summary of last lesson excellent review of techniques for pop gen methods of analysis previous...
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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.
Methods and Materials 2
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)
Methods and Materials 2
• 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.
Methods and Materials 3
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.
Recognition of self vs. non self
• 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
• 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
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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
Recognition of self vs. non self
• 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
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QuickTime™ and aTIFF (Uncompressed) decompressor
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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
0.05 substitutions/site
NJ
Fir-SpruceFir-Spruce
Pine EuropePine Europe
Pine N.Am.Pine N.Am.
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
0.0005 substitutions/site
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…..
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
0.05 substitutions/site
NJ
Fir-SpruceFir-Spruce
Pine EuropePine Europe
Pine N.Am.Pine N.Am.
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
0.0005 substitutions/site
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 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)
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
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– 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
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RAPD DATA W/O COSEGREGATING MARKERS
PCA
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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)