Q: What will the form of the test be?
A: a mixture of multiple choice, T/F, and short answer, problem solving, with an emphasis on experimental approaches. Do not memorize the more obscure protein/gene names (e.g. the individual subunits of the RNA polymerases) but do know the more fundamental ones (e.g. RNA Pol I vs RNA Pol II.).

Q: what will the material on the tests be drawn from (i.e emphasis on the lecture notes, the lecutre notes and the book, etc...)
A: emphasis on the lectures and notes, use the book to interpret confusing statements in lecture or in the notes. Unless I state otherwise, you are not responsible for material in the text that is not covered in the notes or in lecture.

Q: How much of the specifics will we be tested on.(i.e is it vital to know specific examples)
A: little on specifics (e.g. number of introns in DHFR? will not be asked), a lot on concepts (e.g. are the introns or exons in DHFR better conserved? will be asked)

Q: I understand this is a very demanding class, the pace of your lectures is very fast and gives little time for complete absorbtion of the materials. There is so much information I don't even know what to study for. I'd appreciate any positive feedback.
A: In terms of studying, there are a couple of things I would suggest: - 1. scan the lecture notes and do the reading before lecture so that you have a sense of what we will be covering. - 2. look at the 'Basic Concepts' on the Web -- for every lecture, we have a series of general questions that you should feel comfortable discussing -- if you can, you are probably picking up what I consider to be the important material. - 3. look at the sample problems on the Web -- these are designed to mimick the types of questions you may expect on the tests. - 4. look at the FAQ on the Web if you have specific questions -- every time I get a question from a student, I post my response to it here. It is likely that if something is confusing, someone else will have already asked about it. - 5. attend the review sessions.

Overview and Cloning Techniques

Q: What is the point of cloning? I used to think it was to make a lot of copies of the insert, but is it really to get a lot of whatever product is produced after the organism "expresses" the insert? What if the insert is never expressed?
A: There are several reasons why one would be interested in cloning a gene --including the following: 1. Once one has a gene cloned, the gene can be sequenced -- by comparison to the sequences of other genes, this information often helps to define what type of gene product the gene codes for. If genetic studies have isolated mutants of the gene, one can determine the sequence differences between wild type and mutant copies of the gene to infer the molecular basis for misfunction by the mutants. 2. Having the gene greatly facilitates experiments to characterize the biochemical and cellular function of the gene product. With the gene, one can: - a. express large quantities of the gene product, purify it, and carry out in vitro studies on the protein / RNA (e.g. x-ray crystallography) - b. raise antibodies to the gene product, making it possible to define the gene product's function in vitro and in vivo (e.g. immunoprecipitation) - c. introduce tags into the gene product to define when and where it is expressed/localized (e.g. green fluorescent protein fusions) - d. set up genetic screens to identify other gene products that interact with the gene (e.g. yeast two hybrid) - e. analyze what factors are important for expression (e.g. promoter bashing) - f. engineer mutants that reveal how the gene product functions (e.g. site-directed mutagenesis) 3. Having the gene makes it possible to express the gene product under artificial control in non-natural settings. This is generally done in two ways: - a. expression in bacteria makes it possible to obtain large quantities of purified protein which can then be used directly (e.g. as pharmaceuticals or as industrial enzymes). - b. expression in non-natural hosts may confer new properties to the host (e.g. gene therapy, pest-resistant crops, etc.)

Q: What does "transformation efficiency" mean? How well the bacteria accepts and reproduces the vector containing the insert?
A: Transformation efficiency is a measure of the efficiency with which naked DNA is taken up by a host cell. It is generally expressed in terms of transformants / microgram of DNA. Typical values are: - plasmids: 100,000 transformants / ug - phage: 10,000,000 plaque-forming units / ug

Q: How would one go about isolating bacterial mRNA from rRNA and tRNA, since the poly A tail doesn't exist; in addition does the fact that translation and transcription proceed simulataneously cause problems?
A: It is hard to isolate bacterial mRNA for two reasons: - 1. as you note, bacterial messages are not poly-A-tagged, so oligo-dT cellulose is no help. - 2. bacterial cells are full of RNAse, and the half-life for bacterial mRNAs is exceedingly short (e.g. 3 minutes in E. coli versus 6-24 hrs. in humans). As such, many messages at any given instant will be partially cleaved. Undaunted, though, one can run sucrose gradient centrifugation experiments which fractionate macromolecules on the basis of size, density, and apparent radius. The bulk of bacterial mRNA runs faster than the ribosomal RNA and the tRNA, and can thus be purified away from these contaminants (realizing however, that short messages and partially degraded messages will be lost...).

