Chapter 4, Problem 26Q

A.

Summary Introduction

To draw: Graph of the substrate concentration in the X axis versus the reaction rate in the Y axis and the estimate the K_M and V_max values.

Concept introduction: Enzymes are biological protein catalysts that alter the speed of the reaction in the biological system. The chemical reactions that are catalyzed by these enzymes are studied under enzyme kinetics. The reaction rate is measure based on the rate at which the substrate of the enzymes combines with the enzyme and forms the products. K_M is the Michaelis–Menton constant. K_M value is not a direct measure of the affinity of the enzyme toward the substrate, but concentration of S where velocity at maximum is half. V_max is the maximum velocity with which the enzyme catalyzes the reaction.

$\text{The rate of enzyme reaction }\left(\text{V}\right)\text{ =}\frac{{\text{V}}_{\text{max}}\left[\text{S}\right]}{\left({\text{K}}_{\text{M}}\text{+}\left[\text{S}\right]\right)}$

Explanation of Solution

Pictorial representation:

Fig. 1: A graph of the substrate concentration verses the rate of the reaction

Explanation:

[S] µM	Rate µ mole/min
0.08	0.15
0.12	0.21
0.54	0.70
1.23	1.1
1.82	1.3
2.72	1.5
4.94	1.7
10.00	1.8

With the above data, the graph is plotted. The black curve in Fig. 1 is the obtained graph from which the K_Mis evident as 1µM and the V_max evident as 2 µmole/min.

B.

Summary Introduction

To explain: The results agree with the estimates made from the first graph of the raw data.

Concept introduction: Enzymes are biological protein catalysts that alter the speed of the reaction in the biological system. The chemical reactions that are catalyzed by these enzymes are studied under enzyme kinetics. The reaction rate is measure based on the rate at which the substrate of the enzymes combines with the enzyme and forms the products. K_M is the Michaelis–Menton constant. K_M value is not a direct measure of the affinity of the enzyme toward the substrate, but concentration of S where velocity at maximum is half. V_max is the maximum velocity with which the enzyme catalyzes the reaction.

$\text{The rate of enzyme reaction }\left(\text{V}\right)\text{ =}\frac{{\text{V}}_{\text{max}}\left[\text{S}\right]}{\left({\text{K}}_{\text{M}}\text{+}\left[\text{S}\right]\right)}$

Explanation of Solution

Pictorial representation:

Fig. 2: A reciprocal graph of the substrate concentration verses the rate of the reaction

Explanation:

1/[S] (1/ µM)	1/Rate min/ µ mole
12.50	6.7
8.30	4.8
1.85	1.4
0.81	0.91
0.56	0.77
0.37	0.67
0.20	0.59
0.10	0.56

With the above data, the graph is plotted. The straight black line in Fig. 2 is the obtained graph from which the K_Mis evident as 1µM and the V_max evident as 2 µmole/min. The data are easier to interpret in the second graph as the plot gives a straight line.

C.

Summary Introduction

To explain: The reason why the statement that very little product was formed in the given reaction condition is important.

Concept introduction: Enzymes are biological protein catalysts that alter the speed of the reaction in the biological system. The chemical reactions that are catalyzed by these enzymes are studied under enzyme kinetics. The reaction rate is measure based on the rate at which the substrate of the enzymes combines with the enzyme and forms the products. K_M is the Michaelis–Menton constant. K_M value is not a direct measure of the affinity of the enzyme toward the substrate, but concentration of S where velocity at maximum is half. V_max is the maximum velocity with which the enzyme catalyzes the reaction.

$\text{The rate of enzyme reaction }\left(\text{V}\right)\text{ =}\frac{{\text{V}}_{\text{max}}\left[\text{S}\right]}{\left({\text{K}}_{\text{M}}\text{+}\left[\text{S}\right]\right)}$

Explanation of Solution

The consequence will be that the calculated rate of the reaction will be lower than they actually should be because if there are more products, the rate of the reaction will decrease as the substrate is getting used to form the products. Thus products will be accumulated and the substrates will be less. The whole Micheaelis–Menton model of enzyme kinetics is based on the substrate concentration, K_M and V_max values. Look up Concept introduction for the formula.

D.

Summary Introduction

To explain: Plot of the graph for substrate concentration versus rate of the reaction and reciprocal of the substrate concentration versus the rate of the reaction when the enzyme upon regulated with phosphorylation, the K_M value increases by a factor of 3 without any change in the V_max.

Concept introduction: Enzymes are biological protein catalysts that alter the speed of the reaction in the biological system. The chemical reactions that are catalyzed by these enzymes are studied under enzyme kinetics. The reaction rate is measure based on the rate at which the substrate of the enzymes combines with the enzyme and forms the products. K_M is the Michaelis–Menton constant. K_M value is not a direct measure of the affinity of the enzyme toward the substrate, but concentration of S where velocity at maximum is half. V_max is the maximum velocity with which the enzyme catalyzes the reaction.

$\text{The rate of enzyme reaction }\left(\text{V}\right)\text{ =}\frac{{\text{V}}_{\text{max}}\left[\text{S}\right]}{\left({\text{K}}_{\text{M}}\text{+}\left[\text{S}\right]\right)}$

Explanation of Solution

When the K_M value increases, the substrate concentration to give half of the maximum rate is increased.

$\begin{array}{l}\text{Rearranging}\text{the}\text{rate}\text{of}\text{the}\text{enzyme}\text{reaction}\text{equation}\\ K_M=\frac{V_{Max}[S]}{V}-[S]\end{array}$

K_M value is directly proportional to the substrate concentration. Therefore, as the K_M increases, the substrate concentration to give half of the maximum rate increases.

Since more concentration of the substrate is needed to produce the same rate of the reaction; the regulation with phosphorylation inhibits the enzyme-catalyzing reaction. The expected data plots for the same is plotted as dotted curve in Fig 1 for substrate concentration versus rate of the reaction and dotted line in Fig 2 for reciprocal plot of substrate concentration versus rate of the reaction.

[S] µM	Rate µ mole/min
0.24	0.15
0.36	0.21
1.62	0.70
3.69	1.1
5.46	1.3
8.16	1.5
14.82	1.7
30.00	1.8
1/[S] (1/ µM)	1/Rate min/ µ mole
4.16	6.7
2.78	4.8
0.61	1.4
0.27	0.91
0.18	0.77
0.12	0.67
0.06	0.59
0.03	0.56

With the above data, 3[S] is taken for X axis and the rate of the reaction remains the same. The expected data plots for the same is plotted as dotted curve in Fig. 1 for substrate concentration versus rate of the reaction and dotted line in Fig. 2 for reciprocal plot of substrate concentration versus rate of the reaction.