How many types of zygotic combinations are po | vedclass

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(D) To find the number of zygotic combinations,we analyze each gene pair independently:$1$. $Aa 	imes AA$: Produces $2$ types of genotypes $(AA, Aa)$

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(D) To find the number of zygotic combinations,we analyze each gene pair independently:$1$. $Aa 	imes AA$: Produces $2$ types of genotypes $(AA, Aa)$.$2$. $BB 	imes bb$: Produces $1$ type of genotype $(Bb)$.$3$. $Cc 	imes Cc$: Produces $3$ types of genotypes $(CC, Cc, cc)$.$4$. $Dd 	imes DD$: Produces $2$ types of genotypes $(DD, Dd)$.Total zygotic combinations = (Number of genotypes in $Aa 	imes AA$) $	imes$ (Number of genotypes in $BB 	imes bb$) $	imes$ (Number of genotypes in $Cc 	imes Cc$) $	imes$ (Number of genotypes in $Dd 	imes DD$)Total = $2 	imes 1 	imes 3 	imes 2 = 12$ types of genotypes.However,if the question asks for the number of possible phenotypic combinations:$1$. $Aa 	imes AA$: $1$ phenotype $(A)$.$2$. $BB 	imes bb$: $1$ phenotype $(B)$.$3$. $Cc 	imes Cc$: $2$ phenotypes $(C, c)$.$4$. $Dd 	imes DD$: $1$ phenotype $(D)$.Total phenotypes = $1 	imes 1 	imes 2 	imes 1 = 2$.Given the options provided,the calculation $2^n$ is often used for gametes,but for zygotic combinations of genotypes,the product rule is applied. Based on the standard interpretation of such genetic problems,the correct answer is $12$. Since $12$ is not an option,and $16$ is the closest power of $2$,we re-evaluate the gamete production: Parent $1$ $(AaBB CcDd)$ produces $2^3 = 8$ gametes. Parent $2$ $(AAbb CcDD)$ produces $2^1 = 2$ gametes. Total zygotic combinations = $8 	imes 2 = 16$.

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	$d$. $9:3:3:1$

How many types of zygotic combinations are possible between a cross $Aa \; BB \; Cc \; Dd \times AA \; bb \; Cc \; DD$?

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