7.06 Problem Set 7 Answer Key 2003
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7.06 Problem Set 7 Answer Key 2003
Question 1
Use the following critical concentrations of ATP-G-Actin to fill in the tables below:
Plus-end
0.12 mM
Minus-end
0.6 mM
A. Given the critical concentrations shown in the table above, what do you expect to
happen when you add a sample of F-actin to a reaction tube with ATP and the
following G-actin concentrations:
In-vitro Concentration [G-Actin]
0.01 mM
0.9 mM
0.25 mM
Change in F-Actin Filament Length
Decreases because depolymerization
at both ends
Increases because polymerization at
both ends
Stays the same because treadmilling
occurs (polymerization at + end and
depolymerization at – end)
B. You repeat the experiment above, except this time you add a minus-end capping
protein to your reaction tube as well. What do you expect to happen when you add a
sample of F-actin to a reaction tube with ATP, the minus-end capping protein, and
the following G-Actin concentrations:
In-vitro Concentration [G-Actin]
0.01 mM
0.9 mM
0.25 mM
Change in F-Actin Filament Length
Decreases because depolymerization
at + end (- end capped)
Increases because polymerization at
+ end (- end capped)
Increases because polymerization at
+ end (- end capped)
Question 2
A. How does ATP binding affect the interaction of myosin and actin?
ATP binding weakens the affinity of myosin for actin.
B. What is the effect of a non-hydrolyzable analogue of ATP on muscle contraction?
Why?
Non-hydrolyzable ATP prevents myosin from acting as a motor since ATP
hydrolysis allows rebinding to actin and the subsequent release of ATP and
Pi power the movement of actin relative to myosin.
C. How could you use an in-vitro assay to study the effects of non-hydrolyzable ATP
on myosin function?
7.06 Problem Set 7 Answer Key 2003
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Use a sliding filament assay in which myosin molecules are layed down on a
slide with their tails attached to the slide. Then add fluorescent actin to the
slide and also either ATP or non-hydrolyzable ATP. If you added normal ATP,
you would observe actin fibers sliding along the microscope slide. However,
if you added a non-hydrolyzable ATP analogue, you would not observe any
actin movement.
Question 3
Advances in optics and electronics allow the movements of single motor protein
molecules to be analyzed. Using polarized light, it is possible to create interference
patterns that exert a centrally directed force. Individual molecules that enter the
interference patter are rapidly pushed to the center, allowing them to be captured
and manipulated at the experimenter’s discretion.
Using such a technique, single kinesin molecules can be positioned on a microtubule
that is fixed to a glass slide. Although a single kinesin molecule cannot be seen, its
movement can be tracked by attaching a larger silica bead to it and following the
bead (See Figure A). In the absence of ATP the kinesin molecule remains at its
original position, but upon addition of ATP the kinesin “walks” along the microtubule
simulating what happens in a cell. Traces of the movements of two kinesin
molecules along a microtubule are shown in Figure B.
A.
What direction is the kinesin moving and does it move unidirectionally?
Kinesin is a + end directed motor protein so it is moving towards the + end
of the microtubule. It is believed to only be able to move in the + direction
and the data is consistent with this.
B.
What is the average speed of kinesin along a microtubule? Is this a
continuous speed or does kinesin take discrete steps? What are the lengths of the
steps?
The speed of the kinesins is about 10nm/s, it varies greatly between the
two experiments. The kinesin seems to take discrete steps as it stay at one
position for a period of time before moving to a new position almost
instantaneously. The length of each step is very close to 8nm as it took ten
steps for each kinesin to go about 80nm.
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C.
Given what you know about microtubule protofilaments and kinesin what does
the kinesin step size tell you about how it moves?
The kinesin head subunit binds to beta-tubulin in microtubule
protofilaments, and each beta-tubulin is about 8nm apart. Thus this data
suggests that kinesin “steps” from one beta-tubulin subunit to the next
without skipping.
D.
The amount of ATP hydrolyzed by kinesin in each step is not completely
understood and scientists argue over whether one ATP is hydrolyzed for each step or
not. What other information besides that given in this question would you need to
calculate the number of ATP molecules hydrolyzed per step?
This data gives the distance kinesin travels in a given amount of time. If
the rate of ATP hydrolysis was measured (ATP used per time) then the two
values could be combined to give ATP used per distance which could easily
be converted to ATP used per step. The problem with this is ensuring that
when you measure the ATP used per time the ATP is solely being used by
the kinesin.
