Question

For enzyme \(\alpha\)-amylase with \(K_m = 0.005\,M\), the substrate concentration for one-fourth of \(V_{\max}\) (rounded to two decimals, mM) is \(\underline{\hspace{1cm}}\).

Hint:

At \( v = V_{\max}/n \), substrate concentration is \( K_m/(n-1) \) for simple Michaelis–Menten kinetics.

Answer 1

Correct Answer: 1.6

Using Michaelis–Menten equation:
\[ v = \frac{V_{\max} [S]}{K_m + [S]} \]
Given:
\[ v = \frac{V_{\max}}{4} \]
Thus,
\[ \frac{1}{4} = \frac{[S]}{K_m + [S]} \]
Cross-multiplying:
\[ K_m + [S] = 4[S] \]
\[ 3[S] = K_m \]
\[ [S] = \frac{K_m}{3} = \frac{0.005}{3} = 0.00167\,M \]
Convert to mM:
\[ 0.00167 \times 1000 = 1.67\ \text{mM} \]