Using the Michaelis-Menten equation competitive inhibition, the Inhibition constant, Ki of the inhibitor is 53.4 μM.
<h3>What is the Ki for the inhibitor?</h3>
The Ki of an inhibitor is known as the inhibition constant.
The inhibition is a competitive inhibition as the Vmax is unchanged but Km changes.
Using the Michaelis-Menten equation for inhibition:
Making Ki subject of the formula:
where:
- Kma is the apparent Km due to inhibitor
- Km is the Km of the enzyme-catalyzed reaction
- [I] is the concentration of the inhibitor
Solving for Ki:
where
[I] = 26.7 μM
Km = 1.0
Kma = (150% × 1 ) + 1 = 2.5
Ki = 26.7 μM/{(2.5/1) - 1)
Ki = 53.4 μM
Therefore, the Inhibition constant, Ki of the inhibitor is 53.4 μM.
Learn more about enzyme inhibition at: brainly.com/question/13618533
Answer:
ans is (2) 2,4- hexadiene
The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =

Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =


Thus vant hoff factor for sodium chloride in X is 1.9
Answer:
the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature