HW #4 BP401/P475 Fall 2015
Assigned Th 9/17/15: due: Friday 5pm 9/25/15 Paul Selvin
Reading: Optional
“Direct observation of the rotation of F1-ATPase”, Noji, et al, Nature 386:300, 1997.”
I’ve decided to wait on proving that the rotary motor only requires one ATP per turn. We
will revisit the issue when we study translational motors (such as myosin and kinesin).
Instead, I want to make two thing clearer.
1. Keq
Several people made the observation that:
G= -kBT ln Keq
is valid only if Keq is a dimensionless number; otherwise taking the logarithm of a number with
dimension doesn’t make sense. (By changing the dimension, e.g. moles/liter into moles/milliliter,
the number gets multiplied by 1000, or G by ln 1000.) In the case of a number of reactants
going to the equal number of reactants:
nA  nB
Keq = [B]n/[A]n = {[B]/[A]}n is dimensionless.
With unequal nA  mB + pC
Keq = [B]m[C]p/[A]n which in not dimensionless.
(Although this is obvious, I never realized this until it was brought to my attention!)
Not surprisingly, this has been covered
(see http://en.wikipedia.org/wiki/Equilibrium_constant#Definitions )
and as long as you keep all of the terms in the same units (e.g. Moles/liter), you can consider
this to be dimensionless and all works out.
Wikipedia calls this an illusion:
“It must in fact be dimensionless or it would not be possible to take its logarithm. The illusion is
the result of omitting the standard concentrations from the defining expression. Nevertheless it
is not unusual, particularly in texts relating to biochemical equilibria, to see a value quoted with a
dimension.”
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HW #4 BP401/P475 Fall 2015
Assigned Th 9/17/15: due: Friday 5pm 9/25/15 Paul Selvin
2. ΔG and ΔGo: which is which?
There is also confusion about the terms ΔG and ΔGo, which are sometimes used
inconsistently in the literature (and in my lectures!).
I recommend reading a short section on the web, from which this is taken:
http://www.adriandingleschemistrypages.com/ap/making-sense-of-delta-g-and-delta-gstandard-when-it-comes-to-equilibrium/
ΔGo is the Gibbs free energy in standard conditions. Standard conditions are 1 Moles/liter
for all reactants and products, temperature is 25 C (or, equivalently, 298 K) (and if you are
dealing with gasses, which is generally irrelevant for biochemistry, you have atmospheric
pressure equal to 1 atmosphere.)
Then ΔGo = - kBTln Keq
[Recall, if dealing with moles: ΔGo = - RTln Keq, where R = NkB, with N = Avogado’s number, 6
x 1023 molecules/mole.]
For the chemical reaction a[A] + b[B]  c[C] + d[D] :
where Keq is the equilibrium constant and Keq =
[𝐶]𝑐 [𝐷]𝑑
[𝐴]𝑎 [𝐵]𝑏
.
Notice that this is the situation for equilibrium. The amount of products and reaction are
fixed. Keq is a number and it tells how much the free energy of the products are minus the free
energy of the reactants. In particular, ΔGo is not equal to zero at equilibrium. It tell about how
much the products vs. reactants
This does not predict whether a reaction of some different amounts of products and
reactants will spontaneously go. For this, we need to find ΔG, which is:
∆G = ∆G° + RT ln Q
where Q is defined the same way as Keq, but is the values that you have (or values that you
start with), not necessarily the equilibrium values.
(You will use these thoughts later, in this homework.)
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HW #4 BP401/P475 Fall 2015
Assigned Th 9/17/15: due: Friday 5pm 9/25/15 Paul Selvin
1. Energy from Splitting ATP to ADP + Pi
One can crudely calculate the energy released in splitting a phosphate from ATP to make ADP
and phosphate. In ATP, negatively charged oxygen’s are in close proximity. In particular, one of
the negatively charged oxygen’s of the terminal phosphate is near (~2Å) a negatively charged
oxygen of the second phosphate. In ADP, this bond is cleaved, letting the oxygen’s separate.
What is the potential energy released? [What dielectric constant to choose is tricky; it’s in water,
but water can’t completely get in between the oxygen’s – do the calculation assuming  =80o
and =o]. Actually, the energy for splitting the negatively charged oxygen’s is about 12kT.
What would  have to be if the energy released was 12kT? When splitting an ATP one should
get 25kT of energy, where does the other 13kTs of energy come from?
2. Energy from ATP using ΔG:
How Efficient is ATPase?
1st = 3 x 7kT. Practically 100% Efficient
2nd : Free energy in ATP
(Calculate #pN-nm for DG; compare it to ATP)
ATP à ADP + Pi
DG0 = -kbT ln Keq
Keq = exp –DG0 / kbT;
At standard conditions (1M)
[See above for the difference between DG0, and DG]
DG = DG0 + kbT .ln {[ADP][Pi]/[ATP]}
In problem #1, we estimated the energy from splitting ATP. Now we want to show that it
depends on the concentration of the final products, [ADP] and [Pi].
For the reaction A + B ↔ C + D, one can calculate the free energy (ΔG) using the following
equation:
Here ΔGo is the standard free energy change.
ATP + H2O ↔ ADP + inorganic phosphate (Pi)
Note that since H2O is the solvent, [H2O] = 1 M and body temperature is 37°C.
a. After an equilibrium experiment at body temperature, we find that Keq = 130,000.
What is ΔGo?
b. Lets say that [ATP] = 10 mM, [ADP] = 100 µM, and [Pi] = 1 mM. What is the free
energy, or the energy available from the hydrolysis of 1 mole of ATP? For an individual
ATP?
