summary of last lesson

Summary of last lesson

• Excellent review of techniques for pop gen

• Methods of analysis• Previous lesson: density dependence/janzen

connel/red queen hypothesis/type of markers• humnogous fungus

• Testing the marker/testing sample size

Frequency-, or density dependent, or balancing selection

• New alleles, if beneficial because linked to a trait linked to fitness will be positively selected for.– Example: two races of pathogen are present, but

only one resistant host variety, suggests second pathogen race has arrived recently

• Rapid generation time of pathogens. Reticulated evolution very likely. Pathogens will be selected for INCREASED virulence

• In the short/medium term with long lived trees a pathogen is likely to increase its virulence

• In long term, selection pressure should result in widespread resistance among the host

Overview

Armillaria bulbosa (gallica)

• Known as the Humungous Fungus, or honey mushroom

• Form rhizomorphs, which make up much of the “humungous” part

• Basidiocarp: cap 6 cm in diameter, stem is 5-10 cm tall

• Facultative tree root pathogen

Life cycle: Reproduction

• Sexual – Basidiocarps release spores (n) after

karyogamy and meiosis– 2 mating-type loci, each with

multiple alleles in the population– Isolates (n) must have different

alleles at two mating type loci to be sexually compatible

• Asexual – vegetative spreading of rhizomorph

• The large mass of rhizomorph that is genetically isolated is called a clone

Building up the question…

• “By extending the areas sampled in subsequent years, we were finally able to delimit the large area occupied by this genotype and then go on to show that this genotype likely represents and ‘individual’”

- Myron Smith

Researcher’s Question

• The clonal “individual” is especially difficult to define because the network of hyphae is underground

• How do you unambiguously identify an individual fungi within a local population?

Approach

1. Collect samples

2. Check mating type

- Somatic compatibility test

- Distrubution of mating-type alleles

3. Molecular testing

- RFLP

- RAPD

4. Statistics

5. More testing

Methods and Materials 1

1. Collecting samples• Researcher collected samples over a 30 hectare

area by baiting Armillaria with poplar stakes and taking tissues and spores

• They then grew the successfully colonized stakes in soil taken from the study site

• Each fungal colony cultured was called an isolate.


2. Checking mating type

- Somatic incompatibility

For two fungal isolates to fuse, all somatic compatibility loci must be the same.

Fusion means they’re clones

Example (not Armillaria)


• 2. Checking mating type

- Distrubution of mating alleles

- Mating occurs only when coupled isolates have different alleles at two unlinked, multiallelic loci: A and B. (They have an incompatibility system)

- If fruit bodies had the same alleles at A and B, and were collected from the same area, they were assumed to be from the same clone

Result 1

• Somatic compatilbilty:– isolates from vegetative mycelium from a large

sampling area fused

• Mating alleles– They had the same mating type

Result 1

• “Clone 1” was found to exceed 500 m in diameter– Used previously

collected mtDNA restriction fragment patterns

Sensitivity of Approach

• Problem: These tests alone are not enough to distinguish a clone from closely related individuals

Why?

• Q: The first two tests were not sensitive enough to tell a clone from a close relative…Why?

• A: Spores from same point source have the same mating-type alleles, but the offspring they produce after inbreeding are genetically distinct.


3. Molecular Testing

- RFLP analysis at 5 polymorphic, heterozyg. loci of mtDNA from “Clone 1”

- RAPD analysis at 11 loci

RAPDS vs. RFLPs

• Use 1 short PCR primer

• When it finds match on template at a distance that can be amplified (primer binds twice within 50 to 2000 bp) RAPD amplicon

• Dominant, annoymous

• Total genomic, vs single locus

• Use endonuclease to digest DNA at specific restriction site

• Run digest and see how amplicon was cut

• Single locus is co-dominant

Result 2• RFLP

– All 5 loci from Clone 1 were heterozygous and identical (both alleles present at loci: 1,1)

• RAPD – All 11 RAPD products were present in all vegetative

isolates”

Statistical Analysis

• The probability of retaining heterozygosity at each parental locus in an individual produced by mating of sibling monospore isolates…

= 0.0013

• So they were pretty confident that cloning was responsible for their results, not inbreeding

More testing, just in case

• To be completely confident, they tested:– 1) that nearby Clone 2 was different and lacked 5 of

the Clone 1 heterozyg. RAPD fragments,

– 2) more loci, totaling• 20 RAPD fragments

• 27 nuclear DNA RFLP fragments

** all were identical in Clone 1

Sensitivity of RAPDs

• Tested on subset of spores from same basidiocarp

• RAPDs differentiated among full sibs

Conclusions

• Somatic compatibility, mating allele loci, mtDNA, RFLP, and RAPD tests all indicate that a single organism could indeed occupy a 15 hectare area

Conclusions

• The larger individual, Clone 1 was estimated to weigh 9700 kg and be over 1500 years old

Implications

• ?????

