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231 Biol problem set key fall 2009 Q1. E ukaryotic cells have a number of membrane-bounded organelles with specific and essential functions in the life of the cell. For the list of organelles below, match each organelle to the best descriptions of their properties or functions that occur there. For each organelle there is more than one answer (exact number in brackets), and individual answers may apply to more than one organelle. P ossible functions: (1) synthesizes membrane proteins and proteins to be secreted from the cell (2) covalently modifies proteins (3) sorts proteins to many other sites in the cell (4) maintains a pH much more acidic than that of cytoplasm (5) synthesizes sugar molecules using the energy of light (6) produces molecular oxygen by splitting water (7) consumes oxygen and releases carbon dioxide (8) synthesizes ATP from ADP and phosphate (9) site of genomic DNA replication (10) site of genomic DNA transcription into RNA (11) degrades proteins, lipids and nucleic acids (12) probably evolved from bacteria that were engulfed by the early eucaryotic cell ORGANELLE FUNCTIONS NUCLEUS 9, 10 MITOCHONDRION 7 , 8, 12 CHLOROPLAST 5, 6, 12 ENDOPLASMIC RETICULUM 1, 2 GOLGI APPARATUS 2, 3 LYSOSOME 4, 11 [Hint: use your text book's glossary and the figures in chapter 1] Q 2. Y ou have approximately 1013 cells in your body (human cells, that is - we'll come back to the rest in a few weeks!). If all of those human cells arose by successive divisions of your original zygote, how many divisions did it take to produce you? Determine the integral number of divisions that gives the closest result Answer: Log2 of 101 3 is approximately 43 . So it would have taken 43 rounds of division starting with the original zygote to produce you (this would get you closest to 10 13 , with 8.80 x 10 12 cells). Of course, the development of an adult organism from a zygote isn't quite that simple. In addition, things can go awry with cell division - notably the transformation of normal cells into cancer cells that divide in an uncontrolled way. If a single cancer cell in a tissue of the human body divides once every day, how long will it take for that cell to produce a tumor 20 cm3 in size (approximately equivalent to a golf ball)? Assume that the original cell has a volume of 10! 9 cm3 and determine the integral number of divisions that gives the closest result. Answer: 10 -9 cm 3 x 2 n = 20 cm 3 2 n = 2 x 10 10 n = approx 3 4 divisions (resulting in a volume of 1.72 x 101 0 cm3 ), so 34 days total) Q 3. C ells can take up molecules from the outside world by endocytosis, through membrane channels, or by simple diffusion. But regardless of the avenue, the rate at which the cell can acquire material from the outside world is limited by the total surface area of the cell. On the other hand, the amount of material that they need to take up is proportional to the cell volume - i.e., how much biomass is in the cell. Some cells are spherical. (note: if you dont remember the formulae for surfaces and volumes of regular solids, one place to find them is: h ttp://math2.org/math/geometry/areasvols.htm (a) Determine a general expression for the ratio of surface area (for example, in m ) to volume (in m 3 ) for any sphere. 2 Surface:Volume = (4 r 2 )/(4/3 r 3 ) = 3/r (b) Now try a calculation - what is the actual surface-to-volume ratio for a spherical adipocyte (fat cell) with a radius of 1 0 m? F or r = 10 m, S:V = 3/10 m = 0 .3 m ! 1 (c) For a human oocyte (egg cell) with a radius of 100 m? F or r = 100 m, S:V = 3/100 m = 0 .03 m ! 1 (d) For a frog oocyte with a radius of 1000 m? Note that you do not need to calculate the individual surfaces and volumes if you have first determined a general expression for the ratio. F or r = 1000 m, S:V = 3/1000 m = 0 .003 m ! 1 What do these values mean? The units here, m ! 1 , are the same as ( m 2 / m 3 ), which means that they tell you how many m 2 of surface area are supporting each m 3 of volume in the cell in question. So here you can see that for different cells of the same shape cell (spherical), as the radius goes up, the amount of surface area p er unit volume goes down. (e) Do you think that there is an upper limit to how large a spherical cell can be? [ Why or why not?] YES! o ne reason for an upper limit on cell size would be that, for a uniform shape like a sphere, the ratio of surface to volume decreases linearly with the cell radius. As the size of a spherical cell increases, it eventually reaches a point at which the amount of surface area is inadequate to transport the nutrient molecules needed to support the cells volume. Q 4. C ells in the body of a metazoan organism like yourself have many shapes and sizes that are suited to their particular functions. Consider a muscle cell, which has the shape of a long cylinder. A cell in one of the tiny dorsal interosseous muscles of your hand is 10 mm (10,000 m) long, with a radius of 10 m. (A) What is its surface-to-volume ratio? The surface area of a cylinder (with radius = r and length = l) equals the area of the wall plus the area of the two circular ends, so the surface = (2 r x l) + (2 r 2 ). The volume = r2 l. Here, r = 10 m and l = 10,000 m. So for this cell, surface And the volume = = = = = (2 1 0 m 10,000 m) + (2 (10 m)2 ) 628,318 m 2 + 628 m 2 628,946 m 2 (10 m)2 10,000 m 3,141,600 m 3 So the S:V for the muscle cell = (628,946 m 2 ) / (3,141,600 m 3 ) = 0.2002 m ! 1 . (B) The volume of the human oocyte from tjhe previous problem is similar to that of the dorsal interosseous muscle cell in (A), but their S/V ratios are very different. Calculate (Surface-to-volume ratio of the muscle cell)/(Surface-to-volume ratio of the oocyte): C ompared to 0.03 m! 1 for the oocyte, this is a difference of 6.67-fold . In other words, although the muscle cell has the roughly the same volume as the oocyte, it has much more (6.67x more) surface area, because of its shape. Q 5. H uman epithelial cells can be grown in a culture dish, where they undergo many rounds of cell division. During most of the cell cycle, they adhere tightly to the surface of the dish, spreading out into a rectangular solid shape that approximates a White Castle burger, 65 m x 65 m x 1 m thick. But during cell division they lose most of their attachment to the surface of the dish and round up into a spherical shape with a radius of 10 m. In both shapes the cell has about the same volume (check it for yourself), but the surface area of the cell goes down by about 7-fold when it rounds up (you can check this, too). How can a cell quickly decrease its S/V by decreasing its surface area, while keeping its volume about the same? [multi choice, random order] (A) Exocytosis of many small secretory vesicles, since each of these would have a S/V that is much less than that of the cell. (B) Exocytosis of many small secretory vesicles, since each of these would have a S/V that is much greater than that of the cell. ( C ) Endocytosis that takes up many small coated vesicles, since each of these would have a S/V that is much less than that of the cell. **(D) Endocytosis that takes up many small coated vesicles, since each of these would have a S/V that is much greater than that of the cell. During uptake of fluid or specific macromolecules from the outside world by endocytosis, a cell pinches in vesicles from its plasma membrane that are much smaller (by $ 1 00fold) than the cell itself. Remember (from problem 3, above) that as the radius of a sphere goes DOWN, its surface-to-volume ratio goes UP. Thus, the vesicles pinched in by the cell from its membrane during endocytosis contain much more surface area per unit volume than the cell as a whole. So by performing lots of endocytosis, the cell can decrease its surface area much more than it increases its volume. Thus, the surface-tovolume ratio of the flat cell as a whole goes down as it adds a little volume, loses proportionately more surface area, and becomes spherical. As the surface:volume ratio of a solid diminishes, its shape approaches that of a sphere. You can do a test calculation: imagine a spherical cell with radius 100 m how will its S and V change if it endocytoses a vesicle with radius 10 m? Conversely, note also that a lot of the reverse process exocytosis, the fusion of vesicles with the cell surface to release their contents will tend to INCREASE the cells surface area relative to its volume. Q 6. A ssume that a spherical cell, with a diameter of 20 m, has spherical mitochondria, each with a diameter of 0.2 m. If mitochondria occupy 10% of the volume of the cell, how many mitochondria are there? [You don't need a calculator to figure this out.] The ratio of the cell diameter: mitochondrial diameter is 100/1 = 100, so the ratio of the volumes is (100)3 , or 1,000,000. Thus, it would take 1,000,000 mitochondria to equal the volume of the cell, and 10% of the cells volume would be occupied by 10% of 1,000,000, or 1 00,000 mitochondria . Q7. N umber each of the following in order of its surface area, with 1 being the largest. V ery briefly, explain your reasoning: Plasma membrane of an epithelial cell [ 1] Total membrane of a secretory vesicle [ 4] Total membrane of the ER [ 2] Total membrane of a mitochondrion [ 3] Two discrete organelles, a secretory vesicle (4) and a mitochondrion (3) differ in the total area of their membranes because a mitochondrion is generally longer and has both an outer and a highly folded inner membrane. The ER (2) is a large, highly branching and anastomosing (reticulated, thus the term endoplasmic reticulum) organelle that can extend throughout much of the cell, and whose surface area is probably exceeded on this list only by the plasma membrane (1) of an entire epithelial cell, which is itself typically highly folded at one or more surfaces. Q8. N umber each of the following in order of its numerical value, with 1 being the largest. Very briefly, explain your reasoning: ! log(10 ! 14 ) ! log(10 ! 1 ) ! log(10 ! 7 ) log(10 3 ) 103 [2] [5] [3] [4] [1] Its just arithmetic: 2,5,3,4,1 Q9. (A) If a solution is at pH 5, what is the concentration of protons? 10 -5 M Remember that pH = ! log[H + ] (B) How many protons are in 10 mL solution at pH 3? 6.022 x 10 18 protons pH 3 = a concentration of 10-3 M protons. So how many protons in 10mL? Well, if there are 10-3 moles of protons per Liter, then there are 10-5 moles per 10mL. To get from moles to number of protons, you just multiply by Avogadros number: 10-5 moles (6.022 1023 protons/mole) = 6.022 10 18 protons (C) Assuming the solution described in (b) is not buffered, how much water would you need to add to the solution (b) in order to raise the pH to 5? You need to add 9 90 mL of H 2 O to the original 10 mL of solution. Going from pH 3 to pH 5 is a 100-fold dilution of protons, so you need to add 990 mL water to 10 mLs of starting solution. If this isnt obvious to you, a reliable way to work out this kind of problem is to use the dilution relationship: C 1V1 = C 2V2 where C 1 and V 1 are the concentration and volume, respectively, of the starting solution and C 2 and V 2 are the concentration and volume of the diluted solution. Since (concentration of solute) (volume of solution) = (mass of the solute), this is simply an expression of conservation of mass. Q10. Identify by label: a C-C single bond, a C=C double bond, amines, carboxyls, a phosphate, a diphosphate, a sulfhydryl, an ester, a phosphoester, a phosphodiester [possible molecules displayed: phosphatidylcholine; ADP; lysine; phosphoserine] Q11. Identify by label: the 3 and 5 carbons of ribose/deoxyribose; 3', 4', 5' carbons of glucose; the sole structural difference between ribose and deoxyribose; a glycosidic bond [possible molecules: glucose; maltose; deoxyribose] Q12. Identify by label: The atom in the base rings that will be covalently linked to the 1 carbon of deoxyribose if these pairs were part of an DNA molecule OR Identify by label: All Hs involved in H-bonding the base pairs [molecules shown: an A:T and a G:C pair] Q13. Identify by label: Fatty acid tails; ester bond; phosphoester bond; glycerol backbone; polar head group [molecule shown: phosphatidylcholine] Q14. T he Go for the binding of O 2 to ferromyoglobin is ! 31.4 kJ/mol. What is the K eq for this reaction? G = G 0 + RTln([P]/[R]); at equilibrium, G = 0 and [P]/[R] = K eq , so: G0 = ! RTln(K eq ) = ! 5.7 kJ/mole @ log(K eq) log(K eq ) = (! 31.4 kJ/mole)/(! 5.7 kJ/mole) Therefore K e q = 3.23 10 5 What does a K eq>1 tell you about this, or any, reaction? How about a K eq <1? Look carefully at the expression G 0 = ! RTln(K eq). Because G 0 and K eq have a minus log relationship, K e q < 1 results in a large positive G0 and describes a reaction that will NOT proceed spontaneously from left to right u nder standard conditions. But if K e q > 1 then the reaction has a large negative G0 and WILL proceed spontaneously from left to right under standard conditions. But also realize what K e q means: it is a ratio of products to reactants at equilibrium. So, whether or not a reaction will proceed in a particular direction can be changed by a large shift in the ratio of [products]:[reactants] (a departure from standard conditions). Q 15. Y ou have identified found an enzyme in the lab freezer labeled bloggsose kinase/ GTP which catalyzes the reaction: bloggsose + GTP <=> bloggsose-phosphate + GDP You want to determine whether it is still active, so you place in solution in an eppendorf tube a very tiny amount of pure bloggsose kinase and a mixture of bloggsose and GTP and allow the reaction to proceed to equilibrium. Then you carefully assay the concentrations of each reactant and product, and find that they are: [bloggsose] = 10 mM [bloggsose-phosphate] = 0.1 M [GTP] = 10 mM [GDP] = 0.1 M (a) Calculate the K eq for this reaction. K eq = [bloggsose-P][GDP] [ bloggsose][GTP] = (10 -1 M)(10 -1 M) (10-2 M)(10 -2 M) (b) Using the K eq, calculate G 0 for this reaction. At equilibrium, G 0 = ! RTln(K ) eq = ! 5.7 kJ/mol @ log (K eq ) = ! 5.7 kJ/mol @ log(10 2 ) = ! 