Biochem 3

Question 1

1st challenge: oxygen is reactive and typically oxidizes species to which it binds -> How do we bind O2 reversibly without oxidation? -> the only iron that can do this is ferrous +2 iron
-contains iron which normally chemically reacts with O2 -> rust

Answer 1

• bury O2 binding sites away from water (hydrophobic) in the heme pocket to allow for reversible binding of O2 to ferrous iron rather than the chemical rxn of ferrous iron with oxygen and water to form ferric iron
• avoids oxidation of the iron

Question 2

2nd challenge: reversible binding of ligand to tissue is hyperbolic (michaelis-menten behavior) -> how to dissociate O2 at pressures which are not near 0?

Answer 2

-oligomerize binding sites to permit positive cooperative interactions of 4 binding sites (present in hemoglobin tetramer) rather than the hyperbolic binding characteristics of michaelis-menten relationship

Question 3

3rd challenge: normal metabolism results in the generation of metabolic acids and CO2 (2 sources of protons) -> how to facilitate transport of CO2 and protons to the lungs so they can be eliminated

Answer 3

• lower pK of amino acid side chains as O2 is bound and raise pK as oxygen is released
• protons will bind as oxygen dissociates
• hemoglobin carries protons to the lungs -> release protons in lungs -> recombine with bicarbonate ions and form CO2 -> exhale

Question 4

CO2

Answer 4

• anhydrous form of carbonic acid

- carbonic acid (hydrated form) will dissociate into bicarbonate ions and protons

Question 5

hemoglobin

Answer 5

• tetramer
• 50% volume of whole blood
• millimolar (big in comparison)
• we need a lot for proton and O2 transport
• must be packaged in RBC bc if we put it in plasma the osmotic pressure would result in large amount of water to flow out of tissues into blood
• shape gives large surface

Question 6

myoglobin

Answer 6

• 8 helices
• one polypeptide of the hemoglobin (monomer)
• non helical segments- between the helices -> important for transmitting information
• binds 1 heme
• O2 binding sites are buried in a hydrophobic environment to allow reversible binding without oxidation of the iron

Question 7

heme

Answer 7

• tetrapyrrole
• four 5 member rings
• iron in the middle -> porphyrin structure
• iron is in ferrous state

Question 8

hemoglobin

Answer 8

• ferriprotoporphyrin 9 structure

- heme with an iron in it -> iron is in ferrous state

Question 9

hemoglobin chains

Answer 9

• 4 of the 6 binding sites that iron can make in a hexocoordinate structure are occupied by 4 of the Nitrogen of the tetrapyrrole (porphyrin ring)
• additional binding site provided by the protein itself -> histidine
• each of the four chains have a histidine -> F8 histidine
• 8th residue in the 6th of 8 helices
• forms a coordination with ferrous iron
• 6th position is where oxygen will bind
• ferrous irons are linked to protein through a pentacoordinate complex which becomes a hexacoordinate when O2 is bound

Question 10

hemoglobin and myglobin

Answer 10

• ferrous irons are linked to protein through a pentacoordinate complex which becomes a hexacoordinate when O2 is bound
• O2 binding sites are buried in a hydrophobic environment to allow reversible binding without oxidation of the iron

Question 11

oxygenated hemoglobin

Answer 11

• O2 bound in 6th position
• iron is hexacoordinate
• always in ferrous state -> no oxidation of iron takes places

Question 12

hydrophobic environment of hemoglobin

Answer 12

• no oxidation of iron takes place bc there is no access to water in hemoglobin
• hydrophobic residues surround the iron
• oxidation of iron to ferric iron proceeds with an intermediate that makes use of water

Question 13

myoglobin affinity for O2

Answer 13

• hard to get O2 off myoglobin until you reach very low O2 tensions in the surrounding environment of the molecule
• good oxygen buffer -> it will carry a lot of O2 but wont release effectively -> hyperbolic
• binding of O2 is hyperbolic
• high affinity for O2
• native state -> binds too tightly
• even in the lab a myoglobin with lower affinity still wont bind effectively in lungs
• never be able to transport O2 safely to tissue with myoglobin

Question 14

binding of Carbon monoxide

Answer 14

• binds just like O2 to hemoglobin
• affinity is 200-250x greater than the affinity for hemoglobin for O2
• toxic to the body
• does come off hemoglobin
• people with carbon monoxide poisoning affixiate
• carbon monoxide also blocks the function of myoglobin and mitochondrial cytochromes that are involved in oxidative phosphorylation
• Methods to displace the carbon monoxide (least to most effective):
• very high pressures of O2 in a hypobaric chamber -> reversible binding rxn
• bright light can cause carbon monoxide to be photolytically cleaved off the heme
• chemically oxidize the hemes to the ferric state -> carbon monoxide like oxygen will not bind to ferric heme -> immediately reduce iron with methylene and ascorbic acid back to ferrous state
• low levels of carbon monoxide bind to some subunits of hemoglobin instead of oxygen and shift the hemoglobin structure to high affinity form of the tetramer

Question 15

2 conformations of hemoglobin

Answer 15

• one has a low affinity for O2 -> sufficiently low to give up O2
• the other binds O2 much more tightly (in lungs)
• transition from low to high affinity gives the S shaped curve -> called positive cooperative interactions
• 4 binding sites are essential for cooperativity
• must be at least 2 conformations in order for cooperativity to take place!

Question 16

cooperativity: quantitative description

Answer 16

• cooperative proteins have multiple ligand binding sites
• hill equation
• n = hill coefficient (degree of cooperativity)
• k= affinity
• n is measured by making a hill plot
• x-axis = log of pressure of O2
• y-axis = log of the ratio of the % saturation / 100 - % saturation
• S shaped curve plotted using the hill equation gives a value of the slope of the hill plot -> nh
• nh > 1 when there are positive cooperative interactions
• when there is no cooperativity n= 1
• if there is negative cooperativity 0 < n < 1
• n can never be greater than the number of subunits in the protein and never less than 0

Question 17

positive cooperativity

Answer 17

• S shaped binding curve
• n>1 on hill plot
• binding to one site makes it easier to bind to the next (so on)
• can never be more than the number of subunits (4)
• in reality it is typically 2.8

Question 18

negative cooperativity

Answer 18

• 0 < n< 1 on hill plot
• not found in hemoglobin
• binding to first site is easier than the second (so on)

Question 19

no cooperativity

Answer 19

• independent of how many subunits are on the protein

- n=1 on hill plot

Question 20

pH effect O2 binding to hemoglobin: Bohr Effect

Answer 20

• myoglobin is independent to pH
• as pH decreases, proton concentration goes up -> affinity for hemoglobin for O2 goes down -> great for tissues!
• tissues are actively metabolizing generating the protons to facilitate the release of O2 near the lungs (catalyzed by carbonic anhydrase)
• CO2 + H2O HCO3- + H+
• tissues also release CO2 which rxn with H2O -> enzyme carbonic anhydrase converts CO2 to bicarbonate ions and protons (CO2 is a source of protons)
• H+ binds to hemoglobin and stabilizes the T state
• pronates His146 -> forms a salt bridge with Asp94 -> release of O2
• pH different between lungs and metabolic tissues increases efficiency of O2 transport
• Bohr effect

