Consider the following cross: Aa BB Cc Dd EE (male) x Aa Bb Ce Dd EE (female) ()

Question

Consider the following cross: Aa BB Cc Dd EE (male) x Aa Bb Ce Dd EE (female) () How many cells (boxes) would there be on a Pumett square for this cross if you were to make one? (2) What fraction of the offspring from this cross would you expect to have the same genotype as the male parent? (3) What fraction of the offspring from this cross would you expect to be homozygous for all five genes? (5) What fraction of the offspring from this cross would you expect to be homozygous dominant for all five genes? (5) What fraction of the offspring from this cross would you expect to display the dominant phenotype for all five genes?

Explanation / Answer

1:there would be 54 boxes in the punnet square for this cross.

2:From the Forkline method we get AaBBCcDdEE in the ratio 1/2*1/2*1/2*1/2*1=6.2%

Therefore, 6.2%of the offsprings will have genotype similar to male parent.

3:From the Forkline method 0.78% of the offsprings will be Homozygous for all the five genes ie:AABBCCDDEE ;1/4*1/2*1/4*1/4*1=0.78%.

4:In total there will be 54 genotypes generated by this cross. And out of all the offsprings, only 0.78% will be Homozygous dominant.

5:32.7% of the offsprings will show dominant phenotype

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