Download Questions for exam #1

C2006/F2402 ’11 -- Exam #1 – Questions & Answers
Each answer (A-1, A-2, etc.) is worth 2 points, and the explanation to each section (A, B, C etc.) is worth 2 points, unless it
says otherwise. Therefore each part, A, B, C, etc. is usually worth 6 points. Explanations that simply repeated the circled
answer did not receive credit. Additional information had to be provided.
1. Claudins are a family of proteins found only in tight junctions. Claudin 16 has 4 hydrophobic TM segments, each about 25
AA long, and 5 hydrophilic segments.
A. Consider the claudin in an epithelium one cell layer thick.
A-1. Claudin is needed to connect epithelial cells (to the basal lamina) (to other cells) (both) (neither).
A-2. Claudin would be found in the cell membrane (on the BL surface) (on the apical surface) (on both surfaces). (where the
BL and apical surfaces converge) (all over the entire cell) (in none of these – it’s not in the membrane).
Tight junctions are located at the boundary between the apical and BL surfaces. They prevent lateral diffusion of proteins from
one area to the other, and they create a water tight seal that prevents materials from passing between the cells from one side of
the epithelium to the other. (To get full credit, you had to mention one function or the other.)
B. In normal cells, claudin binds to ZO, a peripheral protein that binds to actin.
B-1. ZO should be (entirely in the cytoplasm) (transmembrane, with an intracellular domain) (in the ECM)
(inside a part of the EMS) (in the nucleus) (in mitochondria) (none of these). Circle all reasonable possibilities.
B-2. You can break cells open, and add antibodies to various parts of the cytoskeleton. Using this procedure with the correct
antibodies, you can precipitate either MF, MT or IF. You are most likely to precipitate claudin if you add antibody to (MF) (IF)
(MT) (MT or MF) (IF or MF) (IF or MT) (any of these).
B-1. ZO must be on one side of the membrane or the other (by definition). Since actin is in the cytoplasm, and ZO binds to
actin, ZO must be intracellular. ZO must bind to the intracellular (cytoplasmic) domain of the TM protein claudin, as well as
to actin.
B-2. Since MFs are made of actin, that means MF and claudin are indirectly connected – ZO binds to both and connects them.
Note all connections here consist of weak bonds, not covalent bonds. If Ab cross links and precipitates MF, it will also ‘pull
down’ ZO and claudin as all three proteins are connected.
C. In some patients with kidney disease there is a mutation affecting the last hydrophilic segment of claudin 16, and the mutant
protein does not bind to ZO.
C-1. From these results, the amino end of normal claudin 16 should be (intracellular) (extracellular) (TM)
(intra- or extra-cellular) (any of these) (none of these).
C-2. Suppose you examine another claudin, say claudin 3, which is a member of the same protein family. You expect the
amino end of normal claudin 3 to be (intracellular) (extracellular) (intra- or extra-cellular) (TM)
(can’t predict) (none of these).
C-1. The COOH end of claudin binds to ZO, so the COOH end must be intracellular. Since the protein has 4 TM segments, the
COOH end and the amino end should be on the same side of the plasma membrane.
C-2. Proteins in the same family generally have similar structures and functions. So you would expect the same orientation for
all claudins.
C-3. Assume that each cell type contains a single claudin – claudin 3, or 16, etc, and that claudins are the major TM proteins in
tight junctions. Suppose you examine cells containing claudin 16. Under the conditions you use, claudin 16 is found in the
plasma membrane whether it is normal or mutant. You freeze-fracture membranes & look at both E and P faces.
(i). In membranes containing normal claudin 16, you expect to see continuous ridges of bumps on the membrane’s (E face)
(P face) (both) (neither) (either P or E, but can’t predict which).
(ii). In membranes containing mutant claudin 16, you expect to see more bumps on the membrane’s
(E face)* (P face) (both) (neither – pattern will be the same).
*The intended answer to (ii) is E, because there will be more bumps on the E face in the mutant than in the normal. However
some students chose the answer ‘neither’ because they understood this question to ask whether there were more bumps on the
P face or the E face in the mutant. If you took it that way, and you clearly explained your reasoning, it was okay.
Explanation: Claudin is normally anchored to the cytoskeleton, so when the membrane is split open by freeze-fracture, the
protein is more likely to stay with the P face than with the E. The mutant protein is not anchored, so it is just as likely to go
with the P face as the E face.
Note: Membrin is a made up name. Claudin is real. All the parts of questions 1-3 are derived from a paper about claudin.
