Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A...

29
Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175 103 4 Aa 171 102 5 aa 155 101 6 aa 152 103 Subject Genotype Height Basic Concepts

Transcript of Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A...

Page 1: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Gene, Allele, Genotype, and Phenotype

A pair of chromosomes Father Mother

Gene A,with twoalleles Aand a

Phenotype

1 AA 185 1002 AA 182 104

3 Aa 175 1034 Aa 171 102

5 aa 155 1016 aa 152 103

Subject Genotype Height IQ

Basic Concepts

Page 2: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Bad news: It is very hard to detect such a gene directly.

Genetic Mapping

Marker 1

Marker 2Marker 3

Marker k

QTL

A gene that affects a quantitative trait is called a quantitative trait locus (QTL).

A QTL can be detected by the markers linked with it.

A QTL detected is a chromosomal segment.

.

.

.

Linkage Map

Page 3: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

QTL Mapping in Natural Populations

• Basic theory for QTL mapping is derived from linkage analysis in controlled crosses

• There is a group of species in which it is not possible to make crosses

• QTL mapping in such species should be based on existing populations

Page 4: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Human Chromosomes

Male Xy

X yFemaleXX X XX Xy Daughter Son

Page 5: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Human Difference

Page 6: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

How many genes control human body height?

Page 7: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Discontinuous Distribution

due to a single dwarf gene

Page 8: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Continuous Distribution

due to many genes?

Page 9: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.
Page 10: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Continuous Variation due to

• Polygenes 31=3, 32=9, …, 310=59,049

• Environmental modifications

• Gene-environmental interactions

Page 11: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Power statistical methods are crucial for the identification of human height genes

Page 12: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Subject

Marker (M) Conditional prob M1 M2 … Mm

Phenoty

pe (y)

of QTL genotypeQQ(2) Qq(1) qq(0)

1 AA(2) BB(2) … y1

2|1

1|1

0|1

2 AA(2) BB(2) ... y2

2|2

1|2

0|2

3 Aa(1) Bb(1) ... y3

2|3

1|3

0|3

4 Aa(1) Bb(1) ... y4

2|4

1|4

0|4

5 Aa(1) Bb(1) ... y5

2|5

1|5

0|5

6 Aa(1) bb(0) ... y6

2|6

1|6

0|6

7 aa(0) Bb(1) ... y7

2|7

1|7

0|7

8 aa(0) bb(0) … y8

2|8

1|8

0|8

Data Structure

Page 13: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Association between marker and QTL

-Marker, Prob(M)=p, Prob(m)=1-p

-QTL, Prob(A)=q, Prob(a)=1-q

Four haplotypes:

Prob(MA)=p11=pq+D p=p11+p10

Prob(Ma)=p10=p(1-q)-D q=p11+p01

Prob(mA)=p01=(1-p)q-D D=p11p00-p10p01

Prob(ma)=p00=(1-p)(1-q)+D

Linkage disequilibrium mapping – natural population

Page 14: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

AA Aa aa ObsMM p11

2 2p11p10 p102 n2

Mm 2p11p01 2(p11p00+p10p01) 2p10p00 n1

mm p012 2p01p00 p00

2 n0

MM p112 2p11p10 p10

2 n2

p2 p2

p2

Mm 2p11p01 2(p11p00+p10p01) 2p10p00 n1

2p(1-p) 2p(1-p) 2p(1-p)

mm p012 2p01p00 p00

2 n0

(1-p)2 (1-p)2 (1-p)2

Joint and conditional (j|i) genotype prob. between marker and QTL

Page 15: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Mixture model-based likelihoodwith marker information

L(|y,M)=i=1n[2|if2(yi) + 1|if1(yi) + 0|if0(yi)]

Sam- Height Marker genotype QTL genotypeple (cm, y) M AA Aa aa1 184 MM (2) 2|1 1|1 0|1

2 185 MM (2) 2|2 1|2 0|2 3 180 Mm (1) 2|3 1|3 0|3 4 182 Mm (1) 2|4 1|4 0|4 5 167 Mm (1) 2|5 1|5 0|5 6 169 Mm (1) 2|6 1|6 0|6 7 165 mm (0) 2|7 1|7 0|7 8 166 mm (0) 2|8 1|8 0|8

Prior prob.

Linkage disequilibrium mapping – natural population

Page 16: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Conditional probabilities of the QTL genotypes (missing) based on marker genotypes (observed)

L(|y,M)

= i=1n [2|if2(yi) + 1|if1(yi) + 0|if0(yi)]

= i=1n2 [2|if2(yi) + 1|if1(yi) + 0|if0(yi)] Conditional on 2 (n2)

i=1n1 [2|if2(yi) + 1|if1(yi) + 0|if0(yi)] Conditional on 1 (n1)

i=1n0 [2|if2(yi) + 1|if1(yi) + 0|if0(yi)] Conditional on 0 (n0)

