Questions and Answers for 153A

Question:

Hello Dr. Bates, Do we need to memorize all the reactions of the pyruvate dehydrogenase complex? Thanks

Answer:

Yes.

Question:

Dr. Bates, In the examples you gave in lecture for carbon tracing you used glucose labeled at C2 and C5 and glucose labeled at C1 and C6. Is it wrong to think that you will only see a label show up in the TCA cycle if glucose is labeled at C5 or C6? It seems that the C2 and C1 labels are being lost before the TCA cycle. Is that a correct assumption to make? Thanks for your help.

Answer:

No! Carbons 3 & 4 (the carboxyl of pyruvate) are lost as CO2 in the pyruvate dehydrogenase reaction. The other four carbons enter TCA via acetylCoA, 1 & 6 as the methyl group of AcCoA and 2 & 5 as the carbonyl.

Question:

Dr. Bates, Is there a difference in referring to the reduction potential of say, NAD+ and NADH? For example, taking an arbitrary reaction like succinate + FAD --> fumarate + FADH2, if we wanted to know which direction the reaction would go, would we look up the reduction potentials for fumarate or succinate, FAD or FADH2 or does it make a difference, or am I not making any sense?

Answer:

A reduction potential refers to a redox couple, which is the oxidized and reduced form of a compound that can participate in an oxidation-reduction reaction. It is the measure of a compound's tendency to become reduced (accept electrons). See pages 144 - 147 of the compendium. Thus reduction potentials refer to half-reactions. For example: FAD + 2e- + 2H+ -> FADH2 is a half-reaction and FAD/FADH2 is a redox couple. fumarate + 2e- + 2H+ -> succinate is a half-reaction So to determine the direction or a redox reaction between these compounds, you would compare the reduction potential of the FAD/FADH2 (one redox couple) with the reduction potential of fumarate/succinate (the other redox couple).

Question:

Dr. Bates, Quick question, what are the differences between positive and negative effectors of phosphofructokinase? ie. between ATP and ADP or AMP

Answer:

I am not sure what you are asking here. Of course positive effectors increase the PFK activity and negative effectors decrease it. ATP, ADP and AMP make up the adenylate pool: ATP having three phosphates, ADP two and AMP one. A major purpose of glycolysis is to make ATP, the currency of energy exchange in the cell. When ATP levels are high there is less need for the PFK reaction, which is the rate limiting step of glycolysis. Thus ATP acts as a negative effector (or inhibitor) of PFK. When ATP levels are low, ADP and ADP are elevated so they can act as positive effectors stimulating PFK activity. Each effector binds to its own regulatory binding site on PFK to mediate its effect on PFK activity. See page 117 of the compendium.

Question:

Professor Bates. What do you want us to memorise from glycolisis, TCA cycle and Oxidative phosphorilation?

Answer:

It is all spelled out in the review sheet that was handed out in lecture during the past 2 days and again today. The review sheet is also posted right here in our Chem 153 voh site, under WHAT YOU SHOULD KNOW FOR THE FINAL.

Question:

Dr. Bates, Do you want us to memorize every reaction that takes place in the Electron Transport Chain?

Answer:

NO.

Question:

Dr. Bates, 1) when pyruvate is converted to oxaloacetate, what makes it decide whether it will undergo the pathway of making glucose or keep the TCA cycle going? 2) Also, for naming enzymes, some names contain the name of the substrate exactly (G-6-P isomerase) and some don't contain the name exactly (phosphofructokinase). Would it be wrong if we always put the exact name of substrate in the enzyme name? Thank you.

Answer:

1) The flux of the carbons of OAA into and through these alternative pathways is, of course, controlled by the same regulatory mechanisms we have been studying. This regulation responds to the current needs of the cell. For example the next reaction leading into the TCA cycle is catalyzed by citrate synthase which is regulated by substrate availability and feedback inhibition. The next reaction of gluconeogenesis is catalyzed by PEPCK which is regulated by hormone stimulation. You already know about some other regulated reactions in each of these pathways. 2) In naming some enzymes you could be mistaken in using the exact name of a specific substrate, because some enzymes are not entirely specific for a single substrate. An example of this is hexokinase which phosphorylates some other hexoses in addition to glucose.

Question:

Will you have the seating chart for the final on Voh?

Answer:

Yes, it will probably go up on Thursday this week.

Question:

Dr. Bates, I was wondering if it's too late to do a regrade for Midterm 2. If it's not, can you tell me how to go about doing it?

Answer:

Attach a note to the front of the exam and give it to me or Robin by Tuesday, June 9.

Question:

Prof. Bates, When will the regrades for the second midterm be finished? Will you announce when in class?

