CH 306 TEST 2 Flashcards

1
Q

Why is an assay required for protein purification?

A

An assay, which should be based on some unique biochemical property of the protein that is being purified, allows the detection of the protein of interest.

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2
Q

What physical differences among proteins allow for

their purification?

A

Differences in size, solubility, charge, and the specific binding of certain molecules

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3
Q

What are the two properties of enzymes that make them especially useful catalysts?

A

Rate enhancement and substrate specificity

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4
Q

What are the general characteristics of enzyme active sites?

A

The active site is a three-dimensional crevice or cleft; it makes up only a small part of the total volume of the enzyme. Active sites have unique microenvironments. A substrate binds to the active site with multiple weak interactions. The specificity of the active site depends on the active site’s precise three-dimensional structure.

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5
Q

What does an apoenzyme require to become a holoenzyme?

A

A cofactor

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6
Q

What are the two main types of cofactors?

A

Coenzymes and metals

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7
Q

Why are vitamins necessary for good health?

A

Vitamins are converted into coenzymes that are required for most biochemical reactions.

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8
Q

What is the fundamental mechanism by

which enzymes enhance the rate of chemical reactions?

A

Enzymes facilitate the formation of the transition state.

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9
Q

What is the structural basis for

enzyme specificity?

A

The intricate three-dimensional structure of proteins allows the construction of active sites that will recognize only specific substrates.

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10
Q

What is meant by the term binding

energy?

A

Binding energy is the free energy released when two molecules bind together, such as when an enzyme and a substrate interact.

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11
Q

What is the role of binding energy in enzyme catalysis?

A

Binding energy is maximized when an enzyme interacts
with the transition state, thereby facilitating the formation of the transition state and enhancing the rate of the reaction.

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12
Q

What would be the result of an enzyme having a greater binding energy for the substrate than for the transition state?

A

There would be no catalytic activity. If the enzyme–substrate complex is more stable than the enzyme–transition-state complex, the transition state would not form and catalysis would not take place.

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13
Q

Protein catalyst

A

Enzyme

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14
Q

Reactant in an enzyme-catalyzed reaction

A

Substrate

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15
Q

A coenzyme or metal

A

Cofactor

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16
Q

Enzyme minus its cofactor

A

Apoenzyme

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17
Q

Enzyme plus cofactor

A

Holoenzyme

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18
Q

Small vitamin-derived organic cofactors

A

Coenzymes

19
Q

Function of K′eq

A

ΔG°′

20
Q

The least-stable reaction intermediate

A

Transition state

21
Q

Site on the enzyme where catalysis takes place

A

Active site

22
Q

Change in enzyme structure

A

Induced fit

23
Q

Change in enzyme structure

A

Induced fit

24
Q

Why does the activation energy of a reaction not appear in the final ΔG of the reaction?

A

The energy required to reach the transition state (the activation energy) is returned when the transition state proceeds to product.

25
Q

Proteins are thermodynamically unstable. The ΔG of the hydrolysis of proteins is quite negative, yet proteins can be quite stable. Explain this apparent paradox. What does it tell you about protein synthesis?

A

Protein hydrolysis has a large activation energy. Protein synthesis requires energy to proceed.

26
Q

Suggest why the enzyme lysozyme, which degrades cell walls of some bacteria, is present in tears.

A

Lysozyme helps protect the fluid that surrounds eyes from bacterial infection.

27
Q

Transition-state analogs, which can be used as enzyme inhibitors and to generate catalytic antibodies, are often difficult to synthesize. Suggest a reason.

A

Transition states are very unstable. Consequently, molecules that resemble transition states are themselves likely to be unstable and, hence, difficult to synthesize.

28
Q

Suppose that a mutant enzyme binds a substrate 100 times as tightly as does the native enzyme. What is the effect of this mutation on catalytic rate if the binding of the transition state is unaffected?

A

The mutation slows the reaction by a factor of 100. The activation-free energy is increased by 11.4 kJ mol−1 (2.73 kcal mol−1). Strong binding of the substrate relative to the transition state slows catalysis.

29
Q

Separating proteins on the basis of size differences

A

Molecular exclusion chromatography

30
Q

Allows high resolution and rapid separation

A

High-pressure liquid chromatography (HPLC)

31
Q

Produced by hybridoma cells

A

Monoclonal antibodies

32
Q

An immunoassay technique preceded by gel electrophoresis

A

Western blotting

33
Q

A measure of the rate of movement due to centrifugal force

A

Sedimentation coefficient

34
Q

Separating proteins on the basis of net charge

A

Ion-exchange chromatography

35
Q

Specific site recognized by an antibody

A

Antigenic determinant (epitope)

36
Q

Based on the fact that proteins have a pH at which the net charge is zero

A

Isoelectric focusing

37
Q

Based on attraction to a specific chemical group or molecule

A

Affinity chromatography

38
Q

A means of identifying a protein based on a unique property of the protein

A

Assay

39
Q

Why do proteins precipitate at high salt concentrations?

A

If the salt concentration becomes too high, the salt ions interact with the water molecules. Eventually, there are not enough water molecules to interact with the protein, and the protein precipitates.

40
Q

Although many proteins precipitate at high salt concentrations, some proteins require salt to dissolve in water. Explain why some proteins require salt to dissolve.

A

If there is a lack of salt in a protein solution, the proteins may interact with one another—the positive charges on one protein with the negative charges on another or several others. Such an aggregate becomes too large to be solubilized by water alone. If salt is added, it neutralizes the charges on the proteins, preventing protein–protein interactions.

41
Q

What types of R groups would compete with salt ions for water of solvation?

A

Charged and polar R groups on the surface of an enzyme

42
Q

(a) The octapeptide AVGWRVKS was digested with the enzyme trypsin. Would ion-exchange or molecular exclusion chromatography be most appropriate for separating the products? Explain.
(b) Suppose that the peptide had, instead, been digested with chymotrypsin. What would be the optimal separation technique? Explain.

A

(a) Trypsin cleaves after arginine (R) and lysine (K), generating AVGWR, VK, and S. Because they differ in size, these products could be separated by molecular exclusion chromatography.
(b) Chymotrypsin, which cleaves a er large aliphatic or aromatic R groups, generates two peptides of equal size (AVGW) and (RVKS). Separation based on size would not be effective. The peptide RVKS has two positive charges (R and K), whereas the other peptide is neutral. Therefore, the two products could be separated by ion-exchange chromatography.

43
Q

The detergent sodium dodecyl sulfate (SDS) denatures proteins. Suggest how SDS destroys protein structure.

A

The long hydrophobic tail on the SDS molecule (p. 73) disrupts the hydrophobic interactions in the interior of the protein. The protein unfolds, with the hydrophobic R groups now interacting with SDS rather than with one another.

44
Q

In the course of purifying an enzyme, a researcher performs a purification step that results in an increase in the total activity to a value greater than that present in the original crude extract. Explain how the amount of total activity might increase.

A

An inhibitor of the enzyme being purified might have been present and subsequently removed by a purification step. This removal would lead to an apparent increase in the total amount of enzyme present.