Match the terms in Column I with its phenotyp | vedclass

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(C) $i$. Monohybrid test cross: $A$ cross between a heterozygous individual and a homozygous recessive individual results in a $1:1$ phenotypic ratio.

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$i-c, ii-a, iii-d, iv-b$

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(C) $i$. Monohybrid test cross: $A$ cross between a heterozygous individual and a homozygous recessive individual results in a $1:1$ phenotypic ratio. Thus,$i-c$.$ii$. Incomplete dominance: In plants like $Mirabilis  jalapa$,the $F_2$ generation shows a phenotypic ratio of $1:2:1$ (Red:Pink:White). Thus,$ii-a$.$iii$. Law of independent Assortment: Dihybrid crosses follow this law,resulting in a phenotypic ratio of $9:3:3:1$ in the $F_2$ generation. Thus,$iii-d$.$iv$. Sickle Cell Anaemia: It is a point mutation where the ratio of normal to affected individuals in a cross between two carriers $(Hb^A Hb^S 	imes Hb^A Hb^S)$ is $1:2$ (Normal:Carrier:Affected is $1:2:1$,but often simplified to $1:2$ for phenotypic expression of trait vs disease). Thus,$iv-b$.Therefore,the correct matching is $i-c, ii-a, iii-d, iv-b$.

Column $I$	Column $II$
$i$. Monohybrid test cross	$a$. $1:2:1$
$ii$. Incomplete dominance	$b$. $1:2$
$iii$. Law of independent Assortment	$c$. $1:1$
$iv$. Sickle Cell Anaemia	$d$. $9:3:3:1$

Column $I$	Column $II$
$(a) XX-XO$ method of sex determination	$(i)$ Turner's syndrome
$(b) XX-XY$ method of sex determination	$(ii)$ Female heterogametic
$(c)$ Karyotype $45$	$(iii)$ Grasshopper
$(d) ZW-ZZ$ method of sex determination	$(iv)$ Female homogametic

Similar Questions

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Match the terms in Column $I$ with its phenotypic ratio in Column $II$ correctly and choose the correct option.Column $I$Column $II$$i$. Monohybrid test cross$a$. $1:2:1$$ii$. Incomplete dominance$b$. $1:2$$iii$. Law of independent Assortment$c$. $1:1$$iv$. Sickle Cell Anaemia$d$. $9:3:3:1$