1. 12/04/2017
RNA seq (I)
Edouard Severing

2. A typical heat stress experiment (climate change)
Economically important frog
Heat stress
(convection)
Control
85 minutes
How does the frog adapt and survive?
5 days

3. Coping with heat stress
The frog likely has to change several processes in order to cope with the heat stress.
Adaptation of metabolic pathways.
Prevent water loss through skin
Changing the concentration of several enzymes, other proteins and molecules.
We want to determine these molecule concentration changes
Starting with proteins.

4. Changes at the molecular level
We could measure protein concentration directly
Not often done on a large scale
We could measure changes in the expression of the genes that encode these proteins.
Gene expression can be approximated by measuring the amount of mRNA molecules that are produced by the gene.

5. Gene count and complexity
genes
genes

6. From genes to proteins (I)
Initial assumption N
mRNA
Molecules N
Proteins N
Protein coding
genes
Assumption is based on studies that were performed on bacterial systems

7. From genes to proteins (II)
Current view
X N
mRNA
Molecules
? N
Proteins
What happens here ? N
Protein coding
genes

8. Splicing Pre-mRNA Intron Gene Splicing mRNA 5’- -3’ 5’- -3’ 5’- -3’
Exon
5’-
-3’
Intron
Exon
5’-
-3’
Splicing
mRNA
Gene

9. Alternative splicing Pre-mRNA Splicing Splicing 5’- -3’ 5’- -3’ 5’-

10. Gene count and complexity
90% of genes have AS
60% of genes have AS
The average number of transcripts produced by human genes is also higher than the average number of transcripts produced by plant genes

11. An extreme case
Dscam gene produces over 38,000 different transcripts

12. Major alternative splicing event types
In humans exon skipping is most frequent AS event type
In plants intron retention are the most common AS event type
Humans
Exon skipping
Plants
Intron retention

13. Difficulty: Distinguish genuine RNA-editing from sequencing errors
Primary transcript
(Predicted sequence) C U C
5’- A G
U - 3’ A
RNA-Editing
After editing
(Observed sequence) A C U U
5’- A G
U - 3’ A
Difficulty: Distinguish genuine RNA-editing from sequencing errors

14. Not everything is translated
A large fraction (>30%) of transcripts of protein coding genes are degraded by the nonsense-mediated decay (NMD) pathway.
The position of the stop codon is used to predict whether a transcript is likely to be degraded by the NMD pathway

15. Detecting putative NMD candidates
Pre-mRNA
5’-
-3’
Exon/Exon junctions
mRNA
5’-
-3’
Open reading frame M
5’-
-3’
Stop d
> 50-55nt

16. Remember
The number of unique mRNA molecules is much larger than the number of genes.
A large fraction of the mRNA molecules is degraded by the NMD pathway.
NMD provides a means to regulate gene-expression at the post-transcriptional level

17. Process the frogs into reads for analysis
Sequencing
Grind N2
Prepare for sequencing
>s1
ATCGTAGGGTA
>s2
ATGGCCTAGGT
Bioinformatics

18. Basic transcriptome analysis steps
Many research questions require the following steps:
Reconstruction of the transcriptome
We usually only have fragments
Quantification of the transcriptome
Differential expression analysis
Other fun stuff.

19. de novo transcriptome reconstruction (I)

20. de novo transcriptome reconstruction (II)

21. Genome-guided transcriptome reconstruction
5’-
-3’
mRNA

22. Genome-guided transcriptome reconstruction

23. Genome-guided transcriptome reconstruction

24. Remember de novo transcriptome assembly
When no reference genome is available
Finding features which are not on the reference genome (tDNA insertion)
Programs: Trinity, Trans-ABySS, Velvet Oases
Genome-guided transcriptome reconstruction
Reference genome is available with or without annotation
Mapping programs: TopHat, GSNAP
Transcriptome reconstruction: Scripture, Cufflinks

25. RNA seq (II) Quantification
12/04/2017
RNA seq (II) Quantification
Edouard Severing

26. A typical heat stress experiment
(convection)
Control

27. Raw counts
Counting number of reads/fragments falling with exonic regions of a gene.
Example: HTseq-count
Exon 1
Exon 4
Exon 3
Exon 2

28. The same fragment count yet different expression levels
Exon 1
library
Exon 1
Library size matters
library

29. The same fragment count yet different expression levels.
Exon 1
Exon 4
Exon 3
Exon 2
Exon 1
Transcript/gene length matters

30. Normalizing/correcting for feature length and library size
Reads mapped to region
RPKM ≈ 1.7
300 nt
Feature length
10,000,000
All mapped reads