The mutant ‘membrin’ described here is really claudin that can’t bind to ZO, as in problem 1. See
Http://www.ncbi.nlm.nih.gov/pmc/articles/PMC1180395/ or http://jcem.endojournals.org/cgi/reprint/91/8/3076
2. Membrin is a multipass TM protein. In normal cells, membrin is located in the plasma membrane. In cells from patients with
mutant membrin, most of the membrin ends up in lysosomes. You do experiments 1 to 3 (described on the last page) to locate
normal and mutant membrin using immunofluorescence. Please read the details on the last page and then answer the questions.
(If your exam had another part here, we ignored the answer.)
A-1. Consider the labeled secondary antibody used here to locate membrin. This Ab combines with
(human protein) (rabbit protein) (both) (neither) (either one).
A-2. Antibodies for research are not made in humans. However, if it were ethically feasible, the secondary antibody used here
to detect Ab 1 could be made in a (human) (rabbit) (neither) (either one).
A-1. The primary antibody binds to membrin (which is a human protein). The secondary Ab binds to the constant part of the
primary Ab (which is a rabbit protein).
A-2. An organism does not make Ab to its own proteins, except in cases of autoimmune disease. Organisms only make Ab to
substances (antigens) the individuals considers foreign. (How this works will be discussed later in the term.) So you can’t make
secondary Ab to rabbit protein in a rabbit. You have to inject the rabbit protein into an individual of another species. Humans
would consider rabbit protein foreign, and make Ab to it. However, as a student pointed out, it may not be optimal to inject the
primary Ab into a human, as the primary Ab will bind to a human protein (membrin) and the binding may interfere with the
functioning of membrin.
B-1. If you carry out expt. 1 with cells from patients with mutant membrin, which antibodies should be found in an
‘intracellular vesicular staining pattern?’ (Ab 1) (Ab 2) or (both) (neither) (one or the other but not both).
B-2. For this experiment, the secondary antibodies used to detect Ab 1 and Ab 2 would need to
(bind to different epitopes) (have different colored fluorescent labels) or (both) (either one)
(neither – you can use the same secondary Ab with the same label).
1 pt for each correct answer; 2 pts for each explanation; 8 pts for B.
B-1. Ab2 sticks to a protein found in lysosomes. So it is always found in a ‘vesicular’ staining pattern. Mutant membrin also
ends up in lysosomes (although it goes to the plasma membrane first – see below). Any Ab, such as Ab1, that binds to the
mutant membrin, will be carried with the membrin to lysosomes.
We know that Ab 2 co-localizes with lysosomes. Therefore it is reasonable to assume that Ab 1 & Ab 2 can get into
lysosomes from the cytoplasm. (We know antibodies can get into permeabilized cells, and therefore we assume they can get
into organelles as well.) However the experiment will work and give the observed staining patterns even if the antibodies can’t
cross the lysosomal membrane. Ab 2 could remain in the cytoplasm and stick to the cytoplasmic tail of a lysosomal TM protein;
if it does, it will give the same ‘vesicular’ pattern. If Ab 1 can’t enter lysosomes from the cytoplasm, it will end up in lysosomes
anyway, because it will bind to the membrin on the surface of the cell and be endocytosed with the membrin.
B-2. The two primary antibodies were made in different species, and have different constant ends. A secondary Ab that binds to
one primary Ab will not bind to the other, and vice versa. Each primary Ab has unique epitopes. If you want to see where each
secondary Ab is located, you need to have a different tag on each one.
C. What is the simplest interpretation of expt. #2?
C-1. When there is no inhibition of endocytosis:
(i) Mutant membrin is found (in the plasma membrane) (lysosomes) or (both) (neither).
(ii). Mutant membrin is more likely than normal membrin to (reach the plasma membrane) (get made)
(stay in the plasma membrane) (get removed from the plasma membrane) (get degraded). Circle all that apply.
C-2. When endocytosis is inhibited, mutant membrin is more likely than before to (reach the plasma membrane)
(get made) (stay in the plasma membrane) (get removed from the plasma membrane) (get degraded). Circle all that apply, and
explain briefly as usual.
Each correct answer 1 pt; explanation 2 pts. 7 pts. Total. The experiment shows that most normal membrin is found on the
plasma membrane, but most mutant membrin is found in lysosomes, unless endocytosis is inhibited. The simplest explanation is
that newly made membrin goes to the plasma membrane, whether the membrin is normal or mutant. Ab1 binds to that membrin
from the outside of the cell. After the membrin reaches the membrane, the normal membrin (with bound Ab) stays put,
accounting for the ‘plasma membrane staining pattern.’ The mutant protein is endocytosed (along with Ab1) and sent to
lysosomes to be degraded, accounting for the ‘vesicular staining pattern.’ If endocytosis is blocked, the mutant membrin and
Ab bound to it stays on the surface.
