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What is the ratio of [S] to KM when the velocity of an enzyme catalyzed reaction is 80% of Vmax?
- a)4 : 1
- b)1 : 4
- c)1 : 8
- d)8 : 1
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
What is the ratio of [S] to KM when the velocity of an enzyme catalyze...
since v is 80% of Vmax assume Vmax=1, therefore v=0.80
1/v = Km/Vmax x 1/[S] + 1/Vmax
1/0.80 = Km/1 x 1/[S] + 1/1
1/0.80 - 1/1 = Km/[S] * Km/1 was multiplied with 1/[S]
0.25= Km/[S] but we want [S]/Km ratio so.0.25[S]= Km *remember Km has invisible 1 coefficient
[S]/Km= 1/0.25
[S]/Km= 4
Most Upvoted Answer
What is the ratio of [S] to KM when the velocity of an enzyme catalyze...
Given information:
Velocity of an enzyme catalyzed reaction is 80% of Vmax.
To find:
Ratio of [S] to KM.
Solution:
The Michaelis-Menten equation for an enzyme-catalyzed reaction is given by:
V = Vmax[S] / (KM + [S])
Where,
Vmax = maximum velocity of the reaction
[S] = substrate concentration
KM = Michaelis constant.
At 80% of Vmax, we can write:
0.8Vmax = Vmax[S] / (KM + [S])
Solving for [S], we get:
[S] = (0.8Vmax x KM) / (Vmax - 0.8Vmax)
[S] = (0.8 x KM) / 0.2
[S] = 4KM
Therefore, the ratio of [S] to KM is 4 : 1.
Answer:
Option A. 4 : 1.
Velocity of an enzyme catalyzed reaction is 80% of Vmax.
To find:
Ratio of [S] to KM.
Solution:
The Michaelis-Menten equation for an enzyme-catalyzed reaction is given by:
V = Vmax[S] / (KM + [S])
Where,
Vmax = maximum velocity of the reaction
[S] = substrate concentration
KM = Michaelis constant.
At 80% of Vmax, we can write:
0.8Vmax = Vmax[S] / (KM + [S])
Solving for [S], we get:
[S] = (0.8Vmax x KM) / (Vmax - 0.8Vmax)
[S] = (0.8 x KM) / 0.2
[S] = 4KM
Therefore, the ratio of [S] to KM is 4 : 1.
Answer:
Option A. 4 : 1.
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What is the ratio of [S] to KM when the velocity of an enzyme catalyzed reaction is 80% of Vmax?a)4 : 1b)1 : 4c)1 : 8d)8 : 1Correct answer is option 'A'. Can you explain this answer?
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