Ch 4 - Ch 6 Flashcards

Question

proteostasis

Answer 1

the continual maintenance of cellular protein activity accomplished by the coordination of many different pathways 3 kinds of processes contribute to proteostasis: 1) protein synthesis 2) protein folding 3) protein degradation improperly folded or unfolded protein may aggregate and contribute to disease and aging processes

Answer 2

denaturation: loss of structural integrity with accompanying loss of activity proteins can be denatured by * heat or cold * pH extremes * organic solvents * chaotropic agents: urea and guanidinium hydrochloride

Answer 3

the sequence alone determines the native conformation of proteins Experiment * ribonuclease is a small protein that contains 8 cysteines linked via four disulfide bonds * ribonuclease denatures fully with urea and 2-mercaptoethanol * when urea and 2-mercaptoethanol are removed, the protein spontaneously refolds and the correct disulfide bonds are reformed

Answer 4

it is mathematically impossible for protein folding to occur by randomly trying every possible conformation until the lowest-energy one is found

Answer 5

not random; follow a distinct path toward the native structure that's thermodynamically most favorable 1) secondary structures form first (ionic interactions play an important role in guiding these early folding steps) 2) motifs are formed (hydrophobic effect plays a significant role throughout the process) 3) domains fold (domains near the amino terminus are synthesized first) 4) entire polypeptide is folded

Answer 6

a) multiple folding pathways (folding order somewhat random); no stable folding intermediates; one stable native structure b) represents a more typical protein; multiple possible stable folding intermediates on the multiple folding pathways; possibly several stable native structures c) one stable native structure; no stable folding intermediates; few folding pathways d) stable folding intermediates on every pathway leading to one native state (e.g. motif or domain always folds quickly, but other parts of the protein fold more slowly and in random order)

Answer 7

proteins that interact with partially folded or improperly folded polypeptides; facilitate correct folding pathways or provide microenvironments in which folding can occur

Answer 8

a soluble protein that is normally secreted from the cell is secreted in a misfolded state; most of these proteins have a concentration of aromatic amino acid residues in a core region of beta-sheet or a-helix the core folds into a beta-sheet before the rest of the protein folds correctly, and the beta-sheets from two or more incompletely folded protein molecules associate to begin forming an amyloid fibril other parts of the protein then fold differently, remaining on the outside of the b-sheet core in the growing fibril the fibers are highly ordered and unbranched amyloidoses: diseases like type 2 diabetes, Alzheimer, Huntington disease, and Parkinson disease associated with amyloid fibers

Answer 9

a) disulfide bonds are covalent bonds, which are much stronger than the noncovalent interactions that stabilize most proteins. They cross-link protein chains, increasing their stiffness, mechanical strength, and hardness. b) Cystine residues (disulfide bonds) prevent the complete unfolding of the protein

Answer 10

φ = (f) and Ψ = (e)

Answer 11

a) bends are most likely at 7 and 19; pro residues in the cis configuration accommodate turns well b) cys residues at 13 and 24 can form disulfide bonds c) external surface: polar and charged residues (Asp, Gln, Lys) interior: nonpolar and aliphatic (Ala, Ile) Thr has a hydropathy index near zero thus can be found either on the external surface or in the interior of the protein

Answer 12

a) Because only a single 2,4-dinitrophenyl (DNP) amino acid derivative is found, there is only one kind of amino acid at the amino terminus (i.e. all the polypeptide chains have the same amino-terminal residue) comparing the number of moles of DNP-valine to the number of moles of protein equals the number of amino termini and thus the number of polypeptide chains b) 4 c) different chains would probably run as discrete bands on an SDS PAGE.

Answer 13

Protein B has a higher affinity for ligand X; it will be half-saturated at a much lower concentration of X than will protein A. Protein A has K_a = 10⁶ M^–1 Protein B has K_a = 10⁹ M^–1 | (when [L] = K_d, half of the ligand-binding sites are occupied)

Answer 14

a) decreases b) increases c) decreases d) decreases The affinity of Hb for O₂ is regulated by the binding of the ligands H⁺, CO₂, and BPG. The binding of each ligand shifts the O₂-saturation curve to the right – that is, the O₂ affinity of hemoglobin is reduced in the presence of ligand.