Answer:

I don't know. Sometime this week. I will announce it in class.

Question:

Dr. Bates, I have a question about the reduction potential o f NADH and FADH2. When you were explaing Rx 5 in the pyruvate dehydrogenase complex, I thought I heard you said that FAD has a lower reduction potential than NAD, but when it is bound to an enzyme, the reduction potential gets higher. And then when you were going over the succinate dehydrogenase in the TCA cycle, you said that FAD has a higher reduction potential than NAD. Can you please tell me again which has the higher reduction potential and how this is related to those two reactions. Also, you said that Oxaloacetate will bind first to citrate synthase rather than Acety-SCoA and explained why. Can you tell me the explanation again, because I didn't quite get it in class. Thank You very much.

Answer:

Yes, I think I may have misspoken in that lecture and made it seem as if FADH2 has a lower standard reduction potential than NADH. We should be careful to distinguish between standard and actual reduction potentials just as we do for Gibbs Free Energy change. The standard reduction potential of FADH2 is higher than that of NADH. And it is true that covalently enzyme-bound FAD/FADH2 generally has an even higher standard reduction potential. It happens that under the reaction conditions for E3 the actual reduction potential of FADH2 is lower than that of NADH so that the electrons can be passed spontaneously from FADH2 to NAD+. This is an unusual situation. We would generally find that the transfer of electrons between these two redox couples would go in the other direction under most conditions. The mechanism of citrate synthase is an induced fit mechanism analogous to that of hexokinase, and it serves a similar purpose. If AcCoA could bind first, it could be hydrolysed to acetate and Co A. The free energy of hydrolysis would be wasted, because the acetyl group would not have been transferred to OAA to form citrate.

Question:

Hi Professor, Just a few questions: 1) What exactly is coordinate regulation? 2) What's the difference between an acyl and acetyl group? 3) In our class discussion of phosphoglyerate mutase, did you say that it can also catalyze the formation of 2,3-bisphosphoglycerate? Thanks, Payal

Answer:

1) Coordinate regulation as applied to multienzyme complexes means that the regulation of one enzyme in the complex also affects the activities of other enzymes in the complex. For example, inhibition of E2 of PDH complex by AcCoA binding to the CoA binding site, prevents lipoamide from becoming deacylated. This in turn, prevents hydroxyethyl-TPP of E1 from giving up its hydroxyethyl group, which in turn makes TPP unavailable to accept another hydroxyethyl group from pyruvate so that the reaction of E1 is also inhibited. The term, coordinate regulation, can also be used to describe the regulation of opposing pathways such that the regulated enzymes of the "forward" pathways are inhibited when the regulated enzymes of the "reverse" pathway are active and vice versa. 2) Acyl is a general term, and acetyl is a specific term for a 2 carbon acyl group. 3) Yes. 2,3-bisphosphoglycerate is an enzyme bound intermediate of the mutase reaction. In a certain fraction of the reactions, this intermediate is released into solution leaving the enzyme dephosphorylated. The enzyme then needs to be "primed" by 2,3-bisphosphoglycerate before it can continue to catalyze the mutase reaction.

Question:

What is the difference between a "high-energy bond" and a labile bond? When you talk about ATP you make that distinction but I don't get it. Thanks.

Answer:

This answer is going to be hard to follow without the use of illustrations. So you will probably need to talk to me personally where I can draw out some reactions for you in connection with the explanations. But here goes .... Both terms refer to the same thing. My point is that the term "high energy bond", although in common use, is misleading and less accurate than the term labile bond. The former implies that the bond possesses some kind of potential energy that is released upon the breaking of that bond, and that the energy released is used to drive the formation of other bonds. This is not the case. Actually it requires an input of energy to break bonds, and energy is released in the formation of bonds.. The free energy change in a reaction is negative (and thus the reaction spontaneous) if the energy released in the formation of all new bonds in the reaction is greater than the energy required to break all existing bonds that are broken in the reaction. Thus a net reaction that results in the formation of certain chemical bonds and involves the breaking of a thermodynamically labile bond (i.e. a bond that breaks with relatively low energy input) will proceed more readily than one involving the formation of the same bonds but the breaking of a more thermodynamically stable bond. So when we say that the hydrolysis of ATP, for example, drives the phosphorylation of glucose, we mean that the phosphodiester bond of ATP is easier to break than the bonds that must be broken in the direct phorphorylation of glucose by inorganic phosphate. Now you see why biochemists find it more convenient to talk about hydrolysis "high energy bonds" driving reactions no matter how inaccurate.

