Introduction
Anabaena, a filamentous cyanobacterium, is remarkable in that it has both the abilities to harness the sun's energy and to fix nitrogen, two mutually exclusive processes. The oxygen produced by photosynthesis normally poisons the process of nitrogen fixation by inactivating the protein that catalyzes the reaction. Anabaena, therefore has a mechanism by which it must separate the two reactions-nitrogen fixation only occurs in specialized cells called heterocysts, which are fairly evenly spaced along the filament. These cells, and thus the process of nitrogen fixation, only appear when the filament is starved of nitrogen, when there is none in the medium on which it is growing.

This is the part of Anabaena that is the most interesting. How does the filament of this primitive organism decide which cells should become heterocysts, or in other words, how does the pattern of differentiation come about? If it can be determined exactly how these organisms differentiate, it can eventually lead to how higher organisms, and eventually humans, differentiate from a ball of cells to a working, complex piece of machinery. There is a standard model of how these simple organisms differentiate that has been accepted for the past 25 years. It states that the first cells to start to differentiate are random (random initiation), and that these cells diffuse an inhibitor into the surrounding cells, so that they can't differentiate. This lateral inhibition is caused by a protein and results in heterocysts placed an average of about 10 cells apart.

This explanation was good enough to not be challenged (because of a lack of technique needed to closely investigate it) for quite a while, but there is a problem with it. It does not account for the observation that often clusters of cells will all begin to differentiate at approximately the same time-this defies the random initiation assumption. Also, if the lateral inhibitor is knocked out, all cells that begin to differentiate should eventually become heterocysts, and this was not observed. A new model has been proposed, similar to the original, except for the idea of nonrandom initiation. This model suggests that a cell's ability to become a heterocyst depends on that cell's position in the cell cycle. The lateral inhibition half of the standard model remains in place. The connection of differentiation to the cell cycle indicates that neighboring cells, which are the most closely related, are probably in the same or similar places in the cycle. This can account for the clumps of cells all beginning to differentiate at the same time.

The method of connection of the cell cycle and differentiation is of prime importance. Four methylases have been discovered within the Anabaena genome, dmnA, dmnB, dmnC, and dmnD. These genes encode enzymes that place methyl groups on newly formed strands of DNA. This could be the connection, as methylation occurs in a certain place in the cell cycle and could be the signal to differentiate. Both DmnA and DmnD recognize the sequence GATC in a newly minted strand of DNA, which raises some interesting questions. Are both of the enzymes required, and if so, why would the cell need two proteins that seem to perform the same function? It has already been proven that DmnA is necessary, even in the presence of DmnD. So the next logical question is whether or not DmnD is necessary in the presence of DmnA. I would like to be able to look into DmnD and its specific function and importance in the differentiating cell, as well as its relation to DmnA, to see if it plays any part in deciding which cells become heterocysts.

Methods
In order to accomplish the challenge of figuring out just what DmnD does and is, several experiments will have to be performed. I want to figure out its role in differentiation in Anabaena and whether or not it accomplishes the same things there as in E. coli, where it plays roles in mismatch repair and DNA replication. The first experiment to be performed to see what it might do is to create a dmnD- genome and insert it into Anabaena to see if the cyanobacterium can survive without the gene, and whether it can differentiate without it. A restriction map of the plasmid on which it appears must be constructed using a computer, so a restriction enzyme may be found that only cuts at a point within dmnD. The gene that encodes resistance to kanamycin is then isolated and inserted into the cut in dmnD, so that the kanamycin-resistance gene lies in the middle of dmnD. The dmnD gene with kan is then cut out of the plasmid and inserted into an E. coli plasmid, which confers sensitivity to sucrose through sacB, by cutting both with the same enzyme and mixing them together.