Answer to 2C, cont.
The simplest explanation of these results is that the mutant membrin (like claudin that does not bind ZO) is not anchored in the
plasma membrane, and so is trapped in coated pits, endocytosed and sent to lysosomes. An alternative possibility is that
membrin is a receptor that normally recycles. In that case, the mutation affects sorting in the endosome. The normal membrin
is constantly removed from the plasma membrane by endocytosis, but returned to it by exocytosis. The mutant membrin is
endocytosed, but then targeted from endosomes to lysosomes instead of to membrane bound vesicles. Both normal and mutant
membrin are endocytosed and but only the normal protein is recycled back to the plasma membrane.
D. Normal membrin is made on the RER and inserted into the ER membrane. From there it travels to the plasma membrane.
Assume the fusion protein described in experiment 3 behaves exactly like unmodified mutant membrin. When the fusion
protein is made, which of the following will be the first to fluoresce?
(plasma membrane) (endosomes) (lysosomes) (coated vesicles)* (beats me).
Both normal and mutant membrin travel in vesicles to the plasma membrane, as explained above. The mutant protein does not
go directly to lysosomes. It goes first to the plasma membrane, and is then endocytosed and goes from coated vesicles to
endosomes to lysosomes.
*Note: The term ‘coated vesicles’ is intended to refer to the clathrin coated vesicles of endocytosis. All vesicles have some sort
of coat protein, and it was okay if you took ‘coated vesicles’ to mean the vesicles traveling from the ER/Golgi to the plasma
membrane (if you explained yourself properly).
3. Some tight junctions between kidney epithelial cells allow Mg2+ to pass across the epithelium by going between the cells.
(These junctions are impermeable to water as usual.) Passage of ions through TJ is called the paracellular pathway, and is
limited to certain parts of the kidney. Suppose Mg2+ can use the paracellular pathway in part of the kidney. The passage of ions
by the paracellular pathway is very fast, and the rate of passage increases linearly with [Mg 2+] at all normal (physiological)
values of [Mg2+].
A-1. The proteins of the tight junctions in the paracellular pathway are acting as (channels) (carriers) (pumps)
(co-transporters) (exchangers) (transporters, but can’t tell what kind) (any of these).
A-2. Passage of Mg2+ through the paracellular pathway should be dependent on (splitting of ATP)
(a cation gradient of some other ion – not Mg2+) (either one) (neither).
The properties of passage through the paracellular pathway indicate that the TJ contains channels – because passage is so
fast, and the rate increases linearly with [Mg2+] – the rate doesn’t saturate as it would if there were a carrier or transporter
protein involved.
B-1. In the area where Mg2+ uses the paracellular pathway, you would also expect passage through the tight junctions by (Na+)
(K+) (both) (either one) (neither) (can’t predict).
B-2. There are mutants of the TJ transmembrane proteins that do not allow passage of Mg2+ by the paracellular pathway. The
mutant proteins are correctly localized, but do not function correctly. The mutations are likely to be in the parts of the protein
that are (intracellular = cytoplasmic) (TM) (extracellular) (intra or extra cellular)
(extra cellular or TM) (intracellular or TM) (any of these).
Channels are specific, so a channel that allows passage of a divalent cation like Mg 2+ is not likely to allow passage of
monovalent cations like Na+ or K+. Since the Mg2+ is going between the cells, not through them, the channels must be
extracellular. Therefore mutations that alter the channels must be in the extracellular domain of the TM proteins.
Note: These channels are actually formed by the extracellular domain of claudin 16. The mutation that prevents passage of
Mg2+ affects a different domain of the protein from the mutation that causes mislocalization (as in problems 1 & 2). Both
mutations cause kidney disease due to loss of Ca2+ and Mg2+ (which is normally recovered in the kidney).
C. Because of the action of the various transport proteins in the membrane, there is a voltage difference across the epithelium
where the paracellular pathway occurs. The lumen (apical) side is positive, relative to the BL side.
C-1. Suppose [Mg2+] is the same on both sides of the epithelium. Then you would expect net movement of Mg2+ (from the
lumen side to the BL side) (from the BL side to the lumen side) (neither) (could be either way).
