Handout08ChiSquare

Handout08ChiSquare - expected The expected in this case is...

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BIS101-001/Engebrecht Spring 2010 Handout08ChiSquare A cross is made between AABB and aabb individuals. From the AaBb F1 individuals, the following meiotic products are obtained. Do the data suggest that the A and B loci are linked? AB 280 ab 270 Ab 220 aB 230 1) Calculate the RF: # recombinant products / total = 220 + 230 / 1000 = 450/1000 = .45 2) There are two hypotheses: a. The loci are linked and are approx. 45 mu. apart. b. The loci are assorting independently, and this is just random fluctuation from the expected 1:1:1:1 ratio of meiotic products (50% recombinants). 3) Make a null hypothesis . Make the simplest and most straightforward hypothesis: The loci are not linked. In this case, the null hypothesis is assuming that the numbers represent a random fluctuation from a 1:1:1:1 ratio. 4) Use the Chi-squared formula: Chi-squared = total of (observed-expected ) 2 for all classes.
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Unformatted text preview: expected The expected in this case is a 1:1:1:1 ratio, or 250 for each class Chi-squared = (280-250 ) 2 + (270-250 ) 2 + (220-250 ) 2 + (230-250 ) 2 250 250 250 250 3.6 + 1.6 + 3.6 + 1.6 = 10.4 5) Look up the Chi-squared value in the table (Ch. 4, p.126 in 7 th ; Ch. 2, p.41 in 8 th edition). degrees of freedom (df) = number of classes - 1. Reject the null hypothesis if the chi-square value you calculate is greater than the value given in the p = 0.05 column. For this problem, 3 df, the value of 10.2 is greater than the value for p(0.05) = 7.8 (and lies between the values for p(0.025) and p(0.01) This means that this great a deviation from the expected numbers for independent assortment is only seen 1- 2.5% of the time, rarely. Therefore we reject the hypothesis of independent assortment. The data are thus consistent with linkage (but don't prove it)....
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This note was uploaded on 07/02/2010 for the course BIS 101 taught by Professor Simonchan during the Spring '08 term at UC Davis.

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