• Fungi are one of the oldest and largest organisms on the planet

• Recycle nutrients…very important!• Armillaria bulbosa also a pathogen; its effects on

forest above may be huge as well.

HOST-SPECIFICITY

• Biological species• Reproductively isolated• Measurable differential: size of structures• Gene-for-gene defense model• Sympatric speciation: Heterobasidion,

Armillaria, Sphaeropsis, Phellinus, Fusarium forma speciales

Phylogenetic relationships Phylogenetic relationships within the within the HeterobasidionHeterobasidion complexcomplex

Het INSULARE

True Fir EUROPE

Spruce EUROPE

True Fir NAMERICA

Pine EUROPE

Pine NAMERICA

0.05 substitutions/site

NJ

Fir-SpruceFir-Spruce

Pine EuropePine Europe

Pine N.Am.Pine N.Am.

The biology of the organism drives an epidemic

• Autoinfection vs. alloinfection

• Primary spread=by spores

• Secondary spread=vegetative, clonal spread, same genotype . Completely different scales (from small to gigantic)

Coriolus

Heterobasidion

Armillaria

Phellinus

OUR ABILITY TO:

• Differentiate among different individuals (genotypes)

• Determine gene flow among different areas

• Determine allelic distribution in an area

WILL ALLOW US TO DETERMINE:

• How often primary infection occurs or is disease mostly chronic

• How far can the pathogen move on its own

• Is the organism reproducing sexually? is the source of infection local or does it need input from the outside

IN ORDER TO UNDERSTAND PATTERNS OF INFECTION

• If John gave directly Mary an infection, and Mary gave it to Tom, they should all have the same strain, or GENOTYPE (comparison=secondary spread among forest trees)

• If the pathogen is airborne and sexually reproducing, Mary John and Tom will be infected by different genotypes. But if the source is the same, the genotypes will be sibs, thus related

Recognition of self vs. non self

• Intersterility genes: maintain species gene pool. Homogenic system

• Mating genes: recognition of “other” to allow for recombination. Heterogenic system

• Somatic compatibility: protection of the individual.


• What are the chances two different individuals will have the same set of VC alleles?

• Probability calculation (multiply frequency of each allele)

• More powerful the larger the number of loci• …and the larger the number of alleles per

locus


• It is possible to have different genotypes with the same vc alleles

• VC grouping and genotyping is not the same

• It allows for genotyping without genetic tests

• Reasons behing VC system: protection of resources/avoidance of viral contagion

Somatic incompatibility

QuickTime™ and aTIFF (Uncompressed) decompressor

are needed to see this picture.



More on somatic compatibility

• Perform calculation on power of approach

• Temporary compatibility allows for cytoplasmic contact that then is interrupted: this temporary contact may be enough for viral contagion

SOMATIC COMPATIBILITY

• Fungi are territorial for two reasons– Selfish– Do not want to become infected

• If haploids it is a benefit to mate with other, but then the n+n wants to keep all other genotypes out

• Only if all alleles are the same there will be fusion of hyphae

• If most alleles are the same, but not all, fusion only temporary

SOMATIC COMPATIBILITY

• SC can be used to identify genotypes• SC is regulated by multiple loci• Individual that are compatible (recognize one

another as self, are within the same SC group)• SC group is used as a proxy for genotype, but in

reality, you may have some different genotypes that by chance fall in the same SC group

• Happens often among sibs, but can happen by chance too among unrelated individuals


• What are the chances two different individuals will have the same set of VC alleles?

