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

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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

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