A: The heterozygotic female for haemophilia m | vedclass

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Question

(A) Haemophilia is an $X$-linked recessive disorder. $A$ heterozygotic female (carrier) has one normal $X$ chromosome and one $X$ chromosome carrying

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Assertion and Reason both are correct and Reason is the correct explanation of Assertion.

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(A) Haemophilia is an $X$-linked recessive disorder. $A$ heterozygotic female (carrier) has one normal $X$ chromosome and one $X$ chromosome carrying the haemophilia gene $(X^hX)$.During gametogenesis,she produces two types of eggs: $X$ and $X^h$. If an egg with the $X^h$ chromosome is fertilized by a $Y$-bearing sperm,the resulting son will be haemophilic $(X^hY)$. Thus,the assertion is correct.Haemophilia exhibits criss-cross inheritance,where the trait is passed from an affected father to his daughter (who becomes a carrier) and then to his grandsons. Since $X$-linked recessive traits are transmitted from mother to son and father to daughter,they follow a criss-cross pattern. Thus,the reason is correct and explains why the disease is transmitted in this specific manner.

$A$: The heterozygotic female for haemophilia may transmit the disease to sons.
$R$: Such traits show criss-cross inheritance.

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$A$: The heterozygotic female for haemophilia may transmit the disease to sons.$R$: Such traits show criss-cross inheritance.