Assignment 8.3 Falconer & Mackay, chapter 8

Assignment 8.3Falconer & Mackay, chapter 8 Sanja Franic VU University Amsterdam 2013

What is VD/VG for a single locus with the following degrees of dominance: 1) d = ½ a 2) d = a 3) d = 2a Plots graphs to show the relationships in each case, and find the allele frequency at which the approximate maximum value occurs.

VG = VA + VD VA= 2pq[a + d(q – p)]2 • VG = 2pq[a + d(q – p)]2+ (2pqd)2 VD= (2pqd)2

VG = VA + VD VA= 2pq[a + d(q – p)]2 • VG = 2pq[a + d(q – p)]2+ (2pqd)2 VD= (2pqd)2 • For the general case of d=ca: • VD/VG = (2pqd)2 / ( 2pq[a + d(q – p)]2+ (2pqd)2 ) • = (2pqca)2 / ( 2pq[a + ca(q – p)]2+ (2pqca)2 )

VG = VA + VD VA= 2pq[a + d(q – p)]2 • VG = 2pq[a + d(q – p)]2+ (2pqd)2 VD= (2pqd)2 • For the general case of d=ca: • VD/VG = (2pqd)2 / ( 2pq[a + d(q – p)]2+ (2pqd)2 ) • = (2pqca)2 / ( 2pq[a + ca(q – p)]2+ (2pqca)2 ) • = (2pqca)2 / ( 2pq ([a + ca(q – p)]2+ 2pqc2a2 ) ) • = 2pqc2a2/ ( [a + ca(q – p)]2+ 2pqc2a2 ) • = 2pqc2a2 / ( a2 + 2ca2(q – p) + c2a2(q - p)2 + 2pqc2a2 ) • = 2pqc2a2 / ( a2( 1 + 2c(q – p) + c2(q - p)2 + 2pqc2) ) • = 2pqc2/ ( 1 + 2c(q – p) + c2(q - p)2 + 2pqc2) • = 2pqc2 / ( [1 + c(q – p)]2+ 2pqc2 )

VG = VA + VD VA= 2pq[a + d(q – p)]2 • VG = 2pq[a + d(q – p)]2+ (2pqd)2 VD= (2pqd)2 • For the general case of d=ca: • VD/VG = (2pqd)2 / ( 2pq[a + d(q – p)]2+ (2pqd)2 ) • = (2pqca)2 / ( 2pq[a + ca(q – p)]2+ (2pqca)2 ) • = (2pqca)2 / ( 2pq ([a + ca(q – p)]2+ 2pqc2a2 ) ) • = 2pqc2a2/ ( [a + ca(q – p)]2+ 2pqc2a2 ) • = 2pqc2a2 / ( a2 + 2ca2(q – p) + c2a2(q - p)2 + 2pqc2a2 ) • = 2pqc2a2 / ( a2( 1 + 2c(q – p) + c2(q - p)2 + 2pqc2) ) • = 2pqc2/ ( 1 + 2c(q – p) + c2(q - p)2+ 2pqc2) • = 2pqc2 / ( [1 + c(q – p)]2+ 2pqc2 )

VG = VA + VD VA= 2pq[a + d(q – p)]2 • VG = 2pq[a + d(q – p)]2+ (2pqd)2 VD= (2pqd)2 • For the general case of d=ca: • VD/VG = (2pqd)2 / ( 2pq[a + d(q – p)]2+ (2pqd)2 ) • = (2pqca)2 / ( 2pq[a + ca(q – p)]2+ (2pqca)2 ) • = (2pqca)2 / ( 2pq ([a + ca(q – p)]2+ 2pqc2a2 ) ) • = 2pqc2a2/ ( [a + ca(q – p)]2+ 2pqc2a2 ) • = 2pqc2a2 / ( a2 + 2ca2(q – p) + c2a2(q - p)2 + 2pqc2a2 ) • = 2pqc2a2 / ( a2( 1 + 2c(q – p) + c2(q - p)2 + 2pqc2) ) • = 2pqc2/ ( 1 + 2c(q – p) + c2(q - p)2+ 2pqc2) • = 2pqc2 / ( [1 + c(q – p)]2+ 2pqc2 ) • because (a+b)2=a2+2ab+b2 , so [1 + c(q – p)]2= 1 + 2c(q – p) + c2(q - p)2

Let’s plot it (e.g., in R): • # The ratio of dominance variance to the total genetic variance, as a function of the degree of dominance and allele frequency • c=c(.5,1,2) # the constant c (i.e., the ratio of d to a) • f=function(p,q,c) { 2*p*q*c^2 / ( (1 + c*(q - p))^2 + 2*p*q*c^2 ) } # the function we are plotting (derived in the slides); the curve is a function of allele frequencies and the constant c • p=seq(0,1,by=.001) # a range of frequencies of the value increasing (A) allele; from 0 to 1, with a .001 step • q=1-p # a range of frequencies of the other (B) allele • a=f(p,q,c[1]) # the function evaluated at the full range of allele frequencies, at c=.5 • b=f(p,q,c[2]) # same, at c=1 • c=f(p,q,c[3]) # same, at c=2 • plot(p,a,ylim=c(0,1),xlab='Frequency of A allele',ylab='Dominance variance / Genetic variance',main='The VD/VG ratio') • lines(p,b,type='p',col='coral') • lines(p,c,type='p',col='cornflowerblue') • legend(0,1,c('Incomplete dominance; d=.5a','Complete dominance; d=a','Overdominance; d=2a'),col=c(1,'coral','cornflowerblue'),pch=19)

To find the maximum for each curve (e.g., in R): • ma=max(2*p*q*.5^2 / ( (1 + .5*(q - p))^2 + 2*p*q*.5^2 ));ma • mb=max(2*p*q*1^2 / ( (1 + 1*(q - p))^2 + 2*p*q*1^2 ));mb • mc=max(2*p*q*2^2 / ( (1 + 2*(q - p))^2 + 2*p*q*2^2 ));mc • The maxima are: • 0.1428571 for the case of incomplete dominance (d=.5a) • NaN for the case of complete dominance (or 1, if we look only within the feasible range of allele frequencies (from 0 to 1)) • 1 for the case of overdominance • At what allele frequencies do the maxima occur?

p=.75 p=1

Assignment 8.3 Falconer & Mackay, chapter 8