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by: Keshawn Howell


Keshawn Howell
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This 8 page Class Notes was uploaded by Keshawn Howell on Thursday October 29, 2015. The Class Notes belongs to BPSC 148 at University of California Riverside taught by Staff in Fall. Since its upload, it has received 23 views. For similar materials see /class/231737/bpsc-148-university-of-california-riverside in Botany at University of California Riverside.

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Date Created: 10/29/15
Chapter 2 Population Genetics Review Gene frequencies and genotype frequencies Consider an autosomal locus of a diploid organism with two alleles A1 and A2 There will be three genotypes A1141 A1142 and AzAz Let Nbe the total number of individuals sampled from the population among which N11 are of genotype A1141 N12 of A1142 and N22 of 142142 The genotype frequencies are de ned and estimated from the following table Genotype Count Frequency Estimate A1141 N11 P 13 N A1142 N 12 H 1 N 12 N A2142 N 22 Q A N 22 N N N11 le sz The gene frequencies are defined and estimated from the following table Gene Count Frequency Estimate A1 2M1N12 P 2NllNIZ 15f 2N A2 2sz l39Niz q 2N2N12 QAIfI Hardy Weinberg equilibrium In a large random mating population with no selection no mutation and no migration the gene frequencies and genotype frequencies are constant from generation to generation There is a simple relationship between the gene frequencies and the genotype frequencies The relationship follows the square law HardyWeinberg equilibrium can be reached at one generation of random mating 21 HardyWeinberg law the square law Gene Genotype A1 A2 A1141 A1142 142142 Frequency P 9 p 2 2P9 q 2 Square law pq 2 p2 2179 q 2 Proof of the H W law a From gene frequencies to genotype frequencies Female gamete Male gamete A1p 1429 A1p A1A1p2 A1A2pq A2q AzAmqp Azquz wmwm mw cxrwf wmw2 b From genotype frequencies to gene frequencies A1141 A1142 A2142 P Hhw Qf L 2 1 2 FrequencyofA1 allele P2Hp 22pqp pqppqp Frequency ofAz allele QH qz 2pq q2 pq qpq q 22 If a population is not at HW equilibrium eg male and female populations have different gene frequencies HW equilibrium can be reached in one generation of random mating Female gamete Male gamete A1pf AZWI A1pm A1141 Pmpf A1A2PmQ A2qm A2A1qmpf A2A2me pmqm landpfqfl The frequency ofA1 in the offspring P PmPf Pm9f quf 2 707m Pf The frequency ofAz in the offspring q qmqf Pm9f quf 7 qf When the offspring population is random mated the gene frequencies will remain at p and q because the frequencies of the males and females are the same Multiple alleles Consider n alleles A1 A2 An with frequencies q1 qz qn The array of genotypes and their frequencies are A1QI A2qz Anqn 14191 A1A1H11 A1A2H12 AlAnH1n A2Q2 A2A1H21 A2A2H22 AzAnH2n Anqn AnA1Hn1 AnAzan AnAnHm H i is the genotype frequency of AAj The table is symmetrical ie H U Hji because AAj and AA represent the same genotype 23 In HardyWeinberg equilibrium H U qiqj which re ects random mating The gene frequencies in the next generation will be 9 H ZHU Hj 1 It can be veri ed that Z q l What is the genotype frequency array According to the 1 square law the genotype frequency array is n n71 n 919239quotqn2 qu 22 2an 11 11 x1 There are 71 types of homozygotes and nn 1 types of heterozygotes Applications of the H W law a Estimating gene frequency of the recessive allele The frequency of the recessive genotype is Q which equals qz under HW equilibrium Since Q is observable it can be used to estimate q by using a J b Frequency of carriers Carriers are individuals who look normal but carry one recessive allele ie Act We cannot distinguish Aa fromAA because of dominance The frequency of carriers is defined as the frequency of heterozygotes among normal individuals Frequency of normal P H Frequency of heterozygotes H Frequency ofcarriers fi22i2EF qq2q PH p 2pq l q 2l qq lq 2 4 Test of Hardy Weinberg equilibrium The MN blood group frequencies Genotype Gene frequency MM MN NN Total M N Observed 233 385 129 747 05696 04304 Expected 24236 36626 13838 747 2 Z O E2 233 242362 385 366262 129 138382 E 24236 36626 13838 The test is not signi cant and thus the population is in HardyWeinberg equilibrium 196 