11.4 kJ/mol = 100 Q16. C onsider the reaction: Glucose + P i ] G lucose-6-phosphate + H 2 0 G o = +12.5 kJ/mol (a) What is the equilibrium constant, K eq, for this reaction? G = G 0 + RTln([P]/[R]); at equilibrium G = 0 and [P]/[R] = K eq, so: G 0 = ! RTln(K eq) = ! 5.7 kJ/mol log(K eq) 12.5 kJ/mol = ! 5.7 kJ/mol log(K eq ) log(K eq ) = (12.5 k J/mol/-5.7 k J/mol) Therefore K e q = 6.41 10 ! 3 (b) What does the positive G o for the reaction tell you about whether or not the reaction will go spontaneously from left to right? A large positive standard free energy change means that the reaction will NOT run spontaneously for L to R under standard conditions. To make it run L to R will require a VERY high concentration of reactants relative to products. (c) Under physiological conditions, [glucose] = 5 mM, [g-6-P] = 83 M, and P i = 1 mM. Under these conditions, will the reaction go spontaneously from left to right? [Note: the concentration of water is always taken to be 1 M]. If not, what would the concentration of glucose need to be in order for the reaction to go from left to right, if the concentrations of the other reactants and products are as stated above? Divide the concentration of products by that of reactants under current conditions to get the quantity we call q . Compare q to the K eq. If q > K eq, then the products are in excess and the reaction will not proceed spontaneously; if q < K eq, then the reactants are in excess and the reaction will proceed spontaneously. H ere, q = [ G6P]/[glucose][Pi] = 16.6, which is >> 6.41 10! 3 . The reaction will not proceed spontaneously until: [G6P]/[glucose][Pi] < 6.41 10 ! 3 (83 10! 6 )/(10! 3 )(6.41 10 ! 3 ) < [glucose] [glucose] > 12.9 M!!! Even Coke is not this sweet. Q17. C onsider the reaction: glutamate + ammonia X g lutamine + water G o = +14 kJ/mol (a) What is the K eq for this reaction? Again, what does the K eq describe? G 0 = ! RTln(K eq) = ! 5.7 kJ/mole log(K eq) K e q = 3.50 10 -3 ; K e q is the ratio of [products] : [reactants] at equilibrium! (b) If the concentration of ammonia is 10 mM, what is the ratio of glutamate to glutamine required for the reaction to proceed spontaneously from left to right? Remember, we define the ratio of P to R under current conditions as q . Reactions proceed spontaneously from left to right if q < K eq . S o this reaction proceeds spontaneously from left to right if [gln]/[glu][Amm] < 3.50 10 -3 So, [gln]/[glu] < (3.50 10 ! 3 )(10 ! 2 ) [gln]/[glu] < 3.50 10! 5 , and (invert both sides) [glu]/[gln] > 2.86 10 4 (c) You have probably guessed that in the cell the synthesis of glutamine from glutamate and ammonia does not occur by the reaction described above. The actual reaction c ouples g lutamine synthesis to ATP hydrolysis: glutamate + ammonia + ATP X g lutamine + ADP + P i If the Go for the hydrolysis of ATP is ! 30.7 kJ/mol, what is the G o for the overall reaction? Add the G0 values for the two coupled reactions. The net standard free energy change = ! 30.7 kJ/mol + 14 kJ/mol = ! 16.7 kJ/mol. (d) Suppose that all the reactants and products, except ammonia, are present at 10 mM. What concentration of ammonia would be required to drive the reaction to the right; that is, drive the synthesis of glutamine? First, calculate the K eq for the coupled reaction, using the G 0 of ! 16.7 kJ/mol: G0 = ! RTln(K eq ) = ! 5.7 kJ/mole log(K eq) log(K eq ) = (! 16.7/! 5.7 ) = 2.93 K eq = 851 Then solve for [Amm] under spontaneous L => R conditions with K e q = 10 3 : To proceed L => R, q < K eq (meaning that [P]/[R] < K eq ), so (10! 2 )(10! 2 )(10! 2 ) / (10! 2 )(10! 2 )[Amm] < 851 10 ! 2 /[Amm] < 851 [Amm] > 1.18 10 ! 5 [Amm] must be anything > 1.18 10 ! 5 M, i.e., >11.8 M. Q 18. S uppose you are studying an enzyme that catalyzes the reaction: E+SYP+E After incredibly hard work in the lab, you have developed a simple assay by which you can measure the production of P as a function of time. You obtain the following data at a constant enzyme concentration under standard buffer, pH, and temperature conditions: [S] ( M) 0.00 0.20 0.25 0.33 0.50 1.00 2.00 4.00 (1/S), M ! 1 mol P produced/min 5.0 0 4.00 3.00 2.00 1.00 0.50 0.25 0.0 55.6 62.5 71.4 83.3 100.0 111.1 117.6 (1/V), mol/min ! 1 0.01799 0.0160 0.0140 0.0120 0.0100 0.0090 0.0085 Using these data, determine the V max and K M of the enzyme under these conditions. Use a Lineweaver-Burk plot (1/[S] on the x-axis and 1/V on the y-axis, see Fig. 3-28 in your text). The y-intercept is 0.008 ( moles/min)-1, so the V m ax = 125 moles/min. T he slope is 0.00199 (uM/umols/min), and since slope = K M/V max, we multiply this by V max to get K M = 0.250 M. N OTE that the units are extremely important! I used Excel to do the least-squares fit to the data, but if you used a graphing or statistical calculator you ought to get the right values too. See my graph of these data, below: Q 19. S uppose you are working in the laboratory and have been utilizing a DNA polymerase in an experiment. The polymerase catalyzes the synthesis of DNA from individual nucleotides. In order for the polymerase to function, it requires a single-stranded DNA template, a short DNA primer, and nucleotides (ATP, GTP, CTP, and TTP). If everything is OK, the polymerase will then synthesize a new strand of DNA that is complementary to the template strand. After a bit of trial and error, you find conditions where the synthesis reaction goes well: 1 g of template; 100 pmols of primer; 5 units of polymerase; and 25 M of each of the nucleotides. J ust when the Nobel Prize looms clearly on the horizon, tragedy strikes: the reaction ceases to work well. The only difference between when the reaction was working and now is that the DNA polymerase is from a different source (isolated from a different organism). [NOTE: enzymes like polymerases used in most laboratories are purchased in kits from companies.] The K M of the "old" polymerase was 20 M nucleotide, and the K M of the "new" polymerase is 100 M. What is the first thing you would do to change the reaction conditions (using the new polymerase) in your attempt to get the reaction to work? Briefly explain your answer. Try increasing the [S] to >100 M. With the old enzyme, the rxn was being run with the [S] (nucleotides) slightly > K m (1.25 the K m ) and it was working fine. Since the new enzyme has a higher K m (100 vs 25 M), it will take a higher [S] to get the same rxn rate as before: > 100 M, 1 25 M nucleotides w ould be a good starting point. Q 20. In general, the rate of chemical reactions increases with temperature, but enzymatically-catalyzed reactions have an optimum temperature, above and below which the reaction rate goes down. Which the following statements best explain this observation? (A) At the optimum temperature, the substrate has a higher affinity for the enzymes active site than at higher or lower temperature. (B) Below the optimum temperature, the substrate tends to diffuse out of the active site before the reaction can be catalyzed. ** ( C) As the temperature increases, approaching the optimum, typical features of reactions such as the kinetic energy of the components cause the rate to increase with T. (D) At any temperature above the optimum, the reaction typically occurs efficiently even without the catalytic action of the enzyme. OR ** (A) As the temperature increases above the optimum, the effects of elevated temperature on protein structure cause the enzyme to become less catalytically effective and perhaps ultimately, to denature. (B) At any temperature above the optimum, the reaction typically occurs efficiently even without the catalytic action of the enzyme. ( C) Below the optimum temperature, the substrate tends to diffuse out of the active site before the reaction can be catalyzed. (D) At the optimum temperature, the substrate has a higher affinity for the enzymes active site than at higher or lower temperature. [ each students order of alternatives differs and is randomized, AND there are two different correct answers depending on the set of alternatives that is presented] Q21. C onsider a GTPase that catalyzes the reaction: GTP + H 20 Y G DP + P i. There are simple assays that utilize a color reaction to measure the amount of P i in a solution. Suppose you utilize such an assay to measure the velocity of the release of P i catalyzed by the GTPase. In addition, you have an inhibitor Z that inhibits the GTPase. You vary the concentration of GTP and measure the product released in the presence and absence of an inhibitor and obtain the following data: [GTP] (M) 0.16000 0.19000 0.24000 0.33000 0.49000 0.99000 V no I (mmol/min) 0.99000 1.01000 1.25000 1.54000 1.82000 2.89000 V + I (mmol/min) 0.64000 0.72000 0.85000 1.05000 1.32000 1.75000 Using these data, what is the V max and K M of the GTPase for GTP without inhibitor? Does the inhibitor Z behave as a competitive or noncompetitive inhibitor? As in the previous kinetics problem, use a Lineweaver-Burk plot (note - this is real data, it helps to look at plots in addition to using a linear regression function in your calculator or Excel). The KM is 511 mM and the Vmax is 3.94 mmole/min. In the presence of inhibitor Z the x-intercept of the reciprocal plot is about the same while the y-intercept is higher. Thus with inhibitor present the KM is the same while the Vmax is reduced and therefore Z is a non-competitive inhibitor. 1/[S] (M-1) 6.25000 5.26316 4.16667 3.03030 2.04082 1.01010 1/V no I (min/mmol) 1.01010 0.99099 0.80000 0.64935 0.54945 0.34602 1/V + I (min/mmol) 1.56250 1.38889 1.17647 0.95238 0.75758 0.57143 y-int Vmax slope Km 0.253769 3.94059 mmol/min 0.12974 0.51125 M 0.374761 2.66836 mmol/min 0.19120 0.51019 M T he uninhibited GTPase plot is in blue, the inhibitor plot is in red: Q 22. T he relationship between enzyme velocity, V, and the concentration of substrate, S, can be expressed in the following equation, the Lineweaver-Burk equation, which describes a straight line in a double-reciprocal plot: 1/V = 1/V max + (K M /V max )(1/[S]) A competitive inhibitor, I, affects the slope of the line produced in a double reciprocal plot by increasing the slope by the factor: (1 + [I]/k i), where k i = [E][I]/[EI] Suppose an enzyme has a K M = 10! 4 M and a competitive inhibitor has a k i = 10 ! 3 M. If [I] = 2 10! 3 M, how much will the K M change in the presence of the inhibitor? Use the Lineweaver-Burk equation to solve this problem: ki = 10!3 M [I] = 2 x 10 ! 3 M So, 1 + ([I]/k i) = 3 Therefore the slope, (K M /V max), of the reciprocal plot increases 3-fold. So (K M /V max) increases 3-fold but V max remains the same (because it's a comp inhibitor). Therefore, K M must increase by a factor of 3. Note that you don't have to plot anything this to solve the problem - y ou just need to understand what the elements of the Lineweaver-Burk equation mean. A competitive inhibitor competes with S for the active site, increasing the apparent K M. This is reflected in the plot by an increased slope (because the x-intercept is closer to the origin). Sketch it out and be sure that you understand. Q23. As we have seen, you can analyze the kinetic properties of a Michaelis enzyme by measuring its velocity of catalysis at a range of substrate concentrations and plotting [S] vs V, or 1/[S] vs 1/V. You can also analyze the properties of an enzyme inhibitor by carrying out the same experiment and graphic analysis in the presence of the inhibitor. If you do this for a noncompetitive inhibitor, what difference will you see when you compare the enzyme kinetics to an uninhibited control? (A) Vmax will decrease and KM will increase. (B) Vmax and KM both remain the same with a non-competitive inhibitor. ( C) Vmax and KM will both decrease. (D) Vmax will be unchanged but KM will increase. ***(E) Vmax will decrease but KM will be unchanged. [ each students order of alternatives is randomized] As you increase [S], V will increase, reaching half of its maximum at the same [S] as without inhibitor (K m unchanged). But the V max will be reduced, i.e., in the presence of noncompetitive inhibitor no matter how much S you add the enzyme will not reach the same V max as it does without inhibitor. This is because a non-competitive inhibitor does NOT compete with S for the enzymes active site. Q 24. T he figure below shows the short protein that has the amino acid sequence: GRAKLSS. (A) Click on the side chain of Lysine (4th residue) o r Arginine (2 nd residue) [ 3] (B) What do you expect the charge of this protein will be at pH 2? [ -1] ( C) What do you expect the charge of this protein will be at pH 14? At pH 2, expect all these groups to be protonated: the N-terminal amino group (= +1), the basic residues (K & R) (each = +1), and the C-terminal carboxylic acid group (= uncharged, or 0). So the overall charge at pH 2 = +3. A t pH 14, expect the ionizable groups to be unprotonated: the N-terminal amino (0), the basic residues (each 0), and the C-terminal carboxylic acid group (-1). Thus the overall charge = -1. (7.8) (10.8) (12.5) (3.6) N-term K R C-term net charge 2 +1 +1 +1 0 +3 14 0 0 0 !1 !1 pH Q25. C onsider the peptide: NH 2 -GAPAGPAGTGKTETTKDLAKSMALLCWFNCS-COOH Using Table I below, determine the net charge of the peptide at: pH 3, 6, 7, 9, 12. [ each student gets just 3 of these pHs to solve, randomly] T able I (from Stryer, Biochemistry, 2nd ed., 1981) p K values of some amino acids a mino acid -COOH group 2.3 2.4 1.8 2.1 2.3 2.0 2.2 1.8 1.8 2.2 2.2 1.8 Alanine (A) Glycine (G) Phenylalanine (F) Serine (S) Valine (V) Aspartic acid (D) Glutamic acid (E) Histidine (H) Cysteine (C) Tyrosine (Y) Lysine (K) Arginine (R) peptide amino terminus peptide C terminus -NH 3 group 9.9 9.8 9.1 9.2 9.6 10.0 9.7 9.2 10.8 9.1 9.2 9.0 7.8 side chain 3.9 4.3 6.0 8.3 10.9 10.8 12.5 3.6 pK: 7.8 10.8 4.3 10.8 3.9 10.8 8.3 8.3 3.6 pH N-ter K E K D K C C C-ter total 3 +1 +1 0 +1 0 +1 0 0 0 +4 6 +1 +1 -1 +1 -1 +1 0 0 -1 +1 7 +1 +1 -1 +1 -1 +1 0 0 -1 +1 9 0 +1 -1 +1 -1 +1 -1 -1 -1 -2 12 0 0 -1 0 -1 0 -1 -1 -1 -5 Q 26. N umber each of the following in order of its size, with 1 being the largest. Be sure to reason through it logically, based on what you know and what you can find in the text figures, lecture