• Probability calculation (multiply frequency of each allele)

• More powerful the larger the number of loci• …and the larger the number of alleles per

locus

Recognition of self vs. non self:probability of identity (PID)

• 4 loci• 3 biallelelic• 1 penta-allelic

• P= 0.5x0.5x0.5x0.2=0.025

• In humans 99.9%, 1000, 1 in one million

INTERSTERILITY

• If a species has arisen, it must have some adaptive advantages that should not be watered down by mixing with other species

• Will allow mating to happen only if individuals recognized as belonging to the same species

• Plus alleles at one of 5 loci (S P V1 V2 V3)

INTERSTERILITY

• Basis for speciation

• These alleles are selected for more strongly in sympatry

• You can have different species in allopatry that have not been selected for different IS alleles

MATING

• Two haploids need to fuse to form n+n

• Sex needs to increase diversity: need different alleles for mating to occur

• Selection for equal representation of many different mating alleles

MATING

• If one individuals is source of inoculum, then the same 2 mating alleles will be found in local population

• If inoculum is of broad provenance then multiple mating alleles should be found

MATING

• How do you test for mating?

• Place two homokaryons in same plate and check for formation of dikaryon (microscopic clamp connections at septa)

Clamp connections







MATING ALLELES

• All heterokaryons will have two mating allelels, for instance a, b

• There is an advantage in having more mating alleles (easier mating, higher chances of finding a mate)

• Mating allele that is rare, may be of migrant just arrived

• If a parent is important source, genotypes should all be of one or two mating types

Two scenarios:

• A, A, B, C, D, D, E, H, I, L

• A, A, A,B, B, A, A

Two scenarios:

• A, A, B, C, D, D, E, H, I, L

• Multiple source of infections (at least 4 genotypes)

• A, A, A,B, B, A, A

• Siblings as source of infection (1 genotype)

SEX

• Ability to recombine and adapt

• Definition of population and metapopulation

• Different evolutionary model

• Why sex? Clonal reproductive approach can be very effective among pathogens

Long branches in between Long branches in between groups suggests no sex is groups suggests no sex is occurring in between occurring in between groupsgroups

Het INSULARE

True Fir EUROPE

Spruce EUROPE

True Fir NAMERICA

Pine EUROPE

Pine NAMERICA


NJ




Small branches within a clade indicate Small branches within a clade indicate sexual reproduction is ongoing within that sexual reproduction is ongoing within that

group of individualsgroup of individuals

11.10 SISG CA

2.42 SISG CA

BBd SISG WA

F2 SISG MEX

BBg SISG WA

14a2y SISG CA

15a5y M6 SISG CA

6.11 SISG CA

9.4 SISG CA

AWR400 SPISG CA

9b4y SISG CA

15a1x M6 PISG CA

1M PISG MEX

9b2x PISG CA

A152R FISG EU

A62R SISG EU

A90R SISG EU

A93R SISG EU

J113 FISG EU

J14 SISG EU

J27 SISG EU

J29 SISG EU


NJ

890 bpCI>0.9

NA S

NA P

EU S

EU F

Index of association

Ia= if same alleles are associated too much as opposed to random, it means

sex is not occurring

Association among alleles calculated and compared to simulated random

distribution

Evolution and Population genetics

• Positively selected genes:……• Negatively selected genes……• Neutral genes: normally population genetics

demands loci used are neutral• Loci under balancing selection…..

Evolutionary history

• Darwininan vertical evolutionary models

• Horizontal, reticulated models..

Phylogenetic relationships Phylogenetic relationships within the within the HeterobasidionHeterobasidion complexcomplex

Het INSULARE

True Fir EUROPE

Spruce EUROPE

True Fir NAMERICA

Pine EUROPE

Pine NAMERICA


NJ




Geneaology of “S” DNA insertion into P Geneaology of “S” DNA insertion into P ISG confirms horizontal transfer.ISG confirms horizontal transfer.

Time of “cross-over” uncertainTime of “cross-over” uncertain

11.10 SISG CA

2.42 SISG CA

BBd SISG WA

F2 SISG MEX

BBg SISG WA

14a2y SISG CA

15a5y M6 SISG CA

6.11 SISG CA

9.4 SISG CA

AWR400 SPISG CA

9b4y SISG CA

15a1x M6 PISG CA

1M PISG MEX

9b2x PISG CA

A152R FISG EU

A62R SISG EU

A90R SISG EU

A93R SISG EU

J113 FISG EU

J14 SISG EU

J27 SISG EU

J29 SISG EU


NJ

890 bpCI>0.9

NA S

NA P

EU S

EU F

Because of complications such as:

• Reticulation

• Gene homogeneization…(Gene duplication)

• Need to make inferences based on multiple genes

• Multilocus analysis also makes it possible to differentiate between sex and lack of sex (Ia=index of association), and to identify genotypes, and to study gene flow

Basic definitions again

• Locus• Allele• Dominant vs. codominant marker

– RAPDS– AFLPs

How to get multiple loci?

