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A woman whose blood type is B- is at risk of having a child with hemolytic disease of the newborn based on the Rh factor (also called "D"):

a. with her first pregnancy, if the child is B+
b. with her second pregnancy, if the first was B+ and the second is O+
c. with her second pregnancy, if the first was O+ and the second is B- (neg)
d. both b. and c. are correct (but not a.)
e. none of the above is correct

1 Answer

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Final answer:

A woman with blood type B- is at risk of having a second child with hemolytic disease of the newborn if both her first child and the second are Rh positive. The risk is due to the mother's potential sensitization to Rh+ blood during the first pregnancy, causing her to produce anti-D antibodies that could affect a second Rh+ pregnancy.

Step-by-step explanation:

A woman with blood type B- has Rh-negative blood and she is at risk of having a child with hemolytic disease of the newborn (HDN) based on the Rh factor if her body develops anti-D antibodies. This can happen if she was exposed to Rh-positive blood from a previous pregnancy and subsequently has an Rh-positive baby. Hence, she is at risk:

  • b. with her second pregnancy, if the first was B+ and the second is O+.

During her first pregnancy with an Rh-positive (Rh+) child, a mixing of maternal and fetal blood may sensitize her immune system to Rh+ blood, leading her to form anti-D antibodies. During a subsequent pregnancy with another Rh+ child, these antibodies can cross the placenta and cause HDN. However, if her second child is Rh negative (Rh-), as in option c, there will be no risk of HDN due to Rh incompatibility. Therefore, the correct answer is b.

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