notes etc. lysine glycine CTP CaM kinase cholesterol [4] [5] [3] [1] [2] CaM kinase (1) is a protein comprised of hundreds of amino acid residues and thus much larger than the other molecules. The largest of the others is cholesterol (2), a 27-carbon lipid with 4 characteristic carbon rings bearing one polar -OH group, 2 methyl groups and an 8-carbon tail, all of which makes it slightly larger than CTP (3), which has a pyrimidine ring (6 members) with amino and carboxyl groups, a ribose sugar (5 more carbons), plus 3 phosphates. The smallest are the 2 amino acids, and lysine (4) has a larger R-group than glycine (5), which is the smallest amino acid (R-group = - H). See diagrams in Chapter 2. Q27. N umber each of the following in order of its size, with 1 being the largest. A TP [2] glucose [5] glucose-6-phosphate [ 4] ADP [3] hexokinase [1] Hexokinase (1) is an enzyme (and therefore a protein, and thus comprised of hundreds of amino acid residues) that uses glucose and ATP as substrates. ADP (3) and ATP (2) are purine nucleotides each with a double ring (9 members), a ribose sugar (5 carbons) plus 2 or 3 phosphates, respectively. Glucose (5) and glucose-6-phosphate (4) are identical 6 carbon sugars, without or with a phosphate group, respectively. Q28. m ulti choice on Protein Data Bank study (Alternative #1) Protein Data Bank h ttp://www.pdb.org/pdb/home/home.do Use this web resource to probe the structure of a Nerve Growth Factor. Search the database using "neurotrophin," and select neurotrophin 4, or just enter id# 1 b8m . From the Display Molecule menu on the left, select "WebMol Viewer" to render the structure, then use the various display options and menus that I demonstrated in lecture (check the lecture notes for help) to answer these questions: (A) What is the overall shape of neurotrophin-4? [multi choice] - It is an elongated, roughly dumbbell-shaped protein [ correct] - It is a spherical protein - It is an elongated protein composed mainly of helical coiled-coil (B) How many polypeptide subunits is this protein composed of? Answer: 2 [hint: easiest way to see this is to select "Color/Chain"] (C) How many disulfide bonds does this protein have? Answer: 6 [hint: you can get this info by clicking the [?] button at lower right, and then display the protein using "Backb" and "Color/Chain" to confirm the number by inspection (you'll have to rotate the structure to see them all)] (D) How many alpha helices does each subunit polypeptide have? A nswer: 0 [ hint: use the "Backb" and "Color/Sec" display options and rotate the structure to look for and count helices (gold) or beta strands (blue)] (E) How many beta strands does each subunit polypeptide of the protein have? Answer: 8 [ hint: use the "Backb" and "Color/Sec" display options and rotate the structure to look for and count helices (gold) or beta strands (blue)] (F) What % of the protein is alpha helix? Answer: 0 [ hint: you can get this info by clicking the [?] button at lower right] (G) What % of the protein is beta strand? Answer: 50% [hint: you can get this info by clicking the [?] button at lower right] (Alternative #2) Protein Data Bank h ttp://www.pdb.org/pdb/home/home.do Use this web resource to probe the structure of bovine hemoglobin. Search the database using "bovine hemoglobin," or just enter id# 1 g09 , or just search the id#. From the Display Molecule menu on the left, select "WebMol Viewer" to render the structure, then use the various display options and menus that I demonstrated in lecture (check the lecture notes for help) to answer these questions: (A) What is the overall shape of hemoglobin? [multi choice] - It is an elongated, roughly dumbbell-shaped protein - It is a roughly spherical protein [ correct] - It is an elongated protein composed mainly of helical coiled-coil (B) How many polypeptide subunits is this protein composed of? Answer: 4 [hint: easiest way to see this is to select "Color/Chain." Hemoglobin has 2 alpha subunits (tendered in green and yellow) and two beta subunits (red and blue). It is easy to see that these subunits have very similar structures.] ( C) How many disulfide bonds does this protein have? Answer: 0 [hint: you can get this info by clicking the [?] button at lower right, and then display the protein using "Backb" and "Color/Chain" to confirm the number by inspection (you'll have to rotate to look carefully)] (D) Blink the "Het Atom" check-box to toggle non-protein groups in and out of the image. T his will reveal the heme groups that directly bind oxygen. How many heme groups does the entire protein have? Answer: 4 [one per subunit] (F) How many alpha helices does each subunit polypeptide have? A nswer: 6 (G) How many beta strands does each subunit polypeptide of the protein have? Answer: 0 [ hint: use the "Backb" and "Color/Sec" display options and rotate the structure to look for and count helices (gold) or beta strands (blue); the "Trace" command is good for confirming your count. Note that there are NO unambiguous beta strands in this protein.] (H) What % of the protein is alpha helix? Answer: 77% [hint: you can get this info by clicking the [?] button at lower right] (I) What % of the protein is beta strand? A nswer: 0% [hint: you can get this info by clicking the [?] button at lower right] (Alternative #3) Protein Data Bank h ttp://www.pdb.org/pdb/home/home.do Use this web resource to probe the structure of bacterial flagellin (a component of the rotary tail that drives bacterial swimming). Search the database using "bacterial flagellin," or just search id# 1 ucu . From the Display Molecule menu on the left, select "WebMol Viewer" to render the structure, then use the various display options and menus that I demonstrated in lecture (check the lecture notes for help) to answer these questions: (A) What is the overall shape of flagellin? [multi choice] - It is an elongated, roughly dumbbell-shaped protein - It is a roughly spherical protein - It is a highly elongated protein, several times as long as it is wide. [ correct] (B) How many polypeptide subunits is this protein composed of? Answer: 1 [hint: easiest way to see this is to select "Color/Chain"] ( C) How many disulfide bonds does this protein have? Answer: 0 [hint: you can get this info by clicking the [?] button at lower right, and then display the protein using "Backb" and "Color/Chain" to confirm the number by inspection (you'll have to rotate to look carefully)] (D) How many alpha helices does flagellin have? A nswer: 6 or 7, this is tricky (E) How many beta strands does each subunit polypeptide of the protein have? Answer: 18 [ hint: use the "Backb" and "Color/Sec" display options and rotate the structure to look for and count helices (gold) or beta strands (blue); the "Trace" command is good for confirming your count. Note that there are NO unambiguous beta strands in this protein.] (F) What % of the protein is alpha helix? Answer: 42% [hint: you can get this info by clicking the [?] button at lower right] (G) What % of the protein is beta strand? Answer: 20% [hint: you can get this info by clicking the [?] button at lower right] (Alternative #4) Protein Data Bank h ttp://www.pdb.org/pdb/home/home.do Use this web resource to probe the structure of HIV reverse transcriptase, the enzyme that the AIDS virus uses to copy its genomic RNA into DNA in the host cells. Search the database using "HIV reverse transcriptase," or just type in id# = 1 s1t. With 1s1t selected, from the Display Molecule menu on the left, select "WebMol Viewer" to render the structure, then use the various display options and menus that I demonstrated in lecture (check the lecture notes for help) to answer these questions: (A) What is the overall shape of HIV reverse transcriptase? [multi choice] - It is a relatively flat structure with 3 perpendicular "legs." [ correct] - It is a spherical protein - It is a highly elongated, rope-like protein, many times as long as it is wide (B) What is the identity of residue #316? (Use the single-letter abbreviation) Answer: G (gly) [hint: When you put the arrow cursor on the chain, the residue you are pointing at is displayed in a small box at the upper right, with the residue number and single-letter abbreviation. If you simplify the structure by selecting "Backb" from the first menu, it is very easy to move along the chain and find a particular residue.] ( C) How many polypeptide subunits is this protein composed of? Answer: 2 [hint: The easiest way to see this is to select "Color/Chain." Note that its overall shape makes it look as though it might have >1 subunit.] (D) How many disulfide bonds does this protein have? Answer: 0 [hint: you can get this info by clicking the [?] button at lower right, and then display the protein using "Backb" and "Color/Chain" to confirm the result by inspection (you'll have to rotate the structure to see them all)] (E) How many alpha helices does the entire protein have? A nswer: 23 [12 in one subunit, 11 in the other] (F) How many beta strands does the entire protein have? Answer: 39 [24 and 15] [hint: Use the "Backb" and "Color/Sec" display options and rotate the structure to look for and count helices (gold) or beta strands (blue). Using the "Trace" command with these settings can help you to check on the numbers.] (G) What % of the protein is alpha helix? Answer: 34% [hint: you can get this info by clicking the [?] button at lower right] (H) What % of the protein is beta strand? Answer: 18% [hint: you can get this info by clicking the [?] button at lower right] Q 29. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: human telomere human chromosome an Okazaki fragment DNA polymerase complex nucleotide [2] [1] [4] [ 3] [5] A human chromosome (1) typically has a telomere (2) - a region of repeated DNA sequence, 3,000 - 18,000 nucleotide bases in length, important for faithful replication - at each end. The subunits from which the DNA of a chromosome is constructed are nucleotides (5). When the DNA on the lagging strand is replicated by a large assembly of proteins, the DNA polymerase complex (3). The segments of a few hundred nucleotides (in eukaryotes - but several thousand nucleotides in prokaryotes) between replication primers are called Okazaki fragments (4) (after their discoverer, Reiji Okazaki of Nagoya University). Q30. C onsider this generalized diagram of a nucleotide triphosphate: (A) Identify each of the labeled structures (a-d) in the diagram. (1) is a phosphate; (2) is the 5' carbon; (3) is the 3' carbon; (4) is the 1' carbon; (5) is the base (B) Is it possible that the base attached to the 1' carbon is a purine? N o , purines have a nine-member, two-ring structure; this is a pyrimidine base (A). ( C ) Could this nucleotide triphosphate be incorporated into a growing DNA strand? No, this is a triphosphate and DNA polymerase uses monophosphates [correct] No, this is a ribose-containing nucleotide triphosphate, not a deoxyribose-containing one No, the base is an RNA base (U), not a DNA base It has an -OH at the 2' position this is an RNA subunit. Q 31. T ranslation begins at an initiation codon in the mRNA: AUG. Translation ends at a termination codon in the mRNA: UAA, UAG, or UGA. Consider the following small gene sequence: 5'-TAAGCTACGTGCTGGCGAATTAATTTATTTCGTTATAGTCGTACATCTAT-3' 3 '-ATTCGATGCACGACCGCTTAATTAAATAAAGCAATATCAGCATGTAGATA-5' (A) Which strand is transcribed into mRNA? Do a virtual transcription of each strand, being careful about polarity. b ottom DNA strand: 3'-ATTCGATGCACGACCGCTTAATTAAATAAAGCAATATCAGCATGTAGATA-5' resulting transcript: 5'-UAAGCUACGUGCUGGCGAAUUAAUUUAUUUCGUUAUAGUCGUAGAUCUAU-3' top DNA strand, flipped so it reads 3' to 5' left to right: 3'-TATCTACATGCTGATATTGCTTTATTTAATTAAGCGGTCGTGCATCGAAT-5' r esulting transcript: 5'-AUAG AUG UACGACUAUAACGAAAUAAAU UAA UUCGCCAGCACGUAGCUUA-3' T hen look for which potential transcript has an open reading frame a start and stop codon in the same frame. The DNA strand that you transcribed to get that mRNA sequence is the one that can be transcribed. The transcript of the top strand has an AUG and a UUA in frame, while the transcript of the bottom strand has two stop codons but no AUG at all. Thus, the top strand is the one that will be transcribed into mRNA. (B) What is the mRNA sequence? Write out just the open reading frame, from the start codon to the first stop codon. 5 ' - AUGUACGACUAUAACGAAAUAAAUUAA - 3' ( C) What is the translated protein, using the single letter amino acid abbreviations? Use the codon table from an old exam or the lecture notes: A UG UAC GAC UAU AAC GAA AUA AAU UAA NH 2 - M YDYNEIN - C OOH Q 32. C onsider the following situation. Suppose there is a mutant cell in which there is a mutation in the gene marked by the * in the above sequence so that the T in the top strand is changed to a G. 5'-TAAGCTACGTGCTGGCGAATTAATTTATTTCGTTATAGTCGTACATCTAT-3' 3'-ATTCGATGCACGACCGCTTAATTAAATAAAGCAATATCAGCATGTAGATA-5' * (A) What is the new anticodon sequence for the affected codon? Be sure to label (note) the polarity of the anticodon sequence. M utated DNA: 5'-TAAGCTACGTGCTGGCGAATGAATTTATTTCGTTATAGTCGTACATCTAT-3' 3'-ATTCGATGCACGACCGCTTACTTAAATAAAGCAATATCAGCATGTAGATA-5' previous mRNA transcript, copied from the original top strand (done in previous problem): 5'-AUAG AUG UACGACUAUAACGAAAUAAAU UAA UUCGCCAGCACGUAGCUUA-3' n ew mRNA transcript, copied from the mutated top strand: 5'-AUAG AUG UACGACUAUAACGAAAUAAAU UCA UUCGCCAGCACG UAG CUUA-3' (!) At the point affected by the mutation, the original codon was UAA (a stop codon), so the perfect-fit anticodon would have been 3'-AUU-5' BUT a stop codon has NO anticodon, because it has no tRNA-aa that matches it. However the new, mutated codon is UCA, which encodes serine and thus has an anticodon: [ 3'-AGU-5'] (B) With this mutation, what is the resulting protein sequence, using the single letter amino acid abbreviations? T he mutation has replaced a stop codon with a codon encoding serine. So the amino acid sequence of the protein becomes: NH 2 - M YDYNEINSFAST - C OOH It does terminate, but only because there is another stop codon (which was silent, so to speak, in the original mRNA) just 5 codons downstream. Q33. S uppose you have cloned and sequenced a piece of double-stranded DNA: 5'-CTATGTCGCCTCCCTCCCTTCTCGGGCCGACAAGAGTTCGCAATACAT-3' 3 '-GATACAGCGGAGGGAGGGAAGAGCCCGGCTGTTCTCAAGCGTTATGTA-5' Assuming that this is the entire gene, and that your sequence ends precisely at start and stop codons, which strand is the one that will be transcribed into mRNA? Again, do a virtual transcription of each strand: bottom DNA strand: 3'-GATACAGCGGAGGGAGGGAAGAGCCCGGCTGTTCTCAAGCGTTATGTA-5' resulting RNA transcript: 5'-CUAUGUCGCCUCCCUCCCUUCUCGGGCCGACAAGAGUUCGCAAUACAU-3' top DNA strand, reversed so 3' to 5' is left to right: 3'-TACATAACGCTTGAGAACAGCCGGGCTCTTCCCTCCCTCCGCTGTATC-5' resulting RNA transcript: 5'- AUG UAUUGCGAACUCUUGUCGGCCCGAGAAGGGAGGGAGGCGACA UAG -3' N ext, find which potential transcript has an open reading frame. The virtual transcript of the bottom strand has a potential start codon, but it begins at base 3, not at the beginning of the sequence, and it has no stop codon at the end in fact it has no stop codons at all. So here, the top strand w ill be transcribed, since its transcript begins with a start, ends with a stop, and has them in frame with each other. (B) What will the mRNA sequence be (remember to label the 5' and 3' ends)? 