Answer 15

Θ = the fraction of occupied binding sites a) 0.13 pM b) 0.6 pM c) 7.6 µM

Answer 16

The observations indicate that the cooperative behavior of hemoglobin arises from the interaction between subunits. O₂ cooperative binding is graphically represented as a sigmoid curve. Noncooperative binding like that of myoglobin is hyperbolic. BPG bind in the cavity between the ß subunits in the T state. The four polypeptide chains of Hb communicate with each other about H⁺ binding to His of the ß subunits. When protonated, His forms an ion pair to Asp that helps stabilize the T state (deoxyhemoglobin). CO₂ binds as a carbamate group to the Hb and releases a H+, contributing to the Bohr effect. The bound carbamate also forms additional salt bridges that help to stabilize the T state and promote the release of oxygen.

Answer 17

More tightly. An inability to form tetramers would limit the cooperativity of these variants, and the binding curve would become more hyperbolic. Also, the BPG-binding site would be disrupted. BPG binds in the cleft between the two ß subunits. Oxygen binding would probably be tighter because the default state in the absence of bound BPG is the tight-binding R state.

Answer 18

The effect of pH and CO₂ concentration on the binding and release of oxygen by hemoglobin. The binding of H⁺ (amino acid residues, particularly His¹⁴⁶ in beta subunits) and CO₂ (amino-terminal end of each globin chain) is inversely related to the binding of oxygen. At the relatively low pH and high CO₂ concentration of peripheral tissues, the affinity of hemoglobin for oxygen decreases as H⁺ and CO₂ are bound, and O₂ is released to the tissues.

Answer 19

O₂ binding to one subunit affects the affinity for O₂ in adjacent subunits. the first molecule of O₂ that interacts with deoxyhemoglobin binds weakly because it binds to a subunit in the T state. Its binding leads to conformational changes that are communicated to adjacent subunits, making it easier for additional O₂ molecules to bind. T→R transition occurs more readily in the second subunit. The last O₂ molecule binds to a heme in a subunit that is already in the R state, and thus binds with much higher affinity than the first molecule.

Answer 20

The concentration of ligand at which half the binding sites are occupied is the dissociation constant. A, K_d = 2 µM B, K_d = 6 µM Because A is half-saturated at a lower [L], it has a higher affinity for the ligand.

Answer 21

Quantitative Description of Cooperativity n = Hill coefficient (the degree of cooperativity) n = 1 → no cooperativity (myoglobin) n \> 1 → positive cooperativity (hemoglobin) n \< 1 → negative cooperativity

Answer 22

The K_m measured in the presence of a competitive inhibitor is defined as aK_m. By definition, the value of a' for a competitive inhibitor is 1. a = 2; a' = 1

Answer 23

Strong binding: K_d \< 10 nM weak binding: K_d \> 10 µM

Answer 24

complementary surfaces of the binding site and ligand are preformed

Answer 25

Complementary of the binding sites and the ligand. complementary in size shape charge or hydrophobic/hydrophilic character

Answer 26

conformational changes occur upon ligand binding; both the ligand and the protein can change their conformations allows for tighter binding of the ligand allows for high affinity for different ligands

Answer 27

CO has a similar size and shape to O₂ and fits the same binding site CO binds over 20,000 times better than O₂ because the carbon in CO has a filled lone electron pair that can be donated to vacant d-orbitals on the Fe²⁺ Protein pocket decreases affinity for CO, but it sill binds about 250 times better than oxygen; histidine residue in pocket hydrogen bonds with O2 CO is highly toxic as it competes with oxygen and blocks the function of myoglobin, hemoglobin, and mitochondrial cytochromes that are involved in oxidative phosphorylation

Answer 28

Protein side chains lack affinity for O₂ Some transition metals bind O₂ well but would generate free radicals if free in solution organometallic compounds such as heme are more suitable, but Fe²⁺ in free heme could be oxidized to irreversible Fe³⁺ Capture oxygen with heme that is protein-bound