31. Normalizing/correcting for feature length
FPKM is analogous to RPKM
RPKM = 1
RPKM = 2
FPKM = 1
Different picture emerges from raw counts and RPKM/FPKM values

32. Counting method issues
What to do with reads that map to multiple isoforms (alternative splicing) or genes
Pure Random assignment?
No, expression can differ
Count multiple time?
No, it has been derived from a single transcript
Gene 1
Gene 2
Isoform 1
Isoform 1
Isoform 2

33. Count issues: Back to the gene level (I)

34. Count issues: Back to the gene level (II)

35. Statistical methods: Expression levels of transcripts

36. Fishing in the dark lake experiment
Question: What fraction (t) of the fish in the lake is green?
Method: We catch a number of fish and determine what fraction is green.
Caution: Fish have to be immediately thrown back in the water.

37. Fishing in the dark lake results (I)
Sane people would do:
Sample(X)
Fraction of fish that is green
t = 1/3

38. Fishing in the dark lake results Maximum likelihood estimate of t
Sample(X)
Maximum likelihood estimate of t
The probability of observing our
sample X given a certain t:
𝑃(𝑋)= 3! 2!∙1! ∙𝑡 ∙ (1−𝑡) 2
Find a t that maximizes the probability of our observation
P(t)) t

39. Fishing in a complex dark lake.
Transcript quantification using RNAseq is like fishing in a dark lake with fragmented fish.
We are also forced to determine the possible origin(s) of the fish fragments
Only lost an eye and a vin but not its tail

40. Estimating relative transcript abundances
Target
Fragmentation α1
Transcript 1
Transcript 2 α2
Sequencing
>s1
ATCGTAGGGTA
>s2
ATGGCCTAGGT
Observation
Read mapping
Which values of the α1 and α2 gives the highest probability of observing these reads. (α1 + α2 = 1 )

41. Maximum likelihood alignments
The likelihood of our observation (ʎ) corresponds to the product of observing each of the individual mapped reads (rj ) in our set (R)
𝛌= 𝑗=1 𝑅 𝑃(𝑟 𝑗 ) R

42. Probability of observing a read
Probability of observing a read rj is the sum of the individual probabilities that a read originates from each transcript (t) in our transcript set (T).
𝑃(𝑟 𝑗 )= 𝑡=1 𝑇 𝐾 𝑗𝑡 ∙ 𝛼 𝑡 𝑙 𝑡 𝑖=1 𝑇 𝛼 𝑖 𝑙 𝑖 ∙ 𝑃 𝑗𝑡 (𝑞)
Probability that rj originated from transcript t
Read j

43. Component 1: Compatibility
𝑃(𝑟 𝑗 )= 𝑡=1 𝑇 𝐾 𝑗𝑡 ∙ 𝛼 𝑡 𝑙 𝑡 𝑖=1 𝑇 𝛼 𝑖 𝑙 𝑖 ∙ 𝑃 𝑗𝑡 (𝑞)
Does read j map to transcript t
t=1
Kj1 = 1
t=2
Kj2 = 1
t=3
Kj3 = 0

44. Component 2: Sequencing a read from a specific transcript
𝑃( 𝑟 𝑗 )= 𝑡=1 𝑇 𝐾 𝑗𝑡 ∙ 𝛼 𝑡 𝑙 𝑡 𝑖=1 𝑇 𝛼 𝑖 𝑙 𝑖 ∙ 𝑃 𝑗𝑡 (𝑞)
Probability of “sequencing” a read from transcript t
Product of the relative expression level and length of transcript t
𝛼 𝑡 𝑙 𝑡

45. Component 2: Sequencing a read from a specific transcript
Why and not just ?
𝛼 𝑡 𝑙 𝑡
𝛼 𝑡
Longer transcripts produce more fragments than shorter transcripts at equal expression levels.
Fragments
Fragmentation α1 α2
α1 = α2

46. Component l1 = 200; α1 = 0.3 l2 = 150; α2 = 0.2 l3 = 50; α3 = 0.5
𝛼 1 𝑙 1 𝑖=1 𝑇 𝛼 𝑖 𝑙 𝑖 = 0.3 ∙ ∙ ∙ ∙50 ≈0.52
Adjust for length
normalize

