Q35 The turnover numbers (k₋ₐₜ) for enzymes E1 and E2 are 150 s⁻¹ and 15 s⁻¹ respectively. This means
- A (A) E1 binds substrate with higher affinity
- B (B) velocities could be equal if [E2] = 10×[E1]
- C (C) velocity of E1 > velocity of E2
- D (D) velocity of E1 < velocity of E2