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BIOC1009 Dina > Experiment D: enzyme assay > Flashcards

Experiment D: enzyme assay Flashcards

1
Q

regarding pH, when does an enzyme show maximum activity?

A

at the optimum pH

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2
Q

why does pH affect enzymatic activity?

A
  • substrate binding involves interaction with amino acid side chains at the active site
  • the pH may affect the ionisation of the substrate and also of the side chains at the active site, thus affecting binding
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3
Q

what is the effect of extreme values of pH?

A

extreme values of pH may disrupt the tertiary structure of the enzyme to distort the active site and even denature the protein

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4
Q

what is the effect of having mainly uncharged reactive groups at the active site?

A

the enzyme has activity over a relatively broad range of pH values

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5
Q

what is the plasma or intracellular pH?

A

pH 7.35-7.45

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6
Q

how do enzymes with very different pH optima to the intracellular average maintain their activity?

A

by sub-cellular compartmentation

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7
Q

how should the activity of an enzyme be determined with respect to pH?

A

the activity of an enzyme should be determined near or at the optimum pH to produce meaningful results

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8
Q

what is the relationship between incubation time and amount of product formed?

A

the longer an enzyme is incubated with its substrate, the greater the amount of product formed

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9
Q

true or false: the rate of product formation is a simple function of incubation time

A

false; all proteins suffer denaturation, and hence loss of catalytic activity, over time

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10
Q

are enzyme-catalysed reactions reversible or irreversible?

A

reversible

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11
Q

how does the reversibility of enzyme-catalysed reactions affect the formation of product?

A
  • initially, there is little or no product present, so the reaction proceeds only in the forward direction
  • as the reaction continues, there is a significant accumulation of product and therefore a significant rate of back reaction
  • therefore, the rate of product formation decreases as the incubation proceeds
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12
Q

what is the effect of having too long an incubation time?

A

the measured activity of the enzyme is falsely low

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13
Q

how is the rate of product formation measured regarding incubation time?

A
  • measurements made at short time intervals (eg. 10 seconds) for the first minute or so
  • graph plotted of amount of product formed against incubation time
  • initial rate determined by drawing a tangent to the steepest part of the curve
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14
Q

what is the problem with short incubation times?

A
  • considerable error of timing

- only a small amount of product is formed so analytical errors are magnified

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15
Q

what incubation time should be used for enzyme-catalysed reactions?

A
  • long enough to allow a moderate amount of product to be formed
  • long enough so that an error in timing is insignificant
  • not so long that there is a detectable levelling-off of the curve
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16
Q

what must you ensure when determining the rate of reaction?

A

that the enzyme activity has been constant throughout the incubation

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17
Q

what effect does concentration have on some monomer enzymes?

A

some enzymes form dimers or other aggregates at high concentrations, affecting the catalytic activity

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18
Q

what effect does concentration have on some enzymes that have a natural form of dimers or other aggregates?

A

these enzymes may disaggregate when the preparation is diluted, affecting the catalytic activity

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19
Q

how can the ready denaturation in dilute solution of some enzymes be countered?

A

by adding a relatively large amount of inert protein, such as albumin, to act as a chaperone to the purified complex

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20
Q

what would be the effect of product formation other than that as a result of enzyme activity?

A
  • graph of product formed against enzyme concentration
  • constant gradient
  • y-intercept above zero, since some product is present even when there is no enzyme present
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21
Q

what amount of enzyme should be used?

A
  • enzyme preparation is generally the most valuable constituent of the incubation
  • lowest amount of enzyme that can be pipetted with acceptable precision
  • and that leads to the formation of a readily detectable amount of product
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22
Q

how is the activity of an unpurified enzyme expressed?

A

mol of product formed / unit time / volume of preparation

mol of product formed / unit time / mass of original tissue (g)

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23
Q

how is the activity of a partially purified enzyme expressed?

A

specific activity;

mol of product formed / unit time / mass of protein (mg)

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24
Q

how is the activity of a purified enzyme expressed?

A

mol of product formed / time (s) / mol of enzyme

if the molecular mass is known

25
Q

what is the effect of temperature on an enzyme-catalysed reaction?

A
  • as the temperature increases, the rate increases
  • since more molecules have an energy at/above the activation energy
  • sharp increase in the formation of product 5-50C
    BUT
  • because enzymes are proteins, they are denatured by heat
  • at higher temperatures, there is a rapid loss of activity as the protein suffers irreversible denaturation
26
Q

why is it not useful to determine the ‘optimum’ temperature for an enzyme-catalysed reaction?

A

the longer the incubation time, the lower the temperature at which there is maximum formation of product, because of the greater effect of denaturation of the enzyme

27
Q

what is the conventional temperature used for enzyme-catalysed reactions?

A

30C; this is a compromise between

  • mammalian/clinical biochemists - 37C
  • microbial biochemists - 20C
28
Q

what is the Arrhenius equation?

A

k = AE^-Ea/RT

29
Q

when is k equal to Vmax?

A

for temperatures with significant denaturation, if the enzyme is saturated with substrate

30
Q

what is the relationship between rate of reaction and the concentration of substrate?

A

there is a hyperbolic relationship between the two variables

31
Q

what is the limiting factor at low substrate concentrations?

A

the concentration of available substrate

32
Q

what is the limiting factor at increasing substrate concentrations?

