How does inhibitor affect the reaction rate of enzyme activity?

The graph given below shows the Michaelis-Menten kinetics for two reactions ($R_1$ and $R_2$; represented as solid and dotted lines, respectively). Enzymes $E_1$ and $E_2$ have $V_{\max }$ values, $V_{\max 1}$ and $V_{\max 2}$, respectively. The respective $K_M$ values for $E_1$ and $E_2$ are $K_{M1}$ and $K_{M2}$. ($K_{M1}=K_{M2}$).

The correct statement(s) for reactions $R_1$ and $R_2$ is/are

Given below are two statements :

Statement I : Low temperature preserves the enzyme in a temporarily inactive state whereas high temperature destroys enzymatic activity because proteins are denatured by heat.

Statement II : When the inhibitor closely resembles the substrate in its molecular structure and inhibits the activity of the enzyme, it is known as competitive inhibitor.

In the light of the above statements, choose the correct answer from the options given below : 

The correct match is:

A scientist performed enzyme kinetics and obtained velocity$\mathrm{v}$ as a function of substrate concentration $\left[\mathrm{S}\right]$ and the values are given in the table below.

The ${\mathrm{K}}_{\mathrm{m}}$ of the substrate could

Study the following

The correct combinations are

The following graph shows the concentration of substrate on enzyme activity:

What does the $\mathrm{Y}$-axis represent?

The graph below depicts the velocity of an enzyme-catalyzed reaction vs. substrate concentration. Identify the alphabets with the correct descriptions.

I. $\mathrm{A}=$ Enzyme activity

II. $\mathrm{B}=$ Substrate concentration at maximum velocity

III. $\mathrm{A}=$ Substrate concentration at which half of the maximum velocity

IV. $\mathrm{B}=$ Half of the maximum velocity