Evolution, Trees and HIV

Collection date

Cum

ulat

ive

bran

ch le

ngth Trend line

Random biological fact of the day: Galton and regression

https://commons.wikimedia.org/wiki/File:Francis_Galton_1850s.jpghttps://commons.wikimedia.org/wiki/File:Sweet_Pea-01.jpghttps://commons.wikimedia.org/wiki/File:Sweet_pea_(Lathyrus_odoratus)_seeds.jpg

Mother pea diameter

Daug

hter

pea

dia

met

er

Where we’ve been

• Part 1: mechanisms• 4 forces• Pop gen model (simulator)• Sequence change over longer time scales:

substitutions

• Part 2: methods• Calculating distances (Jukes-Cantor)• Reconstructing trees (Neighbor joining)

• Unrooted trees• Inferring host switch events

• Inferences about selection (Ka/Ks)• Cumulative br len vs. collection date plots

https://commons.wikimedia.org/wiki/File:HIV-budding-Color.jpg

Themes• Evolution as organizing principle• Sequence information has many uses• Molecular shape → molecular function• Importance of stochastic processes

♬

Advice for studying…

• Review problems• Written homework• Coding homework• Recitation problems• Activities from lecture

• Go through lecture slides• Spend time connecting the pieces

Topics for today

• Revisiting the homework (a bit)• Some notes on methods we’ve talked about• Questions?

Homework 2: population size, drift and selection

Homework 2: genetic variation, drift and selection

Mutation vs. substitution rate

GGG

GGG

GGG

GGG

GGG GGG

1 replication

GGG

GGG

GGG

GGG

GGG

GGG GGG

Population size is 4, so we have 4 replications per generation

TGG

GGG

GGG

GGG

! mutationssite∗ replication ∗ 0 replicationsgeneration

10substitutionsmutation*Substitution

rate=

Expected number of substitutions per site per generation

=

Homework 3: mutation vs. substitution rate

! mutationssite∗ replication ∗ 0 replicationsgeneration

10substitutionsmutation*Substitution

rate=

Expected number of substitutions per site per generation

=

Case D (no selection) Case E (selection)

>>> mutProb=0.0003>>> printSummary(subL,".6f")0 0.000299 (0.000266-0.000332)1 0.000266 (0.000235-0.000297)2 0.000289 (0.000255-0.000323)>>> mutProb=0.0006>>> printSummary(subL,".6f")0 0.000575 (0.000531-0.000619)1 0.000585 (0.000539-0.000631)2 0.000630 (0.000578-0.000682)

>>> mutProb=0.0003>>> printSummary(subL,".6f")0 0.000050 (0.000039-0.000060)1 0.000052 (0.000040-0.000063)2 0.000293 (0.000268-0.000317)>>> mutProb=0.0006>>> printSummary(subL,".6f")0 0.000112 (0.000095-0.000130)1 0.000101 (0.000085-0.000117)2 0.000583 (0.000551-0.000614)

Topics for today

• Revisiting the homework (a bit)• Some notes on methods we’ve talked about• Questions?

Methods for reconstructing a tree: how they fit together

TAATTCATGAGAAAGATATTAGTA------GCAGGAAT

TAATTCATGAGAAAGATAT

TAGTAGCAGGAAT

Sequence samples

Align sequences

Calculate proportion of sites that are different

613

Jukes-Cantor correction to get substitutions per site

$ = −34 ln 1 − 43* = 0.717Fill in

entry in

distance matrix

Repeat until we have distances for all pairs of sequences we want to look at

Neighbor-joining

Jukes-Cantor: two applications we’ve seen

TAATTCATGAGAAAGATATTAGTA------GCAGGAAT

Calculate proportion of sites that are different

613

Jukes-Cantor correction to get substitutions per site

$ = −34 ln 1 − 43* = 0.717

Ancestor: GTA

CTA (aa)

GCA (aa)

TTA (aa)

GTT (syn)

GTG (syn)GTA

GTA

GTA

Calculate average proportion of sites that are different

$. = −34 ln 1 − 43 ∗316 = 0.216

Jukes-Cantor correction to get substitutions per site

316

Ka/Ks

“Industrial strength” approaches• More realistic trees• Ancestor seq not

required

Ancestor

Our approach• Ancestor seq• Star phylogeny

Topics for today

• Revisiting the homework (a bit)• Some notes on methods we’ve talked about• Questions?