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Principles of Analysis > Enzymology > Flashcards

Enzymology Flashcards

1
Q

what are enzymes

A

a substance (usually a protein) that increases the rate of a chemical reaction without itself being changed in the process

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2
Q

what does orotidine 5’-phosphate decarboxylase do?

A

enhances the rate of decarboxylation by 1017 fold from once every 100 million years to 40 x a second

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3
Q

how many major classes of enzyme types are there?

A

6

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4
Q

properties of enzymes

A

increase the rate of reaction
do not change the reaction equilibrium
specific for their substrates
they are regulated - genetic (transcription/translation)/ allosteric

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5
Q

why do labs measure enzymes

A

during disease, tissue damage may lead to release of enzymes from tissues
inborn errors of metabolism may be due to a deficiency of a certain enzyme

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6
Q

what does the release of aminotransferases (ALT and AST) indicate?

A

liver damage

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7
Q

what is PKU caused by

A

deficiency of the enzyme phenylalanine hydroxylase

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8
Q

what is an example of using enzymes of reagents within the labs?

A

glucose measurement

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9
Q

what is the equation for glucose?

A

glucose -> (hexokinase) (ATP->ADP) -> Glucose 6 phosphate
Glucose-6-phosphate dehydrogenase (NAD+->NADH) -> 6=Phosphogluconate

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10
Q

how is glucose routinely measured

A

hexokinase and G6PD NAD+->NADH

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11
Q

factors that can affect enzyme activity

A

the conc of substrate
enzyme
pH
temp
type of buffer
cofactor
substrate
incubation time

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12
Q

how do we measure enzyme activity?

A

E+S - > ES -> E+P

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13
Q

What does rate consumption (substrate) =

A

rate formation (product)

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14
Q

what enzymes do a fixed-time method use

A

enzymes with high affinities (Low Km)

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15
Q

how do you measure a fixed-time method?

A

Measure substrate or product concentration at two time points and calculate the difference then divide by time period to obtain rate.

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16
Q

what enzymes does a continuous monitoring method use?

A

enzymes with low affinities (high Km)

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17
Q

how do you measure a continuous monitoring method?

A

Reaction is monitored by continuously taking a rate measurement over a designated time period.

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18
Q

What are the advantages of a continuous monitoring method?

A

assess reaction progress, ensure true initial rate measurement obtained and identify any lag phase

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19
Q

what is the lag phase

A

few seconds for E+S to find each other, not reflective of the true rate

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20
Q

what is the true initial rate

A

is the exponential phase when product is increasing and enzyme working at highest rate of activity

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21
Q

what is happening in the plateau phase

A

the substrate starts to be used up

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22
Q

units for enzyme activity - SI unit of activity - katal

A

the amount of enzyme which will convert 1 mole of substrate in 1 second

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23
Q

units for enzyme activity- International unit - U

A

the amount of enzyme which will convert 1 μmole of substrate in 1Minute

24
Q

1 katal / L =

A

1,000,000 x 60 = 6 x 107 U/L

25
Q

1U/L =

A

1 / 60 x 106 = 16.67 x 10-9
= 16/67 nkatal/L

26
Q

steps in calculating enzyme activity

A

convert physical measurement (e.g. absorbance) made over a timed interval
divide change in absorbance by the time period of measurement (mins) to give rate (ΔA/min).
convert into substrate concentration units and derive rate for the enzyme-catalysed reaction
divide by molar absorptivity (L/mol/cm) and cuvette path length (cm) (Beer-Lambert’s Law)

27
Q

how to convert to the concentration of enzyme units in the clinical samples

A

multiply by 1,000,000 to convert from mol to umol
multiply by total reaction volume (mL) and divide by sample volume (mL) - ‘dilution factor’

28
Q

what is the final equation for calculating enzyme activity in a clinical sample

A

ΔA/min x 1,000,000 x Total vol (mL) /
ε (L/mol/cm) x l (cm) x Sample vol (mL)

29
Q

steady-state kinetics / transition state theory

A

Enzymes accelerate the approach to equilibrium between substrates and products by lowering the activation energy of the reaction.
No effect on position of equilibrium
The ‘standard free energy of activation’ (ΔG°‡) is the additional free energy that substrate molecules must have to attain the transition state.

30
Q

what is the term ES

A

Enzyme - substrate complex

31
Q

what is K+1

A

rate constant for formation of ES

32
Q

what is k-1

A

rate constant for dissociation of ES

33
Q

What is Kcat

A

rate constant for the chemical (catalytic) step (turnover number)

34
Q

what is the total enzyme concentration

A

[E]tot = [E] + [ES]

35
Q

what does the Michaelis-Menten equation do?

A

describes the effcet5 of substrate concentration on rate of enzyme catalysed reactions

36
Q

what is the equation for enzyme activity

A

E+S (k-1)—(k+1)> ES (k-2)—(kcat)> E+P

37
Q

what happens a few miliseconds after mixing enzyme (E) and substrate (S)?

A

the enzyme-substrate complex [ES] builds up and does not change. this is called the steady state [ES] is constant.

38
Q

what is the michaelis-menten equation?

A

v=vmax[S] / km + [S]

39
Q

what is Km

A

the michaelis constant
the substrate conc at half vmax
measure of the strength of the ES complex
a high km indicates weak binding, a low km indicates strong binding

40
Q

what is the V

A

the initial reaction velocity at [S]

41
Q

what is vmax

A

the maximum possible initial reaction velocity

42
Q

what happens at low concentrations of substrate?

A

the initial velocity of the reaction (v) is directly proportional to the substrate concentration. the equation approximates to v=Vmax[S]/Km. this is a linear expression.

43
Q

when a reaction rate is proportional to a single concentration term it is…

A

said to follow first-order kinetics

44
Q

what is the equation of first-order kinetics?

A

v=Vmax[S]/Km

45
Q

what happens at very high S

A

when S»>Km most of the enzyme exists as the ES complex. the enzyme is saturated with substrate. the equation approximates to v=vmax[S] / [S] = Vmax. rate becomes independent of [S] and constant (v=Vmax).

46
Q

when reaction rate is independent of [S] it is said to follow…

A

zero-order kinetics

47
Q

what is the equation of zero-order kinetics

A

v=vmax

48
Q

the lineweaver-burk plot

A

used for determining the types of enzyme inhibition

49
Q

what do enzyme inhibitors do?

A

they lower the reaction velocity, decreasing Vmax and increasing Km

50
Q

what are the 3 main types of enzyme inhibitors

A

competitive inhibitors
non-competitive inhibitors
uncompetitive inhibitors

51
Q

how do competitive inhibitors work

A

usually structurally related to the substrate, bind reversibly with the enzyme at/near active site. The Km is increased therefore the V is reduced. it is theoretically possible to overcome competitive inhibitors at high [S].

52
Q

how do non-competitive inhibitors

A

they bind reversibly to the enzyme at areas other than the active site. binding of the inhibitor and substrate is independent, therefore can form complex between enzyme and inhibitor [EI] and also between enzyme, inhibitor and substrate [EIS]. Km is unchanged, but Vmas is reduced and therefore v is reduced

53
Q

how can non-competitive inhibition be overcome?

A

by increasing the substrate since binding of the 2 sites is independent

54
Q

how do uncompetitive inhibitors work?

A

combine only with the ES complex. vmax is reduced, km is also reduced but this is insufficient to overcome the reduction in vmax therefore overall v is reduced

55
Q
A

Principles of Analysis (8 decks)

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