Homework #7
Physics 498Bio Spring 2012
Prof. Paul Selvin
1. One or Two ATPs Binding?
Molecular motors like myosin or kinesin, which undergo a hand-over-hand stepping, have a
“lag-time” in the fluorescence when one leg is labeled with a fluorophore. After this leg takes a
step, it waits until other leg takes a step. The histogram of the time between a step therefore
has a “lag”, or zero probability immediately after a step. The probability therefore looks like
texp(-kt), where t is the time, and k is the stepping rate. If the motor “walks” by rotating, as in the
F1F0ATPase (lec 6), it will have the same behavior (green or red curve shown on the right-hand
graph, below). Simultaneous arrival of two (or more) ATP molecules are expected to be a rare
event. If, on the other hand, each step (i.e. rotation) is driven by one ATP molecule, the
histogram should be exponential (black line), as shown.
60
50
40
count
ATPase
Myosin or kinesin,
labeled on foot
30
20
10
0
0.0
0.2
0.4
0.6
0.8
1.0
1.2
dwell time (sec)
Let’s first figure out what the graph of ATPase would look like if a rotation were due to a single
ATP binding. As discussed in class, the rate of change in ATP concentration will be directly
proportional to the concentration of ATP
d [ ATP ](t )
= −k1 × [ ATP ](t )
dt
Here k1 is a positive rate constant.
a) Solve for [ATP](t)
b) Note that [ATP] is directly proportional to the probability of an event occurring, P(t).
Solve for P(t), using normalization. This equation will reflect the shape of the graph
of dwell times if only a single ATP binds for each stepping event. This in-fact, fits the
data (black line) for the data.
Now let’s explore the scenario where two ATPs bind for each 120º stepping event. This
involves two separate binding events.
c) Rewrite your solution to part b for each ATP binding event. Each event has a
different rate constant, k1 and k2. We also need new notation for the probabilities
since they will not be the same. Use f(t) and g(t).
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Homework #7
Physics 498Bio Spring 2012
Prof. Paul Selvin
If the combined event takes a total of t seconds, and the first ATP binding takes u
seconds, then the second ATP binding must take t-u seconds. To get the probability of
both events occurring in time t we must multiply and two separate probabilities and
integrate over the total time:
t
P(t ) = ∫ f (u ) g (t − u )du
0
d) Plug in your equations from part c. and solve for the total probability.
e) Simplify the equation by setting the rate constants equal. Be careful, a simple
substitution into part d. will yield zero divided by zero.
f) Sketch a graph of the probabilities vs. dwell time for both one and two ATP bindings.
According to the graph of dwell times for F1-ATPase shown in class, do one or two
ATP bind for each 120º rotation?
g) If myosin V (which has a 37 nm center-of-mass movement, or 74 nm - 0 nm foot
movement) is labeled on the leg (rather than the foot—see blue circle), the answer
will be a single exponential (see data, right-hand-side, below). Why is this?
2. Multiple Steps of an Enzyme. If we assume that each step requires n hidden
irreversible Poissonian steps with identical rate k, the data can be fit with the gamma
distribution, tn–1·exp(–kt). For a single ATP (n = 1), as in the case of F1ATPase, this
decays to exp(–kt). For the case of n = 2, a possibility which was considered in the
ATPase (but did not properly fit the data) it reduces to t·exp(–kt). It fit the data very well
for a kinesin or myosin labeled on the foot.
Derive the most general form, tn–1·exp(–kt), using the n = 1 and n = 2 as examples.
Later in class, we will see how a certain process associated with DNA proteins follows
this more general case.
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Homework #7
Physics 498Bio Spring 2012
Prof. Paul Selvin
3. Why is the excited-state lifetime an exponential? That is, you put in a pulse of light to a
fluorophore, and a fluorophore will decay as exp(-t/tf) (where tf is typically a few nanoseconds). It
turns out that a fluorophore’s emission rate (rate = 1/lifetime) is an example of Poissonian
statics. Example of Poissonian statistics, can be found everywhere. The number of raindrops in
a specific area in a given amount of time; the number of nuclear decay in a given mass of
uranium in a given time; the number of photons hitting a detector in Astronomy….
We want to find the probability that a molecule, X will remain in its excited state over a unit of
time. kF is the probability per unit time that X*, which is already in the excited state at any time
“t”, will leave the excited state through door “F” (fluorescence).
(a) What is the approximate probability that X*, which is already in the excited state at time
“t=0”, will leave the excited state through door “F”, after the “short” time, Δt?
(b) What is the approximate probability per unit time that X*, which is already in the excited
state at any time “t”, will remain in the excited state.
(c) What is the approximate probability for X*, which is already in the excited state at time
“t=0”, will remain in the excited state for the time 2Δt?
(d) If you consider the total time from t=0 to t=T, and divide up this time interval into “n” time
points, so that Δt = T/n, what form does the equation from part c take?
(1-kFΔt)n = (1-kFT/n)n
(e) What is the exact probability of X* staying in the excited state over this interval t = T if you
take the limit of n approaching infinity? [HINT: what is the Taylor expansion of e-x? What if
you take only the first two terms… and what is x here?]
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