LS1a Fall 2014 Problem Set #4 Due Monday 11/3 at 6 pm in the drop boxes on the Science Center 2 nd Floor

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1 LS1a Fall 2014 Problem Set #4 Due Monday 11/3 at 6 pm in the drop boxes on the Science Center 2 nd Floor Note: Adequate space is given for each answer. Questions that require a brief explanation should be answered only in the provided space. I. Basic Concept Questions 1. (5 points) Determine how each change listed below affects the fluidity of phospholipid bilayer. Change Effect on Membrane Fluidity (circle one) Increasing the temperature Increase Decrease Increasing the cholesterol content (assume at a high temperature) Replacing phospholipids with longer fatty acid tails for phospholipids with shorter fatty acid tails Conversion of trans double bonds in phospholipid fatty acid tails to cis double bonds Conversion of saturated phospholipid fatty acids to cis unsaturated fatty acids Increase Increase Increase Increase Decrease Decrease Decrease Decrease 2. (2 points) The conversion of a membrane from solid-like to fluid-like is a temperature dependent reaction: Given what you know about the temperature dependence of membrane fluidity, is ΔS rxn term positive or negative for the phase transition as the phospholipid membrane goes from solid-like to fluid-like? Briefly explain your answer. ΔS rxn is positive because the ΔG rxn for the melting of the membrane becomes more negative (favorable) at higher temperatures. 1

2 3. (10 points) Phosphotidyl ethanolamine is a type of phospholipid commonly found in cell membranes. Shown below are two different molecules of phosphotidyl ethanolamine (named Phospholipid 1 and Phospholipid 2 ). These two molecules differ in the types of fatty acids bound to the glycerol backbone. a. (4 points) Consider a plasma membrane composed entirely of phospholipid 1. Would this membrane be more or less fluid than a plasma membrane composed entirely of phospholipid 2? Briefly explain why. A membrane comprised of Phospholipid 1 would be more fluid than one comprised of Phospholipid 2. Phospholipid 1 has two cis-double bonds that create kinks that would interfere with ordering of the membrane. These kinks would limit the amount of contact surface area shared between neighboring phospholipids, decreasing the amount of van der Waals interactions that can occur between the neighboring phospholipids. Phospholipid 2 has transdouble bonds, which do not create kinks and would affect fluidity much less. b. (4 points) Cells with membranes consisting entirely of EITHER phospholipid 1 OR phospholipid 2 were made to express a membrane protein that was linked to GFP and subjected to a FRAP experiment. The results are shown below. Which curve, A or B, was obtained from a cell membrane composed entirely of Phospholipid 1? Briefly explain your reasoning. 2

3 c. (2 points) Cholesterol was added to the membrane that yielded Curve A. On the curve above, draw the result you would expect if one were to conduct a new FRAP experiment with these cells. Keep in mind that cells live at temperatures where their membranes are fluid (a temperature we therefore consider as high ). The new curve should be to the right of the original curve, which signifies a longer recovery time. If you are artistically inclined, please feel free to draw whatever you like below this line. 3

4 II. Applied Concept Questions 4. (10 points) The hydrophobic interior of a lipid bilayer favors certain intermolecular interactions. a. (2 points) Briefly explain how the three-dimensional structure of an alpha helix allows it to favorably span a phospholipid bilayer. The helix can project nonpolar amino acid side chains outward, which can interact favorably (because of the hydrophobic effect and through van der Waals interactions) with the hydrophobic interior of the membrane while buffering the polar peptide backbone from the hydrophobic interior. In addition, alpha-helices can satisfy all of the potential hydrogen bonds between its amide groups without exposing these polar groups to the hydrophobic interior of the membrane. b. (4 points) Briefly explain why the flipping of a phospholipid from one leaflet of the membrane to the other leaflet is a kinetically slow process. Placing the hydrophilic head groups of phospholipids in the hydrophobic membrane is thermodynamically unfavorable, and as such, creates a very high energy transition state for the flipping reaction. As a result of the high activation energy associated with the flipping reaction, the reaction is slow (the rate constant k e ΔG /RT ). Interactions between the hydrophilic portions of the phospholipid and water must be broken in order for the phospholipid head to enter the hydrophobic membrane interior. This process is enthalpically unfavorable, which contributes to the high activation energy of the flipping process. c. (4 points) Which of the three reaction energy diagrams shown below best describes the flipping of a phospholipid from one leaflet of the membrane to the other leaflet? Justify your choice with a brief explanation. (Assume that the membrane is composed of only one type of phospholipid and that the solution on both sides of the membrane is the same.) 4