Question:

Dr. Bates, What was the class average and standard deviation for the second midterm in the second lecture? There seems to be some confusion among different people. Thanks, John

Answer:

The average was 59, and the standard deviation was 17 for lecture 2.

Question:

Hi Dr. Bates, I was just wondering if there was still a chance to get an "A" since my midterm scores averaged exactly a 70? Thanks

Answer:

Sure. For example, if you score 250 on the final and 50 in discussion, your total score will 440. That is 88% which is sure to be an "A".

Question:

hi,Dr. Bates i have some questions on today's lecture. Can you explain what is the "substrate-levelo phosphorylation" and how does this rxn act as a phosphate donor. and what are differences for the enzymes in Gluconeogenesis and glycolysis i.e. Hexokinase and glucose 6-phosphatase, and which is the limiting steps?? and also, are those regrading for midterm #1 ready yet??thanks

Answer:

The regrades have been ready for two weeks, and I have announced this repeatedly in lecture. I bring them to lecture with me every day. I cannot give a concise answer to your questions, as they don't seem to focus on specifics that can be readily addressed in this format. What you seem to be asking for is a recapitulation of about half of the lecture. I think it would be very inefficient for me to try to type out all of this material (which would fill a couple of pages) without really knowing where your uncertainties lie. Please come to one of the real office hours to get these things cleared up. There are 16 offered per week (2 with me, and 14 with the various TA's).

Question:

Are these both correct examples of lyases? 1. the addition of H2O to break a double bond 2. the elimination of H2O to form a double bond when water is involved in the making or breaking of a double bond, how do we know if it's a lyase or a hydrolase?

Answer:

Yes. A hydrolase reaction cleaves C-O, C-N or C-C single bonds to yield two products (or the reverse).

Question:

Hi Dr. Bates. I was wondering where the other students are getting copies of your old exams. I have not been able to find them on the web. Thanks!

Answer:

Some are still posted on voh sites for Chem 153 in previous quarters.

Question:

Dr. Bates- How much above 0 can the Delta G be for the reaction to still be able to go in either direction? What is the cut off point?

Answer:

We cannot set a cut off point. See answer to previous question.

Question:

Dr. Bates- How much above 0 can the Delta G be for the reaction to still be able to go in either direction?

Answer:

This is not a valid question. While delta G is greater than zero for a given reaction written from left to right, the reaction will only go in that direction. You probably meant to say delta G o'. Even so there is no set answer to your question. We can say that in general (in the absence of other information) the higher the magnitude of a positive delta G o', we would expect a lower probability of reaching a [P]/[R] ratio in the cell that will enable the reaction to proceed in the other direction.

Question:

Hi Dr. Bates. You said in class that the binding kinetics of hemoglobin can be partially explianed by both the koshland and monod models. It seems to me that the sequential pathway better describes hemoblobin's binding behavior. Where does the concerted mechanism come into play?

Answer:

This question has already been answered. Check about 3 or 4 down the list.

Question:

Hi Professor Bates, I missed your review session and was wondering if you could answer a couple of my questions: 1) Does a hydrolase only work by adding water, or can it also eliminate it? 2) Do we not need to worry about any of the structures in glycolysis? Only those pertaining to the first two reactions? 3) Should we memorize the 3 reactions of hemoglobin? 4) What is a mixed inhibitor? Is it the same thing as a noncompetitive inhibitor? 5) Should we memorize all the steps of chymotrypsin activity? Or just know the basic ideas? Thanks, Payal

Answer:

1) Since a any enzyme can catalyze its reaction in either direction, a hydrolase can catalyze the breaking of a single bond by addition of water or the formation of a bond by the elimination of water. e.g. the formation of a glycosidic bond. 2) Yes, know the reactants and products of the first two reactions for this exam. 3) I would prefer that you understand them. 4) Non-competitive is a special case of mixed inhibition. In mixed inhibition the inhibitor can bind to E or E-S, but not necessarily with the same binding constant for both. Non-competitive inhibition is the case when the binding constant is the same for both E and E-S. Non-competitive is the only case of mixed inhibition that we will concern ourselves with in this course. 5) Understand all the steps and the concepts of catalysis that they illustrate. You will not have to draw them from memory, if that is what you are asking.

Question:

Okay you are probably sick of this ?..but to verify. We (both lectures that is) are only responsible for the first two reactions of glycolysis on the 2nd midterm, right???

Answer:

Right on both counts.

Question:

Hi Dr. Bates. Can you explain to me why the binding kinetics of hemoglobin can partially be explained by "both" the Koshland and Monod models? It seems to me that hemoglobin's binding kinetics follows more of the sequential rather than the concerted pathway.

