lecture 12

Question 1

What helps to speed up biochemical reactions?

Answer 1

enzymes

Question 2

T/F, the enyzme concentration is much lower than that of the substrate?

Answer 2

T

Question 3

so when substrate levels are high, what does the rate of the reaction depend? example

Answer 3

the reaction depends of the concentration of the enzyme, also known as the catalyst

Question 4

explain how a basic enzyme catalyzed reaction works in terms of E, S, E x S, P and K1, K2, K3, and K4?

Answer 4

Some come together and some dissociate, depending on the rate constant; so when K1 is bigger, affinity is higher and favor enzyme substrate form. K1 is association, K2 is dissociation; once ES forms, and K3 is bigger number it moves forward and vice versa with K4.

Remember that the enzyme (E) binds substrate (S) and forms a non-covalent complex between E and S (E x S); substrate is converted to product (P) which is simulataneously released; the product formation is usually the rate limiting step of the reaction and the covalent bonds are broken or formed in this step

In cells, the level of P is usually low, therefore can neglect k4 and the back reaction of P to E·S; P formation depends on the rate of ES breakdown, k3 which k3 often called the “turnover number” or kcatalysis or kcat; Thus, the velocity (v) of product formation is determined by v = k3[E·S].

Question 5

what is the michaelis mentor equation?

Answer 5

v=V sub max times concentration of substrate/K sub m plus the concentration of substrate

Question 6

what does the michaelis menton equation mean?

Answer 6

predicts enzyme velocity (v) at any [S] as long as the constants Vmax and Km are known of which both are experimentally derived

Question 7

going back to enzyme kinetics, with respect to enzyme product formation, why can we neglect k4?

Answer 7

Because [P] is usually very small, the term k4 [E][P] can be neglected, so we can neglect the k4 reaction which converts P back to ES.

Question 8

define Vmax?

Answer 8

Vmax is the maximum rate an enzyme can attain.  simply means that 100% of E is making P, each enzyme has its own Vmax and so Vmax = K3[E]total

Question 9

define Km?

Answer 9

a collection of constants, defined as Km= K2 + K3/K1. Recall from the previous slide, we see that k3 is often the slow step, meaning the value of k3 is small. Typically k1 is very large for most substrates that bind enzymes. Because of this, we can neglect k3 from the numerator, which reduces the Km equation to k2 over k1, which is the exact same thing as a dissociation constant, Kd. Because of this, Km is often considered to be a dissociation constant. Km is NOT exactly the same as Kd, but if k3 is smaller than k2, this is approximately true. Dissociation constants are measures of affinity – they tell how likely a substrate is to bind the enzyme. If k1 is very large and k2 is very small, this means the S stays bound to E. If k1 is small and k2 is big, it means S is often not bound to E. Notice, if k1 is big and k2 is small, the Kd will be very small. So a small Kd actually means high affinity, and a big Km means low affinity.

Question 10

cover objectives e-i

Answer 10

refer to notes

Question 11

whats an important multi substrate reaction we should consider?

Answer 11

Glucokinase / Hexokinase – use ATP to phosphorylate glucose

Question 12

how does the aforementioned multi reaction occur?

Answer 12

Hexokinase binds glucose and ATP together to form glucose-6-phosphate and ADP as products. Normally ATP is in large supply in the cell but free glucose is low. So the enzyme will be preloaded with ATP, but nothing will happen until glucose comes in the cell and binds to the active site. Then the EAB complex (E.ATP.glucose) is complete and product formation can occur.

Question 13

in the multi substrate reaction named as the example, why are A and B required?

Answer 13

Both A and B are required for the reaction to proceed, both have an influence on the v vs. [S] curve

Lots of A but no B…no reaction (both A and B required)…and vice versa
Lots of A? Velocity will increase as [B] increases
Lots of B? Velocity will increase as [A] increases

Question 14

explain the coenzymes as they relate to the enzyme active site and enzymatic reaction

Answer 14

without coenzymes, compounds A,B, and CD don’t respond to their enzymes and so with coenzymes available and in place on the enzyme the compounds are attracted to their sites on the enzymes and the reactions proceed. The coenzymes often donate or accept electrons, atoms, or groups of atoms. The reactions are completed with either formation of new product, AB, or the breaking apart of a compound into two new products C and D and release of energy

Question 15

name the characteristics of enzyme cofactors and list examples?

Answer 15

May be small organic molecules (coenzymes) or bound metal atoms or metal ions (essential ions); May be permanently bound to the enzyme or may diffuse in and out of the active site with substrates/products; they can also be derived from vitamin and minerals; examples include Nicotinamide Adenine Dinucleotide, NAD+, derived from Vitamin B2, niacin

Question 16

what is enzyme inhibition?

Answer 16

Any substance that affects ES formation will inhibit product formation

Question 17

what is competitive inhibition?

Answer 17

when substances block the active site, preventing S from binding

Question 18

what is non-competitive inhibition?

Answer 18

Substances may bind to E to change the shape of the active site

Question 19

what is reversible inhibition?

Answer 19

when the inhibitor binds and dissociates freely

Question 20

what is irreversible inhibition?

Answer 20

bind tightly to the enzyme and inactivate it

Question 21

what is the effect of of each inhibitor (competitive and non-competitive) on enzyme Km and Vmax?

