Question

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Question 8: (20 points) Using the following data set, graph a primary plot, a lineweaver burk plot and calculate K and V round to the nearest thousandthfor the enzyme. Also determine what type of inhibitor each inhibitor is and provide a lineweaver burk plot that shows the enzyme and its inhibited rate. There should be one lineweaver burk graph for each inhibitor Enzyme rate 0.2 0.35 Inhibited enzyme rate 1 0.1 0.17 0.22 0.27 0.3 0.5 Inhibited enzyme rate 2 0.14 0.18 0.22 0.25 0.27 0.28 0.3 0.55 0.59 0.63 0.64 0.35 0.4

Answer 1

Answer:

8):

• The lineweaver burk plot for the given values of inhibited enzyme rate 1 is shown below,

rata 1_ Lineweaver burk plot 1/[S] 1/enzyme rate 1/rate 1 0.2 5 10 0.1 2.857142857 5.882353 0.066667 2 4.545455 0.05 1.818181818 3.703704 0.04 1.694915254 3.333333 0.033333 1.587301587 2.857143 0.028571 1.5625 2.5 y = 42.84x + 1.514 RP = 0.997 1/rate y = 20.47x + 0.843 R = 0.992 - 1/enzyme rate 1/rate 1 0.05 0.1 0.15 0.2 0.25 1/[S]

$\rightarrow$The equation for lineweaver burk plot is,

$\rightarrow$

1_ v Km Vmax [S] 1 Vmax

• This equation resembles straight line equation
  y = mx + c
  , where, m is the slope and c is the y-intercept.

$\rightarrow$For enzyme

$\rightarrow$The y-intercept in the above plot is given by.

$\rightarrow$From the above plot y-intercept =0.843.

Vmax = 1. 2 1

$\rightarrow$The slope in the above plot is given by .

$\rightarrow$From the above plot slope =20.47

Km = 20.47 Vmax

$\rightarrow$For inhibited enzyme rate 1,

$\rightarrow$The y-intercept in the above plot is given by.

$\rightarrow$From the above plot y-intercept =1.514.

${V_{max}}=0.66\approx 1$

$\rightarrow$The slope in the above plot is given by .

$\rightarrow$From the above plot slope =42.84

• From the graph it is inferred the two plots does not have same slope or y-intercept and have near about similar value of x-intercept,.
• Thus, inhibition 1 represents non-competitive inhibition.

$\rightarrow$The lineweaver burk plot for the given values of inhibited enzyme rate 2 is shown below,

Line weaver burk plot 00 1/[S] 1/enzyme rate 1/rate 2 0.2 5 7.142857 0.1 2.857142857 5.555556 0.066667 2 4.545455 0.05 1.818181818 0.04 1.694915254 3.703704 0.033333 1.587301587 3.571429 0.028571 1.5625 3.333333 y= 22.21x + 2.904 R? = 0.972 in y = 20.47x + 0.843 R2 = 0.992 1/rate + - 1/enzyme rate -1/rate 2 N O 0.05 0.1 0.15 0.2 0.25 1/[S]

$\rightarrow$The equation for lineweaver burk plot is,

1_ v Km Vmax [S] 1 Vmax

• This equation resembles straight line equation
  y = mx + c
  , where, m is the slope and c is the y-intercept.

$\rightarrow$For inhibited enzyme rate 2,

• The y-intercept in the above plot is given by.
• From the above plot y-intercept =2.904.

$\rightarrow$The slope in the above plot is given by .

$\rightarrow$From the above plot slope =22.21

Km – 22.21 Vmax

$K_{m}=7.6\approx8$

• From the graph it is inferred the two plots does not have y-intercept or x-intercept, but have near about same values of slope.
• Thus, inhibition 2 represents uncompetitive inhibition.

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