Explanations of key for Bi 336 Examination 1 questions:

Question 13:

13.The valence of oxygen is:

The valence or combining capacity of oxygen is 2. The number of valence electrons in oxygen is 6 (needs 2 to fill its outer electron shell).

Question 14:

14.A two-electron reduction of a carboxyl group would produce:

Sorry about this mistake, I should have corrected it on all of the Scantrons, if not, please come and see me.

Question 22:

22: Heterotrophs can use which of the following as energy sources

It was pointed out to me (correctly) that heterotroph in the microbiological sense refers to the carbon source used by the organism, not it's energy source. However, in general (and in this course) heterotroph refers to where the organism gets it's food from (hetero = other, troph = food), other feeder. This is contransted with autotrophs which make their own food. Since it was mentioned in lecture (see Lecture 3, slide 6) that heterotrophs are "Organisms using organic compounds for energy", the only correct answer here is c. Protein since it is the only organic compound.

Question 23:

23: When glucose is phosphorylated in glycolysis it is to

The energy of the phosphate attached to glucose is too low to transfer to ATP (see Figure 3.27), and there is no electron to transfer to NAD. Glucose can be isomerized to fructose in the absence of phosphorylation. Additionally, the reaction glucose 6-phosphate to fructose 6-phosphate has a positive delta-G0, so it should go in the opposite direction. In any case the cell does not want to lose glucose (or any other glycolytic intermediates) through the cell membrane, therefore it keeps all of them charged either by phosphorylation or negatively charged carboxylic acids (see Lecture 6, Slide 9 "Steps in Glycolysis: Hexokinase". Charged molecules have much more trouble crossing the membrane due to the charged head-groups on phospho-lipid bilayers.

Questions 24 and 42:

24: For the complete oxidation of glucose to carbon dioxide and water the deltaG0’ is –686 kcal/mol. The deltaG0’ for the formation of ATP is + 7.3 kcal/mol. If these reactions were directly coupled how many ATPs could theoretically be made?

Theoretically, 686 kcal/mol of energy are available to make ATP which needs 7.3 kcal/mol to make, therefore 93.97 could be made. So many are not made because the reactions are not 100% efficient.

42: For the reaction CO2 + 4H2 -> CH4 + 2H2O what is the deltaG0f, given that the deltaG0f of CO2 is -390 kJl/mol, H2 is 0 kJ/mol, CH4 is –50 kJ/mol and H2O is -240 kJ/mol.

deltaG0 = sum (deltaG0f of products) - sum (deltaG0f of reactants) = -390 + 4 (0) - (-50 + 2(-240)) = -140

The numbers in the answers were not exact, sorry. However it is clear which the best responses were, as stated at the top of page 1 on the test. I told this to everyone who asked me during the examination. I did not say that people should bring calculators and neither of these questions should have required them. Evidently I should have used one when writing the exam.

Question 29:

29: Your student shows you some data on an enzyme assay that she has done with a new enzyme. She has plotted the data on a Lineweaver-Burk plot (1/[S] vrs 1/V) to determine the Vmax and Km of the uninhibited enzyme. Then she added a new compound to the reaction and repeated the experiment and replotted the data. The slope of the line is less than that of the enzyme assay without addition but her X-intercept is the same, is this new compound that she tried a

This is a classic example of an enzyme activator, if this is unclear, please come and see me.

Question 35:

35: Enzymes basically do the same thing as heating up the reactants:

Enzymes are catalysts, catalysts increase the rates of reactions, heating up reactants also increases the rate of the reaction.

Question 41:

41: In the reaction Se4+ + As3+ = Se2+ + As5+ which element is being reduced?

The correct answer to this question is actually "d" since in a reversible reaction both Se and As are adding electrons, however since it was not clear on the examination points will be given for both "a" and "d". This is not reflected on the Scantrons, but will be changed. If you answered "d" and were not given a point and want to make sure that I have added this point correctly, please come and see me or give me a call. My apologies.