In the Lineweaver-Burk plot of (initial rate) ${ }^{-1}$ vs. (initial substrate concentration) ${ }^{-1}$ for an enzyme catalysed reaction following Michaelis-Menten mechanism, the y-intercept is $5000 \mathrm{M}^{-1} \mathrm{s}$. If the initial enzyme concentration is $1 \times 10^{-9} \mathrm{M}$, the turnover number
(a)$2.5 \times 10^{3}$
(b)$1.0 \times 10^{4}$
(c)$2.5 \times 10^{4}$
(d)$2.0 \times 10^{5}$
Answer
Answer: D ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
y-intercept = 1/V_max = 5000 M⁻¹ s ⇒ V_max = 2×10⁻⁴ M s⁻¹; turnover number = V_max/[E]₀ = 2×10⁻⁴/10⁻⁹ = 2×10⁵ s⁻¹.