2. 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

3. 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

4. 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

5. 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?

6. 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

7. 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.

8. 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)

9. 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

10. Result 1 Somatic compatilbilty: –isolates from vegetative mycelium from a large sampling area fused Mating alleles –They had the same mating type

11. Result 1 “Clone 1” was found to exceed 500 m in diameter –Used previously collected mtDNA restriction fragment patterns

12. Sensitivity of Approach Problem: These tests alone are not enough to distinguish a clone from closely related individuals

13. 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.

14. Methods and Materials 3 3. Molecular Testing - RFLP analysis at 5 polymorphic, heterozyg. loci of mtDNA from “Clone 1” - RAPD analysis at 11 loci

15. 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

16. 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”

17. 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

18. 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

19. Sensitivity of RAPDs Tested on subset of spores from same basidiocarp RAPDs differentiated among full sibs

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

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

22. 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.

23. 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

25. Phylogenetic relationships within the Heterobasidion complex Fir-Spruce Pine Europe Pine N.Am.

26. 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

28. OUR ABILITY TO: Differentiate among different individuals (genotypes) Determine gene flow among different areas Determine allelic distribution in an area

29. 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

30. 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

31. 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.

32. 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

33. 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

34. Somatic incompatibility

35. 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

36. 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

37. 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

38. 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

39. 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

40. 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)

41. 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

42. 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

43. 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

44. MATING How do you test for mating? Place two homokaryons in same plate and check for formation of dikaryon (microscopic clamp connections at septa)

45. Clamp connections

46. 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

47. Two scenarios: A, A, B, C, D, D, E, H, I, L A, A, A,B, B, A, A

48. 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)

49. 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

50. Long branches in between groups suggests no sex is occurring in between groups Fir-Spruce Pine Europe Pine N.Am.

51. Small branches within a clade indicate sexual reproduction is ongoing within that group of individuals 890 bp CI>0.9 NA S NA P EU S EU F

52. 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

53. Evolution and Population genetics Positively selected genes:…… Negatively selected genes…… Neutral genes: normally population genetics demands loci used are neutral Loci under balancing selection…..

54. Evolution and Population genetics Positively selected genes:…… Negatively selected genes…… Neutral genes: normally population genetics demands loci used are neutral Loci under balancing selection…..

55. Evolutionary history Darwininan vertical evolutionary models Horizontal, reticulated models..

56. Phylogenetic relationships within the Heterobasidion complex Fir-Spruce Pine Europe Pine N.Am.

57. Geneaology of “S” DNA insertion into P ISG confirms horizontal transfer. Time of “cross-over” uncertain 890 bp CI>0.9 NA S NA P EU S EU F

58. 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

59. Basic definitions again Locus Allele Dominant vs. codominant marker –RAPDS –AFLPs

60. 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!)

61. 60 Example 2: Distinguishing taxa in the Pleurotus eryngii (King Oyster Mushroom) complex using AFLPs (Urbanelli et al., Appl. Microbiol. Biotechnol. (2007) 74:592-600) Photo: The New York Times Photo: Wikimedia Commons

62. 61 Distinguishing taxa in the Pleurotus eryngii (King Oyster Mushroom) complex using AFLPs Goal: to determine, using multilocus genotypes, whether the distinction between Pleurotus eryngii, P. ferulae, and P. eryngii var. nebrodensis is supported by genetic data 90 populations sampled 94 AFLP loci scored P. ferulae (http://steinpilz.up.seesaa.net)

63. 62 Sample AFLP Gel

64. 63 AFLP Data Map from Urbanelli et al. (2007)

65. 64 AFLP Data Map with UPGMA dendogram from Urbanelli et al. (2007)

66. 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)

67. 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)

68. RAPDS can also be obtained with Arbitrary Primed PCR Use longer primers Use less stringent annealing conditions Less variability in results

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

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

72. Ponderosa pineIncense cedar

73. Yosemite Lodge 1975 Root disease centers outlined

74. Yosemite Lodge 1997 Root disease centers outlined

75. 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

82. 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

83. RAPD combination 1 2 1010101010 1010000000 1011101010 1010111010 1010001010 1011001010 1011110101

84. 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

85. 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

86. 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

87. Saturation (rarefaction) curves 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 No Of New alleles

88. Dealing with dominant anonymous multilocus markers Need to use large numbers (linkage) Repeatability Graph distribution of distances Calculate distance using Jaccard’s similarity index

