All five questions were scaled to be worth the same amount of total credit.
Question 2 Final Exam Answer Key
1 – The ecological purposes of chemotaxis in neutrophils and bacteria is location and
destruction of pathogens, and location and consumption of nutrient sources respectively.
In both cases the distribution will be non-uniform through space. This is particularly true
of the formyl-peptide attractants of neutrophils which are presumably only produced at
infected sites. In both cases small-scale local gradients are likely to be smooth.
2 – Both bacteria and neutrophils consume attractant, directly, or by consuming the
source of the attractant (respectively). Attractant molecules can be expected to degrade,
diffuse, or be subject to movement due to the fluid properties of the medium containing
the nutrient. The sources of the nutrients may also change their location in time, as is
certainly the case in the moving bacterial targets of neutrophils.
3 - foodSurf(yactual-Nsize:yactual+Nsize,xactual-Nsize:xactual+Nsize)=
foodSurf(yactual-Nsize:yactual+Nsize,xactual-Nsize:xactual+Nsize)-deplAmount;
This is zero-order depletion. First-order depletion might more accurately reflect
consumption, or something Michaelian in which there was a maximal rate of
consumption that dropped off as nutrient levels fell. Zero-order depletion is inaccurate in
the case of very low nutrient, the cells will deplete it entirely at this point, trying even to
take more nutrient than is available for consumption.
4 – Neutrophil direction is determined by considering a number of points around the cell
(corresponding to points on the membrane of the cell where peptide is detected). The
single point in highest nutrient concentration is identified, and the cell will then proceed
in that direction (plus or minus a random dither of its position). This is more or less
accurate. Neutrophils move roughly in the direction of the steepest nutrient gradient, but
there is a lag in re-orientation not found in our simulation. Since we are simulating long
timescales of chemotaxis, and reorientation happens over short timescales this should not
harm the qualitative accuracy of the simulation. The dither effect on position may or may
not be accurate in its magnitude but corresponds to a Brownian motion effect that
certainly happens in real life.
5- When trying to make a simulation comparison between two mechanisms, there must
certainly be different parameters, otherwise the mechanisms being compared would be
identical. Since all simulation-specific parameters, such as step size, dither, and depletion
rates, are the same, a comparison should be feasible.
6 – Hand-drawn examples of what you should see (Matlab isn’t working on my
computer):
Bacteria Neutrophil
note depletion of nutrient where the cell dwelled. The drawn path must appear
significantly longer than the path at the end of the movie.
7 – Bacteria execute a biased random walk. The move randomly until the reach a good
area and then tend to stay there. Ultimately they may fall off that peak and end up back in
low nutrient regions. When this happens, sometimes they are able to find better peaks.
Generally they do a pretty good job of finding the points of very high nutrient before the
end of the simulation. They rarely completely consume a peak.
Neutrophils migrate directly up the nutrient gradient to the nearest peak. They
stay on that peak until it is essentially all consumed, or until it is optimal to move toward
a different peak. Most of the time by the end of the simulation they have covered one or
two peaks, and rarely do they find the optimal peaks, and never do they fall off peaks.
8 – The Neutrophils should look something like:
Bacteria look like:
Summing the values experienced as a function of time was a fine way to integrate the
curve. Using any better numerical methods earned you bonus points. A lot of people gave
averages but not the under-the-curve values for ten repetitions. Including these areas was
required for full credit. Because there is significant noise in the final areas as a function
of whether or not the cells find the best peaks, a wide range of average answers were
acceptable. Typical answers had bacterial nutrient experienced around 104, and neutrophil
nutrient experienced averaging around 8000.
9 – Since the criteria used to determine “reaching a local peak” varied widely among the
answers, a wide range of answers was suitable, as long as the same criterion was applied
to both cell types. Some strategies included 1) taking the first time in the course of
nutrient experienced in which nutrient dropped, 2) qualitatively assessing it, 3) smoothing
the nutrient experience curved and applying 1, 3) setting an absolute nutrient threshold,
and looking for the time it took to reach any point at that threshold or above, or 4)
looking at the original landscape, and asking how long it took to get within a fixed
distance of one of the original maxima. In all cases the bacteria took longer. Stating that,
and justifying the method used to determine this earned full credit.
10 – Redundant question with #1, sorry about that, but you can be consoled by the fact
that it was a chance to boost your score without having to engage your brain. It was
interesting to see when this answer differed from that in #1 though.
Answers from 8 do make sense, since the goal of bacterium is to use nutrient for
growth, and it consumes more than the neutrophils whose goal is to track down a
pathogen and thoroughly eliminate it.
