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# 27 - Friday Lecture 27 1 Midterm exam mean 92 out of 124 =...

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Friday, October 29, 2010 Lecture 27 1. Midterm exam mean 92 out of 124 = 74.2%, SD 19; Make-up 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, right-hand 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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2. Pyruvate, the final product, is constantly leaving glycolysis. 3. Intermediates in the reaction are connected to other metabolic reactions so
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27 - Friday Lecture 27 1 Midterm exam mean 92 out of 124 =...

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