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

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.

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Linkage Map

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

Human Chromosomes

Male Xy

X yFemaleXX X XX Xy Daughter Son

Human Difference

How many genes control human body height?

Discontinuous Distribution

due to a single dwarf gene

Continuous Distribution

due to many genes?

Continuous Variation due to

• Polygenes 31=3, 32=9, …, 310=59,049

• Environmental modifications

• Gene-environmental interactions

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

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

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

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

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

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

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

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

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)

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)

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.

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

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)

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)

Objective

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

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*

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