Question

A tobacco plant heterozygous for albinism is self pollinated and 1200 seeds are subsequently germinated. how many seedlings would have the parental genotype?

Answer 1

A tobacco plant which is heterozygous for albinism (a ... germinated, how many of the seedlings would have the parental genotype ...

Answer 2

Answer:

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.