Consider a collection of four populations of humans. These populations are each mating at random locally so each population is at HWE. At the ADH locus (alcohol dehydrogenase) there are two alleles &). The frequencies of each allele in each population are given below. + Allele 0.98 0.04 0.69 0.00 Population 1 2 3 4 -Allele 0.02 0.96 0.31 1.00 What is the observed heterozygosity (Hs) across the four populations?

Frequency of heterozygotes :

P1 = 2 x 0.98 x 0.02 = 0.0392

P2 = 2 x 0.04 x 0.96 = 0.0768

P3 = 2 x 0.69 x 0.31 = 0.4278

P4 = 2 x 0.00 x 1 = 0.00

Average of heterozygotes (Hs) = (0.0392+0.0768+0.4278) / 4 = 0.1359

Average allele frequency of the  + allele = (0.98+0.04+0.69) / 4 = 0.4275

Average allele frequency of the - allele = (0.02+0.96+0.31+1) / 4 = 0.5725

1. a tropical dry forest carries out more rapid nutrient cycling