Question

Interpret the data above

1. What is the aporoximate Km and Vmax (units matter)

2.Briefly explain how you found each

3.if you used 0.020 microM enzyme in your studies, what is kcat in units of s^-1? Show your work units matter

4. Does this enzyme appear to display cooperativity? Explain how you came up to that conclusion.

Directions: Below are data from 4 separate experiments that you must analyze/evaluate. Using the information from all 4 experiments, you must then propose a catalytic mechanism for the enzyme. Be sure to include all steps that you identify and explicitly show the movement of electrons (PUSH ARROWS). Problem: You are an emerging team of biochemists that wants to discover a new cure for a key metabolic disease in hopes of winning a Nobel Prize. Your team has identified an enzyme that could be a physiologically important target. The enzyme that you are interested in catalyzes the following dehydration reaction: O=P-o -o-o-t- ozp-o- o ó Id=CH₂ + H2O C-CH2 OH OH Your team has extensive experience in protein purification (thanks to your undergraduate biochem lab!) and has purified 1 kg of the protein (enough to last a scientific lifetime!). Next, your team have developed a sensitive, simple kinetic experiment in order to determine the kinetic parameters (Km and Vmax), which can then be used to study the catalytic mechanism of the substrate. Your team has run 4 separate experiments over the course of the last 8 months, and you are having troubles putting all of the results together into a complete picture. However, you are confident that once you are able to compile the data you can propose a reasonable catalytic mechanism and which you hope will lead to the development of transition state analog inhibitors for the enzyme.

Answer 1

Following is the - complete Answer -&- Explanation: for the first three sub-parts ( i.e. Sub- part : 1 to 3 ), of the given:

Question: in...typed & image format...

$\Rightarrow$Answer:

1. K_m =   slope / y - intercept = ( 2.696 ) / ( 8.151 ) = 0.33 mM
2. V_max =   1/ y - intercept = 1/ ( 8.151 ) = 0.1227 mM / min.
3. k_cat = V_max / [E_t ] = [ (0.1227 mM/min ) x ( 1 min / 60 sec ) ] / ( 0.00002 mM ) = 4.09 x 10^-8 sec^-1

$\Rightarrow$Explanation:

Following is the complete: Explanation: for the above: Answer...

• Given:

Following is the given: data set: slightly modified for building Lineweaver Burk plot: prepared by using MS Excel, presented....here, in....image format...

1/VO [S], mM ( milli-molar) || Vo, mM/min (milli- molar/minute) 0 0 0.1 0.028 10 0.2 0.048 0.3 0.06 3.3333333 0.085 1.6666667 0.09 0.095 0.6666667 0.098 0.5 35.71428571 20.83333333 16.66666667 11.76470588 11.11111111 10.52631579 10.20408163

$\Rightarrow$Where:

1. [S] =  Substrate molar concentration, [=] mM  ( milli-mol/L )
2. V_o = Reaction velocity , [=]  mM / min. ( milli-mol/ min. )
3. [E_t ] = Enzyme concentration [=] 0.020 micro-mol/L = 0.00002 milli-mol/L ( mM )

• Step - 1:

​​​​​​​We know the following : Michaelis Menten equation:

$\Rightarrow$ 1/ V_o = K_m / V_maxx ( 1/ [S] ) +   1/V_max------------------------Equation - 1

$\Rightarrow$Where:

1. K_m =   Michaelis Menten constant [=] mM
2. V_max = Maximum reaction velocity [=] mM / min.

$\Rightarrow$Now, if we plot:  1/V_o  vs. 1/[S] , according to Equation - 1, we will be getting the following : i.e.

1. Slope = K_m / V_max
2. y - intercept = 1/V_max
3. slope / y - intercept = K_m
4. 1/ y - intercept = V_max

• ​​​​​​​Step - 2:

​​​​​​​Following is the graph: of 1/V_o vs. 1/[S]  , prepared using MS Excel...presented in ...image format...

1/Vo vs. 1/[S] graph y=2.696x + 8.151 1/Vo 1/Vo vs. 1/[S] graph —Linear (1/Vo vs.1/[S] graph) + 00 1/[S]

$\Rightarrow$Linear Equation of the TrendLine: y = 2.696x + 8.151 -----------------------Equation - 2

$\Rightarrow$ We get:

1. slope = 2.696
2. y - intercept = 8.151  

• ​​​​​​​Step - 3:

Therefore:

1. K_m =   slope / y - intercept = ( 2.696 ) / ( 8.151 ) = 0.33 mM
2. V_max =   1/ y - intercept = 1/ ( 8.151 ) = 0.1227 mM / min.
3. k_cat = V_max / [E_t ] = [ (0.1227 mM/min ) x ( 1 min / 60 sec ) ] / ( 0.00002 mM ) = 4.09 x 10^-8 sec^-1

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