A tobacco plant heterozygous for albinism is self pollinated and 1200 seeds are subsequently germinated. how many seedlings would have the parental genotype?
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A tobacco plant which is heterozygous for albinism (a ... germinated, how many of the seedlings would have the parental genotype ...
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The correct answer would be 600.
Let "A" and "a" be the alleles of a gene responsible for albinism in a plant.
The genotype of a tobacco plant which is heterozygous for albinism would be "Aa".
The gametes formed would be A and a.
The cross would result in the production of seed with three types of genotype AA, Aa, and aa in 1:2:1.
Thus, the total number of seeds with parental genotype (Aa) would be = 2/4 of 1200.
It comes out to be 600.
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