Title: Mapping Mutations in HIV RNA
1Mapping Mutations in HIV RNA
- By Nimrod Bar-Yaakov nimrod-b_at_orbotech.com
- With co-operation of Dr. Zehava Grossman of the
Israels Multi-Center AIDS Study Group, National
HIV reference Laboratory in Tel-Hashomer.
2Todays Topics
- HIV What is it and how it operates.
- What so important about the HIV DNA mutations?
- Extracting the RNA sequence for analyze.
- Naïve view of the HIV RNA sequences
- Locating the RNA mutations
- Analysis of the RNA mutation interactions
3Virus Overview
- Viruses may be defined as acellular organisms
whose genomes consist of nucleic acid, and which
obligately replicate inside host cells using host
metabolic machinery and ribosomes to form a pool
of components which assemble into particles
called VIRIONS, which serve to protect the genome
and to transfer it to other cells.
4Virus Animation
5Virus Overview
- The concept of a virus as an organism challenges
the way we define life - viruses do not respire,
- nor do they display irritability
- they do not move
- and nor do they grow,
- however, they do most certainly reproduce, and
may adapt to new hosts.
6What is an HIV
- human immunodeficiency virus, A type of
retrovirus that is responsible for the fatal
illness Acquired Immunodeficiency Syndrome (AIDS) - Retrovirus A virus that's carry their genetic
material in the form of RNA rather than DNA and
have the enzyme reverse transcriptase that can
transcribe it into DNA. - In most animals and plants, DNA is usually made
into RNA, hence "retro" is used to indicate the
opposite direction
7How does the HIV infects the body cells?
- HIV begins its infection of a susceptible host
cell by binding to the CD4 receptor on the host
cell - The genetic material of the virus, which is RNA,
is released and undergoes reverse transcription
into DNA, which enters the host cell nucleus
where it can be integrated into the genetic
material of the cell. - Activation of the host cells results in the
transcription of viral DNA into messenger RNA
(mRNA), which is then translated into viral
proteins. - The viral RNA and viral proteins assemble at the
cell membrane into a new virus. - The virus then buds forth from the cell and is
released to infect another cell.
8Treatment related to the active RNA sites
- The HIV DNA generates proteins that are essential
to the virus life-cycle. - Medical treatment interfere or block the
operation of these proteins. - Reverse Transcriptase medicines
- Inhibits the transcription of the HIV RNA into
the cells DNA - The HIV protease protein, is required to process
other HIV proteins into their functional forms. - Protease inhibitors medicines, act by blocking
this critical maturation step. -
9RNA mutations
- Environmental/Biological processes may cause
mutations in the HIV RNA. - The mutated HIV RNA merge into the infected
cells DNA. - The generated Amino-Acids sequence is then
altered. - A different Protein is generated by the cell.
- The altered protein may resist the medical
treatment!
10Mutation families
- The HIV RNA has a high mutation rate (a 1000
times more than a regular cell). - Fast evolutionary processes causes the best
mutated viruses to increase their population in
the infected body. - Well focus on 3 main mutation families
- Resistance mutations
- Clade mutations
- Other noise/random
11The importance of identifying the resistance
mutations
- Selecting the best medicine treatments
- Understanding the way different medicines
interacts with the HIV - Understanding the functional interpretation of
the RNA sequence
12Extracting the RNA Sequence
- The RNA sequences are transcript into DNA
sequences. - The DNA sequences then multiplied several times
- A DNA sequencer read the aligned DNA sequences.
- The decision how to interpret a specific DNA
segment is based over image processing algorithms
(define the segment boundaries and find the best
match for the segment pattern) and isnt
deterministic!
13Sequence Alignment (from Ron Shamirs Course)
14(No Transcript)
15Sequence Alignment
- Before alignment
- AtaaagakagggggacagctaaaagaggctctcTTAGACACAGGAGCAGA
TGATACA - ACTCTTTGGCAGCGaCCCCGTTGTCACaATAAAAATagGGGGACAGCTAA
gGGagGc - TAAAAGAGGCTCTCTTAGCACACAGGMGCAGAYGAYACAGTMCTTASCAA
GAAATAA - ACTCTTTGGCAGCGACCCCTTGTcACAATAAAAGTAGAGGGACAGCTAAG
GGAKGCT - ACTCTTTGGCAGCGaCCCCTTGTCACAATAAAAATAGGGGACAGCTAAGG
GAGGCTC - ACTCTTTGGcAGCGACCCCTtGTCACAATAAAAGtAGGGGGaCAGCTAAA
gGAGGCT - aCTnTTnGRCAGCGaCCCCTTgTCYCARtAAAAATAGGGGGGCAGRTAAR
GGAGGCt - After Alignment
- ------------------------------ATAAAGAKAGGGGG-ACAG-
CTAAAAGAGG - ------------C-GACCCC--TTGTCACAATAARAATAGGGGG-ACAG-
CTAAAAGAGG - ACTCTTTGGCAAC-GACCCC--TTGTCACAATAAGAGTAGGGGG-ACAG-
CTAAAAGAGG - -CTCTTTGGCAAC-GA-CCCC-TTGTCACAGTAAAAATAGRAGG-ACAG-
CTAAAAGAAG - ACTCTTTGGCAAC-GA-CCCC-TTGTCACAGTAAAAATAGGAGG-ACAG-
CTAAAAGAAG - ACTCTTTGGCAAC-GA-CCCC-TTGTCACAGTAAAAATAGGAGG-ACAG-
CTMAAAGAAG - ACTCTTTGGCAAC-GA-CCCC-TTGTCACAGTAAGAATAGGAGG-ACAG-
CTAAAAGAAG - Degapping
16Reduction from Bio problem to CS Problem
- Generation of a consensus RNA sequence.