Q: Why do cosmids need cos sites?
A: The cos sites (cos = 'cohesive') are 12-base overhangs at the 5'-ends of the dsDNA form of the single phage genome. The ends are complementary to each other such that once phage DNA is injected into an E. coli host, the ends mediate circularization of the genome. Once covalently closed by DNA ligase, DNA polymerase catalyses the synthesis of a concatenated polymer of phage DNA. An endonuclease recognizes the cos sites, catalyzes their cutting to release a single copy of phage DNA. Cutting is coupled to packaging, so... to package DNA as a lambda phage, you need to have cos sites. To further clarify what a cosmid is, it is basically a plasmid that happens to have cos sites. The cos sites allow it to be packaged in vitro as a phage virion which can then infect a bacterium (thus you get high efficiency transformation). Once inside a bacterium, however, the cosmid lacks all of the phage structural genes, so it is basically trapped as a plasmid and cannot code for the coat proteins, etc. that are required to generate new virions for further rounds of bacterial infection.

Q: Does the fact that translation and transcription proceed simulataneously [in prokaryotes] cause problems?
A: The directions of translation and transcription are such that the first part of the message synthesized (the 5'-end) is the first part translated -- as such translation proceeds on a message that is still in the process of being transcribed. Bacteria seem to do fine with this arrangment. It obviously presents some limitations, however, including: - 1. if introns were present [which they almost universally are not], one would need to have a mechanism that would block translation until splicing was completed -- the nuclear envelope accomplishes this effectively in eukaryotes. - 2. with this arrangement, virtually all regulation of gene expression must exist at the transcriptional level, eliminating the fine tuning one obtains in eukaryotes from splicing, RNA export, translational efficiency, RNA degradation.

Q: I do not quite understand how the insertion of a reverse-transcribed mRNA would give rise to pseudogenes?
A: cDNA generated from an mRNA lacks the promoter and upstream regulatory element sequences that would be required for the sequence to be expressed (i.e. transcribed). As such, if the cDNA inserts into a random place on the genome, it generally will have no effect and as such is not subject to any evolutionary selective pressure (ultimately, it will incorporate other mutations that would make it non-functional in addition to its lack of promoter sequences). Occassionally, such inserts land at actively transcribed locations in the genome and do exert effects --retrovirus-mediated transformation by viral oncogenes is the prime example.

Q: In making cDNA I don't quite understand what you mean by "running total cellular RNA over an oligo-dT column,"? what is a oligo-dT column, is it just a sepharose column of DNA containing only thymine? And if yes, how does that help isolate mRNA?
A: One takes total cellular RNA (obtained by cracking open cells, digesting them with DNAse and then extracting them with phenol to remove all proteins) and loads it onto a column which contains beads of cellulose that have short oligonucleotides of dT covalently attached to them. The average length of a oligo-dT oligomer is roughly 20 nucleotides. It helps to isolate mRNA because mRNAs are polyadenylated after transcription (a tail of roughly 200 A's are tagged on to the 3'-end of the transcript). Simple base-pairing then directs immobilization of poly-A-containing mRNAs on the column while rRNAs and tRNAs are washed away. Drawing schematically:

column|-TTTTTTTTTTTTTTT-3'
3'-|||||AAAAAAAAAAAAAAAACUGCUACGUAGCAUGCAUCGGCAUGCACGU-5' | |

Q: Is reverse transcriptase used twice when making cDNA? it seems as if you use it once to produce dna from mRNA and then again to produce the loop of DNA back on to itself after degradation of the mRNA strand.?.?
A: yes it can be. It turns out that reverse transcriptase (RT, which comes from retroviruses) normally catalyzes both first strand (using RNA as a template) and second strand synthesis (using DNA as a template) in converting the retroviral genome from a single-stranded RNA to a double-stranded DNA. Old protocols for making cDNA libraries use RT for both steps. More modern protocols often incorporate a second DNA polymerase for second strand synthesis since (by eliminating the RNAse-H activity of RT) it is possible to get longer reads. If you are interested in the nitty-gritty of this process, please look here.

