Aspects of Genetic and Genomics in Cancer Research

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Aspects of Genetics and

Genomics in Cancer Research

Li Hsu

Biostatistics and Biomathematics Program

Fred Hutchinson Cancer Research Center

Outline

• Cancer facts

• Linkage analysis of family studies

• Genome-wide association studies

Etiology of Cancer

• The etiology of cancer is multifactorial, with genetic, environmental, medical, and lifestyle factors interacting to produce a given malignancy.

• The breakthroughs in high throughput genotyping technologies have made it possible for systematically identifying genes that are responsible for disease occurrence.

BRCA1 and Breast Cancer

• BRCA1 (breast cancer 1) is a human gene that belongs to a class of genes known as tumor suppressors, which maintains genomic integrity to prevent uncontrolled proliferation. Variations in the gene have been implicated in a number of hereditary cancers, namely breast, ovarian and prostate. The BRCA1 gene is located on the long (q) arm of chromosome 17 at 38Mb.

Probability of developing breast cancer by age (Chen et al.

2009) carriers

Non-carriers

Age 50

Age 60

Age 70

Age 80

Probability of Developing Breast Cancer for

BRCA1 carriers

Average Person BRCA1 Carrier

2.1%(1.7%-2.7%) 18.8%(8.2%-2.3%)

4.1%(3.4-5.0%)

7.2%(6.0%-9.0%)

31.3%(14.3%-61.2%)

45.4%(22.7%-74.3%)

10.2%(8.4%-12.5%) 54.9%(30.4%-81.4%)

• How was BRCA1 found?

Linkage Analysis

1/2

1/3

3/4

2/4

3/4

3/2

1/4 1/4 1/2 3/2

Assume disease gene (D) is rare with full penetrance

3/4

D/d

1/2 d/d

1/3 d/D

3/2

D/d

3/4

D/d

2/4 d/d

1/4 1/4 d/d D/d

1/2 d/d

3/2

D/d

Linkage Analysis (continued)

• Disease allele (D) originally in chromosome with allele 3

• How often does D co-segregate with allele 3 (non-recombinant)?

Assume disease gene (D) is rare with full penetrance

3/4

D/d

1/2 d/d

1/3 d/D

3/2

D/d

3/4

D/d

2/4 d/d

1/4 1/4 d/d D/d

1/2 d/d

3/2

D/d

Linkage Analysis (continued)

• Disease allele (D) originally in chromosome with allele 3

• How often does D co-segregate with allele 3 (non-recombinant)?

– 5 meiosises

• How often is D separated from allele 3

(recombinant)?

Assume disease gene (D) is rare with full penetrance

3/4

D/d

1/2 d/d

1/3 d/D

3/2

D/d

3/4

D/d

2/4 d/d

1/4 1/4 d/d D/d

1/2 d/d

3/2

D/d

Linkage Analysis (continued)

• Disease allele (D) originally in chromosome with allele 3

• How often does D co-segregate with allele 3 (non-recombinant)?

– 5 meiosises

• How often is D separated from allele 3

(recombinant)?

– 1 meiosis

Likelihood function

• Set a parameter θ which measures the distance between allele 3 and D by how frequently they recombine.

• The likelihood function L(θ) = (1- θ) 5 θ

• The maximum likelihood estimate is 1/6

• LOD = log

10

= 0.63

L(1/6)/L(1/2)

• LOD for 7 families = 7x0.63 = 4.41

Issues

• Linkage analysis has narrowed down to a region about 1Mb. However it took another four years before the BRCA1 gene was mapped.

• Reduced penetrance, phenocopy, and genetic heterogeneity are among the factors that limit the success of the linkage analysis.

• Relevance of the findings to the population at large.

Genome-Wide Association

Studies(GWAS)

• The Human Genome Project began in 1990 and completed in 2003.

