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1. Figure 1

a. (1 pt) Referring to Figure 1, the authors state that "the labeled, replicated DNA was supercoiled, but the unreplicated DNA was not". The authors correlate this with "preferential assembly" of nucleosomes on the replicated DNA strands. Explain in detail how this figure supports this statement, referring to both the autoradiogram and the EtBr-stained agarose gel.

Gel a (the autoradiogram) shows very distinct bands at the bottom of the gel in the region marked "I" for supercoiled. These bands are very dark and visible when present. This gel should show only replicated DNA [because the DNA is visualized in this case via the incorporation during replication of radioactive dATP]. Gel b (the ethidium-stained gel), on the other hand, shows both the replicated DNA from gel a and unreplicated DNA, because all DNA present can be seen in the EtBr staining. Bands are seen in the "I" region on this gel as well, but they could all by due to the replicated DNA and not the unreplicated DNA. This is supporeted by the fact that there are only bands in this region in the lanes that have them present in gel a. Finally, were they to be due to the unreplicated DNA in addition to the replicated DNA, one would expect a very dark, wide band as seen in gel a where the presence of supercoiled DNA is high.

[Ethidium staining allows visualization of all of the SV40 DNA. Most of the ethidium-stained DNA is NOT fully supercoiled, as indicated by the fact that it migrates more slowly than fully supercoiled DNA. In contrast, most of the newly replicated DNA, which is radioactive due to incorporation of labeled dATP during synthesis, is fully supercoiled.]

b. (1 pt) What new information is demonstrated by the data in figure 1?

Figure 1 demonstrates the existence of a factor activated during DNA replication that is necessary to the supercoiling of replicated DNA. This factor is stable on ice, but becomes inactivated when incubated at 37°C.

2. Figure 2

a. (.5 pt) Why did the authors allow the replication reaction to go for 40 minutes as opposed to a shorter time?

To allow time for complete replication of a detectable number of molecules of SV40 DNA. [SV40 is a circular molecule. If replication is incomplete, the replication intermediate consists of a partial template circle linked to two newly-forming circles. This complex structure migrates as a large, slowly moving, smeared band and is not useful in these assays. The large smeared regions that migrate more slowly than nicked circular DNA in these figures are due to incompletely replicated SV40 molecules.]

b. (.5 pt) The authors state that "the dependence of chromatin assembly on DNA replication could be due to a cis-acting mark left on the replicated DNA". What do you suppose they mean by "cis-acting mark"? What kinds of modifications can you think of that might constitute such a mark?

The effects of a "cis-acting mark" are localized to the area of DNA near the "mark." The mark could be a structural modification of the DNA, a protein bound to DNA, or both.

c. (0.5 pt) In testing for the possibility of cis-acting marks on DNA, explain in detail what the authors hoped to accomplish by including a spin column chromatography step and adding fresh S100 extract to the supercoiling reaction.

[The authors had observed that DNA replication activates a signal that causes preferential supercoiling of newly replicated DNA. They wanted to confirm their hypothesis that at least part of this signal was cis-acting. If the putative cis-acting mark is stable, then replicated DNA that is purified through a spin column should retain that mark, and thus retain the ability to be preferentially supercoiled upon re-addition of S100 extract and CAF-1. In other words, this experiment tests the hypothesis that what is changed or activated by DNA replication is a cis-acting mark, and not a trans-acting factor.

Also, if the putative cis-acting mark exists and is stable enough to be purified with the DNA in a spin column, it will then be possible to isolate any trans-acting factors that are also required. This is what the authors proceed to do next.]

d. (.5 pt) The authors state that "the preferential supercoiling observed in Fig. 2 was due to formation of an array of nucleosomes...". Explain the assay they used to demonstrate this point. How and why do the data they obtained (as described in the text of their paper) support this idea?

In order to show that the preferential supercoiling in fig. 2 was due to formation of an array of nucleosomes, the authors digested the products of the experiment with micrococcal nuclease. The resulting bands were 140-160 bp [or multiples thereof]. These data support their hypothesis because the structure of nucleosomes are such that around 146 bp segments are wrapped up in the histone complexes and separated and attached by linker DNA. The nuclease would cut at these easily accessible linker DNA sites, creating the 140-160 bp fragments observed.

3. Figure 3

a. (1 pt) Compare lanes 18-21 to lanes 25-28. What is different about these two sets of lanes, and what do these data suggest?

