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15-12-2025 15:48

Danny Newman Danny Newman

Melanospora cf. lagenaria on old, rotting, fallen

15-12-2025 15:54

Johan Boonefaes Johan Boonefaes

Unknown anamorph found on the ground in coastal sa

15-12-2025 21:11

Hardware Tony Hardware Tony

Small clavate hairs, negative croziers and IKI bb

15-12-2025 07:09

Danny Newman Danny Newman

indet. Rutstroemiaceae sp. on unk. fallen leavesMc

15-12-2025 21:47

Pol Debaenst

Good evening, On 12/11/2025 I found ascomycetes w

15-12-2025 07:05

Danny Newman Danny Newman

Pseudosclerococcum golindoi (det: Zotto)near Cosb

15-12-2025 11:49

Danny Newman Danny Newman

ITS sequences from the following two collections B

15-12-2025 12:34

Danny Newman Danny Newman

indet. Rhytismataceae on oak leafnear Purchase Roa

09-12-2025 12:06

Andgelo Mombert Andgelo Mombert

Bonjour,Je recherche l'article concernant Hypobryo

13-12-2025 17:26

Buckwheat Pete

Hello everyone,I have a rather interesting ascomyc

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Who can help me?
hannie wijers, 26-07-2014 13:48

Hello everyone,


I hope anyone of you can give me a hand with this problem


On a pine cone I found a pair of yellow / orange slices. I'm not quite sure but think that there are only fours pores in the asci. The measure of the asci is between 75-85 x18 µm. The measure of the spores is between 23-26 x 4-7 µm. I thought there were konidiensporen but were appendages of the spores. Parafysen are knobby, J-.
Is there anyone who has an idea in which direction I should think?


Greetings


Gannie

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Guy Marson, 26-07-2014 14:16
Re : Who can help me?
Hi Gannie,

If the asci are IKI+, it might be Pezicula eucrita. It grows right now, after the rains in mid-europe.  
Cheers,
Guy  
hannie wijers, 26-07-2014 14:29
Re : Who can help me?

Hello Guy,


Thank you, I looked for some pictures and I think you're right. Can you explain what IKI+ and how to see this?


Greetings


Hannie

Guy Marson, 26-07-2014 15:38
Re : Who can help me?
Hi Hannie,

IKI is the so called Lugol´s solution (or "J" - but without chloral hydrate). 
With such IKI, your fungus will show a hemiamyloid (= red!) reaction of the AA (apical apparatus).
In Melzer´s reagent, the reaction is negative.
Pre-treated with KOH (5%), the reaction will be blue in Melzer´s and in Lugol´s solution and not turning into red after a little while.. 

Cheers,

Guy
hannie wijers, 26-07-2014 16:07
Re : Who can help me?

Hello Guy,


thanks for explanation. I did it with Lugol but I can't see any reaction.


Greetings


Hannie

Hans-Otto Baral, 26-07-2014 20:03
Hans-Otto Baral
Re : Who can help me?
Your reagent obviously lost its iodine. Is it very pale yellow? It should be red-brown. Iodine is very volatile.
hannie wijers, 26-07-2014 20:10
Re : Who can help me?

Hello Guy,


yes it's pale yellow. It's hard to get it here in Holland, so someone sent it to me. Maybe he put some water to it? So it's better to buy some new?


Hannie

Martin Bemmann, 26-07-2014 22:20
Martin Bemmann
Re : Who can help me?
Dear Hannie,

to my experience there are several varieties of Lugol's solution in trade. I once got one that was almost clear and prepared for lichenologists, since the reaction in lichen is very strong, but gives no sufficiant results with "standard" ascomycetes. On the other hand I experienced false descriptions of the reaction where concentrations of iodine stronger than 2% were used and thus an hemiamyloid reaction was suppressed or overrun.
The standard formula for Lugol's for ascomycetes should be 1% iodine + 3% potassium iodide added to water.

Regards
Martin
Hans-Otto Baral, 26-07-2014 22:37
Hans-Otto Baral
Re : Who can help me?
The concentration must not be very exact, 0.5% is also quite good, but since the iodine disappears with time, it will finally be 0.3% or only 0.1% and that will be too low. 

I have crystalline iodine and potassium iodide and mix it needed.

See http://www.invivoveritas.de/articles/iodine-reaction/


Zotto
hannie wijers, 27-07-2014 10:36
Re : Who can help me?

Martin and Zotto,


thanks for the answer. I'll study your paper Zotto but it will take some time for me.


Cheers


Hannie