5 Curve A best describes the flipping of a phospholipid from one leaflet of the membrane to the other. While this is a slow process, the energies of the reactant and product are equal since the composition of both leaflets of the membrane and the solutions on both sides of the membrane are the same (i.e., there is no energy difference for the phospholipid to be on either side of the membrane). 5. (26 points) Shown below is the transition state for an RNA catalyzed self-splicing reaction that utilizes an exogenous guanine nucleoside (shown in blue). C11 and C24 are specific catalytic bases involved in the self-cleaving reaction (shown in bold). 5

6 a. (6 points) In the space below, draw the reactants of this phosphodiester bond cleavage reaction and include the arrow pushing mechanism for this reaction. b. (6 points) Draw the products of this reaction. c. (4 points) Both the eukaryotic spliceosome and self-splicing introns carry out similar reactions (the excision of an intron), but use slightly different chemical mechanisms. How do the overall structures of the introns released by the selfsplicing mechanism and the eukaryotic spliceosome differ? Briefly describe the reason for this difference. The introns excised by the eukaryotic spliceosome are released in the shape of a lariat because the nucleophile in the first step is an internal, branch-point adenosine 2 OH (this adenosine s 5 and 3 hydroxyls are bound to the upstream and downstream nucleotides in the intron, respectively). No lariat is generated by the self-splicing mechanism since the nucleophile is an exogenous guanosine that has is not covalently attached to any other upstream or downstream nucleosides. 6

7 d. (6 points) Given the transition state shown for this self-splicing reaction, identify whether uracil is a part of the intron or the exon. Is adenine a part of the intron or exon? Is the phosphodiester bond cleaved in this transition state at the 5 splice junction or the 3 splice junction? Uracil is a part of the exon. Adenine is part of the intron. The phosphodiester bond broken in this transition state is at the 5 splice junction because the exogenous guanine is attacking it and because it frees up the exon uracil 3 OH to attack the 3 splice junction, which will happen in the second transesterification reaction, thereby ligating the two exons. e. (4 points) Provide one reason for why a segment of DNA with the same sequence as this self-splicing RNA would not be expected to carry out the self-splicing reaction. [Please be sure to address the molecular differences between RNA and DNA in your answer.] There are at least three reasonable answers: - Because DNA lacks a 2 OH, it is missing a critical chemical moiety that is involved in the transition state of the reaction (the 2 OH of the uracil nucleotide acts as a catalytic base/acid). Without this chemical group, the DNA would be unable to catalyze the reaction. - Because DNA lacks a 2 OH, it would fold into a different three-dimensional conformation than this segment of RNA, and as a result, it would not have a folded structure that is capable of catalyzing the reaction. - Because DNA is always double stranded in the cell, it would form a double helix, and it would be unable to assume the complex, three-dimensional folded structure that is necessary to catalyze the reaction. 6. (10 points) TMP1a and TMP1b are two transmembrane proteins that bind to one another in the membrane. Neither TMP1a nor TMP1b are typically located in lipid microdomains. TMP1a is embedded in membrane by a single transmembrane alpha helix, whose sequence is shown below along with the amino acids that flank it N-terminus MSKARRWVIIVLLAVLVMLFILYGNQEAKDDTA C-terminus a. (2 points) Draw a box around the amino acids that constitute the single alpha helix that spans the membrane. b. (4 points) Which of the mutated sequences shown below would most likely cause TMP1a to localize to lipid microdomains? Briefly explain your selection, including an explanation why each other mutation is a worse choice than the answer you select. A. Deletion of amino acids N25 through E27. B. Insertion of the sequence ALF between amino acids I10 and V11. C. Deletion of amino acids M18 through F20. 7

