Biomedical Engineering Reference
In-Depth Information
states that natural processes of the universe proceed from states that are more or-
dered to states that are more disordered. The movement of any molecule up (to a
higher concentration) or down (to a lower concentration) a concentration gradient
involves a change in free energy (
Δ
G
—down releases energy so
Δ
G
is negative; up
consumes energy so
G
is positive). Ideally, the amount of free energy released or
consumed is calculated using
Δ
C
Δ
GJ
[ /mol]
=
RJ
[ /mol.K] [K]ln
T
in
+
zFC
[
/mol]
Δ
VV
[
]
(2.1)
C
out
G
is the change in Gibbs free energy per mole; R is the universal gas con-
stant (1.987 cal.mol
−1
.K
−1
or 8.314 J.mol
−1
.K
−1
); T is the absolute temperature,
which must be in Kelvin (K
where
Δ
°F);
C
in
is the
concentration inside the cell (mM);
C
out
is the concentration outside the cell (mM);
z
is the electrical charge of the transported species;
F
is Faraday's constant and
=
273
+
°C) or Rankine (°R
=
459.67
+
V
is
the voltage across the membrane (discussed in Chapter 3). Faraday's constant is the
number of units of static charge per mol of charge, which is 96,484.5 Coulomb/mol
(
Δ
23,062 calories/V.mol) released as one mole of ions moves down a voltage gradi-
ent of 1 volt (1,000 mV). Equation (2.1) presumes that concentrations represent
the activity of that component in each compartment and there are no interactions
with other species.
The first term on the right side of (2.1) is due to the chemical gradient across
the membrane and is called chemical potential. It is the amount by which the free
energy of the system would change if an additional particle is introduced without
altering the distributions and volume of the system. The second term on the right
side of (2.1) provides energy due to the electrical gradient, normally referred to as
drift, and is discussed in Chapter 3. For the transport of uncharged molecules, the
contribution from the second term is neglected and the total Gibbs free energy is
calculated by knowing the total change in the number of moles in the system.
=
EXAMPLE 2.2
The concentration of glucose inside the cell is 0.5 mM, and the concentration of glucose
outside the cell is 5 mM. Assuming a body temperature of 37°C, calculate
Δ
G
.
Solution: Using (2.1) without the electrical gradient
C
G T
C
0.5
ln
in
out
8.314 * 310 * ln
5,934.545 J/mol
5.9 kJ/mol
Δ=
=
=−
=
5
As the process proceeds with the release of free energy, it can proceed spontaneously
if the phospolipid bilayer allows glucose to pass through it. However, the phospholipid
bilayer is not permeable to glucose and needs other means of transport (discussed in Sec-
tion 2.3.2).