Q: Why is RNA single stranded, why isn't it double stranded like DNA?
A: This is more of a philosophical question than a biochemical question. There are both double stranded RNAs and single-stranded DNAs in biological situations (e.g. certain viral genomes exist transiently as each). Teleologically thinking though, one might guess that DNA is double stranded because it is the genetic material and by having two complementary strands, an error or mutation to one strand will not necessary effect the genetic information (because there is always a backup copy to code for repair of the other strand). In contrast, most RNAs function either as structural molecules (rRNAs, tRNAs) or as transient messangers (mRNAs), contexts in which it is not essential that high fidelity be maintained (if an mRNA gets altered between its synthesis and its translation, you can always make another one). For the structural molecules, it is hard to do much if you are locked into a double helix. By being single stranded, rRNAs and tRNAs are able to fold into complex three dimensional structures, acting more like proteins than like dsDNA.

Q: What is the YAC transformation efficiency?
A: Relatively low

Q: If packaging is involved with a cosmid, then why is the insert larger than a phage?
A: In a lambda replacement vector, the lysogenic genes have been deleted but the lytic genes are still present. In a cosmid (which again is basically a plasmid with cos sites), all of the phage genes have been deleted. As such, there is simply more room for the insert.

Q: Can you explain gene transfer? You mentioned it when you talked about why you would want to clone genes
A: The idea of gene transfer is that you take a gene from one organism and by introducing that gene in another organism, allow a foreign gene product to be expressed. This could be useful in two kinds of ways: 1) conferring new biological benefits to another organism (e.g. making purple tomatos), 2) correcting a biological defect (e.g. gene therapy, using the correct copy of a gene to replace a defective copy).

Q: Can't phages be both ss & ds DNA or RNA? Or for the purpose of using them as vectors you use the ones with ds DNA.
A: Most phages pass through different forms as a function of their life cycle. A given type of phage, though, will generally spend most of its time in one form (e.g. M13 = ssDNA, lambda = dsDNA). When we talk about phage as vectors for making libraries, we are talking about lambda phage, which is packaged in the form of dsDNA and thus considered as dsDNA phage.

Q: Can you briefly explain the importance of having TRP1 and URA3 in YAC's? And why you want each half of the YAC (before you insert the insert) to have a telomere at the end
A: Selecting with both TRP1 and URA3 forces you to incorporate each half of the YAC plasmid, meaning that you will end up with telomeres on either end and a single centromere. In addition, by having selectable markers at either end of the YAC, you prevent chromosomal deletions from happening that might eliminate one side or the other of your insert.

Q: In class you briefly talked about when you would use a specific vector for cloning. you said you would use a phage for a single gene. I was wondering why you wouldn't use a plasmid for a single gene. What is the range of a typical gene.
A: The average length of a primary transcript (including introns) is roughly 20kb, way too big for most plasmids. The average length of a cDNA is only 2.2kb but there are lots of mRNAs that are substantially bigger. A phage or cosmid vector is a safer way to get cDNAs since larger than average cDNAs are not going to be selected against.

Q: Why would use want to use an RNA probe in a southern blot, if your target was DNA sequences?
A: RNA probes are relatively easy to make (by in vitro transcription with phage RNA polymerases) and RNA:DNA hybrids under many conditions are more stable than DNA:DNA hybrids.

Q: I still do not understand how PCR can be used to clone the intervening sequence directly. Could you give me an example?
A: Lets say that one has already cloned the wild type allele for a gene that is responsible for a disease and one is interested in defining the specific point mutations that are responsible for disease in individual patients. One can design PCR primers that correspond to the 5' and 3' ends of the gene and, starting with a few scraps of tissue from a single patient, use PCR to amplify the sequence between the two primers (the intervening sequence corresponds to the entire gene). The resulting dsDNA can be either ligated into a vector and sequenced or sequenced directly to reveal the point mutations that may be present in the specific individual. p.s. in re-reading your question, I hope the confusion is not that you are equating intervening sequence with intron. In this context, we mean intervening sequence to be the region that intervenes (lies) between the two PCR primers.