Answer:

I said it was partially consistent with each of the models, not that it is explained by the models. It is explained by the actual mechanisms that have been elucidated by experimentation. The consistency with the Koshland Model is that the initial conformational change to the "R" form is induced by oxygen binding as predicted by that model. The consistency with the Monod Model is that the conformational change appears to occur in a concerted manner rather than sequencially (i.e. all subunits of a given molecule change from T or R states simultaneously rather than the subunits changing one at a time from T to R state so that some can be T and some R at the same time.)

Question:

Hi, Dr. Bates, i have a question regarding to this midterm.. are we responsible for whole glycolysis ? or just stage I..b/c i heard people from the lecture 2, they have been covered almost everything in the glycolysis..thanks

Answer:

Would I lie to you? I said you would be responsible for material that was covered in detail in the lecture that was given on Tuesday in section 2 and Wednesday in section 1. Don't worry, be happy. Well, maybe you can't be happy with an exam breathing down your neck, but ... Well, maybe you can't be worry free either, but exam 2 will only cover the first two reactions of glycolysis.

Question:

Dr. Bates, how can you determine which step of a metabolic pathway is going to be the first committed step. From the example of glycolysis, it does not necessarily have to be the first one.Thank you.

Answer:

You may not always be able to tell by just looking at the reaction sequence of the single pathway. If you know all the interconnecting pathways, you would look for the first step after a branch point. e.g. in the sequence A ->B->C->D, if a is known to be utilized in other reactions that do not belong to this sequence, but B, C, and D are not, you would consider the conversion of A to B as the first step committed to this pathway. If you did not know about the other reactions, another possible clue would be the thermodynamics. The first committed step is frequently a highly spontaneous reaction. As a scientist investigating the question of whether compound A is a branch point, you could approach the problem by providing the organism with compound A carrying radioactive atoms such as C14, and checking to see which various compounds contain this radioactive label after a period of metabolic activity. If the radiolabel shows up in other compounds besides B,C, and D, then A must be a branch point.

Question:

Dr. Bates, I have a few questions in the practice problems. 1) First, pg 70 #8b why would the reaction be favored one way or the other when the concentrations of B and C are the same and in the other synarios. 2) Second, pg 66 #3 was the answer of 33 torr derived by multiplying .5 by 65% and qualitatively reading the graph or is there a more systematic quantitative way of finding the answer(a and b). 3) Lastly, when everything is given in the m and m equations plotting a v vs s graph does not exactly correspond to the graph as in pg 48 # 10 b

Answer:

1) Recall the equation, deltaG =deltaGo'+RTln [C]/[B]: This tells you what direction the reaction will go under actual reaction conditions. DeltaG is a function not only of the standard Gibbs free energy but also of the concentrations to which you refer. 2) No. If the transfer of oxygen to the tissues is 65% efficient, then 35% of the oxygen is still bound to Hb in the tissues. So you look at the graph to determine the pO2 at which Hb is 35% saturated. 3) It is not clear to me what you are asking here. I am not even sure this is a question. Maybe we can clarify your problem at the review session.

Question:

Follow-up on question about posting of review sheet for exam 2.

Answer:

I just checked, and found that the review sheet titled, "What you should know for midterm 2" has been posted on voh under "Exams". It will also be handed out in class today and tomorrow.

Question:

Professor Bates, Can you please explain what the limitation is with using the L-B plot in enzyme kinetics? And also why is it better to use the E-H plot? Thanks a lot!

Answer:

It is not necessarily better to use the E-H plot. You just have to be careful, in using the L-B, to put more weight on the points near the origin in drawing your best fit line. The % error in your smallest measurements is generally greater than the % error in your largest measurements. Thus your largest measurements are the most reliable values. The reciprocals of your largest measurements will have the smallest numeric values on your L-B plot . The errors in your small measurements will be magnified by taking the reciprocals.

Question:

I was wondering when the review sheet for midterm 2 will be posted on VOH. Thank You

Answer:

I submitted it for posting early Friday afternoon 5/22. I'll check on it.

Question:

this is a test!

Answer:

test test test test

Question:

hi.. Dr. Bates, i just want to know that the regraded test has been done yet ?? and do we pick them up at your office or from the TA. thanks.

Answer:

I received an unusually large number of regrade requests for this exam (40), and many of them were of exactly the type I asked the class not to submit, i.e. asking for a general regrade of from 3 to 6 out of the 10 questions or in some cases the entire exam. This is extremely time consuming and unproductive for the student (average change in score = one-half a point). In some cases, it is counter-productive as the a significant percentage of the grading errors I am finding were made in the student's favor. Of course I must also correct these errors when I find them. Some regrades are finished but not yet recorded. Others are awaiting consultation with the grader or investigation of answers that appear to have been changed prior to resubmission. I will bring regraded exams to lecture and make an announcement when they have been recorded.

