Chapter 6 Reading Enzymes Flashcards Preview

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Flashcards in Chapter 6 Reading Enzymes Deck (75):

Q # 1 What components besides amino acid residues are sometimes necessary for an enzyme to be active?



Q # 2 What are the functions of co-enzymes? Note that many are derivatives of vitamins.

organic or metalloorganic molecule that act as transient carriers of specific functional groups.



an additional chemical component that aids in enzymatic activity


Prosthetic group

a co-enzyme or metal ion that is bound tightly or covalently bound to the enzyme of a protein.



A complete catalytically active enzyme together with its bound co-enzyme and or metal ions.



aka apopenzyme. The protein part of an enzyme


HOT LILY (International classes of enzymes)

312456; Hydrolyases, oxidoreductases; transferases; Lyases, isomerases, Ligases



EC: 1, Transfer electrons (hydride ions or H atoms)



EC 2; Group transfer reactions



EC 3: Hydrolysis reactions (transfer of functional groups to water)



EC4 ; Cleavage of C-C, C-O, C-N or other bonds by elimination, leaving double bonds or rings, or addition of groups to double bonds



EC 5; transfer of groups within molecules to yield isomeric forms



EC 6: Formation of C-C, C-S, C-O and C-N bonds by condensation reactions coupled to cleavage of ATP or similar cofactor.


Q # 4 Why are catalysts necessary for reactions in living systems to proceed at a useful rate?

Many reactions that occur in cells would not occur (or would occur at incredibly low rates) without catalysts. Most biologically active molecules are stable at cellular pH, temperature, and environment.


Active site

Region of an enzyme where its substrate binds to. The active site provides and environment for reactions to occur.



The molecule bound in the active site and acted upon by its substrate


Q # 5 What is the difference between a transition state and a reaction intermediate?

The transition state is not a chemical species. It is simply a point where decay to product or substrate is equally likely. A reaction intermediate is a chemical species other than the substrate and product individually.


Q # 6 What affects the reaction equilibrium between S and P? What affects the reaction rate of the conversion from S to P? Which aspect of a reaction can an enzyme alter?

a) The activation barrier b) Reaction rates can be increased or decreased by temperature and or pressure c) enzymes decrease the activation barrier for a reaction to occur


What do catalysts actually affect?

Catalysts increase the rate of a reaction, catalysts do NOT affect reaction equilibria. (A reaction is in equilibrium when there is no net change in the concentration of reactants or products)


Activation Energy

The energy difference between the transition state and the ground state of either the products or the reactants.


Rate limiting step

the reaction intermediate(s) that have the highest activation energy


Q # 7 What does a large, positive K'eq (e.g, K'eq = 1,000) for a reaction mean in terms of the final relative concentrations of products and reactants? What does a very small K'eq (e.g, K'eq = 0.001) mean?

A large positive value for K'eq means a large negative value for ∆G'˚ and a higher value of products to reactants. A small K'eq means a large positive value for ∆G'˚ which means a greater concentration of substrate.


Q # 9 In qualitative terms, what is the relationship between the rate constant K and the activation energy for an enzymatic reaction?

k relates to the amount of substrate that will be converted to product in a period of time.


What is K'eq?

K'eq is the equilibrium constant under standard conditions.


What is the relationship between ∆G'˚ and K'eq?

∆G'˚ = -RTlnK'eq; Large negative ∆G'˚ reflects a favorable reaction equilibrium. one in which there is much more product than substrate at equilibrium.


Q # 10 What is the specific source of energy for lowering the activation energy barriers in enzyme-catalyzed reactions?

(Binding energy)Most of the energy comes from weak, non covalent interactions between substrate and enzyme. Covalent interactions also play a role.


Binding energy

∆Gb; the major source of free energy used by enzymes to lower the activation energies of reactions. 1) Catalytic power comes from energy released in forming many weak bonds and interactions between enzyme and substrate. 2) enzymes are complementary to transition states. not to the substrates


Q # 11 Why is it important for an enzyme to be complementary to the reaction transition state rather than to the substrate?

