Consider the following cross AA Bb Dd Ee Aa bb Dd ee a How

Consider the following cross: A/A; B/b; D/d; E/e * A/a; b/b; D/d; e/e a) How many different genotypes could potentially be seen in the offspring of this cross? b) How many different phenotypes could potentially be seen in the offspring? Assume alleles represented by capital letter show complete dominance. c) What is the chance that an offspring from these two parents would have the genotype: A/a; B/b; D/D; E/e ? d) What is the chance that an offspring from these two parents would have the phenotype: A_bb D_ee ?

Solution

a) to solve it, we need to do individual crosses separately.

First one is AA x Aa it bears 2 genotypes

Second is Bb x bb it bears 2 genotypes as well

Third is Dd x Dd it bears 3 different genotypes

Fourth is Ee x ee it bears 2 different genotypes.

In total it would bear 2 x 2 x3 x 2 different genotypes. Therefore 24 different genotypes

b) since the phenotypes are different from genotypes:

The cross AA x Aa gives dominant only

Cross Bb x bb gives half dominant half recessive

Cross Dd x Dd gives dominant and recessive

Cross Ee x ee gives dominant and recessive .

Therefore the combination is 1 x 2 x 2 x 2 = 8 different phenotypes.

c) To have the given combination of genes we again need to multiply the probablities found out through individual crosses.

Probability of Aa= 1/2 , Bb = 1/2 , DD = 1/4 , Ee = 1/2

Therefore 1/2 x 1/2 x 1/4 x 1/2 = 1/32

d) Again we need to proceed in same way as we did in part c . Probability of A_ = 1, bb= 1/2 , D_ = 3/4 and ee= 1/2

Therefore 1× 1/2 × 3/4 × 1/2 = 3/16