47. 𝑃(𝑟 𝑗 )= 𝑡=1 𝑇 𝐾 𝑗𝑡 ∙ 𝛼 𝑡 𝑙 𝑡 𝑖=1 𝑇 𝛼 𝑖 𝑙 𝑖 ∙ 𝑃 𝑗𝑡 (𝑞)
Component 3:
𝑃(𝑟 𝑗 )= 𝑡=1 𝑇 𝐾 𝑗𝑡 ∙ 𝛼 𝑡 𝑙 𝑡 𝑖=1 𝑇 𝛼 𝑖 𝑙 𝑖 ∙ 𝑃 𝑗𝑡 (𝑞)
Probability of originating from position q on transcript t
In the case of no bias:
𝑃 𝑗𝑡 (𝑞)= 1 𝑙 𝑡

48. Components: Fragment comes from a certain position of the transcript (I)
Occurence
More likely

49. Components: Fragment comes from a certain position of the transcript (II)
Frequency
Frequency
Not all regions are equally covered.
Frequency
Frequency

50. Search for abundances that best explain the observed fragments
The method used to find the optimum differs per program.
Trapnell et al. 2010

51. Uncertainty in expression estimate
The statistical methods can also provide an indication of the uncertainty in the expression estimates
One of the sources of that uncertainty are reads that do not map uniquely.
Occurrence
FPKM

52. Remember
The statistical methods calculate the expression level of each transcript
The gene expression can then be obtained by simply summing expression levels of its isoforms
Gene
RPKM = 11
Isoform 1
RPKM = 6
Isoform 2
RPKM = 5

53. Programs employing statistical models
Cufflinks
Genome annotation based
FPKM values
Numerical method for finding the maximum likelihood optimum
RSEM
de novo transcriptome
Counts and RPKM values
Expectation maximization for finding the optimum
BitSeq
Markov chain Monte Carlo for sampling from the posterior distribution.

54. RNA seq (III) Differential expression
12/04/2017
RNA seq (III) Differential expression
Edouard severing

55. A typical heat stress experiment
(convection)
Control
Single measurement
Many measurements
Is this gene really important ?
HSP38
Expression level
Expression level

56. RNA extraction Procedure
Sources of variation
Biological
Technical N2
Grind
RNA extraction Procedure
Sequencing
Bioinformatics
Treatment
Convection
Freezer

57. Determining expression variation
Accurately determining the variation requires many biological samples (replicates).
Unfortunately in most case we only have two or three replicates.
Other methods are needed to approximate/model the variation.

58. Determining within condition fragment count variation modeling (DESeq/cuffdiff)
Main assumption:
Variance depends on the mean.
Objective:
Find a function that best describes the relationship between the mean and variance.
Trapnell et al

59. Building the within condition fragment count distribution (DEseq)
At this stage DESeq and Cuffdiff determined for each transcript/gene
The within condition fragment count mean
The within condition fragment count variance
With these parameters a fragment count probability distribution can be determined. DESeq uses a negative binomial (NB) distribution.
NB: Variance is larger than the mean

60. Building the within condition fragment count distribution (Cuffdiff)
In addition to NB distribution of DESeq Cuffdiff also takes the uncertainty in transcripts fragment assignment into account.
The resulting count distribution is called a beta negative binomial (BNB) distribution
This count distribution is in the end transformed to a distribution of FPKM values.

61. Differential gene expression
Now that the count or FPKM distributions are known we can start testing for significant differences in expression levels.
There are several ways in which one can test whether the gene/transcript levels in two conditions are significantly different.
DESeq: uses an exact test which is analogous to the Fisher exact test.
(The hyper-geometrical is replaced by the NB distribution)
Cuffdiff: uses a t-test
(We will look at that in the next slide)

62. Testing for differential expression (Cuffdiff)
log FC
Expression
control
Heat
Many measurements leads to many fold changes
log FC distribution

63. 𝑇= 𝐸[ log 𝐹𝐶 ] 𝑉𝑎𝑟[ log 𝐹𝐶 ]
Cuffdiff p-value
According to authors of cufflinks
State that the quantity T is approximately
Normally distributed
log FC distribution
𝑇= 𝐸[ log 𝐹𝐶 ] 𝑉𝑎𝑟[ log 𝐹𝐶 ]
Unadjusted p-value of cufflinks indicates what the probability of observing
a T which is more extreme the current T
(Red areas)
-|T|
|T|

64. Samtools view
SRR Chr M * AGCGAGAGAGATCGACGGCGAAGCTCTTTACCCGCT%%"&'"(&&#'$&$)+$#,%83%0&1250'III+$' AS:i:-4 XN:i:0 XM:i:2 XO:i:0 XG:i:0 NM:i:2 MD:Z:34G0A0 YT:Z:UU XS:A:+ NH:i:1
Sw accepted_hits.bam
/mnt/geninf15/work/course_2013/day2/mapped_reads