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

Vedclass · Accepted Answer

$1/20$

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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 = $P(Hh) 	imes P(	ext{disease}|Hh) = (1/2) 	imes (1/10) = 1/20$.

Column-$I$	Column-$II$
$(p)$ Mendel	$(i)$ Chromosomal theory of Inheritance
$(q)$ Morgan	$(ii)$ Recombination map
$(r)$ Alfred Sturtevant	$(iii)$ Linkage
$(s)$ Sutton	$(iv)$ Law of Segregation

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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Match the following and choose the correct option:
Column-$I$ Column-$II$
$(p)$ Mendel $(i)$ Chromosomal theory of Inheritance
$(q)$ Morgan $(ii)$ Recombination map
$(r)$ Alfred Sturtevant $(iii)$ Linkage
$(s)$ Sutton $(iv)$ Law of Segregation

Which of the following is genetically dominant in man?

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