Probability

1. Probability • Dealing with events with mutually exclusive outcomes • Tossing a coin: 2 outcomes - heads or tails • Probability of getting tails: one way to get tails out of 2 possible outcomes • P = 1/2

2. P PP Pp p Pp pp Probability • Applies to Mendelian situations also • Crossing two heterozygous purple pea plants: • 2 possible outcomes: purple or white • probability of purple = 3/4 • probability of white = 1/4 P p

3. Probability • Two basic rules • Sum rule • applies when looking for eitherof two or more outcomes of one event • Product rule • applies when looking at more than one event

4. Sumrule • Probability of rolling a 1 or 2 with a die • 1/6 chance of rolling a 1 • 1/6 chance of rolling a 2 • 1/6 + 1/6 = 1/3 chance of either

5. Product rule • Looking for specific outcomes of more than one event • Clue word: “and” • Probability of tossing a coin twice and getting heads both times • 1/2 chance of heads on first toss • 1/2 chance of heads on second toss • (1/2)(1/2) = 1/4 chance of two heads

6. Phenylketonuria (PKU) • Disease caused by recessive allele • locus on chromosome 12  • P produces enzyme: phenylalanine hydroxylase • p defective; no enzyme • No enzyme; no metabolism of phenylalanine • build-up of neurotoxins • severe mental retardation

7. PKU • PP = normal • Pp = normal, but carrier • pp = no enzyme: PKU • 1/8000 U.S. (white); 1/50,000 U.S. (black); 1/18,000 S. Eur.; 1/4000 Ireland/Scotland • Blood test developed in 1950’s • 1st test for a genetic disease • Symptoms totally preventable w/ diet!

8. Product rule applied to Mendel: • A couple who are both heterozygous for PKU are expecting their first child. What is the probability the child will be a boy with PKU? • Pp X Pp ----> 3/4 normal: 1/4 PKU • 1/2 chance boy; 1/2 chance girl • (1/4)(1/2) = 1/8

9. Product rule applied to Mendel: • A couple who are both heterozygous for PKU are expecting their first child. What is the probability the child will be a boy without PKU?

10. Sum and Product Rules • Sum rule: situations where you are given more options. P should be increasing. • ADD FRACTIONS • Product rule: situations where you have more restrictions. P should be decreasing • MULTIPLY FRACTIONS

11. Sum and Product Rules • Two rules together: • P of a couple having two children; one boy and one girl • two possible ways: • first child boy; second child girl • first child girl; second child boy

12. Sum and Product Rules • P of boy then girl: 1/2 x 1/2 = 1/4 • P of girl then boy: 1/2 x 1/2 = 1/4 • Only one of these two cases can occur, so we add the probabilities! • 1/4 + 1/4 = 1/2

13. Sum and Product Rules • A couple who are both heterozygous for PKU plan on having 2 children. What is the probability only one will have PKU? • Two ways: • 1st child PKU; 2nd no PKU • 1st child no PKU; 2nd PKU • (1/4)(3/4) + (3/4)(1/4) = 6/16 = 3/8

14. The Binomial Expansion • Any easy way to combine the sum and product rules • Use when you have unordered outcomes of more than one event

15. The Binomial Expansion n! P = -------- ps qt s! t! n = number of events s = outcome 1 t = outcome 2 p = prob of outcome 1 q = prob of outcome 2

16. The Binomial Expansion... n! P = ---------- ps qt ru s! t! u! n = number of events s = outcome 1; t = o. 2; u = o. 3 p = prob of outcome 1 q = prob of outcome 2 r = prob of outcome 3

17. The Binomial Expansion A couple who are both heterozygous for PKU plan on having 5 children. What is the probability only one will have PKU? 5! P = ------- (3/4)4 (1/4)1 4! 1! P = 0.40

18. The Binomial Expansion A woman with attached earlobes marries a man with free earlobes whose mother had attached earlobes. They plan on having 3 children. What is the probability they will have two boys with free earlobes and one girl with attached earlobes? A a aa x Aa a Aa aa

19. The Binomial Expansion • 1/2 chance of having attached earlobes • 1/2 chance of having free earlobes • 1/2 chance of having a boy • 1/2 chance of having a girl

20. A woman with attached earlobes marries a man with free earlobes whose mother had attached earlobes. They plan on having 4 children. What is the probability they will have two boys with free earlobes, one girl with free earlobes and one girl with attached earlobes?

