No idea if this is the right logic but: Start with a punnett square. We know there are 4 alleles ass. with alpha-globins. TBH, i'm not sure why, but i think of them as received together. So i used the following nomenclature:
A/ B/ C/ D = normal Ao / Bo = abnormal.
Mom has one mutation, so AoBCD x ABCD Dad had two mutations, one on each chromosome (trans): AoBCD, ABoCD
Crossing them together:
AoBCD x AoBCD = Hit (two mutations, oppisite chromosomes) AoBCD x ABoCD = hit (two mutations, oppisite chromosomes) ABCD x AoBCD = non-hit ABCD x ABoCD = non-hit - therefore, 50% chance of having a trans mutation in the offspring.
submitted by โbwdc(697)
The mom will pass on her deletion in 50%. The father will pass it on in 100% (because both of his copies are affected). Therefore, the child will automatically have at least one deletion and will have the double deletion in 50%.