Biol/Chem 5310

Lecture: 22

November 14, 2002

Transport through Membranes

The structure of biological membranes, with their nonpolar cores, makes them effective barriers to charged and highly polar molecules. This includes most metabolites.

To function, cells need to take up or pump out a variety of molecules, including many polar or charged ones. Specialized mechanisms are necessary. This movement is call transport. Most transport is mediated by integral membrane proteins.

1. Thermodynamic view of Transport.

A. Consider the simple movement of molecules from a region of one concentration (C₁) to a region of a different concentration (C₂)        

The free energy required to move these molecules is   DG = RT ln (C₂/C₁)

If C₂ < C₁, then    DG < 0

This is a favorable process, this is transport down a concentration gradient.

At equilibrium (in this simple system),    C₁ = C₂ ,    so DG = 0

At this point (perhaps after a long time) there is no further net transport.

Important: Transport down a concentration is favorable (DG < 0).

Example:                 at 37 degrees C (310K)

C₁ = 100 mM

C₂ = 10 mM

Convert C to K, convert concentrations to M

DG = RT ln (C₂/C₁)             from C₁ region to C₂ region

DG = (8.31 J mol^-1 K^-1) (310 K) ln (0.1)

DG = - 5.9 kJ mol^-1

B. Charged molecules are also influenced by membrane potential, in addition to concentration. Membrane potential is called DY . It is a measure of the distribution of charged species across a membrane.

Defined mathematically:

DY = Y_in - Y _out              with units of Volts

The free energy to move an ion of net charge z across a membrane potential of DY is DG = zFDY

where F is the Faraday constant which is 96.5 kJ mol^-1 V^-1            ( 1 Coulomb = 1 J/V)

The effects of charge and concentration are additive, So:

DG to move a molecule of charge z from a region of C₁ to C₂ is

DG = RT ln (C₂/C₁) + z FDY

A typical cell has a membrane potential of about -100 mV (negative inside)

This means that an ion can move against its concentration gradient, if the membrane potential is significantly favorable.

For example:

Consider K⁺ transport into a cell with a membrane potential of -100 mV (negative inside)

z = +1

F = 96.5 kJ mol^-1 V^-1

DY = -100 mV = - .1 V

DG_{(membrane potential)} = -9.65 kJ mol^-1

Transport would be favorable as long as the concentration term were less than 9.65 kJ mol-1.

DG_{(Concentration)} = RT ln (C₂/C₁)        

Set  -DG_{(membrane potential)} = DG_{(Concentration)}

and solve for C₂/C₁

9.65 = RT ln (C₂/C₁)

exp(9.65/RT) = C₂/C₁ = exp(3.74) = 42

So K⁺ could move against a concentration gradient as high as 42-fold with this membrane potential.

If the concentration gradient were less than 42-fold, the net DG would be negative, and the transport would go.

If the concentration gradient were more than 42-fold, the net DG would be positive, and the transport would not go.

If going down a concentration gradient, DG <0

If a positive ion is moving towards a negative region, DG < 0

C. When transport is obligatorily coupled to a chemical reaction there is a third term to consider when calculating DG

DG = RT ln (C₂/C₁) + z FDY + DG'

This DG' might be from ATP hydrolysis (very favorable term) or from a second transport process.

In this way an unfavorable transport can be driven by a favorable chemical reaction or a favorable secondary transport.

2. What are the Rates and Mechanisms of Transport?

A. Non-mediated transport

Define J as the rate of transport per unit area (Flux) with units of moles cm^-2 sec^-1

For a membrane of thickness l

From C₁ to C₂

J = -D_A (C₂-C₁)/l

Where D_A is the diffusion coefficient of A in the membrane (units are area/sec).

D_A /l   is defined as P, the permeability coefficient. So,

J = P (C₂-C₁)

We can see that the rate is proportional to the difference in concentration between the 2 regions.

B. Passive, mediated transport

In this case the transport is mediated by another molecule, so it resembles the E + S --> ES situation of Michaelis-Menten kinetics: J vs. [S] yields a hyperbolic plot that saturates at some value Jmax .

