Transmission tree reconstruction by augmentation of ...€¦ · Let Tbe a time-tree (rooted, with branch lengths in units of time). Let V be its node set of size n. Suppose the isolates

Transmission tree reconstruction by augmentation ofinternal phylogeny nodes

Matthew Hall

Li Ka Shing Institute for Health Information and Discovery, University of Oxford

February 2017

Matthew Hall (Oxford) Transmission tree reconstruction February 2017 1 / 29

The relationship of the phylogeny to the transmission tree

Let T be a time-tree (rooted, with branch lengths in units of time).

Let V be its node set of size n.

Suppose the isolates at the tips of T come from a set of H of hosts.

Initial assumptions:

Complete sampling of the epidemic since the TMRCANo superinfection or reinfectionTransmission is a complete bottleneck



The transmission tree N (aDAG whose nodes are themembers of H, depicting whichhost infected which other) canbe represented by a mapd : V → H taking each node toa host (tips to the host theywere sampled from).

Visualised by collapsing thenodes in the preimage of eachh ∈ H under d to a single node.



The transmission tree N (aDAG whose nodes are themembers of H, depicting whichhost infected which other) canbe represented by a mapd : V → H taking each node toa host (tips to the host theywere sampled from).

Visualised by collapsing thenodes in the preimage of eachh ∈ H under d to a single node. J

A

B

G

F

H C

I

E

D


The simplest version

Assume that the phylogeny andtransmission tree coincide;internal nodes are transmissionevents.

This implies no within-hostdiversity and necessitates nomore than one tip per host.

If n is internal with children nC1

and nC2, then eitherd(n) = d(nC1) ord(n) = d(nC2).

Trivially 2n−1 transmission treesfor a fixed T .


Within-host diversity

If within-host diversity isassumed then internal nodes arecoalescences of two lineageswithin a host.

The subgraph induced by thepreimage of d for any host mustbe connected.

An extra set of parameters qrepresent the infection times.

Question: How manytransmission trees for a fixed T ?(Depends on the topology.)

With one tip per host?With ≥ 1 tip per host?(Sometimes 0.)


Simultaneous MCMC reconstruction of phylogeny andtransmission tree

In either case we get an (injective but not surjective) map z from theset of possible ds to the space of transmission trees.

Thus an MCMC method that samples from the posterior distributionof phylogenies with internal node augmentation obeying either set ofrules simultaneously samples from the posterior distribution oftransmission trees.

Not only a method for reconstructing N , but a population model(tree prior) for reconstruction of T that is more realistic for anoutbreak than the standard unstructured coalescent models.


Decomposition

Let S be the sequence data and φ the various model parameters.

Without within-host diversity:

p(T , d , φ|S) =p(S |T )p(T , d |φ)p(φ)

p(S)

p(S |T ) is the standard phylogenetic likelihood and p(T , d |φ) theprobability of observing the augmented tree under a transmissionmodel.

With within-host diversity:

p(T , d , q, φ|S) =p(S |T )p(T |N , q, φ)p(N , q|φ)p(φ)

p(S)

p(N , q|φ) is the probability of the transmission tree and its timings asabove; p(T |N , q, φ) is the probability of the within-hostmini-phylogenies under a coalescent process.


MCMC implementation

Hall et al., 2015 implementedsimultaneous reconstruction ofboth trees in BEAST, withMCMC proposals that respectthe rules of node augmentation.

Several other approaches (e.g.Didelot et al., 2014, Morelli etal., 2012; Ypma et al., 2013;Klinkenberg et al., 2017) withrecent work on the incompletesampling problem (Didelot etal., 2016; Lau et al., 2016).

i)

ii)

iii)

ii)i)

iii)

iv)

50%

50%

Exchange

Subtree slide

Wilson-Balding

50%

50%

Exchange

Subtree slide

Wilson-Balding


43617 tips


The BEEHIVE study

NGS short-read sequence data acquired from samples taken fromEuropean (and one African) HIV cohort studies.

Some cohorts go back to the early epidemic in the 1980s

Current data from 3138 individuals

Epidemiology: age, gender, date of first positive test, countries oforigin and infection, risk group, ART dates, etc.

