231: Cell Biology - Study Guide
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Date Created: 10/20/14
Biol 231 problem set key fall 2009 Q1 Eukaryotic cells have a number of membranebounded 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 Possible 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 Hintr use your text book39s glossary and the gures in chapter I Q2 You have approximately 1013 cells in your body human cells that is we39ll come back to the rest in a few weeks If all ofthose 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 ofI013 is approximately 43 So it would have taken 43 rounds ofdivision starting with the original zygote to produce you this would get you closest to 1013 with 880 x 10 cells Of course the development of an adult organism from a zygote isn39t 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 109 cm3 and determine the integral number of divisions that gives the closest result Answer 109 cm3 x 2quot 20 cm3 2 2 x 10 n approx 34 divisions resulting in a volume of 72 x 1010 cm3 so 34 days total Q3 Cells 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 E of the cell On the other hand the amount of material that they need to take up is proportional to the cell Volume ie how much biomass is in the cell Some cells are spherical note if you don t remember the formulae for surfaces and Volumes of regular solids one place to find them is httpmath2orgmathgeometryareasvolshtm a Determine a general expression for the ratio of surface area for example in umz to Volume in um3 for any sphere Surface Volume 47rr2437rr3 3r b Now try a calculation what is the actual surfacetoVolume ratio for a spherical adipocyte fat cell with a radius of 10 um Forr I0um S39V 3I0um 03 um c For a human oocyte egg cell with a radius of 100 um Forr 100 um S V 3I00 um 003 um d For a frog oocyte with a radius of 1000 um Note that you do not need to calculate the individual surfaces and Volumes if you have first determined a general expression for the ratio For r 1000 um S V 3I000 um 0003 um1 What do these values mean The units here um are the same as umzum3 which means that they tell you how many um2 ofsurface area are supporting each um3 of volume in the cell in question So here you can see thatfor dijferent cells ofthe same shape cell spherical as the radius goes up the amount ofsurface area per 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 one reason for an upper limit on cell size would be that for a uniform shape like a sphere the ratio ofsurface to volume decreases linearly with the cell radius As the size ofa spherical cell increases it eventually reaches a point at which the amount ofsurface area is inadequate to transport the nutrient molecules needed to support the cell s volume Q4 Cells 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 10000 um long with a radius of 10 pm A What is its surfacetoVolume ratio The surface area ofa cylinder with radius r and length l equals the area ofthe wall plus the area ofthe two circular ends so the surface 27rrx l 27rr2 The volume 7rr2 X l Here r I0um andl 10000 um Sofor this cell surface 2 X 75 X I0um gtlt I0000um 2 X 75 X I0um2 628318 um2 628 um2 628946 um2 And the volume 7 X I0um2 X 10000 um 3141600 um3 So the S Vfor the muscle cell 628946 um2 3I4I600 um3 02002 um 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 SV ratios are Very different Calculate SurfacetoVolume ratio of the muscle cellSurfacetoVolume ratio of the oocyte Compared to 0 03 um 1for the oocyte this is a dijference of 667fold In other words although the muscle cell has the roughly the same volume as the oocyte it has much more 667x more surface area because ofits shape Q5 Human 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 um X 65 um X 1 pm 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 um 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 7fold when it rounds up you can check this too How can a cell quickly decrease its SV 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 ofthese would have a SV that is much less than that of the cell B Exocytosis of many small secretory Vesicles since each of these would have a SV 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 SV 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 SV that is much greater than that of the cell During uptake of uid or specific macromoleculesfrom the outside world by endocytosis a cellpinches in vesiclesfrom its plasma membrane that are much smaller by 2 I00 fold than the cell itself Remember from problem 3 above that as the radius ofa sphere goes DOWN its surfacetovolume ratio goes UP Thus the vesicles pinched in by the cellfrom its membrane during endocytosis contain much more surface area per unit volume than the cell as a whole So by performing lots ofendocytosis the cell can decrease its surface area much more than it increases its volume Thus the surfaceto volume ratio ofthe at 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 ofa solid diminishes its shape approaches that ofa sphere You can do a test calculation imagine a spherical cell with radius 100 um how will its S and V change if it endocytoses a vesicle with radius 10 um Conversely note also that a lot ofthe reverse process exocytosis thefusion ofvesicles with the cell surface to release their contents will tend to INCREASE the cell s surface area relative to it s volume Q6 Assume that a spherical cell with a diameter of 20 um has spherical mitochondria each with a diameter of 02 um If mitochondria occupy 10 of the Volume of the cell how many mitochondria are there You don39t need a calculator to figure this out The ratio ofthe cell diameter mitochondrial diameter is 1001 100 so the ratio ofthe volumes is 1003 or 1000000 Thus it would take 1000000 mitochondria to equal the volume ofthe cell and 10 ofthe cell s volume would be occupied by 10 of1000000 or 100000 mitochondria Q7 Number each of the following in order of its surface area with l being the largest Very 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 dijfer in the total area oftheir membranes because a mitochondrion is generally longer and has both an outer and a highlyfolded inner membrane The ER 2 is a large highly branching and anastomosing reticulated thus the term endoplasmic reticulum organelle that can extend throughout much ofthe cell and whose surface area is probably exceeded on this list only by the plasma membrane 1 ofan entire epithelial cell which is itselftypically highlyfolded at one or more surfaces Q8 Number each of the following in order of its numerical Value with l being the largest Very briefly explain your reasoning 10g10 4 2 10g10 5 10g1017 31 10g103 41 103 1 It sjust arithmetic 2534I Q9 A If a solution is at pH 5 What is the concentration of protons U5 M Remember that pH logH B How many protons are in 10 mL solution at pH 3 6022 x I0 8protons pH 3 a concentration ofI0393 Mprotons So how many protons in I0mL Well ifthere are 103 moles ofprotons per Liter then there are 105 moles per I0mL To getfrom moles to number ofprotons youjust multiply by Avogaclro s number 105 moles X 6 022 X 1023 protonsmole 6 022 X 1018 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 990 mL ofH20 to the original 10 mL ofsolution Goingfrom pH 3 to pH 5 is a I00fold dilution ofprotons so you need to add 990 mL water to 10 mLs of starting solution Ifthis isn t obvious to you a reliable way to work out this kind of problem is to use the dilution relationship qnQn where C and V are the concentration and volume respectively ofthe starting solution and C2 and V2 are the concentration and volume ofthe clilutecl solution Since concentration ofsolute gtlt volume ofsolution mass ofthe solute this is simply an expression of conservation of mass Q10 Identify by label a CC single bond a CC double bond amines carboxyls a phosphate a diphosphate a sulfhydryl an ester a phosphoester a phosphodiester possible molecules clisplayecl phosphaticlylcholine ADP lysine phosphoserine Q11 Identify by label the 3 and 5 carbons of ribosedeoxyribose 339 439 539 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 Hbonding the base pairs molecules shown an AT and a GC pair Q13 Identify by label Fatty acid tails ester bond phosphoester bond glycerol backbone polar head group molecule shown phosphatidylcholine Q14 The AG for the binding of O2 to ferromyoglobin is 314 kJmol What is the Keg for this reaction AG AG RTlnPR at equilibriumAG 0 and PR Keg so AGO RTlnKeg 57kJmole logKeg logKeg 314 kJmole 57 kJmole Therefore Keg 323 X 105 What does a Keegt1 tell you about this or any reaction How about a Keg lt1 Look carefully at the expression AGO RTlnKeg Because AG and Keg have a minus log relationship Keg lt 1 results in a large positive AG0 and describes a reaction that will NOTproceed spontaneouslyfrom left to right under standard conditions But ifKeg gt I then the reaction has a large negative AG and WILL proceed spontaneouslyfrom left to right under standard conditions But also realize what Keg means it is a ratio of products to reactants at equilibrium So whether or not a reaction willproceed in a particular direction can be changed by a large shift in the ratio ofproductsreactants a departurefrom standard conditions Q15 You have identified found an enzyme in the lab freezer labeled bloggsose kinase GTP which