VIEWS: 4 PAGES: 30 POSTED ON: 7/7/2011
Population Genetics Lab 2 BINOMIAL PROBABILITY & HARDY-WEINBERG EQUILIBRIUM Last Week : Sample Point Methods: Example: Use the Sample Point Method to find the probability of getting exactly two heads in three tosses of a balanced coin. 1. The sample space of this experiment is: Outcome Toss 1 Toss 2 Toss 3 Shorthand Probabilities 1 Head Head Head HHH 1/8 2 Head Head Tail HHT 1/8 3 Head Tail Head HTH 1/8 4 Tail Head Head THH 1/8 5 Tail Tail Head TTH 1/8 6 Tail Head Tail THT 1/8 7 Head Tail Tail HTT 1/8 8 Tail Tail Tail TTT 1/8 2. Assuming that the coin is fair, each of these 8 outcomes has a probability of 1/8. 3. The probability of getting two heads is the sum of the probabilities of outcomes 2, 3, and 4 (HHT, HTH, and THH), or 1/8 + 1/8 + 1/8 = 3/8 = 0.375. Sample- point method : Example: Find the probability of getting exactly 10 heads in 30 tosses of a balanced coin. Total # of sample points = 230 = 1,073,741,824 Need a way of accounting for all the possibilities Example: In drawing 3 M&Ms from an unlimited M&M bowl that is always 60% red and 40% green, what is the P(2 green)? If one green M&M is just as good as another… Binomial Probability Distribution Where, n = Total # of trials. y = Total # of successes. s = probability of getting success in a single trial. f = probability of getting failure in a single trial (f = 1-s). Assumptions of Binomial Distribution 1. # of trials are independent, finite, and conducted under the same conditions. 2. There are only two types of outcome.(Ex. success and failure). 3. Outcomes are mutually exclusive and independent. 4. Probability of getting a success in a single trial remains constant throughout all the trials. 5. Probability of getting a failure in a single trial remains constant throughout all the trials. 6. # of success are finite and a non-negative integer (0,n) Properties of Binomial Distribution Mean or expected # of successes in n trials, E(y) = ns Variance of y, V(y) = nsf Standard deviation of y, σ (y) = (nsf)1/2 Example: Find the probability of getting exactly 10 heads in 30 tosses of a balanced coin. Solution: We know, n = 30 y = 10 s = 0.5 f = 0.5 Example: Find the expected # of heads in 30 tosses of a balanced coin. Also calculate variance. Solution: E(Y) = ns = 30*0.5 = 15 V(Y) = nsf = 30*0.5*0.5 = 7.5 Problem 1 (10 minutes)(2 points) An allozyme locus has three alleles, A1,A2, and A3 with frequencies 0.947, 0.033, and 0.020, respectively. If we sample 30 diploid individuals, what is the probability of: •Not finding any copies of A2? •Finding at least one copy of A2? •GRADUATE STUDENTS ONLY: Finding fewer than 2 copies of A2? Example: How many diploid individuals should be sampled to detect at least one copy of allele A2 from Problem 1 with probability of at least 0.95? Solutions: Thus, to detect at least one copy of allele A2 with probability of 0.95, one would need to sample at least 90 alleles (i.e., at least 45 diploid individuals). Problem 2 (15 minutes)(2 points) The frequency of red-green color-blindness is 0.07 for men and 0.005 for women. Cities are considering implementing new traffic lights that can be distinguished by people with this disability. However, the investment is large, so President Obama has asked you to calculate the minimum size of a city that would have a probability of 0.9 of having at least 10 men who are red-green color-blind. Assume equal population sizes for men and women. Full credit (2 pts): Solve for at least 1 man +1 Bonus: Set up the equation for at least 10 men +3 Bonus: Solve for at least 10 men Estimation of allele frequency for Co-dominant locus Where, p = Frequency of allele A1 q = Frequency of Allele A2 N11 = # of individuals with genotype A1A1 N12 = # of individuals with genotype A1A2 N22 = # of individuals with genotype A2A2 N = total # of diploid individuals =N11+N12+N22 Estimation of Standard Error Where, p = Frequency of allele A1 q = Frequency of Allele A2 SEp = Standard error for frequency of allele A1 SEq = Standard error for frequency of allele A2 N = total # of diploid individuals =N11+N12+N22 Standard