Answer :
To solve this problem, we need to analyze the genetic cross between a parent that is homozygous recessive for both traits (bbee) and a dihybrid parent (BbEe). Let's break down the process into clear steps.
1. Identify Gametes:
- For the homozygous recessive parent (bbee), we only have one type of gamete: `be`.
- For the dihybrid parent (BbEe), the possible gametes are: `BE`, `Be`, `bE`, and `be`.
2. Construct the Punnett Square:
- We will combine the gametes from each parent to form the offspring genotypes.
- The Punnett Square will represent all possible combinations.
3. Determine Offspring Genotypes:
- Combine each gamete from the homozygous recessive parent with each gamete from the dihybrid parent.
| | BE | Be | bE | be |
|-----------|------|------|------|------|
| be | BbEe | Bbee | bbEe | bbee |
4. Determine Phenotypes:
- The phenotypes are based on the dominant and recessive alleles:
- Fur Color: Dominant B = Black Fur, Recessive b = White Fur
- Eye Color: Dominant E = Black Eyes, Recessive e = Red Eyes
- For each genotype:
- `BbEe`: Black Fur and Black Eyes (dominant B and E)
- `Bbee`: Black Fur and Red Eyes (dominant B and recessive e)
- `bbEe`: White Fur and Black Eyes (recessive b and dominant E)
- `bbee`: White Fur and Red Eyes (recessive b and e)
5. Count Phenotypes:
- We have a total of four different phenotypes and each combination is equally likely.
6. Predicted Fractions:
- Each phenotype occurs 1 out of 4 times in the Punnett Square, representing 1/4 of the total.
Therefore, the table of predicted fractions of each phenotype filled in is:
[tex]\[ \begin{tabular}{|l|c|c|c|c|} \hline & \begin{tabular}{c} Black \\ Fur and \\ Black \\ Eyes \end{tabular} & \begin{tabular}{c} Black \\ Fur and \\ Red \\ Eyes \end{tabular} & \begin{tabular}{c} White \\ Fur and \\ Black \\ Eyes \end{tabular} & \begin{tabular}{c} White \\ Fur and \\ Red \\ Eyes \end{tabular} \\ \hline \begin{tabular}{c} Predicted \\ Fraction \end{tabular} & $\frac{1}{4}$ & $\frac{1}{4}$ & $\frac{1}{4}$ & $\frac{1}{4}$ \\ \hline \end{tabular} \][/tex]
1. Identify Gametes:
- For the homozygous recessive parent (bbee), we only have one type of gamete: `be`.
- For the dihybrid parent (BbEe), the possible gametes are: `BE`, `Be`, `bE`, and `be`.
2. Construct the Punnett Square:
- We will combine the gametes from each parent to form the offspring genotypes.
- The Punnett Square will represent all possible combinations.
3. Determine Offspring Genotypes:
- Combine each gamete from the homozygous recessive parent with each gamete from the dihybrid parent.
| | BE | Be | bE | be |
|-----------|------|------|------|------|
| be | BbEe | Bbee | bbEe | bbee |
4. Determine Phenotypes:
- The phenotypes are based on the dominant and recessive alleles:
- Fur Color: Dominant B = Black Fur, Recessive b = White Fur
- Eye Color: Dominant E = Black Eyes, Recessive e = Red Eyes
- For each genotype:
- `BbEe`: Black Fur and Black Eyes (dominant B and E)
- `Bbee`: Black Fur and Red Eyes (dominant B and recessive e)
- `bbEe`: White Fur and Black Eyes (recessive b and dominant E)
- `bbee`: White Fur and Red Eyes (recessive b and e)
5. Count Phenotypes:
- We have a total of four different phenotypes and each combination is equally likely.
6. Predicted Fractions:
- Each phenotype occurs 1 out of 4 times in the Punnett Square, representing 1/4 of the total.
Therefore, the table of predicted fractions of each phenotype filled in is:
[tex]\[ \begin{tabular}{|l|c|c|c|c|} \hline & \begin{tabular}{c} Black \\ Fur and \\ Black \\ Eyes \end{tabular} & \begin{tabular}{c} Black \\ Fur and \\ Red \\ Eyes \end{tabular} & \begin{tabular}{c} White \\ Fur and \\ Black \\ Eyes \end{tabular} & \begin{tabular}{c} White \\ Fur and \\ Red \\ Eyes \end{tabular} \\ \hline \begin{tabular}{c} Predicted \\ Fraction \end{tabular} & $\frac{1}{4}$ & $\frac{1}{4}$ & $\frac{1}{4}$ & $\frac{1}{4}$ \\ \hline \end{tabular} \][/tex]
