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My answer ( after rounding off) is 9% -100( 0.75 X 0.25 X 0.5) but the answer given is 22 % Am I correct ?

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up vote 2 down vote accepted

I think the quoted answer is correct.

First, the probability of getting a son( and not a daughter) is $P_s=0.5$. Next, since both the parents are heterozygous, the probability of genotype $hh$ is $P_h=0.25$. This genotype will not produce any antigen precursor, and hence there will be no antigens on RBC. As blood group O is defined by the absence of antigens, this can be classified as blood group O.For the expression of the antigen (dominant at the $H$ locus), the probability is $P_H=0.75$.

Again owing to the heterozygosity of both the parents at $I$ locus, the probability of getting a blood group with genotype $I^oI^o$ is $P_I=0.25$. This is the probability, that if the antigen precursor is made, $75%$% of the times the expressed blood group will be O. Therefore the probability of blood group O due to homozygous $I$ locus and dominant $H$ locus is $P_I\times P_H=3/16$.

Net probability for blood group O=(probability due to homozygous h+probability due to homozygous I and dominant H)*probability of a son.

$$P_{net}=P_s(P_h+P_HP_I)\approx 0.22$$

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The blood group without hh antigen is also known as Bombay Blood group ! One of the rarest type in the world ! Phenotypically O but when O blood transfusion is done, it will create a severe reaction. –  biogirl Jan 11 at 4:14
    
@biogirllajja I knew that but I guessed, it was the phenotype of O whose probability was asked. –  Satwik Pasani Jan 11 at 6:54
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I realised my mistake of not considering hh as phenotype O when I saw your working. I just wanted to share the information regarding bombay blood group. –  biogirl Jan 11 at 13:07

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