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Lecture 9 Phylogenetic Prediction. Bioinformatics. Dr. Aladdin HamwiehKhalid Al- shamaa Abdulqader Jighly. Aleppo University Faculty of technical engineering Department of Biotechnology. 2010-2011. Phylogenetic Trees and Dissimilarity estimation. Historical Note. - PowerPoint PPT Presentation
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Bioinformatics
Dr. Aladdin Hamwieh Khalid Al-shamaaAbdulqader Jighly
2010-2011
Lecture 9Phylogenetic Prediction
Aleppo UniversityFaculty of technical engineeringDepartment of Biotechnology
Phylogenetic Trees and Dissimilarity estimation
3
Historical Note• Until mid 1950’s phylogenies were constructed
by experts based on their opinion (subjective criteria)
• Since then, focus on objective criteria for constructing phylogenetic trees– Thousands of articles in the last decades
• Important for many aspects of biology– Classification – Understanding biological mechanisms
Morphological vs. Molecular
• Classical phylogenetic analysis: morphological features: number of legs, lengths of legs, etc.
• Modern biological methods allow to use molecular features– Gene sequences– Protein sequences– DNA markers
5
Rat QEPGGLVVPPTDA
Rabbit QEPGGMVVPPTDA
Gorilla QEPGGLVVPPTDA
Cat REPGGLVVPPTEG
From sequences to a phylogenetic tree
There are many possible types of sequences to use (e.g. Mitochondrial vs Nuclear proteins).
.
Basic Assumptions Closer related organisms have more similar genomes.
Highly similar genes are homologous (have the same ancestor).
Phylogenetic relation can be expressed by a dendrogram (a “tree”) .
Aardvark Bison Chimp Dog Elephant
Dangers in Molecular Phylogenies
• We have to emphasize that gene/protein sequence can be homologous for several different reasons:
• Orthologs -- are genes in different species that have evolved from a common ancestral gene via speciation.
• Paralogs -- sequences diverged after a duplication event
• Xenologs -- sequences diverged after a horizontal transfer (e.g., by virus)
8
Species Phylogeny
Gene Phylogenies
Speciation events
Gene Duplication
1A 2A 3A 3B 2B 1B
Phylogenies can be constructed to describe evolution genes.
Three species termed 1,2,3.Two paralog genes A and B.
9
Types of Trees
A natural model to consider is that of rooted trees Common
Ancestor
10
Types of treesUnrooted tree represents the same phylogeny without the
root node
Depending on the model, data from current day species does not distinguish between different placements of the root.
11
Distance-Based MethodInput: distance matrix between species
For two sequences si and sj , perform a pairwise (global) alignment. Let f = the fraction of sites with different residues. Then
Outline:• Cluster species together• Initially clusters are singletons• At each iteration combine two “closest” clusters
to get a new one
3 4log(1 )4 3ijd f (Jukes-Cantor Model)
Human, Chimp, Gorilla, Orangutan, and Gibbon
UPGMA
Taxa 1 2 3 4 5 6 7OTU-1 T G C G T A TOTU-2 T G G G T A TOTU-3 T G C G C T TOTU-4 T G C T G T GOTU-5 T A G T A G C
Step 1: Generate data (Sequence/ Genotype/ Morphological) for each OTU.
Distance can be calculated by using different substitution models:1. # of nucleotide differences.2. p-distance.3. JC distance4. K2P distance.5. F816. HKY857.GTR etc
Step 2: Calculate p- distance for all pairs of taxa
Taxa 1 2 3 4 5 6 7OTU-1 T G C G T A TOTU-2 T G G G T A T
= 0.142857143
Step 3: Calculate distance matrix for all pairs of taxa and select pair of taxa with minimum distance as new OTU.
Taxa 1 2 3 4 51 0 1 2 4 62 0.1428 0 3 5 53 0.2857 0.4285 0 3 64 0.5714 0.7142 0.4285 0 55 0.8571 0.7142 0.85710.7142 0
OTU-1OTU-2
0.0714
0.0714
Step 4: Recalculate new distance matrix, assuming OTU-1 and OTU-2 as one OTU.
= 0.3571
taxa 1+2 3 4 51+2 0
3 0.35714 0 4 0.64285 0.4285 0 5 0.78571 0.8571 0.7142 0
Taxa 1 2 3 4 51 0 2 0.1428 0 3 0.2857 0.4285 0 4 0.5714 0.7142 0.4285 0 5 0.8571 0.7142 0.8571 0.7142 0
Step 5: Select pair of taxa with minimum distance as new OTU.
OTU-1
OTU-2
0.071
0.071
OTU-30.179
0.107
0.107 + 0.071 + 0.179 = 0.357
Step 6: Again select pair of OTU with minimum distance as new OTU and recalculate distance matrix.
= 0.5714
taxa (1+2)3 4 5(1+2)3 0
4 0.5714 0 5 0.8095 0.7142 0
taxa 1+2 3 4 51+2 0
3 0.35714 0 4 0.64285 0.4285 0 5 0.78571 0.8571 0.7142 0
Taxa 1 2 3 4 51 0 2 0.1428 0 3 0.2857 0.4285 0 4 0.5714 0.7142 0.4285 0 5 0.8571 0.7142 0.8571 0.7142 0
Step 7: Again select pair of taxa with minimum distance as new OTU.
OTU-2
OTU-10.071
0.071
OTU-30.179
0.107
OTU-40.286
0.107
0.107 + 0.107 + 0.071 + 0.286 = 0.571
Step 8: Again select pair of OTU with minimum distance as new OTU and recalculate distance matrix.
