CHEMISTRY 1AA3
April 5 – 8, 2010
TUTORIAL PROBLEM SET 12 – Full SOLUTIONS
______________________________________________________________________________
1
Questions 1  4:
Some practical enzyme kinetics
Our diet contains many complex polysaccharides (polymers of different sugars), which must be
broken down to monosaccharides in order to be converted to energy.
Polysaccharides are broken
down through a variety of routes, which includes significant amounts of disaccharide formation.
Let's consider three disaccharides, cellobiose, gentiobiose and lactose, that must be hydrolyzed to
the corresponding monosaccharides, glucose & galactose.
A hypothetical enzyme, glycosidase X
(gX), which catalyzes this reaction.
gX will be digest each disaccharide with different steady state
rate constants.
1.
(a)
With gentiobiose:
k
cat
= 10 s
1
and
K
M
= 10
4
M.
If:
[gX] = 10
7
M, and[gentiobiose] = 10
5
M, what is
v
0
?
SOLUTION:
We can plug the values into the MichaelisMenten equation:
v
0
= (10
×
10
7
×
10
5
) / (10
4
+ 10
5
)
= 9
×
10
8
M/s
(b)
If [gentiobiose] increases to 2
×
10
5
M, what is the new
v
0
?
We can plug the values into the MichaelisMenten equation:
v
0
= (10
×
10
7
×
2
×
10
5
) / (10
4
+ 2
×
10
5
)
= 1.7
×
10
7
M/s
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentCHEMISTRY 1AA3
April 5 – 8, 2010
TUTORIAL PROBLEM SET 12 – Full SOLUTIONS
______________________________________________________________________________
2
(c)
Why does the rate roughly double when [gentiobiose] doubles?
When [S] <<
K
M
, the reaction is first order w.r.t. [S], which is the case in this question.
2. (a) If
v
0
= 2
×
10
6
M/s with 4
×
10
4
M cellobiose and 1.8
×
10
6
M/s with 2
×
10
4
M
cellobiose, what can we conclude about
K
M
(cellobiose)?
There is almost no change in
v
0
when [S] doubles, so the reaction is roughly zero order w.r.t. [S].
This
occurs when [S] >>
K
M
, so we know that
K
M
(cellobiose) << 2
×
10
4
M.
(b)
If [gX] = 10
7
M, make an estimate of
k
cat
.
When [S] >>
K
M
, (
K
M
+ [S])
≈
[S].
The MM equation reduces to
v
0
=
k
cat
[E]
0
, or
k
cat
=
v
0
/[E]
0
k
cat
= (2
×
10
6
M/s) / 10
7
M
= 20 s
1
3.
(a)
With lactose:
k
cat
= 2 s
1
and
K
M
= 3
×
10
5
M.
What would the
v
0
vs [lactose] profile look like in comparison to the gentiobiose
profile?
The enzyme reaches 1/2 maximal rate at a lower value of [S] with lactose because its
K
M
value is lower.
Its
k
cat
value is also lower, so the maximal rate is lower than gentiobiose.
The maximal rate for gentiobiose is
off the top of the graph shown.
(b)
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '10
 Tammie
 Linear Algebra, Algebra, Reaction, Chemical reaction, ea, Tutorial Problem

Click to edit the document details