Question

Given an enzyme with a Km = 10m M and Vmax = 100 m mol/min. If [S] = 100 m M, which of the following will be true?
A.A 10 fold increase in Vmax would increase velocity 10 fold y
B. A 10 fold decrease in Km would increase velocity
C. Both (a) and (b)
D. A 10 fold increase in Vmax would decrease velocity 20 fold

Answer 1

The data given to us is from Kinetics of Enzymatic activities and the kinetic accounting for enzyme formations in equations for Michaelis-Menten Enzymatic kinetic equations.

Functions of enzymes are to enhance and purport the rates induced with higher rates in the reactions that should be balancing the needs of other organisms. For taking in the Enzymatic function a kinetic activity is needed for a brief overview in how does it actually "increase the reaction rates".

An enzyme can be defined for set rate of catalytic reaction as $\bf{V_0}$ that is the total number of moles for a product in every counted second, which is in a slight differentiation to that of the "S" of the substance or the substrate concentration of the given Enzymatic substance. A substrate concentration is increased when $V_0$ is increasing until it becomes in a constant rate after finishing the stage of Half of Max velocity, it just means that the substrate concentration is adhering with the principles of Michaelis-Menten Enzymatic kinematical equation and exhibits the criteria for reaching the maximum or maximal velocity that is $\bf{V_{max}}$ in a asymptomatic manner. The standard constant of $\bf{K_M}$ or the Michaelis constant is attaining that half max velocity of $\bf{\frac{V_{max}}{2}}$. Basically this means the catalytic rate is in a linear rise because of which the substrate concentration is increasing to and attains the final maximum point in a equally increasing substrate concentration.

Deriving the equation for the Enzyme-Substrate complex is the kinetic property itself in many of the Enzymatic reactions. That is,

$\boxed{\bf{Enzyme + Substrate \rightleftharpoons^{k_1}_{k_{- 1}} \: ES comp. \longrightarrow Enzyme + Product}}$

Here, the Enzyme is in the combination of a Substrate to form a new kinematic complex called as the Enzyme-Substrate complex or the ES complex, assigned with a specific rate constant of $k_1$. Enzyme-Substrate complex can either continue to further indulge in the disassociation process to form an Enzyme and a Substrate with a different rate constant called $k_{- 1}$, or as for the second process form Enzyme and the product with a new rate constant of $k_2$. Here, No products are going to take part in an reversible process to get back to its original form of Substrate, this presumption is critical to attain concentration of the product in final stage after initial reactions have subsided.

An equation is expressed for relating the catalytic rate to the substrate concentration that is, the rate of catalysis is equivalent to it's concentrations for the product that is of the rate constant $k_2$ and the Enzyme-Substrate Complex.

$\bf{Catalysis Rate, \: V_0 = k_2[ES \: comp.]}$

Now, Apply the expressions for the Enzyme-Substrate complex with the help of known quantitative analysis that is,

$\bf{Rate \: of \: Breakdown \: of \: ES = k_1[E] \: [S]}$

$\bf{Rate \: of \: Formation \: of \: ES = (k_{- 1} + k_2) \times [ES]}$

According to the state of steadiness presumption, the intermediate concentration of Enzyme-Substrate complex is not changing even if the concentrations for initial starters and the products at the end are "changing". ES complex (rate of formation) and it's breakdown are considered equal for the Enzyme-Substrate complex.

Therefore,

$\textbf{Rate of formation of ES = Rate of breakdown of ES}$

$\bf{k_1 \times [E][S] = (k_{- 1} + k_2) \times [ES]}$

$\bf{\therefore \quad \dfrac{[E][S]}{[ES]} = \dfrac{(k_{- 1} + k_2)}{k_1}}$

Now, the newly formed Enzyme-Substrate complex is defined as the Michaelis constant that is, $\textbf{K_M}$

$\bf{K_M = \dfrac{k_{- 1} + k_2}{k_1}}$

$\bf{[ES] = \dfrac{[E][S]}{K_M}}$

The enzyme has a total concentration that is $[E]_T$.

$\bf{\therefore \quad [E] = [E]_T - [ES]}$

$\bf{\therefore \quad [ES] = \dfrac{([E]_T - [ES]) \times [S]}{K_M}}$

$\bf{\therefore \quad [ES] = [E]_T \dfrac{[S]}{[S] + K_M}}$

Since, $\bf{V_0 = k_2 [ES]}$

$\bf{\therefore \quad V_0 = k_2 [E]_T \dfrac{[S]}{[S] + K_M}}$

Since, A max velocity rate or the maximal velocity is achieved when the enzyme's sites for catalysation is saturated for the concentration of substrate that is,

$\bf{V_{max} = k_2 [E]_T}$

$\boxed{\bf{\underline{V_0 = V_{max} = \dfrac{[S]}{[S] + K_M}}}}$

Which is the Michaelis-Menten Enzymatic kinematical equation.

Now, just substitute the following values to obtain the final answer.

$\huge{\textbf{CHECK THE ATTACHMENTS}}$

$\boxed{\bf{\underline{OPTION \: \: A) \: Is \: true}}}$