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leva [86]
3 years ago
9

Assuming that the experiments performed in the absence of inhibitors were conducted by adding 5 μl of a 2 mg/ml enzyme stock sol

ution 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.
Chemistry
1 answer:
prohojiy [21]3 years ago
4 0

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!



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Explanation:

Given -

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  • On treatment with alkaline iodine it gives yellow ppt.
  • On oxidation with CrO₃/H⁺ forms an aldehyde (C₂H₄O)

To Find -

  • Name the compound and write the reaction involved

Now,

Let A be the organic compound.

Then,

  1. A + Na → + H₂↑
  2. A + I₂ → CHI₃ (yellow ppt.)
  3. A + CrO₃ + H⁺ → C₂H₄O

Now,

Here we see that compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives aldehyde.

  • Functional group of aldehyde = —CHO

And It forms only 2 Carbon aldehyde it means, It is Ethanal (CH₃CHO).

Compound A reacts with chromic oxide (CrO₃) in the presence of acidic medium gives ethanal.

It means,

We know that 1° alcohol on oxidation gives aldehyde.

Here it gives 2 Carbon aldehyde.

It means,

Here 2 Carbon and 1° alcohol is used.

Now,

Its cleared that Compound A is Ethanol.

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  1. CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + H₂↑
  2. CH₃CH₂OH + I₂ + OH⁻ → CHI₃↓ + HCOO⁻ + HI + H₂O
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