ch5 - | r . 5—54 1. Protein Function Relationship between...

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Unformatted text preview: | r . 5—54 1. Protein Function Relationship between Affinity and Dissociation Constant Protein A has a binding site for ligand X with a Kd of 10—6 M. Protein B has a binding site for ligand X with a Kd of 104’ M. Which protein has a higher affinity for ligand X? Expiain your reasoning. Convert the KC; to Ka for both proteins. Answer Protein B has a higher affinity for iigand X. The lower KG indicates that protein B will be half-saturated with bound ligand X at a much lower concentration of X than wili protein A. Because IQ = l/Kd, protein A has Ka : 106 NI—l; protein B has Ka = 109 M“. . Negative Cooperativity Which of the foilowing situations would produce a Hill plot with RH < 1.0? Explain your reasoning in each case. (a) The protein has muitiple subunits, each with a singie ligand-binding site. Binding of ligand to one site decreases the binding affinity of other sites for the ligand. (b) The protein is a single poiypeptide with two ligand-binding sites, each having a different affinity for the ligand. ((3) The protein is a single polypeptide with a single ligand-binding site. As purified, the protein preparation is heterogeneous, containing some protein molecules that are partially denatured and thus have a lower binding affinity for the ligand. Answer All three situations would produce a}; < 1.0. An 44H (Hm coefficient) of <10 gener— ally suggests situation (a)——the ciassic case of negative cooperativity. However, closer exami— nation of the properties of a protein exhibiting apparent negative cooperativity in ligand bindw ing often reveals situation (1)) or (c). When two or more types of ligand-bmding sites with different affinities for the ligand are present on the same or different proteins in the same solution, apparent negative cooperativity is observed. In (b), the higher-affinity ligandwbinding sites bind the ligand first. As the ligand concentration is increased, binding to the lower-affuuty sites produces an nil; < 1.0, even though binding to the two ligandwbinding sites is completeiy independent. Even more common is situation (c), in which the protein preparation is heteroge- neous. Unsuspected proteoiytic digestion by contaminating proteases and partiai denaturation of the protein under certain solvent conditions are common artifacts of protein purification. There are few well-documented cases of true negative cooperativity. Affinity for Oxygen of Hemoglobin What is the effect of the following changes on the 02 affinity of hemoglobin? (a) A drop in the pH of blood plasma from 7.4 to 7.2. (b) A decrease in the partial pres— sure of 002 in the lungs from 6 kPa (holding one‘s breath) to 2 kPa (normal). (c) An increase in the BPG level from 5 mM (normal altitudes) to 8 not (high altitudes). (d) An increase in CO from 1.0 parts per rniilion (ppm) in a normal indoor atmosphere to 30 ppm in a home that has a maifunctioning or ieaking furnace. Chapter5 Protein Function S~55 Answer The affinity of hemogiobin for 02 is regulated by the binding of the ligands if“, 002, and BPG. The binding of each ligand shifts the 02~saturation curve to the right—that is, the 02 affinity of hemoglobin is reduced in the presence of ligand. (3.) decreases the affinity; (13) increases the affinity; (c) decreases the affinity; (6) decreases the affinity. 4. Reversible Ligand Binding The protein caicineurin binds to the protein calmodulin with an association rate of 8.9 X 103 Mmls“l and an overali dissociation constant, Kd, of i0 nM. Calculate the dissociation rate, ltd, inciuding appropriate units. Answer Kd, the dissociation constant, is the ratio of led, the rate constant for the dissociation reaction, to lea, the rate constant for the association reaction. Kc! = [Cd/kn Rearrange to soive for ha and substitute the known values. a, w 1rd >< a, = (10 x 10""9 M)(8.9 x 103 Mrs-i) = as x 10”“5 s“1 5. Cooperativity in Hemoglobin Under appropriate conditions, hemoglobin dissociates into its four subunits. The isolated a subunit binds oxygen, but the Og-saturation curve is hyperbolic rather than sig- moid. In addition, the binding of oxygen to the isolated or subunit is not affected by the presence of H4", 002, or BEG. What do these observations indicate about the source of the cooperativity in hemoglobin? Answer These observations indicate that the cooperative behavior—mthe sigmoid ngbinding curve and the positive cooperativity in iigand binding—“of hemoglobin arises from interaction between subunits. 