Multiple Alleles & Blood Types - Genetics

A newborn is suspected of having been switched at birth in a hospital. The infant has blood type B. One couple, the Smiths, has blood types A and B. The other couple, the Joneses, both have blood type A. Which statement provides the most accurate analysis of the situation?
Answer: The infant must belong to the Smiths, as the Joneses cannot have a Type B child. 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.
If two parents have the genotypes $I^A I^B$ and $I^B i$ , what is the probability that they will have a child who is phenotypically identical to one of them?
Answer: 75% The parents have phenotypes Type AB (genotype $I^A I^B$ ) and Type B (genotype $I^B i$ ). The cross is $I^A I^B \times I^B i$ . The possible offspring genotypes are: $I^A I^B$ (from $I^A$ and $I^B$ ), $I^A i$ (from $I^A$ and $i$ ), $I^B I^B$ (from $I^B$ and $I^B$ ), and $I^B i$ (from $I^B$ and $i$ ). Each has a 25% probability. The corresponding phenotypes are: Type AB, Type A, Type B, and Type B. The question asks for the probability that the child is phenotypically identical to a parent (i.e., has Type AB or Type B blood). The probability of Type AB is 25%. The probability of Type B (from either $I^B I^B$ or $I^B i$ ) is 25% + 25% = 50%. Since these are mutually exclusive outcomes, the total probability is P(Type AB) + P(Type B) = 25% + 50% = 75%.
A child is born with Type O, Rh-negative blood. His mother has Type A, Rh-positive blood. Which of the following paternal phenotypes is not possible?
Answer: Type AB, Rh-negative The child's genotype is $ii rr$ . The child inherited one $i$ and one $r$ from each parent. The mother is A-positive and has an O-negative child, so her genotype must be $I^A i Rr$ . The father must therefore contribute an $i$ allele and an $r$ allele. Let's evaluate the choices for the father. A) O-neg ( $ii rr$ ): Possible. B) B-pos (e.g., $I^B i Rr$ ): Possible. C) A-pos (e.g., $I^A i Rr$ ): Possible. D) AB-neg ( $I^A I^B rr$ ): Impossible. A man with Type AB blood has genotype $I^A I^B$ and cannot contribute the $i$ allele needed for the child to have Type O blood.
A man with Type A blood and a woman with Type B blood have a child with Type O blood. They are expecting a second child. What is the probability that this second child will have Type AB blood?
Answer: 25% 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%.
A population contains three alleles for the ABO blood group locus. How many different genotypes are possible for this locus, and how many of these genotypes would result in a heterozygous individual?
Answer: 6 genotypes are possible; 3 of these are heterozygous. With three alleles ( $I^A, I^B, i$ ), the number of possible genotypes can be calculated. The homozygous genotypes are $I^A I^A$ , $I^B I^B$ , and $ii$ (3 total). The heterozygous genotypes are $I^A I^B$ , $I^A i$ , and $I^B i$ (3 total). In total, there are 3 + 3 = 6 possible genotypes. The question asks for the total number of genotypes and the number of heterozygous ones. There are 6 total genotypes, and 3 of them ( $I^A I^B$ , $I^A i$ , $I^B i$ ) are heterozygous.
What is the expected phenotypic ratio for ABO blood types from a cross between two individuals with the genotype $I^A i Hh$ ?
Answer: 9 Type A : 7 Type O 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.
A patient with an unknown blood type is brought to the emergency room. A blood test reveals that their serum causes agglutination when mixed with red blood cells from both a Type A donor and a Type B donor. Which of the following statements is correct regarding this patient?
Answer: The patient has Type O blood and can receive red blood cells from a Type O donor only. 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.
A woman with blood type A and a man with blood type B have four children, each with a different blood type (A, B, AB, and O). What is the probability that their fifth child will have blood type A?
Answer: 25% 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%.
A child has blood type AB. Which of the following parental phenotypic pairs is biologically impossible?
Answer: One parent is Type AB, and the other is Type O. A child with Type AB blood has the genotype $I^A I^B$ . This means the child must inherit an $I^A$ allele from one parent and an $I^B$ allele from the other. A parent with Type O blood has the genotype $ii$ and can only pass on an $i$ allele. Therefore, a Type O individual cannot be the parent of a Type AB child, as they cannot provide either the $I^A$ or $I^B$ allele. All other combinations listed can produce an $I^A I^B$ child.