Answer 29

Breaking ion pairs between the a1 - b2 interface

Answer 30

actively metabolizing tissues generate H+, lowering the pH of the blood near the tissues relative to the lungs * CO2 produced by metabolism is exported in the form of a carbamate, which yields a proton * the rest of the CO2 is exported as dissolved bicarbonate, also producing a proton H+ bind to Hb and stabilizes the T state, releasing O2 in the tissues * carbamate forms additional salt bridges, stabilizing the T-state

Answer 31

negative heterotropic regulator of Hb present at mM concentrations in erythrocytes (produced from an intermediate in glycolysis) small negatively charged molecule binds to the positively charged central cavity of Hb, stabilizing the T states allows for O2 release in the tissues and adaptation to changes in altitude; maintains the same relative difference in O2 bound and unbound Hb

Answer 32

antibody composed of 2 heavy chains and 2 light chains light chains: one constant and one variable domain heavy chains: three constant and one variable domain variable domains make up antigen-binding site (two per antibody) that are hypervariable (confers high antigen specificity) antigens bind via induced fit

Answer 33

Glu6 → Val in the beta-chain of Hb the new Val side chain can bind to a different Hb molecule to form a strand, sickling the red blood cells untreated homozygous individuals generally die in childhood heterozygous individuals exhibit a resistance to malaria

Answer 34

column chromatography and ELISA

Answer 35

can cause conformational changes in proteins, generally required for their function (e.g. for motor proteins that control the movement of cells and organelles) allows for spatial and temporal regulation of interactions

Answer 36

muscle fiber: large, single, elongated multinuclear cell each fiber contains about 1,000 myofibrils myofibrils contain thick filaments (myosin) and thin filaments (actin); divided into sarcomeres (contractile unit) I band contain only thin filaments; A band is where parallel thick and think filaments overlap. myosin thick filaments slide along actin thin filaments Z disk serves as an anchor to which the thin filaments are attached; M line is a region of high electron density in the middle of the thick filaments

Answer 37

muscle contraction caused by a series of conformational changes to protein structure

Answer 38

availability of myosin-binding sites on actin is regulated by troponin and tropomyosin (avoids continuous muscle contraction) nerve impulse triggers the release of Ca²⁺; causes conformational changes to tropomyosin-troponin complex, exposing myosin-binding sites

Answer 39

Enzymes are catalysts They increase reaction rates without being used up by lowering the activation barrier (the free energy of the transition state) they do not affect equilibrium (∆G); any enzyme that catalyzes the reaction S → P also catalyzes the reaction P→S. The reaction reaches equilibrium much faster because the rate of reaction is increased.

Answer 40

Most enzymes are globular proteins Some RNA (ribozymes and ribosomal RNA) also catalyze reactions

Answer 41

* Greater reaction specificity: avoids side products; metabolites have many potential pathways of decomposition; enzymes make the desired one most favorable * milder reaction conditions: conducive to the cell environment * higher reaction rates: in a biologically useful timeframe * capacity for regulation: control of biological pathways

Answer 42

defined by the reactions catalyzed 1. oxidoreductases: transfer of electrons (H atoms) 2. transferases: group transfer reactions 3. hydrolases: hydrolysis reactions (transfer of functional groups to water) 4. lyases: cleavage of C-C, C-O, C-N by elimination leaving double bonds or addition of groups to double bonds 5. isomerases: transfer of groups within molecules to give isomeric forms 6. ligases: formation of C-C, C-S, C-O, and C-N bonds by condensation reactions coupled with expenditure of energy from ATP or other similar cofactor

Answer 43

The equilibrium between S and P reflects the difference in the free energies of their ground states. The free energy of the ground state of P is lower than that of S, so ∆G° for the reaction is negative (the reaction is exergonic) and at the equilibrium, there is more P than S. The position and direction of equilibrium are not affected by any catalyst. The transition state indicates a point at which decay to the S or P state is equally probable (it is downhill either way). The difference between the energy levels of the ground state and the transition state is the activation energy, ΔG^‡. Higher activation energy corresponds to a slower reaction.

