# hERG rundown

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lazy
hERG rundown

Hi
When I am recording hERG current, always the current shows rundown. How you guys take it into account for the assay?

Fraser Moss
Some people include PIP2 in

Some people include PIP2 in the patch pipette. You will need to sonicate this into your pipette soln immediately before each patch.

Another way is to run your proptocol for at least 10 sweeps before drug application and get a fit of the tail currents during ths time. Extrapolate the fit over the duration of the whole experiment and normalize you tail currents in the presence of drug to this fit.

There is a useful paper describing a computer algorithm called I_VOC which deals with current run down. Reference: Marrannes - ComputerMethods&Programs in Biomedicine (2004) v74 page 167. You dont need the actual program to do the run down compensation.. I can do it offline easily using pclamp and Origin graphing software.

Many labs state that if the rundown is more than 10% of the starting peak amplitude then they don't use the cell. The problems is that you will have hardly any data if you do this. However, ensure that your cell has a resting membrane potential before and after your drug application and that you have not just killed it.

lazy
Current voltage relationship

Current voltage relationship of hERG peak tail current shows S-shape curve. From I=V/R, current amplitude seems to be liniealy increased at higher voltage, if R does not increase. But the current amplitude reaches to the max around 0 mV and no more increase. Why V/R is constant at higher voltage? Maybe I am completely misunderstanding??
Thanks

Fraser Moss
I'm not sure I totally

I'm not sure I totally understand what you are asking here.
Can you add images of the current voltage relationships with your questions referring to each example please?

Also, are you performing any kind of leak subtraction on your current recordings?

lazy
At the peak of the tail

At the peak of the tail current, the channel must be fully open (as far as I understood). Then, the conductance (g) seems to be max. From I=V*g, the current amplitude (I) must be greater, more positive 2nd voltage step is applied. The reality is, however, as is the shape of conductance. Something that I do not know must be involved. I like to understand what.

lazy
Dear Frasermoss

Dear Frasermoss

You can find the curent/voltage relatiohship in this site. I got the identical data, but I do not understand why the current amplitude does not increase with higher voltage step.

Thanks

http://avivabio.com/herg-electrophysiology-assay.php

Fraser Moss
Ok so you are seeing the

Ok so you are seeing the classic I-V relationship.

the reason that you do not see an increase in the peak tail current amplitude at voltages more positive than 0mV is that at that voltage you have fully activated all the hERG channels in the cell.

Each single hERG channel has a finite single channel conductance. The whole cell current amplitude is the sum of each of the single channel conductances. When you step from 0mV (where you have opened and then inactivated all the channels) to a hyperpolarized voltage like -60mV, then you will open the maximum number of channels in that cell. No matter how high you drive the voltage you will not be able to push any more potassium ions through the open K channels than is dictated by their single channel conductance for that voltage.

You must understand the activation/inactivation scheme of hERG Channels.

The reason that they have such a unique tail current phenotype is that most of the channels HAVE to pass back through the open state from an inactivated state when you hyperpolarize the membrane in order to go back to the closed state. Most Voltage dependent channels have a non-conducting pathway which is energetically more favorable to return to the closed state from an inactivated state, so you do not see any tail currents.

lazy
Thanks.

Thanks.
Maybe I am misunderstanding, but how to be interpretted by Ohm's law V=Ixg?

g : the maxmal level (constant max value over +20 mV)

V : more positive 2nd voltage step (driving force increased)

Then, the current amplitude (I) must increased, if we can simply apply the Ohm's law??

Possible explanation is that g value over +20 mV can be decreased. I do not know if this is a good way of thinking.
Thanks.

Golden ball
The similar question I have.

The similar question I have. I think the question comes from the figure 9 by Sanguinetti which is attached.