In the Caucasian population of the US, 1 in 2500 babies is affected by a recessive condition – cystic fibrosis. In this population, the frequency of the dominant allele is

The correct option is 0.98.

To find the frequency of the dominant allele in the population, we can apply the Hardy-Weinberg principle. According to this principle, the frequency of alleles and genotypes in a population remains constant from generation to generation in the absence of evolutionary influences.

Let:
- $p$ be the frequency of the dominant allele.
- $q$ be the frequency of the recessive allele.

The equation representing the relationship between the allele frequencies is:
$p+q=1$

The equation representing the genotype frequencies is:
$p^2+2pq+q^2=1$

Here, cystic fibrosis is a recessive condition, which means individuals affected by it must have the homozygous recessive genotype ($q^2$). The problem states that 1 in 2500 babies is affected. Therefore, the frequency of the homozygous recessive genotype ($q^2$) is:
$q^2=\frac{1}{2500}=0.0004$

To find the frequency of the recessive allele ($q$), we take the square root of $q^2$:
$q=\sqrt{0.0004}=0.02$

Now, we can find the frequency of the dominant allele ($p$) using the relation $p+q=1$:
$p=1-q$
$p=1-0.02=0.98$

Thus, the frequency of the dominant allele in this population is 0.98.