Questions

1 Outline the main forms of biological free energy and the ways in which they are utilised and inter-converted.

2 A lipid bilayer of thickness t = 5 nm and relative dielectric constant εr = 2.5, incorporating sodium channels (permeable to sodium ions only) and ATP-driven sodium pumps, forms a vesicle of radius r = 100 nm. The initial sodium concentration is 100 mM both inside and outside the vesicle.

(a) Calculate the aproximate capacitance, C, of the bilayer, neglecting the effects of the proteins.

(b) The pumps pump 11sodium ions across the membrane for every 3 ATP molecules hydrolysed, and are fully reversible. When an excess of ATP is added to drive the pumps, with the sodium channels closed, the membrane voltage, φ, rises asymptotically to 150 mV.

(i) Calculate the number of sodium ions (N1) that must be pumped across the bilayer to establish the membrane voltage, and compare this to the total number, N2, initially in the vesicle.

(ii) Estimate the free energy of hydrolysis of ATP, ΔGATP, and the overall efficiency of the process of charging the membrane.

(c) After the membrane voltage is established, the pumps are inhibited (stopping all function) and a total of 10 sodium channels, with a single-channel conductance of 10 pS, open instantaneously. Obtain an expression for the membrane voltage as a function of time, justifying any assumptions you make.

(d) NaCl is added to increase the sodium concentration in the external medium to 200 mM. What is the membrane voltage after reequilibration?

Answers

1 This question is asking for an essay covering the material of Lectures 1 & 2, and any additional knowledge gained by further reading.

2 (a) The bilayer can be regarded as a planar capacitor as its thickness is much less than the vesicle diameter.

C = 4πr2 ε0 εt/t

= 4π (100 x 10-9 m)2 x (8.85 x 10-12 F m-1) x 2.5/(5 x 10-9 m)

= 5.6 x 10-16 F

(b) (i) To charge the capacitor:

N1 = Cϕ/e = (5.6 × 10-16F) × (0.15V) / (1.6 × 10-19 C) = 520

Initially in the vesicle:

concentration c = 0.1 mol L-1 = 0.1 x 1000 mol m-3

[Note that 1 M means 1 mol litre. See the lecture Chemical reaction kinetics and equilbrium.]

Problem Equation 1


(ii) The pump is reversible and stops when it reaches equilibrium, i.e. when there is a balance between the free energy changes corresponding to ATP hydrolysis and sodium pumping. The ions that are pumped make a negligible difference to the internal concentration (see above), so the ion concentrations either side of the membrane remain equal to the initialconcentration, 100 mM, and the free energy change of the ions on crossing the membrane is simply the electrical energy. So:


Problem Equation 2


[The pumps in reality are called ATP-synthases. The Na+ versions in this question are less common than the H+ versions. The process in this problem mimics one at the heart of bioenergetics. Ions (usually H+, but sometimes Na+) are pumped across membranes by light or by reactions coupled to oxidising carbohydrates (usually), storing free energy as an electrochemical potential gradient: the ion-motive force or proton-motive force (see Biological Energy Lecture 1: Biological Free Energy):

IMF = Vm + ( kBT/e ) ln (cin / cout)               (Equation 18)

where c represents ion concentrations on either side of the membrane.

The first term, Vm, is the chemical potential difference (per unit charge) across the membrane, which happens to be zero in this part of this question. ATP-synthase normally uses the ion-motive force as the source of free-energy to make ATP, which is in turn the source offree energy for most other biological processes – but in many organisms it works equally well in reverse.]

Define efficiency as the fraction of free energy of ATP hydrolysis (E­ATP­) converted into free energy of ion separation, which in this problem is, to a good approximation, only electrical energy, Eel stord in the capacitor:

EelCφ2 = 6.3 ×10–18 J

Free energy of ATP used:


Problem Equation 3


So the efficiency is 50%.

Some will wonder at such a round number. Its origin is the same as the half in the expression for the energy stored in a capacitor when charged (out of equilibrium) by a constant-voltage source. Note that biological molecular machines are not heat engines, but work isothermally. They can be 100% efficient when working close to equilibrium, and often come very close to achieving this under natural conditions. However this problem started far from equilibrium, with the free energy of ATP hydrolysis initially being converted to heat while pumping ions across a gradient lacking an opposing ion-motive force.

(c) Again, the number of sodium ions that need to flow through the sodium channels to discharge the membrane capacitance makes negligible difference to the internal sodium concentration. So the final concentrations will remain equal inside and outside, and the membrane will discharge to zero volts. So we have simply discharge of a capacitor:

φ = φ0 exp(t/t0)

where

t0 = RC

= (100 x 10-12)-1 Ω x 5.6 x 10-16 F

= 5.6 x 10-6 s

(d) Nernst Equilibrium: sodium channels are still open, so sodium ions can cross the membrane, driven by the asymmetric concentrations, until they reach equilibrium. The equilibrium voltage is given by the Nernst Equation (Equation 17 in Biological Energy Lecture 1: Biological Free Energy):


Problem Equation 4


Here ΔV is φ, and so (with the sign convention of inside minus outside) we have


Problem Equation 5


Understanding energy storage in a concentration gradient across a membrane, and the Nernst Equation, are the key goals of this problem. The Nernst Equation gives the voltage at which the ion-motive force is zero, i.e. ions are in equilibrium across the membrane. Asymmetric concentrations give rise to an entropic force that causes ions to cross the membrane until the resulting electric field stops them. Across a membrane, each type of ion has its own Nernst Potential (a.k.a. diffusion potential) defined by the internal and external concentrations, and the actual membrane potential is largely determined by the ion type which has the largest conductivity across the membrane. Each species of ion that is out of equilibrium across the membrane acts as an electrical cell, and switching conductivities is a mechanism for inducing active changes in membrane potential, for example in nerve signalling.

Taking this further, consider the question: if the pumps worked with protons (H+ ions) instead of sodium, as they do in most living cells, how would things be different? This opens up a whole can of worms. At pH 7 the number of free protons in the vesicle is 106-fold fewer than for the sodium ions at 100 mM – less than one! This sounds like a problem. The key is buffering. Buffers are ionizable molecules, in equilibrium with free protons, that maintain an approximately constant pH by supplying or removing protons from solution. Most lipid head-groups in biological bilayer membranes have some buffering capacity that augments that of any buffers in solution. Even if these are absent, H+ ionscan be supplied by water. Students can look up pH, acid-base equilibrium, and buffering. For example, a bacterial cell might contain only about 100 free protons, but supports fluxes of >106 protons per second across its membrane.

If more complication is needed, note that the relation between surface (capacitive) and bulk charges is actually rather complicated – students can look up Gouy–Chapman electrical double layer theory, Poisson–Boltzmann equation, Stern layer and much more on the distributions of charges in solutions, at and near solvated surfaces.