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Learning objectives for Wed 11/15/00

1) describe the purpose of replication origins and the features they share in bacteria, yeast, and higher eukaryotic cells

Replication origins provide for an efficient method of DNA replication and serve as a universally recognized site (within a species) for initiation of DNA replication. Features in replication origins shared between bacteria, yeast, and higher eukaryotes include the following:

• the replication origins all are unique, repeating sequences of DNA
• These DNA segments are recognized by multimeric proteins that bind to the origin
• Origin regions are usually AT-rich because AT base pairing required less energy to melt and therefore allows the DNA in this area to be more easily unwound

2) state the three replication problems encountered by DNA polymerases and describe the solutions to these problems

Problem #1: DNA polymerases work to replicate DNA only of a single-stranded template, meaning the DNA must be melted, a process of which DNA polymerases are not capable

Solution: Proteins called helicases such as DnaB in E. coli melt double-stranded DNA [ahead of the relication fork]. DnaB DNA melting is aided by DnaA [which separates the two strands at the origin], and DnaB attaches in a circle around each strand of template DNA.

Problem #2: DNA polymerases can only extend DNA of primers, not initiate DNA chains.

Solution: The primers are made by primase, which makes short RNA strands

Problem #3: DNA polymerases only work 5' to 3', but DNA is [antiparallel. So as the replication fork proceeds, one template strand is revealed in the 3' to 5' direction and can be copied directly, but the other template strand is "backward," or 5' to 3'.]

Solution: Replication bubbles grow in both directions, and each strand is replicated as the bubble grows. As each growing fork consists of DNA of opposite polarities, there is a leading strand and a lagging strand. The leading template strand has [a 3' to 5' polarity when read in the direction of the growing fork], so replication occurs continuously from a single primer. The lagging strand template has a [5' to 3' polarity in the direction of the growing fork], so replication must occur in a 3' to 5' direction overall, so small amounts of DNA are replicated at a time with multiple primers. Each set of primer and section of replicated DNA is called an Okazaki fragment. The RNA primer is then excised [by DNA polymerase, one nucleotide at a time] and replaced with DNA.

3) compare and contrast the coordinated synthesis of leading and lagging strands in prokaryotes and eukaryotes. Identify and describe the roles of the proteins at the growing replication fork.

In prokaryotes, the leading strand is synthesized continuously by a complex of DNA polymerase III and beta clamp. On the lagging strand, a helicase moves continuously in the direction of fork growth, associated with primase. The primase creates short RNA primers for DNA polymerase III, which binds to the DNA in association with a beta clamp. Pol III enlongates the primer in a 5' to 3' direction, which is opposite the direction of the growing fork. Once the polymerase meets the previous RNA primer, [Pol III dissociates and] DNA Pol I binds to the strand and uses its exonuclease activity to remove the RNA and [its polymerase activity to] replace it with DNA. DNA ligase then stitches the fragments together. Each of the separately synthesized DNA fragments are called Okazaki fragments. The core proteins of Pol III on the lagging and leading strands are connected to one another by a protein dimer called tau, which ensures that both strands are syntehsized concurrently.

DNA replication in eukaryotes is carried out in a similar process. Synthesis of the leading strand is initially carried out by polymerase alpha, bound to primase and RFC. This complex synthesizes the 5' end of the leading strand. Following this, a protein called PCNA binds to RFC and displaces pol alpha and primase. DNA polyemrase delta then binds to PCNA and the leading strands are synthesized continuously. The lagging strand is synthesized discontinuously by a complex of pol alpha, primase, and RFC.

Learning objectives for Fri 11/17/00

1) Define telomere and state why it is important. Describe why telomerase is required to maintain telomere ends and the mechanism by which it achieves this.

Telomeres are repetitive oligomeric sequences that allow for the end of the lagging strand to be synthesized. Since the lagging strand requires the addition of a primer upstream and then is filled in by polymerase, it would be impossible for the polymerase to complete the very end of the lagging strand unless some extra material was placed upstream of the end of the DNA strand. This task is accomplished by telomerase, which can elongate the lagging strand from its 3' hydroxyl end and creates enough extra material for a primer to be placed on, from which DNA polymerase will finish the lagging strand.

NOTE that telomerase does not "fill in the gap" by extending the 5' end of the new daughter strand. Instead, it actually extends the gap by lengthening the (already overhanging) 3' template strand. This allows for renewed lagging strand synthesis of the daughter strand, but never resolves the 3' overhang. (Draw this out for yourself and see what I mean!)

2) Define supercoiling, both negative and positive, and explain the advantages and disadvantages of supercoiling. Explain why and how the type I and type II topoisomerases handle supercoils in DNA.

