Class02-ClassSlides-Spring 2011-03102011

# Class02-ClassSlides- - Genetic Analysis II(385 Population Genetics Lecture 02 Date Instructor Professor Derek Gordon Office Life Sciences Building

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1 Genetic Analysis II (385) – Population Genetics Lecture 02 Date: 03/10/11 Instructor: Professor Derek Gordon Office: Life Sciences Building, Room 128 Phone: 445-3386 Phone: 445 3386 E-mail: [email protected] Office Hours: Tues, Thurs: 12:00pm – 1:15pm, and by appt. HOW TO RECOGNIZE A POPULATION IN HWE In Genetics you have to deal with real populations. These may or may not be in or near HWE. may or may not be in or near HWE. Consider a population with three genotypes A 1 A 1, A 1 A 2, & A 2 A 2 and the genotype array of X + Y + Z. 3/10/2011 PopGen 02- 2011 2

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2 1. First calculate the values of p and q . p = X + 1/2Y q = Z + 1/2Y 2. Next calculate the e xpected genotype frequencies: p 2 (A 1 A 1 ) + 2pq(A 1 A 2 ) + q 2 (A 2 A 2 ). 3. Calculate the expected number of each genotype by multiplying the expected genotype frequencies by the total number 3/10/2011 PopGen 02- 2011 3 number. 4. Compare the expected genotype array under HWE to the observed genotype array X + Y + Z . •In natural populations you would use a Chi-square test for goodness-of-fit. •If you are given a frequency distribution assume an infinitely large population. (Must have an exact fit) 5. If the observed and expected distributions are statistically the same then you can assume the population is in 3/10/2011 PopGen 02- 2011 4 , then you can assume the population is in HWE. *
3 Are the following in HWE? Observed Distribution or Freq. pq Expected Distribution or Freq. 2 HWE A 1 A 1 A 1 A 2 A 2 A 2 a. .60 .20 .20 b. 533 734 233 c. 1000 6000 3000 3/10/2011 PopGen 02- 2011 5 d. 2050 3900 2050 2 With 1 df, Critical value is 3.84 at .05 level of significance. I'll step through the first one for you X+ Y + Z = .6 + .2 + .2 = 1 p = X + 1/2(Y) = .6 + 1/2(.2) = .7 q = Z + 1/2(Y) = .2 + 1/2(.2) = .3 In HWE, X + Y + Z = (p+q) 2 3/10/2011 PopGen 02- 2011 6 (p+q) 2 = (.7+ .3) 2 = .49 + .42 + .09 Evaluate: .49 + .42 + .09 does not equal .6 + .2 + .2. Not in HWE

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4 EXAMPLES: Observed Distribution or Freq. pq Expected Distribution or Freq. 2 HWE A 1 A 1 A 1 A 2 A 2 A 2 A 1 A 1 A 1 A 2 A 2 A 2 a. .60 .20 .20 .70 .30 .49 .42 .09 NO b. 533 734 233 c 1000 6000 3000 3/10/2011 PopGen 02- 2011 7 c. d. 2050 3900 2050 2 With 1 df, Critical value is 3.84 at .05 level of significance. Example 2 533 + 734 + 233 = 1500 p = [533 + 1/2(734)]/1500 = 900/1500 = .6 q = [233 + 1/2(734)]/1500 = 600/1500 = .4 At HWE 3/10/2011 PopGen 02- 2011 8 [.6 2 + (2*.6*.4) + .4 2 ] * 1500 = [.36 + .48 +. 16] * 1500 = 540 + 720 + 240 Expected distribution
5 Example 2 533 734 233 Observed distribution Expected distribution Obs-Exp (Obs-Exp) 2 (Obs-Exp) 2 /Exp 3/10/2011 PopGen 02- 2011 9 Example 2 533 734 233 Observed distribution 540 720 240 Expected distribution 3/10/2011 PopGen 02- 2011 10

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6 Example 2 533 734 233 Observed distribution 540 720 240 Expected distribution -7 14 -7 Obs-Exp 3/10/2011 PopGen 02- 2011 11 Example 2 533 734 233 Observed distribution 540 720 240 Expected distribution -7 14 -7 Obs-Exp 49 196 49 (Obs-Exp) 2 3/10/2011 PopGen 02- 2011 12
7 Example 2 533 734 233 Observed distribution 540 720 240 Expected distribution -7 14 -7 Obs-Exp 49 196 49 (Obs-Exp) 2 .09 .27 .20 (Obs-Exp) 2 /Exp 3/10/2011 PopGen 02- 2011 13 Example 2

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## This note was uploaded on 04/04/2011 for the course GENETICS 385 taught by Professor Brennemanandgordon during the Spring '11 term at Rutgers.

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Class02-ClassSlides- - Genetic Analysis II(385 Population Genetics Lecture 02 Date Instructor Professor Derek Gordon Office Life Sciences Building

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