Question 21

Bohr Effect

Answer 21

• oxyhemoglobin is a stronger acid deoxyhemoglobin and gives up protons easier
• hemoglobin binds oxygen in lungs and gives up protons -> bicarbonate ions surround and react with protons -> make carbonic acid -> carbonic anhydrase converts carbonic acid to CO2 -> exhale
• exhaling gets rid of CO2 bound in tissues
• hemoglobin releases O2 it binds protons
• hemoglobin in a low pH -> dissociation of O2 (tissues)
• high pH -> O2 is tightly bound (lungs)
• protons come from the environment (plasma)

Question 22

p50

Answer 22

• pressure of O2 required to half saturate the molecule
• p50 is lower when affinity is raised
• high pH, high affinity, low p50
• low pH, low affinity, high p50
• lower pH the curve shifts right due to higher p50

Question 23

re-emphasis

Answer 23

• hemoglobin helps carry protons in tissues and release them in the lungs
• carries protons and O2 from lungs to tissues
• low pH in tissues -> release O2
• bicarbonate ions are in the plasma
• bicarbonate provides a reactant for the protons coming off the hemoglobin as it is binding O2 in lungs
• CO2 is synthesized using carbonic anhydrase -> exhaled

Question 24

allosteric modulation by 2,3-DPG

Answer 24

• 2,3-diphosphoglyceric acid; small molecule
• aka 2,3-biphosphyglyceric acid -> 2,3-BPG
• negative allosteric effector
• affinity for O2 is decreased in the presence of 2,3-DPG
• decreased affinity -> right shifted oxygen dissociation curve with an increased value of P50
• properties of 1 binding site by a ligand bound to another (allosteric)
• low pH and high DPG -> decrease affinity

Question 25

CO2

Answer 25

• decreases affinity for O2

- through the protons

Question 26

high affinity

Answer 26

• no CO2
• no 2,3-DPG
• high pH
• low p50
• left shifted

Question 27

low affinity

Answer 27

• high CO2
• high 2,3DPG
• low pH
• high p50
• right shifted

Question 28

shifting of F histidine and helix

Answer 28

• when O2 is bound to the heme iron -> pulls on the heme linked histidine (F8) in each of the 4 polypeptide chains closer together towards the protoporphyrin ring -> planar -> tertiary and quaternary changes
• FG loop is narrow -> kicks out penultimate tyrosine which drags the C-terminus with it that was involved in salt bridges btwn alpha 1 and 2
• 4 hemes
• 4 irons
• 4 oxygens
• when O2 is bound to heme the heme linked histidine needs to straighten up (tilted when no O2 bound)
• if tilt was maintained while O2 is bound there would be unfavorable van der walls overlap between the O2, heme linked histidine, heme pyrrole ring
• shift (straighenting) of the heme linked histidine causes the entire helix the histidine is apart (6th) to shift as well -> also for the alpha and beta chains -> extends all the way to the FG corner

Question 29

beta chains of hemoglobin

Answer 29

• when the beta chain is deoxygenated the FG corner accommodates a tyrosine ring -> penultimate tyrosine
• phenolic ring sticks into the FG corner in the deoxygenated chain -> drag with it the C terminal residue (histidine)
• each chain when it is deoxygenated interacts in this way -> involving a penultimate tyrosine that is in the FG corner only when the chain is deoxygenated
• when the chain is oxygenated the chain is spat out and drags the c terminus with -> opportunities for ion pairs (salt bridges?) involving the c termini of both the alpha and beta chains are broken upon oxygenation
• deoxyhemoglobin will see there are 8 ion pairs that are formed -> all 8 will be broken when the molecule is fully oxygenated
• staging of how these are broken are going to account for the bohr effect -> amino acid side chains that participate in the salt bridges give up protons when the salt bridge is broken but hold on to protons when it is formed

Question 30

allosteric proteins that display cooperative interactions: principle

Answer 30

-molecular interactions in one state that are broken when the protein is transformed into another state is a universal principle for allosteric proteins that display cooperative interactions

Question 31

tetramer

Answer 31

• essential to give rise to S shaped binding curve
• critical structural change that can occur only in the tetramer is a change in the size of cavity between the beta chains -> cavity is wider in the deoxygenated tetramer
• O2 bound would shift the equilibrium between the wide state and narrow site -> lower affinity form
• right shift, high p50
• modulates*

Question 32

quaternary conformational change

Answer 32

• mutant hemoglobins do not undergo quaternary conformational change
• change in the relationship of the subunits to each other
• O2 bound tertiary conformation changes in each subunit result in quaternary conformational changes in the tetramer
• once the 2,3-DPG comes out the quaternary conformation is taken on

Question 33

dovetail

Answer 33

• polypeptide chains dovetail with each other
• dovetails between subunits ensure that there are only two quaternary conformations of hemoglobin
• other wise the chains would collide
• if you try to slip into another conformation you will see vander walls overlap -> impossible
• two conformations allow for cooperativity

Question 34

2,3-DPG

Answer 34

• exists between the beta chains
• exists only in the cavity of deoxygenated hemoglobin tetramer
• cavity is lined with positive charges
• there are about 5 negative charges in 2,3-DPG -> high ionic interaction in the cavity
• cavity must be big enough to accommodate the 2,3-DPG protein -> only in deoxygenated state (wide form)
• the more 2,3-DPG you add to hemoglobin the lower the O2 affinity! -> higher p50

Question 35

deoxygenated structure of hemoglobin

Answer 35

• big space in the FG corner of each polypeptide chain
• big space gets occupied by the tyrosine phenolic ring thats next to the c terminus in each of the 4 polypeptide chains
• 4 polypeptides, 4 FG corners, 4 penultimate tyrosines in the pockets -> dragging of the c termini -> forms 8 salt bridges (2 per chain)
• alpha chains - only interchain salt bridges
• beta chains- 1 intrachain salt bridge and 1 interchain salt bridge
• space between the two beta chains is big enough to accommodate the polyanion 2,3-DPG

Question 36

oxygenated hemoglobin structure

Answer 36

-iron moves -> FG corner narrows
-tyrosine is kicked out
-c terminus is dragged with tyrosine -> 2 salt bridges are broken when one O2 is bound
-not as stabilized by salt bridges
-if another O2 binds another 2 salt bridges are broken (4 remaining salt bridges)
-more unstable
-if 2,3-DPG concentration falls a bit -> 2,3-DPG leaves -> beta chains get closer -> quaternary conformational change -> rupture of another 2 salt bridges
-easier for the other 2 oxygens to bind
-3rd oxygen binds -> breaks another 2 salt bridges
-at this point the only thing stabilizing the protein is 1 salt bridge on the beta chain -> cooperativity (easier to bind O2 bc fewer salt bridges resisting)
-