89. Jaccard’s Only 1-1 and 1-0 count, 0-0 do not count 1010011 1001011 1001000

90. Jaccard’s Only 1-1 and 1-0 count, 0-0 do not count A: 1010011 AB= 0.60.4 (1-AB) B: 1001011 BC=0.50.5 C: 1001000 AC=0.20.8

91. Now that we have distances…. Plot their distribution (clonal vs. sexual)

92. 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

93. 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

94. AMOVA groupings Individual Population Region AMOVA: partitions molecular variance amongst a priori defined groupings

95. Example SPECIES X: 50%blue, 50% yellow

96. AMOVA: example v Scenario 1Scenario 2 POP 1 POP 2 v

97. 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

98. Results: Jaccard similarity coefficients 0.3 0.900.920.94 0.960.98 1.00 0 0.1 0.2 0.4 0.5 0.6 0.7 Coefficient Frequency P. nemorosa P. pseudosyringae: U.S. and E.U. 0.3 Coefficient 0.900.920.940.960.981.00 0 0.1 0.2 0.4 0.5 0.6 0.7 Frequency

99. 0.90.910.920.930.940.950.960.970.980.99 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.

100. P. nemorosa P. ilicis P. pseudosyringae Results: P. nemorosa

101. Results: P. pseudosyringae P. nemorosa P. ilicis P. pseudosyringae = E.U. isolate

102. 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)---

106. RAPDS> not used often now

107. RAPD DATA W/O COSEGREGATING MARKERS

109. PCA

110. 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

111. 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

114. White mangroves: Corioloposis caperata Distances between study sites

115. Coriolopsis caperata on Laguncularia 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

118. 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

119. Using DNA sequences Testing alternative trees: kashino hasegawa Molecular clock Outgroup Spatial correlation (Mantel) Networks and coalescence approaches

122. From Garbelotto and Chapela, Evolution and biogeography of matsutakes Biodiversity within species as significant as between species

123. 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)

124. 123 MS18(AC) 38 218 bp (AC) 39 220 bp (AC) 40 222 bp MS43a(CAGA) 70 373 bp MS43a(CAGA) 71 377 bp MS43a(CAGA) 72 381 bp (220-218) 2 2 2 (222-218) 2 4 2 (377-373) 2 4 2 (381-373) 2 8 2 (39-38) 2 1 2 (40-38) 2 2 2 (71-70) 2 1 2 (72-70) 2 2 2 ACACACACACACACACAC AMOVA Analysis of Molecular Variance

125. 124 Example 1: Origins of the Sudden Oak Death Epidemic in California (Mascheretti et al., Molecular Ecology (2008) 17: 2755-2768) Photo: UC Davis Photo: www.membranetransport.orgwww.membranetransport.org Photo: Northeast Plant Diagnostic Network

126. 125 Spatial autocorrelation Geographical distance (m) 10 100 1000 Moran’s I 0 Within approx. 100 meters the genetic structure correlates with the geographical distance

127. 126 Spatial autocorrelation Moran’s I (coefficient of departure from spatial randomness) correlates with distance up to Distribution of genotypes (6 microsatellite markers) in different populations of P.ramorum in California

128. 127 NJ tree of P. ramorum populations in California SC-1 MA-4 NURSERY SC-3 MA-3 SO-1 SO-2 MA-5 SC-2 MO-1 MO-2 MA-2 MA-1 HU-1 HU-2

129. 128 Phytophthora ramorum (Oomycete) –causal agent of Sudden Oak Death (SOD) first reported in California in 1994 –SOD affects tanoak (Lithocarpus densiflora), coast live oak (Quercus agrifolia), Californian black oak (Quercus kelloggii), and Canyon live oak (Quercus chrysolepis) –P.ramorum also cause a disease characterized mostly by leaf blight and/or branch dieback in over 100 species of both wild and ornamental plants, including California bay laurel (Umbellularia cailfornica), California redwood (Sequoia sempervirens), Camellia and Rhododrendron species Example: microsatellites genotyping of P. ramorum isolates Collection of infected bay leaves from several forests in Sonoma, Monterey, Marin, Napa, Alameda, San Mateo

