Lab 11 :Test of Neutrality and Evidence for Selection

Goals:

1.Calculate exp. # of different allele in a population for different marker.

2.Detect departure from neutrality using 1. Ewens- Watterson test.

2. Tajima’s D test.

3. HKA test and 4. Synonymous and Nonsynonymous nucleotide substitution test

Infinite Alleles Model (IAM)

• • • Each mutation produces a new allele At equilibrium, number of alleles and shape of allele frequency distribution remain constant Lost alleles replaced by new mutations

Ewens -Watterson test

E

(

k

) 

i

2

N

   1 0   

i

 1     1     2  ...

    2

N

 1

Where

,   4

Neu

Expected homozygosity under mutation-drift equilibrium and assuming IAM:

f e

 1 4

N e

  1

Expected homozygosity under HWE:

P-value < 0.025: Too even -> Balancing selection or recent bottleneck

f HW

 

p i

2 P-value > 0.975: Too uneven -> Directional selection or population growth

Problem 1.

Estimates of the long-term

effective population size

of human populations vary widely, ranging from as low as ~3,000 to as high as ~100,000. To estimate allele frequencies for a forensic identification study, you are genotyping individuals selected at random from a population with an

estimated N e 0.8



10 -6

and



= 7,500.

You are using one

allozyme a

nd one

microsatellite marker

, with estimated mutation rates

=

9.2



10 -2

,

respectively 

=

. How many different alleles do you expect to find for each marker in a sample of:

•

7 people?

•

12 people?

•

What assumptions were made for these calculations to be valid?

Tajima’s D

• Under neutrality, we expect the following: • 

S



S

 1

n



i

1

i

   

m

Test of the coalescent model – Assumes neutral alleles and constant population size

Under neutrality

d =

  − 

S

= 0

D

=

d SE

(

d

)

D

 0  Purifying/

D

 0  Balancing positive selection or population selection or recent bottleneck growth

(Hamilton 270)

plantsciences.ucdavis.edu

Problem 2.

File

aspen_phy.arp

Arlequin

(which is already in format) contains sequence data from exon 1 of the

phytochrome B2

(

phyB2

) gene of 24 aspen (

Populus tremula

) trees sampled along a wide latitudinal gradient in Europe. Use

Arlequin

to: a.Determine the number of polymorphic sites (

S

) and calculate the nucleotide diversity (  ) based on these sequences.

b.Perform the tests of neutrality developed by Ewens Watterson and Tajima and interpret the results.

c. Provide a statistical and a biological interpretation of the results from the two neutrality tests.

Hudson-Kreitman- Aguade(HKA) test

(Hamilton 266)

Hudson-Kreitman- Aguade(HKA) test

Adh

Polymorphism within 0.101

species (S/m) Divergence between Species(D/m) Ratio (within/between)

χ

2 p-value 0.056

1.80

6.09

0.016

Control locus 0.022

0.052

0.42

Problem 3.

Files

utr_mays.arp

,

utr_par.arp, exon_mays.arp,

and

exon_par.arp

contain sequence data from the 5’ untranslated region and from an exon of the

teosinte branched1

(

tb1

) gene of maize (

Zea mays

ssp.

mays

) and its most likely wild progenitor

Zea mays

ssp.

parviglumis

. For each of these regions of

tb1

•Use

Arlequin

and for each subspecies: to determine the number of segregating sites (

S

) and calculate the nucleotide diversity (  ). What can you infer by comparing nucleotide diversity between the two species for each region? •

Use Arlequin

to perform the tests of neutrality developed by Ewens-Watterson and Tajima. Interpret and discuss the results.

•

Interpret and discuss the results from the following 2 HKA tests:

•

GRADUATE STUDENTS ONLY: Download and read the paper describing this study (Wang et al. 1999), which is uploaded on the lab page of the class website, and provide an extended biological interpretation of the results of a) – c).

File

utr_mays.arp

utr_par.arp

exon_mays.arp

exon_par.arp

Test A Polymorphism within subspecies Divergence between subspecies

χ

2

p

-value Test B Polymorphism within subspecies Divergence between subspecies

χ

2

p

-value Region of

tb1

5’ untranslated region 5’ untranslated region exon exon

tb1

5’ untranslated region 0.00093

Subspecies

mays parviglumis mays parviglumis

Average of control loci 0.01996

0.05255

13.58

0.001

tb1

translated region 0.00243

0.01273

2.70

0.26

0.02242

Average of control loci 0.01996

0.02242

Synonymous and Nonsynonymous Nucleotide Substitution test

 

d N d S

dN = Observed # nonsynonymous substitutions/nonsynonymous site dS= Observed # synonymous subsitutions/synonymous site 5’-ATT GTT CAT CG T 5’-ATT GTT CAT CG C ACC CA

T

ACC CA

A

CGA-3’ CGA-3’ Synonymous site Synonymous mutation Nonsynonymous site Nonsynonymous mutation

Problem 4.

Calculate the

ω

=

d N

/

d S

ratio based on the following 2 DNA sequences: 5’-ATG GTT CAT TTT ACC GGA CGA AGT CGA TTA-3’ 5’-ATG GTT CAC TTG ACC GCA CGA AGT AGA TTA-3’ Seq 1 Codon ATG GTT No. potential synonymous sites (s j ) 0 No. potential nonsynonymous sites (n j ) 3 Seq 2 Codon ATG GTT No. potential synonymous sites (s j ) 0 No. potential nonsynonymo us sites (n j ) 3

Total