6. Comparison of Fetal and Maternal Hemoglobins Studies of oxygen transport in pregnant mammals show that the ngsaturation curves of fetal and maternal blood are markedly different when measured under the sarde conditions. Fetal erythrocytes contain a structural variant of hemoglobin, HbF, consisting of two a and two 7 subunits (c2272), whereas maternal erythrocytes contain HbA (0:252). 01) (b) (c) Which hemoglob'ui has a higher affmity for oxygen under physioiogical conditions, HbA or HbF? Explain. What is the physiologicai significance of the different 02 affinities? When all the BPG is carefully removed from sampies of Him and HbF, the measured Oz-saturation curves (and conseouently the 02 affinities) are displaced to the ieft. However, HbA now has a greater affinity for oxygen than does HbF‘. When BPG is reintroduced, the 02«saturation curves return to normal, as shown in the graph. What is the effect of EEG on the 02 affinity of hemoglobin? How can the above infomation be used to explain the different 02 affinities of fetal and maternai hemogiobin? 5-56 Chapter 5 Protein Function Answer (a) The observation that hemoglobin A (HhA; maternal) is about 60% saturated at 1302 w 4 kPa (the p02 in tissues), whereas hemoglobin F (HbF; fetal) is more than 90% saturated under the same physiological conditions, indicates that HbF has a higher 02 affinity than libA. In other words, at identical 02 concentrations, libF binds more oxygen than does HbA. Thus, HbF must bind oxygen more tightly (with higher affinity) than HbA under physiological conditions. The higher 03 affinity of HbF ensures that oxygen will flow from maternal blood to fetal blood in the placenta. For maximal ()2 transport, the oxygen pressure at which fetal blood approaches full saturation must be in the region where the 03 affinity of libA is low. This is indeed the case. (b) (c) Binding of BPG to hemoglobin reduces the affinity of hemoglobin for 02, as shewn in the graph. The Og-saturation curve for HbA shifts far to the right when BPG binds (solid curves)wthat is, the 02 affmity is dramatically lowered. The (lg-saturation curve for HbF also shifts to the right when BPG binds (dashed curves), but not as far. Because the (Dz-saturation curve of HbA undergoes a larger shift on BPG binding than does that of lle, we can conclude that HbA binds BPG more tightly than does HbF. Differential binding of BPG to the two hemoglobins may determine the difference in their 02 affinities 7. Hemoglobin Variants There are almost 500 naturaliy occurring variants of hemoglobin. Most are the result of a single amino acid substitution in a glohin polypeptide chain. Some variants produce clinical illness, though not all variants have deleterious effects. A brief sample follows: HbS (sickle-cell lib): substitutes a Val for a Glu on the surface lib Cowtown: eliminates an ion pair involved in T—state stabilization lib Memphis: substitutes one uncharged polar residue for another of similar size on the surface lib Bibba: substitutes a Pro for 3. Leu involved in an or helix lib Milwaukee: substitutes a Glu for a Val Hb Providence: substitutes an Ash for a Lys that normally projects into the central cavity of the tetramer Hb Philly: substitutes a Phe for a Tyr, disrupting hydrogen bonding at the 0:351 interface Explain your choices for each of the following: (a) The Eli variant least likely to cause pathological symptoms. (b) The variantCs) most likely to show p1 values different from that of l-IbA on an isoelectric fo« casing gel. 8. 