8 D. Insertion of the sequence GLP between amino acids L12 and L13. Mutation B increases the length of the transmembrane helix, which would increase the protein s chance of being localized to a lipid microdomain (since microdomains are thicker than the rest of the membrane). Mutation A does not affect the length of the transmembrane helix. Mutation C makes the transmembrane helix shorter, which is the opposite of what is needed for localization to the lipid microdomain. Mutation D inserts a glycine and proline into the middle of the transmembrane helix, which would destabilize the transmembrane alpha helix (proline in particular is very unstable within an alpha helix, recall the first PyMOL lab assignment). c. (4 points) The mutation you selected in part (b) causes TMP1a to localize to lipid microdomains. Would you expect cells bearing this mutant TMP1a instead of normal TMP1a to contain more, less, or about the same amount of TMP1a bound to TMP1b as in cells with normal TMP1a? Briefly explain. Less TMP1a would be bound to TMP1b in cells that contain mutant TMP1a because mutant TMP1a and TMP1b would not come in contact with each other: TMP1a would reside in lipid microdomains while TMP1b would not. Lipid microdomains confer heterogeneity to the membrane, as evident from the GFP/RFP experiment Susan showed at the end of lecture (8 points) When bacteria grow, their waste products acidify the environment in which they re growing. One of the problems that this poses to bacteria is that high H 3O + concentrations enable the inter-conversion of the trans and cis double bonds in membrane lipids, as described below. K eq for the overall process is slightly larger than one. Some bacteria have evolved to resist this by replacing double bonds with cyclopropane rings (three-membered carbon rings). An example of such a lipid is shown below: Line through the faux double bond carbons on the same side as faux double bond 8

9 a. (4 points) Does the modified lipid affect membrane fluidity more like a transunsaturated lipid or more like a cis-unsaturated lipid? Briefly explain why. It is more like a cis-unsaturated lipid. The modified lipid prevents one of the carbon-carbon bonds from rotating. This has the effect creating a kink in the carbon chain since it is impossible for the chain to assume an extended conformation that would be possible with a saturated or trans-unsaturated lipid. This interferes with stacking with adjacent carbon chains and increases membrane fluidity, like a cis-unsaturated lipid would. b. (4 points) If a bacterium didn t have the ability to introduce the cyclopropane modification into its lipids, would you expect its membrane fluidity to increase or decrease when exposed to an acidic environment? Briefly explain your answer. Membrane fluidity would decrease. The lipid modification preserves the ability of cis-unsaturated lipids to increase membrane fluidity. Without that modification, the cis-unsaturated lipids would be partially converted to transunsaturated lipids that don t have the same ability to increase fluidity because they pack well with other extended carbon chains. Therefore, this would have the overall effect of decreasing membrane fluidity. 8. (11 points) The RNA world hypothesis has gained wide acceptance over the past 30 years with the discovery of naturally occurring ribozymes, which are catalytically active single-stranded RNA molecules. a. (2 points) While several ribozymes have been discovered that can cleave phosphodiester bonds between ribonucleosides, it has been much more difficult to find a ribozyme capable of synthesizing a phosphodiester bond between ribonucleotides. Since we have not been able to find such a ribozyme, scientists have been trying to artificially synthesize a ribozyme that polymerizes RNA using an RNA template. Briefly describe why it would bolster the RNA world hypothesis to generate a ribozyme uses an RNA template to guide RNA polymerization activity (i.e., a replicase ). A ribozyme that can polymerize RNA and use RNA as a template could be able to copy itself, first as a template, then by copying the template regenerate its own sequence to generate another functional replicase. Geneteric variation of such replicases could allow for better replicases to evolve over time, thereby serving as the linchpin for Darwinian selection for this early form of life. Membranes enable this evolution of early life by keeping the replicases and its templates near each other, and excluding other templates, such that each replicase only copies its own template. b. (3 points) Scientists have tried to artificially evolve an RNA replicase using techniques such as SELEX. To test whether the RNAs produced by these experiments are functional as replicases, they are incubated with an RNA template 9