lt 3571 388 0 stands for observed number E stands for expected number The expected number is calculated as follows Let p be the frequency of allele M and q 1 p be the frequency of allele N 2X233385 385 2gtlt747 747 2gtlt747 2X129385 385 2gtlt747 747 2gtlt747 05696 5 04304 Q The expected number for MM is if X 747 24236 The expected number for MN is 2f7 X 747 36626 The expected number for NN is z X 747 13838 25 Appendix C Statistics Review Linear Model Regression and Correlation Linear model Consider two variables X and Y Assume that the change of Y depends on the change of X then Y is called the dependent variable andX is called the independent variable No matter what the exact relationship betweenX and Y is within a short range of value inX Y can always be expressed by the following linear model Y o lX 8 where 50 is called the intercept 51 is the regression coefficient and s is called the residual error The residual error is the collective contribution made by all other unknown variables The residual error is usually assumed to be normally distributed with mean zero and variance 6 ie 8 N N0 6 Regression The regression coefficient 51 represents the amount of change in Y corresponding to one unit change inX Mathematically the regression coefficient is de ned as C0VXY6XY 1 VarX cg39 Because the covariance can take any real value we have oo S l S oo Sometimes 51 is denoted by by X re ecting the direction of the regression process by X is the regression of Y on X which is different from 3X y the regression ofX on Y CovXY 6 XY 3 VarY 6 The predicted Y value is given by o 1X The line representing the predicted Y is shown in Fig C1 Insert Fig C1 here Correlation Correlation describes the association of two variables The degree of association is measured by the socalled correlation coefficient which is defined by CovX Y 6X 6Y where 6X 46 and 6V 46 In contrast to regression coefficient correlation has no direction In addition correlation coefficient can only take value within i1 ie 1 S rXY S 1 Cl Correlation coefficient can be considered as a standardized measurement of covariance De ne Z X and W X Variables Z and W are said to be 6 6 X Y standardized variables because both have an expectation zero and a variance one This can be shown by EltZgt El W LEGO Ema im uX 0 and VarX 2 1 6X VarZ Var izVadX uX L2VarX VaruX 6X 6X 6X The covariance between Z and W is C0VZ W EZ EZW EW EZW E X X Y My 2 EX anY M CovXY r 6X 6y 6X 6y 6X 6 XY Relationship between regression and correlation coef cients 61 6 rXY YX X XY 39 6V 6X Estimation of regression and correlation coef cients We have learned how to estimate variance and covariance given a random sample from the joint distribution of X and Y The estimated regression and correlation coefficients are obtained by replacing the variances and covariance by their estimated values as shown below a C6VltXYxI rcxyx r W 6 ice ff and C6VXY gal roar XY 5amp5 Multiple regression Consider variable Y that is described as a linear function of k variables the linear model is Y 0 1X1 2X2 kag where 50 is still called the intercept lquot39 k are called partial regression coef cients These partial regression coef cients are de ned collectively by 1 2 1 6X1 6X1X2 6an 6W 2 l 6 6 6 6 2 XX X XX XY 1 2 1 2 2 k 2 cg V C 2 6 k SAA3 SAX quot39 SA Xky where Vis a matrix array of elements with k rows and k columns V1 is called the inverse of matrix Vand C is a column vector with k elements The intercept is defined by e EY 2 REM The jth partial regression coef cient represents the amount of change in Y due to per unit change ian given that the values of all other variables are fixed at a particular level The least square estimates of 50 51 Assume that we have a random sample of n observations x1y1 xnyn Each observation can be described by a linear model y 50 1xj 8 forj l n Rearrange the above equation we have 8 y o B1xjyj J7 De ne Qisj fol e we 1 11 The least squares estimates of 50 and 51 are those values of 50 and 51 which minimize Q To nd the least square solution one needs to nd the partial derivatives of Q with respect to 50 and 31 and set these partial derivatives to zero g n ago 2gb Ho xxx1 0 j Qf 2gb o 1xj xj 0 2y n o ij 1 0 1 1 Zxy Zx 0Zx2 l 0 11 11 1 C3


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