Part of sequence from Chromosome 7

AGACGGAGTTTCACTCTTGTTGCCAACCTGGAGTGCAGTGGCGTGATCTCAGCTCACTGCACACTCCGCTTTC C/T GG

TTTCAAGCGATTCTCCTGCCTCAGCCTCCTGAGTAGCTGGGACTACAGTCACACACCACCACGCCCGGCTAATTTTTG

TATTTTTAGTAGAGTTGGGGTTTCACCATGTTGGCCAGACTGGTCTCGAACTCCTGACCTTGTGATCCGCCAGCCTCT

GCCTCCCAAAGAGCTGGGATTACAGGCGTGAGCCACCGCGCTCGGCCCTTTGCATCAATTTCTACAGCTTGTTTTCTT

TGCCTGGACTTTACAAGTCTTACCTTGTTCTGCCTTCAGATATTTGTGTGGTCTCATTCTGGTGTGCCAGTAGCTAAAA

ATCCATGATTTGCTCTCATCCCACTCCTGTTGTTCATCTCCTCTTATCTGGGGTCAC A/C TATCTCTTCGTGATTGCATTC

TGATCCCCAGTACTTAGCATGTGCGTAACAACTCTGCCTCTGCTTTCCCAGGCTGTTGATGGGGTGCTGTTCATGCCT

CAGAAAAATGCATTGTAAGTTAAATTATTAAAGATTTTAAATATAGGAAAAAAGTAAGCAAACATAAGGAACAAAAAG

GAAAGAACATGTATTCTAATCCATTATTTATTATACAATTAAGAAATTTGGAAACTTTAGATTACACTGCTTTTAGAGAT

GGAGATGTAGTAAGTCTTTTACTCTTTACAAAATACATGTGTTAGCAATTTTGGGAAGAATAGTAACTCACCCGAACA

GTGTAATGTGAATATGTCACTTACTAGAGGAAAGAAGGCACTTGAAAAACATCTCTAAACCGTATAAAAACAATTACA

TCATAATGATGAAAACCCAAGGAATTTTTTTAGAAAACATTACCAGGGCTAATAACAAAGTAGAGCCACATGTCATTT

ATCTTCCCTTTGTGTCTGTGTGAGAATTCTAGAGTTATATTTGTACATAGCATGGAAAAATGAGAGGCTAGTTTATCAA

CTAGTTCATTTTTAAAAGTCTAACACATCCTAGGTATAGGTGAACTGTCCTCCTGCCAATGTATTGCACATTTGTGCCC

AGATCCAGCATAGGGTATGTTTGCCATTTACAAACGTTTATGTCTTAAGAGAGGAAATATGAAGAGCAAAACAGTGCA

TGCTGGAGAGAGAAAGCTGATACAAATATAAATGAAACAATAATTGGAAAAATTGAGAAACTACTCATTTTCTAAATT

ACTCATGTATTTTCCTAGAATTTAAGTCTTTTAATTTTTGATAAATCCCAATGTGAGACAAGATAAGTATTAGTGATGGT

ATGAGTAATTAATATCTGTTATATAATATTCATTTTCATAGTGGAAGAAATAAAATAAAGGTTGTGATGATTGTTGATTA

TTTTTTCTAGAGGGGTTGTCAGGGAAAGAAATTGCTTTTTTTCATTCTCTCTTTCCACTAAGAAAGTTCAACTATTAATT

TAGGCACATACAATAATTACTCCATTCTAAAATGCCAAAAAGGTAATTTAAGAGACTTAAAACTGAAAAGTTTAAGATA

GTCACACTGAACTATATTAAAAAATCCACAGGGTGGTTGGAACTAGGCCTTATATTAAAGAGGCTAAAAATTGCAATA

AGACCACAGGCTTTAAATATGGCTTTAAACTGTGAAAGGTGAAACTAGAATGAATAAAATCCTATAAATTTAAATCAA

AAGAAAGAAACAAACT A/G AAATTAAAGTTAATATACAAGAATATGGTGGCCTGGATCTAGTGAACATATAGTAAAGA

TAAAACAGAATATTTCTGAAAAATCCTGGAAAATCTTTTGGGCTAACCTGAAAACAGTATATTTGAAACTATTTTTAAA

Genome-Wide Association Study

• 550,000 SNPs on an array

• 2000 diseased individuals (colon cancer cases) and 2000 normal individuals

• Genotype all DNAs for 550,000 SNPs

• That is 2 billion genotyping!

GWAS on Type 2 Diabetes (Steinthorsdottir et al., 2007, Nature Genetics)

AA

Aa aa

Cases Controls

751

539

108

3107

1887

277

1398 5271

3858

2426

385

6669

AA

Aa aa

Cases Controls

809

509

81

3049

1917

305

1398 5271

• Expected count for cases if AA is not associated with the disease. First, calculate the frequency of AA genotype in both cases and controls combined: freq = 3858/6669 = 57.85%

• For 1398 cases, we expect to see 1398*57.85%=809 individuals having genotype AA.

3858

2426

385

6669

GWAS on Type 2 Diabetes

• The chi-square statistic is calculated by finding the difference between each observed and expected for each cell, squaring them, dividing each by the expected, and taking the sum of the results.

( 757 809 )^2/ 809 +( 3107 3049 )^2/ 3049 +…

• Compare the value to a standard chi-square distribution with degrees of freedom (# rows-1)*(# col -1) = 2.

• The p-value for this SNP is 6.772e-5.

Issues

• Too many SNPs!

• Identifying gene-gene and geneenvironmental interactions are now possible.

Germline mutations account for only a small portion of cancer cases.

http://envirocancer.cornell.edu/FactSheet/General/fs48.inheritance.cfm

Summary

• The amount of the data that have been generated increases exponentially in the last few years.

• This creates a great demand on efficient and valid computational and statistical methods and tools for picking the needles from a haystack.

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