catalyzes the reaction bloggsose GTP ltgt bloggsosephosphate 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 bloggsosephosphate 01 M GTP 10 mM GDP 01 M a Calculate the Keq for this reaction K bloggsosePGDP I0quotMI0quotM I00 bl0ggs0seGTP I0392 MI0392 M b Using the Keq calculate AG for this reaction At equilibrium AGO RTlnKq 5 kJmol log Keg 57 kJmol l0gI02 114 kJmol Q16 Consider the reaction Glucose Pi ltgt Glucose6phosphate H20 AG l25 kJmol a What is the equilibrium constant Keq for this reaction AG AG RTlnPR at equilibrium AG 0 and PR Kee so AG0 RTlnKee 57kJmol gtlt logKee 125 kJmol 57kJmol gtlt logKee logKee 125 kJmol57 kJmol Therefore Kee 641 X 103 b What does the positive AG for the reaction tell you about whether or not the reaction will go spontaneously from left to right A large positive standardfree energy change means that the reaction will NOT run spontaneouslyfor L to R under standard conditions To make it run L to R will require a VERYhigh concentration ofreactants relative to products c Under physiological conditions glucose 5 mM g6P 83 uM and Pi 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 ofproducts by that ofreactants under current conditions to get the quantity we call q Compare q to the Kee Ifq gt Kee then the products are in excess and the reaction will notproceed spontaneously ifq lt Kee then the reactants are in excess and the reaction will proceed spontaneously Here q G6PglucosePi 166 which is gtgt 641 X 103 The reaction will not proceed spontaneously until G6PglucosePi lt 641 X 103 83 X 10610 3641 X 103 lt glucose glucose gt 129 M Even Coke is not this sweet Q17 Consider the reaction glutamate ammonia S glutamine Water AG 14 kJmo1 a What is the Keg for this reaction Again what does the Keg describe AG0 RTlnKeg 57kJmole gtlt logKeg Keg 350 X 103 Keg is the ratio ofproducts 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 de ne the ratio ofP to R under current conditions as q Reactions proceed spontaneouslyfrom left to right ifq lt Keg So this reaction proceeds spontaneouslyfrom left to right ifglngluAmm lt 350 X 103 So glnglu lt 350 X I0 3I0 2 glnglu lt 350 X 1075 and invert both sides glugln gt 286 X 104 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 couples glutamine synthesis to ATP hydrolysis glutamate ammonia ATP r glutamine ADP Pi If the AG for the hydrolysis of ATP is 307 kJmo1 what is the AG for the overall reaction Add the AG0 valuesfor the two coupled reactions The net standardfree energy change 307 kJmol 14 kJmol 167 kJmol 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 Kegfor the coupled reaction using the AG0 of 167 kJmol AGO RTlnKeg 57kJmole gtlt logKeg logKeg 16 7 57 293 Keg 85 I Then solvefor Amm under spontaneous L gt R conditions with Kee 103 To proceedL gt R q lt Kee meaning that PR lt Kee so I0 2I0 2I02 I0 2I0 2Amm lt 851 102Amm lt 851 Amm gt 118 X 10 Amm must be anything gt 118 gtltI0 5 M ie gtII8uM Q18 Suppose you are studying an enzyme that catalyzes the reaction E S gt P 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 uM IS uM pmol P producedmin IV umolmin 000 00 020 500 556 001799 025 400 625 00160 033 300 714 00140 050 200 833 00120 100 100 1000 00100 200 050 llll 00090 400 025 ll76 00085 Using these data determine the Vmax and KM of the enzyme under these conditions Use a LineweaverBurkplot IS on the xaxis and IV on the yaxis see Fig 328 in your text The yintercept is 0008 umolesminI so the Vmax 125 umolesmin The slope is 000199 uMumolsmin and since slope KMVmax we multiply this by Vmax to get KM 0250 uM NOTE that the units are extremely important I used Excel to do the leastsquaresfit to the data but ifyou used a graphing or statistical calculator you ought to get the right values too See my graph ofthese data below y mn pvt u 8 Lam Hfm rruquot s WW 3 125 um m n 1 lVumofmin391 w mergptz x 4u um quot39 KM 2 i25 lL m 002 F3 9 U1 0005 001 1 us uVl391 Q19 Suppose 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 singlestranded 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 ug oftemplate 100 pmols of primer 5 units of polymerase and 25 14M of each of the nucleotides Just 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 KM ofthe quotoldquot polymerase was 20 uM nucleotide and the KM ofthe quotnewquot polymerase is 100 uM 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 gt100 uM With the old enzyme the rxn was being run with the S nucleotides slightly gt Km I25gtlt the Km and it was working ne Since the new enzyme has a higher Km 100 vs 25 uM it will take a higher S to get the same rxn rate as before gt 100 uM I25 uM nucleotides would be a good starting point Q20 In general the rate of chemical reactions increases with temperature but enzymaticallycatalyzed 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 enzyme s 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 enzyme s active site than at higher or lower temperature each student s order of alternatives differs and is randomized AND there are two different correct answers depending on the set of alternatives that is presented Q21 Consider a GTPase that catalyzes the reaction GTP H20 gt GDP Pi There are simple assays that utilize a color reaction to measure the amount of Pi in a solution Suppose you utilize such an assay to measure the velocity of the release of Pi 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 V no I mmolmin V I mmolmin 016000 099000 064000 019000 101000 072000 024000 125000 085000 033000 154000 105000 049000 182000 132000 099000 289000 175000 Using these data what is the Vim and KM 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 LineweaverBurk 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 5 11 mM and the Vmax is 394 mmolemin In the presence of inhibitor Z the Xintercept of the reciprocal plot is about the same while the yintercept is higher Thus with inhibitor present the KM is the same while the Vmax is reduced and therefore Z is a noncompetitive inhibitor 1S M 1 1V no I minmmol 1V I minmmol 625000 101010 156250 526316 099099 138889 416667 080000 117647 303030 064935 095238 204082 054945 075758 101010 034602 057143 yint 0253769 0374761 Vmax 394059 mmolmin 266836 mmolmin slope 012974 019120 Km 051125 M 051019 M The uninhibited GTPase plot is in blue the inhibitor plot is in red ExE fl39E I 3III EIII 1iE I fIIIII 1iEi 21 p 4 E EaEi IiquotIii 1 Ei l iIrr39I i nafm r39r1 IIiia Q22 The relationship between enzyme Velocity V and the concentration of substrate S can be expressed in the following equation the LineweaverBurk equation which describes a straight line in a doublereciprocal plot P 1Vmax H A competitive inhibitor 1 affects the slope of the line produced in a double reciprocal plot by increasing the slope by the factor 1 Iki Where ki EIE1 Suppose an enzyme has a KM 10 M and a competitive inhibitor has a k 10 M If I 2 X 10 3 M how much will the KM change in the presence ofthe inhibitor Use the LineweaverBurk equation to solve this problem k 10 M 1 2 x 10 M So I 1k 3 Therefore the slope KMVmax ofthe reciprocalplot increases 3fold So KMVmax increases 3fold but Vmax remains the same because it39s a comp inhibitor Therefore KM must increase by afactor of3 Note that you don 39t have to plot anything this to solve the problem you just need to understand what the elements ofthe LineweaverBurk equation mean A competitive inhibitor competes with Sfor the active site increasing the apparent KM This is reflected in the plot by an increased slope because the xintercept is closer to the origin Sketch it out and be sure thatyou 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 lV 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 noncompetitive 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 student s order of alternatives is randomized As you increase S Vwill increase reaching halfofits maximum at the same S as without inhibitor Km unchanged But the Vmax will be reduced ie in the presence ofnon competitive inhibitor no matter how much Syou add the enzyme will not reach the same Vmax as it does without inhibitor This is because a noncompetitive inhibitor does NOT compete with Sfor the enzyme s active site Q24 The figure below shows the short protein that has the amino acid sequence GRAKLSS CHEGIZIOH3 1 G H D 4 1 CH3 H H ClL H H H cH7 H ia sig1i39 J quot3 1 1 1 H I H H 3 I I1 1 n 1 o 1 H C41 0 H CHLO H 0 CHZD H I CEHL DH 3111 f f39IH Wig NH NH3 A Click on the side chain of Lysine 4th residue or Arginine 2 quot residue B What do you expect the charge of this protein will be at pH 2 3 C What do you expect the charge of this protein will be at pH 14 1 AtpH 2 expect all these groups to be protonated the Nterminal amino group 1 the basic residues K amp R each 1 and the Cterminal carboxylic acid group uncharged or 0 So the overall charge at pH 2 3 AtpH I4 expect the ionizable groups to be unprotonated the Nterminal amino 0 the basic residues each 0 and the Cterminal carboxylic acid group 1 Thus the overall charge I 78 108 125 36 pH Nterm K R Cterm net charge 2 1 1 1 0 3 14 0 O 0 1 1 Q25 Consider the peptide NH2GAPAGPAGTGKTETTKDLAKSMALLCWFNCSCOOH Table I from Stryer Biochemistry 2nd ed 1981 pK Values of some amino acids Using Table 1 below determine the net charge ofthe peptide at pH 3 6 7 9 12 each student gets just 3 of these pH s to solve randomly amino acid onCOOH onNH3 