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2020_SS1_Bis2a_Facciotti_Reading_20 - Biology

2020_SS1_Bis2a_Facciotti_Reading_20 - Biology


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Learning objectives associated with 2020_SS1_Bis2a_Facciotti_Reading_20

  • Dissect the chemical reaction responsible for DNA synthesis and tell its energy story.
  • Draw the process of DNA replication for the leading strand, including key sites on the DNA, reactants, products, enzymes, and the depiction of energy requirements for each step.
  • Draw the process of DNA replication for the lagging strand, include the design challenge associated with replicating the lagging strand template, and the solutions afforded by Okazaki fragments and DNA ligase.
  • Describe the challenges of replicating the ends of linear chromosomes. Contrast these challenges with prokaryotic chromosome replication.
  • Create a depiction of the telomerase-mediated telomere-lengthening mechanism that explicitly describes the role of structural RNA in the enzyme complex.
  • Explain how DNA replication is a source of geneticvariation,and describe the mechanisms that can change the frequency of mutation and the trade-offs associated with fixing errors.

The DNA Double Helix and its Replication

The Problem

In this module, we discuss the replication of DNA—one of the key requirements for a living system to regenerate and create the next generation. Let us first briefly consider the problem through a literary analogy.

The human genome comprises roughly 6.5 billion base pairs of DNA if one considers the full diploid genome (i.e., if you count the DNA inherited from both parents). Six point five billion looks like this: 6,500,000,000. That's a large number. To get a better idea of what that number means, imagine that our DNA is a set of written instructions for constructing one of us. By analogy, we can then compare it to another written document. For this example, we begin by considering Tolstoy's War and Peace, a novel many people are familiar with for its voluminous nature. Data from Wikipedia estimates that War and Peace

contains

about 560,000 words. A second written work many are familiar with are the seven volumes of

J.K. Rowling

's Harry Potter. This work checks in at ~1,080,000 words (Referenced Statistics on Wikipedia). If we assume that the length of the average English word is five characters, the two literary works are 2.8 million and 5.4 million characters, respectively. Therefore, even all seven volumes of "Harry Potter" have over 1000x fewer characters than our own genomes. The number of characters in these novels

are

, however, much closer to the number of nucleotides in a typical bacterial genome.

Now imagine for a moment developing a machine or mechanical process (not an electronic process) that reads and copies these books. Or imagine yourself copying these texts. How fast could you do it? How many mistakes are you likely to make? Do you expect there to be a trade-off between the speed at which you can copy and the accuracy? What type of resources does this process need? How much energy does the process require? Now imagine copying something 1000x larger! Oh, and just for good measure, your imaginary mechanical device needs to do its work on text that is ~25Å wide (i.e., 0.0000000025 meters wide). By comparison, a typical ten point font is ~0.00025 meters wide, about 100,000x larger than the width of a DNA base pair.

With that in mind,

it is worth noting that

a human cell can take about 24 hours to divide (DNA replication must therefore be a little faster). A healthy E. coli cell may take only 20 minutes to divide (including replicating its ~4.5 million base pair genome). Both the human and bacterium do this while typically making few enough mistakes that the subsequent generation remains viable and recognizable. That should seem rather amazing! Now consider that we estimate the human body comprises ~10 trillion cells (10,000,000,000,000) and that it may have between two and ten times that number of microbial residents. That's

a lot of cell

division to consider.

Design challenge

If the cell is to replicate—its

ultimate

goal—a copy of the DNA must

be created

. So one clear problem statement/question is "how can the cell

effectively

copy its DNA?" Given the analogy above, here are some relevant sub-questions: What are the chemical and physical properties that enable DNA to

be copied

? With what fidelity must the organism copy its genome? What speed must

it be copied

at? Where does the energy come from for this task and how much is necessary? Where do the "raw materials" come from? How do the molecular machines involved in this process couple the assembly of raw materials and the energy required to build a new DNA molecule together? The list could

, of course,

go on.

In the following discussion and in the lecture, we examine how the process of DNA replication

is accomplished

while keeping in mind some driving questions. As you go through the reading and lecture materials, try to be constantly aware of these and other questions associated with this process. Use these questions as guideposts for organizing your thoughts and try to find matches between the "facts" that you think we might expect you to know and the driving questions.