= 0.7857
taxa ((1+2)3)4 5
((1+2)3)4 0
5 0.7857 0
taxa (1+2)3 4 5(1+2)3 0
4 0.5714 0 5 0.8095 0.7142 0
taxa 1+2 3 4 51+2 0
3 0.35714 0 4 0.64285 0.4285 0 5 0.78571 0.8571 0.7142 0
Taxa 1 2 3 4 51 0 2 0.1428 0 3 0.2857 0.4285 0 4 0.5714 0.7142 0.4285 0 5 0.8571 0.7142 0.8571 0.7142 0
Step 9: Again select pair of OTU with minimum distance as new OTU and make final rooted tree.
OTU-2
OTU-10.071
0.071
OTU-30.179
0.107
OTU-40.286
0.107
OTU-50.393
0.107
0.393 + 0.107 + 0.107 + 0.107 + 0.071 = 0.785
Jukes-Cantor distancethe rate of nucleotide substitution is the same for all pairs of the four nucleotides A, T, C, and G A A
A CA GA TC AC CC GC TG AG CG GG TT AT CT GT T
25% similar (= distance of 0.75). 75% which is what you expect with random assignment of nucleotides to a pair of taxa
طريقة الوراثية UPGMAتفترض القرابة شجرة أفرع طول في ثابتة نسبة
=-(3/4)*LN(1-(((4/3)*0.1594)))
Neighbor-joiningطريقة • - أفرع طول في ثابتة نسبة استخدام على مارغولياش فيتش طريقة تعHتمد ال
طريقة في هي كما الوراثية القرابة UPGMAشجرةبأقل • المدروسة للوحدات أزواج أقرب تحديد على تعتمد الطريقة هذه
( . المقارب الزوج تعريف ويمكن لألفرع قيمة( Pair of neighborاألطوال بأنهجذرية ) غير بعقدة وحدتين بين (.unrooted nodeاالرتباط
والغوريال: • األنسان عكس على وحدة في متحدان والشيمبانزي اإلنسان مثال ) الغHوريال، ) مع تجاور على والشيمبانزي اإلنسان األولى الوحدة ندعو وعليه
باقي مع القرابة عن نبحث والغHوريال األولى الوحدة بين القرابة دراسة وبعد. المدروس المجتمع أفراد
• : نبدأ مدروسة أفراد ثمانية لدراسة مثالمرتبطون جميعا أنهم لو كما المقارنة
إثبات وعند بعدها واحدة، بعقدةبين على 2و 1االرتباط الشجرة تصبح
Neighbor-joiningطريقة
Neighbor-joiningطريقة
A B C D E rA (human) — 0.015 0.045 0.143 0.198 0.4010B (chimp) — 0.03 0.126 0.179 0.3500C (gorilla) — 0.092 0.179 0.3460D (orangutan) — 0.179 0.5400E (gibbon) — 0.7350
Neighbor-joiningطريقة
A B C D EA (human) — 0.0150 0.0450 0.1430 0.1980B (chimp) -0.3605 — 0.0300 0.1260 0.1790C (gorilla) -0.3285 -0.3180 — 0.0920 0.1790D (orangutan) -0.3275 -0.3190 -0.3510 — 0.1790E (gibbon) -0.3700 -0.3635 -0.3615 -0.4585 —
A:B = 0.015-(0.4010+0.35)/2
Example:
A B C D E r r/3A (human) — 0.015 0.045 0.143 0.198 0.4010 0.1337B (chimp) — 0.03 0.126 0.179 0.3500 0.1167C (gorilla) — 0.092 0.179 0.3460 0.1153D (orangutan) — 0.179 0.5400 0.1800E (gibbon) — 0.7350 0.2450
=0.179/2+(0.18-0.245)/2
=0.179-0.057
Human and chimpanzee have the smallest value of Mij and they are replaced by node 2.
dijMij
• PHYLIP (Phylogeny Inference Package)
a = 0.016
3
2
1
b = -0.001
c = 0.006
d = 0.057
e = 0.1221'= 0.0403
2'= 0.024
E
D
A
B
C• UPGMA
• Neighbor-joining (NJ)
Genetic distanceMarker1 Marker2 Marker3 Marker4 Marker5 Marker6 Marker7
Plant1 1 0 1 1 0 1 1
Plant2 1 1 1 0 0 1 0
Plant11 0
Plant2 1 Fa=3 Fb=10 Fc=2 Fd=1
N= Fa+Fb+Fc+Fd
Simple Match distance = Fa/N= 3/7= 0.43Genetic distance (Jaccard) = Fa/(Fa+Fb+Fc) = 3/6= 0.5
Dissimilarity indices – Continuous
Euclidean distance
Euclidean Distance is the most common use of distance. In most cases when people said about distance , they will refer to Euclidean distance. Euclidean distance or simply 'distance' examines the root of square differences between coordinates of a pair of objects.
Example:Point A has coordinate (0, 3, 4, 5) and point B has coordinate (7, 6, 3, -1). The Euclidean Distance between point A and B is
Features
cost time weight incentive Plant A 0 3 4 5 Plant B 7 6 3 -1
Dissimilarity indices – Continuous
Euclidean distance
Manhattan (City-Block)It is also known as Manhattan distance, boxcar distance, absolute value distance. It examines the absolute differences between coordinates of a pair of objects.
Features cost time weight incentive
Plant A 0 3 4 5 Plant B 7 6 3 -1
Thank you
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