BIOL0140 Ecology and Evolution Spring 2008
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The handy cheat sheet to calculating expected counts for use in the Chi–squared test. 1. If the question is of the type “THE NUMBER OF INDIVIDUALS OF THE FOLLOWING GENOTYPES ARE: A1A1=320, A1A2=210, A2A2=100. IS THIS POPULATION IN HARDY-WEINBERG EQUALIBRIUM?”. To determine if observed genotype counts conform to Hardy-Weinberg ratios you use a Goodness of Fit Chi-Squared test. Calculate p and q as p counting alleles method.) Calculate expected genotype counts (not proportions) assuming the population is in the Hardy-Weinberg ratios: p2 : 2pq: q2 [e.g. Number of A1A1 individuals is p2*N, Number A1A2 individuals is 2*p*q*N, etc.] Use the Goodness of Fit Chi-Squared test to determine if observed numbers are different (O E ) 2 from expected numbers. Remember that the test statistic is the sum of for each E cell in your table (i.e. each genotype). For this test df (degrees of freedom) = number of rows–2. The critical value for for =0.05 and df = 1 is 20.05,1 = 3.84.
2 N A1 A1 N A1 A 2 , and the analogous proportion for q. (This is the 2 N total
2. If the question is of the type “THE ALLELE FREQUENCY OF THE A1 ALLELE WAS 0.5 BEFORE AN EL NINO EVENT AND 0.8 AFTER THE EVENT. ARE THE ALLELE FREQUENCIES DIFFERENT? To determine if allele frequency has changed over time you use a Chi-Squared Test of Independence (AKA Contingency Table Chi-Squared Test). Record the allele frequencies in a 2X2 table like this: p 0.5 0.8 q 0.5 0.2
Observed allele frequencies At t=0 At t=1
Calculate the counts of alleles in the population at t=0 and t=1 by multiplying p and q by the number of alleles in the population. Remember that each individual has 2 alleles! For example, if n=200 at both time points, then the table of observations becomes: A1 200 320 520 A2 200 80 280 Row total 400 400 800
Observed number of alleles At t=0 At t=1 Column total
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BIOL0140 Ecology and Evolution Spring 2008
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Calculate expected number of gametes assuming no change in counts by calculating for each cell in the table (row total X column total) / grand total. Repeat for all cells and then use these expected values to set up a Chi–Squared test as described above In this test degrees of freedom (df) = (number of rows –1) X (number of columns –1) =1. If P=0.05 then critical value is 3.84.
3. If the question is of the type ““THE NUMBER OF INDIVIDUALS OF THE FOLLOWING GENOTYPES ARE FOUND BEFORE A DROUGHT A1A1=320, A1A2=210, A2A2=100. AFTER THE DROUGHT THE NUMBERS ARE A1A1=100, A1A2=190, A2A2=50. IS THERE A CHANGE IN GENOTYPE COUNTS?” There are two ways to answer this question: Option 1 is to use the Goodness of Fit Test. Use the counting alleles method to determine pt=0 and qt=0 at t=0. Assume the population is in H-W equilibrium and calculate expected genotype counts for a population of the same size as the t=1 population [e.g. Number of A1A1 individuals is pt=o2*N, Number A1A2 individuals is 2*pt=0*qt=0*N, etc.] Use the Goodness of Fit Chi-Squared test to determine if observed numbers at t=1 are different from expected numbers at t=1. For this test df (degrees of freedom) = number of rows–1 and for p=0.05, df=2 the critical value is 5.99.
Option 2 is to use the Test of Independence. Set up the data into a table like this: Before drought A1A1 320 A1A2 210 A2A2 100 Column total 630 After drought 100 190 50 340 Row total 420 400 150 970
Use the row and column totals to calculate the expected values (as before). Then calculate the Chi-Squared value as before. In this test degrees of freedom (df) = (number of rows –1) X (number of columns –1) =2. If P=0.05 then critical value is 5.99.
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