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Cross a parent that is homozygous recessive for both traits (bbee) with a dihybrid (BbEe).

- First, draw a Punnett square in your notes.
- For each offspring in the Punnett square, determine the phenotype (fur color and eye color).
- Finally, use that information to fill in the predicted fraction of each phenotype in the data table below.

\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}{l}
Predicted \\
Fraction
\end{tabular} & [tex]$\square / 16$[/tex] & [tex]$\square / 16$[/tex] & [tex]$\square / 16$[/tex] & [tex]$\square / 16$[/tex] \\
\hline
\end{tabular}



Answer :

GinnyAnswer
Sure, let's walk through the problem step by step:

1. Identify the Genotypes of the Parents:
- One parent is homozygous recessive for both traits: [tex]\( bbee \)[/tex].
- The other parent is a dihybrid: [tex]\( BbEe \)[/tex].

2. Determine the Possible Gametes for Each Parent:
- The homozygous recessive parent (bbee) can only produce one type of gamete: [tex]\( be \)[/tex].
- The dihybrid parent (BbEe) can produce four types of gametes through meiosis: [tex]\( BE \)[/tex], [tex]\( Be \)[/tex], [tex]\( bE \)[/tex], and [tex]\( be \)[/tex].

3. Set Up the Punnett Square:
- The top row (or column) of the Punnett square will be the gametes from the homozygous recessive parent: [tex]\( be \)[/tex].
- The side row (or column) will be the gametes from the dihybrid parent: [tex]\( BE \)[/tex], [tex]\( Be \)[/tex], [tex]\( bE \)[/tex], [tex]\( be \)[/tex].

Constructing the Punnett square:
```
be
-----------------
BE | BbEe
Be | Bbee
bE | bbEe
be | bbee
```

4. Determine the Genotype of Each Offspring:
- Combining the gametes, we have these possible outcomes:
- [tex]\( BE \times be = BbEe \)[/tex]
- [tex]\( Be \times be = Bbee \)[/tex]
- [tex]\( bE \times be = bbEe \)[/tex]
- [tex]\( be \times be = bbee \)[/tex]

5. Determine the Phenotype of Each Offspring:
- For [tex]\( BbEe \)[/tex]: Black Fur and Black Eyes (1 combination)
- For [tex]\( Bbee \)[/tex]: Black Fur and Black Eyes (1 combination)
- For [tex]\( bbEe \)[/tex]: White Fur and Black Eyes (1 combination)
- For [tex]\( bbee \)[/tex]: White Fur and Red Eyes (1 combination)

Since the homozygous recessive parent always pairs with the same gamete type, each combination will occur 4 times in our resulting 16-box Punnett square.

6. Determine the Phenotype Ratios for the Offspring:
- Black Fur and Black Eyes (BbEe and Bbee): Occurs 4 times out of 16
- Black Fur and Red Eyes: Does not occur
- White Fur and Black Eyes (bbEe): Occurs 2 times out of 16
- White Fur and Red Eyes (bbee): Occurs 2 times out of 16

7. Fill in the Predicted Fraction of Each Phenotype:
```
\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}{l}
Predicted \\
Fraction
\end{tabular} & 4 / 16 & 0 / 16 & 2 / 16 & 2 / 16 \\
\hline
\end{tabular}
```

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