Question

Show the cross for two heterozygous guinea pigs.75% with short hair and 25% with long hair.
Punnet square:

Answer 1

To perform a Punnett square for two heterozygous guinea pigs, we need to define the alleles involved. Let's use:
- **H** = short hair (dominant)
- **h** = long hair (recessive)

Since both guinea pigs are heterozygous, their genotypes will be **Hh** (one allele for short hair and one for long hair).

### Punnett Square

We can set up a 2x2 Punnett square as follows:

| | H | h |
|------|------|------|
| **H** | HH | Hh |
| **h** | Hh | hh |

Now let's analyze the possible genotypes and their corresponding phenotypes:

- **HH**: This genotype will produce a guinea pig with short hair.
- **Hh**: This genotype will also produce a guinea pig with short hair.
- **hh**: This genotype will produce a guinea pig with long hair.

### Genotype Ratios
From the Punnett square, we can see:

- 1 HH (short hair)
- 2 Hh (short hair)
- 1 hh (long hair)

In terms of percentages:
- **HH**: 1 out of 4 (25%)
- **Hh**: 2 out of 4 (50%)
- **hh**: 1 out of 4 (25%)

### Phenotype Ratios
Combining the genotypes that produce short hair (HH and Hh):
- Short hair (HH + Hh): 3 out of 4 (75%)
- Long hair (hh): 1 out of 4 (25%)

### Summary
Thus, when crossing two heterozygous guinea pigs (Hh x Hh), we expect:
- 75% of the offspring to have short hair
- 25% to have long hair

This correlates with your original statement of 75% short-haired and 25% long-haired offspring.