side group group chain Alanine A 23 99 Glycine G 24 98 Phenylalanine F 18 91 Serine S 21 92 Valine V 23 96 Aspartic acid D 20 100 39 Glutamic acid E 22 97 43 Histidine H 18 92 60 Cysteine C 18 108 83 Tyrosine Y 22 91 109 Lysine K 22 92 108 Arginine R 18 90 125 peptide amino terminus 78 peptide C terminus 36 pK 78 108 43 108 39 108 83 83 36 pH Nter K E K D K C C Cter total 1 1 1 0 1 0 0 0 4 1 1 1 1 1 1 0 0 1 1 1 1 1 1 1 1 0 0 1 1 0 1 1 1 1 1 1 1 1 2 12 0 0 1 0 1 0 1 1 1 5 Q26 Number each ofthe following in order of its size with l 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 4 glycine 5 CTP 3 CaM kinase 1 cholesterol 2 CaM kinase 1 is aprotein comprised ofhundreds ofamino acid residues and thus much larger than the other molecules The largest ofthe others is cholesterol 2 a 27carbon lipid with 4 characteristic carbon rings bearing one polar OH group 2 methyl groups and an 8carbon tail all ofwhich 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 Rgroup than glycine 5 which is the smallest amino acid Rgroup H See diagrams in Chapter 2 Q27 Number each ofthe following in order of its size with l being the largest ATP 2 glucose 5 glucose6phosphate 4 ADP 3 hexokinase I Hexokinase I is an enzyme and therefore a protein and thus comprised ofhundreds of amino acid residues that uses glucose andATP 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 3phosphates respectively Glucose 5 and glucose6phosphate 4 are identical 6 carbon sugars without or with a phosphate group respectively Q28 multi choice on Protein Data Bank study Alternative 1 Protein Data Bank httpwwwpdborgpdbhomehomedo Use this web resource to probe the structure of a Nerve Growth Factor Search the database using quotneurotrophinquot and select neurotrophin 4 or just enter id lb8m From the Display Molecule menu on the left select quotWebMol Viewerquot 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 neurotrophin4 multi choice It is an elongated roughly dumbbellshaped protein correct It is a spherical protein It is an elongated protein composed mainly of helical coiledcoil B How many polypeptide subunits is this protein composed oi Answer 2 hint easiest way to see this is to select quotColorChain C How many disulfide bonds does this protein have Answer 6 hint you can get this info by clicking the 7 button at lower right and then display the protein using quotBackbquot and quotColorChainquot to confirm the number by inspection you39ll have to rotate the structure to see them all D How many alpha helices does each subunit polypeptide have Answer 0 hint use the Backbquot and quotColorSecquot display options and rotate the structure to lookfor 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 quotBackbquot and quotColorSecquot display options and rotate the structure to lookfor 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 7 button at lower right G What of the protein is beta strand Answer 50 hint you can get this info by clicking the 7 button at lower right Alternative 2 Protein Data Bank httpWWWpdborgpdbhomehomedo Use this web resource to probe the structure of bovine hemoglobin Search the database using quotbovine hemoglobinquot or just enter id lg09 or just search the id From the Display Molecule menu on the left select quotWebMol Viewerquot 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 dumbbellshaped protein It is a roughly spherical protein correct It is an elongated protein composed mainly of helical coiledcoil B How many polypeptide subunits is this protein composed oi Answer 4 hint easiest way to see this is to select quotColorChain quot 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 7 button at lower right and then display the protein using quotBackbquot and quotColorChainquot to confirm the number by inspection you39ll have to rotate to look carefully D Blink the quotHet Atomquot checkbox to toggle nonprotein groups in and out of the image This 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 Answer 6 G How many beta strands does each subunit polypeptide of the protein have Answer 0 hint use the quotBackbquot and quotColorSecquot display options and rotate the structure to lookfor and count helices gold or beta strands blue the quotTracequot command is goodfor 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 7 button at lower right I What of the protein is beta strand Answer 0 hint you can get this info by clicking the 7 button at lower right Alternative 3 Protein Data Bank httpwwwpdborgpdbhomehomedo Use this web resource to probe the structure of bacterial agellin a component of the rotary tail that drives bacterial swimming Search the database using quotbacterial flagellinquot or just search id lucu From the Display Molecule menu on the left select quotWebMol Viewerquot 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 dumbbellshaped 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 oi Answer I hint easiest way to see this is to select quotColorChain C How many disulfide bonds does this protein have Answer 0 hint you can get this info by clicking the 7 button at lower right and then display the protein using quotBackbquot and quotColorChainquot to confirm the number by inspection you39ll have to rotate to look carefully D How many alpha helices does agellin have Answer 6 or 7 this is tricky E How many beta strands does each subunit polypeptide of the protein have Answer I 8 hint use the quotBackbquot and quotColorSecquot display options and rotate the structure to lookfor and count helices gold or beta strands blue the quotTracequot command is goodfor 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 7 button at lower right G What of the protein is beta strand Answer 20 hint you can get this info by clicking the 7 button at lower right Alternative 4 Protein Data Bank httpWWWpdborgpdbhomehomedo 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 quotHIV reverse transcriptasequot or just type in id lslt With lslt selected from the Display Molecule menu on the left select quotWebMol Viewerquot 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 quotlegsquot correct It is a spherical protein It is a highly elongated ropelike protein many times as long as it is wide B What is the identity of residue 316 Use the singleletter 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 singleletter abbreviation Ifyou simplify the structure by selecting quotBackb quotfrom thefirst menu it is very easy to move along the chain andfind a particular residue C How many polypeptide subunits is this protein composed oi Answer 2 hint The easiest way to see this is to select quotColorChain quot Note that its overall shape makes it look as though it might have gt1 subunit D How many disulfide bonds does this protein have Answer 0 hint you can get this info by clicking the 7 button at lower right and then display the protein using quotBackbquot and quotColorChainquot to confirm the result by inspection you39ll have to rotate the structure to see them all E How many alpha helices does the entire protein have Answer 23 12 in one subunit II in the other F How many beta strands does the entire protein have Answer 39 24 and I5 hint Use the quotBackbquot and quotColorSecquot display options and rotate the structure to lookfor and count helices gold or beta strands blue Using the quotTracequot command with these settings can help you to check on the numbers G What of the protein is alpha helix Answer 34 hints you can get this info by clicking the 7 button at lower right H What of the protein is beta strand Answer 18 hints you can get this info by clicking the 7 button at lower right Q29 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning human telomere 2 human chromosome I an Okazaki fragment 4 DNA polymerase complex 3 nucleotide 5 A human chromosome I typically has a telomere 2 a region ofrepeated DNA sequence 3000 18000 nucleotide bases in length importantforfaithful replication at each end The subunitsfrom which the DNA ofa 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 ofa 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 Consider this generalized diagram of a nucleotide triphosphate 5 NH2 N A Identify each of the labeled structures ad in the diagram 1 is a phosphate 2 is the 5 carbon 3 is the 3 carbon 4 is the I carbon 5 is the base B Is it possible that the base attached to the 139 carbon is a purine N0 purines have a ninemember tworing 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 ribosecontaining nucleotide triphosphate not a deoxyribosecontaining one No the base is an RNA base U not a DNA base It has an OH at the 2 p0siti0n this is an RNA subunit Q31 Translation 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 539TAAGCTACGTGCTGGCGAATTAATTTATTTCGTTATAGTCGTACATCTAT339 339ATTCGATGCACGACCGCTTAATTAAATAAAGCAATATCAGCATGTAGATA539 A Which strand is transcribed into mRNA Do a virtual transcription ofeach strand being careful aboutpolarity bottom DNA strand 339ATTCGATGCACGACCGCTTAATTAAATAAAGCAATATCAGCATGTAGATA539 resulting transcript 5 39 UAAGCUACGUGCUGGCGAAUUAAUUUAUUUCGUUAUAGUCGUAGAUCUAU 339 top DNA strand ipped so it reads 3 to 5 left to right 339TATCTACATGCTGATATTGCTTTATTTAATTAAGCGGTCGTGCATCGAAT539 resulting transcript 539AUAGAUGUACGACUAUAACGAAAUAAAUUAAUUCGCCAGCACGUAGCUUA339 Then lookfor which potential transcript has an open