Q: How can one design a probe?
A: I assume you are talking about nucleic acid probes: one needs to have some idea of the nucleotide sequence. This can come from 1. a cloned insert isolated from a library. For example, in the process of chromosome walking, one uses cloned inserts to generate probes which can then identify neighboring clones. 2. knowledge of the nucleotide sequence of a related gene. For example, the gene for actin expressed in cardic muscle has been determined. Using the sequence of an exon from this gene, we could make a probe that might hybridize to related exons in the actin genes expressed in other muscle types. 3. knowledge of peptide sequence. If some peptide sequence is known (e.g. you have purified a protein biochemically and by microsequencing know the amino acid sequence of some peptides generated by proteolytic digestion), you can use the genetic code to direct the synthesis of oligonucleotides that would code for the peptide sequence. Because of the degeneracy of the genetic code, one typically synthesizes a mixture of all possible oligonucleotides that might code for a given sequence (e.g. genes that code for the amino acid sequence Val-Trp-Ala-Gly would be identified by hybridization with a oligonucleotide mixture with the general sequence GTNTGGGCNGGN -- in reality, one would need a longer sequence to get enough specificity).

Q: I have a question regarding blocking. I dont remember you talking about blocking, however it is mentioned in one of the web pages you linked to your web page. Should we include blocking as one of the steps when doing hybridizations? Would we need to block when hybridizing a probe to a filter when analyzing librarys (case one, two and three mentioned in the reader)?

A:Yes, in real blotting experiments you need to do something to prevent non-specific binding of your probe. In hybridizations you generally add in a large excess of a non-specific DNA to prevent non-specific binding of the probe (the property of the membrane used to initially bind the nucleic acids from your gel). In a western blot, you add a large excess of non-specific protein to prevent your antibody probe from binding the membrane itself (remember the membrane is designed to bind proteins-it has the proteins from the gel bound to it). These blocking reagents don't displace the nucleic acid or protein that was bound to the membrane during the transfer step, they just bind the remaining binding sites left on the membrane prior to the hybridization or antibody probing steps. The upshot of this is that the only way for your probe to bind the membrane is through a specific interaction with molecules that were transferred from the gel onto the membrane.

Just to keep you up on your trivia, the most commonly used nucleic acid blocker is salmon sperm DNA (I have no idea why they chose that!) and Carnation non-fat instant milk for western blots.

Gene structure

Q: How do you deal with intron that have a restriction site in the exon trap? How does the trap select for exons alone? Is there a requirement for the two sides of the exon?
A: I don't have any direct experience with exon trapping vectors but would make the following guesses: 1. as with any kind of library construction, restriction sites can be present in the insert. you can get around problems by doing a partialdigest with a frequent cutter so that at least some of the inserts will not have the insert digested 2. the idea is that the fragments containing exons have the necessary sequences on both the 5'-side and 3'-side to let the ends of the exon be recognized by the splicing machinery. After a transcript is generated, all intron sequences will be spliced out so if you're characterizing the mRNA produced by the vector, all you will see is exon sequences. 3. If you insert a fragment which is lacking the sequences at the 3'-side of the exon (for example), you probably wouldn't get accurate splicing. Either an inappropriate splice site would be chosen or splicing would be inhibited. In either case, your product would be different in size relative to the insert-free construct.

Q: Is it true that the good reasons to have introns are: - 1. limit recombination during meiosis? - 2. yet, good also because we want recombination and introns serve as a buffer region? I am a little confused i thought recombination was good for diversity, but it seems that recombination is good sometimes? Why only sometimes, do you mean when recombination produces a bad mutation, maybe? well, i am curious if the above 1.and 2. are considered good reasons to have introns or bad reasons, in your opinion, of course?!?!?
A: Recombination can happen between genes (resulting in new combinations of pre-existing alleles) or within genes (to generate new alleles representing combinations of other alleles). This second process is good at some evolutionary level (because it allows for new genes to be assembled from mixtures of genes, increasing diversity within the gene pool) but can also generally be bad for an individual (since many recombinants result in defective gene products). The level of recombination is thus effectively set to balance these competing effects -- introns, by rapidly diverging may limit the extent of recombination within genes. It is worth noting that as with most evolutionary arguments, it is hard to construct experiments to test these ideas directly.

Q: You list sequence comparison and restriction mapping as two ways of identifying introns. What are the differences between the two?
A: sequence comparison means you go ahead and sequence both the cDNA and the genomic clone of a gene and directly read out the nucleotide sequences of the introns (missing from the cDNA) and the exons (not missing). Restriction mapping basically means that you figure where the restriction sites are in a cDNA and a genomic clone and you look for conserved patterns of restriction sites in both sequences. This can tell you at low resolution where there are big gaps in the genomic clone (introns) but doesn't tell you what the sequences for each are or exactly where the boundaries between introns and exons lie.