Sequences from one time point only (with a few exceptions)

Rather than making a consensus sequence from each host’s reads, wewant to use everything.


Phyloscanner: phylogenetic analysis of NGS pathogen data

sequencingmapping to references

Idea: align all short reads from all hosts to a reference genome andslide a window across the genome, building a phylogeny for the readsoverlapping each window.


Phyloscanner: phylogenetic analysis of NGS pathogen data

Identical reads from a single host are merged but the duplicate countskept as tip traits

We use RAxML for reconstruction

Tips are not associated with each other across different windows, buthosts are.


The topological signal of transmission

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●●

●

●

●●

●●

●

●

●

●●

●

●

●

●●

●

●

●

●

●

●

●

●

●●

●

●●

●

●

●●

Once we have many tips fromeach host, transmission has atopological signal.

Direct transmission is suggestedwhen the clade from theinfectee is not monophyletic(Romero-Severson et al., 2016)but in general we only see thedirection of transmission fromthe topology.

Starts to look like a parsimonyproblem.



●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●●

●

●

●●

●●

●

●

●

●●

●

●

●

●●

●

●

●

●

●

●

●

●

●●

●

●●

●

●

●●






●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●●

●

●

●●

●●

●

●

●

●●

●

●

●

●●

●

●

●

●

●

●

●

●

●●

●

●●

●

●

●●





Challenges reconstructing transmission from this data

43617 tips

Datasets:

Enormous sizeContaminationpresentCoverage is uneven

Epidemiology:

Sampling isincompleteMultiple infectionspresentBottleneck attransmission may bewide (IDUs)


Transmission tree reconstruction using parsimony

For a fixed tree, we aim to:

Reconstruct hosts from those represented in the tips to internal nodesin the tree

But also allow reconstruction to “a host outside the dataset’ asrequired by incomplete sampling

Minimise the number of infection events amongst hosts in thedataset. . .

. . . except, penalise reconstructions which suggest an unreasonableamount of genetic diversity stemming from a single infection event.

Identify multiple infections and contaminations



Suppose the T has nodes V and we are reconstructing charactersfrom the set of states S .

The cost function c(p, q; i , j) determines the cost of transitioningfrom state i to state j along the branch from p to q (in the directionaway from the root).

We take a node- and edge- dependent c on a known tree; costs fortransitions vary depending on the states involved and the branch onwhich they occur.

The lowest cost reconstruction is found with the Sankhoff algorithm.



If reads are taken from n hosts h1, . . . , hn making up the studypopulation, we use the hi as states along with an “unsampled” stateu.

Assume that the root of the tree was in the unsampled state (usingan outgroup if required). Tips from outside the study population (theoutgroup, other reference sequences, contaminants) are assigned u asa state.

We are interested in minimising the cost of infections of members ofthe study population, but not hosts outside that population, soc(p, q; h, u) = 0 for all p, q, h.

Reconstruction starts by putting a u on the root. (Otherwise it willalways be cheap to reconstruct some hi to the root, plausible or no.)



Let l(p, i) be the sum of thebranch lengths in the subtreerooted at p pruned so that onlytips from i remain, or ∞ if thereare no such tips. Then we takec(p, q; i , j) = 1 + kl(p, j) withk ∈ R+ a tunable parameter.

This penalises reconstructionswith unreasonable amounts ofwithin-host diversity (right).

●

●

●

●

●

●

●

●

●

●

r r

o o

p p

qq

s s

hu

l1

l2

Left: C = 1 + k(l1 + l2)Right: C = 2



●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●

●

●

●

●

●

●

●

●

c(p, q; h, u) = c(p, q, h, u) for all p, q, h. Above trees have equal C .Choose to break ties always towards h (for dense sampling), alwaystowards u (conservative), or based on branch lengths.


Parsimony for detection of contaminants

Small numbers of contaminant reads are frequent, where a virus fromthe wrong host has been sequenced

The parsimony reconstruction does double duty to detect these

Find “multiple introductions” where the tips in one split have verylow read counts, and ignore those tips


Without the assumptions of asingle infection event per hostand complete sampling, the“transmission tree” has multiplenodes per host and unsamplednodes.

