Polymerase e may be used for lagging strand synthesis in vivo. Polymerase a has been implicated in primer formation. Answer 6. Each bidirectional origin generates two replication forks which move 2, bp per min. For the two forks per origin, this is 1. A unidirectional mode of replication would show a monophasic gradient of label, highest at the terminus and lowest at the origin, and decreasing continously around the circle between these two sites.
If the bidirectional origins were in fragments E and H, these fragments would be labeled last in the pulse-labeleing experiment. Assuming equal elongation rates for all four replication forks, the termini would be half-way between E and H on both halves of the SV40 molecule, i.
These would label first in the pulse-labeleing experiment, and fragments between the termini and origin would have progressivly less label. Fragment 3, with a bubble arc, has an origin, and fragment 5, with a double-Y arc, has a terminus, as diagrammed below. The other fragments have Y arcs, indicative of replication forks moving through them. Fork movement in fragment 4 is from left to right, moving from an origin to a terminus. Fork movement in fragment 6 is from right to left, moving into a terminus from an origin not on the map.
If you knew that these were bidirectional origins, then one could conclude that fork movement in fragments 1 and 2 are from right to left. For example, if the linear-equivalent length at the transition point is base pairs, and the unit length were bp, then the two forks traversed bp before one reached the restriction site. The DNA has melted locally and the strands separated in the open complex.
DNA in the open complex will show the opposite effect, i. Gyrase is a member of the Topoisomerase II family, which uses the energy of ATP to introduce negative superhelical turns, changing the linking number in steps of 2 recall this from the section on supercoiling in the chapter on DNA structure. Thus one cycle of the reaction catalyzed by gyrase will counteract the unwinding of 20 bp. One cycle of gyrase action will be needed, using one ATP. We discussed a model for lagging strand synthesis in which the b subunit sliding clamp dissociates from Pol III core when it encounters the 5' end of an Okazaki fragment.
If we apply that to this situation, each leading strand polymerase would stop synthesis and dissociate as soon as it encounters the 5' end of an Okazaki fragment; for a circular chromosome this would be the last Okazaki fragment synthesized by the fork moving in the opposite direction. Action by a 5' to 3' exonuclease and polymerase e.
Telomerase catalyzes the synthesis of one hexanucleotide repeating unit GGGGTT in the case of Tetrahymena and then shifts over to synthesize another repeating unit. If the enzyme dissociates from one telomere after each repeating unit, then its processivity is very low, i.
If it shifts over on the same telomere, then its processivity is higher. Note that the template RNA has at least two copies of the complement of the telomere repeating unit, so that there is still some overlap with the extending DNA strand when the enzyme shifts over to make a new repeating unit.
There is an average of six replication forks per chromosome. New replication must initiate every 20 min to sustain this rate of growth. If you picture a replicating DNA molecule that will complete synthesis in 20 min, it is already half-replicated, and each of the nascent daughter molecules has also initiated synthesis, for a total of 3 origins fired, and 6 replication forks for bidirectional replication.
In the molecule illustrated above, the two older replication forks will meet and terminate in 20 min. This will leave 2 molecules in the cell until the cell divides 20 min later. However, replication re-initiates every 20 min as well, so each molecule will still have 6 replication forks. Note that the pattern for Q is a "bubble arc" and that for P is a "Y arc".
The leading strand extends away from the origin, and in the case of the bi-directional replication, the leading strands will extend divergently from the origin. Since only the leading strand is being synthesized and labeled, the hybridization pattern indicates that the bottom strand is made continuously beginning with fragment K hybridizes to the top strand , and the top strand is made continuously beginning with fragment L hybridizes to bottom strand. Thus a bi-directional origin must exist around the junction between K and L.
You cannot map unidirectional origins by this technique - can you see why? The replication fork moves from right to left through fragments A through K. The top, or nontemplate, strand hybridizes, which tells you that the leading strand is the bottom strand, whose 5' to 3' orientation is right to left. The replication fork moves from left to right through fragments L and M. The bottom, or template, strand hybridizes, which tells you that the leading strand is the top strand, whose 5' to 3' orientation is left to right.
Any enzyme that is specifically involved in lagging strand synthesis is a candidate, e. Perhaps ligase or DNA polymerase I "homolog" could also be considered. In fact, emetine is an inhibitor of protein synthesis.
The fact that it also blocks lagging strand synthesis indicates that some component of the machinery that synthesizes the lagging strand requires constant protein synthesis, suggesting that some component is very unstable. M13 vectors. Placing the restriction fragment in one orientation will produce one strand in the viral progeny, whereas placement in the other orientation will produce the other strand. Another good choice and the one used in this paper is a vector like pBluescript, which has promoters for RNA polymerases from bacteriophage T3 and T7 on the two sides of the insert, so transcription from one promoter generates an RNA with the "top strand" sequence, and transcription from the other promoter generates the "bottom strand" sequence.
