molecular clock i. evolutionary rate xuhua xia [email protected]

22
Molecular Clock I. Evolutionary rate Xuhua Xia [email protected] http://dambe.bio.uottawa.ca

Post on 18-Dec-2015

216 views

Category:

Documents


2 download

TRANSCRIPT

Page 1: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Molecular ClockI. Evolutionary rate

Xuhua Xia

[email protected]

http://dambe.bio.uottawa.ca

Page 2: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 2

Within given gene (or DNA region), mutations (nt or aa sub) accumulateat an approximately equal rate in all evolutionary lineages

Rate constancy concept

Originally based on comparisons of protein sequences for hemoglobin, cytochrome c… from different organisms

Information can be used to estimate divergence times, reconstruct phylogenies…

BUT… Does it hold for all genes, all genomes… ?

How to reconcile with irregular rate of morphological evolution?

Molecular clock hypothesis

Page 3: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 3

Combined data for hemoglobins, cytochrome c & fibrinopeptide

Fig. 4.15

Clock-like substitution rate

Page 4: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 4

To compare rates in lineages A and B, use C as reference species

If constant rate, then “distance” from outgroup to each memberwithin group should be equal

KAC = KOA + KOC (1)

KBC = KOB + KOC (2)

KAB = KOA + KOB (3)

So KOA = (KAC + KAB - KBC ) / 2

KOB = (KAB + KBC - KAC ) / 2

KOC = (KAC + KBC - KAB ) / 2

Fig. 4.16

Relative-rate test

Page 5: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 5

Then according to molecular clock hypothesis:

KOA = KOB so KOA – KOB = 0

and from equations (1) and (2)

KOA – KOB = KAC – KBC

Can compare rates of substitution in lineages A and B directly from KAC and KBC

Fig. 4.16

Relative-rate test

Page 6: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 6

A B CA B C

Equal rates in lineagesleading to A and B

Slower rate in B lineage

KAC = KBCKAC > KBC

Rate difference

Page 7: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 7

Relative-rate test

Critical assumption: KAC, KBC and KAB are estimated without error.

K’AC = KOA + KOC + AC

K’BC = KOB + KOC + BC

K’AB = KOA + KOB + AB

KOA – KOB = K’AC – K’BC + AC - BC

A C

B

O

Page 8: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 8

How do you interpret the data shown in this table?

Page 9: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 9

Nr = number aa positions where human vs. rat different

but human vs. chicken identicalso replacement in rodent lineage Nr = 600

Nh = number of aa positions where human vs. rat different

but rat vs. chicken identical So replacement in human lineage Nh = 416

How do you interpret these data?

Sub. rates between rodent and human

Page 10: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 10

“They used amino acid sequences instead of DNA, because the chicken and mammalian lineages diverged about 300 million years ago…

… so it’d be difficult to obtain reliable estimates of divergence at synonymous sites.”

p.149

When to use AA sequence?

Page 11: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 11

Beta hemoglobin gene cluster

Adult:

22(HbA)

22(HbA-2)

Fatal:

21 (HbF1)

22 (HbF2)

Embryonic:

22 (Hb Gower I)

22 (Hb Gower II)

Page 12: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 12

Can use duplicated genes to test if rates are constant (Table 4.13)

How do you interpret the data in Table 4.13 ?

Cautionary note: there may be gene conversion events (“copy correction”) between sequences in multi-gene families

Page 13: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 13

• Mutation rates– Generation time

– Metabolic rate (e.g., high aerobic respiration leads to mutagenic effects of oxygen free radicals)

– DNA repair

• Purifying selection or positive selection• Different genetic background

Causes of rate differences (p.152)

Page 14: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 14

Martin PNAS 1993

Sub. rate, generation time, metabolic rate

Page 15: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 15

For mammalian mitochondrial genes,

Ks ~ 5.7 x 10 -8 sub/ site/ year

~ 10 x higher than for mammalian nuclear genes

Mitochondrial DNA used extensively in taxonomic, forensic, conservation biology,… studies

But.. in plants, mitochondrial nt sub rate very slow…

Rate difference between nuc and mt DNA

Page 16: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 16

An odd pattern in plants

Page 17: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 17

Positive selection?

Tree based on growth hormone genes, with branch length proportional to the number of nucleotide substitutions

Fig. 4.19

Phase Rate of AA replacement KA/Ks

Slow phase 0.30.1 0.03

Ruminant rapid phase 0.30

Primate rapid phase 0.49

Page 18: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 18

Fig. 4.18

Extant organisms

Ancestor

Lineage which has accumulated fewer substitutions, has retained more “primitive” ancestral state

But not necessarily any correlation between “primitive” appearance (morphological state) and amount of molecular change

“Primitive” vs. “advanced” (p.153)

Page 19: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 19

- very rapid rate of evolution (Table 4.17)

HIV retrovirus ~ 10 6 x higher than mammalian nuclear genes

- error prone reverse transcription (RT)

- sequences may be useful in retracing spread through population

RNA viruses and retroviruses

Page 20: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 20

- HIV virions harvested (blue vertical lines) at various times & sequenced

Freeman & Herron Fig. 1.10

“Each blue tick represents a virion sampled from the patient during the course of the infection; its horizontal position indicates when it was sampled and its vertical position indicates how genetically different it was from the first sample”.

Evolution of HIV population within an individual patient

Page 21: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 21

Who brought HIV-1 to America?

Gilbert, M. T. et al. The emergence of HIV/AIDS in the Americas and beyond. Proc Natl Acad Sci U S A 104, 18566-70 (2007).

Page 22: Molecular Clock I. Evolutionary rate Xuhua Xia xxia@uottawa.ca

Xuhua Xia Slide 22

Who brought HIV-1 to America?

Gilbert, M. T. et al. The emergence of HIV/AIDS in the Americas and beyond. Proc Natl Acad Sci U S A 104, 18566-70 (2007).