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

AB1

B2●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

B

Augmented phylogeny "Transmission tree"

UA


Example

Known HIV transmission chain(Lemey et al., 2005, Vranckenet al., 2014).

Reconstruction on RAxML treesfor env and pol genes

Host True env polA B? U UB A? U AC B B BD C D DE C C CF A U AG F U FH B B BI B B BK E E EL C C C

I

E

F

B A

GL D

C

K

H

?


Full phyloscanner results

Full output from phyloscanner is a separate phylogeny for the reads ineach genome window

We would like to use these in a manner similar to bootstrapping, toindicate support for topological relationships

The phylogenies are not trivially comparable across windows as thetips are not the same

The hosts are the same, so the trasmission trees are morecomparable, but:

Some hosts are absent from some windows due to sequencing problemsOne or more nodes for unsampled regionsPotentially multiple nodes per host

Question: Can we nonetheless give a summary or mediantransmission tree?


Classification of relationships

For the time being, concentrate on classifying pairwise relationshipsbetween hosts on each window

Contiguity: is the subgraph induced by the nodes from the pair, withperhaps some unsampled nodes, connected?

Descent: Are all the nodes from one patient ancestral to those fromthe other?

Mean patristic distance in the phylogeny

Then count relationships across windows


Improved HIV cluster detection

By setting a threshold on patristic distance we can refine the proceduresfor identifying HIV clusters with likely directionality.

