Enzyme kinetics Flashcards

1
Q

Explain how enzyme activity may be measured using spectrophotometry

A

If the one of the product of a reaction has a significant absorbance at a wavelength where the substrate does not, then measuring the absorbance using spectrophotometry can be used to track the progress of the reaction

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2
Q

Explain how experimental values of reaction velocity at different substrate concentrations can be used to derive Km and Vmax for an enzyme

A

By plotting the absorbance against time for each concentration, the initial rate can be determined for each. After converting this to a rate involving volume instead of absorbance using the Beer-Lambert law, 1/S can be plotted on the x axis and 1/V0 on the y axis. The x intercept is equal to -1/Km and the y intercept 1/Vmax

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3
Q

Explain the effects of competitive and non-competitive inhibitors on Km and Vmax

A

Competitive inhibition: no effect on Vmax because if completely outcompeted by genuine substrate, maximum effect of enzyme unchanged, yet Km increased because genuine substrate must compete with inhibitor

Non-competitive inhibition: no effect on Km but decreased Vmax because some active sites have been denatured by allosteric inhibitor bindin

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4
Q

Describe the Lineweaver-Burk Plot

A

Since Vmax can be difficult to determine with accuracy directly, Lineweaver and Burk devised a method of plotting the data as reciprocals. The initial velocity (V0) of a series of reactions are calculated for a range of substrate concentrations [S]. As [S] increases 1/[S] tends towards zero, so that 1/Vmax is determined as the value of 1/V0 when 1/[S] = 0 (i.e. the intercept on the axis). Extrapolating the straight line graph until it crosses the other axis gives a 1/[S] value of –1/KM

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5
Q

What is meant by Km

A

According to Michaelis-Menten kinetics, Vmax is the maximum enzyme velocity which is approached asymptotically as [S] becomes large. KM, also known as the Michaelis constant, is the substrate concentration at which the rate of reaction is exactly ½ Vmax

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6
Q

How can you convert absorbance to concentration to rate

A

A = ε ́ c ́ l, where l = 1 cm (the length of the light path through the cuvette), you can simply rearrange the equation to give: c = A/ ε. For example, one group found that with the 4.0 mM GPNA solution the rate of change in A410 was 0.19 absorbance units per minute. This implies a rate of change in the concentration of p-nitroaniline (Vo) of 0.19/8.8 = 0.022 μmol.ml-1.min-1 and a value of 1/V of 46.3 μmol.ml -1.min-1.

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7
Q

What do the intercepts define

A

The intercept on the x axis defines –1/KM.

The intercept on the y axis defines 1/Vmax.

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8
Q

Describe the errors in this practical

A

In this experiment there can be quite a lot of error in determining the intercepts, and taking the reciprocal has the effect of magnifying errors in the measured rate, especially at the lower values of [S] such as the 0.5mM GPNA.

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