Quick start: QUB Online Fitness X-Means Movie Maker

# 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 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 $$P_C$$ 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

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

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

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

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

$$$$\tag{C7.3} 1 = \sum_{k=0}^n P_k (x; n)$$$$ $$= \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}$$ $$= P_C + P_O$$

The last equality is obtained since

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

and $$$$\tag{C7.3.2} P_O = \text{Prob.(at least one channel opens)}$$$$ $$= \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 $$P_O$$ or $$P_C$$ as

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

or

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

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

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

or $$$$\tag{C7.4.2} Ln(P_O) = Ln(x) + Ln(n)$$$$ $$= \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.