C2006/F2402 ’11 Review Questions for Exam #1 = Exam #1 of ‘10
Note that there is additional information & a list of abbreviations at the end of the exam. Spaces for explanations have been removed to save paper, but all answers had to be explained.
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.
B. You have a red dye that is a small hydrophilic 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 inject 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.
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.
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 alternate 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).
Part B is on the next page.
C2006/F2402 ’11 Review Questions for Exam #1 = Exam #1 of ’10
Problem 2, cont.
B. Suppose you add two antibodies (Ab 2/3 and/or Ab 3/4) to the intact RBC. (The antibodies involved 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.
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 all through the cell. This implies the label they use is in (clathrin) (integrin) (Fe) (apoB) (apotransferrin) (transferrin receptor) (beats me). Explain how you know.
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.
4. The scientists in Dr. What’s lab have been studying RME using antibodies. They have found, as expected, that apotransferrin and transferrin receptor are recycled, but LDL (including apoB) is not.
In Dr. What’s lab, they inject labeled antibodies into cells and follow the label. If they add only transferrin to their cells, the injected antibodies bind to uncoated vesicles. If they add only LDL, the antibodies do not bind to uncoated vesicles.
Suppose a researcher in the lab adds both LDL and transferrin and waits long enough for the added substances to reach uncoated vesicles. Then she breaks open the cells, adds antibody, and purifies the vesicles that have bound antibody.
A-1. These vesicles should contain (transferrin) (LDL) (both) (neither).
A-2. The antibody should be binding to (transferrin) (transferrin receptor) (apoB) (LDL receptor). Explain briefly.
B. Suppose the antibody is found in or on vesicles budding off of endosomes.
B-1. These vesicles should contain (apotransferrin) (apoB) (both) (neither) (can’t predict).
B-2. These vesicles should contain antibody (inside the lumen of the vesicle) (on the outside of the vesicle) (either way). Explain briefly. Please be selective -- do not reproduce the entire class handout on RME.
5. Most cells contain a spectrin web under the plasma membrane. When coated vesicles are first formed, they seem to get trapped in the spectrin web, so it takes a few minutes for added transferrin to get from coated vesicles to endosomes. A type of myosin, myo6, sticks to uncoated vesicles. It does not bind to coated vesicles. If myo6 is missing, transferrin takes much longer to reach endosomes.
A-1. From the information given, it seems likely that vesicles normally move to endosomes by
(simple diffusion) (movement along MT) (movement along MF) (movement along IF) (none of these).
A-2. Myo6 probably binds to (Fe) (spectrin) (clathrin) (actin) (tubulin) (transferrin) (none of these).
Circle all correct answers and explain both parts briefly.
B. If myo6 is defective, what protein(s) could cells make that would be likely to speed up the vesicle traffic to endosomes? (dynein) (a rare kinesin) (ordinary kinesin) (none of these). See info on last page for properties of these motor molecules. Circle all reasonable choices and explain.
C. There are genetically engineered cells that make a fragment of myo6 in addition to the whole protein. The fragment contains the domain of myo6 that binds to uncoated vesicles. The fragment does not include any other domain of the protein. (These cells do not make any additional proteins that can substitute for, or interfere with, myo6.) In these cells, that make both myo6 and the fragment, vesicles should move to endosomes (faster than normal) (slower than normal) (about the same rate as normal) (can’t predict). Explain your prediction.
C2006/F2402 ’11 Review Questions for Exam #1 = Exam #1 of ’10
6. GLUT1 is the only glucose transporter in endothelial cells and RBC. People with GLUT1-DS have low levels of GLUT1. This does not seem to have serious effects on most of the body, perhaps because energy sources other than glucose can be used instead. However, the brain is seriously affected. Glucose levels in the blood are normal, but levels in the CSF (= cerebral spinal fluid = interstitial fluid of the brain) are low.
A. Consider the endothelial cells that surround capillaries. Should GLUT1 be required for glucose to enter the endothelial cells?
