Biochemistry 2: Enzymes

Question 1

Oxioreductase

Answer 1

catalyze redox reactions (electron transfer)

ex. dehydrogenase, reductase, oxidase

Question 2

Transferase

Answer 2

catalyze the movement of a functional group from one molecule to another

ex. aminotransferase, kinase

Question 3

Hydrolase

Answer 3

catalyze cleavage with the addition of water

ex. phosphatase, peptidase, nuclease, lipase

Question 4

Lyase

Answer 4

catalyze cleavage without the addition of water and without the transfer of electrons

can also do the reverse of these reactions!

act as synthases for smaller molecules

ex. ATP –> cAMP

Question 5

Isomerase

Answer 5

catalyze the interconversion of isomers, including both constitutional isomers and stereoisomers

rearrange bonds in a molecule

Question 6

Ligase

Answer 6

catalyze the addition/synthesis reactions between two larger molecules, often of the same type

need ATP

Question 7

How do enzymes work?

activation energy? free energy? enthalpy? rate?

Answer 7

• lower activation energy
• increase rate of reaction
• no change to the equilibrium constant
• no change to overall free energy

Question 8

locke and key theory

Answer 8

hypothesizes that the enzyme and substrate are exactly complementary

Question 9

induced fit model

Answer 9

hypothesizes that the enzyme and substrate undergo conformational changes to interact fully

Question 10

cofactors

Answer 10

inorganic molecules or metal ions that are necessary for enzyme function

Question 11

coenzymes

Answer 11

small organic molecules (like vitamins) that are necessary for enzyme function

Question 12

prosthetic groups

Answer 12

tightly bound cofactors or coenzymes that are necessary for enzyme function

Question 13

saturation kinetics

Answer 13

as substrate concentration increases, the reaction rate increases until a max value is reached (vmax)

Question 14

Michaelis Menten equation

Answer 14

E + S (k_-1)⇔(k₁) ES →(k_cat) E + P

if concentration of enzyme kept constant:

Question 15

K_m

Answer 15

Michaelis constant

equal to the amount of substrate when half of the enzymes are full (at 1/2 vmax)

measure of affinity of the enzyme for substrate

low Km = high affinity

Question 16

k_cat

Answer 16

measures the number of substrate molecules converted into product

vmax = [E]kcat

Question 17

catalytic efficiency

Answer 17

k_cat/K_m

a measure of enzyme efficiency

Question 18

Lineweaver-Burk Plots

x-intercept?

y-intercept?

Answer 18

x-intercept: -1/K_m

y-intercept: 1/v_max

helpful in comparing vmax and Km

Question 19

cooperative enzymes

Answer 19

have multiple active sites and subunits that an be in a low-affinity tense (T) state or high affinity relaxed (R) state

binding substrate encourages other subunits T -> S

sigmoidal kinetics when graphed on v vs. [S] Michealis-Menten plot

Question 20

Hill’s coefficient

Answer 20

quantifies cooperativity

> 1, positive cooperative binding

= 1, no cooperative binding

< 1, negative cooperative binding

Question 21

feedback regulation

Answer 21

regulatory mechanism where enzyme activity is inhibited by high levels of product made later on in pathway

Question 22

feed-forward regulation

Answer 22

enzymes are regulated by intermediates that precede the enzyme in the pathway

Question 23

reversible inhibition

Answer 23

inhibition that, once removed, allows the enzyme it was inhibiting to begin working again, has no permanent effects on the enzyme

can be competitive, noncompetitive, uncompetitive, or mixed

Question 24

competitive inhibition

Answer 24

inhibitor is similar to the substrate and binds at the active site

can be overcome by adding more substrate

vmax unchanged because it enzyme can still perform if more substrate added

Km increases because decreased affinity so more substrate is required

Question 25

noncompetitive inhibition

Answer 25

inhibitor binds with equal affinity to the enzyme and enzyme-substrate complex because it binds at an allosteric site

vmax decreases because less enzyme available to ever reach the normal reaction velocity

Km unchanged because enzyme can still bind to substrate normally

Question 26

mixed inhibition

Answer 26

inhibitor binds with unequal affinity to enzyme and enzyme-substrate complex at allosteric site

vmax decreased because less enzyme available to reach normal reaction velocity

Km increases when binds more to enzyme because it prevents it’s binding to the substrate – there is less affinity for substrate

Km decreases when binds more to enzyme-substrate because the enzyme must bind substrate in order for inhibitor to bind

Question 27

uncompetitive inhibition

Answer 27

inhibitor binds only to enzyme-substrate complex at allosteric site

vmax decreases because eventually there is less enzyme available

Km decreases because the enzyme must bind substrate in order for inhibitor to bind

Question 28

irreversible inhibition

Answer 28

alteration in the enzyme such that the active site is unavailable for a long time or permanently

new enzymes must be synthesized for the reaction to occur again

Question 29

zymogens

Answer 29

enzymes that are secreted in an inactive form and are activated by cleavage

the active site is covered by some regulatory domain until it is cleaved or altered