Practical D Flashcards
You need to be sure that when you determine the rate of reaction (in mol of product formed / minute) the enzyme ….
has been active at a more or less constant rate throughout your incubation.
As long as there is substrate present, the amount of product formed might be expected to increase in a linear fashion as the amount of enzyme increases, as shown in the graph below.
The graph below shows the effect of increasing the concentration of an enzyme that has low activity as the monomer and higher activity when it has formed a dimer at a higher concentration of enzyme.
The graph below shows the opposite - an enzyme that has higher activity as the monomer, and decreased activity as it associates to form a dimer, at higher concentrations.
The graph below shows the results you would observer if there were formation of the product other than as a result of enzyme activity. – nie trzeba myślę, krótko tylko opisać
przypomnienie
It is usual to set up incubations using the lowest amount of enzyme that can be pipetted with acceptable precision, and which leads to the formation of a readily detectable amount of product.
If you are working with a relatively unpurified preparation, or do not know the molecular mass of your enzyme, then you can express the activity of the enzyme as:
- mol of product formed / unit time / volume of preparation
- mol of product formed / unit time / gram of original tissue
- mol of product formed / unit time as the total amount in the original tissue
If you are working with a partially purified enzyme preparation you should express its activity as the specific activity:
specific activity = mol of product formed / unit time / mg protein
If you are working with a purified enzyme preparation, and know its molecular mass, or if there is some other way of determining the molar concentration of the enzyme in the sample, e.g. by measurement of a prosthetic group bound to the enzyme protein, then it is possible to express the activity of the enzyme as the catalytic rate constant:
kcat = mol of product formed / sec / mol of enzyme.
The effect of temperature on the rate of an enzyme-catalysed reaction is the result of two opposing factors:
- As with any chemical reaction, the rate increases as the temperature increases, since the activation energy of the reaction can more readily be provided at a higher temperature. This means, as shown in the graph below, that there is a sharp increase in the formation of product between about 5 - 50°C.
- Because enzymes are proteins, they are denatured by heat. Therefore, at higher temperatures (over about 55°C in the graph below) there is a rapid loss of activity as the protein suffers irreversible denaturation.
Przypomnienie
- 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.
. The equation relating the rate of a chemical reaction to temperature was derived empirically by Arrhenius in 1889:
k = A * e ^ - (E/RT)
where:
k = the rate constant for the reaction
A = a constant for the reaction
E = the activation energy
R = gas constant = 1.987 cal / deg / mol = 8.31434 joules / deg / mol
T = temperature (°K, = °C + 273)
Taking logs gives: log = log A - E / 2.303 * R * T
For temperatures below that at which there is significant denaturation of the enzyme, if the enzyme is saturated with substrate, the rate constant is = Vmax, and a graph of log Vmax against 1 / T will be a straight line with gradient = - E / 2.303 * R
Przypomnienie:
Activation energy
E = -4.576 * gradient (cal / deg / mol)
E = -19.148 * gradient (joules / deg / mol)
A simple chemical reaction with a single substrate shows a linear relationship between the rate of formation of product and the concentration of substrate, as shown below:
For an enzyme-catalysed reaction, there is usually a hyperbolic relationship between the rate of reaction and the concentration of substrate, as shown below:
(A) At low concentration of substrate, there is a steep increase in the rate of reaction with increasing substrate concentration. The catalytic site of the enzyme is empty, waiting for substrate to bind, for much of the time, and the rate at which product can be formed is limited by the concentration of substrate which is available.
(B) As the concentration of substrate increases, the enzyme becomes saturated with substrate. As soon as the catalytic site is empty, more substrate is available to bind and undergo reaction. The rate of formation of product now depends on the activity of the enzyme itself, and adding more substrate will not affect the rate of the reaction to any significant effect.
An enzyme with a low Km
relative to the physiological concentration of substrate, as shown above, is normally saturated with substrate, and will act at a more or less constant rate, regardless of variations in the concentration of substrate within the physiological range.
An enzyme with a high Km
has a low affinity for its substrate, and requires a greater concentration of substrate to achieve Vmax.”
