​An enzyme has been found to efficiently catalyse the following reaction:​

​$$\text{Where,} \\k_f \, : \, \text{Forward rate of the reaction} \\k_r \, : \, \text{Reverse rate of the reaction} \\K_{\text{eq}} \, : \, \text{Equilibrium Constant}$$​

Which one of the following parameters will be increased over the uncatalyzed reaction by the enzyme?

Correct option is A

An enzyme catalyzes a reaction by lowering the activation energy, which increases the rate of the reaction. For the given reaction 

$A\leftrightarrow B$, the forward rate (k_f) and reverse rate (k_r​) define the equilibrium constant <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>K</mi><mrow><mi>e</mi><mi>q</mi></mrow></msub><mo>=</mo><mfrac><msub><mi>k</mi><mi>f</mi></msub><msub><mi>k</mi><mi>r</mi></msub></mfrac></mrow><annotation encoding="application/x-tex"> K_{eq} = \frac{k_f}{k_r} </annotation></semantics></math> K_eq = kfkr\frac{k_f}{k_r}kr​kf​​​

• Enzymes increase reaction rates: An enzyme will increase both <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><msub><mi>k</mi><mi>f</mi></msub></mrow><annotation encoding="application/x-tex"> k_f </annotation></semantics></math>k_f (forward rate) and k_r (reverse rate) by the same factor, as it lowers the activation energy for both directions of a reversible reaction.
• Effect on $K$: Since K_eq = $$\frac{k_f}{k_r}$$​, if both k_f and k_r  increase by the same factor, their ratio (K_eq) remains unchanged. Enzymes do not affect the equilibrium constant; they only speed up the rate at which equilibrium is reached.
• Effect on k_r​ : The reverse rate K_r​ will increase due to the enzyme.
• Effect on $$\frac{1}{k_r}$$​ : If k_r increases, then $$\frac{1}{k_r}$$​  will decrease.
• Effect on $$\frac{1}{K_{eq}}$$​ : Since k_eq remains unchanged, 
  $$\frac{1}{K_{eq}}$$ also remains unchanged.

Now, let's evaluate the options:

1. k_r: This is the reverse rate, which will increase due to the enzyme. This matches our analysis.

2.  k_eq : The equilibrium constant does not change with an enzyme, so this is incorrect.

3.  $$\frac{1}{K_{r}}$$ :  Since k_r​ increases, <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mn>1</mn><mi mathvariant="normal">/</mi><msub><mi>k</mi><mi>r</mi></msub></mrow><annotation encoding="application/x-tex"> 1/k_r </annotation></semantics></math>1/kr decreases, so this is incorrect.

4. $$\frac{1}{K_{eq}}$$​: Since k_eq​ remains unchanged, $$\frac{1}{K_{eq}}$$​ also remains unchanged, so this is incorrect.

Answer: The parameter that will be increased by the enzyme over the uncatalyzed reaction is:

1. k_r​