Q: When you talk about the evolution of introns, and why they exist you mention "limit recombination of coding sequences." Can you explain what you mean by this? Can you also explain what you mean by facilitate swapping between protein domains?
A: Recombination can happen when there are large regions of homologous sequence between two chromosomes. By diverging (evolutionarily) quickly, introns limit the size of regions of homology between different genes that could lead to disruptive recombination. If exons correspond to natural modules of protein structure (early intron hypothesis), one should be able to mix and match them with other modules to create new proteins. By breaking up a protein into discrete modules with introns, when this mixing happens, it is more likely to happen in a way that moves an entire module and is thus more likely to give rise to a functional product.

Q: we've learned several times the idea of "one gene: one polypepetide." How can this be if alternative splicing occurs? Isn't that "one gene: several polypeptides?" This confusion is probably due to a simply misunderstanding of definitions. Could you clear it up?
A: This basic concept was developed before we had any clue that genes corresponded to fragments of DNA and was one of the first observations that ultimately led to the formulation of the central dogma of molecular biology: DNA makes RNA makes protein. This idea remains essentially true for prokaryotes but must be revised for eukaryotes. Instead of 'one gene, one protein,' we can say 'one gene, one set of related protein products...'

Q: When we talk about alternative splicing are we concerned with only eukaryotes or prokaryotes? I thought that eukaryotes were monocistronic and prokaryotes were polycistronic. If so how can alternative splicing occur in eukayotes and not occur in prokaryotes?
A: Prokaryotic genomes essentially lack introns, making alternative splicing impossible since splicing doesn't happen at all... You are right, though, that eukaryotic genes are almost invariably monocistronic while prokaryotic operons often consist of polycistrons.

Q: I was wondering if one function of introns involves developmental regulation via alterantive splicing. If this is the case, is it possible that introns were involved in the evolution of multicellular creatures?
A: Yes and yes. Very often, one finds different patterns of splicing in developmentally distinct tissues, allowing optimization of gene products for specific cell types. It seems obvious that yeast has a relatively simple cell structure, only one cell type, and may thus not require the multiplicity of gene products that can be obtained by alternative splicing, thus explaining the relative absence of introns in this organism.

Genome organization and gene families

Q: I was wondering if another reason for the length of DNA could be that it reduces the overall effect of random mutanagenesis? It seems that if random mutations occur over a larger genome than only the transcribable elements, regulatory elements, etc need to be repaired and the excess can be left alone. Maybe that is another reason why the organism continues to keep pseudogenes, introns, heterochromatin, etc. It seems that evolution forces organisms to be as efficient as possible and that if they are not, they are soon surpassed and survived by another organism which is more efficient. So the question is, why would we have this apparent excess of DNA? Why would the cell/organism waste precious space, energy, and resources unless there were a specific, necesaary function. It seems that the DNA is therefore necessary. Since the active DNA is only a small percentage of the whole length than only a small percentage of the random mutations will occur in active DNA. It also seems that if some organisms had smaller genomes with a larger proportion of active to inactive DNA, that perhaps the rate of mutanagenesis over an extended period of time would eliminate those organisms allowing larger and larger genomes to prosper. And since that would not be the only solution, then other organisms would survive due to other creative solutions, like better, more efficient repair mechanisms and creative ways of controlling the random mutations.
A: The mechanism you suggest for reasons for having a big genome has been suggested by others. The reason I didn't mention it in class is that I don't really see how it would help. Presumably mutations / errors in replication arise in a stochastic way, with a certain probability of happening at any site, let's say 10-8. Increasing the size of the DNA by stuffing in random non-coding sequences is simply going to increase the number of mutations that happen without decreasing the density of mutations within coding regions. Some experiments (which I forget the exact details of) have been done to see if non-coding DNA has any biological function. As I vaguely remember, a yeast strain was engineering in which a major DNA satellite was removed from a chromosome. The resulting strain was seemingly phenotypically identical to the parent strain, but if one grew the two strains in parallel, the satellite conferred a slight growth advantage and eventually, the parent strain took over the culture. The mechanism for this growth advantage, as far as I know, has never been characterized. As far as big genomes being a disadvantage, it really depends on the growth conditions. For something like a bacterium or a yeast which is trying to grow and divide as rapidly as possible, DNA synthesis can be rate-limiting in the cell cycle and you might expect selective pressure to drive towards small genomes. Most eukaryotic cells, however, spend large periods of time in G1 or G2, not replicating their DNA and as such, the time spent copying a genome is not going to drive you to having a smaller genome. The bioenergetic cost of copying DNA is probably not huge compared to all of the other things going on in the cell; eukaryotic cells waste large amountsof energy to ensure regulation / high fidelity of processes, and simple biosynthetic costs may be relatively minor compared to these (just a guess).

