Haemolytic jaundice is caused due to a domina | vedclass

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(D) Let the dominant gene for haemolytic jaundice be $H$ and the normal recessive allele be $h$.The man is heterozygous $(Hh)$ and the woman is homozy

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$1/20$

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(D) Let the dominant gene for haemolytic jaundice be $H$ and the normal recessive allele be $h$.The man is heterozygous $(Hh)$ and the woman is homozygous normal $(hh)$.The cross is: $Hh 	imes hh$.The offspring genotypes are $50\% \ Hh$ and $50\% \ hh$.Only the individuals with the genotype $Hh$ carry the dominant gene.According to the problem, only $10\%$ of those carrying the gene develop the disease.Probability of having the genotype $Hh = 1/2$.Probability of developing the disease given the genotype $Hh = 10\% = 1/10$.Total probability of a child developing the disease = $P(Hh) 	imes P(	ext{disease}|Hh) = (1/2) 	imes (1/10) = 1/20$.

Haemolytic jaundice is caused due to a dominant gene but only $10\%$ of the people who carry the gene actually develop the disease. $A$ heterozygous man marries a homozygous normal woman; what proportion of the children would be expected to develop the haemolytic disease?

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