<span>1.11 g/cm³
Hope this helps! </span>
Answer:
27 min
Explanation:
The kinetics of an enzyme-catalyzed reaction can be determined by the equation of Michaelis-Menten:
![v = \frac{vmax[S]}{Km + [S]}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bvmax%5BS%5D%7D%7BKm%20%2B%20%5BS%5D%7D)
Where v is the velocity in the equilibrium, vmax is the maximum velocity of the reaction (which is directed proportionally of the amount of the enzyme), Km is the equilibrium constant and [S] is the concentration of the substrate.
So, initially, the velocity of the formation of the substrate is 12μmol/9min = 1.33 μmol/min
If Km is a thousand times smaller then [S], then
v = vmax[S]/[S]
v = vmax
vmax = 1.33 μmol/min
For the new experiment, with one-third of the enzyme, the maximum velocity must be one third too, so:
vmax = 1.33/3 = 0.443 μmol/min
Km will still be much smaller then [S], so
v = vmax
v = 0.443 μmol/min
For 12 μmol formed:
0.443 = 12/t
t = 12/0.443
t = 27 min
Answer:
Adding more substrate would overcome the effect of the compound
Explanation:
- Enzymes are biochemical catalysts that speed up chemical reactions. They act on specific substrate to convert them to products.
- Compounds known as inhibitors slow down the rate of enzyme activity.
- Inhibitors are classified as competitive and non-competitive inhibitors.
- Competitive inhibitors will compete with the substrate to bind the active sites on the enzyme. The effect of competitive inhibitors may be reduced by increasing the concentration of the substrate.
- The compound added by the biologist was a competitive inhibitor and therefore adding more substrate would overcome its effect on enzyme catalysis
- Non-competitive inhibitors binds the active site of the enzyme permanently and prevents the substrate from accessing the active sites.
By considering the reaction equation is:
5Br(aq)+BrO3(aq)+6H(aq)= 3Br2(aq)+3H2O(l)
when the average rate of consumption of Br = 1.86x10^-4 m/s
So from the reaction equation
5Br → 3Br2 when we measure the average rate of formation (X) during the same interval So,
∴ 1.86x10^-4/5 = X / 3
∴X = 1.1 x 10^-4 m/s
∴the average rate of formation of Br2 = 1.1x10^-4 m/s