Answer 21

For competitive inhibition, the substrate is blocked from binding to E active site due to presence of inhibitor, so higher S is needed to “compete” against I to form ES. Because of this, it appears the affinity of the enzyme for S is lower, so Km is bigger. However, S can eventually saturate E forming 100% ES, so the enzyme can reach Vmax.

For noncompetitive inhibition, substrate binding is not affected in any way…k1, k2, and k3 are not changed. Therefore Km is not changed. However, when I is bound, the enzyme can’t make product. So any enzyme in the EI form (or ESI form) is inactive. The total amount of “active” enzyme is decreased, so Vmax, which is directly proportional to the enzyme concentration, is decreased.

Question 22

what is a suicide inhibitor?

Answer 22

when the enzyme does the catalysis that leads to its own destruction and so this works by the binding to the active site similar to substrate and so this binds covalently and irreversibly to an amino acid residue in the enzyme active site forming bonds that can’t be broken. This destroys the enzyme active site. Also called mechanism based inhibitors since they depend on normal reaction cycle for action

Question 23

what are the examples of suicide inhibitors?

Answer 23

cyclooxygenase and aspirin; organophosphate inhibitors of the acetylcholine receptor

Question 24

what are the cofactors that can be small organic molecules?

Answer 24

coenzymes

Question 25

what are the cofactors that bound metal atoms or metal ions

Answer 25

essential ions

Question 26

how does competitive inhibition affect Km and Vmax?

Answer 26

Competitive inhibition changes Km, but not Vmax

Question 27

how does non-competitive inhibition affect Km and Vmax?

Answer 27

Non-competitive inhibition does not change Km but Vmax is decreased

Question 28

what is the michaelis menton equation?

Answer 28

velocity = Vmax x [S]/Km + [S]

Question 29

what happens to the MM equation when [S]&raquo_space;Km?

Answer 29

v=Vmax, you can neglect Km in the denominator

Question 30

what happens to the MM equation when [S]=Km?

Answer 30

v=1/2Vmax, substitute [S] in place of Km and solve

Question 31

what happens to the MM equation when [S] <

Answer 31

v=(Vmax x [S])/Km, can neglect [S] in the denominator and this indicates that V is linearly dependent on [S]

Question 32

with respect to the michaelis menton equation, what happens when you increase the substrate concentration?

Answer 32

the velocity increases, increase rate of product formation

Question 33

what is K sub d?

Answer 33

how likely ES will fall apart

Question 34

what is K sub m?

Answer 34

the michaelis constant, K sub m = K sub 2 (likely to fall apart) / K sub 1(likely to form) and K3 is neglected because we only want to know about dissociation to association

Question 35

are K sub m and K sub d the same?

Answer 35

Km is approximately equal to Kd; Km becomes measures for how much affinity there is for enzyme. So if we look at Km=K2/K1 and K1 is really big and K2 is small then the Km/Kd is small indicating high affinity and more association

Question 36

where does hexokinase live?

Answer 36

brain, red blood cells, muscles, most cells

Question 37

hexokinase has low or high affinity for glucose?

Answer 37

it has a small Km which means high affinity so hexokinase is very active at low [glucose]

Question 38

glucokinase has high or low affinity for glucose?

Answer 38

it has a higher Km meaning lower affinity for glucose which means it is less sensitive to glucose

Question 39

what is the line weaver Burk plot?

Answer 39

inverted michaelis menton equation. It can be used to determine the kinetic parameters for an enzyme (KM and Vmax) and also the mechanism of action of enzyme inhibitors

Question 40

in cells, the level of P is high or low? why is this needed to know?

Answer 40

low; we can neglect K4 and the back reaction of P to E x S

Question 41

what is P formation dependent on?

Answer 41

the rate of ES breakdown, K3

Question 42

what is K3, often referred to as?

Answer 42

turnover number or Kcatalysis or Kcat

Question 43

what is the velocity of product formation determined by?

Answer 43

v=k3[ES]

Question 44

with respect to line weaver buck transformation, define the following:

line weaver burk equation
y-intercept
x-intercept
slope

Answer 44

line weaver buck equation:
1/v=Km/Vmax(1/[S]) + 1/Vmax

y-intercept
1/vmax

x-intercept
-(1/Km)

slope
Km/Vmax

Question 45

what is the significance of the line weaver buck plot?

Answer 45

Can be used to determine the kinetic parameters for an enzyme (KM and Vmax) and also the mechanism of action of enzyme inhibitors

Question 46

Define the term catalytic rate (kcat) and turnover number (kcat-1

Answer 46

The catalytic rate is the same thing as k3. It is the rate at which an enzyme can produce product. So you can say 100 per second or 10,000 per second, etc. That is the rate. The turnover number is the number of products that are produced per unit time. That is the same thing. If an enzyme has a kcat of 10 per second, it can produce 10 product molecues per second, which is its turnover number. The -1 subscript is a typo that got in there somehow ?!?! don’t know where that came from… But even so, if you consider the inverse of kcat or the turnover number, that would give you the amount of time it takes to make 1 product molecule. So again, if kcat is 10 per second, then it takes 0.1 sec to make 1 product molecule. That is the extent of this to be covered on the exam (the definition of kcat and turnover number…not the inverse typo thingy…)

Question 47

what is the catalytic rate? define it

Answer 47

same thing as the K3; It is the rate at which an enzyme can produce product.

Question 48

what is the turnover number?

Answer 48

The turnover number is the number of products that are produced per unit time.