Question 37

amino acids on the salt bridges

Answer 37

• 4 out of 8 of them involve nitrogen groups (amino or imidazole)
• 4 bohr protons
• all have a physiological pH around 7
• when the salt bridges form the proton associated with the amino group in on -> weaker acid, relatively high pK
• breaking salt bridge -> proton comes off -> stronger acid, relatively low pK

Question 38

deoxygenated

Answer 38

takes in protons to form salt bridges

• 4 bohr protons are on the amino acid (in reality 3 due to shift in pK)
• 2,3-DPG is present

Question 39

oxygenated

Answer 39

gives up protons when breaking salt bridges

• stronger acid than deoxygenated
• 4 bohr protons come off (in reality 3 due to a shift of pK value)

Question 40

binding of each oxygen molecule results in breakage of two salt bridges and dissociation of one bohr proton

Answer 40

-true

Question 41

the progressive breakage of the salt bridges associated with O2 binding suggests that oxygenation of the protein results in progressive elimination of conformational constraints on the protein that might be conceptually correlated with progressive increases in O2 affinity

Answer 41

-breakage of these salt bridges constitutes the progressive relief of conformational constraints on the protein -> physical basis for the physiological phenomenon of positive cooperativity

Question 42

KNF model

Answer 42

• koshand, nemethy, and filmer model
• induced fit- when a ligand binds to an enzyme the enzyme undergoes (multiple) conformational change
• the specific bound subunit will undergo conformation change but the intersubunit contact will also undergo change
• conformational changes within each subunit when it is bound will produce a change in the intersubunit contacts for a olgomeric protein (bc one changed and one didnt)
• intersubunit will be altered
• changes in the free energy of the intersubunit contacts could contribute to the overall free energy of substrate binding (can be stabilizing or destabilizing)
• explains/you can have positive cooperativity (resulting from stabilizing intersubunit contacts) or negative cooperativity (resulting from destabilizing intersubunit contacts)
• lacks symmetry

Question 43

MWC model

Answer 43

• monod, wyman, and changeux model
• accomodated
• this model only works for positive cooperativity
• wanted to prove there were ONLY 2 crystal forms of hemoglobin -> deoxymolecule and the other associated with oxygenated
• 2 conformations must be in equilibirum all the time
• in absence of O2 the low affinity (of all subunits) deoxygenated quaternary conformation predominates vice versa
• if the conformation changes to high affinity quaternary conformation, ALL subunits must change simultaneously bc there are only 2 conformations
• constrained/tension -> T-state -> nothing is bound -> low ligand affinity
• relaxed conformation -> R-state -> O2 bound -> high ligand affinity
• when O2 binds it shifts the equilibrium (some T-state will switch to R-state to balance)
• % of R-state is not necessarily equal to % of subunits with ligands bound -> bc you can have partially oxygenated molecules that have undergone quaternary conformational change
• amount of conformational change may not be the same as the amount of O2 bound

Question 44

L

Answer 44

• equilibrium between the R and T states in the absence of O2
• c and L define the s shaped binding curve

Question 45

gradual shift the equilibrium balance accounts for

Answer 45

cooperativity

Question 46

gradual shift the equilibrium balance accounts for

Answer 46

cooperativity

Question 47

T-state

Answer 47

• nothing has lower affinity than the completely empty t-state
• low affinity
• constrained state
• no such thing as negative cooperativity in this model because you cant make the affinity lower than this unbound state

Question 48

limitations of MWC model

Answer 48

• only positive cooperativity or no cooperativity are allow (Not negative bc it requires a new quaternary conformation with lower affinity of T-state (which could only be induced))
• fraction of molecules in the R-state may not be the same as the fraction of molecules that have O2 bound (unlike KNF model)

Question 49

mutant MWC model

Answer 49

• there are mutant hemoglobins that have extra stabilization of the low affinity form of the protein bc they have additional salt bridges in them when the molecule is deoxygenated
• high affinity hemoglobin mutants that are missing salt bridges in the T state but are present in the R state
• easy to explain high and low affinity hemoglobins with the two state model

Question 50

calculate values of the intersubunit interactions for all the intermediates of the KNF model and L and c that will fit on hemoglobin O2 binding curve using MWC model

Answer 50

• both models fit the physiological data
• both have aspects that are correct
• two models
• degenerate model

Question 51

degenerate model

Answer 51

• describes every possible permeatation of binding and conformational change
• describes the O2 binding curve
• NWC and KNF are subclasses of the degenerate model
• considers all possibilities

Question 52

MWC accomplishments

Answer 52

• easy to explains the affects of 2,3-DPG bc it binds to only one quaternary confirmation
• consistent with the failure to find intermediate conformation (crystal) forms
• consistent with evidence that there are 2 conformations present even in the absence of ligands (O2)
• consistent with evidence for concerted conformational change (quaternary conformation change in hemoglobin)

Question 53

KNF

Answer 53

• only model to explain negative cooperativity (1>n>0)
• consistent with experimental evidence showing a strict linear relationship between % conformational change and % saturation with O2
• bohr effect in hemoglobin

Question 54

mutations in hemoglobin that destabilize the linkage between the heme and protein

Answer 54

• iron is subject to easy oxidation -> methemoglobinemia develops
• may also cause precipitation of the protein -> induces an inclusion of the RBC -> heinspotting anemia -> macrophages dont like to see lumps in RBC so they will destroy them

Question 55

mutation that destabilizes the T-state

Answer 55

• compromises a salt bridge or hydrogen bond or other kind of interaction that present only in the t state
• increases oxygen affinity

Question 56

mutation that destabilizes the R-state

Answer 56

• stabilizing forces associated with the R-state (hydrogen bonds, salt bridges..) are compromised -> hemoglobin will have a decreased oxygen affinity
• most people are heterozygous with this mutation

Question 57

sickle-cell anemia

Answer 57

• common mutation
• Glu6 -> Val in the beta chain of hemoglobin
• Val is hydrophobic but glu was charged and polar
• val causes aggregation only in the deoxygenated form (only form val is easily accessed)
• aggregation results in deformation of cell
• deoxyhemoglobin S within the RBC aggregates -> deformed into distinctive sickle shape -> tactoids
• people who are homozygous -> generally die in childhood
• shape creates difficulty passing through capillary beds -> occludes circulation -> vaso-occlusive crises
• heterozygous people have half hemoglobin S and half hemoglobin A
• there cells dont sickle unless the hemoglobin is extensively deoxygenated (will develop enough hemoglobin S)
• heterozygous people exhibit a resistance to malaria as the parasite is destroyed when the cells it has invaded sickle and are cleared
• parasite deoxygenates the RBC and causes it to be phagocytized -> eats the parasite with it

