A heterozygous tobacco plant for albinism is | vedclass

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(B) Let the dominant allele be $A$ and the recessive allele for albinism be $a$.The genotype of the heterozygous parent is $Aa$.When $Aa$ is self-poll

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$600$

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(B) Let the dominant allele be $A$ and the recessive allele for albinism be $a$.The genotype of the heterozygous parent is $Aa$.When $Aa$ is self-pollinated $(Aa 	imes Aa)$,the offspring genotypes are $AA$ (homozygous dominant),$Aa$ (heterozygous),and $aa$ (homozygous recessive) in a ratio of $1:2:1$.The parental genotype is $Aa$.In the $F_1$ generation,the proportion of heterozygous offspring $(Aa)$ is $2/4$ or $1/2$.Given that $1200$ seeds are germinated,the number of offspring with the parental genotype $(Aa)$ is $1200 	imes (1/2) = 600$.

$A$ heterozygous tobacco plant for albinism is self-pollinated,and $1200$ seeds are subsequently germinated. How many of the offspring will have the parental genotype?

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