5' A UGUAUUGCGAACUCUUGUCGGCCCGAGAAGGGAGGGAGGCGACAUAG 3 ' ( C) What is the sequence of the protein that this gene encodes? Amino acid sequence: NH 2 - M YCELLSAREGREAT - C OOH Q34. C onsider the following mRNA sequence: 5'-CAUUA AUG GACGCCAAUAAGUGGGCUAUUCUU UAG CGUAUGUG-3' (A) Which base number does the open reading frame begin with? The ORF begins with the 6 th b ase (and runs for 10 codons, including the stop) (B) What is the sequence of the open reading frame (from start through stop codon): mRNA: A UGGACGCCAAUAAGUGGGCUAUUCUUUAG ( C) What is the translated polypeptide sequence? AUG GAC GCC AAU AAG UGG GCU AUU CUU UAG => aa sequence: (N-term) M DANKWAIL (C-term) (D) Suppose there is a mutation in one of the aminoacylsynthases, the enzyme which charges a particular tRNA with the appropriate amino acid. This mutated enzyme takes the tRNA containing the anticodon 5'-GUC-3' and, instead of coupling it to aspartic acid, it couples it to cysteine. If the mRNA above were expressed in a cell containing this mutation in the aminoacylsynthase, would it change the sequence of the polypeptide? Yes! L ook for a codon that would bind the anticodon 3'-CUG-5' a perfect fit would be GAC. This is the second codon of your open reading frame: mRNA: AUG G AC G CC AAU AAG UGG GCU AUU CUU UAG So if the translation system plugs in the wrong amino acid at that codon, it will change the sequence of this protein. (E) Using the single letter code for amino acids, what is the polypeptide sequence in cell with the aminoacylsynthase mutation? In the aminoacylsynthase mutant, the second codon would translated as C instead of D, giving: M CANKWAIL Q35. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: 40S subunit polysome ribosome 60S subunit an aminoacyl tRNA [4] [1] [2] [3] [ 5] A polysome (1) is comprised of a number of ribosomes (2) all translating along an mRNA template. Each ribosome contains a 60S subunit (3) and a 40S subunit (4). An aminoacyl tRNA (5) will fit within the A site formed by the 40S and 60S subunits. Remember that for illustrative purposes, tRNA is often represented in cartoons of translation at an exaggerated size relative to ribosomes and mRNA. Q36. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: tRNA [3] ribosome [2] stop codon [4] endoplasmic reticulum [ 1] nucleotide [5] A stop codon (4) contains just 3 nucleotide (5) residues, and is therefore small relative to an entire tRNA (3), which is approx 90 nucleotide residues in length (see fig 7-28 in ECB). D uring translation, an aminoacyl tRNA molecule fits into a single site within the larger ribosome (2); if the translated protein contains a start transfer sequence, the ribosome will dock on the hulking surface of the ER (1), which as a major organelle in the cell is much larger than any of these macromolecules. Q37. In the extremities of some arctic animals and cold water fowl (e.g., reindeer paws, duck feet) tissues near the surface are maintained at temperatures significantly below that of the body core. Of course this means that the cells in those tissues must function normally at lower temperatures than the cells of the same tissues elsewhere in the animal. An obvious problem for a cell functioning at low temperature is maintaining the fluidity of its membranes. Which changes in the composition of membranes would allow them to maintain fluidity (keep them from solidifying) at low temperatures? ***(A) Decrease the length of the fatty acid chains and increase the % of unsaturated C-C bonds (B) Decrease the length of the fatty acid chains and decrease the % of unsaturated C-C bonds ( C) Increase the length of the fatty acid chains and decrease the % of unsaturated C-C bonds (D) Increase the length of the fatty acid chains and increase the % of unsaturated C-C bonds C hanges in the phospholipids: The acyl chain length could be shorter and there could be more unsaturated bonds in the acyl chains Each of these would decrease the temperature at which the membranes solidify, thus allowing them to be fluid at a lower temperature. A shorter acyl chain does this by providing fewer opportunities for hydrophobic interactions between chains. Increased numbers of cis-unsaturated bonds cause more kinks in the acyl chains, making it less likely that adjacent acyl chains can have interactions along their entire length. Q 38. C onsider the following cell: it has a plasma membrane with surface area of 5000 : m2 ; there are 106 phospholipid molecules per : m2 in each leaflet of the membrane; there are 2000 molecules of flippase enzyme associated with the plasma membrane; each flippase enzyme molecule can catalyze 100 flips per second (well ignore enzyme specificity for the moment). [ note: the numerical input values are generated randomly for each student] (A) What is the maximum rate of inappropriate spontaneous phospholipid flips, in flips/sec for the whole plasma membrane, that can be reversed by the action of the flippases? T he flippases can catalyze: 2 000 molecules x (100 flips /( sec @ m olecule)) = 2 x 1 0 5 flips / sec = 7 .2 x 10 8 flips / hr (B) What fraction of the total phospholipids in the plasma membrane can be flipped per hour? T he cell has: (106 / : m2 ) x (5000 : m2 / leaflet) x 2 leaflets = 1010 phospholipid molecules total So, (7.2 x 108 / 10 10 ) = 0.072 o r 7 .2% o f the total P-lipids that can be flipped each hr ( C ) If each phospholipid in the plasma membrane flips spontaneously on average once every 105 seconds, are the 1000 molecules of flippase adequate to reverse those flips and keep the asymmetry of the lipid bilayer intact? If each phospholipid flips spontaneously every 105 sec, then the number of flips in the whole plasma membrane each sec = (10-5 flips / (sec @ p hospholipid molecule)) x (1010 phospholipid molecules) = 10 5 flips / sec . S ince the flippases can reverse 2 x 10 5 flips / sec twice this number YES, they ARE adequate to maintain the phospholipid asymmetry of the plasma membrane. Q 39. Y ou are investigating the lateral mobility of integral membrane proteins in the plasma membrane of Bloggs cells. You have at your disposal specific antibodies against the extracellular domain of 2 different membrane proteins, Bloggs1 and Bloggs2. You first try a FRAP (fluorescence recovery after photobleaching) experiment. You covalently attach a green fluorescent label onto an antibody against Bloggs1 (anti-Bloggs1) and a red fluorescent label onto anti-Bloggs2. You add these to Bloggs cells growing in culture, and you can see both green fluorescence and red fluorescence distributed over the whole surface of each cell. You then bleach both the red and green fluorescence in one spot using a tightly focused laser beam and follow the recovery of each color of fluorescence in that spot over time. You see a pattern like this in cell after cell: (A) Based on this experiment, which protein, Bloggs1 or Bloggs2, diffuses more quickly in the plane of the bilayer? B loggs 1 . The recovery of fluorescence in a FRAP expt like this is due to diffusion into the bleached region of unbleached molecules. So the rate of fluorescence recovery is proportional to the rate of diffusion the faster the recovery, the more rapid the diffusion. Y ou then do a second experiment: you covalently attach each of your antibodies separately to gold beads 15 nm in diameter and then apply each kind of antibody-bead individually to Bloggs cells in culture. Using video-enhanced light microscopy you can follow the movement of single membrane proteins by tracking the movement of individual gold beads. To keep your observations honest, Ivor, your lab assistant, has encoded the tubes containing the antibody-coupled beads by labeling them A and B instead of Bloggs1and Bloggs2. Only he knows which is which. You follow the movements of many A and many B beads on many cells and plot them as line segments each representing the distance and direction the bead diffused in 1 sec on the cell surface. You find that the following patterns of movement are typical: Although Ivor will not be in until Monday to tell you which beads have anti-Bloggs1 and which have anti-Bloggs2, you have a pretty good idea already. Which preparation, A or B, contains beads coupled to the Bloggs1 antibody? Sample B . Here you are measuring the diffusion rate directly. The membrane protein that bead B is bound to is clearly moving much further each sec than the one that bead A is bound to. Q 40. T he region(s) of a membrane protein that span the membrane have specific primary and secondary structure properties. They are generally alpha helical, and they have a high percentage of hydrophobic amino acid residues and few polar residues. Examine the diagram here [next page] of the amino acid sequence of the single-span protein nerve growth factor, or NGF. This is a simple primary structure (amino acid sequence), with the residues numbered N-terminus to C-terminus, from 1 to 796. The numbering system is simple: the zero in 10" sits directly above the single-letter abbreviation for the 10th residue, the zero in 20" sits above the 20th residue, etc. There are three different 20-residue sequences highlighted (in red, green and blue) that are possible membrane spans. Which do you think is most likely to be the correct one? Blue Hint: look for the sequence with the fewest polar residues NGF sequence (numbers indicate the residue number, starting with the Nterminus) 10 20 30 40 50 60 MLRGGRRGQL GWHSWAAGPG SLLAWLILAS AGAAPCPDAC CPHGSSGLRC TRDGALDSLH 70 80 90 100 110 120 HLPGAENLTE LYIENQQHLQ HLELRDLRGL GELRNLTIVK SGLRFVAPDA FHFTPRLSRL 130 140 150 160 170 180 NLSFNALESL SWKTVQGLSL QELVLSGNPL HCSCALRWLQ RWEEEGLGGV PEQKLQCHGQ 190 200 210 220 230 240 GPLAHMPNAS CGVPTLKVQV PNASVDVGDD VLLRCQVEGR GLEQAGWILT ELEQSATVMK 250 260 270 280 290 300 SGGLPSLGLT LANVTSDLNR KNVTCWAEND VGRAEVSVQV NVSFPASVQL HTAVEMHHWC 310 320 330 340 350 360 IPFSVDGQPA PSLRWLFNGS VLNETSFIFT EFLEPAANET VRHGCLRLNQ PTHVNNGNYT 370 380 390 400 410 420 LLAANPFGQA SASIMAAFMD NPFEFNPEDP IPVSFSPVDT NSTSGDPVEK KDETPFGVSV 430 440 450 460 470 480 AVGLAVFACL FLSTLLLVLN KCGRRNKFGI NRPAVLAPED GLAMSLHFMT LGGSSLSPTE 490 500 510 520 530 540 GKGSGLQGHI IENPQYFSDA CVHHIKRRDI VLKWELGEGA FGKVFLAECH NLLPEQDKML 550 560 570 580 590 600 VAVKALKEAS ESARQDFQRE AELLTMLQHQ HIVRFFGVCT EGRPLLMVFE YMRHGDLNRF 610 620 630 640 650 660 LRSHGPDAKL LAGGEDVAPG PLGLGQLLAV ASQVAAGMVY LAGLHFVHRD LATRNCLVGQ 670 680 690 700 710 720 GLVVKIGDFG MSRDIYSTDY YRVGGRTMLP IRWMPPESIL YRKFTTESDV WSFGVVLWEI 730 740 750 760 770 780 FTYGKQPWYQ LSNTEAIDCI TQGRELERPR ACPPEVYAIM RGCWQREPQQ RHSIKDVHAR 790 LQALAQAPPV YLDVLG Q 41. T hese two diagrams [next page] show two different ways of looking at a membrane protein called the beta-adrenergic receptor, which spans the membrane 7 times. This receptor, which we will talk more about soon, is on the surface of your muscle cells, where it converts the fight or flight signal of adrenaline in your blood stream into enhanced muscle metabolism to power your frantic escape from saber-toothed tigers, angry roommates, etc. The first diagram is a simple primary structure (amino acid sequence), with the residues numbered N-terminus to C-terminus, from 1 to 413, and the membranespanning regions in red. The second is a 2-D representation of how the beta-adrenergic receptor is actually organized in the plasma membrane of your muscle cells, with the amino acid residues labeled at the boundaries between membrane spans and cytoplasmic or extracellular portions. Imagine that you have a panel of different monoclonal antibodies, each specifically binding a small stretch of amino acids. You want to add one or more of these antibodies to live muscle cells in a culture dish in order to bind to the extracellular parts of the receptor so that you can visualize them by immunofluorescence microscopy. Which of the following antibodies would work for you in this experiment? (A) antibody that binds SLIVLAIVFG [ no, membrane span] (B) antibody that binds FERLQTVTNY [ no, cytoplasmic domain] ( C) antibody that binds YQSLLTKNKA [ no, cytoplasmic] (D) antibody that binds AINCYANETC [yes, extracellular] (E) antibody that binds ALKTLGIIMG [ no, cytoplasmic + membrane span] (F) antibody that binds SQGRNCSTND [ no, cytoplasmic] Beta-2 adrenergic receptor, Homo sapiens (i) primary sequence 10 20 30 40 50 60 MGQPGNGSAF LLAPNRSHAP DHDVTQQRDE VWVVGMGIVM SLIVLAIVFG NVLVITAIAK 70 80 90 100 110 120 FERLQTVTNY FITSLACADL VMGLAVVPFG AAHILMKMWT FGNFWCEFWT SIDVLCVTAS 130 140 150 160 170 180 IETLCVIAVD RYFAITSPFK YQSLLTKNKA RVIILMVWIV SGLTSFLPIQ MHWYRATHQE 190 200 210 220 230 240 AINCYANETC CDFFTNQAYA IASSIVSFYV PLVIMVFVYS RVFQEAKRQL QKIDKSEGRF 250 260 270 280 290 300 HVQNLSQVEQ SSKFCLKEHK DGRTGHGLRR ALKTLGIIMG TFTLCWLPFF IVNIVHVIQD 310 320 330 340 350 360 NLIRKEVYIL LNWIGYVNSG FNPLIYCRSP DFRIAFQELL CLRRSSLKAY GNGYSSNGNT 370 380 390 400 410 GEQSGYHVEQ EKENKLLCED LPGTEDFVGH QGTVPSDNID SQGRNCSTND SLL (ii) 2-D arrangement in membrane Q 42. T hese two diagrams [ following pages] again show two different ways of looking at a membrane protein, this time an important signaling protein, adenylate cyclase, or AC. We will talk about AC again soon when we consider how our cells amplify and respond to signals from the outside world. The first diagram is a simple primary structure (amino acid sequence), with the residues numbered N-terminus to C-terminus, from 1 to 1168, and the membrane-spanning regions in red. The numbering system is simple: the zero in 10" sits directly above the single-letter abbreviation for the 10th residue, the zero in 20" sits above the 20th residue, etc. This diagram also colors the three major intracellular portions of the protein in green, and the largest extracellular protein in blue. The second diagram is a simple 2-D representation of how AC is actually organized in the plasma membrane of your cells, with the N- and C-termini labeled. Consider a series of experiments in which you probe the structure of AC by digesting away portions of the protein and looking on SDS polyacrylamide gels to see how many fragments you get. (A) First, what if you added a protease to the outside of the cells, such that any portion of AC exposed on the extracellular side of the membrane was digested away, but any portion in the membrane or cytoplasm was protected. How many fragments of AC would result? [7 fragments - count em] (B) Next, what if a protease in the cytoplasm digested all portions of AC exposed to the cytoplasm, but any portion in the membrane or extracellular side was protected how many fragments would this divide AC into? [6 fragments] ( C ) Finally, what if an unusual protease within the membrane made a single cut midway in each membrane-spanning region how many fragments of AC would this produce? [13 fragments] Adenylate cyclase , type 6 (Homo sapiens) Name: ADCY6 10 20 30 40 50 60 MSWFSGLLVP KVDERKTAWG ERNGQKRSRR RGTRAGGFCT PRYMSCLRDA EPPSPTPAGP 70 80 90 100 110 120 PRCPWQDDAF IRRGGPGKGK ELGLRAVALG FEDTEVTTTA GGTAEVAPDA VPRSGRSCWR 130 140 150 160 170 180 RLVQVFQSKQ FRSAKLERLY QRYFFQMNQS SLTLLMAVLV LLTAVLLAFH AAPARPQPAY 190 200 210 220 230 240 VALLACAAAL FVGLMVVCNR HSFRQDSMWV VSYVVLGILA AVQVGGALAA DPRSPSAGLW 250 260 270 280 290 300 CPVFFVYIAY TLLPIRMRAA VLSGLGLSTL HLILAWQLNR GDAFLWKQLG ANVLLFLCTN 310 320 330 340 350 360 VIGICTHYPA EVSQRQAFQE TRGYIQARLH LQHENRQQER LLLSVLPQHV AMEMKEDINT 370 380 390 400 410 420 KKEDMMFHKI YIQKHDNVSI LFADIEGFTS LASQCTAQEL VMTLNELFAR FDKLAAENHC 430 440 450 460 470 480 LRIKILGDCY YCVSGLPEAR ADHAHCCVEM GVDMIEAISL VREVTGVNVN MRVGIHSGRV 490 500 510 520 530 540 HCGVLGLRKW QFDVWSNDVT LANHMEAGGR AGRIHITRAT LQYLNGDYEV EPGRGGERNA 550 560 570 580 590 600 YLKEQHIETF LILGASQKRK EEKAMLAKLQ RTRANSMEGL MPRWVPDRAF SRTKDSKAFR 610 620 630 640 650 660 QMGIDDSSKD NRGTQDALNP EDEVDEFLSR AIDARSIDQL RKDHVRRFLL TFQREDLEKK 670 680 690 700 710 720 YSRKVDPRFG AYVACALLVF CFICFIQLLI FPHSTLMLGI YASIFLLLLI TVLICAVYSC 730 740 750 760 770 780 GSLFPKALQR LSRSIVRSRA HSTAVGIFSV LLVFTSAIAN MFTCNHTPIR SCAARMLNLT 790 800 810 820 830 840 PADITACHLQ QLNYSLGLDA PLCEGTMPTC SFPEYFIGNM LLSLLASSVF LHISSIGKLA 850 860 870 880 890 900 MIFVLGLIYL VLLLLGPPAT IFDNYDLLLG VHGLASSNET FDGLDCPAAG RVALKYMTPV 910 920 930 940 950 960 ILLVFALALY LHAQQVESTA RLDFLWKLQA TGEKEEMEEL QAYNRRLLHN ILPKDVAAHF 970 980 990 1000 1010 1020 LARERRNDEL YYQSCECVAV MFASIANFSE FYVELEANNE GVECLRLLNE IIADFDEIIS 1030 1040 1050 1060 1070 1080 EERFRQLEKI KTIGSTYMAA SGLNASTYDQ VGRSHITALA DYAMRLMEQM KHINEHSFNN 1090 1100 1110 1120 1130 1140 FQMKIGLNMG PVVAGVIGAR KPQYDIWGNT VNVSSRMDST GVPDRIQVTT DLYQVLAAKG 1150 1160 YQLECRGVVK VKGKGEMTTY FLNGGPSS green = major cytoplasmic domains; blue = major extracellular domain; red = 12 transmembrane domains Q 43. Number each of the following in order of its numerical value, with 1 being the largest. Very briefly, explain your reasoning: [3] the number of solute molecules moved by one cycle of a Na + /glucose symporter [4] the number of membrane-spanning domains in a receptor tyrosine kinase (like the NGF receptor) [1] the number of membrane-spanning domains in a Na+ channel [2] the number of ions moved across the membrane in a single cycle of a Na + /K + ATPase [5] the amount of net charge across the membrane generated by one cycle of a Na + /K + ATPase The numbers are, respectively: s ymporter = 3 moved (2 Na+ plus 1 glucose; would be 2 or 4 with other stoichiometries) NGFR = 1 membrane span Na channel = 24 membrane spans Na/K ATPase = 5 moved (3Na + plus 2K + ) net charge of one cycle of Na/K ATPase = ! 1 Q 44. W e have studied 4 examples of integral membrane proteins, including one composed of a single polypeptide (the NGF receptor) and 3 that have several subunits. For these 4 integral membrane proteins shown below, match them with the function(s) at right: Membrane protein: Function: (1) NGF receptor [A,D] (A) interacts with species form outside the cell (2) ligand-gated Na channel [A,C,E] (3) Na/glucose symporter [A,B,C,E] (B) transports species across the membrane AGAINST their concentration gradient (4) Na/K ATPase [A,B,D,E] ( C) transports species across the membrane DOWN their concentration gradient (D) h as enzyme activity (E) c an affect he magnitude of the Na gradient across the membrane Q 45. Number each of the following in order of its numerical value in mV, with 1 being the greatest. Use as your reference the squid axon that we looked at quantitatively in lecture. [ 4] V m of neuron prior to stimulation of depolarization [1] V m o f neuron in a region X of the axonal membrane where a depolarization has been stimulated, at the time when Na+ permeability is at its peak [2] V m o f neuron in a region X of the axonal membrane where a depolarization has been stimulated, at the time when K + permeability is rising and Na+ permeability is falling [5] V m o f neuron in a region X of the axonal membrane where a depolarization has occurred and the relevant Na+ and K + channels have closed again, but the Cl! leak channels have not yet restored the resting V m. [3] V m o f neuron in axonal region Y that is adjacent to and downstream from X, at the time after its membrane potential has begun to respond to the depolarization at X, but before a depolarization at Y has been triggered At each of these points in the action potential, the values for V m (taken from the widely used experimental system, squid giant axon, that we discussed in lecture) are approximately: prior to depol = ! 70 mV at Na permeability peak = +40-50 mV where K permeability is rising and Na permeability is falling = 0 mV after depol and repol, before Cl! leak channels restore = ! 80 mV adjacent region thats starting to respond but hasnt depold = ! 50 mV >x> ! 70 mV Q 46. If the following four cells had the SAME volume, which would have the SMALLEST surface area? [What is your reasoning?] A cylindrical muscle cell in your foot, with length 1000 times greater than its diameter. A n euron innervating your shoulder, with an axon 10,000 times longer than its diameter A cubiodal epithelial cell in your kidney A spherical adipocyte (fat cell) under your skin. [ correct] The adipocyte takes the prize: A sphere surrounds a given volume using the minimum possible surface area. W hich do you think would have the greatest surface area? Q47. W hich of the following best describes the role of the sodium-potassium ATPase in generating an action potential in neurons? (A) The Na/K ATPase propagates the action potential along the axon. (B) The Na/K ATPase determines the exact resting membrane potential of the neuron. ( C) The Na/K ATPase is responsible for the outward flow of K ion during the action potential. (D) The Na/K ATPase is responsible for the inward flow of Na ion during the action potential. (E) The Na/K ATPase plays no direct role in generating the action potential, but is largely responsible for establishing the initial conditions in which sodium and potassium are out of equilibrium. [ correct] The Na/K ATPase plays no direct role in generating the action potential. However, the Na/K ATPase is largely responsible for establishing the initial conditions in which sodium and potassium are out of equilibrium, and, along with potassium leak channels, helps establish and maintain the resting membrane potential. Q 48. H ere's one experimental approach to finding out how rapidly a membrane transporter can move its solute into the cell: combine biochemical information about how much solute accumulates in the cell over time with structural information about how many transporters are on the surface of the cell. You are studying a spherical eukaryotic cell that has a diameter of 20 m and is engaged in glycolysis. Its energy source is glucose, which it is taking up from its environment via Na/glucose symporters distributed throughout the surface of its plasma membrane. By briefly adding an inhibitor of all glucose breakdown, you are able to determine that glucose accumulates in the cell at the rate of 50 M/hr. (A) Based on this number, how many moles of glucose must be entering the cell per hour? (Assume that one-half of the volume of the cell is composed of organelles, and thus not available for diffusion of glucose) Glucose accumulates at 50 M per hr = 50 mol /( hr @ L ). B ut the cell is much smaller than 1 Liter in volume. The volume of the cytosol here is half the volume of a sphere with radius 10 m, so: 4/3 r3 = 4/3 3.1416 (10 m)3 = 4189 m 3 , [ (1L / 10 15 m 3 )] = 4.18910 ! 12 L a nd half of that = 2 .09410 ! 12 L . S o glucose enters the cytoplasm at a rate of: (2.09410! 12 L) (50 umol / hr*L) = 1.04710 -10 mol / hr (B) What is the flux of glucose into the cells (expressed in moles /sec @cm 2 ) ? Flux is the rate of movement of glucose across the surface area of the cell membrane. You know the rate at which glucose is entering, and you can calculate the surface area of the spherical cell. S urface = 4 r2 = 1257 m2 = [ (1 cm/104 m)2 ] = 1.25710 ! 5 cm 2 . T he amount of glucose accumulating / sec = 1 .04710! 10 mol / hr (1hr / 3600 s) = 2 .90810! 14 mol / s = [ 1 mol / 10 6 mol] = 2.90810 -! 20 mol / s So the flux of glucose is: = (2.90810-20 mol / s) / (1.25710! 5 cm 2 ) = 2 .31310 -15 mol / s @ c m 2 ( C ) If electron microscopy indicates that there is on average 1 Na/glucose symporter per m2 of the plasma membrane, at what rate must each symporter transport glucose, in molecules per second? First convert the flux to moles per sec per m 2 : (2.31310 ! 15 mol / s @ c m 2 ) (1 cm 2 / 10 8 m 2 ) = 2.31310 ! 23 mol / s @ m 2 Next, multiply this by Avogadros number to get the molecules per sec per m 2 : (2.31310! 23 mol / s @ m 2 ) (6.0221023 molecules / mole) = 13.93 molecules / s @ m 2 Since there is 1 symporter / m 2 , it must transport at a rate of 13.93 molecules of glucose / sec. Q 49. A s ea urchin egg is a sphere with a diameter of 100 M. The concentrations of the major cations, K+ , Na+ , and Ca2+ are shown in the table below. [randomized for each student] Assuming that these ions determine the membrane potential, what are the upper and lower limits on the possible membrane potential? cation intracellular concentration potassium sodium calcium 200 mM 20 mM 100 nM extracellular concentration 10 mM 450 mM 10 mM Use the Nernst equation to find out what the equilibrium potential for each ion is: V X = (59 mV/z) @ log([X] out/[X] in) V K = 59mV/1 log (10mM/200mM) = ! 76.8 mV V Na = 59mV/1 log (450mM/20mM) = +79.8 mV V Ca = 59mV/2 log (10mM/0.0001mM) = +147.5 mV The upper and lower limits for V m of the cell are the highest and lowest equilibrium potentials, or + 147.5 mV and ! 76.8 mV. Q 50. S ome marine invertebrates, such as squid, have extracellular fluids that resemble sea water and therefore have much higher intracellular ion concentrations than mammals. For a squid neuron, the ion concentrations are: ion intracellular concentration extracellular concentration K+ Na+ Cl! Ca 2+ pH 410 mM 40 mM 100 mM 210 ! 4 mM 7.6 15 mM 440 mM 560 mM 10 mM 8.0 If the resting membrane potential, V m, is ! 70 mV, are any of the ions in equilibrium? How far out of equilibrium, in mV, is each ion? What will be the direction of net movement of each ion through an open channel for that ion? Use Nernst equation, V x = (59mV/z)log([X] out/[X] in), to calculate equilibrium potential for each ion and compare to the membrane potential, ! 70 mV. Ask, which direction must this ion flow to get from the V M to the ions equilibrium potential, V X ? For K + : V K = 59mV @ log(15mM/410mM) = ! 84.7 mV (will leave c ell to reach equil) For Na + : V Na = 59mV @ log(440mM/40mM) = + 61.4 mV (will e nter c ell) For Cl! : V Cl = (59mV/! 1)@ log(560mM/100mM) = ! 44.1 mV (will leave c ell) For Ca + + : V Ca = (59mV/2)@ log(10mM/0.0001mM) = + 147.5 mV (will e nter c ell) For H + : V H = 59mV @ (! 8 ! (! 7.6)) = ! 23.6 mV (will e nter c ell) Q 51. Intestinal epithelial cells have a Na+-glucose symporter in their plasma membrane on the apical side. The function of this protein is to use the energy contained in the electrochemical gradient of sodium to drive glucose against its concentration gradient from the outside into the cytoplasm. Suppose that Vm = ! 70 mV, [Na+ ]in = 10 mM, [Na + ]out = 100 mM and [glucose]out = 10 M. What intracellular concentration of glucose can be generated by the symporter under these conditions? Assume in this case that the stoichiometry of the symporter is 1 mole glucose to 2 mole Na+ . Note: this is a G problem. The free energy difference in glucose across the membrane cannot exceed the oppositely directed free energy difference for sodium. If you use the Faraday constant (F), assume it is equal to 100 kJ/V*mol. First, determine the G of the Na+ gradient. We will need to use the relationship G Na = zF(V m ! V Na ), so first we will determine V Na using the Nernst equation: V Na = 59 mV @ log ([Na+ ] out / [Na+ ] in) = 59 mV @ log(10) = 59 mV Then, given that V m = -70 mV and, for Na + , z = 1, we solve for G Na: G Na = F(! 70mV ! 59mV) = zF @ ! 129 mV = (102 kJ/V @ m ol)(! 0.129 V) = ! 12.9 kJ/mol If all of the energy in the Na+ gradient can be used to support glucose transport, then: G glucose = ! 