Download LS1a Problem Set #4

LS1a Fall 2014
Practice Problem Set #4
1. Phospholipids typically form bilayer structures when present in aqueous solutions. Briefly
explain how introducing cis-double bonds into saturated fatty acid chains of a phospholipid
will affect the melting temperature (Tm) and the fluidity of that membrane bilayer.
2. Certain molecules such as oxygen (O2) can freely diffuse across cell membranes.
Which of the following molecules do you expect to freely diffuse across cell membranes
(indicate by circling)?
3. Organisms can regulate their lipid composition in response to temperature. Fish are
particularly rich in “Omega-3” and “Omega-6” fatty acids, which are unsaturated fatty acids
containing cis-double bonds (the “omega- #” names refer to the location of the cis-double
bond). On the other hand, poultry and mammalian meat sources (chicken, beef, etc) tend to
be richer in saturated fats, which do not contain any double bonds.
a) In light of what you know about plasma membranes, why is it not surprising that fish
membranes are rich in unsaturated fats while poultry and mammalian membranes are
rich in saturated fats? (Hint: Fish are cold-blooded, whereas birds and mammals are
warm-blooded).
Regulating membrane fluidity is especially important for organisms such as bacteria that
cannot regulate their own temperature.
b) Consider a colony of bacteria that suddenly undergoes a drastic drop in temperature.
What consequence would this have on the fluidity of the membranes of the bacteria?
1
c) Some bacteria have enzymes that can adjust the length and saturation of the fatty acid
chains. If this drop in temperature was gradual, what could the bacteria do to combat
the change in membrane fluidity?
d) If the bacteria were suddenly shifted to a very high temperature after gradually
adjusting to a lower temperature, how would membrane permeability be affected?
4. In your studies of cells, you discover two new transmembrane proteins, A and B, which are
present on the cell surface. To further investigate their distribution in the plasma
membrane, you use fluorescent tags to label them.
a) You first label Protein A with a fluorescent tag and examine its localization in the cell’s
membrane using a microscope. You find it is distributed diffusely across the cell surface.
You then use FRAP (“Fluorescence Recovery After Photobleaching”) to determine the
mobility of protein A in the membrane. You notice that after bleaching an area of the
cell membrane 90% of the fluorescence is recovered in this area within 5 minutes. Draw
the recovery graph you would expect and briefly explain these results.
b) Next you label protein B with a fluorescent tag and find it is also distributed diffusely
across the cell surface. However the FRAP results are very different. After bleaching an
isolated area of the cell membrane, only 25% of the fluorescence is recovered in this
area within 5 minutes. Draw the recovery graph you would expect and briefly explain
these results.
2
c) You further examine the mobility of protein A in cells that have a defect in cholesterol
production. The membranes of these cells have very little cholesterol compared to the
cells used above. If you repeat your FRAP experiments from part (a) using these cells,
how do you predict the mobility of protein A would compare with your previous
observation at both low and high temperatures? Briefly explain your answer.
5. Porins are transmembrane proteins with hollow centers through which small molecules can
diffuse. The structure of one is shown below (in sideview).
a) What is the predominant secondary structure in a porin?
b) When the sequence of a single yellow protein strand is analyzed, it is discovered that
consecutive amino acids alternate between hydrophobic and hydrophilic sidechains.
Does this make sense? Briefly explain.
3
c) If this porin protein is tagged with a fluorescent molecule and FRAP is performed, the
results below on the left are seen. If another transmembrane protein (Protein A) is
tagged with a fluorescent molecule and FRAP is performed, the results below on the
right are seen. Which of these proteins is more mobile in the plasma membrane? Briefly
explain.
d) Brief treatment with proteases (enzymes that catalyze the breaking of peptide bonds)
cleaves some of peptide bonds that are on the surface of a protein and more accessible.
If the membranes containing either tagged Porin or tagged Protein A are briefly treated
with a protease and then FRAP is preformed, the mobility of both proteins are
indistinguishable from the results shown above for Protein A. Based on this information,
provide a plausible explanation for the mobility of Porin in the absence and presence of
protease treatment.
6. An engineered ribozyme is a small type of nucleic acid that can catalyze a chemical reaction.
Shown below is an example of a ribozyme (black RNA) along with its substrate (red RNA).
This particular ribozyme can cleave the phosphodiester bonds of its substrate. The site
where the ribozyme cleaves the substrate is indicated.
