Question

What is the chance of producing a child with the dominant phenotype from each of the following crosses? (pages \(469-70\) ) a. \(A A \times A A\) b. \(A a \times A A\) c. \(A a \times A a\) d. \(a a \times a a\)

Short answer

a. The probability of producing a child with the dominant phenotype from the cross \(A A \times A A\) is 100% or 1. b. The probability of producing a child with the dominant phenotype from the cross \(A a \times A A\) is 100% or 1. c. The probability of producing a child with the dominant phenotype from the cross \(A a \times A a\) is 75% or \(\frac{3}{4}\). d. The probability of producing a child with the dominant phenotype from the cross \(a a \times a a\) is 0% or 0.

Step-by-step solution

Cross a. \(A A \times A A\)

Since both parents are homozygous dominant (having identical dominant alleles), their genotypes are represented as \(A A\). If we create a Punnett square for this cross, all the offspring will have the same genotype, which is \(A A\), and since \(A\) is the dominant allele, they all will show the dominant phenotype. So the probability is 100% or 1.

Cross b. \(A a \times A A\)

In this cross, one parent is homozygous dominant (having identical dominant alleles) \(A A\), and the other is heterozygous (having one dominant and one recessive allele) \(A a\). If we create a Punnett square for this cross, all the offspring will have dominant allele \(A\), which means they all will show the dominant phenotype. So, the probability is 100% or 1.

Cross c. \(A a \times A a\)

In this cross, both parents are heterozygous (having one dominant and one recessive allele) \(A a\). If we create a Punnett square for this cross, we get the following genotypes: \(\frac{1}{4} A A, \frac{1}{2} A a, \frac{1}{4} a a\). The offspring that have at least one dominant allele \(A\) will show the dominant phenotype. So, the probability is \(\frac{1}{4} + \frac{1}{2} = \frac{3}{4}\) or 75%.

Cross d. \(a a \times a a\)

In this cross, both parents are homozygous recessive (having identical recessive alleles) \(a a\). If we create a Punnett square for this cross, all the offspring will have the same genotype, which is \(a a\), and since there is no dominant allele \(A\), none of them will show the dominant phenotype. Therefore, the probability is 0% or 0.