readingframe a start and stop codon in the sameframe The DNA strand thatyou transcribed to get that mRNA sequence is the one that can be transcribed The transcript ofthe top strand has an AUG and a UUA inframe while the transcript ofthe 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 tablefrom an old exam or the lecture notes AUG UAC GAC UAUAAC GAA AUA AAU UAA NH2 MYDYNEIN COOH Q32 Consider 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 39 TAAGCTACGTGCTGGCGAATTAATTTATTTCGTTATAGTCGTACATCTAT3 39 3 39 ATTCGATGCACGACCGCTTAATTAAATAAAGCAATATCAGCATGTAGATA5 39 A What is the new anticodon sequence for the affected codon Be sure to label note the polarity of the anticodon sequence Mutated DNA 5 39 TAAGCTACGTGCTGGCGAATGAATTTATTTCGTTATAGTCGTACATCTAT3 39 3 39 ATTCGATGCACGACCGCTTACTTAAATAAAGCAATATCAGCATGTAGATA5 39 previous mRNA transcript copiedfrom the original top strand done in previous problem 539AUAGAUGUACGACUAUAACGAAAUAAAUUAAUUCGCCAGCACGUAGCUUA339 new mRNA transcript copied from the mutated top strand 539AUAGAUGUACGACUAUAACGAAAUAAAUUCAUUCGCCAGCACGUAGCUUA339 1 At the point a ected by the mutation the original codon was UAA a stop codon so the perfect t anticodon would have been 3 A UU5 BUT a stop codon has NO anticodon because it has no tRNAaa that matches it However the new mutated codon is UCA which encodes serine and thus has an anticodon 3 AGU539 B With this mutation what is the resulting protein sequence using the single letter amino acid abbreviations The mutation has replaced a stop codon with a codon encoding serine So the amino acid sequence ofthe protein becomes NH2 MYDYNEINSFAST COOH 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 Suppose you have cloned and sequenced a piece of doublestranded DNA 539CTATGTCGCCTCCCTCCCTTCTCGGGCCGACAAGAGTTCGCAATACAT339 339GATACAGCGGAGGGAGGGAAGAGCCCGGCTGTTCTCAAGCGTTATGTA539 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 ofeach strand bottom DNA strand 339GATACAGCGGAGGGAGGGAAGAGCCCGGCTGTTCTCAAGCGTTATGTA539 resulting RNA transcript 539CUAUGUCGCCUCCCUCCCUUCUCGGGCCGACAAGAGUUCGCAAUACAU339 top DNA strand reversed so 3 to 5 is left to right 339TACATAACGCTTGAGAACAGCCGGGCTCTTCCCTCCCTCCGCTGTATC539 resulting RNA transcript 539AUGUAUUGCGAACUCUUGUCGGCCCGAGAAGGGAGGGAGGCGACAUAG339 Next find which potential transcript has an open readingframe The virtual transcript of the bottom strand has a potential start codon but it begins at base 3 not at the beginning ofthe sequence and it has no stop codon at the end in fact it has no stop codons at all So here the top strand will be transcribed since its transcript begins with a start ends with a stop and has them inframe with each other B What will the mRNA sequence be remember to label the 539 and 339 ends 5 AUGUAUUGCGAACUCUUGUCGGCCCGAGAAGGGAGGGAGGCGACAUAG 3 C What is the sequence of the protein that this gene encodes Amino acid sequence NH2 MYCELLSAREGREAT COOH Q34 Consider the following mRNA sequence 539CAUUAAUGGACGCCAAUAAGUGGGCUAUUCUUUAGCGUAUGUG339 A Which base number does the open reading frame begin with The ORF begins with the 6 base and runsfor I0 codons including the stop B What is the sequence of the open reading frame from start through stop codon mRNA AUGGACGCCAAUAAGUGGGCUAUUCUUUAG C What is the translated polypeptide sequence AUG GAC GCCAAUAAG UGG GCUAUU CUU UAG gt aa sequence Nterm MDANKWAIL Cterm 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 539GUC339 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 Lookfor a codon that would bind the anticodon 3 CUG5 a perfectfit would be GA C This is the second codon ofyour open readingframe mRNA AUG GAC GCCAAUAAG UGG GCUAUU CUU UAG So ifthe translation system plugs in the wrong amino acid at that codon it will change the sequence ofthis 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 ofD giving M CANK WAIL Q35 Number each of the following in order of its size with l being the largest Very brie y explain your reasoning 40S subunit 4 polysome I ribosome 2 60S subunit 3 an aminoacyl tRNA 5 A polysome I is comprised ofa number ofribosomes 2 all translating along an mRNA template Each ribosome contains a 60S subunit 3 and a 40S subunit 4 An aminoacyl tRNA 5 willfit within the A siteformed by the 40S and 60S subunits Remember thatfor illustrative purposes tRNA is often represented in cartoons of translation at an exaggerated size relative to ribosomes and mRNA Q36 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning tRNA 3 ribosome 2 stop codon 4 endoplasmic reticulum I nucleotide 5 A stop codon 4 containsjust 3 nucleotide 5 residues and is therefore small relative to an entire tRNA 3 which is approx 90 nucleotide residues in length see g 728 in ECB During translation an aminoacyl tRNA molecule ts into a single site within the larger ribosome 2 ifthe translatedprotein contains a start transfer sequence the ribosome will dock on the hulking surface ofthe ER 1 which as a major organelle in the cell is much larger than any ofthese 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 uidity of its membranes Which changes in the composition of membranes would allow them to maintain uidity keep them from solidifying at low temperatures A Decrease the length of the fatty acid chains and increase the of unsaturated CC bonds B Decrease the length of the fatty acid chains and decrease the of unsaturated CC bonds C Increase the length of the fatty acid chains and decrease the of unsaturated CC bonds D Increase the length of the fatty acid chains and increase the of unsaturated CC bonds Changes in the phospholipids The acyl chain length could be shorter and there could be more unsaturated bonds in the acyl chains Each ofthese would decrease the temperature at which the membranes solidify thus allowing them to befluid at a lower temperature A shorter acyl chain does this by providingfewer opportunitiesfor hydrophobic interactions between chains Increased numbers ofcisunsaturated bonds cause more kinks in the acyl chains making it less likely that adjacent acyl chains can have interactions along their entire length Q38 Consider the following cell it has a plasma membrane with surface area of 5000 pmzg there are 106 phospholipid molecules per pmz in each leaflet of the membrane there are 2000 molecules of ippase enzyme associated With the plasma membrane each flippase enzyme molecule can catalyze l00 flips per second We ll 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 ips in flipssec for the Whole plasma membrane that can be reversed by the action of the flippases The ippases can catalyze 2000 molecules x I00 ips sec molecule 2 x 105 ips sec 72 x 108 ips hr B What fraction of the total phospholipids in the plasma membrane can be flipped per hour The cell has 106 pmz x 5000 pmz lea et x 2 leaflets I010phospholipicl molecules total So 72 x 108 I0 0072 or 72 ofthe total Plipids that can be ippecl each hr C If each phospholipid in the plasma membrane flips spontaneously on average once every 105 seconds are the 1000 molecules of ippase adequate to reverse those flips and keep the asymmetry of the lipid bilayer intact Ifeach phospholipicl ips spontaneously every 105 sec then the number of ips in the whole plasma membrane each sec I 0395 ips sec phospholipid molecule x I 0 0phospholipicl molecules 105 ipssec Since theflippases can reverse 2 x I05flips sec twice this number YES they ARE adequate to maintain the phospholipid asymmetry ofthe plasma membrane Q39 You 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 Bloggsl and Bloggs2 You first try a FRAP fluorescence recovery after photobleaching experiment You covalently attach a green fluorescent label onto an antibody against Bloggsl antiBloggsl and a red fluorescent label onto antiBloggs2 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 AntiBlo s1 AntiBoggs2 FLUORESCENCE TIME after photobleaching A Based on this experiment which protein Bloggsl or Bloggs2 diffuses more quickly in the plane of the bilayer Bloggs I The recovery of uorescence in a FRAP expt like this is due to diffusion into the bleached region of unbleached molecules So the rate of uorescence recovery is proportional to the rate of diffusion the faster the recovery the more rapid the diffusion You 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 antibodybead individually to Bloggs cells in culture Using videoenhanced 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 antibodycoupled beads by labeling them A and B instead of Bloggsland 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 W A bead B bead Although Ivor will not be in until Monday to tell you which beads have antiBloggsl and which have antiBloggs2 you have a pretty good idea already Which preparation A or B contains beads coupled to the Bloggsl 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 Q40 The regions 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 singlespan protein nerve growth factor or NGF This is a simple primary structure amino acid sequence with the residues numbered Nterminus to Cterminus from 1 to 796 The numbering system is simple the zero in l0quot sits directly above the singleletter abbreviation for the 10th residue the zero in 20quot sits above the 20th residue etc There are three different 20residue 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 lookfor the sequence with thefewestpolar