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M&MB 2002 "Molecular and Membrane Biology"

UCLA School of Medicine 1st year Course

Offered: Fall 2002


OnLine Dictionaries Stedman's Medical Dictionary
good source for biochemistry terms etc.
  Webster's English Language Dictionary
  Dictionary of cell and molecular biology
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OnLine Cell/Molecular Biology

Jim Berger's list of Cell Biology links (scroll down when you get to top)

  U of Arkansas Cell Biology Page
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Regular Text Books
Champe & Harvey "Lippincott's Illustrated Reviews of Biochemistry"
Good Board Review
  Thomas Devlin "Textbook of Biochemistry, w/ clinical correlations 5th ed.

Good Practice Questions/Bio Med Library has old and new eds.

  Marks, Marks & Smith "Basic Medical Biochemistry"
Bio Med Library has older Ed.
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Responses to 2002 Students Questions
Question #
Question from Student
Dr. Lane'e response [UPDATED 9/5/02]

Q&A from the 2001 course

Amino acid removal Dear Dr. Lane,....Regarding the consequences of changing aa in transmembrane proteins, so that the charge is altered, does loosing a positive charged aa in the juxtamembrane region cause the protein to loose its transmembrane position? Dear [Student].......This is a case where we must think of "likely" vs "relatively unlikely" as components of your answer. Most juxtamembrane regions would have multiple positive charges and the loss of one of them would be unlikely to have a major effect. A bigger question might be if the old aa was substituted for a new one and what the new one would do. Deletion of the codon for the positive aa would be relatively unlikely to cause major problems. But it might cause minor problems. If you substituted the positive aa for a large polar one, this might extend the transmembrane domain and have a bigger effect. Still, the use of terms like likely, very likely, and relatively unlikely, are useful in these types of answers.
Telomeres Dear Dr. Lane,....In DNA replication, wrt to telemores as a remedy to the problem of loosing the ends of chromosome information, is this due to the fact that the RNA segment primer "blankets" a certain section of the DNA and thus that region is not created? Does the primer dislocate after replication and if so, what happens to that short section of DNA? Dear [Student].......I am a bit confused by your question as it seems to wander, but I will make a couple of statements that I hope will help. You can reply to this and we can work it out....OK? Telomeres are created to add new, "non coding" regions to the ends of chromosomes. The "need" for Teleomeres seems to relate to the problem of how to "prime" the reverse strand during DNA replication. DNA Primase, apparently can't bind to the very end of the strand, and so a small amount of DNA would be unreplicated during each cell cycle. Eventually, "important" genes might be lost. In fact, Telomeric DNA IS LOST in every cell cycle, and DNA isolated from young people have longer telomeres than DNA isolated from older people. The enzyme that creates telomeres is not expressed in adults, and the DNA primase problem can be SEEN in the steady shortening of telomeres as we age.
Translation/elongation Dear Dr. Lane,....In protein translation, does only the initial tRNA have to be "charged" with Met (in addition to the other factors that charge the tRNA)? Thereafter, the Met is available within the ribosomal subunit to transfer to incoming tRNA-aa? Dear [Student].......All tRNAa' become charged with an amino acid and a high energy bond. The initial Met tRNAi is special in that the tRNA is special and (in the next step) it associates with an initiation factor to form the tertiary complex that we talked about. All other tRNAs become charged, but then form ternary complexes with an Elongation factor. The last sentence of your question seems to indicate a bigger problem. I am not sure if you are thinking that a MET ever exists as a free amino acid within the Ribosome. That does not occur. The initiating Met is transfered to the next tRNA and the growing chain is always bound to the tRNA that occupies the P position of the ribosome. New aa's are always bound to a tRNA in ternary complex with an elongation factor occupying the "A" position of the ribosome. Take another look at the diagram of elongation and see if this makes more sense. Good luck.
Secretory Pathway Dear Dr. Lane,....In the secretory pathways of proteins, how to transmembrane proteins make it to the membrane of the cell once they are completed translated and positioned within the membrane of the RER? Dear [Student].......The best way to think of it is that the RER membrane is a conveyor belt. What I didn't have time to talk about, is that the RER is connected to the ER which is where new membrane is made. The ER is continually making new membrane and the problem is not so much how to prevent proteins from flowing out of the RER, but how to keep them in! RER proteins, such as signal peptidase, the translocases and the N-glycosylation machinery, all have to be kept in the RER with RER "retention signals". If these RER retention signals are deleted, the proteins that need to stay in the RER would "flow" out to vesicles, which then fuse with the Golgi, which then move into secretory vesicles, which fuse with the cell membrane.
Dear Dr. Lane,....Slide 21 from Lecture 6 says that all eukaryotic proteins are synthesized on cystosalic ribosomes. Aren't those proteins destined to be secreted or deposited in the plasma membrane synthesized on RER ribosomes? How are both of these possible. Dear [Student].......The slide is correct. As you state, it is also true that "cytosolic ribosomes" can be classified as either "free" OR as "membrane-bound". However, both classes of Ribosome are always bathed in cytosol and thus are all "cytosolic".
Primer in replication
Dear Dr. Lane,...if the primer in DNA replication is made up of RNA, and from that point forward the daughter strand is made, how is the section of DNA corresponding to that which the primer is occupying, ever replicated? Dear [Student]......To put your statement into formal language, let me restate it as : DNA primase makes an RNA primer that is then extended by DNA polymerase III. The resulting daughter strand is composed of DNA with a short RNA tag representing the RNA primer made by DNA primase. This short RNA sequence must be repaired (probably by DNA polymerase I), which will replace the RNA primer with DNA. DNA ligase will then link the short pieces of daughter strands together.
DNA primase and telomere shortening
Dear Dr. Lane,... is the RNA primer [created by DNA primase] actually just bound to the telomere and that is how the chromosome gets shorter over time?

