LM2 enzymes pt2 Flashcards

1
Q

define uncompetitive inhibition

A

binds to the ESC preventing release of products

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2
Q

define competitive inhibition

A

binds to the active site preventing substrate binding

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3
Q

define non-competitive inhibtion

A

binds to the allosteric site, changing the shape of the active site so the substrate can no longer bond

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4
Q

how do you calculate velocity of enzyme activity

A

-d[s]/t
d[p]/t

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5
Q

define 1st order kinetics

A

linear increase in velocity with increase in substrate concentration

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6
Q

define 0 order kinetics

A

at plateau where there is no change in velocity with changing substrate concentration

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7
Q

what equation can be used to explain first order kinetics

A

y=mx+c

y= velocity
m= gradient
x=substrate concentration
c= intercept

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8
Q

what does the michaelis menten equation explain

A

explains the plateua mathematically and the aim is to establish a mathematical connection between V0,Vmax and Km

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9
Q

what is the dissociation constant / how do wwe find it

A

k1[E][S] = K-1[ES]
[E][S]/[ES] = K1/K-1
=Kd

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10
Q

what are the assumptions made about the enzyme reations

A

that it is in equilibrium and the concentration of ES remains constant during the enzymatic reaction therefore,
[ES] formation = [ES] breakdown

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11
Q

what does [ES] breakdown =

A

[ES] breakdown = k-1[ES] +kcat[ES]
k1[E][S]=ES

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12
Q

how do we get the Michaelis Menten constant

A

[E][S]/[ES] = (K-1+Kcat)/K1 = Km = MM constant

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13
Q

how can total enzyme concentration be given

A

E0=E+ES

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14
Q

what does Vmax equal

A

Vmax = Kcat[E0] when E0 = ES

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15
Q

What does V0 equal

A

V0=Kcat[ES]

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16
Q

what is the MM equation

A

V0 = Vmax[S]/Km+[S]

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17
Q

what do we assume when the substrate concentration is very large

A

when S»Km the Km can be ignored so the equation become V0=Vmax

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18
Q

how do we get a lineweaver burke plot

A

by inverting the michaelis menten equation to produce a straight line graph
y=1/V0
m=Km/Vmax
x=[S]
c=1/Vmax

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19
Q

what are the limitations of lineweaver burke plot

A

1/[S] and 1/V0 are small values and finding them on graph paper may be difficult

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20
Q

what is y=mx+c on a lineweaver burke plot

A

y=1/V0
m=Km/Vmax
x=[S]
c=1/Vmax

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21
Q

what does the lineweaver burke plot of a competitive inhibitor look like compared to without an inhibitor

A

there is a slope change so Km increases but y-intercept remains the same so Vmax is unaffected
they cross(inhibitor has steeper gradient)

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22
Q

what does the lineweaver burke plot of an uncompetitive inhibitor look like compared to without an inhibitor

A

Km and Vmax are reduced
runs parallel to the original

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23
Q

what does the lineweaver burke plot of a non-competitive inhibitor look like compared to without an inhibitor

A

Km remains the same but Vmax is reduced
both start at the same point but it has steeper gradient with inhibitor

24
Q

what is the purpose of serine in the catalytic triad

A

acts as a nucleophile attacking the carbon in the carbonyl of the peptide bond

25
Q

what are the amino acids in the catalytic triad

A

Asp102
his52
ser195

26
Q

how do aspartate and histidine work together

A

the side chain of serine is an alcohol which is not a very good nucleophile so therefore the negatively charged side chain of aspartate interacts with the slightly positive charged hydrogen on the side chain of the histidine and will form a hydrogen bond, so the nitrogen on the histidine is now partially negatively charge and will interact with the alcohol group on serine transforming it to an alkoxyl group

27
Q

describe the action of the serine alkoxide

A

it acts as a nucleophile and attacks the carbonyls carbon breaking the pi bond and forming a relatively unstable tetrahedral intermediate with a negative charge on the carbonyls oxygen

28
Q

what is the purpose of the oxyanion hole

A

interacts with negative charge on carbonyls oxygen stabilising the tetrahedral intermediate

29
Q

what does the tetrahedral intermediate collapse into

A

an amin and a carbonyl group

30
Q

what happens after the tetrahedral intermediate collapses

A

the serine is acylated and forms an amide molecule that deprotonates the histidines nitrogen

31
Q

what happens after the amide product departs

A

it departs with the hydrogen from histidine and histidine helps to transform water into hydroxide which attacks the carbon in a nucleophilic fashion creating another unstable intermediate to be stabilised by the oxyanion hole

32
Q

describe the end products of the catalytic triad

A

a carboxylic acid product
all of the catalytic triad residues interacting with one another and the cycle repeats

