Environment: Clean Forest
[tex]\[
\begin{tabular}{ccccccc}
& Moths Released & $G_1$ & $G_2$ & $G_3$ & $G_4$ & $G_5$ \\
Typica & 490 & 301 & 387 & 456 & 556 & 878 \\
Carbonaria & 510 & 210 & 190 & 186 & 166 & 54 \\
Total & 1000 & 511 & 577 & 642 & 722 & 932
\end{tabular}
\][/tex]

Phenotype Frequency

[tex]\[
\begin{tabular}{cccc}
& Color & Initial Frequency & \begin{tabular}{c}
Frequency $G_5$ \\
(Round to 2 decimal places)
\end{tabular} \\
Typica & Light & 0.49 & 0.94 \\
Carbonaria & Dark & 0.51 & 0.06
\end{tabular}
\][/tex]

Allele Frequency

[tex]\[
\begin{tabular}{cccc}
& Allele & Initial Allele Frequency & \begin{tabular}{c}
$G_5$ Allele Frequency \\
(Round to 2 decimal places)
\end{tabular} \\
q & d & 0.70 & \\
P & 0.30 &
\end{tabular}
\][/tex]

Genotype Frequency



Answer :

Let's go through the problem step-by-step to calculate various frequencies and understand how they change over the generations in a clean forest environment.

### 1. Initial Phenotype Frequencies
- Initial Typica (light-colored) Moths: 490
- Initial Carbonaria (dark-colored) Moths: 510
- Total Initial Moths: 490 + 510 = 1000

The initial phenotype frequencies are as follows:
- Typica: [tex]\( \frac{490}{1000} = 0.49 \)[/tex]
- Carbonaria: [tex]\( \frac{510}{1000} = 0.51 \)[/tex]

### 2. Phenotype Frequencies in G5
From the final generation (G5) data:
- Typica in G5: 878
- Carbonaria in G5: 54
- Total Moths in G5: 878 + 54 = 932

The phenotype frequencies for G5 are:
- Typica: [tex]\( \frac{878}{932} = 0.94 \)[/tex]
- Carbonaria: [tex]\( \frac{54}{932} = 0.06 \)[/tex]

### 3. Initial Allele Frequencies
We determine the initial allele frequencies under the assumption that the population is in Hardy-Weinberg equilibrium.

Given initial frequency for the recessive allele (q or 'd'):
- [tex]\( q = 0.70 \)[/tex]
Then the frequency of the dominant allele (P):
- [tex]\( p = 1 - q = 1 - 0.70 = 0.30 \)[/tex]

### 4. Allele Frequencies in G5
We use the Hardy-Weinberg equilibrium to calculate the recessive allele frequency (q) from G5.

The frequency of homozygous recessive individuals (carbonaria) in G5 is:
- [tex]\( q^2 \)[/tex] = Frequency of carbonaria in G5 = 0.06

To find q:
- [tex]\( q = \sqrt{0.06} \approx 0.24 \)[/tex]

Then, the frequency of the dominant allele (p) in G5:
- [tex]\( p = 1 - q = 1 - 0.24 = 0.76 \)[/tex]

### Summary Tables:

#### Phenotype Frequency
[tex]\[ \begin{array}{cccc} & Color & \textbf{Initial Frequency} & \textbf{Frequency G_5 (Round to 2 decimal places)} \\ Typica & Light & 0.49 & 0.94 \\ Carbonaria & Dark & 0.51 & 0.06 \\ \end{array} \][/tex]

#### Allele Frequency
[tex]\[ \begin{array}{cccc} & Allele & \textbf{Initial Allele Frequency} & \textbf{G_5 Allele Frequency (Round to 2 decimal places)} \\ q & d & 0.70 & 0.24 \\ p & & 0.30 & 0.76 \\ \end{array} \][/tex]

Through this detailed step-by-step solution, we see how the environment has influenced the moth population, causing a change in both phenotype and allele frequencies over generations.