Question

This is a Biochemistry 200 level question, please answer neatly and as soon as possible. Thanks in advance, and I will like as soon as answer is posted!

The enzyme happyase catalyzes the following reaction: SAD — HAPPY An enzyme kinetics experiment gave the data shown in the table below. The concentration of happyase used in the experiment was 25.00 uM. [SAD] (MM) 1.000 2.000 3.000 4.000 Vo (mM/s) 3.700 6.727 9.250 11.385

Answer 1

Following is the - complete Answer -&- Explanation: for the given Question, in.....typed format...

$\Rightarrow$Answer:

At reaction rate: Vo = 19.8 mM/sec, molar concentration of SAD, will be: [SAD] =  10.338 mM ( milli-mol/L )

$\Rightarrow$Explanation:

Following is the complete Explanation, forr the above Answer....

• Given:

1. balanced chemical reaction:   SAD  $\rightleftharpoons$ HAPPY
2. Name of the substrate: SAD   
3. Name of the product: HAPPY
4. Name of the enzyme: happyase
5. Enzyme concentration throughout the reaction:[Happyase] =  25.0 $\mu$ M = 0.025 mM ( milli-mol/L )
6. Vo =   Reaction Velocity [=] mM/s ( milli-mol/L .sec )
7. [SAD] = Molar concentration of the substrate, in millimol/L ( i.e. mM )
8. The values of [SAD], and the corresponding values of reaction velocity ( Vo ), is given in the question, as prodided below in the section: "Explanation"  

• ​​​​​​​Step - 1:

​​​​​​​Following is the given information, in the form of a table, prepared by using MS Excel, presented here in...image format....

(SAD), IM (millimol/L) Vo, M/s (milli-mol/L. sec ) 3.7 6.7 9.25 11.385

Where:

1. [SAD] = Molar concentration of the substrate, in millimol/L ( i.e. mM )
2. Vo =   Reaction Velocity [=] mM/s ( milli-mol/L .sec ) ​​​​​​​

• Step - 2:

​​​​​​​Since the plot, of Vo vs. [SAD], will not be Linear, we will have to plot the reciprocal values, i.e. the values of 1/Vo vs. 1/[SAD], which should be Linear. Therefore, we will have to prepare the following table, by modifyig the previous, given table,above as following, in....image format...

1/[SAD), mM^-1 ( L/millimol) 1/Vo, s/mM ( sec.1/milli-mol) 0.5 0.333333333 0.25 0.27027027 0.149253731 0.1081081081 0.087834871

• Step - 3:

​​​​​​​Now, let's plot, the values of 1/Vo vs. 1/[SAD], according to the table, given above, using MS Excel, given here in...

....image format....

1/Vo vs. 1/[SAD] plot V = 0.243x + 0.0271 1/Vo, sec/mM 1/Vo vs. 1/(SAD) plot Linear (1/Vo vs. 1/(SAD) plot) 0.2 0.8 1.2 0.6 1/(SAD), mM^-1

We get the following information, from the above plot...

1. Equation of the obtained LinearTrendline: y = 0.243 x + 0.027   -------------------- Eq. (1)
2. y - axis:   1/Vo   
3. x - axis:   1/[SAD]

• ​​​​​​​Step - 4:

​​​​​​​Therefore, the above Equation - (1), can be represented, as the following:

$\Rightarrow$i.e. 1/Vo = ( 0.243 ) x 1/[SAD] + 0.027   -------------------------------------- Eq. (2)

• Step -5:

​​​​​​​Now, we are given:

$\Rightarrow$ Reaction rate:   Vo = 19.8 mM/s  , i.e. under this condition, we will get the following:

$\Rightarrow$1/Vo = 1.0 / ( 19.8 mM/s ) =  0.050  ( approx. )

Plugging in the above obtained value in the given equation: Equation - (2), we will get the following:

$\Rightarrow$ ( 0.050 ) = ( 0.243 ) x 1/[SAD] + 0.027

$\Rightarrow$1/[SAD] = [ (0.050 - 0.027 ) / ( 0.243 ) = 0.09672

Therefore, at that point in time, we will get the following value of [SAD].....

$\Rightarrow$[SAD] = 1.0 / ( 0.09672 ) =  10.338 mM  ( milli-mol/L )

• Answer:

​​​​​​​Therefore, the molar concentration of SAD, will be the following, when the reaction velocity ( Vo ) , will be equal to 19.38 mM/s  

$\Rightarrow$[SAD] = 10.338 mM ( milli-mol/L )

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