Question

When a cross is made between AaBb x AaBb, what fraction of the offspring is able to produce a genotype that is heterozygous for both characteristics?
3/16
9/16
1/16
4/16

Answer 1

To determine the fraction of offspring that are heterozygous for both traits (genotype AaBb) from the cross AaBb x AaBb, we can use a Punnett square.

1. The alleles from each parent are:
- Parent 1: AaBb
- Parent 2: AaBb

2. The possible gametes from each parent are:
- AB
- Ab
- aB
- ab

3. Setting up a Punnett square with these gametes results in the following combinations:

\[
\begin{array}{c|c|c|c|c}
& AB & Ab & aB & ab \\
\hline
AB & AABB & AABb & AaBB & AaBb \\
\hline
Ab & AABb & AAbb & AaBb & Aabb \\
\hline
aB & AaBB & AaBb & aaBB & aaBb \\
\hline
ab & AaBb & Aabb & aaBb & aabb \\
\end{array}
\]

4. Now, we count the genotypes:
- AABB: 1
- AABb: 2
- AAbb: 1
- AaBB: 2
- AaBb: 4 **(this is the genotype we are looking for)**
- Aabb: 2
- aaBB: 1
- aaBb: 2
- aabb: 1

5. The total number of offspring combinations is \(16\) (4 x 4 from the Punnett square).

6. The number of AaBb combinations in the offspring is \(4\).

7. Therefore, the fraction of the offspring that are heterozygous for both traits (AaBb) is:

\[
\frac{4}{16} = \frac{1}{4}
\]

Thus, the answer is **4/16**.