10. Chapter5 Protein Function 8-57 (c) The variantCs) most iikely to show a decrease in BPG binding and an increase in the overall affin- ity of the hemoglobin for oxygen. Answer (a) lib Memphis; it has a conservative substitution that is unlikely to have a significant effect on function. i-IbS, lib Miiwauiree, and lib Providence; all have substitutions that alter the net charge on the protein, which will change the pi. The ioss of an ion pair in lib Cowtown may indicate loss of a charged residue, which wouid also change the pi, but there is not enough information to be sure. Hb Providence; it has an Asn residue in piece of a Lys that normaliy proiects into the central cavity of hemoglobin. Loss of the positively charged Lys that normally interacts with the negative charges on BPG results in Hb Providence having lower affinity for BPG and thus higher affinity for 02. (b) (0) Oxygen Binding and Hemoglobin Structure A team of biocheniists uses genetic engineering to modify the interface region between hemoglobin subunits. The resulting hemoglobin variants exist in soiution primarily as all dimers (few, if any, 0:233 tetramers form). Are these variants likeiy to bind oxygen more weakly or more tightiy‘? Explain your answer. Answer More tightly. Ah inability to form tetramers would limit the cooperativity of these variants, and the binding curve would become more hyperbolic. Also, the BERG-binding site would be disrupted. Oxygen binding would probably be tighter, because the default state in the absence of bound BFG is the tight-binding R state. Reversible (but Tight) Binding to an Antibody An antibody binds to an antigen with a Kd of 5 X 10""8 M. At what concentration of antigen will 6 be (a) 0.2, (b) 0.5, (c) 0.6, (d) 0.8? Answer (a) 1 X 10—8 M, (b) 5 X 10"8 M, (c) 8 X HTS M, ((1)2 X 10””? M. These are calcu- lated from a rearrangement of Equation 5—8 to give {L} = aKd/(i — 8), and for this antigen- antibody binding, {L} m 9(5 X 10“8 M)/(l m 6). For example, for (a) [L] = 0.205 X lflms M)/ (0.8) 2 1 x 10—8M. Using Antibodies to Probe Structure—Function Relationships in Proteins A monoclonal antibody binds to G—actin but not to F~actm What does this tell you about the epitope recognized by the antibody? Answer 'i‘he epitope is likeiy to be a structure that is buried when G-actin polymerizes to form F-actin. . The Immune System and Vaccines A host organism needs time, often days, to mount an inunune response against a new antigen, but memory cells permit a rapid response to pathogens previously encountered. A vaccine to protect against a particular viral infection often consists of weakened or kiiled virus or isolated proteins from a viral protein coat. When injected into a human patient, the vac— cine generally does not cause an infection and illness, but it effectively “teaches” the immune system What the viral particles look like, stimulating the production of memory cells. On subsequent infection, these cells can bind to the virus and trigger a rapid immune response. Some pathogens, inciuding HIV, have developed mechanisms to evade the immune system, making it difficuit or irupossibie to develop effective vaccines against them. What strategy could a pathogen use to evade the immune system? Assume that a host's antibodies and/or T—ceil receptors are available to bind to any structure that might appear on the surface of a pathogen and that, once bound, the pathogen is destroyed. Answer Many pathogens, including iiEV, have evolved mechanisms by which they can repeat— ediy aiter the surface proteins to which immune system components initially bind. Thus the host organism regularly faces new antigens and requires time to mount an immune response to each one. As the immune system responds to one variant, new variants are created. Some 3‘58 12. 13. Chapter 5 Protein Function molecular mechanisms that are used to vary viral surface proteins are described in Part III of the text. HIV uses an additional strategy to evade the immune system: it actively infects and destroys immune system Cells. How We Become a “Stifi‘” When a vertehrate dies, its muscles stiffen as they are deprived of ATP, a state called rigor mortis. Expiain the molecular basis of the rigor state. Answer Binding of ATP to myosin triggers dissociation of myosin from the actin thin fila— ment. in the absence of ATP, actin and myosin bind tightiy to each other. Sarcomeres from Another Point of View The symmetry of thick and thin filaments in a sarcomere is such that six thin filaments ordinariiy surround each thick filament in a hexagonal array. Draw a cross section (transverse out) of a myoiibril at the following points: (a) at the M line; (b) through the I band; (C) through the dense region of the A band; (d) through the less dense region of the A band, adjacent to the M line (see Fig. 5~29b, (3). Answer (a) At M line (b) At I band {0) At dense ((1) At less dense region of region of A band A band The less dense region of the A band, aiso known as the H zone (not shown in Fig. 5—2913), is the region in which the myosin thick filaments do not overlap the actin thin filaments. When the sarcomere contracts (see Fig. 52%), the l-l zone and the I band decrease in width. Biochemistry on the interest 14. Lysozyme and Antibodies To fully appreciate how proteins function in a cell, it is helpful to have a threeadimensional View of how proteins interact with other ceilular components. Fortunately, this is possible using Web—based protein databases and three-dimensional molecular Vieng utilities. Some molecuiar viewers require that you download a program or plug-in; some can be problematic when used with certain operating systems or browsers; some require the use of conunand—line code; some have a more user—friendly interface. We suggest you go to wwumassedufmicrohio/rasmol and look at the information about RasMol, Protein Explorer, and Jmol FirstG-lance. Choose the viewer most com- patible with your operating system, browser, and level of expertise. Then download and install any software or plug-ins you may need. In this exercise you will examine the interactions between the enzyme lysozyme (Chapter 4) and the Fab portion of the anti-lysozyme antibody. Use the PDB identifier lFDL to explore the structure of the lgGl Fab fragmentwlysozyme complex (antibody-antigen complex). To answer the following ques- tions, use the information on the Structure Summary page at the Protein Data Bank (wwrcs’oorg), and View the structure using RasMoi, Protein Explorer, or FirstGlance in Jmol. (a) Which chains in the three~dimensional model correspond to the antibody fragment and which correspond to the antigen, lysozyme? What type of secondary structure predominates in this Fab fragment? How many amino acid residues are in the heavy and light chains of the Fab fragment? in iysozyme? Estimate the percentage of the lysozyme that interacts with the antigen~binding site of the antibody fragment. Identify the specific amino acid residues in lysozyme and in the variable regions of the Fab heavy and light chains that are situated at the antigen-antibody interface. Are the residues contiguous in the primary sequence of the polypeptide chains? 0)) (c) (d) Chapters Protein Function S~59 Answer (a) Chain L is the light chain and chain H is the heavy chain of the Fab fragment of this anti« body molecule. Chain Y is lysozyme. (in) At the PDB, the SCOP and GATE-l data Show that the proteins have predominantly B sec- ondary structure forming inununoglobuiinmlike Bwsandwich folds. Use the Jmol viewing utility at the PDB to view the complex. You should be able to identify the ,8 structures in the variable and constant regions of both the light and heavy chains. (c) The heavy chain of the Fab fragment has 218 amino acid residues, the light chain fragment has 214, and lysozyme has 129. Viewing the structure in the spacefiil mode shows that Iess than 15% of the total iysozyme molecuie is in contact with the combined Vb and VH domains of the antibody fragment. (d) To answer this question you may wish to use FirstGiance in Jrnol (http://firstglance.jmol.org). Enter the PUB 113 lFDL. When the moiecuie appears, check the “Spin” box to stop the molecule from spinning. Next, click “Contacts.” With “Chains” selected as the target, click on the lysozyme portion of the complex (Chain Y). The atoms will have asterisks when they are selected. Click “Show Atoms Contacting Target.” Only the atoms (in the irnrnunogiobulin chains) that are in contact with iysozyme wili remain in space-filling mode. A quick click on each atom will bring up identifying information. Repeat the process with each of the