Question

face color
chaz and bev both have homozygous recessive genotypes for their color.
chaz
bev
phenotype probability (%)
yellow color:
blue color:

Explanation:

Step1: Determine Genotypes

Since both Chaz and Bev have homozygous recessive genotypes, let's assume the recessive allele for their color (say, blue) is represented as \( bb \) for both. So Chaz: \( bb \), Bev: \( bb \).

Step2: Punnett Square Analysis

When we cross \( bb \) (Chaz) with \( bb \) (Bev), all the offspring will inherit one \( b \) from each parent, resulting in \( bb \) genotype for all offspring.

Step3: Phenotype Probability

If the recessive phenotype is, for example, blue (and dominant would be yellow, but since both are homozygous recessive, the only possible phenotype for offspring is the recessive one). Wait, but the problem mentions yellow and blue. Wait, maybe the recessive is blue? Wait, no—if both are homozygous recessive, their offspring will all have the recessive genotype, so the phenotype associated with the recessive allele. Let's correct: if the recessive allele gives blue color, then all offspring will be blue. But wait, maybe the original problem has a typo, but based on the given: both are homozygous recessive. So let's denote: let's say the recessive genotype is for blue, so Chaz (bb) and Bev (bb). Cross: all offspring are bb. So the phenotype probability: if blue is recessive, then blue color probability is 100%, yellow (dominant) is 0%. But wait, maybe the problem had a mix-up, but following the logic: homozygous recessive parents (both) will produce all homozygous recessive offspring. So the phenotype from recessive genotype: if recessive is blue, then blue is 100%, yellow 0%. But let's check the Punnett square:

	b	b
b	bb	bb

All four squares are \( bb \), so 4 out of 4, which is 100% for the recessive phenotype (blue, assuming recessive is blue) and 0% for the dominant (yellow).

Answer:

Yellow color: 0%
Blue color: 100%