Solve this problem: In chimpanzees, straight fingers are dominant to bent fingers. Complete a Punnett square to show the genotypes and phenotypes expected for the following cross: heterozygous straight x homozygous bent.

Answer 1

To solve the problem, we first need to define the alleles involved for the trait of finger shape in chimpanzees:

• Let's denote the allele for straight fingers as \( S \) (dominant).
• Let's denote the allele for bent fingers as \( s \) (recessive).

Genotypes of the parents:

1. The heterozygous straight parent would have the genotype \( Ss \).
2. The homozygous bent parent would have the genotype \( ss \).

Punnett Square:

We can set up a Punnett square to show the possible combinations of alleles from the two parents. The alleles from the heterozygous straight parent (\( Ss \)) will be along the top, and the alleles from the homozygous bent parent (\( ss \)) will be along the side.

          S          s
        ----------------
    s |   Ss      |   ss   |
        ----------------
    s |   Ss      |   ss   |
        ----------------

Results from the Punnett Square:

• The top left cell: \( Ss \) (heterozygous - straight fingers)
• The top right cell: \( ss \) (homozygous - bent fingers)
• The bottom left cell: \( Ss \) (heterozygous - straight fingers)
• The bottom right cell: \( ss \) (homozygous - bent fingers)

Phenotypic Ratio:

Now, let's summarize the genotypes:

• \( 2 \) \( Ss \) (straight fingers)
• \( 2 \) \( ss \) (bent fingers)

The expected phenotypic ratio is:

• Straight fingers (dominant): 2
• Bent fingers (recessive): 2

This can be simplified to a 1:1 ratio.

Conclusion: For the cross of a heterozygous straight finger chimpanzee \( (Ss) \) with a homozygous bent finger chimpanzee \( (ss) \), we expect the genotypes and phenotypes as follows:

• Genotypic ratio: \( 2 Ss : 2 ss \) or 1 \( Ss \) : 1 \( ss \)
• Phenotypic ratio: 1 straight fingers : 1 bent fingers.