Problem Breakdown

Given:

• Initial velocity, v₀ = 1/4 V_max
• Substrate concentration, [S] = 3.0 × 10⁻³ nM

The Michaelis–Menten equation is:

v₀ = (V_max [S]) / (K_m + [S])

Substituting v₀ = 1/4 V_max:

1/4 = [S] / (K_m + [S])

Step-by-Step Solution

Multiply both sides by (K_m + [S]):

(1/4)(K_m + [S]) = [S]

Multiply both sides by 4:

K_m + [S] = 4[S]

Rearranging:

K_m = 3[S]

Unit Conversion

Given substrate concentration:

[S] = 3.0 × 10⁻³ nM

Convert nM to μM:

1 nM = 10⁻³ μM

[S] = 3.0 × 10⁻³ × 10⁻³ μM = 3.0 × 10⁻⁶ μM

Therefore:

K_m = 3 × 3.0 × 10⁻⁶ μM = 9.0 × 10⁻⁶ μM

Hence, the final answer is:

K_m = 9 μM

No Options Provided

Although the question mentions “explain every option,” no multiple-choice options
are given. This format is typical of numerical answer–type questions in
CSIR NET and GATE Life Sciences, where the correct
response is entered directly as a number.

Key Concepts for Exams

• K_m is the substrate concentration at
  V_max/2 and reflects enzyme–substrate affinity.
• At low substrate concentration ([S] ≪ K_m),
  enzyme kinetics are first-order.
• At high substrate concentration ([S] ≫ K_m),
  kinetics become zero-order.

Exam Relevance

Problems based on
enzyme catalyzed reaction Km value initial velocity one fourth Vmax
are common in CSIR NET Life Sciences. Memorizing the shortcut
v₀ = V_max/4 ⇒ K_m = 3[S]
allows quick and accurate problem solving under exam conditions.

SEO Keywords

enzyme catalyzed reaction Km value, initial velocity one fourth Vmax,
Michaelis–Menten numerical CSIR NET, enzyme kinetics GATE Life Sciences,
Km calculation from v0 and Vmax