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Friday, October 29, 2010
Lecture 27
1. Midterm exam mean
92 out of 124 = 74.2%, SD 19;
Makeup midterm exam
mean 111.4 out of 137 = 81.3%, SD 28.
Note: (regular exam) question 14 was actually worth 16 points, not 15, hence the
total point value of 124.
2.
Quiz 6
A: 23.8
B: 25.6
C: 23.0
3.
In case you could not pick up your exams after the lecture today, you will find
yours at the Reserve Desk in the Biology Center in 216 Stimson Hall.
4.
At the end of these lecture notes: How to Estimate Your Grade so far.
5.
Brief advice about study of metabolism: (a) understand at level of seeing a
metabolic step written and then being able to say what is going on; and (b) when
in doubt, see last year’s quizzes.
6.
Note change of homework assignment for this week move to next week:
Ch 15 #4. (
We get to this enzyme on Monday).
Wednesday's lecture:
•
Why all cells need to store fuel, and the special role of hexose
•
Intro to Glycolysis: interconnections to other metabolism; “strategy”
Today's lecture:
First, let's consider an overview of glycolysis:
Page 193
Now for a very important and more subtle point about control of
metabolism:
We have the
∆
G
o
' values for all these reactions.
For some cells
and on this page are shown concentrations for muscle cells we have the
actual
concentrations
of all of the reactants and products inside a cell.
These
concentrations inside living cells are VERY hard to measure, and have not been
found for most cells!
From the
∆
G
o
' together with the concentrations of reactants
and products, we can determine the actual
∆
G for each step.
These are
calculated, and shown on page 193, righthand column.
We see that
∆
G is close to zero (3 to +1 kJ/mol) for almost every step.
In fact,
this number is indistinguishable from zero, because we do not have accurate
measurements for all the concentrations of the intermediates.
This means that
these rxns are not far from equilibrium!
If these rxns speed up, e.g. because
more of the enzymes are synthesized, then overall glycolysis does speed up
somewhat.
But there are three rxns that are far from equilibrium, having large
negative values of
∆
G.
Why is this?
In fact, why shouldn't every step be at
equilibrium?
Here's why:
1. Glucose, the initial reactant, is constantly being fed into glycolysis.
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View Full Document2. Pyruvate, the final product, is constantly leaving glycolysis.
3. Intermediates in the reaction are connected to other metabolic reactions so
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This note was uploaded on 12/10/2011 for the course BCHEM 3350 taught by Professor Feig during the Fall '09 term at Cornell University (Engineering School).
 Fall '09
 feig

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