10 and ribonucleoside triphosphates ( NTPs ) and allowed time to react. Such an experiment was conducted using two promising ribozymes candidates generated by a SELEX. After the two potential ribozymes were incubated with template and NTPs, they reaction products were analyzed by a gel to separate the products according to their sizes, with the results shown below: Replicases that elongated themselves Replicase that didn t elongate themselves RNA used to template replicase elongation A figure of the gel used to separate the RNA polymerization products is shown above. A DNA size ladder was added in lane 1 for reference. In lane 5, identify the bands that represent: i. replicases that elongated themselves ii. replicase candidates that did not elongate themselves iii. RNA used to template replicase elongation. c. (3 point) Which replicase candidate, 1 or 2, is a more robust RNA polymerase? Briefly explain your answer. Candidate 2, because more of its molecules were able to extend themselves to the maximal length (i.e, the longest / highest band on the gel is more populated/concentrated in candidate 2 than for candidate 1). d. (3 points) The diagram below shows two versions of a different replicase selection experiment. In the version on left, the replicase is covalently connected to the RNA that it polymerizes (as in the previous parts of this question). In the version on right, the replicase polymerizes an RNA to which it is not covalently linked. Briefly explain why the system on right would not be useful for the experiment described above. Covalently-connected Not covalently connected 10

11 There are two perfectly correct answers: 1) The covalent linkage attaches the catalytic component of the ribozyme to the substrate component (the 3 end that gets elongated by the enzymatic domain of the replicase). The covalent attachment of the enzymatic component of the ribozyme to the substrate that gets modified (elongated) by the enzyme allows for the most robust enzyme to be selected by selecting the substrate molecule that is most elongated. In other words, picking the longest ribozyme (the one that was elongated most) also selects for the best enzymatic domain, since the enzymatic domain and the substrate are physically attached. In the case where the enzymatic domain and the elongated RNA molecule are not attached, it is much less obvious which replicase generated the longest product; i.e., which replicase molecule was the most robust. 2) If the RNA molecule being extended where not attached to the catalytic component, but instead could only be extended about as long as the length of the template, we could expect our data to look something like this: The prominent band of the template that was included in the reaction would obscure the longest possible products produced by the replicase, making it more difficult to determine which replicase was most robust (synthesized the longest products). 11

12 9. (6 points) Consider four enzymes, A through D, that affect membrane fluidity. If these enzymes modify phospholipids as described below, which two enzymes would most effectively increase membrane fluidity? Briefly explain your answer. Enzyme A because it converts an unsaturated FA into a cis-saturated FA Enzyme B because it converts longer FA tails into shorter ones Enzyme D is incorrect because it converts a cis-fatty acid into a cis-mimic fatty acid (see question 7a): Looking at the rotation of the cis-fatty acid that shows its kink and comparing it to the kink generated by the cyclopropane makes it more evident why they both affect membrane fluidity roughly equivalently. 12

13 % Initial Fluorescence Name: 10. (12 points) The bacterium Helicobacter pylori causes stomach ulcers in humans. During an infection, the bacterium uses a specialized protein to inject a toxin into stomach cells. In order to pierce the membrane with this protein, the pathogen alters the fluidity of the host cell s plasma membrane by converting cholesterol in the membrane into glucosylcholesterol. a. (4 points) Shown below are the results of a FRAP experiment that analyzed the fluidity of membranes from uninfected cells and membranes from infected cells. The experiment was carried out at a relatively high temperature. Based on these results, does infection increase or decrease membrane fluidity? Briefly explain how you can tell. 100 Uninfected Infected 0 Time Infection decreases membrane fluidity. Fluorescence recovers faster in uninfected cells, which means that those membranes are more fluid. b. (4 points) Based on these results, does the glucosyl modification to cholesterol increase or decrease the effect that cholesterol has on membrane fluidity at high temperatures? Briefly explain your answer. The modification increases its effect. When membranes are fluid (as in the case of living cells), cholesterol tends to decrease their fluidity. Glucosylcholesterol seems to decrease fluidity even more, so it seems that glucosylcholesterol is more effective in this role than cholesterol (at least at high temperatures. c. (4 points) Name two changes to the lipid composition of the plasma membrane that could counteract the effects of glucosyl-cholesterol on membrane fluidity. Cells could produce less cholesterol, or shorten the lengths of the fatty acids, or use more unsaturated fatty acids with cis-double bonds, or use more unsaturated fatty acids with trans-double bonds, or cleave off the glycosylgroup. 13