In a population of 2500, how many babies would you expect to have cystic fibrosis, a homozygous recessive condition, if the frequency of the dominant allele is 0.9 and the population is at Hardy–Weinberg equilibrium? a. 0.9×2500=2250 b. 2×0.9×0.1×2500=450 c. 0.9×0.1×2500=225 d. 0.1x0.1x2500=25

Verified step by step guidance

1. Step 1: Understand the problem. We are asked to find the number of babies expected to have cystic fibrosis, a homozygous recessive condition, in a population of 2500. The frequency of the dominant allele is given as 0.9 and the population is at Hardy-Weinberg equilibrium.
2. Step 2: Recall the Hardy-Weinberg principle. This principle states that the frequency of alleles in a population will remain constant from generation to generation in the absence of other evolutionary influences. The equation for Hardy-Weinberg equilibrium is p^2 + 2pq + q^2 = 1, where p is the frequency of the dominant allele, q is the frequency of the recessive allele, p^2 is the frequency of homozygous dominant individuals, 2pq is the frequency of heterozygous individuals, and q^2 is the frequency of homozygous recessive individuals.
3. Step 3: Calculate the frequency of the recessive allele (q). Since p + q = 1, and we know that p = 0.9, we can solve for q: q = 1 - p = 1 - 0.9 = 0.1.
4. Step 4: Calculate the expected number of babies with cystic fibrosis. Since cystic fibrosis is a homozygous recessive condition, we are interested in q^2, which represents the frequency of homozygous recessive individuals. So, q^2 = (0.1)^2 = 0.01. To find the expected number of babies with cystic fibrosis, we multiply this frequency by the total population: 0.01 * 2500 = 25.
5. Step 5: The correct answer is d. 0.1x0.1x2500=25. This means we would expect 25 babies in this population to have cystic fibrosis.