Question:

Dr.Bates, 1)Do we need to know the details of the mechanisms of Papain and Chymotripsin? 2)Does electrostatic catalysis occur only among the charged side chains, or it can also occur via ion-polar interactions? Also, does this type of catalysis involve a transition state? Thank You.

Answer:

1) Yes, you need to recognize and understand them. 2) Any charged group that helps to promote catalysis by virtue of its charge can be designated as electrostatic catalysis, for our purposes. Yes, it can involve the transition state.

Question:

1) On page 53 of the practice problems booklet, question 1a asks about the residue of a papain. The answer is HISTIDINE. Can you please explain the reasoning behind that answer? Why would it not be aspartate? 2) Also, on the same page, the second part of question 1d should be amide nitrogen, since trypsin attacks positive functional groups. But the answer in the back lists Carbonyl carbon. Why is that?

Answer:

1) There isn't even any aspartate at the papain active site. The + charged imidazolium of histidine stabilizes the deprotonated cys side chain electrostatically. The imidazolium form of his is stabilized by the pi electrons of tryptophan. 2) You are confused about the mechanism of trypsin. Trypsin is specific for amide bonds adjacent to lys and arg which do have + charged groups. The nucleophilic attack leading to bond cleavage is on the carbonyl carbon of the amide bond being cleaved. This is a serine protease with a mechanism just like chymotrypsin. Only the specificity for side chains adjacent to the amide bond to be cleaved is different.

Question:

Hello Dr.Bates I was just wodering if there was a mistake on page 88 of the compendium. I believe that the CH2 in Histidine should be bounded to the nearest carbon to the protonated nitrogen of the Imidazole ring. Also, on page 82 of the compendium there are no hydrogens on the Imidazole rings forming H-bonds to the carbonyl oxygens surrounding them. Please correct me if I'm wrong.Thanks

Answer:

Page 88: I think what you are really asking is where the double C=N bond should be and therefore which NH is acidic. The answer is that you can put it in either position. These are delocalized pi electrons, and we can draw two resonance structures of the fully protonated ring putting the double bond and the + charge on either nitrogen. Thus the deprotonated form can leave the double bond at either position. Page 82:You are right. The hydrogens should be shown here. I'll make a notation to have this corrected in the next edition. Thanks for pointing it out.

Question:

Dr. Bates, Can you give us some idea of what material we should study for the quiz this week? Will it be on material from this week's lectures, or will it also cover last week's lectures?

Answer:

It could cover anything we have covered in the course through the most recent lecture. Every section gets a different quiz. Each will consist of 5 easy true/false or multiple choice questions... maybe a few quizzes will have a fill-in-the-blank question or two.

Question:

Do these terms all have the same meaning: Binding site, Positioning site and Active Site.

Answer:

No.

Question:

hi dr. bates. i know that it's impossible for you to definately determine the grade cut-offs for the course, but i am in real need for some advice. i am disappointed to say that i received a 46 on the first midterm. can you please tell me if my chances of receiving an A for the class still exist. i am currently studying really hard for the second midterm and final exam. i am very worried and this has been bothering me lately. thank you so much.

Answer:

It is not impossible for you to get an A, but you will probably need to score between 85 and 90% on your remaining work to have a chance of doing so. Does it have to be an A? I wish your concern was more strongly targeted on learning the course material than on getting an A.

Question:

Hi Dr Bates, I was comparing my exam with the key on the web, I found that some of answers are, but I did not get any credit for it. If I bring my exam to you during office hours, would you please take a look at it. I don't know your policy on regrading. Thank you

Answer:

As I announced in class, you should resubmit the exam by Monday with a note attached specifying the items that you think were misgraded.

Question:

Prof. Bates, I had a question about question 9C on the midterm. The reaction described seems to be a hydrolysis (addition of water to break a double bond), which would in turn make the enzyme a hydrolase. However, the correct answer is listed as "lyase." I am confused because in lecture, she said that lyases apply to bond cleavage and formation reactions done by means other than hydrolysis.

Answer:

Addition of water to an alkene leaving a hydroxylated alkane is not hydrolysis. It is a lyase reaction. Lyase reactions can usually be recognized by virtue of the formation of a double bond or breaking of a double bond leaving a single bond in its place. This can include the cleavage of one compound into two products, one of which has a newly formed double bond. Hydrolysis is the breaking of a single bond by addition of water across that bond to cleave one compound into two compounds without formation of a double bond.