Q: How do gene families disperse? Can you say that unequal crossover also disperses them, besides creating them?
A: It probably has more to do with random chromosomal breaking and repair effects than unequal crossover. Chromosomes can often (in an evolutionary sense) get broken and then fused with other chromosomes or re-fused with the original chromosome in a different orientation / position, leading to the movement of huge blocks of genes. One sees this if one compares the organization of genes on the human and the mouse X chromosomes...

Q: I have a question about the dynamics and overall purpose of chromosome walking. I have read the method outlines in both of the books on reserve, but it still is unclear to me from where the cloned or subcloned fragments are comming and how they are extended accross the genome---are there multiple probes being used or simply one that is repeated throughout. In general, how is it accomplished and can it serve as tool for determining more than gene struture?
A: The cloned fragments are intially generated by fragmenting genomic DNA and are then ligated into an appropriate vector. For chromosome walking people typically use cosmids or YACs because of their large insert sizes. Once one has isolated a clone near a known marker, the ends of the insert in the corresponding clone are used to direct synthesis of a new probe. This new probe can be used to isolate other clones in the library that overlap with the first clone -- effectively walking out from the known marker, hopefully towards the unknown gene.

Q: Can you please explain the logic behind chromosome walking. In what experimental cases would it be better to use chromosome walking than normal sequencing?
A: Chromosome walking allows you to find an ordered set of fragments of DNA that are physically near a known marker. Sequencing provides the order of nucleotides within a defined segment of DNA. As such the two techniques are fundamentally different but may overlap. For instance, (1) one often sequences the fragments one obtains from chromosome walking to identify genes. (2) one can sequence an entire chromosome (a laborius, multi-lab, multi-year task) and obviate the requirement for chromosome walking.

Q: It says in the notes that 40% of all transcribed regions are non-essential. Where did this figure come from? Is it the 14% which cause impeded growth, but are not lethal, and the 20% of mutations which are in transcribed regions?
A: Considering the genome as a whole to be 100%: - 50% of the genome is transcribed (is genes) - 30% of random insertions have a phenotype, thus 70% of the genome is non-essential - Of this 70%, 50% lies in the region not transcribed, leaving 20% of the genome corresponding to transcribed, non-essential genes. Therefore, 20% of the genome is non-essential= 40% of genes are not essential 50% of the genome corresponds to genes

Chromosomes and chromatin

Q: Is it true that under some conditions, Facultative Heterochromatin can become Euchromatin which is potentially transcribed?
A: The Lewin definition of facultative heterochromatin is "the inert state of sequences that also exist in active copies." The basic idea is that some sequences/chromosomes are packaged (e.g. X chromosomes in females) as both heterochromatin and euchromatin (although not simultaneously or reversibly). Heterochromatin is an essentially permanent state. Euchromatin can undergo varying degrees of condensation such that actively transcribed genes are losely packaged while genes that are inducible but which are not actively transcribed are more densely packed. In answer to your direct question, no, facultative heterochromatin cannot be reversibly unpackaged, converted to euchromatin and transcribed. p.s. This is almost completely true. As it turns out, however, in certain tissues (e.g. microvilli in the intestine) the inactive X is reactivated; and certain genes on the inactive X in all tissues (e.g. XIST) are transcribed. How do you like that waffling?

Q: Why are there no polyA tails found on histone mRNA?
A: Duh, I don't know if there's a good reason for it. Plant histones BTW do have poly-A tails so while it's a common feature, it's not universal. As shown in Lewin figure 29.15, the ends of histone messages are generated by a relatively well understood process. One might imagine that histones are a very well expressed protein so you need a lot of message and that this special mechanism of termination / end-processing somehow allows an additional level of gene control...

