Question

Chi square problem need help please show work of how they solved for the chi square value, the correct answer is in red.

**On this page and the next two, report all* calculated non-integer values to the nearest 0.01** 18. x2. As you learned in one or more of your introductory biology courses, the frequencies of the genotypes in a population at "Hardy-Weinberg" equilibrium can be predicted from the frequencies of the individual alleles "A" and "a' at the locus of interest. If "p" represents the frequency of the "A" allele and "q" represents the frequency of the "a" allele, when this occurs the frequency of the genotypes "AA", "Aa", and "aa" would be p2, 2pq and q2. Consider the diploid population shown below. The number of individuals with each of the three possible genotypes is indicated. Genotype #indiv. Aa 220 69 Conduct a X2 goodness-of-fit test assuming Mendelian segregation with these values to determine if the population appears to be at "Hardy-Weinberg" equilibrium eale 5.04 +0.05 (a, 4 pts) What is the X2 value you obtain? (b, 1 pts) How many degrees of freedom does your test have?

Answer 1

a.������ Here we want to test H_0: p = 1/2 then under H_0, P {AA} = 1/2^2 = p^2 = 1/4, since p = 1/2, under H_0 P{Aa} = spq = 21/4 = 1/2, Under H_0: p = 1/2 and q = 1 - p P{aa} = q^2 = 1/2^2 = 1/4, Under H_0: p = 1/2 and q = 1 - p Therefore Genotype the expected frequencies: AA np{AA} = (111 + 220 + 69)/4 = 100 since n = 400 Aa np{Aa} = (111 + 220 + 69)/2 = 200 aa np{aa} = (111 + 220 + 69)/4 = 100 X^2 = f^2_AA/np{AA} + f^2_Aa/np{Aa} + f^2_aa/np{aa} - n = ((111)^2/100 + (220)^2/200 + (69)^2/100) - 400 = 12.82 b. The degrees of freedom = 3 - 1 = 2