If an enzyme was complementary to the substrate, it would actually stabilize the substrate and increase the activation energy of the reaction. ( lower activation energy arises from the sum of activation energy and binding energy)


Q # 12 What is one reason that some enzymes are very large molecules?

Enzymes must have multiple functional groups to provide multiple opportunities for weak interactions. Enzymes also need a larger structure to keep interacting groups properly positioned and to keep the cavity from collapsing.


Q # 13 How does binding energy contribute to the high degree of specificity shown by enzymes?

Each enzyme has functional groups which maximize the non covalent interactions. Specificity arises, because of the difference in interactions within the active site. Specificity arises from weak interactions in the active site.


Q # 14 Describe four physical and thermodynamic barriers to reaction, and explain how enzymatic catalysis overcomes them.

1) Entropy (freedom of motion) reduces the possibility that molecules will interact with one another. 2) The solvation shell of hydrogen bonded water that stabilizes 3) the distortion of substrates that must occur in many reactions 4) the need for proper alignment of catalytical functional groups on the enzyme. Binding energy overcomes all of these barriers.


Q # 15 Explain the lock and key hypothesis and the induced fit mechanism.

Lock and key states that a substrate fits into an enzyme like a lock would a key. induced fit states that when a substrate binds to an enzyme, that both undergo a change to INDUCE a fitting. The conformation changes that come a long


Q # 14 a) Entropy reduction

restriction in the relative motions of two substrates. Binding energy reduces entropy so that substrate and enzyme can react.


Q # 14 b) Desolvation

formation of weak bonds between the substrate and enzyme replace hydrogen bonds from the solvation shell


Q # 14 c) Distortion

non covalent bonding accounts for electron distortion


Q # 16 When is general (as opposed to specific) acid-base catalysis observed?

Specific catalysis that use H+ (H3O+) or OH- ions present in ions only; proton catalysts that use weak acids and bases other than water. General acid-base catalysis is crucial because it allows the transfer of electrons when water is not present, which is likely in the active site.


Q #17 Why must the covalent bond formed between an enzyme and substrate in covalent catalysis be transient?

It must be transient to produce products or substrate. If the covalent bond was not transient, then the Ea maybe greater than the uncatalyzed reaction.


Q # 18 In what ways can metal ions participate in catalysis?

Metals stabilize charged reaction transition states, orient substrate in the correct direction, mediate redox reactions.


Q # 19 What assumption concerning substrate concentration is made in the discussion of enzyme kinetics? Why is this important?

We look only at the beginning of a reaction, because we want [s] to be considered constant, and V0 can be expressed as a function of [s]; We measure initial velocity before significant p has accumulated.


Q # 20 Explain the effect of saturating levels of substrate on enzyme catalyzed reaction?

[s] saturation causes all enzyme to form ES, which gives us information about the maximum initial rate of a reaction (Vmax). Further increases in [S] leads to no effect on the rate.



Concentration of substrate



Initial velocity; determined by the breakdown of ES to E+P; V0 =K2[ES]



Maximum velocity






Enzyme-substrate complex



E+S to ES reaction constant



ES to E+S reaction constant



ES to E + P reaction constant


Q # 22 Why is there no K-2 in the equation describing the reaction E + S to E + P?

K-2 is ignored because it is early on in the reaction and very little product has been formed.


Q #23 What is the steady state assumption?

That the initial rate of reaction reflects a steady state in which [ES] is constant- that is, the rate of formation of ES is equal to the rate of its breakdown.


What are the two definitions of Km? What are its units?

Km is equivalent to the substrate concentration at which V0 = 1/2 Vmax ; The Michaelis constant. Km = (K-1 +K2)/K1


Michaelis-Menten equation

V0 = Vmax [S]/Km + [S]; developed by M&M hypothesis that the rate-limiting step in enzymatic reactions is the breakdown of ES complex to produce product and free enzyme; It is a statement of the quantitative relationship between the initial velocity V0, the maximum velocity Vmax, and the initial substrate concentration.


What happens when V0 is equal to 1/2 of the Vmax?

