SME
Biochemistry
Bioenergetics
Use this for “A muscle cell at 37 °C holds 8 mM ATP, 0.9 mM ADP and 8.1 mM Pᵢ. Find ΔG for ATP hydrolysis in this cell.”
Diagram
K
=
exp(−ΔG°
÷
RT)
=
137,000:
the
minimum
is
at
ξ
=
K
1 + K
=
1
−
7.3 ×
10
-6
,
pressed
against
the
edge:
the
inset
magnifies
it,
and
the
line
below
shows
it
at
log
K.
ΔG = ΔG° + RT ln Q = −50.1 kJ/mol: Q < K, downhill forward, so it runs forward toward more ADP + Pᵢ.
Drawn for a model ATP ⇌ ADP + Pᵢ with ideal mixing; G of pure ATP is 0.
This is an example. Type your own values to start a new problem.
ΔG°′
Standard free energy change
example · kJ/mol
T
Temperature
example · K
[ATP]
ATP concentration
example
[ADP]
ADP concentration
example
[Pᵢ]
Phosphate concentration
example
Q
Reaction quotient
worked out (not typed)
ΔG
Free energy change in the cell
calculated · kJ/mol
Formulas
Q
=
[ADP]
×
[Pᵢ]
÷
(1,000
×
[ATP])
5.00 ×
10
-4
=
0.5
×
5
1,000 × 5
ΔG
=
ΔG°′
+
8.314
×
T
×
ln(
Q
)
1,000
−50.1
=
−30.5
+
8.314
×
310.15
×
ln(5.00 ×
10
-4
)
1,000
Assumptions
•
ΔG°′ is the biochemists’ standard: pH 7 and 1 M for everything else; ATP → ADP + Pᵢ has ΔG°′ = −30.5 kJ/mol.
•
Q = [ADP][Pᵢ] ÷ [ATP] with each concentration in M (mM ÷ 1000); water is left out.
•
ΔG = ΔG°′ + RT ln Q with R = 8.314 J/(mol·K); body temperature, 37 °C, is 310.15 K.
•
A cell keeps ATP high and ADP low, so Q is tiny and ΔG falls well below ΔG°′.
Step-by-step
WE KNOW
[ATP] = 5 mM, [ADP] = 0.5 mM, [Pᵢ] = 5 mM, ΔG°′ = −30.5 kJ/mol, T = 310.15 K
FIND
reaction quotient (Q), free energy change in the cell (ΔG)
Step 1 · Find reaction quotient (Q)
Use Q = [ADP] × [Pᵢ] ÷ (1,000 × [ATP])
Products
over
reactant,
each
in
M
(
mM
1000
):
two
of
the
three
1000
cancel.
Q
=
0.5
×
5
1,000 × 5
Q
=
0.5
×
5
5,000
Q
=
2.5
5,000
Q
=
5.00 ×
10
-4
Step 2 · Find free energy change in the cell (ΔG)
Use ΔG = ΔG°′ + 8.314 × T × ln(Q) ÷ 1,000
Find
RT
ln
Q
in
J/mol,
divide
by
1000
for
kJ/mol,
and
add
it
to
ΔG°′.
ΔG
=
−30.5
+
8.314
×
310.15
×
ln(5.00 ×
10
-4
)
1,000
ΔG
=
−30.5
+
8.314
×
310.15
×
−7.601
1,000
ΔG
=
−30.5
+
2,578.5871
×
−7.601
1,000
ΔG
=
−30.5
+
−19,600
1,000
ΔG = −30.5 + (−19.6)
ΔG = −50.1 kJ/mol → ΔG is below ΔG°′: the cell keeps [ATP] high, so each ATP gives more than the standard value.
Check
Put the numbers back in. Both sides should match.
5.00 ×
10
-4
=
0.5
×
5
1,000 × 5
✓
−50.1
=
−30.5
+
8.314
×
310.15
×
ln(5.00 ×
10
-4
)
1,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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