Dear [Student].......Your question suggests some level of confusion about RNA primers and telomeres. Let me start by restating a few things which may actually be the simplest way to answer your question, email me again if you want to talk this out more......

First, RNA primers are used in all aspects of DNA replication, not just in replication of telomeres. Similarly, telomeres don't have anything to do with the normal "mechanics" of replication, although theu are are important for chromosomal stability. RNA primers will be made to replicate the telomeres, and these primers will still be removed by a repair process. The reason that teleomeres shorten is that DNA primase doesn't seem to bind to the very end of the reverse strand, making it impossible to make an RNA primer for the very end (the tip) of the chromosome.

Let me just say one more thing that may or may not help, but I hope will clear some of this up for you. Replication of DNA (whether in the middle of a chromosome, or at the ends) should be thought of in the context of "the origin of replication" OR "the replication fork". When focusing on what is happening at the replication fork, one sees 2 processes a) leading strand synthesis and b) lagging strand synthesis. At the Origin of replication, one sees Leading strand synthesis AND (once lagging strand synthsis catches up) repair of the RNA primer and ligation of the daughter strand. At telomeres (the ends of linear chromosomes) leading strand synthesis proceeds normally. The polymerase complex starts at an Origin or replication and then heads out toward the ends until it falls off the end. But the other strand can't act as a normal lagging strand. There is always some distance between the true end and where DNA primase actually sits down to start lagging stand synthesis. This is why the telomeres shorten with each adult cell cycle. Embryonic cells (and germ cells) still express telomerase, the enzyme that makes telomeres. So their telomeres don't shorten with every cycle. I hope this helps. Email if you would like to talk more........