- For each sequence, generate a matching binary
sequence, each 1 represents a mismatch between
the consensus and the original sequence, and 0
represents a match. - Now we have a binary feature vector for each
sample. - We can now calculate the correlations between the
mutations to the treatment and between the
mutations to themselves.
17From Sequences to Mutation Matrix
18So where are the problems?
- Curse of dimensionality
- Noisy data
- Sequenced data are of stochastic nature
- Small number of samples
- Clades and sub-clades
- Vague definitions of independent variables
values. - Silent mutations
- Talk Bio language!
19Data Overlook
20Frequencies of Mutations occurrences
21Filtering the Data
- Mutations that occur less than 5 times in a
specific RNA index cannot considered significant
(well see it later in the Chi square slides) - Well filter all the mutations that occur less
than 3 times and replace them with the consensus
value. - Thus filtering much of the noise.
22Naïve clustering of Data
Clustering of 671 RNA samples using Centroid
linkage
Total Cases
A C B
Clade Distribution
Treated Non-Tr
Treatment Distribution
120 9 12 59 8
29 215 65 147 7
Cluster Size
671
23Feature Extraction
- Better to have misdetection than a false alarm.
- Filter the noisy data
- Work within the clades
- Locate the mutations (features) that are highly
correlate with treatment. - Now we have only few dozens of features to work
on.
24Finding mutations and treatment correlation
- We want to find for each RNA index i whether
P(Mut_in_i) is significantly different from
P(Mut_in_ i/ Treatment). - Well use the CHI square distribution test for
each index to find that.
25Chi Square Overview
- We will use the Chi-Square test to check the
probability that our observed results had came
from the same statistical population as the
expected (chance) results. - A probability of less than 0.05 means that the
results are significant, I.e the populations are
significally different .
26Chi Square Calculations
- Calculating the chi-square statistic
- The probability Q that a X2 value calculated for
an experiment with d degrees of freedom (where
dk-1) is due to chance is
27Example Mutation V82A
28Mutation Table
29Calculating the mutations Correlations Matrix
- Because the treatment is a major artifact in all
the treatment mutations, well have to find the
correlations within the treated samples - P(mut_A/Treat.) P(mut_A/mut_b,Treat.)
- Our Chi-Square table will be (all in treated
cases)
Mut B Non -Mut B Total
Mut A
Non Mut A
Total
30Example correlation results
31Example Mutation D30N
- D30N is an important resistance mutation. But it
appears at frequency of 0.0258 in the C clade
compare with 0.0945 in the B clade, Whats the
explanation for this? - Correlation analysis reveals that in clade B,
D30N is highly correlated with other resistance
Mutations. In clade C its not. - One assumption can be that the Clade B structure
can influence the connections between resistance
mutations.
32Using CART to find mutations interactions
- A regression tree is a sequence of questions that
can be answered as yes or no, plus a set of
fitted response values. Each question asks
whether a predictor satisfies a given condition. - In our research we will ask whether a mutation i
(1 value at i index), predicts the existence of
mutation j (1 value at j index). - This way we can identify relationships between
the mutations.
33CART results D95M
34Using clustering to find mutations patterns
- Well cluster the mutation sample vectors in
order to locate mutation patterns. - Our distance function will be the sum of
differences between two samples. - Well use the ward method to cluster nodes.
35Ward Clustering
- Centroid linkage uses the distance between the
centroids of the two groups - Where and Xs defined
similarly. - Ward linkage uses the incremental sum of squares
that is, the increase in the total - within-group sum of squares as a result of
joining groups r and s. It is given by - Where drs is the distance between cluster r and
cluster s defined in the Centroid linkage. The - within-group sum of squares of a cluster is
defined as the sum of the squares of the distance
between all objects in the cluster and the
centroid of the cluster.
36Cluster results
37Using clustering to find mutations patterns
- When we filter the mutation only to significant
ones, we can see mutations pattern as a result of
clustering -
Samples
Mutations
38Whats next?
- Biological interpretation of the findings
- Locating Amino-Acid and protein functional
changes. May lead to better understand of
resistance behavior. - Identifying new resistance mutations and specific
treatment/resistance correlations. - Focus on specific treatments, apply additional
research in order to investigate the efficiency
of such treatment.
39The End!