The simple Y for fragment A is expected, since the replication fork should be elongating through this fragment. The fact that fragment C is also a simple Y-arc tells you that the origin is not in C, but the fact that it labels last in the pulse-labeling experiment tells you that it is very close to the origin.
Thus one can explain the rather complicated pattern in the 2-D gels for fragment D. These can also generate Y-arcs, but there should be a lot of molecules that are being replicated both by new-initiated and previously initiated forks.
Thus the double-Y pattern the straight line on 2-D gels will be generated. Only the outer strands will hybridize to the leading strands, and no information will be gleaned about the origin or terminus. This assay only gives information when there is a transition of leading strand synthesis from one strand to the other, such as at a bi-directional origin.
Since the cells are dividing every 20 min, replication initiates every 20 min. The time to replicate the chromosome is 30 min. Consider initiation event n. Thus the cells cycle between 6 forks and 2 forks, giving an average of 4 forks per chromosome. This is calculated by adding all the time in the cell cycle after replication and dividing by the total time of the cell cycle 24 hr in this case, i.
Both will show as 2 doublets 4 dots for all of G2 4 hr , and for simplicity in arithmetic, we are assuming they will be 4 dots for all of M 1 hr. The fraction of cells will be same as the fraction of the cell cycle occupied, for an asynchronous population. Consideration of the mitotic cells is an interesting complication.
For much of mitosis, the nuclear envelope is disassembled, so technically there is no nucleus. Of course one will still get hybridization to the condensed mitotic chromosomes. Answer 7. We will use arbitrary colors to help in following the fates of the incorporated d C TP and the A in the template strand.
An A :T base pair is changed to an A : C in the initial product of replication. Upon another round of replication, an A :T will be at this position in one daughter molecule the wildtype and a G: C mutation will be at this position in the other daughter molecule.
A T :A base pair is changed to an T : C in the initial product of replication. Upon another round of replication, a T :A will be at this position in one daughter molecule the wildtype and a G: C mutation will be at this position in the other daughter molecule.
First draw the base paired structure with the nucleoside deoxyguanidine in the enol tautomer and the nucleoside thymidine in the keto tautomer. The oxygen attached to position 6 of guanine O 6 is a hydroxyl in the enol tautomer, and the nitrogen at position 1 N1 is fully bonded to the carbons on either side, so it has no hydrogen. Thus the O 6 hydroxyl is a hydrogen bond donor and the N1 imine is a hydrogen bond acceptor. The amino group bonded to position 2 is unchanged by the tautomerization, and it continues to serve as a hydrogen bond donor.
The two keto groups on keto thymidine are hydrogen bond acceptors, one from the O 6 hydroxyl of enol guanidine and one from the N3 amino group of enol guanidine. The N3 amino group of thymidine is a hydrogen bond donor to the N1 imino group of enol guanidine.
The distance between phosphodiester backbones of the complementary strands of B form DNA is sufficient to accommodate a purine base and a pyrimidine base in the anti conformations.
The purine base is larger than the pyrimidine base, and two of them in the anti conformation cannot be accomodated without changing the distance between phosphodiester backbones. By swinging the purine base back over the deoxyribose i. Since pyrimidine bases are smaller than purine bases, two pyrimidine bases can fit between the two phosphodiester backbones without a shift from anti to syn.
The nitrogen at position 1 is in the amino form for both, i. Thus the CpG dinucleotides would be replaced by TpG. Note that if you are looking only at the sequence of one strand of the DNA, a former CpG can become either TpG if the C on the strand you are considering is methylated or CpA if the C on the complementary strand is methylated. This has been observed after the inactivation of a pseudogene for alpha-globin and as repetitive elements such as Alu repeats in humans evolve after transposition.
When adjacent pyrimidines are covalently linked by the cyclobutane or the bond between the bases, the bases are not able to make the 36 o rotation, resulting in a change in the helical structure. An exonuclease requires a free end on linear DNA to cut, whereas an endonuclease cuts within a DNA molecule and hence can use circular DNA as a substrate, whereas an exonuclease cannot. The excinuclease is an excision nuclease used to cut out a segment of single stranded DNA.
It is an type of endonuclease, but it makes two nicks i. Thus a helicase can unwind the DNA between the nicks and remove the damged segment. The glycosylases are specific for particular kinds of damage,. All the glycosylases leave sugar without a base, i. The mismatch recognition MutS and activation of endonuclease MutL functions are conserved from bacteria to mammals.
However, the enzyme that recognizes the sequence distinctive for newly synthesized DNA MutH endonuclease is not conserved.
One possibility is the methylation of CpG dinucleotides. When translesion synthesis occurs, mutations are generated opposite lesions, so defects in translesion synthesis will reduce the number of mutations i. During replication, a T will incorporate opposite the A to replace the original C. This answer is restricted to the strand as written, but C's on the complementary strand are also subject to deamination.