33
Q

define scissile bond

A

cleavage site

34
Q

what polypeptide has a catalytic triad

A

chymotrypsin
subtilisin - similar to chymotrypsin but different residue numbers and slightly different arranged oxyanion hole
carboxypeptidase - different structures to enzymes above but same catalytic triad mechanism and also oxyanion hole

35
Q

what would be proof of convergent evolution

A

all three enzymes evolved differently yet have similar catalytic triads and mechanisms
chymotrypsin
subtilisin - similar to chymotrypsin but different residue numbers and slightly different arranged oxyanion hole
carboxypeptidase - different structures to enzymes above but same catalytic triad mechanism and also oxyanion hole

36
Q

what enzymes use other approaches that dont use activated serine residues

A

cysteine proteases
aspartyl proteases
metalloproteases

37
Q

describe how cysteine proteases cleave a peptide bond

A

papain

uses a histidine activated cysteine residue to generate a nucleophile that attacks the peptide carbonyl

38
Q

describe how aspartyl proteases cleave a peptide bond

A

the central feature of the active site is a pair of aspartic acid residues that act together to allow a water molecule to attack the peptide bond

one aspartic residue in its deprotonated form activates the attacking water molecule by poising it for deprotonation

the other protonated aspartic residue polarizes the peptides carbonyl group so it is more susceptible to attack

for example renin

39
Q

how do metalloproteases cleave a peptide bond

A

the active site of such protein consists of a bound metal ion, almost always zinc, that activates a water molecule to act as a nucleophile to attack the peptide carbonyl group

40
Q

describe the action of carbonic anhydrase

A

catalyses the rapid interconversion of carbon dioxide and water, carbonic acid and the bicarbonate ion and H+

41
Q

describe structure of carbonic anhydrase

A

it contains a bound Zn2+ ion to the imidazole rings of three histidine residues as well as to a water molecule it is also in a cleft near the centre of the enzyme, the three histidines bind the zinc ion while one acts a base

42
Q

what are the optimum conditions for carbonic anhydrase and how are they achieved

A

maximally active at a high pH
the binding of the water molecule to the zinc reduces the pKa of water from 15.7 to 7
with pKa lowered the concentration of OH- increases

43
Q

what is the nucleophile in carbonic anhydrase

A

the zinc bound hydroxyl and is available to attack CO2 more efficiently than water

44
Q

describe the action of the nucleophile in carbonic anhydrase

A

the zinc ion facilitates the release of a proton from water generating hydroxyl ions
the CO2 binds at the active site and is positioned next to the hydroxyl

the hydroxyl group attack the CO2 converting it to a bound bicarbonate ion

the catalytic site is regenerated with the release of HCO3- and the bonding of another water molecule

thus the binding of a water molecule to the zinc ion favours the formation of the transition state by facilitating the release of a proton and positioning the water close to the other reactant

45
Q

what is the paradox associated with carbonic anhydrase

A

CO2 is hydrated at a faster rate than expected

46
Q

what is the answer to our paradox

A

buffer concentrations above 10^-3 may be high enough to support the faster than expected rates of CO2 hydration

47
Q

what is a histidine proton shuffle

A

his64 abstracts a proton from the zinc bound water thus generating nucleophilic hydroxide ion then buffer removes H+and deprotonates/regenerates His64

48
Q

describe nucleoside monophosphate kinases

A

these enzymes catalyse the transfer of the terminal phosphoryl group from a nucleoside triphosphate usually ATP to NMP

49
Q

what is the challenge of NMK

A

promote the transfer without promoting the competing reaction

50
Q

what does the NMK domain consist of

A

central beta pleated sheet surrounded on both side by alpha helices as well as a key loop

51
Q

what amino acid sequence does the p loop have

A

GLY-XXXX-GLY -LYS

52
Q

why is it called the p loop

A

because it interacts with phosphoryl group on bound nucleotides

53
Q

what do NMK require and why

A

they are basically inactive in the absence of a divalent metal ion such as Mg2+ which coordinates to ATP and fixes it in a particular conformation for tighter binding with the enzyme

54
Q

what else does the divalent metal ion do for nucleoside monophosphate kinases

A

provides additional points of interaction between the ATP-Mg2+ complex and the enzyme thus increasing the binding energy

55
Q

What does ATP binding do

A

induces large conformation changes and the p loop closes down on the polyphosphate chain interacting most extensively with the beta phosphoryl group and this movement brings down the top domain of the enzyme to form a lid over the bound nucleotide

56
Q

what do both sets of changes during ATP binding ensure

A

catalytically component conformation is formed only when both the donor and acceptor are bound preventing wasteful transfer of the phosphoryl group to water

57
Q

How do enzymes lower activation energy?

A

Orientating themselves to reduce the energy barrier associated with random movements rotations and vibrations

Enzymes can induce strain on a substrate stretching them and putting them into an unstable transition state

Enzymes can also temporarily add chemical groups to substrates and change their charge