Week 1 - Association Analysis

Question 1

What are the features of complex diseases?

Answer 1

• High incidences so are very common
• They all follow a Non-Mendelian Transmission
• See a clustering within a family group
• Can and often are more than one gene involved.
• Variable Expressivity
• Polygenic – caused by many genes. Environment and Lifestyle are also additional factors

Question 2

Name the types of Polygenic Inheritance?

Answer 2

1. Additive – sim of effects of two or more gene loci
2. Multiplicative – combined effects of two or more gene loci.
3. Epistasis – gene-gene interactions, suppressive or stimulatory
4. Variable penetrance depending on location on the genome

Question 3

Give examples of diseases caused by polygenic inheritance

Answer 3

Asthma, Diabetes, Hypertensiom, Obesity and Cancer

Question 4

Why do we study complex diseases?

Answer 4

To identify genetic risk factors, susceptibility determination, identification of environmental factors and diagnostic tests and gene medicine.

Question 5

What are the effects of genetic drift on allele frequencies?

Answer 5

• Effect of population on genetic drift.
• Allele frequency in a small v large population. In a large population they are usually stabilised. Genetic drift is more likely to occur in a small population.

Question 6

What is association analysis?

Answer 6

• Selection can be detected by analysing sequence variation
• Synonymous substitutions (no change in aa) accumulate faster than non-synonymous substitutions (change in aa) (approx. x10 fold)
• Ratio of synonymous to non-synonymous substitutions (Ka/Ks) is a measure of selection
• Comparisons of rates of substitution can indicate whether selection on a gene has occurred
• Rates between number 1-3 and show any selection pressure.

Question 7

Describe the study that confirmed the presence of genetic factors in diseases at population level

Answer 7

•Prof. Arid and colleagues (1953) observed that O blood group was common in North England and so was Stomach cancer, but True Association is with A blood group
Case-control studies, Odds Ratios and c2 test
•Found true association was with the A blood group.

Question 8

Describe the Candidate Gene Approach

Answer 8

a) Identify specific genes whose known function may influence phenotype
b) Screen such genes for mutations which may affect function

Question 9

Describe the advantages and disadvantages of the Candidate Gene Approach

Answer 9

Advantage: time saver; simple statistical analyses.
Disadvantage: more likely than not that undiscovered genes play important role in disease

Question 10

Describe the Genome Wide Scan

Answer 10

a. Scan entire genome using highly polymorphic DNA markers (i.e. RFLPs, VNTRs, STRs, SNPs)
b. Identify regions of genome which co-segregate with disease phenotype
c. Directly screen known genes in linked regions for mutations
d. Linkage Disequilibrium studies.

Question 11

What is Linkage Disequilibrium

Answer 11

The non-random association of alleles at different loci

Question 12

Linkage Dissociation

Answer 12

• Human genome – look for an association between an alleles frequency and its LD with other genetic markers surrounding it
• LD makes tightly linked variants strongly correlated producing cost savings for association studies – possibility of identifying association with the disease by direct susceptibility marker (direct association) or a marker that is in high LD with susceptibility marker (indirect association)

Question 13

What is Linkage Disequilibrium affected by:

Answer 13

Natural Selection, genetic drift, recombination and mutation

Question 14

What are the features of DNA when it is supercoiled into Chromatin

Answer 14

• In nucleus DNA is associated with proteins (histones)
• DNA helix is tightly coiled around histones to form chromatin
• A nucleosome consists of one histone with 2 turns of DNA around it
• Euchromatin is a lightly packed region of chromatin, often under active transcription
• Chromatin further coils to form solenoids which form loops and minibands (heterochromatin)
• DNA is only visible when chromosomes condense during cell division

Question 15

what is the structure of a chromosome

Answer 15

• Human somatic cells contain 46 chromosomes
• Gametic cells contain 23 chromosomes
• 44 autosomes and 2 sex chromosomes
• Telomeres and centromeres are essential components
• Centromere divides chromosome into 2 arms (p and q) and is where spindle fibres attach during cell division
• Centromere location gives chromosome characteristic shape, useful for identification

Question 16

What is allelic disequilibrium

Answer 16

Particular alleles at two or more neighboring loci show allelic association if they occur together with frequencies significantly different from those predicted from the individual allele frequencies.

Question 17

Case control studies…

Answer 17

are a ratio not a proportion. the mode of inheritance affects the next stage of analysis. it depends on if it is dominant/ recessive or multifactorial

Question 18

Woolfs - the original basic method for association analysis. what are the features?

Answer 18

• Most Common and Simple,
• Can be used with any marker! (genotype, allele, others)
• Robust, Allows combining of data,
• Requires Patients and controls (case control)

Question 19

Steps in association Analysis - Significance. STEP 1

Answer 19

Work out odds ratio = a x d / b x c. anything higher than 1 is very high

Question 20

Steps in association Analysis - Significance. STEP 2

Answer 20

Work out significance of the association. done by:

expected number = row total x column total / grand total

Question 21

Steps in association Analysis - Significance. STEP 3

Answer 21

now you have observed and expected: (observed - expected)^2 / expected.

Question 22

Steps in association Analysis - Significance. STEP 4

Answer 22

o Remember, you have to this for each cell (genotype/allele) and then add these values together to get the chi-square value
o Assess the observed value against the c2 critical value in the table
o Calculate degrees of freedom df = (number of rows -1) * (number of columns – 1)
o For 1 degree of freedom at p value of 0.05, this value is 3.84.
o If observed value is higher than critical value it will indicate significant association

Question 23

Steps in Association Analysis - Confidence Interval

Answer 23

 The confidence interval for the Odds Ratio is calculated on the natural log (Ln) scale and then converted back to the original scale.
 The sampling distribution of the Odds Ratio is positively skewed.
 However, it is approximately normally distributed on the natural log scale.
 The confidence interval is calculated on the natural log scale (Ln)
 After finding the limits on the natural log scale, use the inverse LN function to find the limits on the original scale.