1. Analyzing Data  2 Test….”Chi” Square

2. Forked-Line Method, F2 UuDd x UuDd 1/4 UU 1/2 Uu 1/4 uu 1/4 DD 1/2 Dd 1/4 dd 1/4 DD 1/2 Dd 1/4 dd 1/4 DD 1/2 Dd 1/4 dd 1/4 x 1/4 = 1/16 UUDD 1/4 x 1/2 = 1/8 UUDd 1/4 x 1/4 = 1/16 UUdd 1/2 x 1/4 = 1/8 UuDD 1/2 x 1/2 = 1/4 UuDd 1/2 x 1/4 = 1/8 Uudd 1/4 x 1/4 = 1/16 uuDD 1/4 x 1/2 = 1/8 uuDd 1/4 x 1/4 = 1/16 uudd

3. Genotypes U--D-- 1/4 UU 1/2 Uu 1/4 DD 1/2 Dd 1/4 DD 1/2 Dd 1/4 x 1/4 = 1/16 UUDD 1/4 x 1/2 = 1/8 UUDd 1/2 x 1/4 = 1/8 UuDD 1/2 x 1/2 = 1/4 UuDd If you pick wt F2 worms, what will be the proportion of each class in the F3? Work this out in your notebooks, for Thursday.

4. Chance Deviation  The outcomes of segregation, independent assortment, and fertilization are subject to random fluctuations from predicted occurrences…  The outcomes of segregation, independent assortment, and fertilization are subject to random fluctuations from predicted occurrences… …as a result of chance. …as a result of chance.  As the sample size increases, the average deviation from the expected fraction or ratio is expected to decrease. Experimental Data Does it fit Expected Values? Does it fit Expected Values?

5. Mendel

6. Goodness of Fit Mendel had no way of solving this problem, Mendel had no way of solving this problem, Karl Pearson and R.A. Fisher developed the  2 “chi- square” test for this type of application. Karl Pearson and R.A. Fisher developed the  2 “chi- square” test for this type of application. The chi-square test is a “goodness of fit” test… The chi-square test is a “goodness of fit” test… …it answers the question of how well do experimental data fit expectations.

7. Null Hypothesis null hypothesis: there is no statistical difference between the observed value and the predicted value. The apparent difference can be attributed to chance null hypothesis: there is no statistical difference between the observed value and the predicted value. The apparent difference can be attributed to chance alternative hypothesis : there is a statistical difference between the observed value and the predicted value. The apparent difference can not be attributed to chance alone. alternative hypothesis : there is a statistical difference between the observed value and the predicted value. The apparent difference can not be attributed to chance alone. Important: accepting or rejecting the null hypothesis does not prove anything, it only provides a statistical means of measuring the difference between the observed and expected results. Important: accepting or rejecting the null hypothesis does not prove anything, it only provides a statistical means of measuring the difference between the observed and expected results.

8. Answer these questions…  what is the total number of offspring?  what is the total number of offspring? offspring n = ______ offspring n = ______  how many different classes (phenotypes or genotypes) of offspring did you observe?  how many different classes (phenotypes or genotypes) of offspring did you observe? classes n = _______ classes n = _______  what are the classes, and how many were observed for each class?  based on genotype classes, how many of each phenotype do you expect in each class in each... 290 purple; 110 white flowers Offspring n = 400 Classes n = 2 (purple, white) Observed: 290 purple 110 white Expected: 300 purple 100 white

9. Null Hypothesis  The observed 290 purple flowers and 110 white flowers are statistically no different than the expected values of 300 purple flowers and 100 white flowers.  The apparent difference is due to chance. 290 purple; 110 white flowers Offspring n = 400 Classes n = 2 (purple, white) Observed: 290 purple 110 white Expected: 300 purple 100 white

10. The Formula O: observedE: expected

11. Observed:Expected: 290 purple300 purple 110 white100 white purple white 290 110 300 10010 100 0.33 1 1.33

12. Degree of Freedom  A critical factor in using the chi-square test is the “degrees of freedom”, which is essentially the number of independent random variables involved,  For this analysis (  2 ), the degrees of freedom is simply the number of classes, minus 1.  For our example, there are 2 classes of offspring: purple and white. Thus, degrees of freedom…  (d.f.) = 2 -1 = 1.

13. What does the  2 value mean? Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels... Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels... 22 = 1.33

14. Null Hypothesis  We will use p = 0.05 as the critical value,  if p < 0.05, you “reject the null hypothesis”  if p > 0.05, you “fail to reject the null hypothesis” …that is, you accept that your genetic hypothesis about the expected ratio is correct*. * actually, not not correct...does this make it “TRUE”?

15. What does the  2 value mean? 22 = 1.33 p = ~0.2 One way to think of this, is that if you did this experiment 100 times, you’d get similar results, do to chance, roughly 20 times.

16. Null Hypothesis: There is no statistical difference between the measured values, and the expected values (with details). Chi Square Analysis: “ What is the probability that the null hypothesis is correct?” Reject Null Hypothesis: p = < 0.05

17. Wednesday 1.Pick L4, cross progeny, PUUDD x uudd F1 UuDd (L4) 2.Bioinformatics #2, 1.WormBase WormBase TO DO before Friday! TO DO before Friday! Look up phenotype info. Look up phenotype info.

18. Friday Wormbase Entrez Determine phenotype(s) for each mutant (assignment, due at start of class Friday), and then ID worms in class (on Friday).

19. ?