Question 4
You are studying microtubule polymerization as a UROP by warming a solution of
tubulin and GTP from 4°C to 37°C, and then following polymerization by measuring
the light scattering from a small sample in a cuvette. Your results are seen below.
A.
What three phases do you see, and why do you get a different result with and
without the seed?
The three phases are the lag phase, elongation phase, and steady state.
The lag period is the time required for a seed to form, which explains why
when a seed is added at the beginning the lag phase does not occur.
Your fellow UROP is studying microtubule polymerization by warming a solution of
fluorescently labeled tubulin and GTP from 4°C to 37°C, and then following the
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polymerization of one microtubule using video microscopy. She uses a total
concentration of tubulin that is slightly above the critical concentration. Her results
are seen below.
B.
These results are very different. Why are they different and what is the
explanation for the behavior that your labmate is observing?
This data reflects the behavior of one microtubule as opposed to the entire
population. The percentage of total tubulin polymerized in a population
reaches a steady state, whereas a single microtubule displays dynamic
instability with sharp drops in the length with subsequent repolymerization.
The graduate student that you work for wants to repeat your experiments, but he
uses ATP instead of GTP. His results are not shown since he is too embarrassed.
C.
What would the graduate student see in his assay? What role does GTP play
in microtubule polymerization, catastrophe, and rescue?
The graduate student did not see any polymerization of the microtubules
because GTP is required for microtubules to polymerize. GTP binds to both
alpha-tubulin and beta-tubulin. The GTP bound to alpha-tubulin always
stays in the GTP state. Beta-tubulin requires GTP bound to polymerize.
When in a microtubule it slowly hydrolyzes the GTP to GDP. Beta-tubulin
bound to GDP is unstable, and if it is exposed at the end of a microtubule it
will dissociate and return to the monomeric form. Thus catastrophe occurs
when the GTP hydrolysis catches up to the end of the microtubule.
Recovery occurs when tubulin can be added to the elongated microtubule
faster than GTP hydrolysis occurs.
Question 5
During a mutant screen to identify genes involved in controlling meiotic chromosome
segregation you identified a mutant that appears to arrest with metaphase I
spindles. Subsequent studies lead you to the conclusion that this mutant is defective
in meiosis I kinetochore co-orientation and that kinetochores are instead bi-oriented.
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Why is this mutant arresting in metaphase I?
In this mutant, the sister kinetochores are bi-oriented meaning that each
sister kinetochore is attached to opposite poles of the cell rather than the
same pole. The mutant arrests with metaphase I spindles because even
though arm cohesion is lost at the metaphase I – anaphase I transition,
centromeric cohesion is retained which resists the pulling force of the
meiosis I spindle.
Would the elimination of chiasmata by mutating proteins required for meiotic
recombination allow chromosome segregation in this kinetochore co-orientation
mutant?
No, because even if the physical linkages between chromosome arms were
eliminated centromeric cohesion would still be present and prevent
segregation.
Would the elimination of proteins important for maintaining centromeric cohesion
allow chromosome segregation to occur in this mutant?
Yes, assuming that chiasmata are resolved with wild-type kinetics,
chromosome segregation would occur.
What type of chromosome segregation pattern would you observe during the first
meiotic division in cells that have kinetochores that are bi-oriented instead of cooriented and lack centromeric cohesion protector function?
The first chromosome segregation phase would be an equational, meiosis
II-like segregation. The second chromosome segregation would be random
(or distributive), as sister-chromatids are no longer linked to each other.
You can visualize individual chromosomes by marking them with GFP. When you
GFP-marked both homologs of chromosome V in a co-orientation-, protectormutant, you obtained the following three classes of segregation patterns within the
four meiotic products:
Class I: the four meiotic products have one chromosome V each
Class II: the four meiotic products have two germ cells with two copies of
chromosome V and two with no chromosome V
Class III: the four meiotic products have two germ cells with one chromosome V,
one germ cell without chromosome V and one germ cell with two copies of
chromosome V.
7.06 Problem Set 7 Answer Key 2003
CLASS I
CLASS II
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CLASS III
What would be the frequencies for these individual classes and why?
Class I: 25 percent
Class II: 25 percent
Class III: 50 percent
Although the first meiotic division is equational, both meiosis I products end
up with two sister chromatids (one from each homolog). However in the
second meiotic division these two sister chromatids would segregate
randomly, which generates the above probabilities.