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

Question

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.