The DNA double helix

To build some extra context, we also need a little of empirically determined knowledge. Perhaps one of the best-known and popular features of the hereditary form of the DNA molecule is that it has a double helical tertiary structure. Our appreciation of the double-helical structure of DNA dates to the 1950s. For more on this story, see the short film here.

Models of the structure of DNA revealed that molecule comprises two strands of covalently linked nucleotides that

are twisted

around each other to form a right-handed helix. In each strand, nucleotides

are covalently joined

to two other nucleotides (except at the very ends of a linear strand) via phosphodiester bonds that link the sugars via the 5' and 3' hydroxyl groups (panel

b

in Figure 1). Recall that the labels 5' and 3' refer to the carbons on the sugar molecule. These sugars and phosphate chains form a contiguous set of covalent links that

are often referred

to as the "backbone" of the structure. In a linear molecule, each strand has two free ends. We call one free end the 5' end because the unlinked functional group that

is typically involved

in joining nucleotides is the phosphate linked to the 5' carbon. We call the other end of the strand the 3' end because the unlinked functional group that

is typically involved

in joining nucleotides is the hydroxyl group linked to the 3' carbon of the sugar. Since the two ends of the strand are not symmetrical, this makes it easy to designate a direction one the strand—one can, for instance, say that they are reading from the 5' end to 3' end to

indicate

that they are "walking" along the strand starting at the 5' end and moving towards the 3' end. This direction (5' to 3') is the convention used by most biologists. One can read in the opposite direction (3' to 5') provided we make the direction explicit. We find the two strands of covalently linked nucleotides to be anti-parallel to one another in the double-helix;

that is, the

orientation/direction of one strand is opposite to that of the other strand (panel

b

in Figure 1). The backbone is structurally on the "outside" of the double helix, creating a band of negative charges on the surface. The nitrogenous bases of each of the antiparallel strands stack on the inside of the structure and oppose one another in a way that allows hydrogen bonds between unique purine/pyrimidine pairs (A pairing with T and G pairing with C) to form. We call these specific base pairings complementary base pairs. Thus, we refer to the paired strands of a double helix as complementary strands.

Complementary strands carry redundant information. Because of the strict chemical pairing, if you know the sequence of one strand, you obligatorily know the strand of its complement. Take, for example, the sequence 5′- C A T A T G G G A T G - 3′. Note how the sequence

is annotated

with the orientation (

indicated

by 5' and 3' labels). The complement of this sequence—written according to the 5' to 3' convention is: 5′- C A T C C C A T A T G - 3′. If you

aren't convinced

, write these two sequences out across from one another in your notes, writing them as antiparallel strands. Note that the twisting of the two complementary strands around each other results in the formation of structural features called the major and minor grooves that will become more important when we discuss the binding of proteins to DNA (panel

c

in Figure 1).

Most of the BIS2A instructors will expect you to recognize key structural features depicted in the figure below and that you will

be able to

create a basic figure of the structure of DNA yourself.

Figure 1. DNA has (a) a double helix structure and(b) phosphodiester bonds. The (c) major and minor grooves are binding sites forDNA bindingproteins during processes such as transcription (the creation of RNA from a DNA template) and replication.


Possible NB Discussion Point

Take a moment to review the nitrogenous bases in Figure 1. Identify functional groups as described in class. For each functional group identified, describe what type of chemistry you expect it tobe involvedin. Try to identify whether the functional group can act as either a hydrogen bond donor, acceptor, or both?


At around the same time, three hypotheses for the modes of DNA replication were being considered.

The models for replication were knownas:

the conservative model, the semi-conservative model, and the dispersive model.

1. Conservative: The conservative model of replication postulated that each whole double-stranded molecule could act as a template for the synthesis of a new double-stranded molecule. If one were to put a chemical tag on the template DNA molecule after replication,

none of that tag would be found

on the new copy.
2. Semi-conservative: This hypothesis stipulated that each individual strand of a DNA molecule could serve as a template for a new strand to which it would now associate with.