immunoglobuiin chains selected to find the lysozyme residues at the interface. In the H chain these residues include Gly31, Tyraz, Aspwo, and rl‘yrlm; in the L chain, 'l‘yrgz, Tier”, Tyrw, and ’I‘rpgz. in lysozyme, residues Asnlg, Glyza, Tyrgg, Serz“, Lysm, Giym, Thrns, Aspng, Ginm, and Arg225 appear to be situated at the antigen- antibody interface. Not all these residues are adiacent in the primary structure. in any antibody, the residues in the VL and VH domains that come into contact with the antigen are located primarily in the loops connecting the [3 strands of the [3—sandwich supersec— ondary structure. Folding of the polypeptide chain into higher levels of structure brings the nonconsecutive residues together to form the antigen—binding site. 15. Exploring Reversible Interactions of Proteins and Ligands with Living Graphs Q Use the living graphs for Equations 5—8, 5—31, 5-i4, and 5-36 to work through the foilowing exercises. (a) Reversibie binding of a ligand to a simple protein, without cooperativity. For Equation 58, set up a plot of 6 versus {L} (verticai and horizontai axes, respectively). Examine the piots generated when Ed is set at 5, 10, 20, and 100 ,uM. Higher affinity of the protein for the ligand means more binding at lower ligand concentrations. Suppose that four different proteins exhibit these four different Kd values for ligand L. Which protein would have the highest affinity for L? Examine the plot generated when Kg = 10 on. How much does 6 increase when [L] increases from 0.2 to 0.4 ,(LM? How much does 6 increase when {L} increases from 40 to 80 ,uM? You can do the same exercise for Equation 5—11. Convert [L1 to p03 and Ed to P50. Examine the curves generated when P50 is set at 0.5, 1, 2, and 10 kPa. For the curve generated when P50 : I kPa, how much does 8 change when the p02 increases from 0.02 to 0.0% kPa? From 4 to 8 kPa? (b) Cooperative binding of a ligand to a mifltisubunit protein. Using Equation 5—14, generate a bind“ ing curve for a protein and ligand with Kd = 10 p.M and n = 8. Note the aitered definition of Kd in Equation 5—16. On the same plot, add a curve for a prote'ni with Ed == 20 nM and n = 3. Now see how both curves change when you change to 73 m 4. Generate Hiil plots (Eqn 5—16) for each of these cases. For Kd = 10 nM and it = 3, what is 6 when {L} = 20 am? {(3) Explore these equations further by varying all the parameters used above. Answer (a) The piots should be a series of hyperbolic curves, with B = 1.0 as the limit. Each curve passes through 8 m 0.5 at the point on the x axis where [L] : Kg. The protein with Kd = 5 MM has the highest affinity for ligand L. When Kg = 10 ptM, doubling [Li from 0.2 to 0.4 nM (values weli beiow Kd) nearly doubles 6 (the actuai increase factor is 1.96). 3-60 Chapter 5 Protein Function This is a property of the hyperbolic curve; at low ligand concentrations, 0 is an almost linearfunction of [L]. By contrast, doubling [L] from 40 to 80 ,uM (well above Kg, where the binding curve is approaching its asymptotic limit) increases 6 by a factor of only i.l. The increase factors are identical for the curves generated from Equation Emil. (b) The curves generated from Equation 5—14 should be sigmoidal. increasing the Hill coeftb cient (rt) increases the slope of the curves at the inflection point. Using Equation 5—14, with [L] m 20 aM, Kg = 10 ,uM, and ’n = 3, you will find that 6 = 0.998. (c) A variety of answers will be obtained depending on the values entered for the different parameters. Data Analysis Problem 16. Protein Function During the 1980s, the structures of actin and myosin were known only at the resolu- tion shown in Figure 5—28a, b. Although researchers knew that the Si portion of myosin binds to actin and hydrolyzes ATP, there was a substantial debate about where in the myosin molecule the contractile force was generated. At the time, two competing models were proposed for the mechanism of force gen— eration in myosin. In the “hinge” model, 31 bound to actin, but the pulling force was generated by contraction of the “hinge region” in the myosin tail. The hinge region is in the heavy meromyosin portion of the myosin molecule, near where trypsin