130. 129 Microsatellites (I) mating type A1 (EU) and mating type A2 (US) A2 (US)A1 (EU) Locus 29325/ - 325/337 -/337 Locus 33315/337325/337 Locus 65234/252 236/244 220/222

131. 130 Microsatellites (II) mating type A2 (US) MS39a(GA) 11 129 bp MS39b(GA) 4 (GATA) 31 234, 242, 246, 250, 254 bp MS43a(CAGA) 69 309, 329, 349, 353, 357, 361, 365, 369, 373, 377, 381 bp MS43b(CAGA) 75 (…)(CAGA) 16 416, 420, 466, 470, 474, 476, 478, 482, 486, 490, 494, 498 bp MS45(TCCG) 11 167, 183, 187 bp MS18(AC) 39 218, 220, 222, 254, 264, 272, 274, 276, 278, 282 bp MS64(CT) 16 342, 374, 376, 378 bp

132. 131 Ind.MS39aMS39bMS43aMS43bMS45MS18MS64 Mating type 1129-129246-246369-369486-486167-187220-278342-374A1 2129-129246-246369-369486-486167-187220-278342-374A1 3129-129246-246373-373486-486167-187220-274342-374A1 4129-129246-246373-373 486-486167-187220-274342-378A1 5129-129246-246373-373 486-486167-187220-274342-378A1 6129-129246-246373-373 486-486167-187220-274342-378A1 7129-129246-246373-373 486-486167-187220-278342-378A1 8129-129246-246373-373 486-486167-187220-278342-374A1 9129-129250-250369-369486-486167-187220-278342-374A1 10129-129250-250 369-369486-486167-187220-278342-374A1 11129-129250-250369-369486-486167-187220-278342-374A1 12129-129250-250 377-377490-490167-187220-278342-374A1 13129-129250-250 377-377 490-490 167-187220-278342-381A1 14129-129250-250 377-377 490-490 167-187220-278342-381A1 15129-129250-250 377-377 490-490 167-187220-278342-381A1 16129-129246-246377-377 490-490 167-187220-278342-374A1 17129-129246-246377-377 486-486167-187220-278342-374A1 18129-129246-246369-369486-486167-187220-278342-374A1 19129-129246-246381-381486-486167-187222-null342-374A2 20129-129246-246381-381494-494167-187222-null342-374A2

133. 132 Ind.MS39aMS39bMS43aMS43bMS45MS18MS64 Mating type 111-1132-3269-6917-1711-1639-6818-29A1 211-11 32-32 69-69 17-17 11-16 39-68 18-29 A1 311-11 32-32 70-7017-17 11-16 39-6618-29 A1 411-11 32-32 70-70 17-17 11-16 39-66 18-30A1 511-11 32-32 70-70 17-17 11-16 39-66 18-30 A1 611-11 32-32 70-70 17-17 11-16 39-66 18-30 A1 711-11 32-32 70-70 17-17 11-16 39-6818-30 A1 811-11 32-32 70-70 17-17 11-16 39-68 18-29 A1 911-11 33-3369-69 17-17 11-16 39-68 18-29 A1 1011-11 33-33 69-69 17-17 11-16 39-68 18-29 A1 1111-11 33-33 69-69 17-17 11-16 39-68 18-29 A1 1211-11 33-33 71-7118-1811-16 39-68 18-29 A1 1311-11 33-33 71-71 18-18 11-16 39-68 18-31A1 1411-11 33-33 71-71 18-18 11-16 39-68 18-31 A1 1511-11 33-33 71-71 18-18 11-16 39-68 18-31 A1 1611-11 32-3271-71 18-18 11-16 39-68 18-29A1 1711-11 32-3271-71 17-17 11-16 39-68 18-29 A1 1811-11 32-32 69-69 17-1711-16 39-68 18-29 A1 1911-11 32-32 72-7217-17 11-16 40-null18-29 A2 2011-11 932-32 72-72 19-1911-16 40-null18-29 A2

134. 133 China Camp State Park (Marin)n=24 Bean Creek Forest (Santa Cruz)n=24 Nurseriesn=14