Answer 44

No The transition state is not a chemical species with any significant stability and should not be confused with a reaction intermediate (such as ES or EP). A transition state is simply a fleeting molecular moment (that happens after ES forms) in which events such as bond breakage, bond formation, and charge development have proceeded to the precise point at which decay to substrate or decay to product are equally likely.

Answer 45

when several steps occur in a reaction, the overall rate is determined by the step (or steps) with the highest activation energy for enzymes with several steps that have similar activation energies, all these steps would be partially rate-limiting

Answer 46

without energy barriers, complex macromolecules would revert spontaneously to much simpler molecular forms, and the complex and highly ordered structures and metabolic processes of cells could not exist

Answer 47

1. uncatalyzed bimolecular reaction * converting two reactants into a single restricted transition state is entropically unfavorable 2. uncatalyzed unimolecular reactions * combining two ends of a molecule that's super flexible into a rigid transition state is also entropically unfavorable 3. Catalyzed reactions * **binding energy is a major source of free energy used by enzymes to lower the activation energies of reactions** * binding energy (∆G_B) comes from the formation of the ES complex by forming weak, noncovalent interactions (H-bonds, ionic interactions, and hydrophobic effect) * each weak interaction is accompanied by the release of a small amount of free energy that stabilizes the interaction * entropy cost (decreased entropy) from organizing reactive groups into close proximity and proper orientation is paid by the binding energy * weak intereactions between enzyme and susbstrate are optimized in transition state

Answer 48

acid-base catalysis: transfer of protons covalent catalysis: change reaction pathway metal ion catalysis: use redox cofactors, pKa shifters electrostatic catalysis: preferential interactions with TS the electrostatic mechanism involves binding energy, whereas the others involve transient covalent interaction with a substrate or group transfer to or from a substrate

Answer 49

transition state enzyme active sites are complementary to the transition state of the reaction; therefore, bind better to TS than substrates Sticakse * before we get to the product (broken sticks), the substrate (metal stick) needs to be bent (transition state) * lock-and-key model: stickase with magnet-lined pocket complementary in structure to the substrate stabilizes the substrate; the result is an increase in activation energy (∆G_M) * induced-fit model: an enzyme with a pocket complementary to the reaction TS helps destabilize the stick, contributing to catalysis of reaction. The binding energy of the magnetic interactions compensates for the increase in free energy required to bend the stick (lowering the activation energy - ∆G_M)

Answer 50

amino acid side chains (titratable groups) take on the role of proton donors and acceptors also can be prosthetic groups and cofactors

Answer 51

a transient covalent bond between the enzyme and the substrate changes the reaction pathway requires a nucleophile on the enzyme: serine, thiolate, amine, or carboxylate

Answer 52

involves a metal ion bound to the enzyme interacts with substrate to facilitate binding; stabilizes negative charges participates in oxidation reactions

Answer 53

determines the rate of a reaction and how it changes in responses to changes in experimental parameters: enzyme, substrate, effectors, and temperature we typically look at the appearance of the product

Answer 54

1) test tube: mix enzyme + substrate 2) start taking samples where we measure the absorbance of the product (the appearance of the product), using a spectrophotometer, to get a continuous line 3) measuring the slope (initial velocity) of the beginning of the reaction (where substrate concentration changes very little). This is where we can get a linear response to the substrate concentration. The slope shows us how fast the enzyme is working at that particular [s]; with increasing substrate concentration, the slope (initial velocity) increases 4) change substrate concentration and repeat to get different initial rates 5) once you have a whole bunch of initial rates, you plot them on the Michaelis-Menten plot, which shows us how the initial rate changes with an increasing substrate concentration; also allows us to determine Vmax and Km

Answer 55

you do a single experiment and get a ton of information out of it: a quantitative description of biocatalysis determine the order of binding of substrates elucidate acid-base catalysis understand catalytic mechanism find effective inhibitors understand regulation of activity