C-2. Suppose [Mg2+] is higher on the BL side of the epithelium. Then you would expect net movement of Mg 2+
(from the lumen side to the BL side) (from the BL side to the lumen side) (neither) (could be either way).
Answer to 3C, cont. 2 pts each answer and explanation, 8 total.
C-1. Ions move through channels down their electrochemical gradients. If there is no chemical gradient, as in C-1, the voltage
difference will drive the positive ion away from the positive side toward the negative side of the membrane, from the lumen (+)
side to the BL (-) side.
C-2. Which way the ions will move will depend on both the electrical forces (voltage differences) and the chemical gradient.
The size of the chemical gradient (or the voltage) is not specified here. If the chemical gradient is large enough to overcome
the charge difference across the membrane, then Mg2+ could move from the BL side to the lumen. If the two forces are
balanced, the ions won’t move at all. Which way the ions will move depends on the relative magnitude of the chemical and
electrical forces.
4.
A. Which is the smallest? (a micron) (a nanometer) (a millimeter) (beats me).
B. Which of the following is (are) usually measured in microns? (length of monkey’s tail)
(width of monkey’s tooth) (diameter of monkey cells) (width of microtubules in monkey cell)
(longest dimension of mitochondrion in monkey cells) (diameter of molecules of monkey tubulin – αβ dimers)
(none of these). Circle all choices that are usually measured in microns, and explain B.
1 pt each answer; explanation 2, total 5 pts. Cells are usually measured in microns, although some prokaryotic cells are
smaller. Mitochondria, which have evolved from independent cells, are much larger than most organelles, and are about the
size of their bacterial ancestors (usually a few microns). MTs and tubulin are much smaller; teeth and tails are much longer.
5. Arp is a transporter that catalyzes transport of GABA and glutamate across the E. coli plasma membrane. The reaction it
promotes (reaction 1) is shown on the last page.
A-1. Arp is a (uniporter) (antiporter) (symporter) (none of these) (beats me).
A-2. Suppose Arp is part of a gene super family. (Families of proteins are grouped into ‘super families.’) Then Arp is most
likely to be in the same super family as the transporter(s) responsible for entry into RBC of
(CO2) (glucose) (Bicarbonate = HCO3-) (K+) (Na+) (Cl-). Circle all correct choices and explain your reasoning.
Arp transports GABA in one direction and glutamate in the other; it is an antiporter or exchanger. The Bicarb/Cl - exchanger
of the RBC (band 3, anion exchanger) acts in a similar way. Both facilitate the passage of two substances in opposite
directions, without additional input of energy. The energy to move the substances comes from their respective gradients. Either
both can go down their gradients, or one up and one down, as long as the net ΔG is negative. Some call this facilitated
diffusion (since both substances can go down their gradients) and others call it secondary active transport (since sometimes
one substance goes up its gradient and the other goes down).
No transporter is required for CO2. It is a gas, and can pass through a membrane without assistance.
B. In E. coli, the reaction catalyzed by Arp goes only in one direction, say from side A of the plasma membrane to side B.
Assume there is no other transporter for GABA in the plasma membrane. You can make vesicles of E. coli membrane that
contain Arp, and use them to measure transport of GABA. When you start your experiment, the [glutamate] on both sides of
the membrane is equal, and there is no voltage difference across the membrane.
B-1. To get GABA transported from side B to side A, which of the following do you need?
__(vesicles with the membrane turned inside out) __(high [GABA] on side A) X (high [GABA] on side B).
Check off all necessary conditions.
B-2. Now suppose you set it up right, and GABA is transported from side B to side A. In your experiment, the [glutamate] on
side A will initially (increase) (decrease) (stay the same) (can’t predict). Explain briefly.
This exchanger is reversible, just like the RBC anion exchanger. There is no need to turn the membrane around to reverse the
transport -- it is sufficient to reverse the gradients of GABA and/or glutamate. If the [glutamate] is equal on both sides, which
way the GABA goes depends on the direction of the GABA gradient. GABA can go in or out, but whichever way it goes (down
its chemical gradient), glutamate goes the other way. If one GABA goes from side B to A, one glutamate must go from side A to
side B, and vice versa. In normal E. coli, GABA builds up in the cell, so it always goes out in exchange for glutamate. Under
experimental conditions, GAGA can go either way. Note that for this problem, it doesn’t matter whether side A is the inside or
the outside.
