Question

The condition for erythroblastosis foetalis to occur is

• A. Rh -ve mother and Rh +ve baby
• B. Rh -ve mother and Rh -ve baby
• C. Rh +ve mother and Rh -ve baby
• D. Rh +ve mother and Rh +ve baby

Answer 1

The correct option is A Rh -ve mother and Rh +ve baby

Erythroblastosis foetalis results from Rh incompatibility, which may develop when a woman with Rh -ve blood conceives a foetus with Rh +ve blood. In this, the red blood cells (erythrocytes) of a foetus are destroyed in a maternal immune reaction resulting from blood group incompatibility between the foetus and the mother. This starts when an Rh- mother with no Rh antigens has an Rh+ baby who has Rh antigens. As long as she is pregnant, there is no mixing of blood and nothing bad happens. During delivery, some blood cells of the baby usually enter the mother’s bloodstream. After delivery, mother’s blood will produce antibodies against Rh factor. Months/years after delivery, antibodies against Rh antigen remain in mother’s blood. During the next pregnancy, mother’s antibodies attack foetal RBCs having the Rh factor. This is erythroblastosis foetalis and can even be fatal for the foetus.