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Contents

Population Genetics

Using the Hardy-Weinberg Law

Problems

Problems

As we saw in the previous section, a population must meet many conditions before it can reach Hardy-Weinberg equilibrium. Natural populations rarely meet all of these conditions. Large populations rarely occur in isolation, all populations experience some degree of random mutation, mating is seldom random, but rather is the result of careful selection of mates. Most importantly, selective pressures favor individuals whose alleles give them the greatest fitness, so survival and reproductive success are never random. Because of these factors inherent in natural selection, allelic frequencies do not remain constant and evolution occurs.

So why study the Hardy-Weinberg Law? The answer is twofold. First, the law proves that natural selection is necessary for evolution to occur. Darwin's theory of evolution states that for evolution to occur, populations must be variable, there must be inheritance between generations, and natural selection must make survival and reproductive success non-random. The conditions set up by the Hardy-Weinberg Law allow for variability (the existence of different alleles) and inheritance, but they eliminate natural selection. The fact that no evolution occurs in a population meeting these conditions proves that evolution can only occur through natural selection.

Second, and just as important, the Hardy-Weinberg Law allows us to estimate the effect of selection pressures by measuring the difference between actual and expected allelic frequencies or phenotypes. In order to make such measurements, we must first write a Hardy-Weinberg equation for the frequencies of alleles in the population.

Hardy-Weinberg Equations

Let us start with a given trait that has two alleles T and t. To find the frequency of allele T(p) in the population, we would use the following formula:

p = (# of copies of T in the population) / (total # of copies of all alleles in the population); or p = (2(# of individuals homozygous for T) + (# of individuals heterozygous for T)) / 2(# of individuals in the population). We can find the frequency of allele t (q) in the same way.

Once we have the values of p and q, we can set up an equation based on the Hardy-Weinberg equilibrium. If the frequency of each allele is constant, then the following equation, called a Hardy-Weinberg Equation, is always true:

(p + q)2 = 1 ; or p 2 +2pq + q 2 = 1

The second form of the above equation is the most useful, since each of its factors represents a portion of the population. P 2 and q 2 represent those individuals homozygous for T and t, respectively, while 2pq represents all heterozygous individuals. Using these equations, we can determine the frequency of a given phenotype given the frequency of the two alleles.

Let's look at a sample system. In a population of pea plants, the gene for flower color has two alleles, F and f. Individuals with two F alleles have white flowers, those with two f alleles have purple flowers, and heterozygotes (having one F and one f allele) have red flowers. A scientist calculates the allele frequencies in one generation of the population. The frequency of F is p = 0.9 and the frequency of f is q = 0.1. Based on these findings, we know that if the system is at Hardy-Weinberg equilibrium the frequency of heterozygotes in the next generation should be 2pq or 2(0.9)(0.1) or 0.18. This means that 18% of the population should have red flowers. If we then allow a selection pressure, such as pollinator preference for flower color, to enter the system, we will see that the percentage of the population that has red flowers is not 18%. It has changed due to selection pressure. The effect of this selection pressure can be quantified by comparing the new frequency of red flowers with the expected 18%.

Hardy-Weinberg equations can be written for traits with multiple alleles in the same way we just wrote one for two alleles. For example, the Hardy-Weinberg equation for a trait with three alleles whose frequencies were p, q, and r would be (p + q + r)2 = 1 .

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