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!
<span>This shows that, in most species, a characteristic is controlled by a suite of genes instead of having only one gene doing all the work. Polygenic traits are much more common than monogenic, which leads to an entire range of outcomes being possible, instead of having an either-or outcome.</span>
Answer:
it creates new combinations of genetic material in the 4 daughter cells
I think it is c
Hope this helps!
