H) Invertase mediates the conversion of sucrose into glucose and fructose. You p
ID: 188345 • Letter: H
Question
H) Invertase mediates the conversion of sucrose into glucose and fructose. You place 0.66 mg of invertase in a beaker of 500 mL buffer initially containing 20 mM sucrose and no glucose and fructose. The value of Kw is 2.5 mM and the value of kat is 1550 s. The invertase you are using has a molecular weight of 65000 i) What is the initial reaction rate (mmol sucrose/Lmin)? i) What will be the concentration of sucrose (mM) after 5 min? ii) What will be the reaction rate (mmol sucrose/Lmin) after 5 min? iv) What will be the concentration of sucrose (mM) after 10 min? By what percentage has the reaction rate decreased after 5 min? What will be the concentration of glucose after 10 min? ) What will be the reaction rate (mmol sucrose/Lmin) after 10 min? By what percentage has the reaction rate decreased after 10 min? If in the 500 mL solution you used 1.32 mg of the enzyme (instead of 0.66 mg) what will be the concentration of sucrose after 10 min? vi)Explanation / Answer
Sucrose = 20mM in 500ml
20 = no. of moles/500/1000
no. of moles= 10mmol
Concentration of enzyme = 0.66*10-3/65000/500/1000
= 0.66/65000/1/2
= 2.03 x 10-8mM
V= Kcat [Enzyme] [S] / (Km + [S])
V = 1550[2.03 x 10-8mM][20mM]/(2.5)+20)
=2796*10-8
=0.2796x10-4mmol/sec
Km is the concentration when the substrate concentration reaches the half.
Initial rate of reaction = 0.13702 mmol/sec
rate of reaction = concentration of product formation per unit time
So,
.2796x10-4mmol/sec = [P]/300…………….(5 min = 300sec)
[P] = 0.0083mmol
So after 5 min sucrose = 10-0.0083 = 9.9917mmol = 19.992mM
So the reaction rate after 5 minutes will be
V= Kcat [Enzyme] [S] / (Km + [S])
V = 1550[2.03 x 10-8mM][19.992]/(2.5+19.992)
= 0.2797 x 10-4 mmol/sec
This is similar to the initial rate of reaction. So no decrease in reaction rate after 5 minutes.
Similarly it can be calculated after 10 minutes.
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