A

the enzyme activity

33
Q

what is the rate of reaction when the enzyme is saturated with substrate?

A

Vmax

34
Q

what is Km?

A
  • the concentration of substrate that allows the enzyme to achieve half Vmax
  • the concentration of substrate at which the enzyme is at half-saturation
35
Q

what does it mean if an enzyme has a relatively high Km?

A
  • low affinity for its substrate
  • requires a greater concentration of substrate to achieve Vmax
  • not normally saturated with substrate
  • activity will vary as the substrate concentration varies
36
Q

what does it mean if an enzyme has a relatively low Km?

A
  • high affinity for its substrate

- acts at an approximately constant rate

37
Q

under which conditions can the amount of enzyme in a sample be determined?

A
  • the limiting factor must be the enzyme activity
  • and not the amount of available substrate
  • concentration of substrate must be high enough to ensure that the enzyme is acting at Vmax
38
Q

what is the amount of enzyme generally used to determine the activity for an enzyme-catalysed reaction?

A

about 10-20-fold higher than the Km

39
Q

under which conditions can the concentration of substrate be determined for an enzyme-catalysed reaction?

A
  • the substrate must be the limiting factor
  • the concentration of substrate must be below Km
  • so that the rate of product formation increases steeply with increasing substrate concentration
40
Q

outline the features of a Lineweaver-Burk plot

A 
1/V = 1/Vmax + Km/Vmax (1/[S])
plotting 1/V against 1/[S] ,
y-intercept = 1/Vmax
x-intercept = - 1/Km
gradient = Km/Vmax
41
Q

what is a disadvantage of the L-B plot?

A
  • places undue weight on points obtained at low concentrations of substrate,
  • at which precision of determining the rate of reaction is lowest
  • because of the small amount of product that has been formed
42
Q

outline the features of a Eadie-Hofstee plot

A 
V = Vmax - Km (V/[S])
plotting V against V/[S] ,
y-intercept = Vmax
x-intercept = Vmax/Km
gradient = - Km
43
Q

what are the dis/advantages of the Eadie-Hofstee plot?

A

advantage: overcomes the problem of uneven spacing of points and undue weight given to low values of [S]
disadvantage: V (the dependent variable) is used on both axes; errors in measuring rate of reaction are multiplied, resulting in lower precision of Km and Vmax estimates

44
Q

outline the features of a Hanes plot

A 
[S]/V = Km/Vmax + [S] (1/Vmax)
plotting [S]/V against [S] ,
y-intercept = Km/Vmax
x-intercept = - Km
gradient = 1/Vmax
45
Q

what are the dis/advantages of a Hanes plot?

A

advantage: overcomes the problem of uneven spacing of points and undue weight given to low values of [S]
disadvantage: [S] (the dependent variable) is used on both axes; errors in measuring rate of reaction are multiplied, resulting in lower precision of Km and Vmax estimates

46
Q

explain competitive inhibition and its effect on an enzyme-catalysed reaction

A
  • inhibitor competes with the substrate for the active site
  • Vmax is unchanged
  • Km is increased
  • gradient is increased
47
Q

explain why Vmax is unchanged but Km appears increased on addition of a competitive inhibitor

A

Vmax - if the enzyme concentration is sufficiently high, the enzyme will still reach full saturation

Km - more substrate is required to reach half saturation; though the affinity of the enzyme is not affected

48
Q

explain non-competitive inhibition and its effect on an enzyme-catalysed reaction

A
  • inhibitor binds to both the enzyme and to the enzyme-substrate complex
  • Vmax is decreased
  • Km is unchanged
  • gradient is increased
49
Q

explain why Km is unchanged but Vmax is decreased on addition of a non-competitive inhibitor

A

Km - the enzyme has the same affinity for the enzyme

Vmax - but the reaction cannot reach Vmax

50
Q

explain uncompetitive inhibition and its effect on an enzyme-catalysed reaction

A
  • inhibitor binds only to the enzyme-substrate complex
  • Vmax is decreased
  • Km is decreased
  • gradient remains constant
51
Q

explain why both Vmax and Km are decreased on addition of an uncompetitive inhibitor

A

Vmax - the reaction cannot reach Vmax

Km - the affinity of the enzyme is increased

52
Q

explain why the gradient of a L-B plot remains the same on addition of an uncompetitive inhibitor

A

Vmax and Km decrease to the same extent

53
Q

what is Ki?

A

the inhibition constant;

  • indication of how potent an inhibitor is
  • the concentration of substrate required to produce half-maximum inhibition
54
Q

what is a Dixon plot?

A
  • plot 1/V against concentration of inhibitor at each substrate concentration
  • gives a family of intersecting lines
55
Q

what is the point of convergence of a Dixon plot for a competitive inhibitor?

A
  • convergence above the x-axis

- at an x-value of -Ki

56
Q

what is the point of convergence of a Dixon plot for a non-competitive inhibitor?

A
  • convergence on the x-axis

- intersect is -Ki

57
Q

what incubation time was used in the experiment?

A

10 minutes

58
Q

what amount of enzyme was used in the experiment?

A

100 μL

59
Q

what is the effect of an enzyme on the energy of the reaction?

A
  • lowers the energy of the transition state

- raises the energy of the ground state

BIOC1009 Dina (5 decks)

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