Chapter 6, Problem 8P

(a)

Summary Introduction

To determine: The substrate concentration at which an enzyme with a Kcat of 30.0s^-1 and Km  of 0.0050M operate at one-quarter of its maximum rate.

Introduction:

Kcat is defined as a constant that is used to describe turnover rate of enzyme-substrate complex into product and enzyme. It is a turnover number of an enzyme which is used to indicate the conversion of maximum number of substrate into product per active site per unit time.

Explanation of Solution

Explanation:

The formula for calculating concentration of substrate $\left(V_{°}\right)=\frac{V\max \left[\text{S}\right]}{Km+\left[\text{S}\right]}$ (1)

The values for Kcat, Km and V are substituted in the formula for calculating concentration of substrate,

$\begin{array}{c}{V}_{°}=\frac{V\max \left[\text{S}\right]}{Km+\left[\text{S}\right]}\\ 0.25V\max =\frac{V\max \left[\text{S}\right]}{Km+\left[\text{S}\right]}\\ 0.25=\frac{\left[\text{S}\right]}{\left[\text{Km}\right]+\left[\text{S}\right]}\\ 0.25\left[Km\right]+0.25\left[\text{S}\right]=\left[\text{S}\right]\\ 0.25\left[Km\right]=\left[\text{S}\right]-0.25\left[\text{S}\right]\\ 0.25\left[Km\right]=0.75\left[\text{S}\right]\\ \frac{0.25\left[Km\right]}{0.75}=\left[\text{S}\right]\\ \left[\text{S}\right]=0.33\left[Km\right]\\ =0.33\times 0.0050\\ =0.00165\\ =1.65\times {10}^{-3}\text{M}\end{array}$

Conclusion

Conclusion:

The concentration of substrate is $1.65\times 10^{-3}M$

(b)

Summary Introduction

To determine: The fraction of Vmax which is obtained at following substrate concentrations $\left[\text{S}\right]=\frac{1}{2}Km,2Km$,and 10 Km.

Introduction:

Enzyme kinetics is defined as the study of chemical reactions which is catalyzed by enzyme. It indicates a rate of reaction as a function of substrate concentration with Vmax and Km of enzyme.

Explanation of Solution

Explanation:

The formula for calculating fractions of Vmax at different substrate concentrations is,

$V_{°}=\frac{V\max +\left[\text{S}\right]}{\left[\text{S}\right]}$(Michaelis-Menten equation ) (2)

Substituting the value of  $\left[\text{S}\right]=\frac{1}{2}Km$

$\begin{array}{c}\frac{V_{°}}{V\max }=\frac{\left[\text{S}\right]}{Km+\left[\text{S}\right]}\\ =\frac{\frac{1}{2}Km}{Km+\frac{1}{2}Km}\\ =\frac{\frac{1}{2}Km}{\frac{2+1}{2}Km}\\ =\frac{\frac{1}{2}}{\frac{3}{2}}\\ =\frac{0.5}{1.5}\\ =0.33\end{array}$

Substituting the value of $\left[\text{S}\right]=2Km$ in equation (2),

$\begin{array}{c}\frac{V_{°}}{V\max }=\frac{\left[\text{S}\right]}{Km+\left[\text{S}\right]}\\ =\frac{2Km}{Km+2Km}\\ =\frac{2}{3}\\ =0.66\end{array}$

Substituting the value of $\left[\text{S}\right]=10Km$ in equation (2),

$\begin{array}{c}\frac{V_{°}}{V\max }=\frac{\left[\text{S}\right]}{Km+\left[\text{S}\right]}\\ =\frac{10Km}{Km+10Km}\\ =\frac{10}{11}\\ =0.90\end{array}$

Conclusion

Conclusion:

The fractions of Vmax at different substrate concentration is 0.33, 0.66 and 0.90.

(c)

Summary Introduction

To determine: The different curves corresponds to which enzyme.

Introduction:

Enzyme is a biological macromolecule that acts as a catalyst and enhance rate of chemical reactions. Various biological processes are dependent on enzyme. All enzymes have an active site on to which amino acids of proteins get binds and form an enzyme-substrate complex.

Explanation of Solution

Explanation:

The upper curve indicates enzyme B as [X] is greater than Km of enzyme while lower curve indicates enzyme A as the [X] is smaller than Km of enzyme.

It can be concluded that if the rate of initial concentration of substrate is greater than $Km$ then rate of chemical reaction is less sensitive to conversion of substrate at initial stage of reaction. So rate of reaction is linear for longer period of time.

Conclusion

Conclusion:

In the given graph, upper curve corresponds to enzyme B whereas lower curve corresponds to enzyme A.