Answer 56

at relatively low [S], initial velocity increases almost linearly with an increase in [S] (the more substrate, the faster the reaction goes) at higher substrate concentrations, initial velocity increases by smaller and smaller amounts in response to increases in [S] eventually reaching a plateau-like V₀ region (changes in V₀ are super small as [S] increases) V₀ approaches but never quite reaches maximum velocity (V_max) the substrate concentration at which V₀ is half-maximal is K_m (Michaelis constant)

Answer 57

substrate prep contains inhibitors enzyme prep contains inhibitors substrate inhibition limitation of measurements

Answer 58

The V_max for a catalyzed reaction is observed when the enzyme is "saturated" with its substrate. Increases in [S] has no effect on the initial rate. Essentially all free enzymes have been converted to the ES form. After one enzymatic turnover, P is generated at the same rate that S is consumed; [ES] remains constant. The measured V₀ generally reflects the steady-state and analysis of these initial rates is referred to as steady-state kinetics.

Answer 59

the rate-limiting step is the breakdown of the ES complex to product and free enzyme (k₂); highly simplified assumption (one reactant, one product, no inhibitors) * at low [S], K_m\>\>[S], and the [S] term in the denominator of the Michaelis-Menten equation becomes insignificant; V₀ exhibits a linear dependence on [S] * at high [S], where [S] \>\> K_m, the K_m term in the denominator becomes insignificant and the equation simplifies to V0=Vmax

Answer 60

linearized double-reciprocal plot good for analysis of two-substrate data or inhibition also allows a more accurate determination of V_max, which can only be approximated by the Michaelis-Menten plot

Answer 61

K_m can vary greatly from enzyme to enzyme, and even for different substrates of the same enzyme K_m only represents the binding affinity (dissociation constant K_d) when the rate-limiting step is k₂ (ES → E + P) more common are cases in which the reaction goes through several steps after formation of ES; K_m then becomes a very complex function of many rate constants V_max reflects the rate-limiting step in a reaction

Answer 62

turnover number (unitless) how many substrate molecules can one enzyme molecule convert per second; equivalent to the rate constant for the rate-limiting step k_cat = V_max / [E_tot]

Answer 63

the best way to compare the catalytic efficiencies of different enzymes or the turnover of different substrates by the same enzymes high velocity vs high affinity; you can't have both k_cat / K_m diffusion-controlled limit (ideal efficiency): the rate at which E and S can diffuse together in an aqueous solution (10⁸ to 10⁹ M^-1s^-1); many enzymes have a specificity constant near this range; such enzymes are considered to be perfect

Answer 64

Km = 10 µM

Answer 65

[S] = 10 µM

Answer 66

most enzyme reactions are not simple reactions; they involve 2 substrates * sequential reaction: form a ternary complex * random order: both active sites already open * ordered: S₁ binds before S₂ (conformational change for binding site 2) * ping-pong reaction: no ternary complex is formed * the first substrate is converted to product and dissociates before the second substrate binds * part of the S1 gets left behind in the active site; the altered enzyme interacts with S2 * ex: transaminase moves an amino acid to E to form a covalently modified enzyme intermediate and pyruvate (p1) is released; alphaketoglutarate (s2) binds to modified enzyme; amine left behind is transferred to s2, which results in free enzyme and glutamic acid

Answer 67

yes with Lineweaver-Burke plot the concentration of substrate 1 is varied while the concentration of substrate 2 is held constant. This is repeated for several values of [S2], generating separate lines sequential reaction: lines intersect Ping pong reaction: parallel lines

Answer 68

inhibitors are compounds that decrease the enzyme's activity 1. irreversible inhibitors (inactivators) * bind covalently with or destroy a functional group on an enzyme that's essential for enzyme activity * form a highly stable noncovalent association * often powerful toxins; also used as drugs 2. reversible inhibitors bind to and can dissociate from the enzyme * often structural analogs of substrates or products * used as drugs to slow down a specific enzyme * competitive, uncompetitive, and mixed inhibitors

