Age discrimination: Erasing young DNA

Despite the diligence and proofreading of DNA polymerase, replication is inherently error-prone. Reasons include the bases' tendency to occasionally to visit their tautomeric forms, and the simple fact that the bases aren't sufficiently dfferent to allow fault-free choices by polymerase (at least, not 100 million times out of 100 million, which is the ACHIEVED error rate of DNA repliction and repair!!!). So errors WILL creep in. Cells have a variety of mechanisms for further policing the work of DNA Polymerase. One of the coolest operates in E. coli and involves an enzyme that physically modifies adenines--the so-called dam (for deoxyadenine modifying) methylase.

The 'idea' behind the operation of this system is simple--since new strands of DNA are made by copying (actually, complementing) old strands, there is a logical way to resolve situations where the strands end up with non-matching partners--go with the old strand! More concretely, let us say there is an A-C mismatch in the DNA. If the A is on the original (template) strand and the C is the newly-incorporated nucleotide, the smart money deduces that the C was put in erroneously. Thus it should be replaced with a T to match the A. So far, so good. The trick, of course, is figuring out which strand is new and which old! That's where the dam methylase comes in.

The methylase is a very simple-minded machine--it wanders along DNA and covalently modifies adenines in the sequence -GATC- whenever it finds them. One important thing to note at this point is that this is a well-chosen sequence for the task of marking strands--its complement (the sequence on the other strand) is also -GATC- (don't forget that the strands are complementary and anti-parallel). Thus there are anywhere this sequence appears, BOTH strands have the potential to be 'marked' by the methylase. It is the timing of the marking that is central to our story. The marking itself is shown in green below. Note that addition of the methyl group would NOT alter the ability of A to basepair with T, but WOULD create a 'flag' on the A that could be read in the major groove (added methyl group shown in green).

They say that timing is everything, and that is certainly true for DNA marking. Here's the poop--the dam methylase is a relatively SLOW enzyme. Old DNA is fully methylated, so when replication occurs, the old DNA strand has all its GATC sequences 'marked' by adenine methylation. The new DNA is synthesized using plain ole dATP, so it is unmethylated. This creates the asymmetry we need! For a limited time after DNA replication, E. coli machines can 'count on' the fact that new DNA looks different than old DNA--it lacks methylated adenines! This situation won't persist forever--the dam methylase is always chugging along--but it creates a 'window of opportunity' for repair machinery to spot and fix basepairing errors that slipped past the proofreading function of DNA polymerase.

So how does E. coli actually READ and USE the information provided by the dam methylase system? It uses three proteins that cooperate to identify which strand of a mismatched basepair needs to be replaced. The first player in the game is called MutS. MutS is a protein that travels along the DNA looking for the odd kinks and bends that signal incompatible basepair partners. Upon discovering a mismatch (and it's pretty good at this--apparently it can identify all mismatches except C:C) it joins forces with MutL and MutH. MutL is thought to be a 'matchmaker' that brings the active players MutS and MutH together. When MutH gets 'turned on' by MutS it starts traveling along the DNA 'looking for' GATC sequences. Note that it isn't letting go of MutL which is still holding on to MutS which is still in touch with the offending mismatch!

When MutH finds a GATC sequence it identifies the new DNA strand (the one NOT methylated an the A!) and breaks the phosphate backbone of that strand. From this point, other repair machinery takes over. DNA Polymerase III (For most DNA repair, Pol I is used, but in this special case, the nick is likely DISTANT from the site of mismatch--and Pol I is unlikely to 'go the distance). Further, since PolIII is being used, there is NO associated 5' -> 3' exonuclease activity, so a special exonuclease is 'called in' to chew up the unmethylated strand. If the team traves travels far enough (and in the correct direction! See below) it will eventually come to the mismatch that started this whole process. Since it has been working at replacing the NEW strand, it will inevitably replace the new (and presumed erroneous!) part of the mismatch with a matching nuceleotide.

Voila! By discovering a 'timestamp machine' (the slow DNA modifying enzyme dam methylase) E. coli has bought itself further insurance against the unavoidable errors that arise during DNA replication. Not a bad trick for a 'simple' prokaryote, however smelly!

The polarity issue: only nicks on ONE SIDE of the mismatch will be productive! Here's why:

Let's take a look at the starting situation: a mismatched pair of G's and a hemimethylated GATC sequence: (note: I have used black dots to represent 5' ends or the 5' direction, while arrowheads represent 3' ends or 3' direction)

Here, the methyl group is indicated by the green dot over the A on the same strand as the BLUE G. Since the methyl is on this strand, it must be the older of the two. Thus it was the TEMPLATE, and the likely error is the orange G on the lower strand. So let's proceed: MutS lands at the mismatch site, recruits MutL and MutH, which track down the DNA and come to the hemimethylated (= 'half-methylated', i.e. methylated on one strand and not the other) GATC sequence. MutH dutifully identifies the unmethylated strand and nicks it:

Now, along come Pol III and Exo VII and beginning chewing up (Exo) and replacing (Pol) the nicked strand in the 5' to 3' direction:

The replaced DNA, as indicated by the dashed lines, will eventually include the erroneous orange G as long as the process proceeds far enough.
But what if the BLUE G had been the new strand? The methylation (and subsequent nicking) would be on the UPPER strand:

If Exo VII and Pol were to perform replacement synthesis, but (to quote Indiana Jones) "They're digging in the wrong place!". So nicks on the 'right' side would non-productive in terms of repairing the upper strand; in this case we would want to nick on the 'left' side...

Instead, in cases where the nick is made 3' of the mismatch, a different exonuclease is recruited! The exo is called Exo I, and it has a 3' to 5' exonuclease activity (similar to that used as the 'proofreading' exonuclease on DNA polymerases). Chewing in this direction will eventually reach the Blue G:

When synthesis starts (after completion of Exo I activity, which has presumably proceeded beyond the G-G mismatch) replacement synthesis will put in a fresh nucleotide (hopefully a C this time!) and re-fill the gap left by Exo I.

Obviously, when MutHLS get started, all they 'know' is that they have a mismatch--they CANNOT know whether to go 'right' or 'left'; the correct direction to go is determined by which strand is 'old' and which 'new'--and THAT can only be known once a GATC is found! So MutHLS must proceed at random, and in the end Exo VII or Exo I is 'chosen' (by what mechanism I do not know--it must be reasonably intricate, in that it requires 'knoweledge' of whether the mismatch lies in the 5' direction (choose Exo I) or 3' direction (choose Exo VII) of the mismatch!