 Solution

ID: past-exam-of-the-mathematical-physiology-course-of-the-university-of-oxford/2023/1/f/solution

The two identical

M

gates are independent Bernoulli random variables, so their open count has a binomial distribution. Multiplying by the independent

H

factor gives the normalized channel densities

S_{ij} = (i 2) m^{i} (1 - m)^{2 - i} h^{j} (1 - h)^{1 - j} .

Explicitly,

S_{00} S_{10} S_{20} = (1 - m)^{2} (1 - h), = 2 m (1 - m) (1 - h), = m^{2} (1 - h), S_{01} S_{11} S_{21} = (1 - m)^{2} h, = 2 m (1 - m) h, = m^{2} h .

(1 2) = 2

counts which of the two distinguishable

M

gates is open. At open count

i

, the total

M

-opening rate is

(2 - i) α

and the total closing rate is

i β

. Dividing the expected value of that count by two therefore recovers the same single-gate ordinary differential equation:

\overset{m}{˙} = α (1 - m) - β m, \dot{h} = γ (1 - h) - δ h .

For the specified initial conditions and common rate sum

λ

, the all-open fraction is

S_{21} (t) = \frac{α ^{2} γ}{λ ^{3}} (1 - e^{- λ t})^{3} .

Thus, assuming positive opening rates,

t_{1/2} = \frac{- lo g ( 1 - 2 ^{- 1/3} )}{λ} .

Three gates take longer to reach half their final all-open fraction than two gates: all three activation factors must build up simultaneously.

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