Question 58

sickle cell anemia treatment

Answer 58

• try to reduce the hydrophobic interactions (val) with chemical modifying reagents like urea
• try to reverse switchover from fetal hemoglobin, in which gamma chains (present in the fetus) have replaced the beta chains, to adult hemoglobin (the hydrophobic patch created by valine at position 6 on the beta chains facilitates the polymerization of hemoglobin S)
• lentiviruses to introduce another form of hemoglobin into patients -> this hemoglobin has a beta chain with substitution at a position that prevents aggregation -> people began to develop leukemia

Question 59

biconcave shape

Answer 59

• high SA

- extensive and rapid gas exchange

Question 60

deoxygenation

Answer 60

• high altitude
• low oxygen
• parasitic

Question 61

exposed valine

Answer 61

• exposed valine finds a hydrophobic pocket with Phe and Leu on another beta chain
• causes aggregation
• orientation of val is only exposed when it is deoxygenated
• leads to tactoid formation -> sickling -> anemia

Question 62

hemoglobin as an enzyme

Answer 62

• its only similar to enzyme in that it binds ligands reversibly
• no catalysis

Question 63

enzymes are catalysts

Answer 63

• no effect on the difference in free energy between the substrate and product
• no affect on the equilibrium between substrate and product
• do not drive the rxn forward
• lowers the activation energy in a rxn required to get to the substrate to the transition state -> does this by stabilizing the transition state (not too low that it would never leave transition state)
• transforms the reactant (substrate) to an intermediate state -> transition state before converting to product
• reduction in activation energy is achieved by a much better fit (tighter) of the transition state to the enzyme than either the substrate or product (tight enough that it can come off though)
• affects the rate at which equilibrium can be reached!
• if there is any effect on substrate or product they must be equal

Question 64

ΔΔG‡

Answer 64

• free energy to get to from the reactant to the transition state
• activation energy
• lowers free energy of transition state
• enzymes dont affect ΔG only ΔΔG‡ -> bc of this it doesnt matter if the enzyme is working forwards or backwards (concentrations and which you start with (product or substrate) doesnt matter)

Question 65

enzyme-substrate complex

Answer 65

• has a lower free energy than the enzyme and free substrate
• favorable rxn
• proceeds to form a transition state -> activation energy barrier that must be overcome in order to form product
• the enzyme-substrate complex is itself not the transition state -» there is a chemical structural modification which results in the formation of the transition state
• transition state has a higher free energy than the enzyme-substrate complex and the enzyme free substrate (highest)

Question 66

uncatalyzed rxn

Answer 66

• higher activation energy barrier

- higher energy transition state

Question 67

transition state analogs

Answer 67

• molecules resembling the transition state
• they are not actually the transition state (they look alike)
• bound much more tightly than either substrate or product
• act as powerful competitive inhibitors
• competitive because it holds on more tightly than substrate or product
• not all competitive inhibitors are transition state analogs
• transition state analogs are always competitive inhibitors

Question 68

v0

Answer 68

• the initial rate of product formation at the beginning of the rxn
• at equilibrium -> no further product is formed
• at the beginning of rxn product concentration is usually zero
• velocity of product formation is linear with time
• at first rate is fast
• as we approach equilibrium rate slows down

Question 69

Koshland analysis of catalytic mechanisms

Answer 69

• distortion and strain
• induced fit
• approximation and orbital steering

Question 70

distortion and strain

Answer 70

• imposed on the substrate by the enzyme
• enzyme doesnt move but substrate is distorted and strained
• pushed the enzyme closer to transition state

Question 71

induced fit

Answer 71

• substrate isnt distorted or strained -> enzyme is moving
• reactive amino acids in side chains of enzyme move (when substrate binds) to form groups that facilitate rxns to allow formation of transition state

Question 72

approximation and orbital steering

Answer 72

• bringing reactants of substrate rxn together facilitated by binding to enzyme
• exclude other competing reactants (like hydrolysis -> exclude water)
• also can move reactants close together so problem of diffusion that involves collision of 2 reactants is eliminated (too close)
• orbital steering- 2 substrates are positioned so their molecular orbitals perfectly overlap so the rxn can occur spontaneously
• requires extremely specific arrangement of amino acids on the protein to hold the reactants together perfectly

Question 73

lysozyme

Answer 73

• endoglycosidase (polysaccharide that cleaves in middle) that cleaves the polysaccharide component of the peptidoglycan that constitutes part of the cell wall of the bacteria
• substrate is a sugar, polysaccharide
• in tears
• alternating copolymer of N-acetylglucosamine and N-acetylmuramic acid (NAG and NAM)
• can only cleave on one side of NAM (btwn NAM and NAG -> never NAG and NAM)
• cleaves glycosidic linkage between anomeric carbon of NAM and oxygen of glycosidic link
• catalyzes rxn
• NAG and NAM are alternating in cell wall
• rxn happens without enzyme but a lot slower
• depth of the cleft (binding sites) vary -> specificity
• substrate is distorted (chair to half chair) -> enzyme doesnt move
• must be capable of acid-catalysis and stabilizing the transition state
• requires the addition of water

Question 74

space filling model of lysozyme

Answer 74

• in the middle running horizontally is a groove and two carboxylic side chains of asp52 and glu35
• substrate binding site is longer -> extended substrate binding site
• subsites function to stabilize binding sufficiently so that 1 of the sugar rings could be forced into a half chair
• binds 6 contiguous sugars in the polysaccharide NAG NAM
• only one bond is cleaved -> specificity
• longer polymers bind with polysaccharide sticking out at either end (only 6 can bind)
• NAM restricts it to sites B, D, and F
• NAG fits into ACE
• sites A,B,C bind sugars very strongly and favorable
• site D cannot take a sugar in its normal conformation and bind -> binding to favorable sites like A,B,C overcomes unfavorable binding at D
• sugar must be distorted to fit into D site -> cyclohexane like molecules (hexoses) -> form 2 structures -> boat or chair
• must distort the chair or boat into the hair chair (one side is flattened) -> the transition state -> this will fit into the D site

Question 75

consequences of distorting the sugar in the D site

Answer 75

• cyclohexane-like molecules (hexoses) are distorted into half chair -> this is the transition state
• half chair can be cleaved easily is acid-base catalyzed mechanism
• product of the cleavage is a sugar that remains in half chair -> transition state for cleaving lysozyme
• mechanism of catalysis is considered to be acid-base catalysis (Asp52 and Glu35)
• has to be N-acetylmuramic acid (NAM) you cant put N-acetylglucosamine (NAG) in

Question 76

asp52 and glu35

Answer 76

• asp and glu are deprotonated at physiological pH -> surround glu with hydrophobic residues -> protonated state
• surround asp with polar residues -> un/deprotonated
• acid-catalysis
• participate in the NAM NAG cleavage involving the sugar in the D site
• Glu35- functions primarily as an acid -> adds proton to facilitate cleavage
• forms a tetrahedral intermediate through substitution attack -> covalent catalysis
• Asp52- basic -> proton acceptor
• Asp52 creates a dipole-dipole interaction with D site
• always present in this conformation
• no induced fit of the enzyme -> its already in the “correct’ structure (substrate is distorted)
• conspire together to form transition state (carbocation)

