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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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/1.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/2.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/3.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/4.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/5.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/6.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/7.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/8.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/9.jpg)
![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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/10.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/11.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/12.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/13.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/14.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/15.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/16.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/17.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/18.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/19.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/20.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/21.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/22.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/23.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/24.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/25.jpg)
![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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/26.jpg)
![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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/27.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/28.jpg)
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.](https://reader037.fdocuments.in/reader037/viewer/2022110209/56649e565503460f94b4e50c/html5/thumbnails/29.jpg)
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