Cross a heterozygous running, heterozygous black mouse with a homozygous running, homozygous black mouse
Parental _
Possible gametes _
Offspring phenotypic ratio _
The correct answer and explanation is :
Parental Genotypes:
- The first mouse is heterozygous for both traits (running and black coat color). The genotype would be Rr Bb, where:
- R represents the dominant allele for running (with the recessive r for non-running),
- B represents the dominant allele for black coat color (with the recessive b for brown).
- The second mouse is homozygous for both traits (running and black coat color). The genotype would be RR BB, where:
- R represents the dominant allele for running,
- B represents the dominant allele for black coat color.
Possible Gametes:
- The first mouse (Rr Bb) can produce four types of gametes: RB, Rb, rB, rb.
- The second mouse (RR BB) can only produce two types of gametes: RB, RB (since it is homozygous for both traits).
Punnett Square:
We can now perform the cross using a Punnett square. The first mouse will contribute one of four gametes (RB, Rb, rB, rb), and the second mouse will contribute one of two gametes (RB, RB).
| RB | RB | |
|---|---|---|
| RB | RRB, BB | RRB, BB |
| Rb | RRB, Bb | RRB, Bb |
| rB | RrB, BB | RrB, BB |
| rb | RrB, Bb | RrB, Bb |
Offspring Genotypic and Phenotypic Ratios:
- Genotypic Ratio:
- RR BB: 2 offspring (homozygous for both traits)
- RR Bb: 2 offspring (homozygous for running, heterozygous for black)
- Rr BB: 2 offspring (heterozygous for running, homozygous for black)
- Rr Bb: 2 offspring (heterozygous for both traits)
- Phenotypic Ratio:
- Since R (running) and B (black) are dominant traits, all offspring will exhibit both the running phenotype and the black coat color.
- Therefore, 100% of the offspring will be running and black.
In conclusion, the phenotypic ratio is 100% running, black mice.