You want to understand your body in a more meaningful way so...

You want to understand your body in a more meaningful way, so over break you purify pepsin and conduct a kinetic analysis on it 2nM Pepsin v0 (uM/sec) 300 200 100 0 20 40 60 80 [S], mM a. (7 points) What is $k_{cat}$ for pepsin? 2nM Pepsin 1/v0 (sec/uM) 0.04 0.03 0.02 0.01 0 0 0.1 0.2 y = 0.0529x + 0.0029 0.3 1/[S] (1/mM) 0.4 0.5 0.6 b. (13 points) What concentration of substrate is present when you have 20 nM/sec velocity from 2nM enzyme? c. (10 points) Since many people drink alcohol during the holidays, you decide to see what the effect of ethanol is on pepsin activity. Surprisingly, you find that it is a competitive inhibitor with a $K_D$ of 0.129M. If you have a blood alcohol of 0.005M, sketch new lines on the graphs above to show the effect of the competitive inhibitor on pepsin activity. Give values for the new $V_{max, app}$ and $K_{M, app}$ on your graph.

Transcript

00:01 Hello student, here the question is given and the question is about these enzyme kinetics.

00:07 So, here the tables is given and we have to find out the value of vmax, we have to calculate.

00:16 So, here by using the given value the graph of the substrate concentration, so the graph of substrate concentration and the rate of reaction is plotted by using the value in the given table.

00:43 So, here we are using the inverse of this substrate concentration and inverse of the rate of reaction and we know between these two the rate of reaction and this is the inverse of this rate of reaction is going to plotted with the inverse of the substrate concentration at the x -axis and this is the y -axis.

01:07 So, by plotting these values this is a line weaver -burk plot.

01:18 So, according to this the value is given which is this 0 .2, 0 .4, 0 .6, 0 .8, so we are taking the inverse value and here similarly the negative values of these 0 .4, 0 .6, so all are the negative 0 .8 and intercept which are going to cut at this particular place is 1 by minus 1 by km if it is cut at that particular place.

01:54 So, here so we are taking getting these kind of graph in the line weaver -burk plot.

02:00 So, intercept at this y -axis intercept we are getting at this y -axis is 1 by vmax and here we are getting the value of 0 .4 and similarly this could be the 0 .2 and all and we are knowing this slope is comes as the kmax upon vmax.

02:28 So, by taking the value from the graph here the intercept which is 0 .4 into 10 to the power 9 and this is equal to 1 by vmax.

02:42 So, from here we are simply calculating the value of vmax which is 1 by 0 .4 into 10 to the power 9.

02:53 So, it could be equal to 2 .5 into 10 to the power minus 9 molar per minute.

03:01 So, this is the value of vmax and if from the graph it is visible that x -axis is the intercept which is going to lead to the km value.

03:13 So, for the 1 by km which is a minus value so here we are getting the 0 .7.

03:19 So, if we are calculating this km so it could be 0 .07 inverse...

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