studies on a large mixed population, such as a racial group, and use statistical tests to detect the presence of substructure; (2) derive theoretical principles that place bounds on the possible degree of population substructure; and (3) directly sample different groups and compare the observed allele frequencies. The third offers the soundest foundation for assessing population substructure, both for existing loci and for many new types of polymorphisms under development.

In principle, population substructure can be studied with statistical tests to examine deviations from Hardy-Weinberg equilibrium and linkage equilibrium. Such tests are not very useful in practice, however, because their statistical power is extremely low: even large and significant differences between subgroups will produce only slight deviations from Hardy-Weinberg expectations. Thus, the absence of such deviations does not provide powerful evidence of the absence of substructure (although the presence of such deviations provides strong evidence of substructure).

The correct way to detect genetic differentiation among subgroups is to sample the subgroups directly and to compare the frequencies. The following example is extreme and has not been observed in any U.S. population, but it illustrates the difference in power. Suppose that a population consists of two groups with different allele frequencies at a diallelic locus:




Group I



Group II



If there is random mating within the groups, Hardy-Weinberg equilibrium within the groups will produce these genotype frequencies:





Group I




Group II




Suppose that Group I is 90% of the population and Group II is 10%. In the overall population, the observed genotype frequencies will be

AA = (0.9)(0.25) + (0.1)(0.81) = 0.306

Aa = (0.9)(0.50) + (0.1)(0.18) = 0.468

aa = (0.9)(0.25) + (0.1)(0.01) = 0.226

If we were unaware of the population substructure, what would we expect under Hardy-Weinberg equilibrium? The average allele frequencies will be

A = (0.9)(0.5) + (0.1)(0.9) = ).54

a = (0.9)(0.5) + (0.1)(0.1) = 0.46

which would correspond to the Hardy-Weinberg proportions of

The National Academies | 500 Fifth St. N.W. | Washington, D.C. 20001
Copyright © National Academy of Sciences. All rights reserved.
Terms of Use and Privacy Statement