C. Arp stands for ‘Acid Resistance Protein’, because reactions 1 & 2 are used together to help E. coli raise their intracellular
pH and survive in an acid environment, such as the stomach. In normal E. coli growing at low pH,
C-1. Reaction 2 should go (to the right) (to the left) (can’t predict).
C-2. GABA should be transported (into the cells) (out of the cells) (either way).
C-3. the ΔG for reaction 1 as written should be (<0) (>0) (about 0) (can’t predict). Explain all your choices briefly.
Answer to 5C.
Bacteria grown in acid medium need to expel protons to raise their internal pH. They soak up protons by converting glutamate
to GABA. (Reaction 2.) Then they expel the GABA in exchange for more glutamate, and repeat the cycle. Reaction 2 is driven
in the desired direction -- to the right -- by Le Chatelier’s principle, because of the high level of protons and low level of CO2
(it’s a gas that bubbles off). The high level of GABA that builds up drives reaction 1 to the left as written; therefore the ΔG is
positive, or favorable to the left. Reaction 1 has a ΔG o of zero, but the ΔG is not zero – it is either negative or positive
depending on the relative concentrations of glutamate and GABA.
D. E. coli grow in minimal medium (glucose & salts) at neutral pH. If the pH of the medium is lowered to 2.5, the bacteria die.
E. coli will grow in rich medium (supplemented with nucleotides, amino acids, vitamins, etc.) at either pH. For E. coli to grow
in minimal medium at pH 2.5, the medium should be supplemented with (glutamate) (GABA) (both) (either one) (neither –
neither will help; you’ll have to try something else). Explain why the bacteria can grow at low pH in your stomach or in rich
medium, but not in minimal medium.
Explanation 3 pts. To resist acid, the cells need glutamate to soak up protons in reaction 2. Glutamate is an amino acid, and
all bacteria contain it – they need it to make protein. Either they make glutamate themselves or take it up from the medium. So
they can carry out reaction 2 and start the cycle. They may exhaust their supply of glutamate, but if there is glutamate in the
medium, the GABA that builds up can be exchanged for more glutamate and the cycle can continue. Rich medium or your
stomach will contain glutamate that can be exchanged for GABA. The E. coli don’t need GABA in the medium – the cells need
to get rid of GABA, not take it up.
Experiments for Problem 2
All experiments.
You have Ab 1 from rabbit (not labeled) to the first extracellular loop of human membrin. The loop is not changed by the
mutation. You also have unlabeled mouse Ab 2 to a (human) lysosomal enzyme. You have ‘suitable fluorescently labeled
secondary antibodies.’ (Quotes are from the actual research paper.) You use the antibodies to locate membrin and lysosomes in
both normal cells and cells from patients. Each time you add an antibody, you wash off unattached antibody before going on.
At the end of each experiment you examine the samples in a microscope.
Experiment 1: You permeabilize your cells, and then add Ab 1 and Ab 2 to your samples. Then you wash off unattached Ab
and add the secondary antibodies. All primary and secondary antibodies were added to each sample.
If you carry out expt. 1 with normal cells, Ab 1 is mostly in a ‘plasma membrane staining pattern,’ and Ab 2 is in an
‘intracellular vesicular staining pattern.’
Experiment 2: You add Ab 1 to the outside of cells. After you wash off unattached Ab, you permeabilize the cells, and add Ab
2. Then you add the appropriate secondary antibodies.
The results with normal cells are the same as in expt. 1. Ab 1 is mostly in a ‘plasma membrane staining pattern,’ and Ab 2
is in an ‘intracellular vesicular staining pattern.’
With cells from patients, both antibodies are found in intracellular vesicles.
If you inhibit clathrin-mediated endocytosis, the pattern for cells from patients is the same as for normal cells -- Ab 1 is
mostly in a ‘plasma membrane staining pattern,’ and Ab 2 is in an ‘intracellular vesicular staining pattern.
Experiment 3: You use genetic engineering to get a cell that makes a mutant membrin-GFP fusion protein; the fusion protein
includes all of both mutant membrin and GFP.
For Problem 5:
Reaction 1
Glutamatein + GABAout ↔
Reaction 2
Glutamate + H+ ↔
Glutamateout + GABAin
GABA + CO2
NH3+-CH-COO- + H+ ↔ NH3+-CH2 + CO2
|
|
R
R
Reactions 1 & 2 are partially responsible for the acid resistance of E. coli. Without acid resistance, E. coli could not survive
passage through the stomach and cause disease. (Most E. coli strains are harmless, but some strains can cause food poisoning.)
Or with formulas: