Question

In a dihybrid cross (AaBb × AaBb) in pea, what is the probability of offsprings to have a pure homozygous dominant genotype for the first trait, and heterozygous for the second trait.

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Answer 1

The gametes produced are AB , Ab , aB , ab.

So we get 16 offsprings.

Required genotype = AABb

2 offsprings of this genotype are formed ; one at 2nd and the other at 5th position of Punnet Square.

So probability = 2/16 = 1/8

Answer 2

In a dihybrid cross $\bold{(\mathrm{AaBb} \times \mathrm{AaBb}})$ in pea, the probability of offspring to have a pure homozygous dominant genotype for the first trait, and heterozygous for the second trait is 2/16 or otherwise 1/8

EXPLANATION:

This can be got easily by drawing a punnet square. A punnet square is a mathematical square that was introduced for predicting the gene ratios in a breeding experiment. The dominant trait will be expressed in heterozygous and homozygous conditions whereas the recessive trait gets expressed only when it is in homozygous condition.