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(B) To find the proportion of the $AAbbccDD$ genotype in the self-pollination of $AABbCcDD$,we analyze each gene pair independently using the law of i

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$1/16$

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(B) To find the proportion of the $AAbbccDD$ genotype in the self-pollination of $AABbCcDD$,we analyze each gene pair independently using the law of independent assortment:$1$. For the $AA$ genotype: $AA 	imes AA$ results in $100\%$ $AA$ (probability = $1$).$2$. For the $Bb$ genotype: $Bb 	imes Bb$ results in $1/4$ $BB$,$2/4$ $Bb$,and $1/4$ $bb$. The probability of $bb$ is $1/4$.$3$. For the $Cc$ genotype: $Cc 	imes Cc$ results in $1/4$ $CC$,$2/4$ $Cc$,and $1/4$ $cc$. The probability of $cc$ is $1/4$.$4$. For the $DD$ genotype: $DD 	imes DD$ results in $100\%$ $DD$ (probability = $1$).Multiplying these independent probabilities: $1 	imes 1/4 	imes 1/4 	imes 1 = 1/16$.

$A$ plant with the genotype $AABbCcDD$ is self-pollinated. Assuming independent assortment of the four genes,what will be the proportion of the $AAbbccDD$ genotype in the progeny?

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