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HW #4 BP401/P475 Fall 2015
Assigned Th 9/17/15: due: Friday 5pm 9/25/15 Paul Selvin
c. Now let [ATP] = 2 mM, [ADP] = 1 mM, and [Pi] = 2 mM. What is the free energy,
or the energy available from the hydrolysis of an individual ATP? (Use units on pN-nm.)
d. Calculate 100 pN-nm (roughly the amount of energy released with splitting one
ATP) in units that allow you to compare this value to the free energy. Are they
consistent?
As you can see, the amount of energy released by ATP depends on the cell’s [ADP] and
[Pi], which vary from cell to cell, hence ATP is 80-100pN-nm.
3. In class, we discussed F0F1-ATPase, a biological rotor that uses a proton gradient to create
ATP1. The α and β subunits are proteins that form a barrel, which holds the γ subunit, another
protein (see below). Protons traversing the membrane cause the γ subunit to rotate, which
drives ATP synthesis, if there is an excess of ADP and Phosphate (see figure below).
But Kinosita et al, in his original 1997 article in Nature, used only a part of the enzyme and
stuck it to a glass slide.
a. Which part did he use, F1 or Fo?
b. He added ATP, instead of ADP and Phosphate. Did he form ATP, or hydrolyze it (into
ADP + Pi)?
c. Where did the energy come from to rotate the actin filament?
d. Where is the membrane and the proton gradient? Why did this experiment work?
e. The actin filament ended up going in a counter-clockwise fashion. Is there a way to
make it move in a clockwise fashion?






Left: The structure of ATPase; Right: The  barrel structure.
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Figure copied from “Direct observation of the rotation of F1-ATPase”, Noji, et al, Nature 386:300, 1997
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HW #4 BP401/P475 Fall 2015
Assigned Th 9/17/15: due: Friday 5pm 9/25/15 Paul Selvin
4. In 1996, Kinosita, et al attached an actin filament to the γ subunit, and watched it rotate. The
actin filament may be approximated as a rod of length 2.6 m, and radius 5 nm. They observed
that, at high concentrations of ATP, the rod rotated at a maximum speed of 0.5 revolutions per
second. They calculated that the torque on a rigid rod held at one end and rotated in a solution is
4
L3 /[ln( L / 2r )  0.447] , where r is the radius of the rod, L is its length,  is the
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rotation rate in radians per second, and  is the viscosity of water, 10-3 N s m-2.
a) Calculate the torque on the rod at this speed, in pN·nm (picoNewton nanometers.) (These
values are as much as 3x too low because the actin is moving near the glass slide, which ends up
increasing the drag considerably.)
b) Kinosita et al. found that the rod didn’t rotate continuously, but would make a series of
discrete “jumps”. Considering the symmetry of the structure of the  barrel shown in the figure
on the right above, what is a likely value for how big each jump is, in degrees?
c) After each jump, it synthesizes a single ATP (See
https://en.wikipedia.org/wiki/ATP_synthase#Binding-change_model for how each of the
binding sites in the 3-fold symmetric structure is occupied). Also note that the ratio of
protons to ATP’s is still 3:1, as the synthase has 10-15 proton binding sites, depending on
the species. These protons supply the energy to bring the ADP + Pi close together.
d) Considering the amount of energy in a single molecule of ATP (~100 pNnm), what is the
efficiency of this process? If your answer is unreasonable, (e.g. significantly above 100%) try
reconsidering your answer to part b.
e) Kinosita also tested the motors with longer filaments (see below). He found that the longer the
filament, the slower it goes. Why is this? Assuming it stays rigid, what range of lengths could he
have used and still gotten the motor to rotate?
f) In part 4a, you calculated that the rotary motor can generate a certain amount of torque, If you
assume that the torque point corresponds to the - interface (see figure above), you might
estimate the torque point to be ~1 nm away. How much force can the rotary motor generate? Is
this more or less than translating motors, like myosin (3-6 pN), kinesin (5-7 pN) or RNA
polymerase (14 pN)?
5. Replication Fork
(This is related to what we have covered in that it involves a rotary type of motor involving
the relative rotation of DNA and an unwinding protein.)
(I have to give you something since I cut out the more difficult problem involving ATP!)
As soon as Watson and Crick announced their double-helix model of DNA structure, both
physicists and chemists were quick to point out some troubling features of the model. For
example, how are the two strands separated for replication, when they’re wound around each
other? Below is a simplified figure showing a replication complex (e.g. DNA polymerase and/or
helicase) (triangle) unwinding DNA. If the replication machinery cannot rotate, because, e.g. it’s
tied down to cellular components, the DNA must rotate about its long axis 360 degrees every 10
bases unzipped.
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HW #4 BP401/P475 Fall 2015
Assigned Th 9/17/15: due: Friday 5pm 9/25/15 Paul Selvin
We will now calculate the frictional drag associated with replication (i.e. the drag caused
by the DNA rotating about its axis) and compare the energy lost due to friction to the energy
required to form the phosphate-sugar backbone bonds.
Model a length of DNA as a cylinder, which is rotated with constant angular velocity 
during replication. The frictional torque is given by,
f = 4R2L, where  is the viscosity of water (at 37 °C), L is the length of DNA, and R is the radius
of DNA.
a) Write the equation for work done against the frictional torque for one helical turn.
b) Consider a DNA piece 6000 helical turns long (still assuming it’s a straight line – which is not
really true). What is L? What is R?
c) If it takes 100 seconds to replicate this DNA, what is ?
d) Calculate the value for work done to overcome friction per helical turn.
e) Let’s assume the energy of each phosphate-sugar bond in the backbone is about 15kT.
How does the energy required to synthesize these bonds per helical turn compare with the
energy required to overcome the rotational drag? Is the rotational drag a significant fraction
of the cost of replicating DNA?
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