In a population of heffalumps, there is a single gene that controls the presence of hair. The H allele, which is a complete dominant, makes a heffalump hairy, whereas the h allele codes for the absence of hair. If 36% of the heffalump population is hairless, what is the frequency of the h allele?

Explanation:

Hardy Weinberg Equilibrium is used to calculate the allelic as well as genotypic frequency

Allelic frequency of dominant and recessive allele is represented by p and q respectively whereas genotypic frequency of dominant genotype is represented by and respectively

Given:

H allele (p) = hairy heffalump (dominant)

h allele (q) = hairless heffalump (recessive)

36% of heffalump population is hairless represents the % of recessive genotype, hh () =36%

Calculation of frequency of the h allele (q) :

Frequency of genotype hh () will be: 36/100=0.36 or 0.6*0.6

Frequency of h allele (q) will be 0.6

explanation:

for  example, forests have higher evapotranspiration than other ecosystems,  such  as grasslands, because of their deeper roots and greater leaf area. thus forests have a net moistening effect on the atmosphere and become a moisture source for downwind ecosystems.