Answers from 9 also make sense, since the bacterium wants to find the best
source, and not necessarily the closest one, and so it will take potentially time to find any
source. In the case of the neutrophil, it wants to find the target bacteria as quickly as
possible, and finding even a single bacterium is important, so it might as well start with
the closest one.
11- Increasing the ingestion rate had little effect on the neutrophil except to allow it to
clear a peak faster and therefore advance to other peaks more easily. The effect on the
bacterium was very different, since increasing the ingestion rate very quickly reduces the
information that the bacterium has available to navigate the landscape. This caused the
behavior to change into a less-biased (and therefore more random) random walk. People
who commented on the paradoxical conclusion that eating too fast (even when eating is
your ecological goal) is counter-productive (when the food source itself is the source of
localization information) earned bonus points.
Changing the step size did not greatly affect the bacterium, which now simply
chemotaxes over smaller scales. That is, its swimming bouts are shorter since in order to
be of the same length, they would have to be composed of more steps, which becomes
less likely. The neutrophil is crippled by this modification since it now cannot step
outside out of the region in which it is eating. It will generally therefore become stuck in
a much smaller region of a single peak. People who commented on the fact that this
corresponds to a requirement that consumption happen on a slow timescale compared to
that of movement earned bonus points.
12 – Even assuming it is true that bacteria are too small to detect spatial differences,
hopefully working through this problem has illustrated that given the ecological goal of
consuming as much nutrient as possible, it is not in the evolutionary interest of a
bacterium to use spatial chemotaxis of the sort used by neutrophils. So it is quite possible
therefore that temporal-based chemotaxis is optimal for bacteria regardless of their size.
Manduca and C elegans both employ temporal-chemotaxis and they are clearly large
enough to evolve spatial sensing. The latter result may make particular sense because C
elegans also chemotaxes for the purpose of consuming nutrient in a complex environment
– compost heaps.
For problems 3 and 4 all parts are worth up to five points.
Problem 3
1.The two positive feedback loops are able to maintain elevated polymerized actin levels even after has returned to basal.
This can be thought of as bi-stable switch where the basal concentration of is above the threshold to maintain the on state but below the threshold to convert the off state to the on state
2.There is no a priori reason why you should need two positive feedback loops.
Two potential reasons are:
a.Temporal.One is quicker and one is slower. The slower feedback helps maintain persistence in actin polymerization while the quick feedback help set up the gradient.
b.Spatial.The actin based feedback appears to tighten the response (as mentioned in lecture). Or in other words the two feedbacks working together can make a much tighter foci of activation than either working alone.
3.PREX is an AND gate. As mentioned in lecture, the activity of PREX depends synergistically on and PIP3. This makes sense in the circuitry because the system needs positive feedback from PIP3 to maintain itself in the on state when returns to basal (due to adaption) and it need input from to ensure that the pathway can turn off if signal goes away. This means it needs to depend on and PIP3 and must be some form of AND gate.
4.There must be an ultrasensitive arrow in each positive feedback loop. This can be just the PREX to Rac arrow or both the Rac to PIP3 and Rac to polymerized actin or just the PIP3 to PREX.
5.The PI3K to PTEN ratio is most critical for setting the threshold from off to on. PTEN sets the amount of activity of which PI3K needs to rise before there is enough PIP3 to trigger the positive feedback loop.
The amount of PREX can also affect the threshold, but it does that in a way that can be thought of as distinct from PI3K to PTEN because it is actually in the positive feedback loops.
6.'or' gate:The pathway could turn on but not off.
'xor' gate:The pathway would turn on than off than on…
If adaptation had similar kinetics to the Rac positive feedback
loops the system would switch its state (turn from on to off or from on to off).
If adaptation was slower than the Rac positive feedback loops
The system would be randomized every time there was chemotactic ligand exposure.
7.Receptor adaptation. Because of receptor adaptation every time the chemotactic ligand concentration is lowered will drop below basal levels. If the positive feedback loop is active, basal levels of are sufficient to maintain the positive feedback. Basal levels of are no sufficient to trigger the positive feedback. If lowering the chemotactic ligand concentration break the positive feedback loop the system needs an increase in chemotactic ligand before chemotaxis will restart.
8a.The system could not be turned on. This is because the positive feedback loop requires both and PIP3, but without PI3K there is no increase in the PIP3 and therefore no positive feedback loop will be triggered.
8.The system will stay on. Once the positive feedback is triggered the system should not require PI3K to remain active.
9.The PIP3 distribution will be broader (or minimal effect). May have temporal or spatial problems (part 2).
10.It could have a minimal effect (if Rac is saturating) or it could make the system harder to turn on and easier to turn off by weaken the positive feedback loop flux two-fold.
