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국진 유
4 years ago
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2 Y 1 Y

6 β α β

9 Independence p qp pq q if X and Y are independent then E(XY)=E(X)*E(Y) so Cov(X,Y) = 0 Covariance can be a measure of departure from independence q

10 Conditional Probability if A and B are independent qp q then p P(A B)=P(A,B)/P(B)=P(A)*P(B)/P(B)=P(A) pq Pr( Ph) = Pr( Ph, G) G Pr( Ph, G) = Pr( G) Pr( G) G = Pr( Ph G) Pr( G) G q

11 p pq qp Pr( disease DD) = Pr( q disease D+ ) = 1 Pr( disease ++ ) = 0, Pr( unaffected DD) = Pr( unaffected D+ ) = 0 Pr( unaffected ++ ) = 1, q Pr( disease DD) = 1 Pr( disease D+ ) = Pr( disease ++ ) = 0,

13 100 y 100 y p (1 p) y

14 α Let H : α = 0, Pr(do not reject H H is true) = 1 α H : α = 0, Pr(do not reject H H is true) = 1 α then Pr(do not reject H H ) where H = H and H = Pr(do not reject H 01 and do no 0 0 = (1- α) (1- α) = (1- α) α α α α k α = = 3 = = = 0 t reject H H ) k (1 α) (1 α) = = (.95).95

15 m = α (1 ) 0.95 = α

16 Terminology Gene DNA segment that codes for a functional unit made up of exons interspersed with introns Locus Location of a gene on a chromosome ABO blood type is on Ch. #9 Allele Different forms of a gene that occupy the same locus.

17 New Gene Discovery Phenotype Segregation Association study Gene? Linkage analysis (LD, sibpair et al) Putative gene (locus)

18 Hardy-Weinberg Equilibrium Statistical Independence of an individual s two genes at a locus Deviations from HWE can be caused by many factors, including inbreeding, selection, population

19 Chi-square Test H 0 : p pq qp ( - ) ~ χ q = -1- q

20 HWE 1 - p = ( )/( 1000) = q = ( )/( 1000) = P(AA) = p = (0.545) = P(A a) = p q = (0.545) (0.4575) = P( aa ) = q = (0.4575) = (expected frequency) AA = P(AA) 1000 = Aa = P(Aa) 1000 = aa = P(aa) 1000 =

21 χ ( ) = ( ) =3-1-1=1 ( ) = p Ho (HWE ). HWE.

22 Test of association (Odds ratio, Chi-square test) p pq R = qp q p /(1 p ) ( n / n ) /( n / n ) n / n n n = = = p /(1 p ) ( n / n ) /( n / n ) n / n n n Chi-square test with df=(#col =(#col-1)(#row-1) : Ho: OR=1 Expected cell freq is bigger than 5, if not use Fisher s s Exact q

23 Chi-square 1 : Genotype-based p qp OR = MM/mm OR = Mm/mm = = pq q n n n n A 0O O 0A n n n n 1A 0O 1O 0A q

24 Chi-square : Allele-based Penetrance of Mm is often intermediate between the two homogyzous genotypes p pq Allele-based model often increase power and qp q precision Each individual now contributes two counts Note: assumption that alleles are sampled independently (random mating) is often not reasonable. q

28 θ θ θ

29 θ θ θ Z = Z( θˆ ) = log [ L( θ ˆ )/ L(1/ )] max 10

31 θ χ χ α

33 4 θ (1 θ) Z( θ) = log10 = log 4 10 θ (1 θ) θ Z (1/ 3) =

35 1 1 L( θ) = θ (1 θ) + θ (1 θ) 1 θ θ θ + θ 4 4 = (1 ) [ (1 ) ] θ

37 L( θ; data) = Pr( Ph )Pr( Ph ) offspring ma pa Pr( Ph Ph, Ph ; θ ) offs ma pa and we know that Pr( Ph ) = Pr( G) Pr( Ph G) ma G

39 Sib-Pair Markers sib1 sib Disease Status d1 d # of Shared i.b.d C µ µ α β ε

42 1 1 Not transmitted Allele1 (A1) Allele (A) Allele 1 0 (a) (b) Transm (A1) 1 1 itted Allele (A) 0 (c) 0(d) - χ ( b c) χ 1 = b + c