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BIOC 3021 Exam 2 Questions and Answers 100% Solved | Graded A+
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BIOC 3021 Exam 2 Questions and Answers 100% Solved | Graded A+ In addition to the basic ES intermediate in an enzyme reaction, what other types of intermediates are formed and why are these intermediates so important? - Transition states occur in enzyme reactions that are important in product ...

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  • December 7, 2024
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  • 2024/2025
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BIOC 3021 Exam 2 Questions and

Answers 100% Solved | Graded A+


In addition to the basic ES intermediate in an enzyme reaction, what other

types of intermediates are formed and why are these intermediates so

important? - ✔✔Transition states occur in enzyme reactions that are

important in product stabilization and changes in energy.

Why would researchers try to determine the initial rate of enzyme activity

(V0) at a very early time point in the reaction? What happens later in the

reaction? Why does this happen? - ✔✔The initial velocity should be

determined early in the reaction because the velocity will change later due

to changes in substrate concentration.

Why does a plot of V vs S taper off and eventually reach a plateau at

higher S levels? - ✔✔Substrate has filled all enzyme active sites and

adding more will not increase the rate of reaction.

What is the Michaelis-Menten Equation? The M-M Equation uses the terms

(V), ( Vmax), (S) and (Km). What do each of these terms mean? - ✔✔The

velocity of the reaction equals maximum velocity times substrate

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,concentration divided by the sum of the substrate concentration and the

Michaelis constant, which has units of concentration.

Why can the rate constant K-2 be ignored in the derivation of the M-M

Equation? - ✔✔There is little product formed early in the reaction, so the

reverse reaction can be ignored.

Why is a V vs S plot linear at low S concentrations? Why does the plot

curve off at intermediate S levels, and why does it plateau at high S levels?

- ✔✔The V vs S plot is linear at low S concentrations because the M-M

equation reduces to V = Vmax * [S] / Km, which is a linear equation. The

plot then curves at intermediate levels with increasing substrate

concentration, and plateaus at high S levels due to the M-M equation

approaching Vmax.

What does the Km term tell us about an enzyme? What does it signify if an

enzyme has a low Km or a high Km? What is the enzyme rate when the

concentration of S = Km? - ✔✔Km tells us about an enzyme's efficiency; a

low Km indicates an efficient enzyme, while a high Km indicates a less

efficient enzyme. When the Km and substrate concentration are equal, the

velocity of the reaction is equal to half the maximum possible velocity of the

reaction.



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,What is the definition of turnover number and how does that constant relate

to the Vmax of an enzyme? What is the absolute value of the turnover

number for the enzyme catalase and what does that mean in terms of the

amount of substrate used by a single molecule of the enzyme per second?

Why doesn't the turnover number of an enzyme change as the enzyme is

purified? - ✔✔The turnover number is the number of molecules of substrate

that can be converted per second per molecule of enzyme. It is equal to

Vmax divided by the concentration of the enzyme. The enzyme catalase

the absolute value of the turnover number is forty million, so it can still

function efficiently even with a poor Km. Turnover number doesn't change

as the enzyme is purified because it's an intrinsic property of enzymes.

What is the Lineweaver-Burk equation? How does it relate to the M-M

Equation and how is it used to determine the Km and Vmax of an enzyme?

- ✔✔1/v = 1/vmax + Km/(vmax*[S]). It is essentially the M-M equation

flipped. -1/Km = 1/[S] intercept, 1/vmax = 1/v intercept.

What is the difference between a reversible and an irreversible inhibitor?

What is the mode of action for the two most common reversible inhibitors,

competitive and noncompetitive? When examined using a Lineweaver-Burk

plot, how do each of these inhibitors change the plots? What information

can be gained by examination of the intersection of these plots with the X

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, and Y intercepts? - ✔✔Reversible inhibitors cause covalently-modified

changes to enzymes that cannot be undone, while reversible inhibitors do

not. Competitive inhibitors reversibly bind at an enzyme's active site. This

increases the Km and thus shifts the LB plot to the right. Noncompetitive

inhibitors reversibly bind at a non-active site of an enzyme and exert their

effects allosterically. This decreases the Vmax of the reaction and shifts the

LB plot up the y-axis (1/v axis).

What is the mode of action of irreversible inhibitors? Why are they

irreversible? What is the specificity of action of TPCK, DIFP, and

iodoacetamide? How can these inhibitors help determine the character of

an enzyme's active site? - ✔✔Irreversible inhibitors cause covalently-

modified changes to enzymes, usually in the active sites. TPCK binds at

the active site of chymotrypsin with a histidine residue. DIFP reacts with

active sites of serine proteases like chymotrypsin and modifies the active

site serine residue of acetylcholinesterase. Iodoacetamide reacts with the

active site of cysteine proteases by covalently bonding with the sulfur atom.

These inhibitors can help determine the character of an enzyme's active

site, since they will only inhibit if certain residues exist.

What are the common mechanisms used by enzymes to achieve catalysis?

How do these mechanisms reduce or increase the energy level of an

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