Q: Does the nuclesome relax the DNA 0.8 of a turn around the nuclesome, and by that you can say that the nucleosome absorbs the supercoil?
A: The basic point is that roughly one of the two superhelical turns that one would predict to be generated by winding the DNA roughly twice around the nucleosome is absorbed on the nucleosome itself by partially unwinding the DNA. In the absense of a nucleosome, the DNA would prefer to be tightly wound (10.5 bp/ helical turn) and superhelical turns appear as kinking of the DNA and a plasmid winds up like a rubber band. On the nucleosome, though, the DNA is stable in a partially unwound state (10 bp/ helical turn) and as such, one of the two superhelical turns is not free to lead to kinking / rubber-banding the DNA.

Q: On page 162 of the reader you explain Nucleosome phasing. I don't understand the test of combining digestion with micrococcal nuclease and restriction enzymes. I understand why you would use micrococcal nuclease, but I don't see how EcoRI is showing specificity as nucleosomes are made. What does the digest tell you?
A:If you consider the fragments that result from digestion with both EcoRI and micrococcal nuclease, one end (digested with MNase) corresponds to the end of the DNA that is wrapped around the nucleosome and the other end (digested with EcoRI) corresponds to a specific site in the sequence (the EcoRI site). If nucleosomes are binding at random to the DNA (i.e. the fragment protected from MNase digestion is different in every different molecule of DNA), we expect there to be no relationship between the position of the two ends (the MNase-generated end and the EcoRI-generated end) and we expect to see a ladder of bands. If the nucleosomes are binding at a specific site on the DNA, the fragment of DNA protected >from MNase will be the same in every different molecule of nucleosome-wrapped DNA, and the two ends generated by the two different enzymes will always be a defined distance apart, giving rise to a limited number of bands (exactly 2: corresponding to the two halves generated by the EcoRI cut of the core particle DNA). The following schematic may help.

Q: The main function of the nucleosome seems to be in efficiently packaging DNA, though I see no reason why this nifty cylindrical wrap is necessary. The nucleosome also stabilizes DNA in a partially unwound (negatively supercoiled?) state- does this play a role in replication and transcription? In bacteria, DNA gyrases maintain the DNA in a negatively supercoiled state, which allows easier progress by the replication/transcription complexes. Is it possible that nucleosomes have a similar role?
A: As we have discussed, naked DNA is roughly 8,000-times longer than it is when it is packaged as a mitotic chromosome. Without this extent of packaging, it would be impossible to segregate chromosomes (too much viscous drag and intertangling of strands) or to fit the DNA into the nucleus. A nucleosome is a very local structure -- the DNA that gets packaged is contained entirely within 200 nts. This makes it possible to keep all of the DNA in the form of condensed nucleosomes and to then unwind only those sequences that one needs (for replication or transcription). As with prokaryotes, eukaryotes have an assortment of enzymes that maintain the overall superhelical tension of the genome at a relatively constant value. For both replication and transcription, negative supercoiling makes it easier to open up the DNA duplex and gain access to the individual template strands. The net effect of nucleosome removal is the release of one negative supercoil -- a supercoil that can be used to facilitate replication / transcription...

Q: As far as supercoiling and histones goes, can you please explain the nomenclature of -1 superhelical turns/ nucleosome. Should we assume that without histones you would have -2 superhelical turns if DNA wrapped around itself? And if so what conformational structure would that look like?
A: The DNA duplex is right-handed. One negative superhelical turn corresponds to underwinding this helix by one turn. To visualize this: imagine holding a long duplex of DNA -- holding the top of the helix still with one hand, rotate the bottom of the duplex one clockwise turn. The act of wrapping a bit of DNA twice around the nucleosome in a left-handed direction is topologically equivalent to introducing two negative superhelical turns. In reality, one of these turns is absorbed by the nucleosome by binding the DNA in an underwound conformation (if the DNA on the nucleosome were as tightly wound as DNA in solution, the superhelical turns would be free in the linker DNA and cause the DNA to contort upon itself). As such, the net effect of introducing a nucleosome is thus to introduce one (-2 + 1 = -1) negative supercoil.

Q: It is my understanding that heterochromatin is not usually transcribed and the reason it is not is due to its structure. I am assuming that it is tightly packed therefore not accessible. How then is it replicated? Does heterochromatin not exist in dividing cells?
A: Heterochromatin is present in all cells -- both actively dividing cells and quiescent cells. While tightly packaged, the replication machinery is able to transiently unpack localized regions of the DNA in the course of DNA synthesis. So: (1) heterochromatin is replicated by transient unpacking, (2) heterochromatic states are maintained from mother to daughter cell, even though the replication machinery unpacks it in the course of replication.

