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(B) Colorblindness is an $X$-linked recessive trait. $1$. The woman has normal vision but her father was colorblind, so she must be a carrier $(X^C X^

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Either colorblind or normal in vision

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(B) Colorblindness is an $X$-linked recessive trait. $1$. The woman has normal vision but her father was colorblind, so she must be a carrier $(X^C X^c)$, where $X^C$ is the normal allele and $X^c$ is the colorblind allele. $2$. The man is colorblind, so his genotype is $(X^c Y)$. $3$. The cross is: $(X^C X^c)$ $	imes$ $(X^c Y)$. $4$. The possible genotypes for their children are: - $X^C X^c$ (Carrier daughter) - $X^C Y$ (Normal son) - $X^c X^c$ (Colorblind daughter) - $X^c Y$ (Colorblind son) $5$. Since the question asks about the fourth child being a boy, the probability of the son being colorblind is $50\%$. However, the question asks what he 'will be' in a definitive sense based on the options provided. Given the options, the boy has a $50\%$ chance of being colorblind or normal. Thus, he will 'either be colorblind or normal in vision'.

Column $I$	Column $II$
$(1)$ Albinism	$(q)$ Recessive gene represented by $cc$
$(2)$ Phenylketonuria	$(s)$ Recessive gene represented by $pp$
$(3)$ Sickle cell anaemia	$(p)$ Recessive gene represented by $Hb^sHb^s$
$(4)$ Alkaptonuria	$(r)$ Recessive gene represented by $aa$

$A$ woman with normal vision, whose father was colorblind, marries a colorblind man. If the fourth child of this couple is a boy, then he will be:

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Match the columns:

Column $I$ Column $II$

$(1)$ Albinism $(q)$ Recessive gene represented by $cc$

$(2)$ Phenylketonuria $(s)$ Recessive gene represented by $pp$

$(3)$ Sickle cell anaemia $(p)$ Recessive gene represented by $Hb^sHb^s$

$(4)$ Alkaptonuria $(r)$ Recessive gene represented by $aa$

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