Certainly! Let's determine the predicted phenotype fractions based on a Punnett square.
Assuming the parental genotypes are heterozygous for the traits of fur color and eye color, we'll denote:
- [tex]\( B \)[/tex] for Black fur (dominant),
- [tex]\( b \)[/tex] for White fur (recessive),
- [tex]\( E \)[/tex] for Black eyes (dominant),
- [tex]\( e \)[/tex] for Red eyes (recessive).
Given these combinations, the Punnett square for a dihybrid cross (BbEe x BbEe) will result in:
\- 9 Black Fur and Black Eyes: The phenotype combines the dominant genes for both fur color and eye color.
\- 3 Black Fur and Red Eyes: The fur color is black due to the presence of at least one dominant [tex]\( B \)[/tex] allele, but the eye color is red because it contains two recessive [tex]\( e \)[/tex] alleles.
\- 3 White Fur and Black Eyes: The fur color is white due to having two recessive [tex]\( b \)[/tex] alleles, but the eye color is black because of the presence of at least one dominant [tex]\( E \)[/tex] allele.
\- 1 White Fur and Red Eyes: Both the fur color and eye color are recessive because it has two recessive [tex]\( b \)[/tex] alleles and two recessive [tex]\( e \)[/tex] alleles.
Filling in the fractions in the table:
[tex]\[
\begin{tabular}{|c|c|c|c|c|}
\hline & \begin{tabular}{l}
Black Fur and \\
Black Eyes
\end{tabular} & \begin{tabular}{l}
Black Fur and \\
Red Eyes
\end{tabular} & \begin{tabular}{l}
White Fur and \\
Black Eyes
\end{tabular} & \begin{tabular}{l}
White Fur and \\
Red Eyes
\end{tabular} \\
\hline Predicted Fraction & $9 / 16$ & $3 / 16$ & $3 / 16$ & $1 / 16$ \\
\hline
\end{tabular}
\][/tex]
For each phenotype, the predicted fractions are:
- Black Fur and Black Eyes: [tex]\(\frac{9}{16}\)[/tex]
- Black Fur and Red Eyes: [tex]\(\frac{3}{16}\)[/tex]
- White Fur and Black Eyes: [tex]\(\frac{3}{16}\)[/tex]
- White Fur and Red Eyes: [tex]\(\frac{1}{16}\)[/tex]
