Question

Which of the following conditions cause erythroblastosis foetalis?

• A. ${\text{Both mother and foetus Rh}}^{+ve}$
• B. ${\text{Mother Rh}}^{+ve}{\text{and foetus Rh}}^{–ve}$
• C. ${\text{Both mother and foetus Rh}}^{–ve}$
• D. ${\text{Mother Rh}}^{–ve}{\text{and foetus Rh}}^{+ve}$

Answer 1

The correct option is D ${\text{Mother Rh}}^{–ve}{\text{and foetus Rh}}^{+ve}$

Erythroblastosis foetalis results from Rh incompatibility, which may develop when a woman with Rh -ve blood group bears a foetus with Rh +ve blood group.

In this, the red blood cells (erythrocytes) of foetus are destroyed in a maternal immune reaction resulting from blood group incompatibility between the foetus and the mother.

This starts when an Rh-ve mother with no Rh antigens has an Rh+ve baby who has Rh antigens. As long as she is pregnant, there is no mixing of blood and nothing bad happens. The Rh +ve child is born safe. However, during delivery, some blood cells of the baby usually enter the mother’s bloodstream. After delivery, mother’s blood will produce antibodies against Rh
antigen. Months/years after delivery, antibodies against Rh antigen remain in mother’s blood.

During the next pregnancy (if the foetus is Rh +ve), mother’s antibodies attack foetal RBCs having the Rh antigen. This is erythroblastosis foetalis and can even be fatal for the foetus.

This can be avoided by administering anti-Rh antibodies to the
mother immediately after the delivery of the Rh +ve child.