residues NGF sequence terminus 10 MLRGGRRGQL 70 HLPGAENLTE 130 NLSFNALESL 190 GPLAHMPNAS 250 SGGLPSLGLT 310 IPFSVDGQPA 370 LLAANPFGQA 430 AVGLAVEACL 490 GKGSGLQGHI 550 VAVKALKEAS 610 LRSHGPDAKL 670 GLVVKIGDFG 730 FTYGKQPWYQ 790 LQALAQAPPV numbers 20 GWHSWAAGPG 80 LYIENQQHLQ 140 SWKTVQGLSL 200 CGVPTLKVQV 260 LANVTSDLNR 320 PSLRWLFNGS 380 SASIMAAFMD 440 FLSTLLLVLN 500 IENPQYFSDA 560 ESARQDFQRE 620 LAGGEDVAPG 680 MSRDIYSTDY 740 LSNTEAIDCI YLDVLG indicate the residue number 30 SLLAWLILAS 90 HLELRDLRGL 150 QELVLSGNPL 210 PNASVDVGDD 270 KNVTCWAEND 330 VLNETSFIFT 390 NPFEFNPEDP 450 KCGRRNKFGI 510 CVHHIKRRDI 570 AELLTMLQHQ 630 PLGLGQLLAV 690 YRVGGRTMLP 750 TQGRELERPR 40 AGAAPCPDAC 100 GELRNLTIVK 160 HCSCALRWLQ 2 2 0 VLLRCQVEGR 280 VGRAEVSVQV 340 EFLEPAANET 400 IPVSFSPVDT 460 NRPAVLAPED 520 VLKWELGEGA 580 HIVRFFGVCT 640 ASQVAAGMVY 700 IRWMPPESIL 760 ACPPEVYAIM 50 CPHGSSGLRC 110 SGLRFVAPDA 170 RWEEEGLGGV 230 GLEQAGWILT 290 NVSFPASVQL 350 VRHGCLRLNQ 410 NSTSGDPVEK 470 GLAMSLHFMT 530 FGKVFLAECH 590 EGRPLLMVFE 650 LAGLHFVHRD 710 YRKFTTESDV 770 RGCWQREPQQ starting with the N 60 TRDGALDSLH 120 FHFTPRLSRL 180 PEQKLQCHGQ 240 ELEQSATVMK 300 HTAVEMHHWC 360 PTHVNNGNYT 420 KDETPFGVSV 480 LGGSSLSPTE 540 NLLPEQDKML 600 YMRHGDLNRF 660 LATRNCLVGQ 720 WSFGVVLWEI 780 RHSIKDVHAR Q41 These two diagrams next page show two different ways of looking at a membrane protein called the betaadrenergic 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 sabertoothed tigers angry roommates etc The first diagram is a simple primary structure amino acid sequence with the residues numbered Nterminus to Cterminus from 1 to 413 and the membrane spanning regions in red The second is a 2D representation of how the betaadrenergic 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 n0 membrane span B antibody that binds FERLQTVTNY n0 cytoplasmic domain C antibody that binds YQSLLTKNKA n0 cytoplasmic D antibody that binds AINCYANETC yes extracellular E antibody that binds ALKTLGIIMG n0 cytoplasmic membrane span F antibody that binds SQGRNCSTND n0 cytoplasmic Beta2 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 DGRTGHGLRR SSKFCLKEHK 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 2D arrangement in membrane 1 Ext139ace11L11a1 35 1 7 4 1 97 BDB P1as111a 1 11e111b1 a1391e 1 2 9 2 7 5 32 9 C Jt0p1as111 413 Q42 These two diagrams following pages again show two different ways oflooking 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 Nterminus to Cterminus from 1 to U68 and the membranespanning regions in red The numbering system is simple the zero in l0quot sits directly above the singleletter abbreviation for the 10th residue the zero in 20quot 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 2D representation of how AC is actually organized in the plasma membrane of your cells with the N and Ctermini 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 Extracellular 7 fragments count em P1b I ZI Cytoplasm V 1N 1668 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 Extracellular A n ragm en ts Plasma membrane I Q Cytoplasm C Finally what if an unusual protease within the membrane made a single cut midway in each membranespanning region how many fragments of AC would this produce Extracellular 1 n A n r Plasma membrane M I I I N I I H I l H 13 fragments C U V 1 1C Ix68 N Adenylate cyclase type 6 Homo sapiens Name ADCY6 10 MSWFSGLLVP 70 PRCPWQDDAF 130 RLVQVFQSKQ 190 VALLACAAAL 250 CPVFFVYIAY 310 VIGICTHYPA 370 KKEDMMFHKI 430 LRIKILGDCY 490 HCGVLGLRKW 550 YLKEQHIETF 610 QMGIDDSSKD 670 YSRKVDPRFG 730 GSLFPKALQR 790 PADITACHLQ 850 MIFVLGLIYL 910 ILLVFALALY 970 20 KVDERKTAWG 80 IRRGGPGKGK 140 FRSAKLERLY 200 FVGLMVVCNR 260 TLLPIRMRAA 320 EVSQRQAFQE 380 YIQKHDNVSI 440 YCVSGLPEAR 500 QFDVWSNDVT 560 LILGASQKRK 620 NRGTQDALNP 680 AYVACALLVF 740 LSRSIVRSRA 800 QLNYSLGLDA 860 VLLLLGPPAT 920 LHAQQVESTA 980 30 ERNGQKRSRR 90 ELGLRAVALG 150 QRYFFQMNQS 210 HSFRQDSMWV 270 VLSGLGLSTL 330 TRGYIQARLH 390 LFADIEGFTS 450 ADHAHCCVEM 510 LANHMEAGGR 570 EEKAMLAKLQ 630 EDEVDEFLSR 690 CFICFIQLLI 750 HSTAVGIFSV 810 PLCEGTMPTC 870 IFDNYDLLLG 930 RLDFLWKLQA 990 40 RGTRAGGFCT 100 FEDTEVTTTA 160 SLTLLMAVLV 220 VSYVVLGILA 280 HLILAWQLNR 340 LQHENRQQER 400 LASQCTAQEL 460 GVDMIEAISL 520 AGRIHITRAT 580 RTRANSMEGL 640 AIDARSIDQL 700 FPHSTLMLGI 760 LLVFTSAIAN 820 SFPEYFIGNM 880 VHGLASSNET 940 TGEKEEMEEL 1000 50 PRYMSCLRDA 110 GGTAEVAPDA 170 LLTAVLLAFH 230 AVQVGGALAA 290 GDAFLWKQLG 350 LLLSVLPQHV 410 VMTLNELFAR 470 VREVTGVNVN 530 LQYLNGDYEV 590 MPRWVPDRAF 650 RKDHVRRFLL 710 YASIFLLLLI 770 MFTCNHTPIR 830 LLSLLASSVF 890 FDGLDCPAAG 950 QAYNRRLLHN 1010 60 EPPSPTPAGP 120 VPRSGRSCWR 180 AAPARPQPAY 240 DPRSPSAGLW 300 ANVLLFLCTN 360 AMEMKEDINT 420 FDKLAAENHC 480 MRVGIHSGRV 540 EPGRGGERNA 600 SRTKDSKAFR 660 TFQREDLEKK 720 TVLICAVYSC 780 SCAARMLNLT 840 LHISSIGKLA 900 RVALKYMTPV 960 ILPKDVAAHF 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 Extracellular 0 1 1 r P er Plasma membrane K Cytplasm Q43 Number each of the following in order of its numerical Value with l being the largest Very brie y explain your reasoning 3 the number of solute molecules moved by one cycle of a Na glucose symporter 4 the number of membranespanning domains in a receptor tyrosine kinase like the NGF receptor I the number of membranespanning domains in a Na channel 2 the number of ions moved across the membrane in a single cycle of a NaK ATPase 5 the amount of net charge across the membrane generated by one cycle of a NaK ATPase The numbers are respectively symporter 3 moved 2 Na plus I glucose would be 2 or 4 with other stoichiometries NGFR I membrane span Na channel 24 membrane spans NaKATPase 5 moved 3Na plus 2K net charge ofone cycle ofNaKATPase I Q44 We 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 functions at right Membrane protein 1 NGF receptor AD 2 ligandgated Na channel ACE 3 Naglucose symporter ABCE 4 NaK ATPase ABDE Function A interacts with species form outside the cell B transports species across the membrane AGAINST their concentration gradient C transports species across the membrane DOWN their concentration gradient D has enzyme activity E can affect he magnitude of the Na gradient across the membrane Q45 Number each of the following in order of its numerical Value in mV with l being the greatest Use as your reference the squid axon that we looked at quantitatively in lecture 4 Vm of neuron prior to stimulation of depolarization I Vm of 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 Vm of neuron in a region X of the axonal membrane Where a depolarization has been stimulated at the time when Kl permeability is rising and Na permeability is falling 5 Vm of 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 Vm 3 Vm of 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 ofthese points in the action potential the valuesfor Vm 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 4050 mV where Kpermeability is rising and Na permeability isfalling 0 mV after depol and repol before Cl leak channels restore 80 mV adjacent region that s starting to respond but hasn t depol d 50 m V gtxgt 70 m V Q46 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 neuron innervating your shoulder with an axon 10000 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 Which do you think would have the greatest surface area Q47 Which ofthe following best describes the role of the sodiumpotassium ATPase in generating an action potential in neurons A The NaK ATPase propagates the action potential along the axon B The NaK ATPase determines the exact resting membrane potential of the neuron C The NaK ATPase is responsible for the outward flow of K ion during the action potential D The NaK ATPase is responsible for the inward flow of Na ion during the action potential E The NaK 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 NaKATPase plays no direct role in generating the action potential However the NaKATPase is largely responsiblefor establishing the initial conditions in which sodium andpotassium are out ofequilibrium and along with potassium leak channels helps establish and maintain the resting membrane potential Q48 Here39s 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 um 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 uMhr A Based on this number how many umoles of glucose must be entering the cell per hour Assume that onehalf of the volume of the cell is composed of organelles and thus not available for diffusion of glucose Glucose accumulates at 50 uMper hr 50 umol hr L But the cell is much smaller than I Liter in volume The volume ofthe cytosol here is half the volume ofa sphere with radius 10 um so 437tr3 43 X 31416 gtlt I0um3 4I89um3 gtlt ILI0 um3 4I89gtltI0 L and hal7 ofthat 2094gtlt10 12L So glucose enters the cytoplasm at a rate of 2094gtlt10 L x 50 umolhrL 1047gtlt101quot um0lhr B What is the flux of glucose into the cells expressed in moles seccmz Flux is the rate ofmovement ofglucose across the surface area ofthe cell membrane You know the rate at which glucose is entering and you can calculate the surface area ofthe spherical cell Surface 47cr2 I257um2 X I cmI04 um2 I257gtltI05 cm2 The amount ofglucose accumulating sec I047gtltI0 10 umolhr gtlt Ihr3600 s 2908gtltI0 umols X I molI06 umol 2908gtlt10quot20 molS So the ux ofglucose is 2908gtltI03920 mols I257gtltI0 5 cm2 2313gtlt10quot5 m0ls cmz C If electron