Question 77

lysozyme end steps

Answer 77

• one the bond between NAM and NAG is cleaved -> NAM reacts chemically with Asp52
• this forms a covalent intermediate which is hydrolyzed by the Glu35 (acting as a base)
• released both ends of the polysaccharide
• glutamic acid starts as an acid and Asp starts as a base -> conspire to form transition state -> forms oxonium or carbocation -> reacts chemically with asp and glu hydrolyzes covalent intermediate to yield 2 products
• oxonium ion is resonance stabilized by double bond (changes from planar to not) -> stabilizes the transition state
• we go into half chair so they are in the same plane to form a double bond and stabilize the transition state
• oxonium ion is stabilized by negative charged asp52 -> electrostatic catalysis

Question 78

titration of histidine

Answer 78

• histidine is a reactor in the bhor effect in hemoglobin
• great participant bc its pK is around 7
• the enzyme ribonuclease cleaves the phosphodiester bond that links the sugars in RNA together -> ex. of microenvironments
• cleavage is achieved by acid-base catalysis -> facilitated by 2 histidine’s
• if you put the histidine into a hydrophobic environment it will lose its proton (uncharged imidazole) and in polar -> holds on proton (charged imidazole)
• microenvironments:
• polar = protonated = acid = charged imidazole
• hydrophobic= unprotonated = base = uncharged imidazole
• acid-base catalysis
• recall: beta chain of hemoglobin the histidine imidazole in the c terminus is shoved up against an Asp (in deoxyhemoglobin) (high polarity) -> histidine holds onto its proton (in deoxyhemoglobin)

Question 79

why does only NAM fit in the D site

Answer 79

• C site has 2 tryptophan’s (bulky)
• NAM has a lactyl side chain
• 2 tryptophans exclude the lactyl side chain of NAM -> cant fit in the C site
• NAM can fit in B, D, F NOT A, C, and E
• A,C, and E can only accommodate for NAG
• NAG occupies E site
• NAM occupies D
• cleavage only occurs between D and E bc thats where the asp and glu that perform the acid-base catalysis are located

Question 80

induced fit

Answer 80

-reactive groups on amino acid side chains will be positioned by that conformational change in the enzyme so they can participate in the catalysis while the substrate is bound

Question 81

carboxypeptidase A

Answer 81

• exopeptidase
• induced fit
• substrate is a peptide
• cleaves only the peptide bond of c terminal amino acid
• prefers nonpolar amino acids (especially tyr and phe) which move closer while hydrophobic residues move away to form catalytic center for enzyme mechanism
• when substrate is bound there are a lot of conformational changes in the enzyme
• tyr* and phe move all the way down when substrate is bound
• arg move away from binding pocket until when substrate is bound and then it moves up against the substrate
• conformational change creates a catalytic center that facilitates the cleavage of the c terminal
• glu moves -> carboxylate end is tucked away normally but when substrate binds it moves to where the substrate binds -> allows for carboxylate cleavage
• cleaves residues off one at a time
• wont touch lys or arg
• used in protein sequencing -> finds the c-terminus
• 1 problem-> it will keep cutting off the next residue until it hits a lys or arg
• has a great stable crystal structure -> soak crystal with oligopeptide substrate to observe movement
• tyr moves a lot

Question 82

carboxypeptidase A catalysis

Answer 82

• pocket in which the hydrophobic aromatic c terminus tends to stick
• pocket is made more hydrophobic and better for binding by the tyr residue
• tyr- stabilizes hydrophpobic environment but also acts as an acid
• c terminal carboxylate forms an ion pair with arg guanadinal group
• glu participates in acid-base catalysis by accepting a proton (base)
• tyr participates in catalytic rxn by helping cleave the peptide bond involving the c terminal residue (donates a proton- acid)
• metalloenzyme- obligate need for zinc -> zinc binds and tends to stabilize the binding of the substrate
• none of these residues are in the correct position until the substrate is bound
• uses acid, base, and metal to stabilize and achieve catalysis

Question 83

peptide bond cleaving for carboxypeptidase A

Answer 83

• as soon as the peptide bond is cleaved a transient covalent intermediate is formed between the enzyme and substrate
• covalent intermediate is a anhydride formed between the carboxylic acid of glu270 on the enzyme and the new carboxylic acid that was created from cleavage of c terminus of substrate
• zinc is helping out
• transition state that involves clustering of amino acids of the ENZYME (in lysozyme it was the substrate)
• subsites functions to have enough of a cluster of side chains to carryout both acid base catalysis and formation of anhydride as a transient covalent intermediate
• products= new c terminus, free c terminal amino acid that diffuses out of the enzyme

Question 84

serine proteases

Answer 84

• endopeptidase
• use for protein sequencing
• peptide substrate
• extended binding site that involves multiple subsites - each subsite has its own function
• next to the catalytic site there is a specificity site
• 3 residues involved in catalysis-> serine, his, asp -. catalytic triad
• chymotrypsin likes to cleave long hydrophobic side chain (phe, leu NOT val, ala)
• 2 unique features -> catalytic triad (in all serine proteases) and specificity pocket (unique to each serine protease)

Question 85

specificity pocket: serine proteases

Answer 85

• next to the catalytic site there is a specificity site
• determines which amino acid will be attacked by the process of catalytic hydrolysis of peptide bond
• enodpeptidases:
• if chymotrypsin has wide pocket with small hydrophobic amino acids -> gly and ser -> allows for bulky hydrophobic side chains
• trypsin has a deep pocket and likes positive amino acids -> there are negative amino acids at the bottom of the pocket -> asp
• elastase has shallow pocket, likes to cleave carboxyl side of small residues like ala -> so bulky side chains are lining the pocket -> lys, arg, leu, thr, val
• cleaving asp (neg and small) -> use lys (pos and large)

Question 86

catalytic triad

Answer 86

• serine (next to a)
• histidine (which N is next to)
• His is acid and base catalyzer
• aspartic acid
• imidazole is a diamine
• one nitrogen of his accepts a proton and the other donates
• the accepting N pulls the proton off the serine (serine’s OH will donate and have a neg charge)
• aspartic acid helps this process by donating a proton to the other N on the same his (ion pair)
• facilitate the ionization of the proton from serine (usually impossible but the position of the triad makes this possible)
• catalysis of the hydrolysis of the peptide bond
• form the active site (not necessarily next to each other but fold near)

Question 87

serine protease catalytic mechanism

Answer 87

• His is a acid and base (physiological pH)
• cleave an internal peptide bond
• substrate binds -> peptide bond becomes a tetrahedral intermediate
• form a covalent intermediate involving the serine hydroxyl (deprotonated)
• carboxyl group of the peptide bond that is going to be cleaved and the product is going to be a acyl enzyme (ester between the carboxyl of the peptide bond thats going to be cleaved and that serine hydroxyl)
• form another tetrahedral intermediate -> breaks down and releases the products of hydrolysis
• tetrahedral intermediates are examples of covalent catalysis
• tetrahedral intermediate is also high energy transition states -> needs to be stabilized

