C2006/F2402 ’10 Key to Exam #1 Corrections made during the exam are blocked in blue.
Unless it says otherwise, each answer was worth 2 pts and each explanation 2 pts. Please email Dr. M if you notice any typos or other errors. Last update: 2/17/2010 5:59 PM Additions made to the explanations after 2/15 are blocked in grey.
1. PFE1 is a transmembrane protein. This question is about the location and function of PFE1.
A-1. PFE1 is best described as (integral) (peripheral on the inside of the cell) (peripheral on the outside)
(peripheral, but can’t tell which side) (peripheral or integral).
A-2. Given the information so far,
(a) PFE1 could be (a ribosomal protein) (clathrin) (a channel protein) (collagen) (tubulin) (none of these).
(b). PFE1 could be part of (the basal lamina) (the cytoskeleton) (a desmosome) (a gap junction) (none of these).
For each part, circle all reasonable possibilities. Then explain the basic principle very briefly.
PFE1 must be embedded in the lipid bilayer. All the proteins and structures in (a) are in the cytoplasm, except collagen and the basal lamina, which are part of the ECM. None are firmly connected to the bilayer. The items chosen in (b) contain transmembrane proteins. (If you thought parts (a) and (b) had to match, and just picked ‘gap junction,’ that was okay if clearly explained.) For this question, 1 pt each correct choice.
B. You have a red dye that is a small molecule. There are no transport proteins for this dye in animals. The dye does not cross the intestinal epithelium (IE). If you inject the dye into a cell of the IE, it spreads throughout the epithelium, but it does not enter the interstitial fluid. If you put dye into the lumen, and PFE1 is missing, the dye enters the interstitial fluid. Given this information, where do you expect to find PFE1?
B-1. In normal epithelial cells, PFE1 is most likely to be bound to (an integrin) (connexin) (cadherin)
(PFE1 from a neighboring cell) (PFE1 in the ECM) (an integrin or cadherin) (any of these).
B-2. If PFE1 is missing, should the injected dye spread through the epithelium? (yes) (no) (can’t predict).
Explain your reasoning – about how normal PFE1, works and the effects of missing protein.
Simplest solution, given the proper corrections:
B-1. PFE1 is part of a tight junction, and that PFE1 is the major transmembrane protein that connects the membranes of neighboring cells. (The protein in tight junctions is not cadherin. Cadherins are restricted to adhesive junctions.) Tight junctions, not adhesive junctions, form a seal between the cells of the IE so liquid cannot pass across it. When PFE1 is defective, the tight junctions become leaky. Then liquid containing dye can pass between the cells of the epithelial layer and reach the interstitial fluid.
B-2. The dye spreads within the IE through gap junctions. Gap junctions and subsequent spread of the injected dye should be unaffected by changes in tight junctions.
An alternative solution is that PFE1 is part of a gap junction, and is bound to (or is) connexin. (In this case it could also be bound to PFE 1from a neighboring cell = connexin from a neighboring cell). When PRE1 is missing, the injected dye leaks out of the cells into the interstitial fluid and the dye does not spread (or does not spread as well) from cell to cell. This answer does not explain how dye added to the lumen could reach the interstitial fluid if PFE1 is missing. It does explain how dye injected into a cell could reach the interstitial fluid (if PFE1 is missing).
Note: If the protein in a gap junction is defective, it might leave a hole where material can leak into the interstitial fluid. However, holes in lipid bilayers tend to seal up if there is no protein holding them open. A gap junction without a critical component is far more likely to fail to connect properly to the next cell, and to prevent dye from spreading from cell to cell.
Because many students were confused by the wording of this problem and/or the corrections, sensible alternative answers were given close to full credit. For grading of this question, no points were given for the answers alone. Credit depended on your reasoning – whether you took into account the roles of both tight junctions and gap junctions in spreading the dye, and correctly explained their structure and function.
Answer to prob. 1, cont.