microscopy indicates that there is on average 1 Naglucose symporter per lII12 of the plasma membrane at what rate must each symporter transport glucose in molecules per second First convert the ux to moles per sec per um239 2313 X101 mols cm2 X 1 cm2108um2 2313gtlt10 23 mols um2 Next multiply this by Avogaclro s number to get the molecules per sec per um239 2313 X10123 mols um2 gtlt 6022 X1023 molecules mole 1393 molecules s um2 Since there is 1 symporter um2 it must transport at a rate of1393 molecules ofglucose sec Q49 A sea urchin egg is a sphere With a diameter of 100 uM The concentrations of the major cations K Na and Ca 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 i intracellular concentration extracellular concentration potassium 200 mM 10 mM sodium 20 mM 450 mM calcium 100 nM 10 mM Use the Nernst equation to ncl out what the equilibrium potentialfor each ion is VX 59 mVZ l0gXouXn VK 59mVI gtlt log I0mM200mM 768 mV VN 59mVI gtlt log 450mM20mM 798 mV 1 Va 59mV2 gtlt log I0mM0000ImM I475 mV The upper and lower limitsfor Vm ofthe cell are the highest and lowest equilibrium potentials or 1475 m Vand 768 m V Q50 Some 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 extracellular concentration concentration K 410 mM 15 mM Na 40 mM 440 mM Cl 100 mM 560 mM Ca 2gtltlO 4 mM 10 mM pH 76 80 If the resting membrane potential Vm 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 Vx 59m Vz gtltlogX0uXl to calculate equilibrium potential for each ion and compare to the membrane potential 70 mV Ask which direction must this ion flow to getfrom the VM to the ion s equilibrium potential VX For K VK 59m V39 logI5mM4I0mM 847 m V will leave cell to reach equil For Na VM 59m V log440mM40mM 614 mVwill enter cell For Cl VC 59m V I log560mMI00mM 441 mV will leave cell For Ca Vca 59m V2 logI0mM0000ImM 1475 mV will enter cell For H VH 59m V 8 76 236 m V will enter cell Q51 Intestinal epithelial cells have a Naglucose 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 Nain 10 mM Naout 100 mM and glucoseout 10 uM 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 kJVmol First determine the AG of the Na gradient We will need to use the relationship AGM zFVm VM sofirst we will determine VM using the Nernst equation VN 1 59 mV log Na0uNa 59 mV log10 59 mV Then given that Vm 70 mVand for Na z 1 we solveforAGNa39 AGM F 70mV59mV zF 129 mV 102 kJV mol0129 V 129 kJmol Ifall ofthe energy in the Na gradient can be used to support glucose transport then AG 2AGN glucose So AGguc 2 129 kJmol RTlngluc gluc0u 258 kJmol RTlngluc gluc0u 258 kJmol 5 7 kJmol loggluc gluc0u log glucmgluc0u 336gtlt104 Thus if glucjom 10uM 001 mM then glucjm 001 mM gtlt 336gtlt104 336 mM Q52 The 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 Vm VK VNa2 If Kin 140 mM and Nain 10 mM for the muscle cell and Naout 150 mM and Kout 5 mM input values randomized what will be the membrane potential of the endplate region of an acetylcholinestimulated muscle cell Use the Nernst equation to calculate the individual equilibrium potentials forpotassium VK 59m V log5mMI40mM 854 mV for sodium VN 59m V logI50mMI0mM 694 m V Ifthese are the only two ions and ifeach contributes equally then the Vm average ofthe two or 80 mV Q53 Consider the following cell it has a transmembrane potential of 0 mV and has the following intracellular and extracellular ion concentrations Nain 20 mM Naout 200 mM Kin 150 mM Kout 50 mM Number each of the following in order ofits absolute Value with 1 being the largest Very brie y explain your reasoning 3 potential energy stored in the K gradient 2 potential energy stored in the Na gradient 1 energy ofthe ATP used by the NaK ATPase to generate the Na and K gradients 4 potential 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 Kl gradient 3 because the Vm is zero and the concentration gradient outin is I0for Nal vs 033 for K No enzyme certainly not the NaK ATPase is perfectly e icient so the amount ofenergyfrom ATP used to generate the Na and K concentration gradients I will always be greater than the potential energy stored in those gradients Since it is greater than the sum ofthe two gradients it is certainly greater than either of them taken separately Finally the existence ofa concentration gradient and therefore the potential energy stored in it depends entirely on the integrity ofthe membrane the potential energy ifyou poke holes in the membrane 4 will be zero Q54 You are studying a spherical eukaryotic cell that has a diameter of 20 um 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 brie y adding an inhibitor of all glucose breakdown you are able to determine that glucose accumulates in the cell at the rate of 50 uMhr 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 of50 uMhr so since 2 Na ions enterfor each glucose molecule Na must be entering at 100 uMhr Normal intracellular Na is 50mM so the symporters would have to run for 5 hours without any other adjustment in order to change the intracellular Na just 1 B By what mechanism could the cell run its Naglucose symporters all of the time WITHOUT increasing the Na in the cell It could also operate the NaK 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 NaK ATPase correct The energy of the glucose disequilibrium across the membrane Q55 A protonophore 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 Ap 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 Q56 If you profoundly restricted the supply of oxygen to the mitochondrion how would each ofthe 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 1 II and 111 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 ofthe 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 CO2 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 Q58 Suppose 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 O2 they ll 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 Ap that will drive the ATP synthase D Lower the pH of the solution creating a proton gradient and Ap that will drive the ATP synthase Electron transport consumes 02 and pumps F out ofthe matrix into the intermembrane space Thus depriving the mitochondria of02 lessens the H7 gradient across the MM and lessens Ap Ifyou lower the pH ofthe bathing solution this willproduce 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 acidifyfurther the bathing solution What happens ifyou get carried away with this approach eg you lower the pH ofthe bathing solution to 3 or 2 or I Hint don t putyourfinger in there Q59 When 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 mitochondria s 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 Suppose that you are able to manipulate the membrane potential Vm of a preparation of isolated mitochondria You measure the pH of the mitochondrial matrix and find it to be 80 You measure the bathing solution and find its pH to be 70 You clamp the membrane potential at 5 9 mV that is you force the matrix to be 5 9 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 kJV mol A What is the ApH for these mitochondria pN PHIMS pHmatrix 78 1 B What is the protonmotive force Ap Vm 59mV ana39ApH 1 so Vm 59 mV gtlt ApH 59mV 59mV I 0 A19 C What is the AG for protons AG FAp 0 D Under these circumstances can the mitochondria use the proton gradient to drive the synthesis of ATP N0 There is NO protonmotiveforce 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 Ap is always 0 Q61 Consider a mitochondrion with a pH in the matrix 84 and the pH of the intramembrane space 65 The matrix is 100 mV more negative than the IMS or cytoplasm If the ratio of ATP to ADP and Pi is such that AG for ATP hydrolysis 60 kJmol 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 Ap across the mitochondrial membrane pHIMS pHmatrix 0 Ap Vm 59mVApH I00mV 59mV gtlt I9 212 mV B If the ratio of ATP to ADP and Pi is such that AG for ATP hydrolysis 60 kJmol 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 AG FNA19 102 kJV mole02I2V 212 kJmole protons this is the energy available note thatfor clijfusion ii the mitochonclrion which e ectively means out ofcytoplasm AG is e ectively 212 kJmole Ifit takes 60 id to make one mole ofATP then the movement into the matrix of 23 moles of protons will be required to synthesize one mole ofATP Q62 A very 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 lquot being the earliest and 4quot 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 ATPdependent proton pump was freed to run in reverse and synthesize ATP respiration 3 B Anaerobic procaryotes ruled the earth They used abioticallygenerated 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 I C An early anaerobic eucaryote engulfed a respirationdependent 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 quotstartquot sequences or signal sequences which initiate the start of insertion into the ER membrane and there are also quotstopquot sequences which halt the insertion into the ER Consider an mRNA that has the following organization randomized numbers for each student 539AUG50 codonsstart sequence75 codonsstop sequence25 codonsstart sequence100 codonsUAG339 A If the start and stop sequences are each 20 amino acids in length how long is the completed protein 311 residues The completeprotein is I 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 Nterminus and the Cterminus of the protein and the