2 G Na So, G gluc = ! 2 @ (! 12.9 kJ/mol) = RTln([gluc] in / [gluc] out) 2 5.8 kJ/mol = RTln([gluc] in / [gluc] out) 2 5.8 kJ/mol = 5.7 kJ/mol@ log([gluc] in / [gluc] out) log ([gluc] in / [gluc] out) = 4.53 ([gluc] in / [gluc] out) = 3.3610 4 Thus, if [gluc] out = 10 M (0.01 mM), then [gluc] in = 0.01 mM 3.3610 4 = 336 mM Q 52. T he membrane potential of a cell is determined by the relative permeability of the membrane to various ions. When acetylcholine binds to its receptors on the muscle endplate, it causes a massive opening of channels, giving equal membrane permeability to sodium and potassium. Under these conditions, V m = (V K + V Na )/2 If [K+ ]in = 140 mM and [Na+ ] in = 10 mM for the muscle cell, and [Na + ] out = 150 mM and [K+ ] out = 5 mM, [input values randomized] what will be the membrane potential of the endplate region of an acetylcholine-stimulated muscle cell? Use the Nernst equation to calculate the individual equilibrium potentials: for potassium: V K = 59mV @ log(5mM/140mM) = ! 85.4 mV for sodium: V Na = 59mV @ log(150mM/10mM) = + 69.4 mV If these are the only two ions, and if each contributes equally, then the V m = average of the two, or ! 8.0 mV. Q 53 . Consider the following cell: it has a transmembrane potential of 0 mV and has the following intracellular and extracellular ion concentrations: [Na + ] in = 20 mM; [Na+] out = 200 mM; [K+ ] in = 150 mM; [K + ] out = 50 mM. Number each of the following in order of its absolute value, with 1 being the largest. Very briefly, explain your reasoning: [3] p otential energy stored in the K + gradient [2] p otential energy stored in the Na + gradient [1] e nergy of the ATP used by the Na+ /K + ATPase to generate the Na + and K + gradients [4] p otential energy of the K + gradient after holes are made in the plasma membrane The potential energy stored in the Na+ gradient (2) should be greater than that in the K + gradient (3) because the V m is zero and the concentration gradient ([out]/[in]) is 10 for Na+ vs 0.33 for K+ . No enzyme certainly not the Na+ /K + ATPase is perfectly efficient, so the amount of energy from ATP used to generate the Na + and K + concentration gradients (1) will always be greater than the potential energy stored in those gradients. S ince it is greater than the sum of the two gradients, it is certainly greater than either of them taken separately. Finally, the existence of a concentration gradient (and therefore the potential energy stored in it) depends entirely on the integrity of the membrane; the potential energy if you poke holes in the membrane (4) will be zero. Q 54. Y ou are studying a spherical eukaryotic cell that has a diameter of 20 m and is engaged in glycolysis. Its energy source is glucose, which it is taking up from its environment via Na/glucose symporters distributed throughout the surface of its plasma membrane. By briefly adding an inhibitor of all glucose breakdown, you are able to determine that glucose accumulates in the cell at the rate of 50 M/hr. The cell normally maintains an internal [Na + ] of 50 mM. (A) If the cell makes no other adjustments, how long will the symporter be able to function at the current glucose uptake rate before the intracellular [Na+ ] has increased by 1% [this value randomized for each student, between 1 and 5%]? The cell is taking up glucose at a rate of 50 M/hr, so since 2 Na+ ions enter for each glucose molecule, Na+ must be entering at 100 M/hr. Normal intracellular [Na + ] is 50mM, so the symporters would have to run for 5 h ours w ithout any other adjustment in order to change the intracellular [Na + ] just 1%. (B) B y what mechanism could the cell run its Na+ /glucose symporters all of the time WITHOUT increasing the [Na+ ] in the cell? It could also operate the Na/K ATPase constantly, maintaining the Na disequilibrium across the membrane. [correct] It could run half of its Na/glucose symporters forward, and half of them in the reverse direction, moving glucose and Na both out of the cell. It could also open Na channels periodically to let Na leak out of the cell. It could also open Cl channels, allowing Cl ion out of the cell, which would be followed by Na ion. ( C) Which of these is the most fundamental source of the energy used by the Na/glucose symporter to take up glucose? The energy of the K disequilibrium across the membrane The energy of the Na disequilibrium across the membrane The energy of ATP hydrolysis that drives the Na/K ATPase [correct] The energy of the glucose disequilibrium across the membrane Q 55. A p rotonophore is a drug or toxin that allows protons to leak easily and rapidly across membranes, including the mitochondrial inner membrane. Protonophores have been very useful tools to mitochondrial biologists studying the mechanisms of electron transport and ATP synthesis. If you added a protonophore to mitochondria, how would each of the following processes be affected? ATP synthesis would increase or decrease? [decrease] The proton gradient across the inner membrane would become greater, smaller, or stay about the same? [smaller] p will become larger, smaller or stay about the same? [smaller] The proton permeability of the outer mitochondrial membrane will become larger, smaller, or stay about the same? [same] Q 56. If you profoundly restricted the supply of oxygen to the mitochondrion, how would each of the following processes be affected? (A) Electron transport complex IV would: become stuck in the reduced state, become stuck in the oxidized state, or stay in the same redox balance as before? [ reduced] (B) Electron transport complexes I, II and III would: become stuck in the reduced state, become stuck in the oxidized state, or stay in the same redox balance as before? [ reduced] ( C) Proton pumping from the matrix to the intermembrane space would: increase, decrease or stay about the same? [ decrease] (D) ATP synthesis would: increase, decrease or stay about the same? [ decrease] (E) pH of the matrix would: increase, decrease or stay about the same? [ decrease] Q57. If you were growing mammalian cells in culture using glucose as their sole carbon source, what would be the effect of profoundly reducing the glucose concentration on each of the following processes? (A) The rate of pyruvate production would: increase, decrease or stay about the same? [decrease] (B) The rate of lactate production would: increase, decrease or stay about the same? [decrease] ( C) The rate of CO 2 production by the cells mitochondria would: increase, decrease or stay about the same? [ decrease] (D) The rate of O2 consumption by the cells mitochondria would: increase, decrease or stay about the same? [ decrease] Q 58. S uppose you isolate mitochondria from cells and maintain them in a buffer solution that mimics the cytoplasm. If you are maintaining them in a tube or chamber that you have deprived of O 2 , theyll stop making ATP. What could you do to the composition of the buffer solution in order to get the mitochondria to resume production of ATP? (A) Increase the pyruvate concentration in order to stimulate the TCA cycle (B) Increase the glucose concentration in the solution ( C) Raise the pH of the solution, creating a proton gradient and p that will drive the ATP synthase *** (D) Lower the pH of the solution, creating a proton gradient and p that will drive the ATP synthase Electron transport consumes O 2 and pumps H + out of the matrix into the intermembrane space. Thus, depriving the mitochondria of O 2 lessens the [H + ] gradient across the IMM and lessens p. If you lower the pH of the bathing solution, this will produce a proton gradient across the inner membrane and ATP can be made. However, this is a temporary solution because after the pH gradient is dissipated, there will be no way to regenerate it, unless you acidify further the bathing solution. (What happens if you get carried away with this approach e.g., you lower the pH of the bathing solution to 3 or 2 or 1? Hint: dont put your finger in there!) Q 59. W hen cell biologists perform experiments on mitochondria in living cells, they employ a variety of drugs and poisons that inhibit specific mitochondrial functions. For example, the rat poison rotenone inhibits mitochondrial complex I, diminishing the mitochondrias capacity to use the electron transport chain to pump protons and generate a potential across the inner membrane. In order to completely eliminate the potential across the inner membrane, it is usually necessary not only to inhibit the electron transport chain, but also to inhibit the inner membrane ATP synthase. What is the most likely reason for this? (A) When electron transport alone is inhibited, the mitochondrial ATP synthase can run in reverse, consuming ATP, pumping protons out of the matrix, and maintaining at least a partial membrane potential. [ correct] (B) When the electron transport chain alone is inhibited, the inner mitochondrial membrane becomes leaky to H + ( C) When both electron transport and the ATP synthase are inhibited, the inner mitochondrial membrane becomes leaky to H + (D) When the mitochondrial ATP synthase is inhibited, electron transport can run in reverse, pumping protons into the matrix, and maintaining at least a partial membrane potential. Q60. S uppose that you are able to manipulate the membrane potential (V m) of a preparation of isolated mitochondria. You measure the pH of the mitochondrial matrix and find it to be 8.0. You measure the bathing solution and find its pH to be 7.0. You clamp the membrane potential at ! 59 mV that is, you force the matrix to be 59 mV positive with respect to the bathing medium. In the following calculations, if you use the Faraday constant (F), assume it is equal to 100 kJ/V mol. (A) What is the pH for these mitochondria? pH = pH IMS ! p H matrix = 7!8 = !1 (B) What is the protonmotive force ( p)? V m = ! 59mV , and pH = ! 1, so: p = V m ! 5 9 mV pH = ! 59mV ! (59mV)(! 1) =0 ( C) What is the G for protons? G = F @ p = 0 (D) Under these circumstances, can the mitochondria use the proton gradient to drive the synthesis of ATP? NO! T here is NO protonmotive force, so mitochondria cannot use the proton gradient to generate ATP. [note that all 3 input values are randomized, but scaled with each other so that p is always = 0] Q 61. C onsider a mitochondrion with a pH in the matrix = 8.4 and the pH of the intramembrane space = 6.5. The matrix is 100 mV more negative than the IMS or cytoplasm. If the ratio of ATP to ADP and P i is such that G for ATP hydrolysis = ! 60 kJ/mol, how many mols of H+ will have to move and in what direction in order to generate a mol of ATP? (A) What is the proton motive force ( p) across the mitochondrial membrane? pH = pH IMS ! p H matrix = (6.5 ! 8.4) = ! 1.9 p = V m ! (59mV)( pH) = (100mV) ! (59mV ! 1.9) = 2 12.1 mV (B) If the ratio of ATP to ADP and P i is such that G for ATP hydrolysis = ! 60 kJ/mol, how many mols of H+ will have to move through the ATP synthase in order to generate a mol of ATP? [The answer should be a whole number.] The energy available: G = (F)( p) = (102 kJ/V A m ole)(0.212V) = 2 1.2 kJ/mole protons (this is the energy available; note that for diffusion into the mitochondrion, which effectively means out of cytoplasm, G is effectively ! 21.2 kJ/mole) If it takes 60 kJ to make one mole of ATP, then the movement into the matrix of $ 3 moles of protons will be required to synthesize one mole of ATP . Q 62. A v ery active area of research at present is the attempt to reconstruct the history of early life on earth, including the evolution of the aerobic eucaryotic cell. This field makes use of a wide range of approaches that combine physical, geological, genetic and other biological data. Based on our consideration of these data, place the following events in temporal order, with 1" being the earliest and 4" being the most recent: (A) Strong selective pressure led to evolution of increased efficiency in bacterial electron transport chains, such that electron transport could pump out so much H + that the ATP-dependent proton pump was freed to run in reverse and s ynthesize A TP (respiration). [3] (B) Anaerobic procaryotes ruled the earth. They used abiotically-generated fermentable carbon compounds as an energy source and excreted organic acids. This made their environment acidic and required them to use membrane H + ATPases to maintain a neutral cytoplasmic pH. [ 1] ( C) An early anaerobic eucaryote engulfed a respiration-dependent bacterium by phagocytosis or endocytosis; the endosymbiont bacterium thus acquired an outer membrane similar to the plasma membrane while retaining an inner procaryotic membrane. This situation became stable, giving the cell its mitochondria. [ 4] (D) Strong selective pressure led to evolution of electron transport chains in anaerobic bacteria, simple systems that could exploit the differences in oxidative state between different molecules in the cell as a source of energy for H + pumping. This reduced the amount of ATP these cells had to expend just to pump out H + . [2] Q63. Suppose you are studying the insertion of a transmembrane protein into the lipid bilayer. There are "start" sequences (or signal sequences) which initiate the start of insertion into the ER membrane, and there are also "stop" sequences which halt the insertion into the ER. Consider an mRNA that has the following organization [randomized numbers for each student]: 5'--AUG----50 codons----start sequence----75 codons----stop sequence---25 codons---start sequence---100 codons---UAG--3' (A) If the start and stop sequences are each 20 amino acids in length, how long is the completed protein? 