Deviation v. Standard Error We expect ~68% of the data to fall within 1 standard deviation of the mean. Standard error is the standard deviation of the mean. We expect sample means to fall within 1 standard error of the true mean ~68% of the time. Example: What are the allele frequencies of alleles A1 and A2, if the following genotypes have been observed in a sample of 50 diploid individuals? Genotype Count A1A1 17 A1A2 23 A2A2 10 Solution: N11 = 17, N12 = 23, and N22 = 10 q = 1 – p = 0.43 Problem 3 (10 minutes) (2 pts) Estimate the allele frequencies (include their respective standard errors) for alleles A1, A2, and A3 if the following genotypes have been observed in a sample of 200 individuals Genotype Count A1A1 19 A2A2 17 A3A3 14 A1A2 52 A1A3 57 A2A3 41 Problem 4 (Time 10 min.)(2 pts) Tay Sachs disease is an autosomal recessive genetic disorder causing the death of nerve cells in the brain due to the steady accumulation of gangliosides. Extensive genotyping has determined that approximately 1 in 30 of the 5 million Ashkenazi Jews within the United States is a carrier. a) Assuming HWE and Mendelian inheritance of the disease, what is the frequency of the recessive allele in this population? b) What is the SE of this estimate? (Assume 1,000 people were sampled) c) How many affected children would you expect to be born in this population? d) What are the assumptions of these estimates? Hypothesis Testing Hypothesis: Tentative statement for a scientific problem, that can be tested by further investigations. 1.Null Hypothesis(Ho): There is no significant difference in observed and expected values. 2.Alternate Hypothesis(H1): There is a significant difference in observed and expected values. Example: Ho = Fertilized and unfertilized crops have equal yields H1 = Fertilized and unfertilized crops do not have equal yields Remember: In final conclusion after the experiment ,we either – "Reject H0 in favor of H1" Or “Fail to reject H0”, Type I error: Error due to rejection of a null hypothesis, when it is actually true (False positive). Level of significance(LOS) (α) : Maximum probability allowed for committing “type I error”. At 5 % LOS (α=0.05), we accept that if we were to repeat the experiment many times, we would falsely reject the null hypothesis 5% of the time. P- value: Probability of committing type I error If P-value is smaller than a particular value of α, then result is significant at that level of significance Testing departure from HWE In a randomly mating population, allele and genotype frequencies remain constant from generation to generation. Ho= There is no significant difference between observed and expected genotype frequencies (i.e. Population is in HWE) H1= There is a significant difference between observed and expected genotype frequencies (i.e. Population is not in HWE) HWE Assumptions 1. Random mating 2. No selection a. Equal numbers of offspring per parent b. All progeny equally fit 3. No mutation 4. Single, very large population 5. No migration χ2 - test Where, Example: A population of Mountain Laurel at Cooper’s Rock State Forest has the following observed genotype counts: Genotype Observed number A1A1 5000 A1A2 3000 A2A2 2000 Is this population in Hardy-Weinberg equilibrium ? Genotype Expected frequency Expected number under HWE under HWE A1A1 p2 = 0.652 = 0.4225 0.4225 10000 = 4225 A1A2 2pq = 0.455 0.455 10000 = 4550 A2A2 q2 = 0.1225 0.1225 10000 = 1225 Genotype Obs. #(O) Exp. #(E) (O-E) (O-E)^2 (O-E)^2/E A1A1 5000 4225 775 600625 142.1598 A1A2 3000 4550 -1550 2402500 528.022 A2A2 2000 1225 775 600625 490.3061 χ2 1160.488 The critical value (Table value) of χ2 at 1 df and at α=0.05 is approx. 3.84. Conclusion: Because the calculated value of χ2 (1160.49) is greater than the critical value (3.84), we reject the null hypothesis and accept the alternative (Not in HWE). Problem 5 (Time 10 min) (2 pts) Based on the observed genotype counts in problem 3, test whether the population that had been sampled is in HWE. What are some possible explanations for the observed results? Genotype Count A1A1 19 A2A2 17 A3A3 14 A1A2 52 A1A3 57 A2A3 41