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QUESTION 21 Consider the following graph and answer questions 21 through 24 below: 10 Cat N-2 1/v(min/mm) -5 - 1 0 1 2 3 4 5
QUESTION 22 What is the approximate Vmax for [1] = 4 mM? OA. 0.25 uM/min, OB.0.50 MM/min, OC.2.00 uM/min, OD.3.00 uM/min, OE.
What type inhibitor is I? A.competitive, B.uncompetitive, C.pure-noncompetitive, OD. mixed ( Kj > Ki), O E. all of the above
What is the ki or Kj? O A. 0.25 mm OB.0.5 mm O C. 1 mM OD.2 mm O E. 4 mM OF. 8 MM
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Answer #1

22.- Remember that Vmax is equal to the inverse of the value where the graph crosses the Y axis, it crosses at a value of 4 mM, then the answer is 1/4

1/4 = 0.25 mM

The answer is option A. 0.25 mM/min

23.- We can now that by the position of the intersection of the 3 different lines. The intersection occurs in none of the axes, that means it is a mixed inhibitor.

The answer is option D. Mixed

24.- Remember that Ki is the inhibitor concentration required to bring the Vmax to half its regular value. The regular value here is 1/2 = 0.5, that means we require to bring the Vmax to 0.5/2 = 0.25. We already did that in question 22. The required concentration of inhibitor is 4 mM.

The answer for this question is option E. 4 mM

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