Responses to 2001 Students Questions
Question #
Question from Student
Dr. Lane'e response
5' to 3' terminology.
Dear Dr. Lane,.....You stated, that DNA is not only synthesized 5' to 3', but also read in the 5'to 3' direction. But, it looks like although the daughter DNA strand is being synthesized 5' to 3', the parent strand is being read 3' to 5'. Can you clarify this, please? Dear [Student].......You are correct. We read and write DNA or RNA sequences 5' to 3'. That is also the way all polymerases synthesize DNA and RNA. You correctly observe that in order to do this, the "polymerase" has to "read" DNA 3' to 5'. We will talk about this in detail in lecture 5.
Leading vs Lagging Strand Synthesis
Dear Dr. Lane,..... I was going over the notes for DNA replication and noticed the following definitions: Leading strand = daughter strand that is moving toward the replication fork Lagging strand = daughter strand that must be synthesized away from the replication fork Assuming the point of origin for replication is somewhere in the middle of the DNA double helix and knowing that DNA replication occurs bidirectionally, won't both daughter strands have leading and lagging parts? Dear [Student].......Exactly right! I only showed you the first half of the bubble that is present at the origin. Why? Because the other half looks exactly the same, only the inverse. The origin is bidirectional, therefore, the origin will have Leading strand synthesis proceeding away from it on each parental strand. Leading strand synthesis "creates" the problem of lagging strand synthesis. One can either look at if from the point of view of the origin, OR from the point of view of the replication fork. I think the view from the replication fork allows a more unified presentation.
Leading vs Lagging Strand Synthesis
Dear Dr. Lane,..... I suppose my question is just a clarification: a daughter strand as a whole cannot be termed leading or lagging, but parts of the daughter strand can be termed leading/lagging. Is this true or am I missing something? Dear [Student].......You have it right. A completed daughter stand would be composed of sequence that was created by both processes.
Differences in DNA polymerases
Dear Dr. Lane,...., From the notes, (Lecture 3) I understand that DNA polymerase III elongates the strand and DNA polymerase I removes the RNA primer put down by DNA primase. What does DNA polymerase II do? Dear [Student].......First, let me explain that DNApol I, II, and III refer to the bacterial proteins. The eukaryotic polymerases are called alpha delta, epsilon, beta and gamma. There is a good discussion of all these enzymes in Devlin if you want to check it out of the library. From what I understand, prokaryotic DNApol I, II and III all have similar activities in vitro, however, mutations in polI or polII have very little effect on replication. Mutations in polI and polII genes affect bacterial DNA repair! Therefore, it is concluded that polII (as with polI) have roles in DNA repair. You are not required to know this for our course, but it is very interesting stuff, and very relevant to problems of DNA repair and the fidelity of the DNA replication. Thanks for your question.
Dear Dr. Lane,.....What is the function of the post-translational modification "glycosylation"? How does it change or regulate the way a protein functions? Dear [Student]......In that slide (Lecture 2) I mentioned that there were two major types of protein glycosylation i. enzymatic (sequence specific) glycosylations (on S, T, or N), versus ii. "non" enzymatic glycosylations. We will talk about sequence specific enzymatic glycosylations next Wednesday or Thursday. The example that I talked about in lecture was non enzymatic glycosylation of HbA. This glycosylation is the consequence of a disease (diabetes) which causes high blood sugar. When blood glucose levels reach a high enough level, glucose attacks free amines and so it gets incorporated into proteins. There is no "beneficial function" of this, it is a biproduct of the disease. Futhermore, glycosylations of this type can cause complications. e.g. when collagen gets non enzymatically glycosylated, it becomes immunogenic. As a consequence, older diabetics can have autoimmune responses that can be tragic.
sickle cell
Dear Dr. Lane,.....In your discussion of sickle cell anemias (Lecture 2) you described three different amino acid substitutions that can cause this disorder. I understand why in HbS the substitution of Val for Glu would cause a lot of problems-- a polar charged amino acid is substituted with an aliphatic one. I am less clear about the effects in HbC and HbO where the substitution involves one polar charged amino acid for another. How and why do these second two substitutions affect the hemoglobin structure? Dear [Student]......There is, in this case, a relatively simple answer. But sometimes, it is not obvious (remember, the size and shape of a side chain may also be critical). In this case, however, you remembered that Glu and Lys are polar, but forgot to take into account that they are also charged! The "simple" answer, is that Glutamic Acid (HbA) and Lysine (HbC/HbO) have opposite charges! It is likely that Glu6 and Glu121 are involved salt-bridges with amino acids of opposite charge. When mutated to Lysine (which is positively charged) this salt bridge would not form AND the new Lysine would actually repel the part of the protein that it needs to associate with.
Dear Dr. Lane,.....What is the definition of a "supercoil" (lecture 4) and what is the distinction between a positive and a negative one? (Is this a directional distinction like right- and left-handed alpha helices?) Dear [Student]......OK, there are a couple of ways to explain this. Of course, your sense seems to be correct. Supercoils are one step up from B-helix (regular coil) DNA. The term "Negative" supercoiling is given, by definition, to DNA that has been cut, then unwound, then religated. This is easiest to picture if you have a closed circular DNA like a plasmid, but the same is true in linear chromosomes, where protein scaffolding holds the supercoils in place. Positive would be what would happen if you cut DNA, overwound it, then religated. It turns out that negative super coiling is VERY useful in biological systems. First, it allows for compaction of genomes which is very useful, but by maintaining negative pressure on B-form DNA, it allows proteins like RNA polymerase to melt short sequences of DNA with "very little" effort.

amino acids and the effect of mutations

Dr. Lane, ..... I am having...difficulties predicting the outcome of particular changes in amino acid sequence on protein function. I don't think that I have a clear understanding of which particular regions of a protein are innocuous if they are altered and which are particularly deleterious. In more obvious examples I seem to be fine: for example that mutations occuring mid-intron are probably not going to alter protein sequence or function, whereas if even one amino acid is substituted at a binding site there is generally a problem with protein function. With more complicated, nuanced questions, however, I find myself getting stuck. Are there any general guidelines that I can follow regarding the amino acid specificity of particular regions of proteins? I was a non-science major and perhaps I just need more background information, in which case maybe you could direct me to the appropriate resource (text, website, etc.)

