Respuesta :
when a haemophilic carrier female marries a haemiphilic man then due to ther genotypes
Xc=carrier of haemophilia
Xn=normal
Y=dont carry haemophilic genes
so here we go!
XcY cross with XcXn
genotypes of ther offsprings
1=XcXc
2=XcXn
3=XcY
4=XnY
so from whole 4 there is 50% parobability
Xc=carrier of haemophilia
Xn=normal
Y=dont carry haemophilic genes
so here we go!
XcY cross with XcXn
genotypes of ther offsprings
1=XcXc
2=XcXn
3=XcY
4=XnY
so from whole 4 there is 50% parobability
The mother inherited her father's X chromosome so her genotype is XHXh and she produces XH and Xh gametes.
The father's genotype is XhY and he produces Xh and Y gametes.
The possible outcomes are XHXh normal female (carrier), XhXh (affected female), XHY (normal son) XhY (affected son)
So the probability of an affected daugher is 1.4
The father's genotype is XhY and he produces Xh and Y gametes.
The possible outcomes are XHXh normal female (carrier), XhXh (affected female), XHY (normal son) XhY (affected son)
So the probability of an affected daugher is 1.4