cleaves off light meromyosin (see Fig. 52730). This is roughly the point labeled “Two supercoiled o: helices" in Figure 5w27a. in the “81" model, the pulling force was gener- ated in the 81 “head” itself and the tail was gust for structural support. Many experiments had been performed but provided no conclusive evidence. In 1987, James Spu- dich and his colleagues at Stanford University published a study that, although not conclusive, went a long way toward resolving this controversy. Recombinant DNA techniques were not sufficiently developed to address this issue in vivo, so Spudich and colleagues used an interesting in vitro motility assay. The alga Nicolai has extremely long cells, often sevm eral centimeters in length and about 1 mm in diameter These cells have actin fibers that run along their long axes, and the cells can be cut open along their length to expose the actin fibers. Spudich and his group had observed that plastic beads coated with myosin would along these fibers in the presence of ATP, just as myosin would do in contracting muscle. For these experiments, they used a more well-defined method for attaching the myosin to the beads. The “beads” were clumps of killed bacterial (Staphylococcus oureus) celis. These cells have a protein on their surface that binds to the Fc region of antibody molecules (Fig. 5w21a). The antibod~ ies, in turn, bind to several (unknown) places along the tail of the rnyosin molecule. When bead~ antibody-myosin complexes were prepared with intact myosin molecules, they would move along Nttelloi actin fibers in the presence of ATP. . (a) Sketch a diagram showing what a bead~antibody-myosin complex might look like at the molecu~ lar level. (b) Why was ATP required for the beads to move along the actin fibers? (c) Spudich and coworkers used antibodies that bound to the myosin tail. Why would this experi— ment have failed if they had used an antibody that bound to the part of 81 that normally binds to actin? Why would this experiment have failed if they had used an antibody that bound to actin? To help focus in on the part of myosin responsible for force production, Spudich and his col— leagues used trypsin to produce two partiai myosin molecules (see Fig. 527): (1) heavy meromyosin (HMM), made by briefly digesting myosin with trypsin; HMM consists of SI and the part of the tail that includes the hinge; and (2) short heavy meromyosin (SHMM), made from a more extensive digestion of HMM with trypsin; SHMM consists of 81 and a shorter part of the tail that does not include the hinge. Brief digestion of myosin with trypsin produces HMM and light meromyosin (Fig. 5-27), by cleavage of a single specific peptide bond in the myosin molecule. Chapter5 Protein Function 8431 (6) Why might trypsin attack this peptide bond first rather than other peptide bonds in myosin? Spudich and coileagues prepared bead-antibody-myosin compiexes with varying amounts of myosin, HMM, and SHMM, and measured their speeds along Nttelia actin fibers in the presence of ATP. The graph below sketches their results. Speed of beads (Jim/S) DD Q 0 Density of myosin or myosin fragment bound to beads (e) Which model (“82” or “hinge”) is consistent with these resuits‘? Expiain your reasoning. (1") Provide a plausible explanation for why the speed of the beads increased with increasing myosin density. (g) Provide a piausibie explanation for why the speed of the beads reached a plateau at high myosin density. The more extensive trypsin digestion required to produce SHMM had a side effect: another spe- cific cleavage of the myosin poiypeptide backbone in addition to the cleavage in the tail. This second cleavage was in the 81 head. (11} Based on this information, why is it surprising that SHMM was stiil capabie of moving beads along actin fibers? (i) As it turns out, the tertiary structure of the 81 head remains intact in SHMM. Provide a plausible explanation of how the protein remains intact and functional even though the polypeptide back- bone has been cleaved and is no ionger continuous. Answer (a) The drawing is not to scale; any given cell would have many more myosin molecules on its surface. (1)) ATP is needed to provide the chemical energy to drive the motion (see Chapter 13). (c) An antibody that bound to the myosin taii, the actin—binding site, would biock actin bind— ing and prevent movement. An antibody that bound to actin would also prevent actin- myosin interaction and thus movement. 