Dear [Student]......I know that it can be a challenge to master this material and the task is especially difficult when it is all new. The 20 amino acids seem very complicated at first, and it takes some effort to understand differences that might appear subtle to the untrained eye. On top of that you must also learn to appreciate differences in the types of environments that proteins encounter (aqueous, membrane, etc). You might not have thought about these issues previously, this is why we have given you so many tools to help your learning process.

That said, I am rather impressed with a couple of your observations, and it sounds like you are getting the basics. Maybe you aren't in as much trouble as you think! Specifically, given the areas that you are comfortable with, it is hard for me to guess what you are having problems with. Please tell me the precise question you missed so that I can help you understand the problem.

.....In general, you should look for clues in the question. If you know that a change leads to a disease, then you can bet it causes some change in structure, stability, binding etc. (for example, the sickle cell problem). Otherwise, you have to make "educated" guesses. Your sense of the importance of introns vs coding sequences, and of binding sites is a start. Then we have internal vs external surfaces of proteins, active sites, and trans membrane domains. If you go to the www site,

you can download my "Review slides 2001". There is one of them (slide 7 ) devoted to summarizing some of this. Let me know if it helps.


Ordering of events associated mRNA processing

Dr. Lane,, .....I have been trying to ascertain the order of the following modification: poly A tail addition, 5' G-cap, and splicing of introns. I've attempted to find the answer from several texts and so far I've got 2 texts and 2 different answers.

a second student adds:

Dr. Lane, In last year's exam, on a question that you asked for a numbered sequence of events, part of the question was to line up these events: . . . . A. Intronic sequences removed . . . .. . . . . . . . . . B. poly-adenylation added . . . . . . . . . . . . . . . . . C. spliceosomes bind to pre-mRNA. . . . . . . . . The correct answer was C, A, and then B,which is what the supplemental notes said, but, this year's notes does not. It actually contradicts last year's notes. According to this year's notes, poly-adenylation occurs BEFORE intronic sequences are removed. I am still unsure where spliceosomes would come in. Can you please clarify this point?

Dear Student... The problem lies in how different books order splicing and poly A addition. I imagine that 5'CAP formation is always listed first and nuclear export is always the last event in pre mRNA processing.

The short answer is that our course has historically placed a bit more emphasis on the order in which the events start (see figures 5:16-21 of this yearsLecture notes)

[ transcription start, 5' CAP, splice formation, poly A addition, nuclear export ].

Some books (and figure 5:15 of this yearsLecture notes) present the order in which the events finish:

[ transcription start, 5' CAP, polyA addition, splice formation, nuclear export.]

the order of polyA addition and splice foramtion is switched becasue splicing can take a long time to complete.

I will give full credit for either order However, some people find it simpler to think of the first list because it is the same as the order in which substrates become available inside the nucleus.

The problem of ordering splicing is that it gets started early but usually ends late and that the timing can be different for different mRNA's.

As to the order of answers in last years exam (please refer to the students question to understand this). Splicing factors start to assemble as soon as introns become visible so C goes first, in some mRNA's introns are removed quickly so A is next. PolyA addition happens soon after the AAUAAA signal is transcribed so B could go last. Last year we presented the order C A B, but I would have accepted B in any position as long as C came before A. This year I presented the order as C A B in lecture 5 slides 16-21, but I also gave an overview slide, Figure 5:15 which some students have interpreted as suporting an answer of B C A. Again, I will accept B in any position, as long as C is listed before A.

Summary Ordering the events in mRNA processing.

In eukaryotic mRNA's transcribed by RNA pol II

i. transcription always begins at an RNA pol II promoter with a correctly assembled transcriptional complex and the RNA pol II holoenzyme.

ii. 5'CAP is assembled soon after the 5' end of the nascent RNA is generated.

iii. splicing factors begin to bind the transcript as soon as introns become available and splicing factors remain attached to the hnRNA until all splicing reactions are complete

iv. RNA cleavage and poly adenylation occur very soon after the AAUAAA signal is reached. Polyadenylation will not wait for splicing to be completed, and so is often completed before splicing is completed.

v. splicing continues until all introns are removed (some introns are processed with fast, others with slow kinetics).

vi. Only when all splicing has occurred, and polyadenylation has occurred, can nuclear export of mRNA take place.

