Cystic fibrosis is an autosomal recessive disease . In an island having a population of 200 people, 98 people suffer from cystic fibrosis. How many people are carrier of this disease ?

Options

A. 42
B. 84
C. 18
D. 64

Answer

To solve the problem of how many people are carriers of cystic fibrosis in a population of 200 people, where 98 people suffer from the disease, we can follow these steps:

### Step 1: Understand the Genetic Basis
Cystic fibrosis is an autosomal recessive disease. This means that an individual must inherit two copies of the mutated gene (one from each parent) to express the disease. Individuals with one copy of the mutated gene are carriers but do not exhibit symptoms.

### Step 2: Determine the Frequency of the Disease
Given that 98 out of 200 people have cystic fibrosis, we can calculate the frequency of the disease in the population.

\[ \text{Frequency of affected individuals (q²)} = \frac{\text{Number of affected individuals}}{\text{Total population}} = \frac{98}{200} = 0.49 \]

### Step 3: Calculate q
To find the frequency of the recessive allele (q), we take the square root of the frequency of affected individuals:

\[ q = \sqrt{0.49} = 0.7 \]

### Step 4: Calculate p
Using the Hardy-Weinberg principle, we know that \( p + q = 1 \). Therefore, we can find p (the frequency of the dominant allele):

\[ p = 1 - q = 1 - 0.7 = 0.3 \]

### Step 5: Calculate the Carrier Frequency
The frequency of carriers (heterozygous individuals, represented as 2pq) can be calculated as follows:

\[ 2pq = 2 \times p \times q = 2 \times 0.3 \times 0.7 = 0.42 \]

### Step 6: Calculate the Number of Carriers
To find the number of carriers in the population, multiply the carrier frequency by the total population:

\[ \text{Number of carriers} = 2pq \times \text{Total population} = 0.42 \times 200 = 84 \]

### Conclusion
Thus, the number of people who are carriers of cystic fibrosis in the population is **84**.

---