Supercoiling is a DNA superstructure that occurs when duplex DNA twists around itself. There are two major kinds of supercoils: positive and negative. In negative supercoils the "front" strand of DNA (when the DNA is viewed standing on its end) falls from right to left [i.e. is right-handed--the same handedness as the DNA double helix itself]. In a positive supercoil the front strand falls left to right. The advantage of supercoiling is that it relieves stress caused by unwinding of the DNA during transcription or replication. In some organisms, supercoils may also perform various structural functions, including gene expression control. [Supercoils also condense DNA, allowing it to fit into a much smaller space.] The disadvantage of supercoils is that they form compact structures [that are relatively] inaccessible to different classes of polymerase. Therefore, something must regulate supercoils in order for protein expression and genomic replication to proceed.

Enzymes called topoisomerases can regulate supercoil formation in DNA. Type I topos relax supercoils by binding to duplex DNA and making a nick in one strand. A tyrosine residue in the active site of the enzyme binds to the free 5' phosphate on the nicked strand. The other side of the nicked strand is then passed under the other strand and the nick is resealed. This process requires no input of energy. Thus type I topos can quickly and efficiently remove supercoils.

NOTE: E. coli topo I can only remove negative supercoils, but eukaryotic topo I's can remove either positive or negative supercoils. Neither can induce supercoils. Both are ATP-independent.

Type II topo changes DNA topology by breaking and rejoining doubled-stranded DNA in an ATP-requiring process. By this mechanism, topo II can change a positive supercoil into a negative supercoil or increase the number of negative supercoils by two. [Topo II cannot create positive supercoils, or remove negative ones.] Top II breaks double-stranded DNA that has already been coiled and passes one end of the broken duplex around the intact duplex and reseals the strands.

SUMMARY OF TOPO I vs TOPO II: Topo I cuts one DNA strand, topo II makes a double-stranded cut. Topo I is ATP-independent and cannot induce supercoils, but only relieve them. Topo II requires ATP and can either create negative supercoils or remove positive ones. In E coli, topo I cannot remove positive supercoils ahead of the replication fork, so topo II (gyrase) is required for replication.

3) State the major forms of DNA damage and their source (exogenous vs endogenous processes). Explain the roles of polymerases and mismatch repair in correcting replication errors.

(see 11/20, #1)

Learning objectives for Mon 11/20/00

1) Describe the main types of DNA damage and their sources.

spontaneous/endogenous sources of DNA lesions

• mismatches: DNA replication is prone to errors every 100,000 to 1 million basepairs, and many times this is fixed by proofreading activities of the DNA polymerases themselves. [Mismatches that are not immediately fixed by DNA polymerase are repaired via mismatch repair.]
• DNA repair synthesis: [DNA repair frequently requires new DNA synthesis, which can be error-prone]
• Deamination of bases with exo-cyclic amino groups: repaired by mismatch repair and base excision repair
• Insertions or deletions of bases during recombination and/or replication: fixed by mismatch repair
• Incorporation of U during replication: fixed by base excision repair
• Loss of bases, either purines or pyrimidines: fixed by base excision repair
• Oxidative damage caused by radicals and peroxides [produced as byproducts of mitochondrial respiration] results in mispairing during replication (base excision repair) as well as chromosomal breaks (no good fix)

Induced/environmental sources of DNA damage

• altered bases result from radiation and alkylating chemicals: fixed by base excision repair
• pyrimidine dimers, caused by UV: fixed by nucleotide excision repair
• double-strand breaks, caused by radicals [and ionizing radiation] are difficult to fix: recombination or end-joining
• chemicals which intercalate between bases in the double helix: nucleotide excision repair

2) Identify what repair pathway would be used to correct each type of damage, and describe the basic mechanisms.

For mismatches that occur during the replication process, the primary repair mechanism is carried out by DNA polymerase. When an incorrect base is inserted into a growing strand, it causes the template/copy duplex to "melt." This triggers the DNA polymerase to flip the growing strand from a polymerase site on the enzyme to an exonuclease site. The exonuclease action removes a short stretch of nucleotides (including the mismatch) and then the strand is flipped back to the polymerase site and replication continues (the removed bases are replaced).

Mismatches in bases can also occur in replication and by the process of deamination. In these cases, another mechanism carries out mismatch repair. In this mechanism, short stretches of newly syntehsized DNA are first excised from duplex DNA. Of course, this necessitates that newly synthesized DNA be differentiable from the template strand. In E. coli, this is accomlished by differential methylation of DNA strands. On the template DNA, GATC sequences are methylated, while newly syntehsized NDA is not. The process is carried out by an enzyme that takes some time to work. Thus duplex DNA comprised of a tempate and a new strand is hemimethylated at GATC sequences. A protein called MutH binds to hemimethylated GATC sequences. Another protein, MutS, recognizes and binds to abnormally paired DNA segments. Then a protein called MutL binds to both proteins, connecting them. This triggers exonulcease activity of MutH, which cleaves the new DNA strand. Following this, the segment of DNA between MutS and MutH containing the mismatch is removed by helicase [and a single-strand exonuclease] and the resulting gap is repaired by DNA polymerase and ligase. Finally the GATC sequence adjacent to the repair is methylated.

