A population in Hardy-Weinberg equilibrium has certain individuals expressing a rare autosomal recessive disease. The frequency of affected individuals in the population is 1 in 90,000. What is the frequency of carriers in this population?

**Core Concept**
In a Hardy-Weinberg equilibrium population, the frequency of alleles and genotypes remains constant from generation to generation, assuming no mutation, gene flow, or natural selection. The frequency of carriers for a rare autosomal recessive disease can be calculated using the Hardy-Weinberg principle.

**Why the Correct Answer is Right**
The frequency of affected individuals (q^2) is given as 1 in 90,000. To find the carrier frequency (2pq), we need to first find the frequency of the recessive allele (q). Since q^2 = 1/90,000, we can take the square root of both sides to get q = √(1/90,000) = 1/300. Now, we can use the formula 2pq to find the frequency of carriers: 2(1/300)(1 - 1/300) = 2(1/300)(299/300) = 598/90,000 = 1 in 150.

**Why Each Wrong Option is Incorrect**
**Option A:** This option is incorrect because it does not take into account the frequency of the recessive allele (q).
**Option B:** This option is incorrect because it assumes the frequency of carriers is equal to the frequency of affected individuals, which is not true.
**Option C:** This option is incorrect because it uses the wrong formula to calculate the frequency of carriers.

**Clinical Pearl / High-Yield Fact**
When dealing with rare autosomal recessive diseases, it's essential to remember that the frequency of carriers is approximately 2-5 times higher than the frequency of affected individuals.

**Correct Answer:** . 1 in 150.

✓ Correct Answer: B. 1 in 150