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!
B. sheet of connective tissue that attaches a muscle to another muscle or bone.
Explanation:
The aponeurosis are mainly made of collagen fibers and works as an insertion to some skeletal muscles. The aponeurosis fibers can connects muscles to the bone or with another muscles connecting the aponeurosis fibers among themselves.