[In both of these sets of lanes, replicated (and spin-column purified) DNA was preincubated with ATP and increasing amounts of purified RFC for 25 min. After the preincubation, a supercoiling reaction was initiated by the addition of CAF-1 and S100. In lanes 18-21, the preincubation was at 4°C, while in lanes 25-28 the RFC preincubation was at 37°C. These data suggest that the cis-acting mark present on the replicated DNA is stable at 4°C but unstable at 37°C.]

b. (.5 pt) The authors added a large excess of CAF-1 over RFC to exclude the "the possibility that RFC might be a competitive inhibitor of CAF-1 or that RFC might block CAF-1 activity by direct binding". Choose one of these possibilities and explain how addition of excess CAF-1 led them to exclude that possibility.

If RFC were a competitive inhibitor of CAF-1, an increase in RFC concentration would produce an increasingly negative effect on chromatin formation [as was observed], as the RFC competed with CAF-1 for the same binding sites. In the same way, a large increase in CAF-1 concentration would be expected to nullify this inhibitory effect. However, inhibition of supercoiling by RFC was independent of CAF-1 concentration, indicating that RFC and CAF-1 are not competitive.

c. (1 pt) The authors state that "RFC...did not increase nicking of DNA (compare lanes 27 and 28 with 22, 23, or 24)." How do they know this? In other words, what would they have expected to see if RFC had nicking activity?

All lanes contain at least a small amount of nicked or relaxed circular DNA, indicated by the II/I0 symbol. If RFC increased nicking of DNA, then more nicked DNA should be visible in the gel. The bands next to the II/I0 symbol would be thicker than those in lanes with less (or no) RFC: 22, 23, and 24. The II/I0 bands in lanes 27 and 28 are not thicker than the II/I0 bands in lanes 22, 23, and 24. In fact, they look thinner.

[Two points are particularly important here. First, nicked DNA has a single-strand break. It is not smaller than unnicked DNA--nicking does not mean chopping up into smaller fragments. Second, nicked DNA cannot be supercoiled. As the DNA double helix is over- or under-wound, the nicked region can spin around to compensate. Supercoils are thus never induced.]

4. Figure 4

a. (.5 pt) Why is supercoiling not seen in Figure 4? (It's not because they ran the supercoiled bands off the gel!)

The DNA wasn't treated with CAF-1 or cell extract to promote supercoiling.

b. (1 pt) What data support the authors' contention that PCNA was being removed by RFC in this experiment?

The DNA was collected by immunoprecipitation with antibodies directed against PCNA. When the DNA was first allowed to react with RFC (at 37°C with ATP) and then subjected to immunoprecipitation, less DNA was collected. These low levels of isolated DNA suggest that the DNA lacked PCNA after treatment with RFC. This suggests that RFC either has the ability to mask the PCNA or to remove it.

5. (1 pt) Of the two experiments represented in Figures 5 and 6, which one demonstrated that PCNA is necessary for assembly of nucleosomes? Explain how.

In figure 5, chromatin assembly reactions were performed in the presence or absence of anti-PCNA antibodies. In the absence of anti-PCNA antibodies, nucleosome arrays formed (see lanes 2, and 5-8). In the presence of anti-PCNA antibodies, nucleosome arrays did not form, suggesting that the "PCNA antibodies inhibited CAF-1-mediated chromatin assembly by masking PCNA that was associated with the replicated DNA." The results of this are seen in lanes 3 and 4, where nuclease digestion appears to have run amok in the absence of protective nucleosome arrays.

6. (1 pt) Based on the data, the authors propose that PCNA serves as the DNA "imprint" that attracts CAF-1 to newly-replicated DNA strands, thereby facilitating replication-coupled chromatin assembly. Figure 8 illustrates the fact that there should be more PCNA on the lagging vs leading strands, implying that more nucleosomes could be loaded onto lagging vs leading strands. They speculate that asymmetric assembly of chromatin on the two sister chromatids might be involved in cell specialization during development. Clearly, though, such asymmetry would not always be desirable. What molecule discussed in this paper might regulate this effect? Explain.

RFC is known to load and unload PCNA from DNA. It is possible that after replication, RFC unloads "extra" PCNA from the lagging strands. I supppose this could help to make the chromatin assembly symmetrical.

7. (1 pt extra credit) Draw a double-stranded chromosome with three origins of replication spaced along it, both before and after replication. Indicate the 5' and 3' ends of each strand. Now draw in the distribution of PCNA that you might expect to see on each replicated chromosome, assuming this distribution has not been altered in the regulated fashion described above.

The key point here is that each new double-stranded molecule represents alternating leading and lagging strand regions. It is not the case that one new double-stranded molecule is the "leading" strand and the other is the "lagging" strand. (If you're not convinced, try drawing it out. If you're still not convinced, come and see me.)

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