Question

A heterozygous dominant brown mouse is crossed with a heterozygous brown mouse (tan is the recessive color.)

Make a Punnett square to calculate the possible gametes they can have and the possible genotype of their offspring.

Answer 1

When two heterozygous dominant brown mice are crossed (Bb x Bb), the Punnett square predicts a genotypic ratio of 1 BB: 2 Bb: 1 bb, and a phenotypic ratio of 3 brown to 1 tan mice.

Step-by-step explanation:

To determine the genotype of the offspring when a heterozygous dominant brown mouse (Bb) is crossed with another heterozygous brown mouse (Bb), we use a Punnett square.

Step-by-step Explanation:

1. Identify the allele for dominant (brown color) which is 'B' and the allele for recessive (tan color) which is 'b'.
2. Assign the parental genotypes. One parent is Bb (heterozygous dominant) and the other is also Bb (heterozygous).
3. Create the Punnett square, placing one parent's alleles across the top and the other's down the side. This represents the meiotic segregation of alleles into gametes.
4. Fill in the Punnett square by combining the top and side alleles for each square, representing the genotype of potential offspring.
5. Analyze the results to determine the genotypic and phenotypic ratios. With both parents being Bb, the possible genotypes are BB, Bb, and bb.

The Punnett square follows:

Parent 1 Gametes
B b
B BB Bb
b Bb bb
Parent 2 Gametes
This results in the following genotypic ratio: 1 BB : 2 Bb : 1 bb. Since brown is dominant over tan, the phenotypic ratio will be 3 brown: 1 tan mice.