Q: I was reading in an older book that in replicating DNA the new and old histones do not mix and that the histones remained attached. I thought that you said that histones come off and rebuild themselves soon after it is transcribed. Is there a difference between replication and transcription of DNA in re: to the removal of histone? I would think that the proteins involved in replication are big thus interfere with the histone.
A: The experiment that we discussed that addresses this question is described in Lewin in figure 28.25. As you remember, a wide range of densities of octamers are observed after switching from light nitrogen to heavy nitrogen, suggesting that the components of the octamers are mixing with each other. The reassembly we are looking at in this experiment is that that occurs as a function of DNA replication (not just transcription). While this experiment suggests that mixing can occur, it (1) does not indicate that at least some histone octamers remain as stable units, and (2) does not preclude partial dissassembly (for instance, if the H3-H4 tetramer and H2A-H2B dimers remain as stable units but mix with each other, we would expect a wide range of densities). Fundamentally, replication and transcription have the same requirements -- single stranded DNA must be made available for nucleic acid synthesis by a large molecular machine. As such, we do not necessarily expect any fundamental difference in the process of nucleosome reassembly (I am unaware of any experiments that would suggest it).

DNA replication and chromosomal elements

Q: I am very confused on the terminology of the book, when they say elongation does that mean the length growth? and when they say synthesis does that mean as you add one nucleotide at a time 5' to 3'.
A: Elongation consists of the individual steps to macromolecule synthesis that come after initiation (i.e. not the first nucleotide/amino acid) and before termination (not the last nucleotide/amino acid). In the process of both DNA synthesis and RNA synthesis, nucleotides are added to the 3'-hydroxyl of the 3'-terminal nucleotide of the growing chain. As such, polynucleotides are synthesized in a net 5'-to-3' direction.

Q: I thought the lagging strand (okazaki fragments) synthesized with primosome and the leading strand synthesizes opposite of it. however the book says the reverse, i guess i am just completely confused? but i've tried drawing it out and still i come up with the same result, maybe you can shed some light and demistify the darkness in which i lay.?.?.?
A: The process is confusing and the figure in Lewin is not especially helpful. If you have access to the Lodish Cell Biology book, you might consult figure 12-11 in which they model the replication fork being broken open by the replisome. The primosome is off to one side (associating intermittently with the lagging strand) to make the short RNA primers, which are then brought through the replisome in essentially the same orientation as the leading strand.

Q: you mentioned some experiments to identify MCM and CDC in class, I did not quite follow. Could you explain those experiments again?
A: The basic idea is to: (1) mutate yeast at random (e.g. treating them with DMS, a chemical mutagen). (2) look for mutants that tend to lose a marker chromosome (a non-essential chromosome that carries a reporter gene that generates a color signal). The mutants presumably have acquired a mutation in a protein that is involved in maintaining chromosomes (but could involve DNA replication, origin recognition, centromere function, telomere function). After repeated rounds of cell division and DNA replication, inefficiency in any of these processes cause the marker chromosome to be lost and the corresponding color signal to be lost. (3) isolate the mutants and see if they are able to maintain a marker chromosome that carries multiple copies of an ARS. The idea here is that if the mutant protein is involved in processes other than initiation of DNA replication, providing multiple copies of the ARS should have no effect -- the marker chromosome will be lost as readily as the first marker. On the other hand, consider what would happen if the mutant protein was involved in recognizing the ARS / ORC? If the protein that has been mutated is partially functional (presumably it is since it is not a lethal mutation), by providing multiple interaction sites for it, we should make it less likely that the marker chromosome will be lost. (If the first marker contains only 1 ARS, the ARS has to be recognized with 100% efficiency in order to be maintained during each S phase. If the second marker contains 5 ARS, each ARS need only be recognized with 20% efficiency to be maintained.) The mutants that lose the first marker but maintain the second are thus likely to be involved in origin recognition.

Q: Normal somatic cells can divide up to 50 times, at this point telomere length is around 0. Is it true that only up to this 50th division, somatic cells cannot not make telomerase? Is telomerase expressed constitutively in transformed cells?
A: Regardless of telomere length, unless transformed, somatic cells don't express telomerase -- once their telomeres get sufficiently short, the cells die. In many transformed cells, telomerase is constitutively expressed and telomeres are maintained.

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