Y 1 Y

β α β

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

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

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,

100 y 100 y p (1 p) y

α Let H : α = 0, Pr(do not reject H H is true) = 1 α H 01 1 01 01 : α = 0, Pr(do not reject H H is true) = 1 α 0 0 0 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 0 0 0 01 0 1 = = 3 = = = 0 t reject H H ) k (1 α) (1 α) 4 1 0.1855 = 0.8145 = (.95).95

m 0.05 10 = α 0.05 4 4 (1 ) 0.95 = 1 0.05 0.005 α

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.

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

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

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

HWE 1 - p = ( 98+489)/( 1000) = 0.545 q = (489+ 13)/( 1000) = 0.4575 - P(AA) = p = (0.545) = 0.943 P(A a) = p q = (0.545) (0.4575) = 0.4964 P( aa ) = q = (0.4575) = 0.093 - (expected frequency) AA = P(AA) 1000 = 94.3064 Aa = P(Aa) 1000 = 496.3875 aa = P(aa) 1000 = 09.3063

χ (98-94.3063) = 94.3063 (13-09.3063) + 09.3063 =3-1-1=1 (489-496.3875) + 496.3875 = 0.15 1 p 0.6379 Ho (HWE ). HWE.

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 1 1 11 1+ 1 1+ 11 1 11 1 + + 1 1 1 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

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

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

θ θ θ

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

θ χ χ α

4 θ (1 θ) Z( θ) = log10 = log 4 10 θ (1 θ) 0.5 0.5 6 4 θ Z (1/ 3) = 0.1475

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

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

3 + 3 3 + + 3 4 + 3 + 3 4 + 4 Sib-Pair Markers sib1 sib 3 3 3 3 3 3 3 4 3 3 3 3 3 4 3 4 3 4 3 4 3 3 4 4 3 3 3 4 4 4 Disease Status d1 d + + + - + - + - - - - - - - - - - - - - # of Shared i.b.d. 1 1 0 0 1 1 C 1 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 µ µ α β ε

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