G
A
CGCGG
GCGCC
GA
UAG
U
C
AGCUCG
UCGAGC
C
A A U
Cleavage
C G
site
A U
5'
3'
4
3'
5'
a) The ribozyme catalyzes the cleavage of the red substrate by forcing it to assume a
conformation that increases the likelihood of a so-called transesterification reaction in
which the 3’-5’ phosphodiester bond at the cleavage site between two adjacent
ribonucleotides within the substrate is cut in an intramolecular reaction where the 2’
OH group attacks the phosphodiester backbone. Using the drawing shown below, draw
in the products of this reaction.
b) Because RNA is difficult to work with, scientists sometimes attempt to make ribozymes
out of DNA. If you made the entire ribozyme (black portion) out of DNA and it adopted
the same shape, would it be able to cleave the substrate? Briefly explain.
c) If the ribozyme (the black portion) were still made of RNA, but the substrate (the red
portion) was made of DNA, could the ribozyme still catalyze the cleavage of the
substrate’s phosphodiester bond?
5
Answers
1) Introducing fatty acids with cis-double bonds will increase the space between adjacent fatty
acid chains and diminish the van der Waal’s forces that exist between nearby chains, thereby
lowering the melting temperature and making the membrane more fluid.
2)
Small uncharged, polar molecules are permeable to the membrane.
3a) In order for membranes to function properly they need to maintain a certain fluidity. The
fluidity of a membrane depends both on temperature and on its composition. One component
that can be varied in membrane composition is the degree of saturation of the fatty acid tails.
Fewer cis-double bonds will allow more van der Waals interactions between the fatty acid
tails, and the membrane will have a higher Tm. Since fish are cold-blooded, their cellular
temperature depends on their environment. If they are living in relatively cooler waters
(compared to the average mammal’s body temperature) it is not surprising that their
membranes would be rich in unsaturated fats to decrease the van der Waal’s interactions and
subsequently decrease their plasma membrane Tm. Thus their membranes resist becoming
more solid-like at this cooler temperature.
3b) The fluidity of the membranes would decrease as they would become more solid-like.
3c) In response to cold the bacteria could shorten their fatty acids chains, synthesize more
fatty acids with cis-double bonds, or increase the concentration of cholesterol in the
membrane. Each of these processes would reduce phospholipid packing and therefore the van
der Waal’s interactions between the fatty acids, increasing the fluidity of the membrane.
3d) Membranes would become too fluid which could cause them to become “leaky” and allow
ions to cross.
4a)
Photobleach
100
90
percent
fluorescence
time
(min)
6
5
The rate of recovery of fluorescence in the bleached area is a measure of the rate of the
mobility (lateral diffusion) of Protein A. Protein A is mobile in the lipid bilayer so after
bleaching the non-fluorescent Protein A moves out of the area and fluorescent Protein A
diffuses in. Since some of the proteins will have permanently lost their fluorescence the
recovery is not 100%.
4b)
100
percent
fluorescence
20
time
(min)
5
Protein B has limited mobility in the lipid bilayer so after bleaching much less of the
fluorescence is recovered after 5 minutes. The majority of the bleached Protein B is relatively
immobile; it might be anchored to proteins inside the cell. Therefore, bleached Protein B
molecules are unable to diffuse out and be replaced by other fluorescent Protein Bs.
4c) At high temperatures cholesterol decreases membrane fluidity. Cholesterol is present in
high concentrations in the membranes of eukaryotic cells. If cholesterol was depleted from
membranes of otherwise similar composition, it is likely that the absence of cholesterol would
cause the mobility of protein A to be greater than what was previously observed.
AND
At low temperatures cholesterol increases membrane fluidity. If cholesterol was depleted
from membranes of otherwise similar composition, it is likely that the absence of cholesterol
would cause the mobility of protein A to be less than what was previously observed.
5a) Beta sheets
5b) Yes, the side chains in a beta sheet alternate between above the plane of the sheet and
below the plane of the sheet. For porin the hydrophobic side chains are in contact with the
hydrophobic fatty acid tails of the membrane phospholipids and the hydrophilic side chains
point into the aqueous environment of the channel.
5c) Protein A is more mobile. The fluorescence in the bleached area of the membrane for
Protein A recovers fairly quickly whereas there is virtually no recovery of fluorescence in the
bleached area of the membrane where porin is labeled.
7
5d) Porin could be anchored to another protein restricting its mobility in the membrane.
When the membranes are treated with protease, it breaks the association between porin and
the fixed protein and allows protein to be mobile in the membrane.
6a)
6b)
Yes. So long as the DNA is able to contort the RNA into the correct conformation, it will
accelerate the reaction.
6c)
No. The 2’ hydroxyl, which is present in RNA but not DNA, has to be in the substrate for this
reaction to occur.
8