Question:

Dr. Bates, What are the approx. grade cut offs for the midterm? What was the standard deviation? Thanks.

Answer:

I don't assign any letter grades until I have the total of all scores earned for the quarter so there are no cut-offs. The stadard deviation on exam 1 was 16 and the average was 62. If you got the average, you are in the C+/B- range. If you got above 80 and continue to perform at that level in both exams and discussion, you are probably on your way to an A. If you scored below 50, you should definitely seek extra help. Come to as many as you can of the 16 available office hours per week and the 21 discussion sections per week. You may also benefit by getting some tutoring, either private or from the tutoring services available through UCLA.

Question:

hi Dr. Bates. i am confused on the nomenclature of glycosides. what is the difference between a pyranoside and a pyranose? when would we use either suffix? thanks.

Answer:

A pyranose has a reducing end, and a pyranoside does not.

Question:

Professor Bates, Are we allowed any calculators (for Henderson-Hasselbach, etc)? If so, are we allowed to use either scientific or graphing? Thanks. David

Answer:

Only simple calculators are allowed. The most complex function they should have is taking logs and anitlogs.

Question:

hi.. Dr. Bates, are we allowed to use any graphing calculator during the exam.. b/c i think most of people are using those kind of calculator rather than simple function one, and it's hard for me to find it by tomorrow.. thanx

Answer:

No

Question:

will you be posting the answers to the questions on page 69 of the compendium???

Answer:

No, I you are uncertain about them, ask about it in discussion or office hours.

Question:

Hi. My question is concerning Vmax. 1) What affects it? I know that [E] does. In lecture you stated that it is not dependant on [S]. However, you also said that if we [S]/2 that Vmax will also reduce by 1/2. (You did this while illustrating that Km would remain constant.) Also, does it matter that [E] remain constant? 2) My other question is in regards to page 67 of the compendium. I know the derivation of Km=[S] is not to long, but will we be responsbile for it on the exam?

Answer:

Vmax = [E] x turnover number 1) I said that if you use 1/2 as much enzyme, [E], (not [S]) that the Vmax would also reduce by 1/2. You can see that this would be the result by evaluating the above equation. 2) Yes

Question:

Dr. Bates, In lecture, you said that cholesterol disrupts the close packing of the fatty acid side chains of the phospholipid bilayer and that this increases fluidity of the membrane. In the v and v textbook, it says that cholesterol decreases membrane fluidity because its rigid steroid ring system interferes with motion of the fatty acid side chains. How can we take both of these views into consideration when examining membrane fluidity?

Answer:

As I said in lecture, this does seem contradictory. However, I think the statement in V & V is somewhat misleading. If you look carefully at figure 11-18 and read the legend, you will see that the presence of the rigid ring structure and relatively short hydrophobic tail of cholesterol tends to immobilize and stiffen the outer region of the membrane leaflet, while it also tends to increase the fluidity of the interior of the membrane by disrupting close packing of the fatty acid chains. This diminishes van der Waals interactions just as unsaturated fatty acids do.

Question:

Dear Dr. Bates, I was wondering if we are responsible for knowing all of the examples for protein structure (i.e. collagen, alpha-keratin, etc.) on page 31 of the compendium. Thanks, Katrina

Answer:

You need not memorize such structures precisely, but you should have a grasp of their characteristics and the applicable concepts that we discussed in lecture.

Question:

Hello Dr Bates Is the amphipathic alpha helix favored or stable if it has like-charge sidechains on one side of the helix?

Answer:

The juxtaposition of charged groups on one side of the helix will tend to destabilize or disfavor the formation of the helix. The net effect of all stabilizing and destabilizing forces will ultimately determine whether a helix will form or not. You should also note that, since the formation of a helix shortens the distance between amino acid residues that are adjacent in the primary sequence, adjacent like charges will also tend to destabilize a helix.

Question:

Dr. Bates, why does water freeze from the top to the bottom as opposed to bottom to top? thank you! karen

Answer:

The hydrogen bonded crystal lattice that forms in the transition from the liquid to solid form of water keeps the atoms of ice further apart than in water so that an equivalent mass will occupy more volume. Thus, ice is less dense than water and therefore floats on top of liquid water. As ice crystals form, they will rise to the top of the body of water ultimately forming a sheet-like covering on the surface that insulates the water beneath allowing much of it to remain liquid at low ambient temperatures.

Question:

Dr. Bates, how can you determine which amino acid is good for the function of glycoprotein? Thank you.

Answer:

If you are asking which amino acid side chains are likely to be glycosylated, look for side chains with hydroxyls and amines. These are the groups that can react with sugar residues to form glycosidic bonds between the protein and the saccharide.