11.It should lower the threshold required to turn the system from off to on.
12.An inhibitor arrow from anything in the positive feedback loop to anything else in the shared positive feedback loops (Rac, PREX, or PIP3). The arrows in the positive feedback need to be local. The inhibitor arrow needs to be global.
Problem 4
0.Ligand inhibits receptor activity. Active receptor phosphorylates CheY. CheY-P increases CW bias. CW bias means more tumbling. Active receptor means more tumbling.
1.See attached.
2.Times: (0,) 200, 400, and 600.
% Occupancy:10, 20, 30, and 10
Not really obvious that this when the concentrationis being changed.
There is just too much noise on the level of an individual bacterium.
3.It is obvious when the concentration is being changed.
Averaging over many bacteria reduces the noise and makes the response visible.
4.It lower the average methylation state and the average activity of a receptor.
5.Even lower average methylation and average activity of a receptor.
6.More methylations per receptor but the same percent methylation and average activity of a receptor.
7.The demethylase only works on active receptors.
The methylase is saturated.
8.The system does not perfectly adapt.
It has trouble at high level of receptor occupancy.
9.dA=R-B*A/(A+K)
A=RK/(B-RK)
Where A stands for active receptor.
R stands for the rate of methylase (CheR).
B stands for the rate of demethylase (CheB).
K stands for the Km of the receptor for CheB.
10.The methylase only binds to unmethylated sites (or only works on receptors with unmethylated sites).
11.R is really dependant on the chance of finding an unmethylated site.
R* is a new variable which is the rate of a methylase that only binds to unmethyl sites.
A=R*K/(B-RK)
For those of you who think that this is kind of silly remember that Barkai changed L to A (L being all receptors and A being just active receptors).
What the letters stand for is really important.
12.This system does not perfectly adapt.
When number of unmethylated sites < the number of methylase the system is no longer saturated and one of the Barkai assumptions breaks down.
13.For the most part it is important. As percent occupancy went up the system would have a harder and harder time finding which way the gradient increased. The cells would be good at finding any source up to a certain concentration but could not distinguish well at concentrations higher than this. It would be difficult to compensate (the advantage of perfect adaptation over fine tuning).
14.The more methylation sites the longer it takes to perfectly adapt. This means it
move in a straight line for longer when sensing a change.
15.Even longer to perfectly adapt.
Less percent methylation, which means potential for a broad dynamic range and
more sensitive response
16.The lower percentligandcycle the tighter binding the ligand.
The tighter binding the receptor the less the maximal response to a change in
ligand concentration. This makes chemotaxis worse.
You want a receptor that binds ligand poorly.
This makes sense because the point of a chemotactic receptor is to sample the
environment. The weaker the binding, the more often it samples the
environment.
17.The time scale of coupling needs to be quicker than the time scale of receptor-ligand cycling. If coupling was slower than by the time a receptor ligand cycling by the time an active receptor coupling had a chance to affect the receptor next to it, the active receptor might no longer be active. This just raises the activity but won't affect the gain. If the time scale is quicker than a broad range of receptor can "work together" affectly increasing the gain of the system.
18. 90% to 91% or 89%.
50% to 52% or 48%.
10% to 15% or 3%.
Problem 5
This problem was very open ended.
You could receive quite a bit of credit by just attempting different approaches.
Approaches
4 points of credit were given for just attempting an analytical approach.
Up to 6 points of credit were given for try a numerical approach (2 points for setting up a 1D system, 2 points for setting up a point source in this system, and 2 points for titrating the level of Mos in that point source).
4 points were also given to a recruitment approach (almost know one did this).
4 points were given for a discussion of an analytical approach that analytically delved into changing the diffusion constant and its affect on the system separate from the concentration effects.
Answers
2 points could be earned for discussing the temporal properties of the system
2 points could be earned for discussing the spatial distribution of the answer
0.5-2.5 points could be earned for discussing the hill coefficient of the answer
Hill coefficient isn’t affected (much)
0.5-2.5 points for a discussion of the concentration offset of the system
The system acts the same just in a different concentration regime.
0.5-2.5 points for a discussion of the scaling offset of the system
The system is less induced because the intermediate can diffuse away
0.5-2.5 points for a discussion of the shape of the curves
The system acts the same just in a different concentration regime.
If you didn’t discuss any of the previous three points but talked about the general shape of the resulting curves
Overall
10 more points could be earned for a synthesis of the answer. How well integrated was your response? How well did you tie together the different aspects of your answer? How thoughtfully did you think about the differences between the case where the system is localized versus where it is not localized?