(multiple answers) This strain of HIV uses a different chemokine coreceptor for entry into cells.

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1 LS1a Fall 06 Problem Set #4 100 points total all questions including the (*extra*) one should be turned in TF 1. (20 points) To further investigate HIV infection you decide to study the process of the viral recognition of host cells. The strain of HIV you have available, HIV-1 NL4-3, normally infects T cells that have the CD4 receptor protein and the T-cell chemokine coreceptor on their surfaces. To dissect the requirements for the entry process, you decide to test whether HIV can infect cheek cells. a) First you add HIV-1 NL4-3 to the cheek cells growing in culture. You find that the HIV viral particles don t bind to the cells. Is this what you would expect? Briefly explain why. (5 points) Yes, it is unlikely the cheek cells express the proteins necessary for viral entry. The fact that the HIV-1 viral particles do not bind or enter the cells is consistent with lack of expression of the CD4 receptor and/or the chemokine co-receptor. b) Next, you engineer your cheek cells to express the CD4 receptor and the T-cell chemokine co-receptor. Do you think that HIV-1 NL4-3 will bind to these engineered cheek cells? (5 points) Yes, if this strain of HIV uses the chemokine co-receptor CXCR4 it is likely that HIV will be able to bind to CD4 and CXCR4 expressed on the engineered cheek cells, and enter these cells. c) (*extra*) Your friend gives you a second strain of HIV-1, HIV-1 YU-2, which normally infects macrophages instead of T-cells. When you try to infect your engineered cheek cells that express CD4 and the T-cell co-receptor with this strain of HIV, you find that HIV-1 YU-2 does not bind to the cells. Give one possible explanation for this result. (multiple answers) This strain of HIV uses a different chemokine coreceptor for entry into cells. d) Since you know that the membrane is a heterogeneous and dynamic structure, you are interested in determining if membrane components are important in the binding of HIV to the cell. You deplete your engineered cheek cells of glycosphingolipids. Why have you chosen to target this type of lipid for depletion? (5 points) Glycosphingolipids with their long fatty acid tails are a component of lipid rafts. If you deplete this type of lipid you may disrupt the formation or structure of the lipid rafts. 1

2 TF e) Interestingly, you find that HIV-1 NL4-3 can no longer bind to and infect the glycosphingolipid-depleted cheek cells, although these cells still express CD4 and the T-cell chemokine receptor on their cell surface. Give one possible explanation for these results. (5 points) CD4 and CXCR4 are normally clustered in lipid rafts and receptor density is important for HIV binding of the receptors and entry into cells. or It is possible that glycosphingolipids act as a coreceptor along with CD4 and CXCR4 to facilitate viral binding and entry into cells. 2. (32 points) 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 membrane, you use fluorescent tags to label and observe these proteins. You first examine Protein A. a) You 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 Photobleach percent fluorescence (4 points for graph) time (min) 5 (6 points) 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 so the recovery is not 100%. 2

3 TF 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 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. 100 percent fluorescence 20 (4 points for graph) time (min) 5 (6 points) 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. 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? Briefly explain and include any assumptions. (12 points: 6 for high, 6 for low temp) 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. 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. 3

4 TF 3. (17 points) Organisms can regulate their lipid composition in response to temperature. Fish are particularly rich in the healthful 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). (6 points) 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 T m. 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 VdW interactions and subsequently decrease their plasma membrane T m. Thus their membranes resist becoming more solid-like at this cooler temperature. 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? (3 points) The fluidity of the membranes would decrease; they would become more solid-like or gel-like. c) If this temperature shift were gradual, what could the bacteria do to combat the change in membrane fluidity? (Hint: some bacteria have enzymes that can adjust the length and saturation of the fatty acid chains) (4 points) In response to cold the bacteria could shorten their fatty acids chains, or they could incorporate double bonds into fatty acids. Either process would reduce the packing and therefore the VdW interactions between the fatty acids. This would increase the fluidity of the membrane. d) If the bacteria were suddenly shifted to a very high temperature, how would the membrane permeability be affected? 4

5 TF (3 points) Membranes would become too fluid which could cause them to become leaky and allow ions to cross. 4. (23 points) Transport of cations, such as Na +, K + and Ca 2+, across membranes is necessary for many processes that occur inside a cell. However, the phospholipid bilayer that comprises the cell membrane is impermeable to these cations. a) Why are cations unable to cross the phospholipid bilayer in the absence of channels? (4 points) The cation s charge and their strong electrostatic attraction to water inhibit them from entering the hydrophobic fatty acid tail phase of the lipid bilayer. The Na + -K + ATPase is an ion pump that is present in the plasma membrane of most animal cells. It pumps 3 Na + out of the cell and 2 K + into the cell. The force that drives an ion across a membrane is made up of two components: one due to the electrical membrane potential and one due to the concentration gradient. b) Are Na + and K + ions being pumped against their concentration gradients? (3 points) Yes c) Are Na + and K + ions being transported with or against the membrane potential? (Hint: a typical membrane potential for an animal cell is -60mV {inside negative}) (3 points) Na + is being transported against the membrane potential. K + is being transported with the membrane potential. The Na + -K + ATPase uses the energy from the hydrolysis of ATP to drive the thermodynamically unfavorable process of ion transport. The energy ( G) from the hydrolysis of ATP to ADP and Pi is roughly 12 kcal/mole. d) What is the sign of G for ATP hydrolysis? (3 points) negative e) What is the sign and value of G for the unfavorable ion transport that one mole of ATP can drive? (3 points) +12 kcal 5

6 TF For transport into the cell the free energy change ( Gin) per mole of K + moved across a plasma membrane with a membrane potential of -60mV at 37 C is: G Kin = (kcal/mole) log Co kcal/mole Ci f) What is the maximum concentration gradient that can be achieved by ATP driven active transport of K + into the cell, assuming that one ATP is hydrolyzed for each solute molecule that is transported? (4 points) Co/Ci= 3.0 X 10-8 Thus a transport system that uses the hydrolysis of 1 ATP to transport 1 Na+ could drive a concentration difference across the membrane of more than 8 orders of magnitude. g) How does this maximum compare with the actual concentration gradient observed in animal cells? (Hint: see your Lecture notes!!) (3 points) Much greater than actual 6

7 TF 5. (8 points) Nerve impulses are transmitted across neurons through the release of neurotransmitter molecules. The neurotransmitter molecules diffuse across the space between the neurons (synapse) and bind to neurotransmitter receptors that in turn propagate the nerve impulse into the neurotransmitter receiving neuron. Neurons are often pre-labeled with lipophilic dyes to visualize all of their processes. The structure for one such dye, DiI is shown below. These dyes are specifically introduced to very small regions of a cells membrane, however the dye will eventually diffuse to label the entire plasma membrane. N + N DiI a) Given DiI s chemical properties and the features of the plasma membrane, explain this observation. (4 points) DiI is an amphipathic molecule like the phospholipids in the plasmid membrane. It has mobility similar to the phospholipids. DiI can therefore diffuse laterally to label the entire plasma membrane. b) Given an infinite amount of time, DiI fluorescence is never observed in the cytosol of a DiI labeled cell. Explain. (4 points) DiI just like the phosopholipids has two long hydrocarbon tails that are hydrophobic. It cannot leave the membrane and enter the aqueous environment of the cytosol. 7