Population Genetics. Matthew B. Hamilton
= √fb2 = √0.717 = 0.847. The other allele frequency at that locus is then determined by subtraction fB = 1–0.847 = 0.153. Similarly, for the second locus fa2 = (148 + 103)/502 = 0.50 and fa = √fa2 = √0.50 = 0.707, giving fA = 1–0.707 = 0.293 by subtraction.
For the hypothesis of one locus with three alleles (hypothesis 2), we estimate the frequency of any of the alleles by using the relationship that the three allele frequencies sum to 1. This basic relationship can be reworked to obtain the expected genotype frequency expressions into expressions that allow us to estimate the allele frequencies (see Problem Box 2.2). It turns out that adding together all expected genotype frequency terms for two of the alleles estimates the square of one minus the other allele. For example, (1 − fB)2 = fO2 + fA2 + 2fAfO; and, checking in Table 2.7, this corresponds to (148 + 212)/502 = 0.717. Therefore, 1 − fB = 0.847 and fB = 0.153. Using similar steps, (1 − fA)2 = fO2 + fB2 + 2fBfO = (148 + 103)/502 = 0.50. Therefore, 1 – fA = 0.707 and fA = 0.293. Finally, by subtraction, fO = 1 − fB − fA = 1–0.153 − 0.293 = 0.554.
Problem box 2.2 Proving allele frequencies are obtained from expected genotype frequencies
Can you use algebra to prove that adding together expected genotype frequencies under hypotheses 1 and 2 in Table 2.7 gives the allele frequencies shown in the text? For the genotypes of hypothesis 1, show that f(aa bb) + f(A_ bb) = fbb. For hypothesis 2, show the observed genotype frequencies that can be used to estimate the frequency of the B allele starting off with the relationship fA + fB + fO = 1 and then solving for fB in terms of fA and fO.
Table 2.7 Expected numbers of each of the four blood group genotypes under the hypotheses of 1) two loci with two alleles each, and 2) one locus with three alleles. Estimated allele frequencies are based on a sample of 502 individuals.
| Blood | Observed | Expected number of genotypes | Observed – Expected | (Observed – Expected)2/Expected | |
|---|---|---|---|---|---|
| Hypothesis 1: fA = 0.293, fa = 0.707, fB = 0.153, fb = 0.847 | |||||
| O | 148 | 502(0.707)2(0.847)2 = 180.02 | −32.02 | 5.69 | |
| A | 212 | 502(0.500)(0.847)2 = 180.07 | 31.93 | 5.66 | |
| B | 103 | 502(0.707)2(0.282) = 70.76 | 32.24 | 14.69 | |
| AB | 39 | 502(0.500)(0.282) = 70.78 | −31.78 | 14.27 | |
| Hypothesis 2: fA = 0.293, fB = 0.153, fO = 0.554 | |||||
| O | 148 | 502(0.554)2 = 154.07 | −6.07 | 0.24 | |
| A | 212 | 502[(0.293)2 + 2(0.293)(0.554)] = 206.07 | 5.93 | 0.17 | |
| B | 103 | 502[(0.153)2 + 2(0.153)(0.554)] = 96.85 | 6.15 | 0.39 | |
| AB | 39 | 502[2(0.293)(0.153)] = 45.01 | −6.01 | 0.80 | |
The number of genotypes under each hypothesis can then be found using the expected genotype frequencies in Table 2.6 and the estimated allele frequencies. Table 2.7 gives the calculation for the expected numbers of each genotype under both hypotheses. We can also calculate a chi‐squared value associated with each hypothesis based on the difference between the observed and expected genotype frequencies. For hypothesis 1, χ2 = 40.32, whereas, for hypothesis 2, χ2 = 1.60. Both of these tests have one degree of freedom (4 genotypes −2 for estimated allele frequencies −1 for the test), giving a critical value of χ20.05,1 = 3.84. Clearly, the hypothesis of three alleles at one locus is the better fit to the observed data. Thus, we have just used genotype frequency data sampled from a population with the assumptions of Hardy–Weinberg equilibrium as a means to distinguish between two hypotheses for the genetic basis of blood groups.
Problem box 2.3 Inheritance for corn kernel phenotypes
Corn kernels are individual seeds that display a wide diversity of phenotypes (see Figure 2.10 and Plate 2.10). In a total of 3816 corn seeds, the following phenotypes were observed:
Purple, smooth 2058
Purple, wrinkled 728
Yellow, smooth 769
Yellow, wrinkled 261
Are these genotype frequencies consistent with inheritance due to one locus with three alleles or two loci each with two alleles?
Figure 2.10 Corn cobs demonstrating yellow and purple seeds that are either wrinkled or smooth.
2.5 The fixation index and heterozygosity
The fixation index (F) measures deviation from Hardy–Weinberg expected heterozygote frequencies.
Examples of mating systems and F in wild populations.
Observed and expected heterozygosity.
The mating patterns of actual organisms frequently do not exhibit the random mating assumed by Hardy–Weinberg. In fact, many species exhibit mating systems that create predictable deviations from Hardy–Weinberg expected genotype frequencies. The term assortative mating is used to describe patterns of non‐random mating. Positive assortative mating describes the case when individuals with like genotypes or phenotypes tend to mate. Negative assortative mating (also called disassortative mating) occurs when individuals with unlike genotypes or phenotypes tend to mate. Both of these general types of non‐random mating will impact expected genotype frequencies in a population. This section describes the impacts of non‐random mating on genotype frequencies and