A similar type of repair mechanism operates to fix damage caused by chemicals that intercalate between basepairs or bind to basepairs and change their shape. The same mechanism can also fix some damage caused by UV radiation (pyrimidine dimers) and mispairing produced in replication. This kind of damage locally distorts DNA topography, and can thus be recognized by certain enzymes. One form of this mechanism, nucleotide excision repair, acts via an ATPase complex that binds to and slides along duplex DNA. When the complex encounters a distortion in the DNA, it produces a kink in the duplex at the error site. At this point, part of the protein complex dissociates and another protein attaches. The bending of the DNA is thought to open space in the duplex for the catalytic site of the enzymes to work. The newly attached enzyme then cleaves the damaged strand and removes a small number of nucleotides in either direction from the damaged site. DNA polymerase and ligase then fill the resulting gap.

A similar mechanism, called base excision repair, works on the same principle. However, instead of removing nucleotides, an enzyme called DNA glycosylase recognizes damage and removes single altered bases. Following removal of the base, an exonuclease removes the corresponding sugar and phosphate. DNA polymerase and ligase then fill the gap. [but NOTE that the "gap" in this case is a single missing nucleotide, whereas an entire region is removed and replaced in NER]

One final method of repairing DNA damage is used when double-stranded breaks have been fromed in a DNA duplex. In this case, non-homologous end joining or homologous recombination may be used to repair the damage. See next week's objectives for descriptions of these mechanisms.

3) Describe the themes common to all forms of DNA repair.

General themes include recognition of the damages dite; excision of these nucleotides, and resynthesis with correct nucleotides. In mismatch repair the MutHLS system is responsible for recognizing the damaged strand and distinguishing it from the parent strand. Excision occurs when MutH's latent endonuclease activity is triggered by linkage of MutS to MutH via MutL. DNA Pol III and DNA ligase facilitate resynthesis and ligation of the new segment. In base excision repair DNA glycosylase recognizes chemically modified DNA and breaks bonds between the base and the sugar. AP endonuclease then excises the sugar and phosphate group, leaving a single-nucleotide gap which is filled with an undamaged nucleotide by DNA polymerase and ligase. In end-joining the proteins Ku and DNA-dependent protein kinase recognize and bind to the ends of double-strand breaks. After a region of microhomology anneals, an exonuclease [or an endonuclease, it's not clear yet] "excises" single strand ends and ligase reseals the free ends.

Learning objectives for Wed 11/22/00

Students should be able to…

1) Diagram and explain the mechanisms involved in: nonhomologous end-joining (fig 12-28), a simple Holliday model of recombination involving single-strand nicks (fig 12-29), and meiotic recombination via double-strand breaks (figure 12-31).

These models are well-explained in your text. See me with questions or concerns. NOTE that what you need to be able to do is to draw and explain these mechanisms on your own, without reference to your text. You also need to be able to answer questions about them. To meet these objectives, you need to understand the processes, not simply memorize them.

2) In reference to the above, use the following terms precisely and correctly: end resection, strand invasion, branch migration, exo vs. endonuclease cleavage, Holliday junction, resolution (of a Holliday junction), recombinant vs. nonrecombinant, heteroduplex.

• end resection: the trimming of DNA strand ends. It occurs through exonuclease activity during the process of end-joining of nonhomologous DNA's and during meiotic recombination.
• strand invasion: is when the 3' end of one strand invades an intact duplex region, displacing one strand from the duplex. It's facilitated by RecA (in E coli) or its homologs, and it occurs during meiotic recombination. The invading 3' end is then extended by DNA polymerase, using the invaded strand as a template.
• branch migration: occurs in the Holliday model of recombination when nicked strands on each chromosome cross and join with 5' ends of homologous strands. The branch point migrates in such a way as to create a ...
• heteroduplex region, which contains a strand from each of the parental DNA molecules, and thus is not perfectly basepaired.
• Holliday structure: is the crossed-strand structure formed during recombination in which one strand from each of a pair of homologs has been exchanged. The two homologs are connected via these crossing strands.
• Resolution of a Holliday structure: is achieved by the cleavage of the chromosomes either at crossover points or at other sites on the linear DNA strands. The juntion is resolved when no crossovers remain and all stands are ligated.
• Recombinant vs nonrecombinant: when the Holliday structure(s) has(ve) been resolved, the new chromosomes are recombinant if they comprise a fusion of one homologous chromosome onto the other. If the genetic material in each new chromosome is derived exclusively from one homolog (with the exception of a short heteroduplex region), they are nonrecombinant.
• Endonuclease activity is needed whenever DNA must be cut in the middle of the molecule, as in Holliday resolution, whereas an exonulcase would be utilized for cutting the ends of DNA, as with end resection.

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