Discussion Section Questions 11.15.11
1) Nicotine adenine dinucleotide (NAD) is cellular redox reagent. The reduced form
of NAD is abbreviated NADH and the oxidized for is NAD+. In the cell, oxygen
O2 is reduced by NADH according to :
.
Assume the two reduction potentials:
a) Calculate the standard cell potential and standard free energy change for
the reduction of oxygen by NADH:
Solution:
Note you do not adjust for stoichiometry
because cell voltages are not intensive (not per mole quantities).
b)Calculate the cell voltage Ecell and ΔG assuming [NADH]=0.035M,
[NAD+]=0.004M, pH=7, and PO2=0.05 bars. Assume also all standard
concentrations are 1M and the standard oxygen pressure is 1 bar. Is this
reaction endergonic or exergonic under these conditions? Is work required or
produced in order to oxidize NADH and reduce oxygen?
Solution:
The cell voltages is:
So we have to calculate the reaction quotient:
Therefore:
Work is produced because ΔG<0. The reaction is exergonic.
c)For every mole of NADH oxidized, three moles of ATP are produced
according to
For this reaction the standard free
energy change is
at pH=7. If [ATP]=0.003M,
[ADP]=0.001M, and [Pi]=0.01M, calculate the work required to produce 3
moles of ATP.
Solution:
Then
d) What fraction of the free energy produced by oxidation of NADH is
used to produce ATP? This is the efficiency of the respiratory system.
Then
Note how once again we need the negative sign in the denominator. Efficiency is
a positive number and equals the fraction of free energy derived from oxidation of
NADH that produces ATP.
2) A protein complex called the sodium/potassium pump uses the free energy of
hydrolysis of ATP to pump sodium ions Na+ out of the cell and potassium ions K+
into the cell. The net reaction for active transport of sodium and potassium ions is
thought to be:
The diagram below shows the concentrations of sodium and potassium ions inside
and outside a cell. The electrical potential E inside and outside the cell is also
given.
a) Calculate the change in the free energy (i.e. ΔG) involved in transporting 1
mole of sodium ion from inside to outside the cell. Assume the activity
coefficients of sodium ion inside and outside the cell are unity. Assume
the temperature is 310K. Is this an endergonic or exergonic process? Is
work required or produced?
Solution:
Endergonic because ΔG>0 so work is required for this transport.
b) Calculate the free energy change involved in transporting 1 mole of
potassium ion into the cell from outside the cell. Assume the activity
coefficients of potassium ion inside and outside the cell are unity. Assume
the temperature is 310K. Is this an endergonic or exergonic process? Is
work required or produced? Compare the results from parts a and b.
Account for the difference between transporting a mole of sodium ion
versus a mole of potassium ion.
Solution:
The transport ndergonic because ΔG>0 so work is required for this transport to take
place. It takes a lot less work to transport potassium than sodium because the nFΔE term
is negative for potassium but positive for sodium.
c) Calculate the total free energy change involved in transporting 3 moles of
sodium ion out of the cell and two moles of potassium into the cell at
T=310K. what is the work required to accomplish this transport?
Solution:
Work=42,800J
d) The standard free energy change for the hydrolysis of ATP i.e.
at 310K is
at pH=7. If
the total concentration of inorganic phosphate Pi is 0.01M, calculate the
ratio of ATP to ADP (i.e. in the reaction quotient Q) which will furnish the
work required to accomplish the transport described in part c.
Solution:
Therefore,
.
e) Using your value of the standard free energy of hydrolysis of ATP i.e.
, calculate K at 310K. Assuming at equilibrium the inorganic
phosphate concentration is still [Pi]=0.01M, calculate the [ADP]/[ATP} ratio.
Explain why a living organism needs to keep the hydrolysis reaction
as far from equilibrium as possible. For the organism to
remain viable and capable of accomplishing ion transport and other tasks, should Q
be greater than or less than K? Assume [Pi]=0.01M. Explain.
Solution:
Then at equilibrium:
If we reach equilibrium there is virtually no ATP left. It has been almost entirely
hydrolyzed to ADP and ion transport, and any other endergonic reaction cannot proceed.
Death occurs quickly. Given that fact, we want Q<<K so there will be abundant ATP.