Multi-channel Popen

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(excerpted from Characterization of Stretch-Activated Ion Channels in Xenopus Oocytes (1989) dissertation by Xian-Cheng Yang, pp 128-131, Appendix C.7

See also: Kijima S, Kijima H.: Statistical analysis of channel current from a membrane patch. II. A stochastic theory of a multi-channel system in the steady-state )

N-channels in the patch

A complication is introduced when there are more than one channel in a patch. Binomial analysis is necessary in this case. One of the ways for carrying this analysis is the method of maximum likelihood (Korn et al., 1981; Sachs et al., 1982), that can be used for determining both the number of channels, N, and the probability of a single-channel being open, from a record containing more than one channel. The number of activatable channels in a patch can also be estimated form experiments as the maximum level observed when applying a saturating dose of pressure. This always gives a lower estimate for N since Pmax < 1. If N (the number of activatable channels) is known in this way, an alternative method for determining the probability of single-channel being open can be developed.

Let PC the probability of channel being closed, and PO the probability of at least one channel being open (or total PO) in a patch containing N-channels. We have

(C7.1)
PC + PO = 1

Assuming that N-channels behave independently, the probability distribution of a channel being at level k (= 1, 2, .... , N) is thus binomial:

(C7.2)
P_k(x;n) = \frac{n!}{k! (n-k)!} x^k (1-x)^{n-k}

with x being the probability of a single-channel begin open. The normalization condition is

(C7.3)
1 = \sum_{k=0}^n P_k (x; n)
= \sum_{k=0}^n \frac{n!}{k! (n-k)!} x^k (1-x)^{n-k}
= (1-x)^n + \sum_{k=1}^n \frac{n!}{k! (n-k)!} x^k (1-x)^{n-k}
= PC + PO

The last equality is obtained since

(C7.3.1)
PC = Prob.(no channel opens)
= Pk = 0(x;n)
= (1 − x)n

and

(C7.3.2)
PO = Prob.(at least one channel opens)
= \sum_{k=1}^n P_k (x; n)

By rearranging (C7.3), the probability of a single-channel being open, x, is solved in terms of PO or PC as

(C7.4)
x = 1 - (1 - P_O)^{\frac{1}{n}}

or

x = 1 - P_C^{\frac{1}{n}}

Both PO and PC can be easily measured from amplitude histograms from (C2.5). If PO << 1 , (C7.4) reduces to

(C7.4.1)
x = \frac{P_O}{n}

or

(C7.4.2)
LnPo = Ln(x) + Ln(n)
= Θp2 + constant

This result shows that sensitivity to membrane tension, Θ, can be directly determined from PO when the probability of the channel being open is very low, say, smaller than 0.1.


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