HW #10 BP401/P475 Fall 2015 Assigned Th 11/05/15: due: Thursday 11/12/15 Paul Selvin
1.
On page 6 of lecture 12, we have the following diagram:
Basics of fluorescence
Shine light in, gets absorbed, reemits at longer wavelength
2. Thermal Relaxa on
[Picosec]
3. Stays at lowest
excited vibrational
states for a “long”
time (nsec)
E
1. Absorp on
[Femtosec]
4. Fluorescence (krad)
& Non-radia ve (knonrad)
[Nanosec]
5. Thermal relaxa on
[Picosec]
kT = knr+ krad
tf = 1/ kT
Quantum yield =
prob fluorescing”
krad /(knr+ krad)
a. Draw an absorption spectrum i.e. the probability of being excited vs. wavelength. (Recall
that E and wavelength or reciprocal, i.e., E =hc/
b. Draw an emission spectrum, i.e. the probability of being excited vs. wavelength.
c. Why does the emission spectrum more-or-less look like the mirror image of the
absorption spectrum?
Now look on lec 18, page 15. Does the mirror image approximately hold true?
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HW #10 BP401/P475 Fall 2015 Assigned Th 11/05/15: due: Thursday 11/12/15 Paul Selvin
2. Signal to noise: Background = fluctuations signal + background light, where most of
fluctuations are in the signal! See pg 5, lec 19
There was some confusion in
Signal to Noise: key to seeing
class, as to what “Background”
More signal, can see it
means. From now on, we define
background as the number of
More noise, harder to see signal from noise
photons entering the detector (your
When signal is about ≈ noise, can’t see it.
eye) from sources not arising from
Noise = background + fluctuations.
the signal of interest. Fluctuations,
we describe as the fluctuations in the
Can subtract off background, but always left with fluctuations.
signal and the background.
If you have some signal N photons, there is always a fluctuation:
What counts is the signal divided
√N.
by the fluctuations. (Unfortunately,
Signal/Noise = N/√N = √N
this, is often called the signal-toWhat about something really bright (like Sun).
background), where the fluctuations
Signal is a lot. Is noise a lot too?
Yes
equals the square root of the number
Why can you see it? Is S/N high?
Yes Yes
Measure some small fluctuation (like star nearby?)
of photons, from the signal +
No
background.
So, for example, say you shine a
light at a fluorophore attached to a protein. It puts out 10,000 photons/sec, which strikes your
detector. If you don’t have the fluorophore present, imagine two cases: 1) one where there is no
background; 2) where there is some background, 1000 photons/sec (from dirt that is laying
around); 3) where there is a ton of dirt around 10,000.
In case 1:
Signal = 10,000 photons
Background = 0 photons
Fluctuations = fluctuation on 10,000 photons = √10,000 = 100.
Your Signal/Fluctuations = 10,000/100 = 100.
In case 2:
Signal = 10,000 photons
Background = 1000 photons
Fluctuations = fluctuation on 10,000 + 1000photons = √11,000 = 105.
Your Signal/Fluctuations = 10,000/105 = 95:1.
In case 3:
Signal = 10,000 photons
Background = 10,000 photons
Fluctuations = fluctuation on 10,000 + 10,000 = √20,000 = 141.
Your Signal/Fluctuations = 10,000/141 = 71:1.
Notice even in this case, you can clearly “see” the signal. The reason is you can subtract off the
background (10,000 photons), which only leaves the fluctuation in the background (√10,000 =
100), which is a lot less. Of course, you still have the fluctuation coming from the signal
(√10,000).
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HW #10 BP401/P475 Fall 2015 Assigned Th 11/05/15: due: Thursday 11/12/15 Paul Selvin
The situation with a light-background experiment, is much worse.
Discuss what the signal, background, fluctuation is. Assume that you a light source which
puts out 10,000 photons. (In fact, this would be an incredibly dim light
source—more likely it puts out a million photons/sec, or bigger.)
Case 1 where a molecule which absorbs 1% of the light (which would
actually be a huge amount);
Case 2, where it absorbs 0.1%;
Case 3, where is absorbs 0.01% of the incident light.
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HW #10 BP401/P475 Fall 2015 Assigned Th 11/05/15: due: Thursday 11/12/15 Paul Selvin
3. 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
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)
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.
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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HW #10 BP401/P475 Fall 2015 Assigned Th 11/05/15: due: Thursday 11/12/15 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?
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HW #10 BP401/P475 Fall 2015 Assigned Th 11/05/15: due: Thursday 11/12/15 Paul Selvin
4. Why is the excited-state lifetime an exponential? See lec 18, pg 15. If you put in a pulse
of light to a fluorophore, the 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 statistics. 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, are all
examples.
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, t?
(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 2t?
(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 t = T/n, what form does the equation from part c take?
(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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