Amino Acid substitutions
Dr. Lane, . . . .I am still having trouble with the review questions at the end of your first lecture packet. I feel as though I have a solid understanding of the concepts presented in the notes, but when I attempt to answer the review questions I feel as if I'm missing something. The answers to the questions suggest that when considering the effects of substituting one amino acid for another in a protein, the location of the substitution is somewhat more crucial than the characteristics of the amino acids being exchanged. Is this correct? And are there any general rules or guidelines regarding which areas of a protein are more likely to result in function alteration should a substitution of an amino acid occur? (ie if the characteristics of the two amino acids involved in a substitution are similar are there areas of the protein that typically would not be affected versus areas that would be dramatically affected?)

Dear Student.....You hit it right on the head, you have to balance what you know about amino acid R groups with what you know about protein structure.

-regions of protein associated with solvent are the least affected by substitutions, but will still be fairly intolerant of a large aliphatic side chain (Trp, Ile, Leu). Small aliphatic amino acids (Ala, Val) are tolerated on protein surfaces, but large aliphatic side chains (Trp, Ile) would not.

-the interiors of proteins tend to be hydrophobic but when a polar group is found it has to be balanced with another polar group.

- Charged residues are usually found on protein surfaces or involved in salt-bridges. Changing a charged amino acid on the surface of a protein can be tolerated if the charged residue is being used to contact water, but changing charge is almost never tolerated on the interior of proteins or at binding surfaces because of the disruption of a salt bridge.

-active sites of enzymes and protein-protein, or protein ligand interfaces are usually very sensitive to non-conservative substitutions.

-regions of proteins binding lipid (eg membrane spanning domains), tend to require aliphatic amino acids and would likely respond poorly to the introduction of a strongly polar or charged amino acid. I wonder if you have read Supplement #1 from last year. It might help. I also have a slide on this in the Review which I will provide on Oct 4th. You can down load this from the www page as well. Finally, there is at least one question on this FAQ related to amino acids.

changing the size of an amino acid
Dr Lane....When does changing the size of an aa make a difference. For example, in question 3a of the practice set at the end of the lecture notes, why doesn't the difference is size between alanine and phenylalanine have an effect on the protein? It seems so much bulkier. In one sections of the tertiary structure does size diff. matter and in what parts does it not?

[ The question was as follows: 3. An alanine residue at the center of a single helix that spans a membrane is replaced by: (a) Phenylalanine?]

Dear Student.....Think about it this way. We have a membrane spanning section of a protein. As the helix crosses the membrane, each of the side chains is being projected into a relativley large sea of lipid. In this case, there are no specific space/size constraints. The protein doesn't care if it is putting a large side chain (Phe) or small side chain (Ala) into the lipid. The only thing it cares about is that the amino acids interact fairly well with lipid. Since both Phe and Ala are aliphatic, they will both work in this microenvironment.

In contrast, if this was a question about the inside of a globular protein where spacing is much more critical, it might be more difficult to predict what an Ala/Phe substitution would do. In this case, it would be safest to assume that the size of the two side chains would definately affect the spacing and shape of the protein!

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GEE WHIZ ....(Facts in Scinence)
A viral genome (about 3000 bp) would take about 50 minutes to read outloud from a printed page.
  A Bacterial genome (about 3 million bp), would take about 34 days to read outloud from a printed page.
  A human genome (about 3 billion bp), would take about 9.5 years to read outloud from a printed page.
  Over 35% of the human genome is made up of transposon ("jumping") DNA.
  One million bases (called a megabase and abbreviated Mb) of DNA sequence data is roughly equivalent to 1 mega byte of computer data storage space. Since the human genome is 3 billion base pairs long, one would need 3 gigabytes of computer data storage space to store the entire genome. This includes nucleotide sequence data only and does not include data annotations and other information that can be associated with sequence data.
* *Contributions to this fact were made by Morey Parang and Richard Mural of Oak Ridge National Laboratory. (3/99)
** **This is based on a rate of 10 bases per second (A, T, C, G, A, T, C, G, A, T), which equals 600 bases/minute, 36,000 bases/hour, 864,000 bases/day, 315,360,000 bases/year. [Contributions to this answer were made by Mark Adams of The Institute of Genome Research.]
*** ***Answer adapted from Analysis of Human Genetic Linkage by Jurg Ott.
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General information about the course

M&MB 2001 "Molecular and Membrane Biology" . .. . (Fall, 2001)
Registration: Pat Anaya ext 5-2866  
Course Co-Chairs: Dr. Judith Gasson, (Biological Chemistry) Dr. Ernest Wright (Physiology)

Professor Lane: Dr. Lane delivers 8 lectures covering the basic features of protein, DNA and RNA biochemistry.

Dr. Timothy F. Lane (Biological Chemistry)

Office: 13-262 Factor


Location and Times

Tuesdays, Wednesdays, and Thursdays 10-11am

73-105 CHS (7th floor Lecture Room)

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