Answer:
The turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1.
Explanation:
Given:
The concentration of bovine carbonic anhydrase = total enzyme concentration = Et = 3.3 pmol⋅L^–1 = 3.3 × 10^–12 mol.L^–1
The maximum rate of reaction = Rmax (Vmax) = 222 μmol⋅L^–1⋅s^–1 = 222 × 10^–6 mol.L^–1⋅s^–1
The formula for the turnover number of an enzyme (kcat, or catalytic rate constant) = Rmax ÷ Et = 222 × 10^–6 mol.L^–1⋅s^–1 ÷ 3.3 × 10^–12 mol.L^–1 = 67,272,727.27 s^–1
Therefore, the turnover number of the enzyme molecule bovine carbonic anhydrase = 67,272,727.27 s^–1
the final velocity will be 0
(v-u)=(0-0)
=0
Answer:
the correct answer is option A
Explanation:
given,
Load = 6000 lbf
depth = 3.5 ft
distance from the load = 4 ft
vertical stress = 
= 
= 28.96 lbf/ft²
= 29 lbf/ft²
hence, the correct answer is option A
Answer:
10 km/s^2
Explanation:
Acceleration = velocity/time
a= 70km/7s
= 10km/s^2