C. Suppose you have unlabeled rabbit antibodies to human PFE1. You want to use them to locate PFE1 that is immobilized (on a blot or gel). Antigen-antibody binding can be detected using a swine antibody linked to the enzyme peroxidase. Peroxidase has a substrate that generates a luminescent product. (Emits light like a firefly or glow stick.)
You want to see if you have PFE1 in your gel or blot. You have (1) the appropriate swine Ab linked to peroxidase, (2) rabbit antibody to PFE1, & (3) peroxidase substrate. You can add them all at once, or add them in any combination. After each addition, you wash the sample to remove unattached material before adding anything else.
C-1. The swine antibody should react with (human Ab) (rabbit Ab) (PFE1 epitopes) (swine Ab).
C-2. Which item(s) should you add first? (swine Ab) (rabbit Ab) (substrate) (both antibodies together) (either Ab) (swine Ab & substrate together) (all 3 at once) (any one -- doesn’t matter which)
C-3. Which item(s) should you add last? (swine Ab) (rabbit Ab) (substrate) (both antibodies together) (either Ab) (swine Ab & substrate together) (all 3 at once) (any one -- doesn’t matter which)
Explain how this works.
The rabbit Ab binds specifically to PFE1, and not to other proteins. (The swine Ab has no affinity or specificity for PFE1). You remove unattached rabbit Ab. Then you add swine Ab. It binds to the constant part of the rabbit Ab. You remove unattached swine Ab. You now have peroxidase located at the site of PFE1. You use the enzymatic activity of the trapped peroxidase to generate a ‘signal’ – a luminescent glow. To do that, you add the substrate, and it is converted to a luminescent product where ever the PFE1(& subsequently the peroxidase) is. If there is no PFE1 present, you won’t get any glow. (No signal.) If PFE1 is present, you get a lot of ‘signal’ – the spot glows brightly -- because each trapped molecule of enzyme can generate many molecules of glowing product.
Notes on procedure: If you add both antibodies at the same time, you will need much more swine antibody, and you will waste most of it, because most of your swine antibody will stick to unattached rabbit antibody. In addition, there may not be enough swine Ab to stick to the limited amount of rabbit antibody attached to PFE1. Therefore you remove all soluble rabbit antibody before adding the swine Ab.
If you add substrate before removing unattached swine Ab (including peroxidase) you will get a glow everywhere – the peroxidase will be all over, not localized, and so you will generate glowing product everywhere.
This is a case of using a primary (rabbit) and secondary (swine) Ab, but you are not using fluorescence as a tag on the secondary Ab. You are using enzymatic activity instead.
2. PFE1 is a transmembrane protein. This question is about its structure. (Its function is not relevant to this question.) PFE1 has 8 hydrophobic segments that are each about 25 amino acids (AA) long. The hydrophobic sections alternative with hydrophilic sections. Both ends of the protein (carboxyl and amino) are hydrophilic.
You have several different antibodies to different regions of the protein. You have intact RBC and unsealed RBC ghosts. You treat either RBC or ghosts with an enzyme that partially degrades proteins, and use the antibodies to find out what part of the protein is left. (See the details of experiment on the last page of the exam.)
A-1. The results of the experiment described on the last page indicate that the carboxyl end of the protein is (extracellular) (intracellular) (either way) (neither).
A-2. You want to be sure of the location of the amino end of the protein by binding of the appropriate Ab. You could probably get the Ab to bind to the amino end by adding it to (intact RBC) (unsealed ghosts) (either way). **
Draw a rough picture that shows how the protein is oriented in the RBC membrane. Stick to the simplest, most obvious orientation, and explain how it fits your answers (& the experimental data).
The antibody to the carboxyl end won’t bind to PFE1 if the protein has been digested with carboxypeptidase Y. The epitope the antibody recognizes will be degraded, and there will be nothing for the Ab to bind to. Binding of Ab to other regions of the protein will be unaffected. Binding to Ab C is a test to see if the C end is still there or whether it has been degraded. Binding to Ab 2/3 is a control to be sure the Ab-antigen reaction is working (and the rest of the PFE1 is still there). No binding means the C end of PFE1 is degraded.