Question:

Hi Dr. Bates, I have a question about page 12, #4 part 1. Why is it that C is also not the answer. Doesn't it have as much charge as B? Payal

Answer:

No, B has 4 predominantly charged groups at pH 7, consisting of the amino terminus, lys side chain, arg side chain, and the carboxyl terminus (about 10% of the his side chains would also be charged, but this will not affect your choice). C only has 3 predominantly charged groups at pH 7, consisting of the two terminii and the glu side chain.

Question:

Are there any old practice exams from previous quarters that we can find on the web, so we can get a feel for the midterm next friday?

Answer:

Yes, look under Chem 153A for previous quarters or in the voh archives.

Question:

Dr. Bates, i would like to know that where does the midterm cover up to?? thanks.

Answer:

Enzyme Kinetics

Question:

Dr. Bates, 1) I understand that gluconic acid is where the aldehyde becomes oxidized to a carboxyl group, but what is glucuronic acid? 2) In alpha helices, is there a specific reason why their side chains stick out, whether they're polar or non polar? Thanks. Thanks, Payal

Answer:

1) Glucuronic acid is the result of oxidation of C-6 of glucose from and alcohol to a carboxyl, while gluconic is the result of oxidation of C-1 of glucose from and aldehyde to a carboxyl. 2) This is due to steric considerations. There is no room for side chains inside this tightly packed structure. This becomes obvious when you look at a space-filling model of an alpha-helix.

Question:

Hi Dr. Bates. 1) I know like charges in the side chains of amino acids destabilize the alpha helix. Do they also destabilize the beta pleated sheet? 2) How far apart do like charges need to be before they do not destabilize the alpha helix and 3) is this number the same for the beta pleated sheet? I predict 4 residues. Thanks in advance.

Answer:

1) No, in fact adjacent side chains of like charge would favor the formation of a beta-strand, because this extended form of the peptide chain would place the two adjacent side chains at their maximal distance from eachother. 2) There is no absolute answer to this question. It is a question of degree. You know that the more distance there is between like charges, the weaker the electrostatic repulsion and the further apart the less destabilization. This is a continuous scale. We cannot say that there is destabilization at one point and none at the next. 3) See answer to (1) above. 4) Don't forget that side chains of like charge which are four residues apart would be fairly close to each other on the same face of the alpha-helix and thus might tend to destabilize the helix.

Question:

When proline forms a peptide bond in a polymer, does the nitrogen lose both hydrogens?

Answer:

Yes

Question:

Dr. Bates,my question is about beta bends. I know that beta bend has four amino acid residues and the second one is usually proline. For the other three, are there usually specific amino acids also or does it vary from protein to protein? If it does vary, does the bend usually consist of amino acids with smaller side chains like alanine or glycine or it does not matter? Thank you. Tatiana.

Answer:

Glycine is frequently found as the "third" residue of the turn following the proline (especially in type II). You can expect residues with small side chains in a beta bend due to steric considerations.

Question:

Dr Bates, I have 2 questions from your lecture today. 1.I don't understand about the difference between glycoside and glucoside. 2. From all the sugar derivatives' structures and disaccharides' structures in the compendium, which ones that you expect us to memorize? I really appreciate your response and thank you very much.

Answer:

1. Glycoside is a general term applied to a sugar compound formed by reaction of the anomeric carbon of a cyclic sugar with an alcohol converting the hemiacetal to an acetal. The term Glucoside is a more specific term used to denote that the sugar involved in the above reaction is glucose. In naming compounds, terms such as glucoside are generally reserved for compounds lacking a reducing end. 2. You should be able to draw glucose, mannose, galactose, glyceraldehyde, dihydroxyacetone, fructose, maltose, cellobiose, lactose, sucrose, segments of cellulose, amylose, amylopectin, and glycogen.

Question:

Dear Dr.Bate Are you going to have old exam posted? If you do, could you please post the exams early? Thank you very much!!!

Answer:

You can find old exam keys under previous quarters of voh, or in the voh archives. Most of the questions in the practice problem book are from old exams. Answers are in the back of the book.

Question:

Dr. Bates Do we need to know the structures of the reagents used in N-/C- terminus ID methods? Names of all the proteolytic enzyme and where they cleave? Vivian

Answer:

You should be able to recognize and name the reagents as well as understand how they are used. You should know the specificities of trypsin and chymotrypsin (as well as cyanogen bromide), but not the other proteases (endopeptidases).

