Genetics: inheritance of blood groups

9 ASCP MLS practice questions on Genetics: inheritance of blood groups, part of Blood Banking. The first 9 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

1. Question 1 · Easy

   A group O mother and a group AB father can have children of which ABO groups?

   1. AA, B, AB or O
   2. BAB only
   3. CA or B
   4. DO or AB

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   Answer: C · A or B

   The mother is genotype OO and passes an O allele to every child. The father is AB and passes either A or B. The children are therefore AO (group A) or BO (group B). A group O or group AB child would need a different explanation, such as a rare cis-AB allele.

   • A: Not all four groups are possible: the father always gives A or B, and the mother always gives O.
   • B: An AB child would need both A and B from the father, which is not possible.
   • D: An O child needs an O allele from each parent, and the AB father has none.

2. Question 2 · Medium

   A D-negative woman's partner is D-positive and heterozygous (Dd). What is the chance that their baby is D-positive?

   1. A75%
   2. B50%
   3. C25%
   4. D100%

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   Answer: B · 50%

   The woman (dd) always passes on no RHD gene, and the father passes RHD to half of his children. So each pregnancy has a 50% chance of a D-positive baby. If the father were homozygous (DD), every baby would be D-positive. Fetal RHD typing on maternal plasma cell-free DNA now shows the actual fetal type.

   • A: 75% applies to two heterozygous parents having a D-positive child.
   • C: 25% would apply if both parents were heterozygous and the question were about D-negative babies.
   • D: 100% applies only if the father is homozygous DD.

3. Question 3 · Hard

   The Xga antigen is inherited as an X-linked dominant trait. If the father is Xg(a+) and the mother is Xg(a-), which children will be Xg(a+)?

   1. AAll of the children
   2. BHalf of the sons and daughters
   3. CAll sons and no daughters
   4. DAll daughters and no sons

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   Answer: D · All daughters and no sons

   A father passes his only X chromosome, carrying Xga, to every daughter and his Y chromosome to every son. All daughters are therefore Xg(a+). Sons get their X chromosome from the Xg(a-) mother, so they are Xg(a-). This father-to-daughter pattern is the signature of an X-linked dominant trait.

   • A: Sons cannot inherit Xga from this father.
   • B: A father passes his X chromosome to all daughters, not half.
   • C: Sons get their X chromosome from the mother, who lacks Xga.

4. Question 4 · Hard

   A child is group O. The mother is group A and the alleged father is group AB. What do the ABO results show?

   1. AThe mother is excluded
   2. BThe alleged father is excluded
   3. CThe results cannot be interpreted
   4. DThe alleged father cannot be excluded

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   Answer: B · The alleged father is excluded

   A group O child must have inherited an O allele from each parent. An AB father has no O allele, so he cannot be the father by ABO, a direct exclusion. The group A mother can be AO, so she is not excluded. Rare alleles such as cis-AB can mislead, which is why parentage is now tested by DNA.

   • A: A group A mother can carry an O allele.
   • C: ABO results alone can give a direct exclusion here.
   • D: An AB man has no O allele to pass on, so he is excluded.

5. Question 5 · Medium

   A man types Jk(a+b-) and his partner types Jk(a-b+). Assuming no silent alleles, their children will all be:

   1. AJk(a+b-)
   2. BJk(a-b-)
   3. CJk(a-b+)
   4. DJk(a+b+)

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   Answer: D · Jk(a+b+)

   Without silent alleles, the man is JK*A/JK*A and the woman is JK*B/JK*B. Every child receives one JK*A and one JK*B allele. Because the alleles are codominant, every child types Jk(a+b+). Such children show dosage: their cells react more weakly with Kidd antibodies than homozygous cells.

   • A: Jk(a+b-) would need JK*A from both parents.
   • B: Jk(a-b-) needs silent alleles, which the stem excludes.
   • C: Jk(a-b+) would need JK*B from both parents.

6. Question 6 · Hard

   In a parentage dispute, a mother, her child and four men are typed. Mother: group O, M+N-, K-, Jk(a+b+). Child: group B, M+N+, K+, Jk(a+b-). Man 1: group AB, M+N-, K+, Jk(a+b-). Man 2: group A, M+N+, K+, Jk(a+b+). Man 3: group B, M-N+, K-, Jk(a+b+). Man 4: group B, M+N+, K+, Jk(a+b+). Maternity is certain, the typings were repeated with fresh samples, and there are no silent or rare alleles in the family. Using direct exclusions only, which man remains a possible biological father?

   1. AMan 4
   2. BMan 3
   3. CMan 2
   4. DMan 1

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7. Question 7 · Medium

   Refer to the family shown. What is the chance that this couple's next child will be group A?

   

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   1. A1 in 4
   2. B1 in 2
   3. C1 in 8
   4. D3 in 4

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8. Question 8 · Easy

   In the family shown, the children's Kidd phenotypes show that the Jka and Jkb alleles are:

   

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   1. ADominant
   2. BRecessive
   3. CCodominant
   4. DX-linked

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9. Question 9 · Hard

   In the family shown, II-2's plasma strongly agglutinates group O red cells at 37 °C, and her red cells do not react with Ulex europaeus lectin. The best explanation for III-1's blood group is:

   

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   1. AIII-1 has an acquired B antigen
   2. BII-2 carries a cis-AB allele
   3. CII-3 is not III-1's biological father
   4. DII-2 has the Bombay (Oh) phenotype

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