(yes) (no) (depends – GLUT1 should be required in cells surrounding some capillaries but not others) (beats me).
B. Consider the capillaries in the liver and in the brain. Should GLUT1 be required for glucose to exit these capillaries? (yes) (no) (depends – GLUT1 should be required in some capillaries but not others) (beats me).
Explain both your answers.
C. You can diagnose GLUT-DS by measuring uptake of glucose by intact RBC. Can you replace RBC with any of the following for diagnostic purposes? Circle the letters of all reasonable choices and explain briefly.
(a) unsealed RBC ghosts. (b) resealed ghosts.
(c) vesicles made from fragments of ghost membrane – resealed randomly either right side or wrong side out
(d) fragments of ghost membrane – not resealed. (e) none of these
D. You measure uptake of glucose by RBC. You measure glucose uptake with time at a fixed concentration of glucose added outside. You plot uptake (vs. time) for RBC from a normal person and for RBC from a person with GLUT1-DS. You would expect a difference in (the plateau value of the curves) (in the initial slope of the curves) (both) (neither – the difference would be one curve would be linear and one wouldn’t).
Note: Plot of ‘uptake vs time’ means you plot the concentration of glucose inside vs time.
E. You have a patient with GLUT1-DS. He makes about half the normal amount of GLUT1, but all the GLUT1 he makes is normal. You measure the uptake of glucose by RBC (vs. time) as above, but you do it repeatedly, using different starting concentrations of glucose added outside. Then you plot the rate of uptake (vs. concentration) for RBC from a normal person and for RBC from the patient. What will the two curves look like? Draw the graph of initial rate of uptake vs. [glucose]outside to start for both the normal person & the patient, and explain.
List of abbreviations and acronyms (A similar list will be provided on each exam this term.)
AA = amino acids
Ab = antibody
ApoB = protein part of LDL
C = carboxyl end of protein
CSF = cerebral spinal fluid; is equivalent to interstitial fluid of brain
ECM = extracellular matrix
Fe = iron
GLUT1 = glucose transporter #1
GLUT1-DS = GLUT1 deficiency syndrome or De vivo disease
IE = intestinal epithelium
IF = intermediate filaments
LDL = low density lipoprotein particle
MF = microfilaments
MT = microtubules
N = amino end of protein
PFE1 = protein invented for exam 1.
RBC = red blood cells
RME = receptor mediated endocytosis
Details for Question 2
Remember protein is always written with its amino (N) end on the left.
You have different antibodies to different epitopes (short peptides) in different regions of the protein.
Ab C – antibody to an epitope (section) near the carboxyl end of protein, after the last hydrophobic domain.
Ab N – antibody to an epitope (section) near the amino end of protein, before the first hydrophobic domain.
Ab 2/3 – antibody to an epitope located in the section between hydrophobic domains 2 and 3.
Ab 3/4 -- antibody to an epitope located in the section between hydrophobic domains 3 and 4.
You have intact RBC and unsealed ghosts.
Expt 1: Unsealed ghosts or intact RBC were treated (or not) with carboxypeptidase Y. Carboxypeptidase Y removes AA one at a time from the carboxyl end. Then PFE1 was isolated, and antibody was added (either Ab-C or Ab 2/3).
Results – Does Ab bind to PFE1?
Peptidase Y Purified PFE Purified PFE
Sample Treatment of Sample? Binds to Ab- C? Binds to Ab 2/3?
Intact RBC yes yes yes
Unsealed ghosts yes no yes
Intact RBC no yes yes
Unsealed ghosts no yes yes
Info for Problem 5:
There are many types of myosin, dynein, and kinesin, and not all cells make all kinds. Each type of myosin, dynein, or kinesin moves in one direction along its respective fiber. All types of myosin except myo6 go toward the plus end. Myo6, and all types of dynein, go toward the minus end of their respective fiber. Most types of kinesin go toward the plus end, but rare kinesins go toward the minus end.