Answer to prob. 2A, cont.
If you add carboxypeptidase Y to intact RBC, it cannot digest the C end of PFE1. If you add carboxypeptidase Y to unsealed ghosts, the peptidase can digest the C end. This means that the C end of the protein is inside the RBC, in the cytoplasm. Since there are an even number of transmembrane segments, so the other end of the protein must also be in the cytoplasm.
Antibodies are proteins, and do not cross membranes. They cannot enter intact cells if they are added from the outside unless the cells have been specially treated to make them permeable to large molecules. That’s why antibodies are sometimes injected directly into cells. The description of this experiment says that antibodies are added to the purified protein, not to whole cells. The cells or ghosts listed in the table indicate what was treated with carboxypeptidase, not what was treated with antibody. However, the table summarizing the results may have misled some students into thinking that the antibodies in this case could enter the intact RBC. If antibodies can enter intact RBC (which they can’t), the answer to A-2 is ‘either way’. Since the table wasn’t as clear as it should have been, full credit was given for a clear explanation of the answer ‘either way’, but not for the answer itself. (Additional headings have been added to the table, and blocked in green, to clarify the intended meaning.)
Grading: 2 pts each answer; 2 pts for picture, and 2 pts each for explaining A-1 & A-2. (10 pts total)
B. Suppose you add two antibodies (Ab 2/3 and/or Ab 3/4) to the intact RBC. (Antibodies are described on the last page.) Which ones would be expected to agglutinate the RBC? (Ab 2/3) (Ab 3/4) (both) (neither) (one of the other – but can’t predict which). Agglutinate = link cells into large clumps. Explain how your answer follows from your picture and antibody structure.
Any Ab that binds to an extracellular domain of the protein will be able to agglutinate the cells. Many molecules of the protein should be present on the surface of the cell, providing many points of attachment for Ab. Each Ab has two identical variable domains that can bind to the cell surface, so each molecule of Ab can link two cells, and each cell can link to many other cells. So you’ll get a giant clump as shown on handout 2-C (antigen-antibody complex).
3. Dr. Who’s lab has been studying RME of transferrin, using pulse-chase experiments.
A. Using their standard pulse-chase methods, the Who lab can follow labeled transferrin to endosomes, but not farther. Once the transferrin reaches the endosomes, the label spreads diffusely around the cell. This implies the label they use is in (clathrin) (integrin) (Fe) (apoB) (apotransferrin) (transferrin receptor) (beats me). Explain how you know.
Fe separates from the apotransferrin-receptor complex in endosomes. The Fe is transported out of the endosomes and reaches many different target enzymes throughout the cell. The apotransferrin-receptor complex stays together and travels through vesicles back to the plasma membrane. If one of the proteins were labeled, the label would remain in spots over the vesicles, not spread around.
‘Around’ here means ‘all throughout the cell’ not ‘all around the cell edges.’ I don’t think there is any way the label could end up around the periphery of the cell. The label can’t be in clathrin, which is the only peripheral protein mentioned here, because the chlathrin separates off from vesicles when they uncoat, before they join endosomes.
B. Over the course of an entire pulse-chase experiment, the amount of label in coated endocytic vesicles should (increase steadily) (increase, and level off) (increase, reach a peak, and decline) (stay constant)
(stay constant at zero) (change, but can’t predict how). Explain the principle briefly.
The labeled material enters the cell, and reaches coated vesicles. As more and more of the labeled material reaches the vesicles, the label in them increases. However the labeled material doesn’t stay in the vesicles. It passes on to the next compartment. Therefore the amount of label in the vesicles starts to decline as the labeled material passes on to endosomes. An ‘entire pulse-chase experiment’ includes one ‘pulse’ of labeled material, not many.
This method is in contrast to a continuous label. In that case, the labeled material would constantly feed into and flow out of coated vesicles. At first the amount of label in the vesicles would increase, but the amount would reach a steady state, as flow in would be counteracted by flow out.