Cross a homozygous running, homozygous black mouse with a heterozygous running, brown mouse
Parental genotypes _
Possible gametes _
Offspring phenotypic ratio _
The correct answer and explanation is :
Let’s break down the genetic scenario involving the two mice with different traits:
Parental Genotypes:
- Homozygous running, homozygous black mouse: This mouse is RRBB. The capital “R” represents the dominant allele for the running trait, and the capital “B” represents the dominant allele for black fur color. Since it’s homozygous, both alleles are the same.
- Heterozygous running, brown mouse: This mouse is Rrbb. “Rr” means that it is heterozygous for the running trait (one dominant allele and one recessive allele), and “bb” means it is homozygous recessive for brown fur color.
Possible Gametes:
To find the possible gametes produced by each parent, we will use a basic Punnett square approach.
- Homozygous running, black mouse (RRBB) can produce the following gametes: RB. Since both alleles are the same, the only gametes it can produce are RB.
- Heterozygous running, brown mouse (Rrbb) can produce the following gametes: Rb and rb. Since “R” and “r” are different alleles, the possible combinations are Rb (dominant running trait, recessive brown fur) and rb (recessive for both traits).
Punnett Square:
Let’s combine the gametes from both parents:
| RB | RB | |
|---|---|---|
| Rb | RRBb | RRBb |
| rb | RrBb | RrBb |
Offspring Phenotypic Ratio:
- Running trait (R) is dominant over the recessive (r), so any offspring with at least one R will be running.
- Fur color: B (black) is dominant over b (brown). So, only bb individuals will have brown fur, while any other combination (such as Bb) will result in black fur.
From the Punnett square, we see:
- RRBb (2 offspring) = Running, Black
- RrBb (2 offspring) = Running, Black
Phenotypic ratio: All 4 offspring will be Running, Black.
Conclusion:
- Parental genotypes: RRBB × Rrbb
- Possible gametes: RB × Rb, rb
- Offspring phenotypic ratio: 100% Running, Black