If frequency of an autosomal recessive lethal gene is `0.6` then what will be the frequency of carrier phenotypes in progeny among 2000 individuals ?

Options

A. 1680
B. 1280
C. 960
D. 320

Answer

To solve the problem step by step, we will use the principles of population genetics, specifically the Hardy-Weinberg principle.

### Step 1: Understand the given information
We are given the frequency of an autosomal recessive lethal gene (q) as 0.6. This means that the frequency of the recessive allele (q) is 0.6.

### Step 2: Calculate the frequency of the dominant allele
Using the Hardy-Weinberg principle, we know that the sum of the frequencies of the dominant allele (p) and the recessive allele (q) must equal 1:
\[ p + q = 1 \]
Substituting the value of q:
\[ p + 0.6 = 1 \]
Thus, we can find p:
\[ p = 1 - 0.6 = 0.4 \]

### Step 3: Apply the Hardy-Weinberg equation
The Hardy-Weinberg equation states:
\[ p^2 + 2pq + q^2 = 1 \]
Where:
- \( p^2 \) = frequency of homozygous dominant individuals
- \( 2pq \) = frequency of heterozygous (carrier) individuals
- \( q^2 \) = frequency of homozygous recessive individuals

### Step 4: Calculate the frequencies
We already have p and q:
- \( p = 0.4 \)
- \( q = 0.6 \)

Now, we can calculate \( 2pq \):
\[ 2pq = 2 \times 0.4 \times 0.6 \]
\[ 2pq = 2 \times 0.24 = 0.48 \]

### Step 5: Find the frequency of carrier phenotypes in a population of 2000 individuals
To find the number of carriers in a population of 2000 individuals, we multiply the frequency of carriers (2pq) by the total number of individuals:
\[ \text{Number of carriers} = 2pq \times \text{Total individuals} \]
\[ \text{Number of carriers} = 0.48 \times 2000 \]
\[ \text{Number of carriers} = 960 \]

### Conclusion
The frequency of carrier phenotypes in the progeny among 2000 individuals is **960**.

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