Linkage disequilibrium mapping – natural population

Page 17: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Normal distributions of phenotypic values

for each QTL genotype group

f2(yi) = 1/(22)1/2exp[-(yi-2)2/(22)],

2 = + a

f1(yi) = 1/(22)1/2exp[-(yi-1)2/(22)],

1 = + d

f0(yi) = 1/(22)1/2exp[-(yi-0)2/(22)],

0 = - a

Linkage disequilibrium mapping – natural population

Page 18: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Differentiating L with respect to each unknown parameter, setting derivatives equal zero and

solving the log-likelihood equationsL(|y,M) = i=1

n[2|if2(yi) + 1|if1(yi) + 0|if0(yi)]log L(|y,M) = i=1

n log[2|if2(yi) + 1|if1(yi) + 0|if0(yi)]

Define2|i = 2|if1(yi)/[2|if2(yi) + 1|if1(yi) + 0|if0(yi)] (1)1|i = 1|if1(yi)/[2|if2(yi) + 1|if1(yi) + 0|if0(yi)] (2)0|i = 0|if1(yi)/[2|if2(yi) + 1|if1(yi) + 0|if0(yi)] (3)

2 = i=1n(2|iyi)/ i=1

n2|i (4)1 = i=1

n(1|iyi)/ i=1n1|i (5)

0 = i=1n(0|iyi)/ i=1

n0|i (6)2 = 1/ni=1

n[2|i(yi-2)2+1|i(yi-1)2+0|i(yi-0)2] (7)

Linkage disequilibrium mapping – natural population

Page 19: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Complete data Prior prob

QQ Qq qq ObsMM p11

2 2p11p10 p102 n2

Mm 2p11p01 2(p11p00+p10p01) 2p10p00 n1

mm p012 2p01p00 p00

2 n0

QQ Qq qq ObsMM n22 n21 n20 n2

Mm n12 n11 n10 n1

mm n02 n01 n00 n0

p11=[2n22 + (n21+n12) + n11]/2n,

p10=[2n20 + (n21+n10) + (1-)n11]/2n,

p01=[2n02 + (n12+n01) + (1-)n11]/2n,

p11=[2n00 + (n10+n01) + n11]/2n, =p11p00/(p11p00+p10p01)

Page 20: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Incomplete (observed) data

Posterior prob

QQ Qq qq ObsMM 2|i 1|i 0|i n2

Mm 2|i 1|i 0|i n1

mm 2|i 1|i 0|i n0

p11=[i=1n2(22|i+1|i)+i=1

n1(2|i+1|i)]/2n, (8)

p10={i=1n2(20|i+1|i)+i=1

n1[0|i+(1-)1|i]}/2n, (9)

p01={i=1n0(22|i+1|i)+i=1

n1[2|i+(1-)1|i]}/2n, (10)

p00=[i=1n2(20|i+1|i)+i=1

n1(0|i+1|i)]/2n (11)

Page 21: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

EM algorithm

(1) Give initiate values (0) =(2,1,0,2,p11,p10,p01,p00)(0)

(2) Calculate 2|i(1), 1|i

(1) and 0|i(1) using Eqs. 1-3,

(3) Calculate (1) using 2|i(1), 1|i

(1) and 0|i(1) based on

Eqs. 4-11,

(4) Repeat (2) and (3) until convergence.

Page 22: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Hypothesis Tests

• Is there a significant QTL?

H0: μ2 = μ1 = μ1

H1: Not H0

LR1 = -2[ln L0 – L1]

Critical threshold determined from permutation tests

Page 23: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Hypothesis Tests

• Can this QTL be detected by the marker?

H0: D = 0

H1: Not H0

LR2 = -2[ln L0 – L1]

Critical threshold determined from chi-square table (df = 1)

Page 24: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

A case study from human populations

• 105 black women and 538 white women;

• 10 SNPs genotyped within 5 candidates for human obesity;

• Two obesity traits, the amount of body fat (body mass index, BMI) and its distribution throughout the body (waist to hip circumference ratio, WHR)

Page 25: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.
Page 26: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.
Page 27: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Objective

Detect quantitative trait nucleotides (QTNs) predisposing to human obesity traits, BMI and WHR

Page 28: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

BMISNP Chrom. Black WhiteADRA1A 8p21 q 0.20

D 0.04a 11.40d -2.63LR 3.90* NS

WHRADRB1 10q24 q 0.83

D -0.07a -0.15d -0.24LR 5.91* NS

ADRB2 5q32-33 q 0.16D 0.07a 0.16d -0.20LR 5.88* NS

ADRB2- 5/20 q 0.83 0.78GNAS1 D 0.02 0.03

a -0.18 -0.15d -0.10 -0.16LR 8.42* 8.06*

Page 29: Gene, Allele, Genotype, and Phenotype A pair of chromosomes Father Mother Gene A, with two alleles A and a Phenotype 1 AA 185 100 2 AA 182 104 3 Aa 175.

Shape mapping meets LD mapping

Mapping Body Shape Genes through Shape Mapping

Ningtao Wang, Yaqun Wang, Zhong Wang, Han Hao and Rongling Wu*

Center for Statistical Genetics, The Pennsylvania State University, Hershey, PA 17033, USA

J Biom Biostat 2012, 3:8