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Sagot :
Certainly! Let's work on filling in the table based on the provided information and assumed genetic principles. The simplified example we are using here assumes a straightforward Mendelian inheritance with complete dominance for the fur color trait. We will assume black (B) is dominant over white (b).
Given that the predicted percentage for black fur is 50%, we can deduce the following:
1. Genotype Configuration: For black fur to appear in 50% of the offspring, the parent genotypes must be such that there is an equal probability of inheriting a pair of alleles resulting in black or white fur.
- One possible scenario is that both parents are heterozygous (Bb). This means each parent has one allele for black fur (B) and one allele for white fur (b).
2. Punnett Square Analysis:
- Cross two heterozygous (Bb) parents:
```
Parent 1: Bb
| B | b |
----------------
P B | BB | Bb |
a ----------------
r b | Bb | bb |
e ----------------
n
t
2
```
- The resulting genotypes in the offspring are:
- BB: 1 (Black fur)
- Bb: 2 (Black fur)
- bb: 1 (White fur)
3. Phenotype Ratio:
- Black Fur (BB and Bb): 3 out of 4 possibilities (75%)
- White Fur (bb): 1 out of 4 possibilities (25%)
However, if our starting point is 50% as given:
- Then the parents must have one black-furred heterozygous (Bb) and one white-furred homozygous (bb). Cross Bb with bb:
```
Parent 1: Bb
| B | b |
----------------
P b | Bb | bb |
a ----------------
r b | Bb | bb |
e ----------------
n
t
2 (bb)
```
- The resulting genotypes in the offspring are:
- Bb: 2 (Black fur)
- bb: 2 (White fur)
4. Phenotype Ratio:
- Black Fur (Bb): 2 out of 4 possibilities (50%)
- White Fur (bb): 2 out of 4 possibilities (50%)
Given these conclusions, we can now fill in the table:
\begin{tabular}{|c|c|c|}
\hline
& Black Fur & White Fur \\
\hline
\begin{tabular}{c}
Predicted \\
Percentage
\end{tabular} & 50 \% & 50 \% \\
\hline
\begin{tabular}{c}
Simulated \\
Number
\end{tabular} & (To be filled in later) & (To be filled in later) \\
\hline
\begin{tabular}{c}
Simulated \\
Percentage
\end{tabular} & (To be filled in later) & (To be filled in later) \\
\hline
\end{tabular}
Given that the predicted percentage for black fur is 50%, we can deduce the following:
1. Genotype Configuration: For black fur to appear in 50% of the offspring, the parent genotypes must be such that there is an equal probability of inheriting a pair of alleles resulting in black or white fur.
- One possible scenario is that both parents are heterozygous (Bb). This means each parent has one allele for black fur (B) and one allele for white fur (b).
2. Punnett Square Analysis:
- Cross two heterozygous (Bb) parents:
```
Parent 1: Bb
| B | b |
----------------
P B | BB | Bb |
a ----------------
r b | Bb | bb |
e ----------------
n
t
2
```
- The resulting genotypes in the offspring are:
- BB: 1 (Black fur)
- Bb: 2 (Black fur)
- bb: 1 (White fur)
3. Phenotype Ratio:
- Black Fur (BB and Bb): 3 out of 4 possibilities (75%)
- White Fur (bb): 1 out of 4 possibilities (25%)
However, if our starting point is 50% as given:
- Then the parents must have one black-furred heterozygous (Bb) and one white-furred homozygous (bb). Cross Bb with bb:
```
Parent 1: Bb
| B | b |
----------------
P b | Bb | bb |
a ----------------
r b | Bb | bb |
e ----------------
n
t
2 (bb)
```
- The resulting genotypes in the offspring are:
- Bb: 2 (Black fur)
- bb: 2 (White fur)
4. Phenotype Ratio:
- Black Fur (Bb): 2 out of 4 possibilities (50%)
- White Fur (bb): 2 out of 4 possibilities (50%)
Given these conclusions, we can now fill in the table:
\begin{tabular}{|c|c|c|}
\hline
& Black Fur & White Fur \\
\hline
\begin{tabular}{c}
Predicted \\
Percentage
\end{tabular} & 50 \% & 50 \% \\
\hline
\begin{tabular}{c}
Simulated \\
Number
\end{tabular} & (To be filled in later) & (To be filled in later) \\
\hline
\begin{tabular}{c}
Simulated \\
Percentage
\end{tabular} & (To be filled in later) & (To be filled in later) \\
\hline
\end{tabular}
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