If the methylated strand were cleaved and degraded, the information in the parental strand would be lost. That leads to a GC original base pair to AT mutant base pair transition. If this altered base is replicated, you get a CG original base pair to TA mutant base pair transition. Other pyrimidine dinucleotides can also form dimers, such as the CT at positions 5 and 6 top strand and any of the several CC dinucleotides.
AP endonuclease will then nick on the 5' side of the AP site, and DNA polymerase can fill in the sequence directed by the opposite strand as the template, thereby repairing the damage. The U will be removed by uracil-N-glycosylase, to leave an apyrimidinic site. In order to show the cleavage of the N-glycosidic bond more clearly, a layer of has been added to denote the sugar-phosphate backbone. The vertical lines are N-glycosidic bonds between the bases and the deoxyribose.
Now the AP endonuclease sees the hole in the helix and cleaves the phosphodiester bond just to the 5' side, leaving a nick with a 3'-OH and a 5' phosphate:. DNA polymerase I can "nick translate" through the AP site and beyond, adding nucleotides as directed by the template top strand indicated by the underlining. Polymerase could fill in the resultant gap with the correct sequence, followed by sealing with ligase.
To illustrate that, lets add another 10 bp to the left of the sequence 3' to the damage on the bottom strand. UvrD catalyzes the breaking of the base pairs to "lift out" the damaged segment.
Thus in this situation, the top strand is the parental presumably correct strand, since it has the methyl on the A in the GATC. The bottom strand presumably incorporated an A erroneously at position The mismatch will be excised in a patch that starts at the GATC and extends past the mismatch.
DNA polymerase plus ligase will restore the wild-type sequence. New DNA is underlined. Note that if the A is not corrected, it will direct the incorporation of the tumorigenic T in the next round of replication. The following are not supported by the data:. FA is a disease resulting from deficiencies in mismatch repair.
However, direct experimental tests not presented here show no change in ligase activity compared to wildtype cells. Specific sites on plasmid DNAs can be mutated by denaturing the plasmid and annealing with an oligonucleotide that has the desired nucleotide substitution. If this heteroduplex plasmid with the plus strand parental and the minus strand newly synthesized and containing the mutation is transformed into E. Increasing the frequency of plasmids containing the mutated sequence is desirable, and a mutant strain defective in dut and ung can be used to decrease the frequency of plasmids with the parental sequence.
Replication of the plasmid is more efficient than repair of the AP sites, thus the replicative polymerase will preferentially use the mutated minus strand as the template, thereby increasing the frequency of the mutated plasmid. This strategy was developed by T. Kunkel Rapid and efficient site-specific mutagenesis without phenotypic selection. Answer 8. The chromosomes would simply be assorted with two to each haploid cell.
Since there is only one of each chromosome in the diploid cell, each chromosome cannot assort with itself. This leaves 12 combinations of two chromosomes, i. Of these 12 combinations, 8 of them have both an A-type and a B-type chromosome.
The result is that every replication fork has a twin replication fork, moving in the opposite direction from that same internal location to the strand's opposite end. It works by coating the unwound strands with rigid subunits of SSB that keep the strands from snapping back together in a helix.
The SSB subunits coat the single-strands of DNA in a way as not to cover the bases, allowing the DNA to remain available for base-pairing with the newly synthesized daughter strands.
As you can see in , when the two parent strands of DNA are separated to begin replication, one strand is oriented in the 5' to 3' direction while the other strand is oriented in the 3' to 5' direction.
DNA replication, however, is inflexible: the enzyme that carries out the replication, DNA polymerase, only functions in the 5' to 3' direction. This characteristic of DNA polymerase means that the daughter strands synthesize through different methods, one adding nucleotides one by one in the direction of the replication fork, the other able to add nucleotides only in chunks.
The first strand, which replicates nucleotides one by one is called the leading strand; the other strand, which replicates in chunks, is called the lagging strand. Since DNA replication moves along the parent strand in the 5' to 3' direction, replication can occur very easily on the leading strand. As seen in , the nucleotides are added in the 5' to 3' direction.
Triggered by RNA primase, which adds the first nucleotide to the nascent chain, the DNA polymerase simply sits near the replication fork, moving as the fork does, adding nucleotides one after the other, preserving the proper anti-parallel orientation.
This sort of replication, since it involves one nucleotide being placed right after another in a series, is called continuous. Whereas the DNA polymerase on the leading strand can simply follow the replication fork, because DNA polymerase must move in the 5' to 3' direction, on the lagging strand the enzyme must move away from the fork.
But if the enzyme moves away from the fork, and the fork is uncovering new DNA that needs to be replicated, then how can the lagging strand be replicated at all? The problem posed by this question is answered through an ingenious method. The lagging strand replicates in small segments, called Okazaki fragments.
These fragments are stretches of to nucleotides in humans to in bacteria that are synthesized in the 5' to 3' direction away from the replication fork. Yet while each individual segment is replicated away from the replication fork, each subsequent Okazaki fragment is replicated more closely to the receding replication fork than the fragment before.
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