Multi-channel Popen

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

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 $P m a x < 1$ . 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 $P_C$ the probability of channel being closed, and $P_O$ the probability of at least one channel being open (or total $P_O$) in a patch containing N-channels. We have

$$\begin{equation} \tag{C7.1} P_C + P_O = 1 \end{equation}$$

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

$$\begin{equation} \tag{C7.2}P_k(x;n) = \frac{n!}{k! (n-k)!} x^k (1-x)^{n-k} \end{equation}$$

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

$$\begin{equation} \tag{C7.3} 1 = \sum_{k=0}^n P_k (x; n) \end{equation}$$ $$ = \sum_{k=0}^n \frac{n!}{k! (n-k)!} x^k (1-x)^{n-k}$$ $$ = (1-x)^n + \sum_{k=1}^n \frac{n!}{k! (n-k)!} x^k (1-x)^{n-k}$$ $$ = P_C + P_O$$

The last equality is obtained since

$$\begin{equation} \tag{C7.3.1} P_C = \text{Prob.(no channel opens)} \end{equation}$$ $$ = P_{k = 0}(x;n)$$ $$ = (1 - x)^n$$

and $$\begin{equation} \tag{C7.3.2} P_O = \text{Prob.(at least one channel opens)} \end{equation}$$ $$ = \sum_{k=1}^n P_k (x; n)$$

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

$$\begin{equation} \tag{C7.4} x = 1 - (1 - P_O)^{\frac{1}{n}} \end{equation}$$

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

Both $P_O$ and $P_C$ can be easily measured from amplitude histograms from (C2.5). If $P_O \lt\lt 1$, (C7.4) reduces to

$$\begin{equation} \tag{C7.4.1} x = \frac{P_O}{n} \end{equation}$$

or $$\begin{equation} \tag{C7.4.2} Ln(P_O) = Ln(x) + Ln(n) \end{equation}$$ $$ = \Theta p^2 + \text{constant}$$

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

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