Question 88

serine protease catalytic mechanism: step 1- formation of the tetrahedral intermediate

Answer 88

• serine hydroxyl and histidine amino group can conspire with the substrate to forma tetrahedral intermediate
• his deprotonates serine (aided by polarization created via H-bond with asp) -> electrostatic catalysis
• enzyme uses serine and histidine imidazole to convert the peptide bond (usually planar) into a tetrahedral intermediate -> 1st transition state

Question 89

serine protease catalytic mechanism: step 2- break down of 1st tetrahedral intermediate

Answer 89

• break down of the tetrahedral intermediate into an ester
• ester forms between the serine hydroxyl and the freed carboxyl of the substrate from the peptide bond we just broke
• true, relatively stable acyl-enzyme intermediate
• substrate is covalently linked to the enzyme

Question 90

serine protease catalytic mechanism: step 3- attack on ester bond

Answer 90

• attack on ester bond results in formation of another tetrahedral intermediate
• second transition state

Question 91

serine protease catalytic mechanism: step 4- break down of 2nd tetrahedral intermediate

Answer 91

• regenerates the active enzyme
• his and asp are helping his extract proton from serine hydroxyl
• true catalysts are not made or destroyed before or after a true catalytic rxn -> this can happen during the rxn though

Question 92

tetrahedral intermediate: serine protease

Answer 92

• 2 are formed during rxn (both stabilized by oxyanion hole)
• both involve a negative charge
• negative charge comes from acyl group on the carboxyl of the bond being cleaved
• negative charge is stabilized by confluence of appropriate counter anions -> oxyanion hole
• carbonyl group of substrate is incorporated into the backbone through H bond-> planar -> stabilizes
• planar product of tetrahedral stabilization is acyl enzyme
• another H bond is formed as well -> stabilizes
• peptide bond is converted to negative charge oxyanion
• oxyanion hole- negative charge on the carbonyl group of substrate
• oxyanion stabilizes the transition state and allows for preferential binding to transition state
• many residues help stabilize the negative charge -> ser, asp, gly, his
• these residues not only participate in catalytic rxn but also stabilize the oxyanion on the substrate of each of the tetrahedral intermediates
• enzymes prefers the transition state

Question 93

issues of serine proteases

Answer 93

• dangerous
• digest a lot of proteins
• body deals with this by synthesizing these enzymes in the form of inactive zymogens (proenzymes)
• made in pancreas
• if it were made as inactive zymogens the pancreas would digest itself -> acute pancreatitis
• activating zymogens ->

Question 94

activating zymogens

Answer 94

• in case of carboxypeptidase -> residues at the c terminus that are cleaved off
• in the case of the serine proteases -> a single peptide bond is cleaved in trypsinogen to allow the swinging of amino acids to create the orientation you need for the catalytic triad
• catalytic triad is formed by cleaving the polypeptide chain trypsinogen- the inactive enzyme
• enteropeptidase in the intestine enterokinase cleaves trypsinogen into trypsin
• trypsin cleaves proelastase and chymotrypsinogen to their active forms

Question 95

enzyme kinetics

Answer 95

• binding and catalysis
• concentrations of substrate and product will reach equilibrium regardless of where you start
• at the beginning of the rxn the concentration of product is typically 0 and the velocity of product formation is linear with time -> V0
• one thing we want to avoid when measuring enzyme kinetics is that overtime the accumulation of product will facilitate the reverse rxn -> so we assume product is 0 at beginning

Question 96

michaelis-menten model

Answer 96

• subunits do not communicate
• if you measure V0 as a function of the substrate concentration…
• at low substrate concentration -> V0 increases linearly
• at high substrate concentration -> V0 reaches a maximal level -> levels off
• define hyperbolic shape of the dependence of velocity on substrate concentration
• two things that quantitate the behavior of enzymes -> maximal velocity and substrate concentration required to achieve half maximal velocity
• measurement of these values must be done at appropriate substrate concentration
• problem with this model: purely phenomenological

Question 97

maximal velocity (Vmax)

Answer 97

• one of the things that quantitate enzyme behavior
• never actually achieved
• asymptotically approached as you increase substrate concentration

Question 98

Km: Michaelis constant

Answer 98

• one of the things that quantitate enzyme behavior
• substrate concentration required to achieve half maximal velocity
• not the half maximal velocity
• substrate concentration that is 10x them Km -> pretty much at Vmax

Question 99

measuring Km and Vmax at substrate concentrations that are too high

Answer 99

• approach Vmax and not get any dependence of velocity on substrate concentration in that range
• dependence of V upon substrate will appear to be zero’th order in substrate
• good way to estimate Vmax but youll never be able to measure Km
• never intersects the x axis
• rate of rxn is independent of the substrate concentration at saturated conditions -> zeroth order
• hexokinase

Question 100

measuring Km and Vmax at substrate concentrations that are too low

Answer 100

• dependence of V upon substrate will appear to be first order
• neither Km or Vmax can be estimated
• but you can get a ratio of them
• goes through the origin -> cant determine Km or Vmax independently
• glucokinase
• high Km
• Vmax x [S]
• enzyme is working directly proportional to substrate
• 1st order
• right shift

Question 101

briggs-haldane model

Answer 101

• mechanistic approach
• psuedo-equilibrium
• an equilibrium between an enzyme and substrate with their enzyme substrate complex with the rate constant K1
• if you throw a substrate into a solution with an enzyme there will be a rate at which the substrate binds to the enzyme as well as reverse rate (dissociation rate)
• as enzyme binds and undergoes catalytic transformation but also some dissociation back to free enzyme and substrate -> eventually the concentration of the enzyme-substrate complex will rapidly ready a steady state (not fixed)
• there will be a magic catalytic event where the substrate converts to product
• once it is at the enzyme-substrate complex stage it can either dissociate or form product
• enzyme-substract concentration is at steady state
• the velocity of the rxn increases until most of the free enzyme is consumed
• K1- association rate constant
• K2 or K-1- dissociation rate constant from the enzyme substrate complex
• K3 or K2 or Kcat- enzyme-substrate catalysis (product)

Question 102

enzyme-substrate complex

Answer 102

• comes before the transition state

- Michaelis complex

Question 103

steady state

Answer 103

• rate of enzyme-substrate production is equal to rate of enzyme-substrate consumption
• the enzyme-substrate complex concentration remains pretty much unchanged
• as enzyme binds and undergoes catalytic transformation but also some dissociation back to free enzyme and substrate -> eventually the concentration of the enzyme-substrate complex will rapidly ready a steady state (not fixed)

Question 104

Km = K-1 + Kp / K1

Answer 104

• as the value of Km increases the enzyme tends to be less involved in generating product -> low apparent affinity
• if Km is small the enzyme is hungry for substrate and grabs it -> high apparent affinity