• Random genomic markers:– RAPDS– Total genome RFLPS (mostly dominant)– AFLPS

• Microsatellites• SNPs• Multiple specific loci

– SSCP– RFLP– Sequence information

Watch out for linked alleles (basically you are looking at the same thing!)

RAPDS use short primers but not too short

• Need to scan the genome

• Need to be “readable”

• 10mers do the job (unfortunately annealing temperature is pretty low and a lot of priming errors cause variability in data)

RAPDS can also be obtained with Arbitrary Primed PCR

• Use longer primers

• Use less stringent annealing conditions

• Less variability in results

Result: series of bands that are present or absent (1/0)

Root disease center in true fir caused by Root disease center in true fir caused by H. annosumH. annosum

Ponderosa pine Incense cedar

Yosemite Lodge 1975 Root disease centers outlined

Yosemite Lodge 1997 Root disease centers outlined

WORK ON PINES HAD DEMONSTRATED INFECTIONS ARE

MOSTLY ON STUMPS• Use meticulous field work and genetics

information to reconstruct disease from infection to explosion

• On firs/sequoia if the stump theory were also correct we would find a stump within the outline of each genotype

Are my haplotypes sensitive enough?

• To validate power of tool used, one needs to be able to differentiate among closely related individual

• Generate progeny

• Make sure each meiospore has different haplotype

• Calculate P

RAPD combination1 2

• 1010101010

• 1010101010

• 1010101010

• 1010101010• 1010000000

• 1011101010

• 1010111010

• 1010001010

• 1011001010• 1011110101

Conclusions

• Only one RAPD combo is sensitive enough to differentiate 4 half-sibs (in white)

• Mendelian inheritance?

• By analysis of all haplotypes it is apparent that two markers are always cosegregating, one of the two should be removed

If we have codominant markers how many do I need

• IDENTITY tests = probability calculation based on allele frequency… Multiplication of frequencies of alleles

• 10 alleles at locus 1 P1=0.1

• 5 alleles at locus 2 P2=0,2

• Total P= P1*P2=0.02

Have we sampled enough?

• Resampling approaches

• Saturation curves

– A total of 30 polymorphic alleles– Our sample is either 10 or 20– Calculate whether each new sample is

characterized by new alleles

Saturation (rarefaction) curves

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

NoOf Newalleles

Dealing with dominant anonymous multilocus markers

• Need to use large numbers (linkage)

• Repeatability

• Graph distribution of distances

• Calculate distance using Jaccard’s similarity index

Jaccard’s

• Only 1-1 and 1-0 count, 0-0 do not count

1010011

1001011

1001000

Jaccard’s

• Only 1-1 and 1-0 count, 0-0 do not count

A: 1010011 AB= 0.6 0.4 (1-AB)

B: 1001011 BC=0.5 0.5

C: 1001000 AC=0.2 0.8

Now that we have distances….

• Plot their distribution (clonal vs. sexual)



• Analysis: – Similarity (cluster analysis); a variety of

algorithms. Most common are NJ and UPGMA



• Analysis: – Similarity (cluster analysis); a variety of

algorithms. Most common are NJ and UPGMA– AMOVA; requires a priori grouping

AMOVA groupings

• Individual

• Population

• Region

AMOVA: partitions molecular variance amongst a priori defined groupings

Example

• SPECIES X: 50%blue, 50% yellow

AMOVA: example

v

Scenario 1 Scenario 2

POP 1

POP 2v

Expectations for fungi

• Sexually reproducing fungi characterized by high percentage of variance explained by individual populations

• Amount of variance between populations and regions will depend on ability of organism to move, availability of host, and

• NOTE: if genotypes are not sensitive enough so you are calling “the same” things that are different you may get unreliable results like 100 variance within pops, none among pops

Results: Jaccard similarity coefficients

0.3

0.90 0.92 0.94 0.96 0.98 1.00

00.10.2

0.40.50.60.7

Coefficient

Fre

quen

cy

P. nemorosa

P. pseudosyringae: U.S. and E.U.

0.3

Coefficient0.90 0.92 0.94 0.96 0.98 1.00

00.10.2

0.40.50.60.7

Fre

quen

cy

Fre

quen

cy

0.9 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.99

Pp U.S.

Pp E.U.