5

7

15

13

2

3

311

12

1

BEE0221-1

BEE0239-1

BEE0220-1

BEE0260-1


Improved HIV cluster detection

7

6

5

12

2

1

7

7

9

7

1

3

23

6

2

8

9

9

5

3

8

1

42

15

5

3

11

6 6

1

20

24

6

13

1

11

3

11

6

7

1

1

7

2

6

2

1

2

16

62

4

5

6

3

1

7

3

13

99

3

6

6

1

7

2

1

1

12

8

7

3

6

42

22

1

73

8

14

3

69

10

1

1

6

6

9

14

1

23

2

11

3

7

5

3

3

6

4

18

2

9

3

2

2

4

512

5

6

2

3

3

63

4

7

1

5

10

2

5

2

7

83

11

7

2

51

9

10

6

15

6

3

4

6

15

1

2

10

5

1

3

11

4

3

13

7

2

13

9

6

1

9

2

2

27

20

1

116

2

4

6

7

1

2

1

6

2

7

1

4

6

4

10

14

6

1

7 10

1

8

1

14

6

13

7

12

3

2

1

7

16

4

3

8

10

216

1

1

65

1

1 115

14

6

6

13

7

1

8

9

11

6

2

2

6

3

2

5

6

57

2

5

21

4

3

5

4 5

36

3

11

7

3

5

7

1

14

12

7

7

18

3

2

8

3

12

6

44

2

3

3

4

2

10

3

5

5

1

2

10

1

1

7

2

3

12

4

3

10

218

2

2

2

4

1

1

14

2

9

8

215

2

4

1

4

54

5

2

4

4

11 12

2

15

11

1

7

6

5

4

12

13

4

10

1

7

7

24

8

1

1

2

4

4

5

3

2

4

3

62

15

4

1

4

2

18

9

3

1 4

13

9

7

12

7

1

2

6

1

1 19

8

5

2

14

1 10

1

3

6

14

4

3

6

711

1

13

1

8

BEE0795-1

BEE0254-1

BEE0248-1

BEE0726-1

BEE0093-1

BEE0015-1

BEE0201-1

BEE0197-1

BEE0095-1

BEE0098-1

BEE0110-1

BEE0577-1

BEE0327-1

BEE0442-1

BEE0323-1

BEE0394-1

BEE0037-1

BEE0036-1

BEE0198-1

BEE0843-1

BEE0840-1

BEE0240-1

BEE0826-1

BEE0728-1

BEE0032-1

BEE0175-1

BEE0009-1

BEE0215-1BEE0804-1

BEE0527-1

BEE0564-1

BEE0738-1

BEE0023-1

BEE0112-1BEE0351-1

BEE0321-1

BEE0427-1

BEE0331-1

BEE0426-1

BEE0809-1

BEE0109-1

BEE0102-1

BEE0205-1BEE0100-1

BEE0259-1BEE0042-1BEE0529-1

BEE0234-1

BEE0096-1

BEE0241-1

BEE0243-1BEE0193-1

BEE0200-1 BEE0146-1

BEE0117-1

BEE0174-1

BEE0735-1

BEE0043-1

BEE0062-1 BEE0115-1

BEE0018-1

BEE0347-1

BEE0567-1

BEE0122-1

BEE0542-1

BEE0541-1BEE0019-1

BEE0141-1BEE0803-1

BEE0260-1 BEE0791-1

BEE0237-1

BEE0592-1BEE0181-1

BEE0017-1 BEE0758-1

BEE0139-1

BEE0057-1

BEE0107-1

BEE0014-1 BEE0767-1

BEE0114-1

BEE0108-1

BEE0345-1

BEE0206-1

BEE0395-1

BEE0799-1

BEE0368-1

BEE0223-1BEE0012-1

BEE0464-1BEE0474-1

BEE0341-1

BEE0160-1

BEE0161-1

BEE0164-1

BEE0573-1

BEE0154-1

BEE0454-1

BEE0532-1

BEE0130-1

BEE0736-1 BEE0807-1

BEE0290-1

BEE0322-1

BEE0470-1

BEE0213-1

BEE0662-1

BEE0305-1

BEE0031-1

BEE0534-1

BEE0273-1

BEE0555-1

BEE0794-1 BEE0808-1

BEE0131-1

BEE0092-1

BEE0222-1

BEE0545-1 BEE0162-1

BEE0144-1BEE0572-1BEE0877-1 BEE0065-1BEE0231-1BEE0097-1

BEE0121-1

BEE0465-1

BEE0780-1

BEE0398-1

BEE0539-1

BEE0797-1

BEE0385-1

BEE0473-1

BEE0221-1BEE0050-1

BEE0220-1

BEE0560-1

BEE0734-1

BEE0040-1

BEE0258-1

BEE0284-1

BEE0849-1

BEE0867-1

BEE0053-1

BEE0301-1

BEE0353-1

BEE0696-1

BEE0789-1 BEE0709-1

BEE0168-1

BEE0547-1 BEE0723-1BEE0148-1

BEE0574-1 BEE0016-1BEE0851-1BEE0230-1BEE0225-1

BEE0725-1

BEE0182-1

BEE0178-1

BEE0552-1

BEE0669-1

BEE0059-1

BEE0710-1

BEE0657-1

BEE0219-1

BEE0285-1

BEE0330-1

BEE0798-1

BEE0120-1 BEE0214-1 BEE0584-1BEE0538-1BEE0733-1

BEE0686-1

BEE0034-1

BEE0543-1BEE0557-1

BEE0253-1

BEE0147-1

BEE0335-1

BEE0318-1

BEE0599-1 BEE0263-1

BEE0049-1

BEE0247-1

BEE0255-1

BEE0233-1

BEE0406-1

BEE0428-1

BEE0086-1

BEE0085-1

BEE0886-1

BEE0381-1

BEE0865-1

BEE0854-1BEE0212-1

BEE0885-1

BEE0563-1

BEE0287-1

BEE0386-1

BEE0409-1

BEE0437-1

BEE0821-1

BEE0741-1

BEE0815-1 BEE0811-1

BEE0196-1

BEE0142-1 BEE0227-1BEE0153-1 BEE0265-1


Conclusions

The transmission tree can be viewed as an augmentation of theinternal nodes of a phylogeny with host information, subject tovarious sets of rules.

Considerable recent work in reconstructing it for smaller datasets,usually using MCMC.

Phyloscanner allows reconstruction with big, NGS datasets.

Work remains to be done for rigorous treatment of the output.



Acknowledgements

Oxford

Christophe Fraser

Chris Wymant

Imperial

Oliver Ratmann

Edinburgh

Andrew Rambaut

Mark Woolhouse


Documents

Transmission tree reconstruction by augmentation of ...€¦ · Let Tbe a time-tree (rooted, with branch lengths in units of time). Let V be its node set of size n. Suppose the isolates