 Solution

ID: past-exam-of-the-biological-physics-course-of-the-university-of-oxford/2018/3/a/solution

Represent one permissive voltage-gated potassium channel gate by an available closed state and an activated state, with voltage-dependent transition rates

α_{n} (V)

and

β_{n} (V)

. Gain minus loss gives

\frac{d n}{d t} = α_{n} (1 - n) - β_{n} n = (α_{n} + β_{n}) (n_{\infty} - n), n_{\infty} = \frac{α _{n}}{α _{n} + β _{n}}, τ_{n} = \frac{1}{α _{n} + β _{n}} .

During a voltage clamp step the imposed voltage and hence these rates are constant. Integration with

n (0) = n_{0}

gives

n (t) = n_{\infty} - (n_{\infty} - n_{0}) e^{- t / τ_{n}} .

The fourth power in the electrical conductance describes four permissive activation variables in the Hodgkin-Huxley model phenomenology. If all must be permissive and are statistically independent, their joint probability is

n^{4}

. Even a simple exponential rise in

n

then gives a delayed-looking macroscopic potassium electric current. The second panel of the model figure in solution illustrates this. A continuously changing voltage makes the rates time dependent and requires solving the differential equation along that voltage trajectory; one fixed exponential is not the general action-potential solution.

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