Question 105

lineweaver-burk transformation

Answer 105

• linearizes the hyperbolic curve
• double reciprocal plot
• tells you about Km (affinity) and Vmax (saturation point)
• the data should be in the middle of the michaelis menten hyperbolic plot -> substrate concentration is not too high or low
• invert the michaelis-menten equation -> linear equation for the inverse velocity (1/V) as a function of inverse substrate concentration (1/S)
• do not use evenly spaced value of substrate concentration (bc the plot is double reciprocal)
• as you increase Km -> slope increase
• if you increase Km -> right shift
• y-intercept = 1/Vmax
• x-intercept = -1/Km
• slope= Km/Vmax

Question 106

do not use evenly spaced value of substrate concentration for lineweaver-burk plot

Answer 106

• bc the plot is double reciprocal
• high values of substrate concentration will be bunched
• low values of substrate concentration will be spaced -> if you make a mistake it will scew the slope value
• work with midline values of substrate concentrations -> do this by first making a michaelis menten plot -> meaningful data

Question 107

a test for cooperativity

Answer 107

• tests for interaction among subunits
• make a michaelis menten plot and observe the substrate concentration that gives you 90% Vmax and the substrate concentration that gives you 10% Vmax
• take the ratio of these concentrations
• if there is no interaction (regardless of affinity) -> the ratio will always be 81
• if ratio is <81 -> binding of substrate to subunits is positively cooperative
• if ratio is >81 -> binding of substrate to subunits is negatively cooperative

Question 108

bisubstrate reactions

Answer 108

• 2 substrates getting together
• if enzyme has separate binding sites for 2 substrates there is no interaction between the sites
• each substrate is bound with michaelis-menten kinetics (monomeric)
• doesnt care which binds first-> random order

Question 109

obligate order

Answer 109

• substrate binding order matters
• binding site for the second enzyme is created when the first one binds (conformational change)
• variations: A binds to enzyme and catalyses the transformation into 2 molecules (splits A) -> one stays bound and the dissociates -> that may facilitate the binding of B

Question 110

inhibtors

Answer 110

• molecules that bind and do not undergo catalytic transformation
• all or none -> either bind to enzyme and totally block the substrate from binding or bind to enzyme and totally block catalytic event
• 2 ways of binding:
• competitive inhibitor- interfering with substrate binding (inhibitor looks like the substrate)
• non-competitive inhibitor- binds to somewhere other than the binding site -> compromising catalysis

Question 111

uncompetitive inhibitor

Answer 111

• may prevent catalysis
• obligate order rxn
• may prevent binding of second substrate
• molecules that inhibit catalysis but can only be bound once the first substrate is bound
• doesnt block second substrate -> it blocks catalysis
• facilitates binding of first substrate
• binds only to the enzyme-substrate complex -> depletes and pulls association of the first substrate onto the enzyme
• apparent Km is lower (affinity if higher)
• vmax lower, Km increase, left shift, high affinity

Question 112

hexokinases: blood glucose

Answer 112

• michaelis menten plot for 2 hexokinase -> hexokinase 1 and hexokinase 4 (glucokinase)
• glucokinase- in liver
• Km for hexokinase is in micromolar range (.2mM)
• Km for glucokinase is in submilimolar range (10mM) -> within physiological concentration of glucose in blood
• hexokinase is always saturated -> rate of synthesis of glucose x phosphate by hexokinase is determined by how much hexokinase you have
• if you double amount of enzyme you double Vmax and you double synthesis rate
• function of hexokinase is to trap glucose in cells (by phosphorylation)
• glucokinase can speed up to metabolize glucose when levels get too high and slow down when levels are low -> regulation of blood glucose

Question 113

glucokinase

Answer 113

• can speed up to metabolize glucose when levels get too high and slow down when levels are low -> regulation of blood glucose
• this is become of its Km
• Km for glucokinase is in submilimolar range (10mM) -> within range of physiological concentration of glucose in blood
• does an okay job

Question 114

catalytic efficiency

Answer 114

• Kcat/Km
• when Kcat is much smaller than K-1 -> then Km is virtually an equilibrium constant
• sometimes Kcat is so large it converts substrate into product instantly
• carbonic anhydrase- very efficient -> high value of Kcat and normal value of K-1 and K1 -> approaches diffusion control rate

Question 115

diffusion controlled rate

Answer 115

• catalytic effiency is very high
• carbonic acid
• very efficient -> high value of Kcat and normal value of K-1 and K1 -> approaches diffusion control rate
• diffusion controlled limit is about 10^8-10^9 M^-1sec^-1
• represents close to the upper limit for an enzyme catalyzed rxn

Question 116

competitive inhibition

Answer 116

• binds to an enzymes instead of substrate
• inhibitor resemble the substrate -> accounting for exclusive binding
• either or event -> either binds substrate or inhibitor (not both)
• transition state analogs- competitive inhibitors which are bound much more tightly than substrate or product -> any substrate analog may be a competitive inhibitor
• only binding is affected
• Km increase, slope of lineweaver burk increase, vmax is the same

Question 117

transition state analogs

Answer 117

• competitive inhibitors which are bound much more tightly than substrate or product
• any substrate analog is a competitive inhibitor -> not every competitive inhibitor is a transition state analog
• never forms product -> dissociates back to free inhibitor and free enzyme
• affinity for transition state analogs is always much higher than affinity for substrate
• CANNOT be converted to product -> what all transition state analogs have in common
• Kcat in presence of inhibitor is 0

Question 118

binding of substrate and inhibitor is

Answer 118

• mutually exclusive

- never bind both

Question 119

michael menten plot for competitive inhibitors

Answer 119

• Vmax/2 is the michaelis constant -> increases as concentration of inhibitor increases
• all approach Vmax in hyperbolic fashion
• in the presence of a competitive inhibitor the slope of the lineweaver burk plot is increased -> value of x-intercept is decreased, and value of y-intercept (1/Vmax) is unchanged

Question 120

lineweaver burk plot for competitive inhibitors

Answer 120

• in the presence of a competitive inhibitor the slope of the lineweaver burk plot is increased
• as concentration increases so does slope
• value of x-intercept is decreased (-1/km) as concentration increases
• value of y-intercept (1/Vmax) is unchanged
• Km increase, slope of lineweaver burk increase, vmax is the same

Question 121

important concepts for competitive inhibitors

Answer 121

• the degree of inhibition is dependent on both the inhibitor concentration and the substrate concentration
• also dependent of the relative values of Km and Ki
• if you keep increasing substrate concentration and hold inhibitor concentration constant -> degree of inhibitor decreases (vice versa)
• bigger value of Ki the easier it is for the inhibitor to fall off the enzyme (dissociate)
• smaller Ki the most tightly bound the inhibitor is
• Ki is NOT directly related to degree of inhibition
• Ki- the concentration of inhibitor that doubles the slope of the linweaver berk plot
• inhibited velocity in presence of competitive inhibitor is the same as the uninhibited velocity as the substrate concentration approaches that which would give you Vmax

