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Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock solution to an assay mixture with a 1 ml total volume, and taking into account that xyzase is a monomeric enzyme with a molecular mass of 45,000 daltons, what is the kcat of xyzase given in s-1 (give the answer in two significant figures)? to see a sample kcat calculation, click on hint in the lower left corner to open the lower panel.

Respuesta :

Hey there!:

From the given data ;

Reaction  volume = 1 mL , enzyme content = 10 ug ( 5 ug in 2 mg/mL )

Enzyme mol Wt = 45,000 , therefore [E]t is 10 ug/mL , this need to be express as "M" So:

[E]t in molar  = g/L * mol/g

[E]t  = 0.01 g/L * 1 / 45,000

[E]t = 2.22*10⁻⁷

Vmax = 0.758 umole/min/ per mL

= 758 mmole/L/min

=758000 mole/L/min => 758000 M

Therefore :

Kcat = Vmax/ [E]t

Kcat = 758000 / 2.2*10⁻⁷ M

Kcat = 3.41441 *10¹² / min

Kcat = 3.41441*10¹² / 60 per sec

Kcat = 5.7*10¹⁰ s⁻¹

Hence   kcat of   xyzase is  5.7*10¹⁰ s⁻¹


Hope that helps!



Cricetus

The [tex]k_{cat}[/tex] of XYZase will be "76 s⁻¹".

As we know,

Line Weaver burk equation,

  • [tex]\frac{1}{V} = (\frac{K_m}{V_{max}.S} )+(\frac{1}{V_{max}} )[/tex]

Series 2 is:

  • [tex]XYZase[/tex]

Here,

→ [tex]1.317 = \frac{1}{V_{max}}[/tex]

→ [tex]V_{max} = \frac{1}{1.317}[/tex]

           [tex]= 0.76[/tex]

Also,

→ [tex]291.4 = \frac{K_m}{V_{max}}[/tex]

→ [tex]291.4 = \frac{K_m}{0.76}[/tex]

→   [tex]K_m = 291.4\times 0.76[/tex]

           [tex]= 221.5[/tex]

hence,

The [tex]k_{cat}[/tex] of [tex]XYZase[/tex] will be:

→ [tex]K_{cat} = \frac{V_{max}}{Et}[/tex]

By substituting the values, we get

→          [tex]= \frac{0.76}{0.01}[/tex]

→          [tex]=76 \ s^{-1}[/tex]

Thus the above answer is right.

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https://brainly.com/question/4745688

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