Info

(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.

The number of genotypes under each hypothesis can then be found by 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 %2 = 43.52, whereas X2 = 1.60 for hypothesis 2. 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 x0 os1 = 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.

0 0

Post a comment