Question 122

noncompetitive inhibition

Answer 122

• inhibitor doesnt bind near the substrate
• inhibitor cant tell if substrate is bound or not -> neither can substrate
• both bind independently
• no catalysis is inhibitor is bound
• affinity of enzyme for the inhibitor is unaffected by presence of absence of substrate in pure noncompetitive inhibition
• only catalysis is affected -> not binding
• Km is unaffected, Vmax decreases, upward shift -> increase in y-intercept, slope increase

Question 123

mixed inhibition

Answer 123

-presence of substrate affects affinity for inhibitor

Question 124

lineweaver burk plot for noncompetitive inhibitor

Answer 124

• intersects on x-axis
• Km for the substrate is unaffected by how much inhibitor you added -> independent
• Vmax goes down if the inhibitor is bound

Question 125

important concepts for noncompetitive inhibitors

Answer 125

• when you have a concentration of an inhibitor that gives you 50% inhibition -> that concentration is equal to Ki
• inhibited velocity in presence of noncompetitive inhibitor is always a constant fraction of the uninhibited velocity regardless of substrate concentration
• Km is unaffected, Vmax decreases, upward shift -> increase in y-intercept, slope increase

Question 126

lineweaver burk plot for inhibitor that is not bound equally by the free enzyme and enzyme substrate complex

Answer 126

• does not intersect with either axis
• gives information about which quadrant the graph is in
• this isnt important for test really

Question 127

bisubstrate rxn

Answer 127

• A and B are bound in random order
• independent
• binding of A doesnt affect B (vice versa)
• competitive inhibitor for A would be strictly noncompetitive for B (vice versa)
• competitive inhibitor for A may allow B to bind but there is no catalysis -> prevents product formation without affected binding of the other substrate)

Question 128

obligate binding order

Answer 128

• cannot bind B unless A is bound first
• in some cases A is protein to be cleaved and B is water that hydrolyze the enzyme
• competitive inhibitor of A could induce a conformational change that would not allow product to form -> would permit binding of second substrate -> functions as a noncompetitive inhibitor for the second substrate
• if you have a competitive inhibitor for the second substrate -> ties up the complex between the enzyme and first substrate
• A= oxyloacetate
• B= acetyl CoA

Question 129

competitive inhibitor of the 2nd substrate for obligate binding orders

Answer 129

• competitive inhibitor for 2nd substrate is an uncompetitive inhibitor for the first
• if you have a competitive inhibitor for the second substrate -> removes the complex between the enzyme and first substrate (not the free enzyme)
• gobbles up the enzyme substrate complex but not the free enzyme
• enhances binding of 1st substrate by mass action (by selectively removing the complex)
• competitive inhibitor of the 2nd substrate ignores the free enzyme
• perturbs the equilibrium between free enzyme and the complex between the enzyme and 1st substrate -> bc only the complex between the enzyme and 1st substrate is tied up by the competitive inhibitor of the 2nd substrate
• equilibrium of free enzyme and the complex with the 1st substrate is shifted in the presence of the competitive inhibitor for the 2nd substrate
• looks like the enzyme affinity for the 1st substrate has increased
• competitive inhibitor of the 2nd substrate increases the affinity for the 1st substrate
• no benefit for catalysis -> catalysis goes down bc inhibitor is bound
• apparent affinity for 1st substrate goes up

Question 130

law of mass action or lachatile principle

Answer 130

• equilibrium of free enzyme and the complex with the 1st substrate is shifted in the presence of the competitive inhibitor for the 2nd substrate
• looks like the enzyme affinity for the 1st substrate has increased
• competitive inhibitor of the 2nd substrate increases the affinity for the 1st substrate
• no benefit for catalysis -> catalysis goes down bc inhibitor is bound
• apparent affinity for 1st substrate goes up
• 2nd substrate binding site is induced by binding of 1st substrate

Question 131

noncompetitive inhibitor of the 2nd substrate for obligate order rxn

Answer 131

-just like a noncompetitive inhibitor for the 1st substrate
-interferes with catalysis (decreases)
-noncompetitive inhibitors of the 2nd substrate are also noncompetitive inhibitors of the 2nd substrate
-

Question 132

lineweaver burk plot for uncompetitive inhibitor

Answer 132

• if you vary concentration of 1st substrate and hold the 2nd substrate concentration constant -> in presence of increasing concentration for competitive inhibitor for 2nd substrate -> series of parallel lines
• as you add competitive inhibitor of 2nd substrate…
• y-axis- decreasing values of Vmax (increasing values of 1/Vmax)
• Km- appears to decrease (-1/Km appears to increase)
• Km (for first substrate) decreases
• looks like affinity for 1st substrate has increased
• apparent Km is lower (affinity if higher)
• vmax lower, Km increase, left shift, high affinity

Question 133

apparent affinity

Answer 133

-the reason the apparent affinity for the 1st substrate is increasing is bc the inhibitor is gobbling up the enzyme substrate complex and not the free enzyme

Question 134

smallest reduction in rate for lysozyme

Answer 134

• correct: addition of a metal-chelating agent such as EDTA to remove any free metal ions from the solution
• false: dissolving both the lysozyme and peptidoglycan in a non-polar solvent to remove water molecules from the rxn -> wrong bc we need water for the mechanism
• false: addition of a strong acid to lower the pH of the solution to below pH 2 -> this would cause both asp and glu to be protonated but we want one protonated and one deprotonated
• false: replacement of the peptidoglycan substrate with a polysaccharide composed of only nacetylmuramic acid sugar linked together by glycosidic bonds -> NAM wont fit into all the sites
• false: addition of high concentration of urea to the rxn

Question 135

microenvironment

Answer 135

• can change everything
• asp nearby can create polar microenvironment
• histidine:
• polar -> protonated
• non-polar -> deprotonated

Question 136

affinity

Answer 136

• due to multiple subunits -> cooperativity

- this is why myoglobin is unaffected by environmental changes

Question 137

which rxn condition is most compatible with making michaelis-menten kinetic measurements

Answer 137

• studying a rxn where the rate of dissociation of the substrate from the enzyme is significantly higher than the rate of conversion of the enzyme-substrate complex to enzyme and product -> understanding the equilibrium
• using enzyme concentrations that are significantly higher than the substrate concentration -> this wouldnt tell us anything about the enzyme
• measuring the rate of rxn at different initial enzyme concentrations
• takes rate measurements before each rxn has reached equilibrium -> you want to study the enzyme-substrate complex @ equilibrium
• continually adding fresh enzyme to the rxn mixture during the course of reach rxn

Question 138

Mur2Ac

Answer 138

NAM

Question 139

GlcNac

Answer 139

NAG

Question 140

CO2 transport

Answer 140

• vast bulk is transported by initial reaction with carbonic anhydrase to convert it to carbonic acid, which has a pK of around 6, and will therefore completely dissociate at physiological pH.
• bicarbonate ions remain in the plasma while the protons bind to four of the 8 ion pairs stabilizing the low affinity conformation of hemoglobin