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Analyzing Data

Analyzing Data. c 2 Test….”Chi” Square. Chance Deviation. The outcomes of segregation, independent assortment, and fertilization are subject to random fluctuations from predicted occurrences…  …as a result of chance .

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Analyzing Data

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  1. Analyzing Data c2 Test….”Chi” Square

  2. Chance Deviation • The outcomes of segregation, independent assortment, and fertilization are subject to random fluctuations from predicted occurrences…  …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?

  3. Mendel

  4. Goodness of Fit • Mendel had no way of solving this problem, • Karl Pearson and R.A. Fisher developed the c2 “chi-square” test for this purpose. • The chi-square test is a “goodness of fit” test… …it answers the question of how well do experimental data fit expectations.

  5. Null Hypothesis • null hypothesis: there isno difference between the observed value and the predicted value. The apparent difference can be attributed to chance • alternative hypothesis: there is a 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.

  6. Answer these questions… • what is the total number of offspring?   offspring n = ______  • how many different classes (phenotypes or genotypes) of offspring did you observe?   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

  7. 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

  8. The Formula O: observed E: expected

  9. Observed: Expected:290 purple 300 purple110 white 100 white purple 290 300 10 100 0.33 white 110 100 10 1 100 1.33

  10. 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 (c2 ), 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.

  11. What does the c2value mean? • Critical values for chi-square are found on tables, sorted by degrees of freedom and probability levels... c2 = 1.33

  12. 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 theory about the expected ratio is correct*. * actually, not not correct...does this make it “TRUE”?

  13. What does the c2value mean? 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. p = ~0.2 c2 = 1.33

  14. Null Hypothesis: There is no 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

  15. 1/4 x 1/4 = 1/16 UUDD 1/4 DD 1/2 Dd 1/4 x 1/2 = 1/8 UUDd 1/4 UU 1/4 dd 1/4 x 1/4 = 1/16 UUdd 1/4 DD 1/2 x 1/4 = 1/8 UuDD 1/2 Dd 1/2 x 1/2 = 1/4 UuDd 1/2 Uu 1/4 dd 1/2 x 1/4 = 1/8 Uudd 1/4 x 1/4 = 1/16 uuDD 1/4 DD 1/2 Dd 1/4 x 1/2 = 1/8 uuDd 1/4 uu 1/4 dd 1/4 x 1/4 = 1/16 uudd Forked-Line Method, F2UuDd x UuDd

  16. 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 Genotypes U--D-- If you pick wt F2 worms, what will be the frequency of each class in the F3? Work this out in your notebooks, for Thursday. 1/4 DD 1/4 UU 1/2 Dd 1/4 DD 1/2 Uu 1/2 Dd

  17. Tomorrow • Pick L4, cross progeny, P UUDD xuudd F1 UuDd (L4) • Bioinformatics #2, • WormBase

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

  19. ?

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