Question

With incomplete dominance, if a black furred animal (B) mates with a tan animal (b), the resulting phenotype is gray fur (Bb), a blend of the two. Set up a Punnett square to determine the phenotypes and genetic ratios for a cross between two heterozygous gray animals. Please note that is a simple monohybrid cross in which the genes exhibit incomplete dominance.

Answer 1

When two heterozygous gray animals (Bb) are crossed, you set up a Punnett square with 'B' and 'b' representing black and tan alleles, respectively. The resulting phenotypic ratio is 1:2:1, with 25% black (BB), 50% gray (Bb), and 25% tan (bb) due to incomplete dominance.

Step-by-step explanation:

To set up a Punnett square for the cross between two heterozygous gray animals, we first establish the alleles for the fur color trait. Let's say 'B' stands for black fur and is incompletely dominant, while 'b' stands for tan fur. A heterozygous gray animal has the genotype 'Bb'. Here's how you would set up the Punnett square:

Parental genotypes (both gray): Bb x Bb

• Gametes from one parent: B, b
• Gametes from the other parent: B, b

Then we fill in the Punnett Square.

BbBBB (black)Bb (gray)bBb (gray)bb (tan)

From this Punnett square, we can see that there is a 1:2:1 ratio for the offspring's genotypes, which translates to a 1:2:1 phenotypic ratio:

1. 25% black (BB)
2. 50% gray (Bb)
3. 25% tan (bb)

This ratio occurs because of incomplete dominance, where the heterozygous phenotype (Bb) is an intermediate between the two homozygous phenotypes.