In the ABO blood system, the relationship between the $I^A$ and $I^B$ alleles is best described as codominance. The relationship between the $I^A$ allele and the $i$ allele is best described as:
Answer: Complete dominance 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.
In the human ABO blood group system, the allele $i$ is recessive to both $I^A$ and $I^B$ , which are codominant. If a man with genotype $I^A i$ and a woman with genotype $I^B i$ have children, what is the expected phenotypic ratio among their offspring?
Answer: 1 Type A : 1 Type B : 1 Type AB : 1 Type O 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.
In a paternity case, a mother with blood type A has a child with blood type O. A man with blood type AB is alleged to be the father. Based solely on this ABO blood typing evidence, what is the most definitive conclusion?
Answer: The man is definitively excluded as the father. 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.
In a hypothetical inheritance system that mimics the human ABO system, an epistatic gene $H$ is required for the expression of blood type antigens. The recessive allele $h$ prevents expression, resulting in a Type O phenotype regardless of the ABO genotype. What is the probability of a child having the Type O phenotype from a cross between parents with genotypes $I^A i Hh$ and $I^B i Hh$ ?
Answer: 7/16 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.
Refer to the pedigree, which shows the inheritance of ABO blood types in a family. Individuals with unknown blood types are indicated by a question mark. Individual II-2 marries individual II-4, who has Type B blood and whose mother had Type O blood. Based on the pedigree, what is the probability that their first child (III-1) will have Type B blood?
Answer: 1/6 First, determine the genotypes of the parents (III-1's parents). Individual II-4 has Type B blood and a Type O ( $ii$ ) mother, so her genotype must be $I^B i$ . Individual II-2 has Type A blood. His parents (I-1 and I-2) are Type A and Type B, respectively, but they have a Type O child (II-1, genotype $ii$ ). This means the grandparents' genotypes must be $I^A i$ and $I^B i$ . The possible Type A offspring from this cross are $I^A I^A$ and $I^A i$ in a 1:2 ratio. So, there is a 1/3 probability that II-2 is $I^A I^A$ and a 2/3 probability he is $I^A i$ . Now, calculate the probability of a Type B child (genotype $I^B i$ ). Case 1: II-2 is $I^A I^A$ (Prob = 1/3). Cross: $I^A I^A \times I^B i$ . The probability of a $I^B i$ child is 0. Case 2: II-2 is $I^A i$ (Prob = 2/3). Cross: $I^A i \times I^B i$ . The probability of a $I^B i$ child is 1/4. The total probability is the sum of probabilities of each case: (1/3 0) + (2/3 1/4) = 0 + 2/12 = 1/6.
A couple, both with Type A blood, have two children. Their first child has Type O blood. What is the probability that their third child will also have Type A blood?
Answer: 3/4 The fact that two Type A parents have a Type O ( $ii$ ) child reveals that both parents must be heterozygous ( $I^A i$ ). The blood types of previous children do not influence the probability for subsequent children. For any child of the cross $I^A i \times I^A i$ , the genotypic probabilities are 1/4 $I^A I^A$ , 1/2 $I^A i$ , and 1/4 $ii$ . The phenotypes are Type A (from genotypes $I^A I^A$ or $I^A i$ ) and Type O (from $ii$ ). The probability of a child having Type A blood is the sum of the probabilities for the corresponding genotypes: P(Type A) = P( $I^A I^A$ ) + P( $I^A i$ ) = 1/4 + 1/2 = 3/4.
A patient with Type AB-positive blood has lost a significant amount of blood and requires an emergency transfusion of fresh frozen plasma (FFP), not packed red blood cells. Which of the following donors would be the most suitable source for the plasma?
Answer: Type AB-positive, because their plasma lacks anti-A and anti-B antibodies. 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.
A man has blood type A-positive. His father had type O-negative blood. He marries a woman with blood type B-positive, whose mother had type O-negative blood. Assuming the genes for ABO blood group and Rh factor are unlinked, what is the probability that their first child will have type AB-negative blood?
Answer: 1/16 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: $1/4 \times 1/4 = 1/16$ .
Given this information, what is the complete genotype of the woman regarding the ABO and H loci?
Answer: $I^B i hh$ 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 $I^B i hh$ .

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