HARDY-WEINBERG EQUILIBRIUM 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: 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. Outcome Toss 1 Toss 2 Toss 3 Shorthand Probabilities 1 Head HHH 1/8 2 Tail HHT 3 HTH 4 THH 5 TTH 6 THT 7 HTT 8 TTT

Total # of sample points = 230 = 1,073,741,824 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… Great example!

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 Number of trials are independent, finite, and conducted under the same conditions. There are only two types of outcome.(Ex. success and failure). Outcomes are mutually exclusive and independent. Probability of getting a success in a single trial remains constant throughout all the trials. Probability of getting a failure in a single trial remains constant throughout all the trials. Number of success are finite and a non-negative integer (0,n)

Properties of Binomial Distribution Mean or expected # of successes in n trials: Variance of y: Standard deviation of y, σ (y) = (nsf)1/2 E(y) = ns V(y) = nsf SD(y) = 𝑽(𝒚)

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 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) Problem 1: A nuclear allozyme locus has three alleles, A1 and A2, and A3, with frequencies 0.847, 0.133, and 0.020, respectively. If we sample 30 diploid individuals, what is the probability of: a) Not finding any copies of A2? b) Finding at least one copy of A2? c) 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: Emphasize flipping of the inequality when dividing by negative number. 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) Problem 2: The frequency of red-green color-blindness is 0.07 for men and 0.005 for women. You are designing a survey to determine the effect of color blindness on educational success. How many males and females would you have to sample to ensure that the probability including at least one color blind individual of each sex would be 0.90 or greater?

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 Standard deviation: variability of individuals around the sample mean Standard error: variability of sample means around the population mean We expect ~68% of the data to fall within 1 standard deviation of the mean.

A1 A11 A12 … A1j A2 A21 A22 … A2j … Ai Ai1 Ai2 …Aij SUM 1 𝑗 𝐴1𝑗 Genotype1 Genotype2 …Genotype j SUM A1 A11 A12 … A1j 1 𝑗 𝐴1𝑗 A2 A21 A22 … A2j 1 𝑗 𝐴2𝑗 … Ai Ai1 Ai2 …Aij 1 𝑗 𝐴𝑖𝑗 Aij ≥𝟎, 𝒊,𝒋=𝟏,𝟐,𝟑,…

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 If p = 0.57 -- what is q?

q = 1 – p = 0.43 What do you notice about the standard error of q?

Problem 3 (10 minutes) 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

Estimation of allele frequency for dominant locus For dominant loci, we only know the genotype of homozygous recessive individuals (in absence of sequence data). (Note: this is an estimate) Where, q = Frequency of Allele A2 N22 = # of individuals with genotype A2A2 N = total # of diploid individuals = N11+N12+N22

Problem 4 (15 minutes) Go to the “Genetics Home Reference” website (http://ghr.nlm.nih.gov) and use the search feature to find a condition caused by a dominant allele in humans. On the main description page, find the frequency of the condition in human populations. Assuming HWE and Mendelian inheritance of the disease, what is the frequency of the recessive allele in this population? What is the standard error of this estimate? How many affected children would you expect in the next generation? (Global/U.S/etc) What are the assumptions of these estimates? Eg. http://ghr.nlm.nih.gov/condition/cornelia-de-lange-syndrome

Hypothesis Testing Hypothesis: Tentative statement for a scientific problem, that can be tested by further investigations. Null Hypothesis(H0): There is no significant difference in observed and expected values. Alternate Hypothesis(H1): There is a significant difference in observed and expected values. Example: H0 = 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. Ho is TRUE Ho is FALSE Accept Ho Fail to reject Ho β 1-α Type II Error α 1-β Reject Ho Type I Error

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. H0= 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) “disturbing factors” is a bit of strange phrasing. I don’t think that part is needed. You just list the assumptions of the model separately.

HWE Assumptions Random mating No selection Equal numbers of offspring per parent All progeny equally fit No mutation Single, very large population No migration

χ2 - test Where,

Is this population in Hardy-Weinberg equilibrium ? 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 ?

χ2 Genotype Obs. #(O) Exp. #(E) (O-E) (O-E)^2 (O-E)^2/E A1A1 5000 4225 Expected frequency under HWE Expected number 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 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). Explain how to determine df

Problem 5 (10 minute) Based on the observed genotype counts in Problem 3, test whether the population that had been sampled is in HWE. Critical Chi square values are given in the table to the right. Think carefully about which one you should use (Hint: How many parameters are estimated from the data when the allele frequencies of 3 alleles are estimated?). What are some possible explanations for the observed results?   d.f. Critical value of χ2 1 3.8415 2 5.9915 3 7.8147 4 9.4877 5 11.0705 6 12.5916