With a Km that equals [S] when J = one half Jmax.

Competitive inhibition exists.

The mediators can be proteins, or other molecules.

They can be categorized functionally as

i) channels or pores (e.g. Gramicidin A)

ii) carriers (e.g. Valinomycin)

Ionophores are organic molecules, usually synthesized by microbes. They increase the transport of ions by acting as carriers or pores.

Examples:

Valinomycin: K+ specific carrier, cyclic polypeptide or mixed D and L form amino acids. One K+ binds inside, coordinated by 6 ligands that come from C=O. Net charge with K+ is plus 1.

Another example: Monensin, a Na+ carrier

Gramicidin A: Monovalent cation channel, 15 amino acids, alternating D and L form. It dimerizes in a head to head fashion, to make a channel across a membrane. It is called a beta-helix, with 6-7 residues per turn. {see Chime LINK}

Porins form channels through membranes.

Glucose transporter: integral membrane protein, with 12 likely transmembrane spans. It is a "gated pore". It has 2 conformations. One binds glucose on the inside, and one binds glucose on the outside . (Figs. 10-34 & 10-35)

C. ATP-driven active transport

Some cells use most of their ATP to pump ions

Na-K-ATPase

Catalytic part (a-subunit) is about 110 kD, found in plasm membrane of red blood cells, many other cells. ATP is hydrolyzed to miantain the desired concentrations of Na+ and K+ inside the cell {See images of natural inhibitors that are used as cardiac drugs in text Box 10-3.}

Ion	outside cell	inside cell	ratio (in/out)
Na	140 mM	10 mM	.071
K	5 mM	100 mM	20.0

ATP must be hydrolyzed continually inside the cell to maintain the high concentration of K⁺.

Steps of the reaction:

1) 3 Na⁺ ions bind inside the cell, with ATP

2) ATP is hydrolyzed, an Asp residue is phosphorylated, ADP is released

3) Na⁺ ions are driven out of the cell, by the proteins conformational change.

4) 2 K⁺ ions rapidly bind on the outside of the cell, to replace the Na⁺ ions.

5) The protein is de-phosphorylated, driving the K⁺ ions to be released inside the cell, P_i is also released inside.

This requires that both Na⁺ and K⁺ are both driven against their concentration gradients. The membrane potential is -70 mV (negative inside)

Will this work?

Check the calculation at 37 C

First calculate the Na⁺ term, then calcculate the K⁺ term, then sum and compare to the DG for ATP hydrolysis.

1) Na⁺ --> out

DG = RT ln (C_out/C_in) + z F DY        

 (the sign for the second term must be positive because this step is unfavorable)

DG = (8.31) (.001) (310) ln (140/10) + (1) (96.5) (.07)

DG = 13.55 kJ mol-1

2) K⁺ --> in

DG = RT ln (C_out/C_in) + z F DY

 (the sign for the second term must be negative because this step is favorable)

DG = (8.31) (.001) (310) ln (100/5) - (1) (96.5) (.07)

DG = 0.97 kJ mol^-1

3) Consider stoichiometry (per mole of ATP hydrolyzed)

DGnet = 3 DG_(Na⁺) + 2DG_(K⁺)

= 42.59 kJ mol^-1

4) DG standard state for ATP is -30 kJ mol^-1, so that would not be enough

but due to the concentrations of ATP, ADP, P_i, and the pH of the cell, the actual DG will be about -50 kJ mol^-1

If [ATP]=3.0 mM

[ADP]=0.8 mM

and [Pi]=4.0 mM

So DGnet = 42-50 < 0, and the process goes.

Other ATPase's:

Ca⁺⁺ ATPase pumps Ca⁺⁺ out of the cytosol or particular organelles, at the expense of ATP.

 This is the only member of this family to have its structure determined at atomic resolution: LINK

Last updated Tuesday, November 12, 2002

Comments/questions: svik@mail.smu.edu

Copyright 2002, Steven B. Vik, Southern Methodist University