311 residues : T he complete protein is 1 + 50 + 20 + 75 + 20 + 25 + 20 + 100 = 311 aa residues long. (B) Draw the protein drawn as it would reside in the plasma membrane. Determine which labels correspond to the outside and inside of the cell, the N-terminus and the C-terminus of the protein, and the amino acid residue number at the ends of each membrane span: Q 64. F or each of the general membrane protein arrangements (A-E) shown below, choose the mRNA sequence that, when translated on the ER, is most likely to produce it. Each grey box indicates a stretch of 20-30 codons that are mainly hydrophobic; a box under an AUG indicates an N-terminal hydrophobic sequence. If an mRNA matches none of the proteins, choose none. [order randomized for each student] C B A none E D none Q 65. Y ou have isolated and sequenced an mRNA for a membrane protein and determined that it has six stretches of 20-30 codons that are hydrophobic enough to serve as membrane spans/signal sequences. You make the diagram below to illustrate your analysis of the sequence information: Of the membrane protein structures shown below, which is the most accurate representation of how the protein translated from this mRNA will be arranged in the plasma membrane? [D] N ow, in order to study the potential importance of your putative membrane-spanning domains, you make a deletion in the gene that eliminates the first 30 residues when the protein is expressed in cells. Which of the structures below most accurately represents the result of this? [ F] Finally, you make deletions that eliminate the last 2 signal sequences, leaving the rest of the sequence intact and in order. Which of the structures below most accurately represents the result of this? [ B] Q 66. C onsider the 5-span integral membrane protein shown below. The sides of the membrane, N- and C-termini, and residue numbers at the boundaries of the membrane are shown. One way to determine a structure like this would be to purify and translate its mRNA in vitro using a preparation of microsomes. You could then treat the microsomes by adding to the buffer solution a mixture of proteases that would digest away any part of the protein that was exposed on the solution side of the microsomal membrane. If you carried out that procedure and then analyzed the resulting pattern of polypeptide fragments on an SDS polyacrylamide gel, which of the patterns below would you expect to see? [ B] If you expressed the protein in cells, digested them using the same protease mixture, but from outside the intact cell, what pattern of fragments of this protein would you expect? [ D] Q 67. C onsider an integral membrane protein with a single membrane span. It is not glycosylated. Suppose you perform the following experiments and obtain the results shown in the SDS polacrylamide gels, drawn below. (i) You translate the protein in vitro in the absence of microsomes. You divide the protein into two portions, treat one portion by adding a mixture of proteases to the solution, and analyze the results on SDS-PAGE: (ii) Next, you translate the protein in vitro in the presence of microsomes. You divide the resulting protein/microsome solution into two portions, treat one portion by adding the same mixture of proteases to the solution, and again analyze the results: (iii) Finally, you study the protein as it actually resides in the plasma membrane of live cells. You treat some of the cells by adding the same mixture of proteases to the culture medium, recover the protein from the membrane, and again analyze the resulting fragments on SDS gels: T o analyze these results, assume that one amino acid has the molecular weight of 100 D. The numbers denote the residue numbers at the boundaries between membrane spans and cytoplasmic or extracellular portions of the chain. (i) Does this protein have a signal sequence removed during cotranslational insertion? [ No - c ompare the size of the protein translated with or without microsomes] (ii) Which of the following diagrams [order randomized for each student] most accurately depicts the arrangement of this protein in the plasma membrane? [ D] A B C D Q 6 8. Y ou have seen that the hydrophobic signal sequences in proteins can be either Nterminal, beginning with the first few residues of the protein, or internal, occurring farther along in the primary sequence. Lets define: X = number of membrane spans in the mature protein S = total number of signal sequences in the protein that emerge form the ribosome during translation I = number of internal signal sequences N = number of N-terminal signal sequences (always 0 or 1) Which of the following is an accurate general formula for how many membrane spans a mature integral membrane protein will have? [order randomized] (a) X = S (b) X = S - N [correct] ( c) X = I - N (d) X = S - I (e) X = 2 rS Q 69. C onsider a transmembrane protein that spans the membrane 3 times. Your goal is to determine the orientation of the protein in the plasma membrane. You have two antibodies, one specific for the N-terminus of the protein and the second specific the C-terminus. You obtain the following data: (1) A SDS PAGE gel of intact undigested protein gives the pattern shown in lane 1 of the gel pictured below. (2) A SDS PAGE gel of the protein after the intact cell is digested with a protease is shown in lane 2 of the gel pictured below. (3) A SDS PAGE gel of the intact protein after in vitro translation in the presence of ER microsomes is shown in lane 3 of the gel pictured below. (4) A SDS PAGE gel of the protein after in vitro translation in the presence of ER microsomes followed by protease treatment is shown in lane 4 of the gel pictured below. (5) The corresponding Western blots of the same four samples (lanes 1-4) are shown below. The blots were probed with either the anti-N-terminus antibody or with the anti-C-terminus antibody (1) W hich terminus (A or B) is the N terminus? [A] (2) W hich (A or B) is the C terminus? [B] (3) Determine the residue number at location (assume 1 aa residue = 100 Da): C: 400 D: 420 E: 920 F: 940 G: 1940 H: 1960 I: 2250 Detailed solution: Gel lane 1 shows that the protein has a MW of 225kD; lane 3 shows that there is no posttranslational cleavage of the protein in the secretory pathway. This means that there was no N-terminal signal sequence, and the mature protein contains the whole translation product. Next, figure out on which side of the membrane the N- and C-termini are located. Lane 4 of the anti-N-terminus western blot shows that the N-terminus of the protein is on the cytoplasmic side of the membrane (protease digestion from that side destroys the Nterminus), while lane 2 of the anti-C-terminus blot shows that the C-terminus is on the extracellular side (digestion from that side destroys the C-terminus). Next, determine how the fragments are orientated in the protein. Lane 4 of the anti-Nterminus blot also shows you that the N-terminus is at the end of the 42kD fragment produced by extracellular proteolysis. Lane 2 of the anti-C-terminus blot shows that the Cterminus is at the end of the 31kD fragment produced by cytoplasmic side proteolysis. G el lane 2 shows that if you proteolyze the portion(s) of the protein exposed on the extracellular surface, 79kD is eliminated and two fragments (104kD & 42kD) remain protected by the membrane. We already know that the 42kD fragment runs from the Nterminus to the end of the first membrane span. Therefore, the 104kD fragment must run from the beginning of 2nd membrane span to the end of the third membrane span. Gel lane 4 shows that if you proteolyze the intracellular portion of the protein, 140kD is eliminated and two fragments (54kD & 31kD) remain protected by the membrane. We already know that the 31kD fragment runs from the beginning of the 3 rd membrane span to the C-terminus. Therefore, the 54kD fragment must run from the beginning of the 1 st membrane span to the end of the 2nd membrane span. Altogether, we have this: Q70. Y ou know that the amino acid sequence of a protein contains targeting information that determines where the protein will be transported after synthesis. You are analyzing a set of cultured cell lines that have highly deleterious temperature-sensitive mutations in genes that encode proteins involved in organelle traffic in the cell. The cellular phenotype of each mutation is described in the table below. Match each phenotype with the change in targeting information that is most likely to have produced it. Possible changes in proteins: (A) Change in a single amino acid critical for the addition to this protein of a sugar chain ending in mannose-6-phosphate. (B) Deletion of this proteins C-terminal KDEL sequence. (C) Deletion of this proteins C-terminal 4-6 amino acid hook motif. (D) Change in the mannose-6-phosphate receptor so that it cannot recognize its ligand. (E) Deletion of an N-terminal hydrophobic signal sequence from this protein. (F) Deletion of the LDL receptor. (G) Addition of an N-terminal hydrophobic signal sequence to this protein. (H) Deletion of an internal hydrophobic signal sequence from this protein. Complete cellular phenotype Change most likely responsible (A - H) Failure of one protease, cathepsin B, to reach the lysosome A Secretion of actin, normally a cytoplasmic protein G Failure of a receptor, the transferrin receptor, to be internalized by endocytosis C Secretion of signal peptidase, which normally resides in the ER B Failure of any degradative enzymes to reach the lysosome D Failure of signal peptidase to be inserted into the ER E Q 71. S uppose you perform the following experiment. You stimulate a cell with a ligand at time = 1. At time = 2, the cell responds to the ligand. All throughout this period of time, you measure the concentrations of five different molecules (called A-E). The cytoplasmic concentrations of the molecules, as a function of time, are plotted below: Evaluate each of the molecules in terms of whether or not it is a likely second messenger in this system. Select the one that is most likely. [ C] (B),(D),(E) do not change concentration between times 1 and 2 and thus are not likely to be 2nd messengers. (A) does change concentration between 1 and 2 and is a possible 2 nd messenger. (C) changes concentration between 1 and 2 and THEN returns to its original concentration after time 2 it is the best candidate. Q 72. C onsider a hypothetical cell whose function is to proliferate when a wound occurs in order to fill in the wound with new tissue. Well call this cell a pre-epithelial cell. At rest, this cell is poised to begin mitosis. When the cell is stimulated with the appropriate ligand, a signal transduction pathway is triggered whose end result is mitosis. Suppose you have isolated this cell and have established it as a slow-growing cell line. You discover that if you add the molecule OUCH to the cells, the cells rapidly undergo mitosis. (Note: OUCH is a small molecule isolated from wounded epithelia. It can be purchased from Sigma Chemical Co.) The pre-epithelial cell is large enough so that you can easily microinject it with molecules. Assuming that there is a cell surface receptor-for OUCH, and that OUCH is the ligand, consider carefully the following experimental data. Note: each is a separate experiment. Match each conclusion below (1-11) with the experiment (A-H) that provides the evidence for it [order of conclusions is randomized]: A. microinject 100 ng of the phosphatase, microinject calcium, do not add OUCH..result: rapid mitosis B. microinject 100 ng of the phosphatase, add OUCH..result, no mitosis C. microinject free calcium ion, do not add OUCH..result: rapid mitosis D. Bathe cells in calcium-free medium, add OUCH...result: no mitosis. E. microinject 100 ng of a phosphatase, do not add OUCH..result, no mitosis F. microinject the catalytic subunit of protein kinase A (cAMP-dependent kinase), do not add OUCH..result: rapid mitosis. G. microinject IP3, do not add OUCH...result: mitosis but it takes a long time for the cell to begin mitosis. H. microinject cGMP, do not add OUCH....result: no mitosis I. microinject 100 ng of the phosphatase, microinject cAMP, do not add OUCH..result: you get different results depending on how much cAMP is injected according the following graph (constant amount of phosphatase): J. microinject cGMP, add OUCH...result: rapid mitosis K. microinject cAMP, do not add OUCH...result: rapid mitosis CONCLUSION EXPT (1) cGMP is probably NOT part of the pathway [H] H (2) cAMP can overcome the phosphatase effect, suggesting that if PKA or a downstream kinase is hyper-stimulated it can overcome the phosphatase effect. [I] I (3) Phosphatase INHIBITS the pathway. [B] B (4) Ca2+ is part of the pathway, after ligand binding. [C] C (5) cGMP does NOT block the normal pathway. [J] J (6) Protein kinase A IS part of the pathway, after ligand binding. [F] F (7) Phosphatase is NOT part of activating the pathway. [E] E (8) The Ca2+ involved in signaling probably comes from outside the cell, rather than from intracellular stores. [D] D (9) cAMP is part of the pathway, after ligand binding. [K] K (10) IP3, which releases internal Ca2+ stores, can substitute for this pathway but is probably not part of the normal pathway. [G] G (11) Ca2+ is involved AFTER the phosphorylation step. [A] A Q 73. W e have discussed the signaling pathway that regulates the availability of glucose in the cytosol when muscle cells require more glucose during heavy exercise or stress. Recall that in that pathway the hormone epinephrine triggers a response that includes increased [cAMP], stimulated activity of protein kinase A (PKA), and net release of glucose from carbohydate stores. A 2nd branch in the pathway inhibits the function of glycogen synthase, thus halting production of glycogen. Read carefully the following 12 statements (A - N) about signaling pathways and states and answer the questions about them below. (A) The membrane protein, Calcium Release Channel (also known as the ryanodine receptor or IP3 receptor) , stimulates the conversion of ATP to cAMP. (B) The concentration of free DAG (diacylglycerol) in the cell membrane rises. (C) Higher concentrations of pyrophosphate (PPi) in the cytoplasm. (D) Adenylyl cyclase activity is balanced by phosphodiesterase activity, maintaining low [cAMP]. (E) G-protein is bound to GTP. (F) G-protein is bound to GDP. (G) Higher concentrations of glucose in the cytoplasm. (H) Lower concentrations of glucose in the cytoplasm. (I) High probability that cAMP will be bound to the catalytic subunit of protein kinase A (PKA). (J) High probability that cAMP will be bound to the regulatory subunit of protein kinase A. (K) Increased phosphorylation of glycogen synthase. (L) The -adrenergic receptor is in the plasma membrane with no bound epinephrine. (M) The cell is converting glucose into the complex carbohydrate, glycogen. (N) The calcium release channel opens up, releasing the stores of intracellular calcium from the endoplasmic reticulum. W hich 5 of the statements above (A - N) are true in muscle cells at rest? LIST THEIR LETTERS CLEARLY in the box: D, F, H, L, M W hich 5 of the statements above (A - N) true in muscle cells at the peak of their response to stimulation by epinephrine? LIST THEIR LETTERS CLEARLY in the box: C, E, G, J, K W hich 2 of the statements above (A - N) are never true? LIST THEIR LETTERS CLEARLY in the box: A, I Which 2 of the statements above (A - N) are not part of this pathway but are part of a different signaling pathway? LIST THEIR LETTERS CLEARLY in the box: B, N Q 74. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: phosphorylated tyrosine residue [ 4] phospholipase C [ 1] diacylglycerol [ 3] phosphatidyl inositol [ 2] Ca2+ ion [ 5] P hospholipase C (1) is a protein; phosphatidyl inositol (2) is a phosphoglyceride that comprises diacylglycerol PLUS an inositol head group; (3) diacylglycerol (glycerol backbone plus two long acyl chains) is larger than a phosphotyrosine (4), which has standard amino acid structure with the R-group = a methyl residue, an aromatic 6-C ring, & a phosphate group on the substituted -OH of the ring. All of these organic molecules are larger than a single calcium ion (5). Q75. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: GTP [ 3] intracellular vesicle [ 1] cAMP [ 5] G protein [ 2] GDP [ 4] An intracellular vesicle (1) is an organelle made of thousands of lipid and protein molecules (think of the surface area of even the smallest vesicles in the cell, with a diameter of 30 nm; if the surface of a sphere = (4 r2 ) and a single phospholipid occupies roughly 2 square nm, how many lipid molecules are there in the two leaflets of the bilayer?). A G protein (2) is larger than the soluble small molecules. GTP (3), GDP (4) and cAMP (5) are all purine nucleotides (9 member ring) and on the question of size differ by a small amount, mainly in the number of phosphates. Q76. Below are schematic diagrams of the protein building blocks of the three cytoskeletal filament systems, just as we saw them in lecture. As arranged below, protein building block 1 is a monomer, 2 is a dimer, and 3 is a tetramer. For each description of a site on or property of the building blocks listed below, match the letter or number that corresponds to it: SITE OR PROPERTY LETTER (A - E) or NUMBER (1 - 3) Forms filaments that are major components of muscle cells 1 GTP binding site, GTP cannot be easily exchanged C Forms bipolar filaments 3 Forms polar filaments 1, 2 GTP binding site, GTP can be hydrolyzed and rapidly exchanged B Forms filaments that serve as tracks for kinesin and dynein motors 2 Interacts via long alpha helical coiled-coils 3 Forms filaments that are major components of the ciliary axoneme 2 ATP binding site A Q 77. If a cell increases the activity of its actin severing proteins, will it lead in the short term to MORE or LESS actin polymer mass? Chose the most accurate answer. *** (A) If the concentration of free actin monomers is above the Cc for both the plus and minus ends, then MORE polymer will result. If the concentration of free actin monomers is below the Cc for both the plus and minus ends, then LESS polymer will result. (B) Increased severing activity will result in LESS polymer under any conditions. ( C) Increased severing activity will result in MORE polymer under any conditions. (D) If the concentration of free actin monomers is below the Cc for both the plus and minus ends, then MORE polymer will result. If the concentration of free actin monomers is above the Cc for both the plus and minus ends, then LESS polymer will result. (E) Increased severing activity will result in NO CHANGE in the total amount of polymer under any conditions. It simply results in a larger number of smaller filaments. Q 78. Y ou are assembling actin filaments in vitro u sing purified actin with or without accessory, actin-binding proteins. In each experiment, you allow assembly to proceed to steady state, then you measure the extent of polymerization (as the % of the total actin that is foundin polymer) and also observe the state of the actin filaments by electron microscopy (how long are they and how are they arranged?). Match each experimental result below with the ONE CLASS of actin binding protein on the list that is most likely to give the result. The first row is a control, no answer there. start with: progress of polymerization: appearance in EM class of accessory protein added? 1 uM pure actin 3' lag phase, then steady polymerization until 80% of actin is in filaments average length = 2 um NONE 1 uM pure actin plus accessory protein 1 3' lag phase, then steady polymerization until 80% of actin is in filaments average length = 2 um, most filaments intersect other filaments cross-linking 1 uM pure actin plus accessory protein 2 no lag phase, steady polymerization until 80% of actin is in filaments average length = 2 um nucleating 1 uM pure actin plus accessory protein 3 3' lag phase, then steady polymerization until 75% of actin is in filaments average length = 1 um severing 1 uM pure actin plus accessory protein 4 3' lag phase, then steady polymerization until 60% of actin is in filaments average length = 2 um monomer-seq Seven possible classes of actin-binding proteins: capping depolymerizing (e.g. cofilin) monomer-sequestering (e.g. thymosin) nucleating c ross-linking (e.g. filamin) bundling (e.g. villin or fimbrin) severing protein (e.g. gelsolin) Q 79. C ytoskeletal filaments are in equilibrium with their free subunits. It is convenient to describe a term, critical concentration (C c), which is the concentration of the free subunits necessary for assembly. When the concentrations of subunits exceeds C c, assembly occurs; when the concentration of free subunits is below the C c, no assembly occurs. Consider the following data from an in vitro a ssembly experiment in which G-actin (globular or unassembled actin) is at different concentrations and its assembly is monitored: Based on this graph, what is the approximate critical concentration of actin for microfilament assembly? 4 0 M. Q 80. T he graph below plots the results of three different test tube experiments on the kinetics of actin polymerization. In each experiment the test tube contained a total actin concentration of 1 m M and an excess of ATP. Polymerization was initiated by addition of salt at time zero and monitored for 50 minutes by light scattering. The course of each experiment was plotted as the % of the total actin subunits in the test tube that was assembled into filaments versus the time of the reaction: Experiment #1: The solid curve shows the result of a control experiment showing the normal time course of actin polymerization. The reaction tube contained pure actin and no additional proteins or factors. E xperiment #2: The open-circles curve shows the time course of actin polymerization in the presence of 0.15 m M of Factor A. E xperiment #3: The closed-diamonds curve shows the time course of actin polymerization in the presence of 1 m M of Factor B. Use these three curves and what you know about the regulation of actin assembly to answer the following questions. (a) What is the critical conc for actin assembly in the control experiment (Expt #1)? 0 .2mM A t steady-state, 80% of the actin is assembled; therefore, 20% of the actin remains as monomer. Total actin concentration = 1 mM; C c = % g-actin at steady state = 1 mM 0.2 = 0.2 mM (b) You can tell from these data that Factor A has two different actin-regulating activities. Which are they? Five choices: ** Nucleating activity, because it eliminates the lag phase of assembly Severing activity, because it generates more filaments at steady-state Minus-end capping activity, because there is more assembled actin at steady-state Cross-linking activity, because actin filaments form a meshwork in its presence ** Plus-end capping activity, because there is less assembled actin at steady-state Factor A has actin nucleating activity. It eliminates the lag phase of assembly as soon as salt is added actin polymer appears. It also has capping (end-binding) activity. You can tell this because the % of assembled actin at steady-state is different in expt 2. Specifically, Factor A is probably a plus end capping protein, because there is less assembled actin at steady state. (Note also that the slope of the assembly curve is shallower, so the assembly rate is slower.) ( c) Factor B increases the lag time and reduces the overall rate of actin polymerization. Based on the data gathered in this experiment, what is the effect of Factor B on critical concentration? Decreases, because the slope of the assembly curve appears to be steeper Increases, because the lag phase is longer *** We don't know from these data expt #3 had not reached steady state yet when we stopped our measurements. Q 81. Rigor e xists when all of the myosin crossbridges are bound to the actin filaments simultaneously and without displacement. Suppose you begin with a skinned muscle preparation in rigor. This is a muscle in which the plasma membrane has been gently permeabilized, causing the cytoplasm to assume the condition so the experimental buffer, which you control. What is the effect of adding to this preparation of each of the following? Treatments: the nonhydrolyzable analog of ATP, AMPPNP [ B] P i [ C] ATP [ A] calcium ion [ C] ADP [ C] Possible effects: A - release of actin filaments by the myosins, probably allowing contraction again if the muscle is stimulated B - release from rigor but no subsequent contraction C - little or no effect ATP will cause the release of actin filaments by the myosins, thus disrupting rigor and allowing the muscle to contract again. ADP will have little or no effect, P i will have little or no effect. A non-hydrolyzable ATP analogue will cause release from rigor but no subsequent contraction. Ca ion will have little or no effect because the myosin heads are in rigor and cannot interact with the sites exposed by Ca-troponin interactions (see ECB, Fig 17-43 and related text about the actomyosin cross bridge cycle) Q 82. T he table below lists a number of molecules (1-12) that we have studied. For each of the statements about cytoskeletal protein function (a-f), list the numbers of all of the molecules from the list that are correct responses. Each statement has at least one correct answer; some statements have more than one correct answer. Items from the list may be used more than once or not at all. (1) ATP (2) GTP (3) troponin (4) kinesin (5) actin (6) myosin (7) dynein (8) cyclin (9) tropomyosin (10) IF protein (11) $ -tubulin (12) " -tubulin STATEMENT ABOUT FUNCTION MOLECULE (a) Regulates the interactions of myosin with actin in the muscle sarcomere: 1 , 9 1, 3, 9 (b) A plus end directed molecular motor: 4 , 6 4, 6 (c) A minus end directed molecular motor: 7 7 (d) Binds and hydrolyzes ATP: 4 , 5, 6, 7 4, 5, 6, 7 (e) Binds and hydrolyzes GTP: 1 1 11 (f) Assembles into bipolar filaments: 6 , 10 6, 10 Q 83. T his figure from your text that we discussed in lecture shows a cross-section of a cilium or flagellum. Match each of the labeled structures with its correct name or the correct statement about its function. Radial spoke (A) Central singlet mictotubule (B) Outer dynein arm (C) Nexin link (D) Inner dynein arm (E) A microtubule (F) B microtubule (G) An outer doublet microtubule (H) A motor protein (C, E) Was probably cut by protease treatment in the sliding disintegration experiment (A, D) Provides the surface along which the motor protein translocates (G) During ciliary beating, slides longitudinally relative to its neighbors (H) Q 84. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: epithelial cell myosin dimer actin monomer typical actin filament microvillus A microvillus (2) is a projection from the surface of an epithelial cell (1) that contains a large bundle of actin filaments (3). A myosin dimer (4), at approx 400 kD, is considerably larger than an actin monomer (5), approx 43 kD (see pp. 709-713 in your text). Q85. N umber each of the following in order of its diameter, with 1 being the largest. V ery briefly, explain your reasoning: actin filament muscle myosin filament sarcomere muscle cell intermediate filament The contractile apparatus within a single muscle cell (1) is comprised of numerous myofibrils running in parallel, each of which is divided into many stacked contractile units, or sarcomeres (2). Within each sarcomere are interdigitated myosin thick filaments (3) and actin thin filaments (5). Intermediate filaments (4) were so named because their diameter is between that of thick and thin filaments. See chap 17 in your text. Q 86. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: kinesin molecule microtubule cilium ciliated cell ciliary axoneme Kinesin (5) is a motor protein that can generate movement along a microtubule (4). A cilium (2) contains a characteristic arrangement of microtubules - usually 11 - and is a beating appendage of a ciliated cell (1). When the plasma membrane and any soluble proteins are extracted from a cilium, the remaining core of protein filaments, cross-linking proteins and motor proteins is called the ciliary axoneme (3). Q87. N umber each of the following in order of its size, with 1 being the largest. Very briefly, explain your reasoning: tubulin h eterodimer microtubule -tubulin monomer flagellum spermatozoon A mature sperm cell, or spermatozoon (1), has a flagellum (2) with which it propels itself. T he flagellum has typically 11 microtubules (3) in a 9+2 arrangement. The microtubules are polymerized from tubulin heterodimers (4) that each consist of one alpha tubulin (5) and one beta tubulin protein monomer. ... View Full Document

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