Chapter 20: Problem 3

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

Expert verified

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

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

Step by step solution

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

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

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

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Biology Textbooks

Chemistry of Life

Biological Structures

Ecology

Reproduction

Genetic Information

Heredity

Study anywhere. Anytime. Across all devices.