Question:

Dr. Bates I have three questions:1) what are the important points of H20 exclusion from the alpha helix core with respect to its stability?2)which of the three solutions that you drew in class(insoluble particles,bolus,and thin layer) is more stable with regard to the system's total entropy?3) With respect to the alpha helix structure again, you drew glu-asp-glu-asp and said that this conformation destabilizes the helix. My question is since the n=3.6, shouldn't the like charges be 3 to 4 amino acids apart rather than two?

Answer:

1) The H-bonds between backbone atoms stabilizing the helix are very strong in an anhydrous environment due to lack of competition from water for H-bonding to the backbone atoms. 2) The system having minimum area of contact between the hydrophobic solute and the water would have the highest entropy and, therefore, the greatest stability with regard to entropy. This would be either the bolus or the layer, depending on the volume of the hydrophobic substance and the size of area occupied by the entire solution. Compare the surface areas in contact with water. Small particles have greater suface to volume ratio than large particles. In actual practice, most hydrophobic compounds will be either more dense or less dense than water and thus would form a layer below or above the water minimizing the contact.

Question:

Will you please explain the different reactions responsible for increasing and decreasing pH upon the addition of OH to solution? OH + H ---> H20 HA + OH ---> A + H2O A + H ---> HA Also, question number 5d. in the buffer section of your practice problems did have an answer for why the addition of H+ to solution would have a greater resistance to change in pH compared to the addition of OH. Why is this so?

Answer:

You'd better come to real office hours for this one. I see that I did not succeed in getting the fundamental concepts across to you in lecture. I think that a face to face discussion is needed so that I can get feedback from you in order to clear things up for you.

Question:

Dr. Bates, I was going through the practice problems when I had some questions. In the answer key, you said that His is an imidazole and Arg has a guanido group. What is an imidazole and a guanido group?

Answer:

The side chain of histidine has an imidazole ring. This is the common name for the heterocyclic amine, 1,3-diazacyclopenta-2,4-diene, with which you should be familiar from your organic chemistry. The side chain of arginine has a guanido group. This is a strongly basic organic functional group consisting of two primary amines and an imine all bound to a central carbon. If you do not recognize these structures and/or their names, you need to review your organic chemistry.

Question:

Are your old office hours still okay? If not, please post or tell us in class when and where your new ones are.

Answer:

Old office hours are cancelled. New office hours are Mondays at noon and Fridays at 11 am.

Question:

Dr. Bates, I know that we have to memorize the symbols of all twenty amino acids but I was wondering if we have to memorize just the three-letter symbol or the one-letter symbol, too.

Answer:

Three letter symbols

Question:

Professor Bates, I'm wondering if Chem 153A honor course is offered this quarter?

Answer:

No, only contracts. This has been discussed about five times in various lectures.

Question:

Dr. Bates, I was going through your practice problems when I came to a question that stumped me. What are amide bonds?

Answer:

You learned about amide bonds in general and organic chemistry. The "peptide bonds" of proteins are amide bonds between the carbonyl carbon of one amino acid residue and the NH group of the next.

Question:

Dear Dr. Bates, I would like to email you however I do not have your email. Can you send it to me?

Answer:

bates@chem.ucla.edu or you can make a confidential submission via voh. That will also come to my e-mail address.

Question:

Prof. Bates: I couldn't make it to class on Monday, 4/13/98. I'm just wondering when we can meet with you and discuss about Honors Contract. I have some ideas in mind (ie. diabetes - insulin dependent, cataracts - what kind of proteins involved, ion channels in the Central Nervous System/muscle tissue). I'm most interested in the ion channels - is this a possible topic? Thank you. --Sherry :-)

Answer:

All of these topics sound promising. You can choose any of them, and then we will need to narrow the topic down to address a specific aspect of the subject.

Question:

Prof. Bates, when water evaporates or when water is heated, what actually happens? What happen to the H-bonds? Thank you. --Sherry p.s. I think the days of the week on your syllabus in the companion book (C3) are wrong. Shouldn't they be MWRF and not MTWF?

Answer:

1. H-bonds are not static. They are constantly forming and breaking. At higher temperatures fewer H-bonds will be intact at any given instant. So there is a greater chance that some of the water molecules will have no H-bonds to other molecules at a given instant. Those molecules are free to escape. 2. Yes, as I announced in class, the schedule in the C3 applies to lecture 2 only. I handed out a syllabus with the correct schedule for lecture 1. If you didn't get one, you may pick one up in Young 5040.

Question:

Prof. Bates, On the syllabus you list April 13th as a holiday, but I dont think that this day is a university holiday. Does this mean we dont have to go to lecture? How about discussion sections on Monday? Thanks Joe

Answer:

That is a mistake in the syllabus, as you may have heard me announce in lecture by now. We will have a lecture on Monday. Thanks for checking.