In a monohybrid cross involving two heterozygous guinea pigs (let's say the allele for hairless is represented as "H" and the allele for hair with fur as "h"), the genotype of the parents would be:
- Parent 1: Hh
- Parent 2: Hh
When we set up a Punnett square for this cross, we get the following combinations for the offspring:
- HH (homozygous hairless)
- Hh (heterozygous hairless)
- Hh (heterozygous hairless)
- hh (homozygous with fur)
The results show:
- 1 HH
- 2 Hh
- 1 hh
Total offspring genotypes:
- Hairless (HH + Hh): 3 (HH + 2 Hh)
- With fur (hh): 1
The percentage of hairless offspring (HH + Hh) is 3 out of 4, which is 75%.
So, the correct answer is 75%.