Answer:
The one from E coli
Explanation:
One way of determining catalytic efficiency is to calculate the ratio of Vmax to Km
For beef heart,
[tex]\dfrac{V_{max}}{K_{m}} = \dfrac{290 \times 10^{-9}\text{ mol/min }}{1.9 \times 10^{-3}\text{ mol$\cdot$L}^{-1}} = 1.5 \times 10^{-5} \text{ L/min} = \mathbf{15 \, \mu \text{L/min}}[/tex]
For E coli ,
[tex]\dfrac{V_{max}}{K_{m}} = \dfrac{18 \times 10^{-9}\text{ mol/min }}{24 \times 10^{-6}\text{ mol$\cdot$L}^{-1}} = 7.5 \times 10^{-4} \text{ L/min} = 750 \, \mu\text{L/min}[/tex]
Thus, the enzyme from E. coli is 50 times as efficient as the one from beef hea rt.