Answer 69

resembles the substrate and both bind to the active site (increase in K_m) does not affect catalysis (no change in V_max) Lineweaver-Burke: lines produced by different inhibitor concentrations intersect at the y-axis

Answer 70

* only binds to ES complex; does not affect substrate binding * inhibits catalytic function; a decrease in V_max * anapparent decrease in K_m; because the [S] required to reach one-half V_maxdecreases by the factor a' * no change in K_m/V_max * Lineweaver-Burke: lines are parallel

Answer 71

* binds enzyme at a site distinct from the active site, but binds to either E or ES * inhibits catalysis by lowering the [E] (decrease in Vmax) * the Km may increase or decrease depending on which enzyme form E or ES the inhibitor binds to most strongly * Lineweaver-Burke: lines intersect left of the y-axis * noncompetitive inhibitors are mixed inhibitors such that there is no change in Km; these are rare

Answer 72

stable molecules that are designed to resemble reaction transition states bind to an enzyme more tightly than does the substrate in the ES complex because they fit into the active site better (form a greater number of weak interactions)

Answer 73

in many allosteric enzymes, the substrate-binding site and the modulator-binding sties are on different subunits (catalytic and regulatory subunits) homotropic regulation: hemoglobin allosteric modulators should not be confused with uncompetitive and mixed inhibitors; although inhibitors bind at a second site on the enzyme, they do not necessarily mediate conformational changes between active and inactive forms usually found at the top of the metabolic pathway and at branch points

Answer 74

zymogen: inactive precursor; cleaved to form the active enzyme (via conformational change) many proteases of the stomach and pancreas are regulated this way active trypsin in the pancreas results in pancreatitis and possible pancreas cancer; trypsinogen is secreted into the lumen of our intestine and cleaved (proteolytic modification); trypsin modifies chymotrypsinogen to active chymotrypsin this type of activation is irreversible; inactivated by inhibitor proteins that bind very tightly to the enzyme active site

Answer 75

noncovalent modification (allosteric effectors) covalent modification (phosphorylation, ubiquitination) irreversible reversible gene expression (making more/fewer enzymes) protein degradation

Answer 76

catalyzes an early step in the biosynthesis of pyrimidines highly regulated by the end-products of pyrimidine synthesis (CTP); binding of CTP results in T state of the enzyme (feedback inhibition); at low levels of CTP, CTP comes off the enzyme, and the enzyme undergoes a T to R transition

Answer 77

myoglobin: high-activity enzyme hemoglobin: high activity state to low-activity state (R to T transition from lungs to tissues) ATCase: from CTP bound (low affinity), as we have a decreasing amount of CTP in the cell, the population of active ATCase increases (more and more high-affinity R state)

Answer 78

9.5 x 10⁸ years

Answer 79

The ES complex is more stable than the free enzyme. The ground state for the ES complex is at a lower energy level than that for the free enzyme, thus increasing the height of the energy barrier to be crossed to get from the native to the denatured state.

Answer 80

a) 1.7 x 10^-3 M b) 0.33; 0.67; 0.91 c) The upper curve = enzyme B; the lower curve = enzyme A. When the initial concentration of the substrate is greater than Km, the rate of the reaction is less sensitive to the depletion of the substrate at the early stages of the reaction and the rate remains approximately linear for a longer time.

Answer 81

a) 0.2 μM s⁻¹ b) 0.6 μM s⁻¹ c) 0.9 μM s⁻¹

Answer 82

a) Calculate the reciprocal values and draw a double-reciprocal plot to determine the kinetic parameters V_max = 51.5 mM/min K_m = 0.59 mM b) competitive inhibitor; in the Lineweaver-Burke plot, the V_max is unchanged and K_m is increased

Answer 83

at pH 5.2, Asp⁵² is mainly deprotonated, whereas Glu³⁵ is protonated. The pH activity profile suggests that maximum catalytic activity occurs at a pH midway between the pK_a values of the two acidic groups when Glu³⁵ is protonated and Asp⁵² is deprotonated.

Ch 4 - Ch 6 Flashcards

4) The three-dimensional structure of proteins 5) protein function 6) enzymes (116 cards)