amino acid residue number at the ends of each membrane span M v E1 1 EH5 pp 0S 1 W 3 fhEZDE FT i EH muisir ei 311 q Q64 For each of the general membrane protein arrangements AE 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 2030 codons that are mainly hydrophobic a box under an AUG indicates an Nterminal hydrophobic sequence If an mRNA matches none of the proteins choose none order randomized for each student N Cytoplasm A I Outside J C Cytoplasm quot1 r C Cytoplasm C I Outside J C D Cytoplasm N B I I Outside KJ C C 1 Cyteplasm I I E Outside 710716 710716 AUG Outside AUG AUG AUG U G l Ijli lei AUG Q65 You have isolated and sequenced an mRNA for a membrane protein and determined that it has six stretches of 2030 codons that are hydrophobic enough to serve as membrane spanssignal 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 Now in order to study the potential importance of your putative membranespanning 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 ofthis 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 ofthis B N cytoplasm 09 Pu FC cytoplasm m A y B l I I utsida L p D T C C Outside cytoplasm K t ltDgt ll l Outside 3 J I J cytoplasm l l N 0 Outside C a C N cytoplasm cytoplasm V e ltEgt l H ll l I outside T W Outside l U C C N o cytoplasm C CYT0l3quot35m 1 t N r K as I ll l ltHgt l l outside U U q Outside J Q66 Consider the 5span integral membrane protein shown below The sides of the membrane N and Ctermini and residue numbers at the boundaries of the membrane are shown N Cytoplasm quot5039Uquot 0 50 320 400 500 600 Lumen or 70 300 42oU48o 620 extracellular 750 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 MW A B C D E expect D I 25kD I 20kD 15kD 3 t 1 I 10kD C 3 I 1 I 5kD 1 Q67 Consider 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 20quot 1 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 SDSPAGE lt smear No trypsin trypsin ii Next you translate the protein in vitro in the presence of microsomes You divide the resulting proteinmicrosome solution into two portions treat one portion by adding the same mixture of proteases to the solution and again analyze the results 20 kD gt j 16 KD gt L No trypsin trypsin iii Finally you study the protein as it actually resides in the plasma membrane of live cells You treat some of the cells by 2okDgt 2 adding the same mixture of proteases to the culture medium recover the protein from the membrane and again analyze the 6 kDgt 1 resulting fragments on SDS gels No trypsin trypsin To 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 N0 compare the size ofthe protein translated with or without microsomesj 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 C 200 N 200 N Cytoplasmo Cytoplasm 60 Cytoplasm 180 CytQpaSA4D O t d 60 160 Outside 80 Outside 160 u SI e C N Outside GUM 200 200 Q 68 You have seen that the hydrophobic signal sequences in proteins can be either N terminal beginning with the first few residues of the protein or internal occurring farther along in the primary sequence Let s 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 Nterminal 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 XS b X s N correct c XIN d XSI e X 2HrS Q69 Consider 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 Nterminus of the protein and the second specific the Cterminus 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 14 are shown below The blots were probed with either the antiNterminus antibody or with the antiCterminus antibody 1 2 3 4 1 2 3 4 1 2 3 4 PG C 1 rjn 1 1 in 1 104 KD 54 KD bl 42 P 1 4 31 kt E E quotquot SDS gel AntiNterm antibody AntiCterm antibody western blot western blot 1 Which terminus A or B is the N terminus A 2 Which A or B is the C terminus B 3 Determine the residue number at location assume 1 aa residue 100 Da B I C 400 D 420 extraceillllutar E 920 D E H F 940 I G 1940 3 F G H 1960 cytoplasm I 2250 A Detailed solution Gel lane I shows that the protein has a MW of225kD lane 3 shows that there is no posttranslational cleavage ofthe protein in the secretory pathway This means that there was no Nterminal signal sequence and the mature protein contains the whole translation product Next gure out on which side ofthe membrane the N and Ctermini are located Lane 4 ofthe antiNterminus western blot shows that the Nterminus ofthe protein is on the cytoplasmic side ofthe membrane protease digestion from that side destroys the N terminus while lane 2 ofthe antiCterminus blot shows that the Cterminus is on the extracellular side digestionfrom that side destroys the Cterminus Next determine how thefragments are orientated in the protein Lane 4 ofthe antiN terminus blot also shows you that the Nterminus is at the end ofthe 42kD fragment produced by extracellular proteolysis Lane 2 ofthe antiCterminus blot shows that the C terminus is at the end ofthe 3IkD fragmentproduced by cytoplasmic side proteolysis Gel lane 2 shows that ifyou proteolyze the portions ofthe protein exposed on the extracellular surface 79kD is eliminated and two fragments I 04kD amp 42kD remain protected by the membrane We already know that the 42kD fragment runsfrom the N terminus to the end ofthefirst membrane span Therefore the I04kDfragment must run from the beginning of2 d membrane span to the end ofthe third membrane span Gel lane 4 shows that ifyou proteolyze the intracellular portion ofthe protein I40kD is eliminated and twofragments 54kD amp 3IkD remain protected by the membrane We already know that the 3IkDfragment runsfrom the beginning ofthe 3 membrane span to the Cterminus Therefore the 54kDfragment must run from the beginning ofthe I membrane span to the end ofthe 2quotd membrane span Altogether we have this Q70 You 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 temperaturesensitive 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 mannose6phosphate B Deletion of this protein s Cterminal KDEL sequence C Deletion of this protein s Cterminal 46 amino acid hook motif D Change in the mannose6phosphate receptor so that it cannot recognize its ligand E Deletion of an Nterminal hydrophobic signal sequence from this protein F Deletion of the LDL receptor G Addition of an Nterminal 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 Failure of a receptor the transferrin receptor to be internalized by endocytosis Secretion of signal peptidase which normally resides in the ER Failure of any degradative enzymes to reach the lysosome Failure of signal peptidase to be inserted into the ER rm U mom Q71 Suppose 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 AE The cytoplasmic concentrations of the molecules as a function of time are plotted below A B Concentration O I I O I I 0 1 0 2 0 0 1 O 2 0 Time Time C D Concentration 0 I I 0 I I 0 1 O 2 0 0 1 O 2 0 Time Time E Concentration 0 I I 0 10 2 0 Time 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 I and 2 and thus are not likely to be 2 messengers A does change concentration between I and 2 and is a possible 2 messenger C changes concentration between I and 2 and THEN returns to its original concentration after time 2 it is the best candidate Q72 Consider a hypothetical cell whose function is to proliferate when a wound occurs in order to fill in the wound with new tissue We ll call this cell a preepithelial 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 slowgrowing 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 preepithelial cell is large enough so that you can easily microinject it with molecules Assuming that there is a cell surface receptorfor OUCH and that OUCH is the ligand consider carefully the following experimental data Note each is a separate experiment Match each conclusion below 11 1 with the experiment AH that provides the evidence for it order of conclusions is randomized A microinject 100 ng ofthe phosphatase microinject calcium do not add OUCHresult rapid mitosis microinject 100 ng of the phosphatase add OUCHresult no mitosis microinject free calcium ion do not add OUCHresult rapid mitosis Bathe cells in calciumfree medium add OUCHresult no mitosis microinject 100 ng of a phosphatase do not add OUCHresult no mitosis F550 microinject the catalytic subunit of protein kinase A cAMPdependent kinase do not add OUCHresult rapid mitosis G microinject IP3 do not add OUCHresult mitosis but it takes a long time for the cell to begin mitosis H microinject cGMP do not add OUCHresult no mitosis I microinject 100 ng of the phosphatase Cells microinject cAMP do not add OUCHresult you stimd get different results depending on how much cAMP is injected according the following graph constant amount of phosphatase J microinject cGMP add OUCHresult rapid mitosis CAMP injected K microinject cAMP do not add OUCHresult 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 I downstream kinase is hyperstimulated it can overcome the phosphatase effect I 3 Phosphatase INHIBITS the pathway B B 4 Ca 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 Ca involved in signaling probably comes from outside the cell rather D than from intracellular stores D 9 cAMP is part of the pathway after ligand binding K K 10 IP3 which releases internal Ca stores can substitute for this pathway G but is probably not part of the normal pathway G 11 Ca is involved AFTER the phosphorylation step A A Q73 We 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 