Rr \times Rr$ cross is 1/4. Since the genes are unlinked, multiply the probabilities: $1/4 \times 1/4 = 1/16$.

Question 12

1. 1 Type A : 1 Type B
2. 1 Type A : 1 Type B : 1 Type AB
3. 1 Type A : 1 Type B : 1 Type AB : 1 Type O (correct answer)
4. 3 Type A : 1 Type B

Explanation: A cross between $I^A i$ and $I^B i$ can be visualized with a Punnett square. The possible genotypes of the offspring are $I^A I^B$, $I^A i$, $I^B i$, and $ii$, each with a probability of 1/4. The corresponding phenotypes are Type AB, Type A, Type B, and Type O. Therefore, the expected phenotypic ratio is 1:1:1:1.

Question 13

1. The patient has Type AB blood and can receive red blood cells from any donor.
2. The patient has Type O blood and can receive red blood cells from a Type O donor only. (correct answer)
3. The patient has Type A blood and their serum contains anti-B antibodies.
4. The patient has Type B blood and their serum contains anti-A antibodies.

Explanation: Agglutination occurs when antibodies in the recipient's serum bind to antigens on the donor's red blood cells. The patient's serum agglutinates both Type A and Type B cells. This means the patient's serum contains both anti-A and anti-B antibodies. The only blood type with both of these antibodies is Type O. Individuals with Type O blood can only receive packed red blood cells from other Type O donors to avoid an immune reaction.

Question 14

1. The infant must belong to the Smiths, as the Joneses cannot have a Type B child. (correct answer)
2. The infant must belong to the Joneses, as they are both Type A.
3. The infant could belong to either couple, so blood typing is not useful here.
4. The infant cannot belong to either couple, indicating a mix-up with a third family.

Explanation: Let's analyze the possibilities. The Joneses are both Type A. Their genotypes could be $I^A I^A$ or $I^A i$. To have a Type B child (genotype $I^B I^B$ or $I^B i$), a parent must contribute an $I^B$ allele. Since neither Mr. nor Mrs. Jones has an $I^B$ allele, they cannot have a Type B child. The Smiths are Type A and Type B. If their genotypes are $I^A i$ and $I^B i$, they can have children with Type A, B, AB, or O blood. Therefore, the infant with Type B blood could belong to the Smiths but could not belong to the Joneses. This allows for a definitive conclusion based on exclusion.

Question 15

1. 0%
2. 25% (correct answer)
3. 50%
4. 100%

Explanation: The fact that the couple has children with all four blood types reveals their genotypes. To have a Type O ($ii$) child, both parents must carry the $i$ allele. To have a Type AB ($I^A I^B$) child, one parent must have the $I^A$ allele and the other must have the $I^B$ allele. Therefore, the parents' genotypes are $I^A i$ and $I^B i$. Each birth is an independent event, and the outcomes of previous births do not affect the probabilities for the next. For the cross $I^A i \times I^B i$, the probability of having a child with Type A blood (genotype $I^A i$) is 1/4 or 25%.

Question 16

1. 9 Type A : 3 Type O : 4 Type Bombay
2. 3 Type A : 1 Type O
3. 9 Type A : 7 Type O (correct answer)
4. 12 Type A : 4 Type O

Explanation: This is a dihybrid cross with epistasis. First, analyze the ABO cross: $I^A i \times I^A i$ yields genotypes in a 1 $I^A I^A$ : 2 $I^A i$ : 1 $ii$ ratio. This corresponds to a 3 Type A : 1 Type O phenotypic ratio. Next, analyze the H-locus cross: $Hh \times Hh$ yields genotypes in a 1 $HH$ : 2 $Hh$ : 1 $hh$ ratio. This corresponds to a 3 H_ (normal expression) : 1 $hh$ (Bombay phenotype) ratio. Now, combine the two. The individuals with H_ genotype (3/4 of offspring) will show their normal ABO phenotype. The individuals with $hh$ genotype (1/4 of offspring) will be phenotypically Type O.

• P(Type A) = P(I^A_) * P(H_) = 3/4 * 3/4 = 9/16.
• P(Genetically Type O) = P($ii$) * P(H_) = 1/4 * 3/4 = 3/16.
• P(Bombay Phenotype O) = P(any ABO) * P($hh$) = 1 * 1/4 = 4/16. Total phenotypic ratio: 9/16 are Type A. The remaining (3/16 + 4/16) = 7/16 are phenotypically Type O. The ratio is 9 Type A : 7 Type O.

Question 17

1. 1/4
2. 3/16
3. 4/16
4. 7/16 (correct answer)

Explanation: The Type O phenotype can result from two genetic conditions: 1) the genotype $ii$ with at least one dominant $H$ allele (iiH_), or 2) any ABO genotype with the homozygous recessive $hh$ genotype (the Bombay phenotype). From the cross $I^A i \times I^B i$, P($ii$) = 1/4. From the cross $Hh \times Hh$, P(H_) = 3/4. So, P(iiH_) = 1/4 * 3/4 = 3/16. From the cross $Hh \times Hh$, P($hh$) = 1/4. This $hh$ genotype will mask any ABO genotype, resulting in a Type O phenotype. The probability of this is 1/4. Since these are mutually exclusive events that both result in the O phenotype, we add their probabilities: P(Type O) = P(iiH_) + P($hh$) = 3/16 + 1/4 = 3/16 + 4/16 = 7/16.

Question 18

1. 0%
2. 25% (correct answer)
3. 50%
4. 75%

Explanation: The fact that a Type A parent and a Type B parent have a Type O (genotype $ii$) child reveals that both parents must be heterozygous carriers of the recessive $i$ allele. Therefore, the father's genotype is $I^A i$ and the mother's genotype is $I^B i$. A Punnett square for the cross $I^A i \times I^B i$ yields the following genotypic probabilities for their offspring: 25% $I^A I^B$ (Type AB), 25% $I^A i$ (Type A), 25% $I^B i$ (Type B), and 25% $ii$ (Type O). Thus, the probability of their second child having Type AB blood is 25%.

Question 19

1. Incomplete dominance
2. Complete dominance (correct answer)
3. Epistasis
4. Pleiotropy

Explanation: In complete dominance, the heterozygote's phenotype is indistinguishable from that of the homozygous dominant individual. A person with genotype $I^A i$ has blood type A, the same phenotype as a person with genotype $I^A I^A$. Therefore, the $I^A$ allele shows complete dominance over the $i$ allele. Incomplete dominance results in a blended intermediate phenotype. Epistasis is when one gene masks the effect of another. Pleiotropy is when one gene affects multiple traits.

Question 20

1. The man is definitively the father because he could have contributed the necessary allele.
2. The man is definitively excluded as the father. (correct answer)
3. The results are inconclusive; more genetic testing is required to determine paternity.
4. The man has a 50% chance of being the father, assuming no other men are involved.

Explanation: The child has blood type O, which corresponds to the genotype $ii$. This means the child inherited one $i$ allele from the mother and one $i$ allele from the father. The mother is Type A and has an O child, so her genotype must be $I^A i$. The alleged father has blood type AB, which corresponds to the genotype $I^A I^B$. A man with this genotype can only pass on an $I^A$ or an $I^B$ allele to his offspring. He cannot pass on the $i$ allele required for the child to be Type O. Therefore, he is definitively excluded as the biological father.