Lecture 16 – February 22, 2010

Review For Midterm Exam

Question 1

Genotype Number BB 22 Bb 16 bb 6 44a) What are allele frequencies?

p(B) = (22 + 8)/44 = .68

Question 1

Genotype Number BB 22 Bb 16 bb 6 44b) Expected numbers for genotypes?

p(BB) = .68 x .68 x 44 = 20.34 p(Bb) = 2 x .68 x .32 x 44 = 19.15 p(bb) = .32 x .32 x 44 = 4.51

Question 1

Genotype Number Expected BB 22 20.34 Bb 16 19.15 bb 6 4.51 44b) HWE? LR = -2[ 22 ln(20.34/22) + 16 ln(19.15/16) + 6 ln(4.51/6) ] = 1.1266 with 1 df < 3.84 = NS

Question 2

About 4% of people express an autosomal recessive genotype.

What is the frequency of the recessive allele? q = sqrt(.04) = .2

What percentage of people would be heterozygous? 2 p q = 2 (.2)(.8) = .32 OR 32%

Question 2

About 32% of people express a heterozygous genotype for some locus.

What is the frequency of the recessive allele? 2 p q = .32 p ( 1-p ) = .16 -p(p) + p -.16 = 0 (quadratic equation) a = -1, b = 1, c = -.16

p = [-1 +/- sqrt( b x b – 4 a c ) ]/ 2 a = -.2 or .8 (take positive result) q = 1 - .8 = .2

Question 3

Assume a sex linked locus with 4 alleles,Females, p1=.1, p2=.2, p3=.3, and p4=.4Males, p1=.5, p2=.3, p3=.2, and p4=0

What are the expected progeny genotypes? .1 .2 .3 .4

.5 .05 11 .10 12 .15 13 .20 14 .3 .03 12 .06 22 .09 23 .12 24 .2 .02 13 .04 23 .06 33 .08 34

Males .1 1Y .2 2Y .3 3Y .4 4Y

Question 3

.1 .2 .3 .4

.5 .05 11 .10 12 .15 13 .20 14 .3 .03 12 .06 22 .09 23 .12 24 .2 .02 13 .04 23 .06 33 .08 34

Males .1 1Y .2 2Y .3 3Y .4 4YWhat are new allele frequencies?

Females, p1 = (.05 + .5(.10+.15+.20+.03+.02))= .3 p2 = .25, p3 = .25, p4=.2Males, p1 = .1, p2=.2, p3=.3, p4=.4

Question 4

Genotype Freq. v(ij) GG .04 8 Gg .32 -4 gg .64 -6

What type of gene action is this?

What is the mean? mean = .04(8) - .32(4) - .64(6) = -4.8What are the allelic effects? (G) = 3.2 (g) = -0.8

Question 4

Genotype Freq. v(ij) GG .04 8 Gg .32 -4 gg .64 -6mean = -4.8 (G) = 3.2 (g) = -0.8

D(GG) = 8 – mean – 2 (G) = 6.4 D(Gg) = -4 – (-4.8) - (G) - (g) = -1.6 D(gg) = .4

Question 4

Genotype Freq. v(ij) A(ij) D(ij) GG .04 8 6.4 6.4 Gg .32 -4 2.4 -1.6 gg .64 -6 -1.6 .4

Var(A) = .04(40.96) + .32(5.76) + .64(2.56) = 5.12

Var(D) = .04(40.96) + .32(2.56) + .64(.16) = 2.56

Question 5

Number of matings to detect carriers, if

matings to carriers, or to general population.

Question 6

Migration, mixing of populations with different allele frequencies – new frequencies.

Question 6b

Two loci, UNLINKED, expected frequencies of gametes.

A B = p(A) x p(B)A b = p(A) x p(b)a B = p(a) x p(B)a b = p(a) x p(b)

Linkage disequilibrium = gametic phase disequilibrium

Question 7

Two loci, LINKED, expected frequencies of gametes.

A B = .5 (1 - r(AB))A b = .5 r(AB)a B = .5 r(AB)a b = .5 (1 – r(AB))

Haldane mapping function

Kosambi mapping function

Question 8

Loci interactions additive dominance add by add dom by dom add by dom add by add by add add by dom by dom add by add by dom, etc.

Question 9

Linkage disequilibrium, between linked loci or unlinked loci.

LD(n) = (1 – r)n LD(0)

If LD(0) = .24, and LD(3) = .12, what is r ?

.12 = ( 1 – r )3(.24), solve for r

Question 10

Three loci, estimate recombination rates between each pair.

What is the crossover interference?

Question 11

Path coefficient method,

Determining the covariance between variables,

and inbreeding coefficients.

Tabular Method of computing relationships.

Wright’s relationship coefficient versus a(i,j), range of values.

A B C D

EFG H I J

L M N OK

Q R S T P

Cov(P,Q) = ??

What was the error on page 3 of section 3.3 of notes?

What are the colour genotypes of Prof’s two yellow labrador retrievers?

What is the map distance between Guelph and Vancouver in Morgans?

Who hit the most home runs in their career in professional baseball? (868)

Who was the greatest cricket player in Australian history?

What race horse was mentioned in class? (not Seabiscuit)

Why was Sewall Wright unable to buy life insurance?

Mid term Exam

February 24, 2010

Rozanski 104, 4:30 PM

I will be in Axelrod 200 at 3:30 pm to help anyone