Final answer:
The frequency of the dominant allele for cystic fibrosis is approximately 99,980,001/100,000,000. The frequency of the recessive allele is approximately 1/10,000. The percentage of heterozygous individuals (carriers) in the Hispanic population is approximately 2%.
Step-by-step explanation:
The frequency of the dominant allele (F) can be calculated using the Hardy-Weinberg formula, which is p^2 + 2pq, where p represents the frequency of the dominant allele. In this case, since cystic fibrosis is a homozygous recessive condition, the recessive allele (f) has a frequency of 1 in 10,000, so q = 1/10,000. To find p, we can subtract q from 1: p = 1 - q. Once we have p, we can calculate p^2 to get the frequency of the dominant allele. The percentage of heterozygous individuals (carriers) can be calculated using the formula 2pq, where p and q represent the frequencies of the dominant and recessive alleles, respectively.
Here's the step-by-step calculation:
p = 1 - 1/10,000 = 9,999/10,000
p^2 = (9,999/10,000)^2 = 99,980,001/100,000,000
The frequency of the dominant allele (F) is approximately 99,980,001/100,000,000.
q = 1/10,000 = 0.0001
2pq = 2 * (99,999/10,000) * 0.0001 = 0.019998 (approximately 2%)