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Sagot :
To solve the problem, we need to use the information from a Punnett square for two traits: fur color and eye color. For simplicity, let’s assume the traits are controlled by two pairs of alleles:
1. Fur color: Black (B) is dominant over white (b).
2. Eye color: Black (E) is dominant over red (e).
We need the genotypes of the parents to complete the Punnett square. Let’s assume the parents are heterozygous for both traits: [tex]\( BbEe \times BbEe \)[/tex].
We can set up a 4x4 Punnett square to illustrate the combinations of these alleles:
[tex]\[ \begin{array}{c|cccc} & BE & Be & bE & be \\ \hline BE & BBEE & BBEe & BbEE & BbEe \\ Be & BBEe & BBee & BbEe & Bbee \\ bE & BbEE & BbEe & bbEE & bbEe \\ be & BbEe & Bbee & bbEe & bbee \\ \end{array} \][/tex]
Now let’s determine the phenotypes resulting from each genotype and count their occurrences:
1. Black Fur and Black Eyes (BBEE, BBEe, BbEE, BbEe):
- [tex]\( BBEE \)[/tex]: 1
- [tex]\( BBEe \)[/tex]: 2
- [tex]\( BbEE \)[/tex]: 2
- [tex]\( BbEe \)[/tex]: 4
- Total: 9
2. Black Fur and Red Eyes (BBee, Bbee):
- [tex]\( BBee \)[/tex]: 1
- [tex]\( Bbee \)[/tex]: 2
- Total: 3
3. White Fur and Black Eyes (bbEE, bbEe):
- [tex]\( bbEE \)[/tex]: 1
- [tex]\( bbEe \)[/tex]: 2
- Total: 3
4. White Fur and Red Eyes (bbee):
- [tex]\( bbee \)[/tex]: 1
To fill in the predicted fractions for each phenotype in the data table, we consider the total number of offspring combinations in the Punnett square, which is 16:
[tex]\[ \begin{array}{|l|c|c|c|c|c|} \cline{2-6} \multicolumn{1}{c|}{} & \begin{tabular}{c} Black Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} Black Fur and \\ Red Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Red Eyes \end{tabular} \\ \hline Predicted Fraction & \frac{9}{16} & \frac{3}{16} & \frac{3}{16} & \frac{1}{16} \\ \hline \end{array} \][/tex]
So, the completed data table with predicted fractions for each phenotype is as follows:
[tex]\[ \begin{tabular}{|l|c|c|c|c|} \cline { 2 - 5 } \multicolumn{1}{c|}{} & \begin{tabular}{c} Black Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} Black Fur and \\ Red Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Red Eyes \end{tabular} \\ \hline Predicted Fraction & \frac{9}{16} & \frac{3}{16} & \frac{3}{16} & \frac{1}{16} \\ \hline \end{tabular} \][/tex]
1. Fur color: Black (B) is dominant over white (b).
2. Eye color: Black (E) is dominant over red (e).
We need the genotypes of the parents to complete the Punnett square. Let’s assume the parents are heterozygous for both traits: [tex]\( BbEe \times BbEe \)[/tex].
We can set up a 4x4 Punnett square to illustrate the combinations of these alleles:
[tex]\[ \begin{array}{c|cccc} & BE & Be & bE & be \\ \hline BE & BBEE & BBEe & BbEE & BbEe \\ Be & BBEe & BBee & BbEe & Bbee \\ bE & BbEE & BbEe & bbEE & bbEe \\ be & BbEe & Bbee & bbEe & bbee \\ \end{array} \][/tex]
Now let’s determine the phenotypes resulting from each genotype and count their occurrences:
1. Black Fur and Black Eyes (BBEE, BBEe, BbEE, BbEe):
- [tex]\( BBEE \)[/tex]: 1
- [tex]\( BBEe \)[/tex]: 2
- [tex]\( BbEE \)[/tex]: 2
- [tex]\( BbEe \)[/tex]: 4
- Total: 9
2. Black Fur and Red Eyes (BBee, Bbee):
- [tex]\( BBee \)[/tex]: 1
- [tex]\( Bbee \)[/tex]: 2
- Total: 3
3. White Fur and Black Eyes (bbEE, bbEe):
- [tex]\( bbEE \)[/tex]: 1
- [tex]\( bbEe \)[/tex]: 2
- Total: 3
4. White Fur and Red Eyes (bbee):
- [tex]\( bbee \)[/tex]: 1
To fill in the predicted fractions for each phenotype in the data table, we consider the total number of offspring combinations in the Punnett square, which is 16:
[tex]\[ \begin{array}{|l|c|c|c|c|c|} \cline{2-6} \multicolumn{1}{c|}{} & \begin{tabular}{c} Black Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} Black Fur and \\ Red Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Red Eyes \end{tabular} \\ \hline Predicted Fraction & \frac{9}{16} & \frac{3}{16} & \frac{3}{16} & \frac{1}{16} \\ \hline \end{array} \][/tex]
So, the completed data table with predicted fractions for each phenotype is as follows:
[tex]\[ \begin{tabular}{|l|c|c|c|c|} \cline { 2 - 5 } \multicolumn{1}{c|}{} & \begin{tabular}{c} Black Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} Black Fur and \\ Red Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Black Eyes \end{tabular} & \begin{tabular}{c} White Fur and \\ Red Eyes \end{tabular} \\ \hline Predicted Fraction & \frac{9}{16} & \frac{3}{16} & \frac{3}{16} & \frac{1}{16} \\ \hline \end{tabular} \][/tex]
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