21. A man and woman are both heterozygous for cleft chin (Cc). One has dimples (Dd) and the other does not (dd). They plan to have five children. What is the probability that only 3 will have dimples but no cleft chin?

22. Statistics: The Chi Square Test • Tests for goodness of fit • i.e. do your results match what you would expect based on your hypothesis of mode of inheritance? • If you expected a 3:1 ratio, is that what you have?

23. Tomatoes P: red x yellow F1: red F1 x F1 F2: 3665 red: 1139 yellow

24. Chi Square Test Is this a 3:1 ratio? Is this a simple case of one gene with two alleles; red dominant and yellow recessive? 3665:1139 = 3:1?

25. Chi Square Test 1. Convert expected ratios to expected NUMBERS Multiply TOTAL by expected fractions: total tomatoes observed: 4804 3/4 x 4804 = 3603 1/4 x 4804 = 1201

26. Chi Square Test red yellow Observed numbers: 3665 1139 Expected numbers: 3603 1201 Are the observed numbers close enough to the expecteds?

27. Chi Square Test (observed - expected)2 c2 = S ----------------------------------- expected

28. Chi Square Test red yellow obs: 3665 1139 exp: 3603 1201 (3665 - 3603)2 (1139 - 1201)2 c2 = ------------------- + ------------------- 3603 1201 c2 = 4.27

29. Chi Square Test The c2 value is a measure of the deviation between your observed numbers and the expected numbers. The larger the c2 value, the greater the deviation. The larger the deviation, the less chance your results match the expecteds!

30. Chi Square Test How much deviation is too much? You must consult a c2 table. First we need to know one more thing: the degrees of freedom

31. Chi Square Test Degrees of freedom Number of other options possible once an organism is classified as one phenotype. d.f. = number of phenotypic classes - 1

32. Chi Square Test Degrees of freedom: phenotypic classes: red yellow d.f. = 1 If you have 8 phenotypic classes, there are 7 degrees of freedom, etc...

33. Chi Square Test Our c2 = 4.27 d.f. = 1 Probability _______________________________________ d.f. .95 .50 .20 .05 .01 1 0.004 0.46 1.64 3.84* 6.64 2 0.10 1.39 3.22 5.99 9.21 3 0.35 2.37 4.64 7.82 11.35 4 0.71 3.36 5.99 9.49 13.28 5 1.15 4.35 7.29 11.07 15.09

34. Chi Square Test Our deviation has a less than 5% chance of occurring due to random causes. There is a statistically significant difference between our observed and expected numbers. We must reject this as a good 3:1 ratio.

35. Chi Square Test Drosophila star x dumpy ---> wild type F1 F1 x F1 ---> 575 wild type 180 dumpy 197 star 48 star, dumpy What is the mode of inheritance of star and dumpy?

36. If star and dumpy are both recessive, we would expect a 9:3:3:1 ratio in the F2. Test for goodness of fit: • wild type • 180 dumpy • 197 star • 48 star, dumpy Total: 1000 Expecteds:

37. Obs.: 575 180 197 48 Exp.: (575 - )2 (180 - )2 (197 - )2 (48 - )2 c2 = ----------- + ----------- + -----------+ -------- c2 = d.f. =

38. Chi square test summary: • Find expected ratios. (from Punnett squares, etc.) • Find expected numbers. • Use c2 formula to find deviation. • Find degrees of freedom. • Consult table to find probability.