Answer:
0.08, because for an X-linked trait the genotype frequencies among males are the same as the allele frequencies. Therefore, you would expect about q2 = 0.0064 (or 0.64%) of the female population to be affected and 2pq =0.147 (or 14.7%) of the female population to be carriers.
Explanation:
The dam would add more support… instead of using dirt to hold up the rainfall running down from the mountain the dam would hold the water back and keep the water away from where it is headed
D makes the most sense, the rest of them aren’t really logical.
Q = recessive allele frequency = 0.3, and thus in H-W equilibrium there are ONLY two alleles, q (recessive) and
p (dominant). Therefore all of the p and q present for this gene in a population must account for 100% of this gene's alleles. And 100% = 1.00.
So p, the dominant allele frequency, must be equal to 1 - q --> p = 1 - q
p = 1 - 0.3 = 0.7.
Since heterozygotes are a combination of the p and q, we must again look at the frequencies of each genotype: p + q = 1, then (p+q)^2 = 1^2
So multiplying out (p+q)(p+q) = 1, we get: p^2+2pq+q^2 = 1 (all genotypes), where p^2 = frequency of homozygous dominant individuals, 2pq = frequency of heterozygous individuals, and q^2 = frequency of homozygous recessive individuals.
Therefore if the population is in H-W equilibrium, then the expected frequency of heterozygous individuals = 2pq = 2(0.7)(0.3)
2pq = 2(0.21) = 0.42, or 42% of the population.
Hope that helps you to understand how to solve population genetics problems!