5432 Reference W WW Chapter 5 Protein Function (d) There are two possible explanations: (1) TIypsin cleaves only at Lys and Mg residues (e) (f) (a) (h) (i) (see Table 3—37) so would not cleave at many sites in the protein. (2) Not all Arg or Lys residues are equally accessible to trypsin; the most-exposed sites would be cleaved first. The Si model. The hinge model predicts that bead—antibody—HMM complexes (with the hinge) would move, but bead-antibody—SHMM complexes (no hinge) would not. The SI model predicts that because both complexes include SI, both would move. The finding that the beads move with SHMM (no hinge) is consistent only with the Si model. With fewer myosin molecules bound, the beads could temporarily fall off the actin as a myosin let go of it. The heads would then move more slowly, as time is required for a second myosin to bind. At higher myosin density, as one myosin lets go another quickly binds, leading to faster motion. Above a certain density, What limits the rate of movement is the intrinsic speed with which myosin molecules move the beads. The myosin molecules are moving at a maxi- mum rate and adding more will not increase speed. Because the force is produced in the 81 head, damaging the 81 head would probably in~ activate the resulting molecule, and SHMM would be incapable of producing movement. The 81 head must be held together by noncovalent interactions that are strong enough to retain the active shape of the molecule. Hynes, 113%., Block, S.M., White, B.'1‘., 8: Spudich, J .A. (1987) Movement of myosin fragments in vitro: domains involved in force production. 0911MB, 9534368. Enzymes 1. Keeping the Sweet Taste of Com The sweet taste of freshly picked corn (maize) is doe to the high level of sugar in the kernels. Store-bought corn {several days after picking) is not as sweet, because about 50% of the free sugar is converted to starch within one day of picking. To preserve the sweetness of fresh corn, the husked cars can be immersed in boiling water for a few minutes (“blanched”) then cooied in cold water. Corn processed in this way and stored in a freezer maintains its sweetness. What is the biochemical basis for this procedure? Answer After an ear of corn has been removed from the plant, the enzy'rneucatalyzed conver- sion of sugar to starch continues. inactivation of these enzymes slows down the conversion to an imperceptible rate. One of the simplest techniques for inactivating enzymes is heat denature- tion. Freezing the corn lowers any remaining enzyme activity to an insignificant level. Intracellular Concentration of Enzymes To approximate the actual concentration of enzymes in a bacterial cell, assume that the celi contains equal concentrations of 1,000 different enzymes in solution in the cytosol and that each protein has a molecular weight of 100,000. Assume also that the bacteria} cell is a cylinder (diameter 1.0 pm, height 2.0 are), that the cytosol (specific gravity 1.20) is 20% solu— ble protein by weight, and that the soluble protein consists entirely of enzymes. Calculate the average molar concentration of each enzyme in this hypothetical cell. Answer There are three different ways to approach this problem. (i) ri‘he concentration of total protein in the cytosoi is (1.2 g/mL) (0.20) 100,000 g/mol Thus, for i enzyme in 3,000, the enzyme concentration is 2.4 x 10—3 M 1000 2 0.24 x 3.0_5 moi/rah m 2.4 X 10”“3 M = 2.4 >< 10—61% moles of each enzyme in cell (11) The average molar concentration m volume of gel: in liters Volume of bacterial cytosol = waft m (3.14) (0.50 rim)2(2.0 am) = 1.6 am3 2 1.6 x 10"” cm3 = 1.6 x i0”12mL m 1.6 >< 10"“15 L Amount (in moies) of each enzyme in ceil is (canine g/cmBXIfi am3)(10”12 ems/amS) {100,000 g/mol)(1000) 3.8 X 10"“23 mol 1.6 x 10‘15 L = 3.8 X 10""21 mol Average molar concentration = = 2.4. x 10‘6 Incl/L = 2.4 x 10"“6 M ...
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