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Erythroblastosis foetalis is caused when :
A
$Rh^-$ female and $Rh^+$ male
B
$Rh^+$ female and $Rh^-$ male
C
$Rh^+$ female and $Rh^+$ male
D
$Rh^-$ female and $Rh^-$ male
Detailed Solution
If an $Rh^-$ mother carries an $Rh^+$ foetus (the foetus gets the Rh antigen from an $Rh^+$ father), some foetal RBCs can enter the mother's blood at delivery.
The mother then forms anti-Rh antibodies. In a later pregnancy with an $Rh^+$ foetus, these antibodies cross the placenta and destroy the foetal RBCs, causing severe anaemia and jaundice (haemolytic disease of the newborn).
It can be prevented by giving anti-Rh antibodies (anti-D) to the mother after the first delivery.
So erythroblastosis foetalis occurs when the female is $Rh^-$ and the male is $Rh^+$.
The mother then forms anti-Rh antibodies. In a later pregnancy with an $Rh^+$ foetus, these antibodies cross the placenta and destroy the foetal RBCs, causing severe anaemia and jaundice (haemolytic disease of the newborn).
It can be prevented by giving anti-Rh antibodies (anti-D) to the mother after the first delivery.
So erythroblastosis foetalis occurs when the female is $Rh^-$ and the male is $Rh^+$.