M= #moles / L
4.35/.75 = 5.6
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!
Complete question from other source attached
Answer:
Explanation:
Catalyzed by DNA polymerase - both. DNA polymerase catalyzes DNA replication in the cell. However, purified versions of the enzyme are also used to synthesise DNA as part of PCR reactions
involves leading strand synthesis only - PCR. In PCR, lagging strand synthesis is not carried out because the DNA is denatured (rendered single stranded). Therefore, each strand is replicated independently by leading strand synthesis
duplicates a small fragment of the genome - PCR. Usually, to carry out PCR, small sequences called primers are used that specify the region of DNA to be replicated
duplicates the entire genome - in vivo replication - when the DNA is replicated in vivo, the entire genome is replicated. This is carried out prior to cell division so that two daughter cells can each inherit a copy of the entire genome
Let the mass of CaO = x grams
So mass of BaO = 5.14 -x grams
moles of CaO = mass / molar mass = x / 56
Moles of BaO = mass / molar mass = 5.14-x / 153
initial moles of CO2 = PV / RT = 750 X 1.50 / 760 X 0.0821 X 303 = 0.06
final mole sof CO2 = PV / RT = 230 X 1.50 / 760 X 0.0821 X 303 = 0.018
So moles of BaCO3 and CaCO3 formed = 0.06 - 0.018 = 0.042 moles
x / 56 + (5.14-x) /153 = 0.042
on solving
x = 0.68
So mass of CaO = 0.68 g
So percentage of CaO = 0.68 X 100 / 5.14 = 13.4 %
Percentage of BaO = 86.6%