1/2 = [s]/ Km + [S]; Km = [s] when V0 = 1/2Vmax


Q # 27 Under what conditions does Km (in this case Kd) represent a measure of affinity of the enzyme for the substrate?

Originally (on its own), Km is a poor indicator for affinity, because Km varies greatly from enzyme to enzyme and even different substrates for the same enzyme. Vmax also is also different for different reactions.


Q # 28 What does a high Kcat value mean? What does a high Km value mean?

Larger Kcat values mean high turnover rate which means greater formation of product from reactant. A high Km value is associated with low affinity of substrate binding.


Q # 29 Why is Kcat/Km, the specificity constant, the most useful parameter for discussing catalytic efficiency?



Dissociation Constant

Kd; conditions K2 is the rate limiting step, K2



general rate constant; describes the limiting rate of any enzyme-catalyzed reaction at saturation; in the Michaelis-Menten equation, Kcat= Vmax/[Et] (for mixed steps) therefore V0 = Kcat [Et][S]/Km + [S]; if there is a definite rate-limiting step, the Kcat = k of that step.


Turnover number

the number of substrate molecules converted to product in a given unit of time on a single enzyme molecule when the enzyme is saturated with substrate.


Specificity Constant

Kcat/Km; the rate constant for the conversion of E + S to E + P; when [S] less than km, equation reduces to V0 = (Kcat/km) [Et][S]; second order reaction.


What limits Kcat/Km?

Diffusion rates


Q # 32 What kinds of information can be obtained by the study of pre-steady state kinetics that cannot be obtained by steady state kinetics alone?

To understand enzyme catalyzed reactions, we must be able to measure the rates of individual reaction steps. This is best done at the pre-steady state.


Q # 33 Why are inhibitors necessary as enzymatic control mechanisms in biological systems?

Inhibitors are necessary to stop process that are producing too much product or are driving a system from equilibrium. Over production can lead to the depletion of resources, or other unwanted physiological effects.


What are the two broad classes of enzyme inhibitors

Reversible and irreversible


Reversible inhibition

Competitive inhibition, uncompetitive inhibition, mixed inhibition,


Competitive inhibitor

Type of reversible inhibition; Competes with the substrate for the active site. An inhibitor (I) binds to the active site preventing the substrate from binding; Vmax becomes normal if we increase the amount of [S]; Km increases in the presence of an inhibitor (factor alpha)


Uncompetitive inhibition

Type of reversible inhibition; inhibitor binds to site distinct from the active site, and unlike a competitive inhibitor, binds only to the ES complex; uncompetitive inhibitors lowers Vmax and Km, because the [S] required to reach one-half Vmax decreases by the factor alpha; Km reduction occurs because uncompetitive inhibitors bind after substrate already binds; Vmax decreases, because it binds the ES complex and lowers the amount forming product.


Mixed inhibitor

Type of reversible inhibition; binds at a site other than the substrate active site. It binds either E or ES; mixed inhibitors effects both Km and Vmax


-1/Km or -alpha/Km

Used for inhibition and double-reciprocal plot; the more negative the value, the greater the binding affinity


Competitive Lineweaver-Burke plot

Slope: Km/Vmax increases
Y-int: 1/Vmax stays constant
Km: increases
Vmax: stays the same


Uncompetitive Lineweaver-Burke plot

Slope: Km/Vmax stays the same
Y-int: 1/Vmax increases (Vmax decreases)
Km: decreases
-1/km: becomes more negative (higher affinity, because the [S] to reach Vmax decreases by factor alpha)


Mixed (noncompetitive) inhibition

Slope: Km/Vmax (increase)
Y-int = 1/ vmax (increases)
Km: increases
decrease: Vmax


Increasing [S] will only overcome which type of inhibition



Irreversible Inhibition

Bind covalently with or destroy a functional group on an enzyme that is essential for the enzyme's activity, or form a particularly stable noncovalent association


Q # 38 Explain How alterations in the surround pH can affect enzyme activity

Amino acid side chains within the enzyme's active site may act as acids or bases which cause critical changes in enzymes. pH changes can change the charge of those aars, or can ruin an enzymes shape.

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