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!
Answer:
a 80% b50% ctwo billionths one one- hundredth
Answer: Acid-fast and Gram
Explanation:
The gram stains peptidoglycan and the acid-fast stain stains mycolic acid, which are both in the cell wall.
Answer:there different because genetics are feactures pass on from your past decendents that you know own and heath factors are represent those things we can modify to improve the length and quality of life for residents.
