Answer:
Can you send the picture of this question
Answer:
O Option 1
Explanation:
IF ENERGY IS RELEASED, THEN ENERGY RELEASED SHOULD BE SUBTRACTED FROM ORIGINAL.
(16.32 X 10^-19) - (5.4 X 10^-19)
10.92 X 10^-19
Answer:
I don't know how to get the answer sorry
Explanation:
db
<span>False. Foods that allow microorganism to grow are not called parasites. Parasites are organisms that feeds on the nutrients of its host.
For example leeches. They suck on our blood.
</span>Foods that allow microorganism to grow are called <span>potentially hazardous foods. They are high in protein and moisture and are slightly acidic.
For example hamburgers. They are high in protein and moisture.</span>
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!
