Use the attached figure to answer the following FOUR questions, including this o

Question

Use the attached figure to answer the following FOUR questions, including this one Trypanothione Reductase, which reduces the disulfide form of the compound trypanothione using NADPH, and is a target for the treatment of trypanosomiasis sleeping sickness). In this example, the inhibitor is a natural product known as "TNQ." (Note that the reaction rate was measured in mol product formed per minute.) The V m(app) for Trypanothione Reductase inhibited by 5 M TNQ is approximately A. 30.0 umol/min B.-0.033 M-1 C. 0.002 min/umol D. 500 mol/min E. 1.0 mol/min Attachment: 0,008 Increasing concentrati on of TNQ 0004 No Inhibitor 0.05 -005%,n /(Trypanothione!,pM Figure 5 pmg

Explanation / Answer

In this question Vamx and Km values are not provided but this indicates the non competitive inhibition curve where Km is not changing, because all the inhibitor concentration graphs are meeting at the same point of X-axis. But in these graphs cut at different points at the the Y-axis, hence, all the graphs are cutting at different Vmax. All the graphs follow having different intercept. If any values of straight line is given than Km and Vmax can be calculate. For non competitive graph we need below equation that is derived in presence of non-competitive inhibitor.

Vma app= Vmax/ (1+ I/Ki) -------------------1

For this equation we don’t have any value.

Hence we have to closely observe the Y-axis graph and see the intersecting point. If we look carefully than find that Y-axis is divided by 0.001 interval scale. In this graph 5uM TNQ is cutting at the intercept position of 0.002, hence, this represents the 1/Vmax

0.002=1/Vmax

Vamx= 1/0.002= 500umol/ min

In this question Vmax without inhibitor

1/0.001= 1000umol/min

In the question no 7 they asked for the Vmax of non competitive. It seems typing error in that question the answer is 1000 umol/min. (question 7)

Question 8.

It is non competitive inhibition.

Km and Ki will be the same.

Now the keep these values in equation 1.

500= 1000/ (1+ 5/Ki)

(1+ 5/Ki) 500=1000

500Ki+2500= 1000ki

2500/km= 1000-500= 500ki

5uM= Ki=Km.

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