0.0

0.1

0.2

0.3

0.4

0.5

0.6

Jaccard coefficient of similarity

0.7

P. pseudosyringae genetic similarity patterns are different in U.S. and E.U.

0.1

4175A

p72

p39

p91

1050

p7

2502

p51

2055.2

2146.1

5104

4083.1

2512

2510

2501

2500

2204

2201

2162.1

2155.3

2140.2

2140.1

2134.1

2059.2

2052.2

HCT4

MWT5

p114

p113

p61

p59

p52

p44

p38

p37

p13

p16

2059.4

p115

2156.1

HCT7

p106

P. nemorosa

P. ilicisP. pseudosyringae

Results: Results: P. nemorosaP. nemorosa

Results: Results: P. pseudosyringaeP. pseudosyringae

0.1

4175A2055.2p44

FC2DFC2E

GEROR4 FC1B

FCHHDFCHHCFC1A

p80FAGGIO 2FAGGIO 1FCHHBFCHHAFC2FFC2CFC1FFC1DFC1Cp83p40

BU9715 p50

p94p92

p88p90

p56Bp45

p41p72p84p85p86p87p93p96p39p118p97p81p76p73p70p69p62p55p54

HELA2HELA 1

P. nemorosaP. ilicis

P. pseudosyringae

= E.U. isolate

The “scale” of disease

• Dispersal gradients dependent on propagule size, resilience, ability to dessicate, NOTE: not linear

• Important interaction with environment, habitat, and niche availability. Examples: Heterobasidion in Western Alps, Matsutake mushrooms that offer example of habitat tracking

• Scale of dispersal (implicitely correlated to metapopulation structure)---

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RAPDS> not used often now





RAPD DATA W/O COSEGREGATING MARKERS

PCA



AFLP

• Amplified Fragment Length Polymorphisms• Dominant marker• Scans the entire genome like RAPDs• More reliable because it uses longer PCR

primers less likely to mismatch• Priming sites are a construct of the sequence

in the organism and a piece of synthesized DNA

How are AFLPs generated?

• AGGTCGCTAAAATTTT (restriction site in red)• AGGTCG CTAAATTT• Synthetic DNA piece ligated

– NNNNNNNNNNNNNNCTAAATTTTT

• Created a new PCR priming site– NNNNNNNNNNNNNNCTAAATTTTT

• Every time two PCR priming sitea are within 400-1600 bp you obtain amplification

White mangroves:Corioloposis caperata

Coco Solo Mananti Ponsok DavidCoco Solo 0Mananti 237 0Ponsok 273 60 0David 307 89 113 0

Distances between study sites

Coriolopsis caperataCoriolopsis caperata on on Laguncularia racemosaLaguncularia racemosa

Forest fragmentation can lead to loss of gene flow among previously contiguous populations. The negative repercussions of such genetic isolation should most severely affect highly specialized organisms such as some plant-parasitic fungi.

AFLP study on single spores

Site # of isolates # of loci % fixed alleles

Coco Solo 11 113 2.6

David 14 104 3.7

Bocas 18 92 15.04

Distances =PhiST between pairs ofpopulations. Above diagonal is the ProbabilityRandom distance > Observed distance (1000iterations).

Coco Solo Bocas David

Coco Solo 0.000 0.000 0.000

Bocas 0.2083 0.000 0.000

David 0.1109 0.2533 0.000

Using DNA sequences

• Obtain sequence

• Align sequences, number of parsimony informative sites

• Gap handling

• Picking sequences (order)

• Analyze sequences (similarity/parsimony/exhaustive/bayesian

• Analyze output; CI, HI Bootstrap/decay indices

Using DNA sequences

• Testing alternative trees: kashino hasegawa • Molecular clock• Outgroup• Spatial correlation (Mantel)

• Networks and coalescence approaches

From Garbelotto and Chapela, From Garbelotto and Chapela, Evolution and biogeography of matsutakesEvolution and biogeography of matsutakes

Biodiversity within speciesBiodiversity within speciesas significant as betweenas significant as betweenspeciesspecies

Microsatellites or SSRs

• AGTTTCATGCGTAGGT CG CG CG CG CG AAAATTTTAGGTAAATTT

• Number of CG is variable• Design primers on FLANKING region, amplify DNA• Electrophoresis on gel, or capillary• Size the allele (different by one or more repeats; if number

does not match there may be polimorphisms in flanking region)

• Stepwise mutational process (2 to 3 to 4 to 3 to2 repeats)

summary of last lesson

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