analysis of gene expression

Question 1

focus on mRNA

Answer 1

• easier to purify and measure than proteins
• easier to make high throughput
• protein expression controlled by PTMs
  
  • activity not proportional to mRNA expression

Question 2

experimental considerations

Answer 2

• uniform growth conditions
  
  • comparable setup without confounding variables
• homogeneous cell population
  
  • heterogeneity produces misleading data
• control choice
  
  • must be relevant

Question 3

microarrays

process of image to numbers

Answer 3

• scan
• correct background noise
• data transformation
• normalisation
• data analysis

Question 4

microarray types

Answer 4

• single channel:
  
  • 2 samples analysed separately
  • 2 sets of expression values
• 2 channel:
  
  • hybridise labelled cDNA from 2 samples to same microarray
  • extratc values
  • different colour tags

Question 5

process of microarrays

Answer 5

• isolate and reverse transcribe mRNA
• label cDNA with labelled nucleotides
• hybridise
• scan and analyse
• calculat eintensity relative to normal background
  
  • some chips have inherently higher intensity than others

Question 6

microarray data transformation

Answer 6

• difference between or ratio of 2 samples
  
  • ratios for larger differences
  • depends what you are comparing
• be wary of small numbers
  
  • differences more likely due to chance
• to identify differential expression:
  
  • plot of log ratios → focus on tails of gaussian distribution
  • plot of log intensity → above or below diagonal line

Question 7

microarrays

data normalisation methods

Answer 7

• global intensity
• houskeeping genes
• exogenous/spiked control
• intensity dependent linear
• loess

Question 8

experimental variables affecting microarray intensity levels

Answer 8

• number of mRNA copies (want to be the only variable)
• hybridisation efficiency
• cross-hybridisation
• efficiency of reverse-transcription
• marker incorporaiton into cDNA
• scanning efficiency
• activity of fluorescent dyes
• equipment differences

Question 9

global intensity normalisation

Answer 9

• assumption:
  
  • same total expression for sample and control
• method:
  
  • calculate sum of intensities for all genes for sample and control
  • use ratio of intensity sum to calculate normalisation factor, k

Question 10

housekeeping gene normalisation

Answer 10

• assumption:
  
  • housekeeping gene mRNA expression is constant
    
    • tubulin, actin
  • their expression does not depend on cell condition/status
• method:
  
  • use ratio of means of hk gene intensities to calculate k
    
    • m(sample)/(control)

Question 11

spiked control normalisation

Answer 11

• add mRNA from foreign genes
  
  • 5-10 B. subtilis genes with artifical polyA tails
• equal amounts of exogenous controls
  
  • use average intensities to compute k
  • whole normalise data sets
• can detect large changes in genome-wide mRNA levels

Question 12

intensity dependent linear normalisation

Answer 12

• predicts experimental expression using control expression minus intercept for normalisation
• assumptions:
  
  • constant mRNA levels
  • all genes on the microarray have the same variance

Question 13

intensity dependent linear normalisation

method

Answer 13

• fit data to linear regression model
• y_i = α + βx_i + ε_i where:
  
  • y_i = background-subtracted intensity of gene in set 1 (experimental)
  • x_i = background-subtracted intensity of gene in set 2 (control)
  • α = y-intercept of microarray data
  • β = normalisation factor
• normalise y_i:
  
  • y’_i = (y_i - α)/β

Question 14

intensity dependent linear normalisation

advantages vs disadvantages

Answer 14

• advantages:
  
  • each gene contributes equally to nromalisaiton factor
    
    • prevents high expression bias
  • can correct skew in data with y-intercept correction
• disadvantages:
  
  • linear data only
    
    • can you be sure of linearity
  • assumes constant mRNA levels

Question 15

M vs A plot

Answer 15

• detemrines whether non-linear normalisation is needed or not
• minus vs add
• difference in log intensity of each channel
  
  • average log intensity of each halved
  • plot against each other
• quick overview
• if no change in expression:
  
  • points lie straight on x=0
  • otherwise loess normalisation needed

Question 16

loess normalisation

Answer 16

• intensity dependent non-linear normalisation
• local running average subtracted
• applies different correction to different parts of data set as appropriate
• apply loess function to MA plot
  
  • applies rules of logs to M and A to correct for M
• good for skewed or non-linear data
• more complex and difficult to calculate

Question 17

RNAseq

Answer 17

• resolves intensity conversion problem
• sequence counts only
• prevents bias against non-coding RNAs/microRNAs and RNAs without probes
• single cell resolution
• calculate reads per nt and get average score for each gene
  
  • repeat with 3’ end
  • sum of reads within the gene divided by length of gene