The resulting plasmid is then placed into Anabaena by conjugation with the E. coli in which it was formed. This is accomplished by introducing a broad host range plasmid into E. coli, making it able to conjugate with a wide variety of cells. Next, the mixture of cells is plated and the Anabaena with mutated dmnD in their genome are selected by their ability to grow without carbon (E. coli cannot) and in the presence of kanamycin (unmutated Anabaena cannot). The resulting cyanobacteria all have one copy of mutated dmnD, and most have a second, wild type copy as well. Single recombination is how most of the mutated genes got into Anabaena, and if this is the case, the rest of the plasmid, along with a second copy of dmnD, is also present. Within the rest of the plasmid is a copy of sacB, which confers sucrose sensitivity. The protein it encodes splits sucrose into its monosaccharide units inside the cell, resulting in a higher solute concentration inside the cell than out, which then swells and bursts the cell. A lot of cells must be grown for this part, since double recombination is very rare. The maintenance of the dmnD- is selected for by the continuing presence of kanamycin in the medium. Double recombination is then selected for, because any cells that have undergone the desired double recombination should have only the mutated copy of dmnD. Once a lot of cells are present, sucrose is added, so any single recombination cells lyse and die because they all have the whole plasmid containing sacB. All of the cells that are left have only one copy of dmnD, a mutated one, because in a double recombination event, the mutated copy actually replaces the wild type, rather than adding before or after it. So in effect, they are all dmnD-.

Once we have some Anabaena with a modified dmnD gene, the actual tests to see what dmnD does can begin. The first test will be whether or not Anabaena can still differentiate if it lacks DmnD. To accomplish this, Anabaena will be grown on a nitrogen-rich medium, and the nitrogen will then be washed out of the medium to see if it will still develop heterocysts. DmnD has a sequence comparable to methylases, so the next test will be for whether or not DNA gets methylated without dmnD there, especially since it performs the same function as DmnA. The question is whether both are needed in order to methylate the sequence GATC. To test for methylation, the dmnD- cyanobacteria will be split and the DNA isolated for cutting with DpnI and DpnII. DpnI requires methylation of the sequnce to cut (GMeATC), whereas DpnII cannot cut when the same sequence is methylated. If DpnI succeeds in cutting the DNA, then DmnD has the same function in Anabaena as in E. coli. The final test is for whether or not the protein produced by dmnD functions in mismatch repair of DNA. In a normal group of Anabaena cells, mutations occur very rarely. To test the mismatch repair, the approximate frequency of normal cells that are usually mutated has to be obtained and compared to the frequency of dmnD- cells mutated.

Results
Ideally, when I select for the double recombinants, there will be some there to use. They should be fairly rare, as double recombination in a particular spot is a rather difficult even to orchestrate. However, there is the possibility that dmnD is a vital gene, and that mutations to it do not allow Anabaena to survive. They phenotype would then obviously be that no cells would grow at all-since everything else would die because of the medium, the double recombinants are the only cells that have any chance of surviving. So if they can't survive at all, no cells will grow. If that is the case, the entire experiment will need to be completely redone so that no dmnD- cells have to be used. But one valuable piece of information would be gleaned from this result: that dmnD is actually vital for cell life.

In the next part, the test for differentiation, the mutated cells are grown on a nitrogen-rich medium (assuming they can grow at all) and when enough cells are present, the nitrogen is washed out and differentiation is noted. If none of the cells differentiate, it can be concluded that DmnD is probably needed for differentiation to occur, despite the fact that it seems to perform the same function as DmnA. That is what occurred with DmnA, cells could not differentiate without it, so it is entirely possible that they cannot differentiate without DmnD. If DmnD is not vital, however, some cells will be able to become heterocysts and fix nitrogen. In this case, it may also be interesting to note patterns of differentiation: do they follow the same pattern as that of wild type Anabaena (about one heterocyst for every ten photosynthetic cells)? If they don't, that is evidence of DmnD's role in differentiation, even if the role isn't vital.

The other tests are completed to pinpoint more possible functions of DmnD in Anabaena, based on known functions of comparable genes in E. coli. In the test for methylation, to see whether the protein product places methyl groups on the same sequence as the bacterium (GATC), DNA cutting by one enzyme of the dmnD- organism represents that Anabaena's DNA is methylated at that sequence by only dmnD-produced protein. If DpnI succeeds in cutting the DNA, then the dmnD- cyanobacteria are unmethylated at the given sequence, implicating DmnD as vital to the methylation, along with DmnA. This leads to the question of why both methylases are needed to perform one function where one suffices in most situations. However, if DpnII cuts, the cyanobacterium is methylated despite the absence of DmnD, so DmnD is not necessary for methylation. This could lead to several different paths, such as: why is DmnA required, but not DmnD? and, does DmnD actually participate in methylation at all? The test for mismatch repair is quite simple. If DmnD plays a role in mismatch repair, the rate of mutation should be significantly higher in the dmnD- group of organisms than in the wild type. If the rates are similar, then DmnD does not help repair problems during replication.