2 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 Gprotein is bound to GTP F Gprotein 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 Badrenergic 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 Which 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 Which 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 JK Which 2 of the statements above A N are never true LIST THEIR LETTERS CLEARLY in the box AI 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 BN Q74 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning phosphorylated tyrosine residue 4 phospholipaseC I diacylglycerol 3 phosphatidylinositol 2 Ca ion 5 Phospholipase C I is aprotein phosphatidyl inositol 2 is aphosphoglyceride that comprises diacylglycerol PL US an inositol head group 3 diacylglycerol glycerol backbone plus two long acyl chains is larger than aphosphotyrosine 4 which has standard amino acid structure with the Rgroup a methyl residue an aromatic 6C ring amp a phosphate group on the substituted OH ofthe ring All ofthese organic molecules are larger than a single calcium ion 5 Q75 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning GTP 3 intracellular Vesicle I cAMP 5 G protein 2 GDP 4 An intracellular vesicle I is an organelle made ofthousands oflipid andprotein molecules think ofthe surface area ofeven the smallest vesicles in the cell with a diameter of30 nm ifthe surface ofa sphere 47572 and a single phospholipid occupies roughly 2 square nm how many lipid molecules are there in the two leaflets ofthe bilayer A G protein 2 is larger than the soluble small molecules GTP 3 GDP 4 and cAMP 5 are allpurine nucleotides 9 member ring and on the question ofsize di quoter by a small amount mainly in the number ofphosphates Q76 Below are schematic diagrams of the protein building blocks of the three cytoskeletal lament 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 1 2 3 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 laments that are major components of 1 muscle cells GTP binding site GTP cannot be easily exchanged C Forms bipolar laments 3 Forms polar laments 1 2 GTP binding site GTP can be hydrolyzed and B rapidly exchanged Forms laments that serve as tracks for kinesin 2 and dynein motors Interacts via long alpha helical coiledcoils 3 Forms laments that are major components of the 2 ciliary axoneme ATP binding site A Q77 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 Q78 You are assembling actin filaments in vitro using purified actin with or without accessory actinbinding 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 Seven possible classes of actinbinding proteins capping monomersequestering eg thymosin crosslinking eg filamin severing protein eg gelsolin start progress of appearance class of accessory with polymerization in EM protein added 1 uM pure 339 lag phase then steady average length actin polymerization until 80 of 2 um NONE actin is in laments 1 uM pure 339 lag phase then steady average length crosslinking actin plus polymerization until 80 of 2 um most accessory actin is in laments laments protein 1 intersect other laments 1 uM pure no lag phase steady average length nucleating actin plus polymerization until 80 of 2 um accessory actin is in laments protein 2 1 uM pure 339 lag phase then steady average length severing actin plus polymerization until 75 of 1 um accessory actin is in laments protein 3 1 uM pure 339 lag phase then steady average length monomerseq actin plus polymerization until 60 of 2 um accessory actin is in laments protein 4 depolymerizing eg cofilin nucleating bundling eg villin or fimbrin Q79 Cytoskeletal filaments are in equilibrium with their free subunits It is convenient to describe a term critical concentration Cc which is the concentration of the free subunits necessary for assembly When the concentrations of subunits exceeds Cc assembly occurs when the concentration of free subunits is below the Cc no assembly occurs Consider the following data from an in vitro assembly experiment in which Gactin globular or unassembled actin is at different concentrations and its assembly is monitored Extent of assembly actin uM Based on this graph what is the approximate critical concentration of actin for microfilament assembly 40 M Q80 The 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 LM 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 so a tc c e 2 er quot c eeEeF3A egeg tee39 2wemA Factor a a a b of aetm pg assembled T 3 into filaments a go P 1 V V 1 0 g Factor E 0 20 30 40 50 time after addition of salt minutes 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 Experiment 2 The opencircles curve shows the time course of actin polymerization in the presence of 015 uM of Factor A Experiment 3 The closeddiamonds curve shows the time course of actin polymerization in the presence of 1 uM 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 02mM At steadystate 80 ofthe actin is assembled therefore 20 ofthe actin remains as monomer Total actin concentration 1 mM39 Cc gactin at steady state I mM gtlt02 02 mM b You can tell from these data that Factor A has two different actinregulating activities Which are they Five choices Nucleating activity because it eliminates the lag phase of assembly Severing activity because it generates more filaments at steadystate Minusend capping activity because there is more assembled actin at steadystate Crosslinking activity because actin filaments form a meshwork in its presence Plusend capping activity because there is less assembled actin at steadystate Factor A has actin nucleating activity It eliminates the lag phase ofassembly as soon as salt is added actin polymer appears It also has capping endbinding activity You can tell this because the ofassembled actin at steadystate is dijferent in expt 2 Speci cally Factor A is probably a plus end capping protein because there is lei assembled actin at steady state Note also that the slope ofthe 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 don39t know from these data expt 3 had not reached steady state yet when we stopped our measurements Q81 Rigor exists 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 Pi 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 ofactin laments by the myosins thus disrupting rigor and allowing the muscle to contract again ADP will have little or no e ect Pl will have little or no e ect A nonhydrolyzable ATP analogue will cause releasefrom rigor but no subsequent contraction Ca ion will have little or no e ect because the myosin heads are in rigor and cannot interact with the sites exposed by Catroponin interactions see ECB Fig 1 743 and related text about the actomyosin cross bridge cycle Q82 The table below lists a number of molecules 112 that we have studied For each of the statements about cytoskeletal protein function af 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 5 actin 9 tropomyosin 2 GTP 6 myosin 10 IF protein 3 troponin 7 dynein 11 Btubulin 4 kinesin 8 cyclin 12 octubulin STATEMENT ABOUT FUNCTION MOLECULE a Regulates the interactions of myosin with actin in the 1 3 9 muscle sarcomere 1 9 b A plus end directed molecular motor 4 6 4 6 c A minus end directed molecular motor 7 7 d Binds E hydrolyzes ATP 4 5 6 7 4 5 6 7 e Binds hydrolyzes GTP 11 11 f Assembles into bipolar filaments 6 10 6 10 Q83 This figure from your text that we discussed in lecture shows a crosssection of a cilium or agellum 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 Q84 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning epithelial cell myosin dimer actin monomer typical actin filament microvillus A microvillus 2 is a projectionfrom the surface ofan epithelial cell 1 that contains a large bundle ofactin laments 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 Number each of the following in order of its diameter with l being the largest Very 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 ofnumerous myo brils running in parallel each ofwhich is divided into many stacked contractile units or sarcomeres 2 Within each sarcomere are interdigitated myosin thick filaments 3 and actin thin filaments 5 Intermediatefilaments 4 were so named because their diameter is between that ofthick and thin laments See chap 17 in your text Q86 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning kinesin molecule microtubule cilium ciliated cell ciliary axoneme Kinesin 5 is a motorprotein that can generate movement along a microtubule 4 A cilium 2 contains a characteristic arrangement ofmicrotubules usually 11 and is a beating appendage ofa ciliated cell 1 When the plasma membrane and any soluble proteins are extracteclfrom a cilium the remaining core ofprotein laments crosslinking proteins and motorproteins is called the ciliary axoneme 3 Q87 Number each ofthe following in order of its size with l being the largest Very brie y explain your reasoning tubulin a heterodimer microtubule atubulin monomer agellum spermatozoon A mature sperm cell or spermatozoon I has a agellum 2 with which itpropels itself The agellum has typically 11 microtubules 3 in a 92 arrangement The microtubules are polymerizeclfrom tubulin heteroclimers 4 that each consist ofone alpha tubulin 5 and one beta tubulin protein monomer
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