In this case

, if

a chemical label were placed

on a double-stranded DNA molecule, one strand on each of the copies would keep the label.
3. Dispersive: This model proposed that a copied double helix would piecewise combine continuous segments of "old" and "new" strands.

If a chemical label were placed

on a DNA molecule that

were copied

using a dispersive mechanism, one would find discrete segments of the resulting copy that

were labeled

on both strands separated by

completely

unlabeled parts.

Meselson and Stahl resolved the issue in 1958 when they reported results of a now famous experiment (describe on Wikipedia) which showed that DNA replication is semi-conservative (Figure 2), where each strand

is used as

a template for the creation of the new strand. To learn more about this experiment, watch The Meselson-Stahl Experiment.

Figure 2. DNA has an antiparallel double helix structure, the nucleotide bases are hydrogen bonded together, and each strand complements the other.DNA is replicatedin a semi-conservative manner,each strand is usedas the template for the newly made strand.

DNA replication

Having established some basic structural features and the need for a semi-conservative mechanism, it is important to understand some of what we know about the process and to think about what questions one might want to answer if they are to better understand what is going on.

Since DNA replication is a process, we can invoke the energy story rubric to think about it. Recall that the energy story rubric is there to help us think systematically about processes (how things go from A to B). In this case the process in question is the act of starting with one double-stranded DNA molecule and ending up with two double-stranded molecules. So, we will ask a variety of questions: What does the system look like at the beginning (matter and energy) of replication? How are matter and energy transferred in the system, and what catalyzes the transfers? What does the system look like at the end of the process? We can also ask questions regarding specific events that MUST

happen

during the process. For instance, since DNA is a long molecule

and

it is sometimes circular, we can ask basic questions like, where does the process of replication start? Where does it end? We can also ask practical questions about the process like, what happens when a double-stranded structure

is unwound

?

We consider some of these key questions in the text and in class and encourage you to do the same.

Requirements for DNA replication

Let's start by listing some basic functional requirements for DNA replication that we can infer just by thinking about the process that must happen and/or

be required

for the replication to happen. So, what do we need?

• We know that DNA

is composed

of nucleotides. If we want to create a new strand, we will need a source of nucleotides.
• We can infer that building a new strand of DNA will require an energy source—we should try to find this.
• We can infer that that there must be a process for finding a place to start replication.
• We can infer that there will be one or more enzymes that help

catalyze

the process of replication.
• We can also infer that since this is a biochemical process, that it will make some mistakes.

Nucleotide structure review

Recall some basic structural features of the nucleotide building blocks of DNA. The nucleotides start off as nucleotide triphosphates.Nucleotides are composedof a nitrogenous base, deoxyribose (five-carbon sugar), and a phosphate group. We name the nucleotide according to its nitrogenous base, purines such as adenine (A) and guanine (G), or pyrimidines such ascytosine(C) and thymine (T). Recall the structures below. Note that the nucleotide Adenosine triphosphate (ATP) is a precursor of the deoxyribonucleotide (dATP) whichis incorporatedinto DNA.

Figure 3. Each nucleotide is madeup of a sugar (ribose or deoxyribose depending on whether it builds RNA or DNA, respectively), a phosphate group, and a nitrogenous base. The purines have a double ring structure with a six-membered ring fused to a five-membered ring. Pyrimidines aresmaller in size; they have a single six-membered ring structure. The carbon atoms of the five-carbon sugarare numbered1', 2', 3', 4', and 5' (1'is readas “one prime”).The phosphate residue is attachedto the hydroxyl group of the 5' carbon of one sugar of one nucleotide and the hydroxyl group of the 3' carbon of the sugar of the next nucleotide,therebyforming a 5'-3' phosphodiester bond.

Initiation of replication

Where along the DNA does the replication machinery start DNA replication?

With millions, if not billions, of nucleotides to copy how does the DNA polymerase know where to start? This process turns out not to be random. There are specific nucleotide sequences called origins of replication along the DNA at which replication begins. Once this site

is identified

, however, there is a problem. The DNA double helix

is held

together by base stacking interactions and hydrogen bonds. If each strand must

be read

and copied individually, there must be some mechanism responsible for helping to dissociate the two strands from one another. Energetically, this is an endergonic process. Where does the energy come from, and how is this reaction

catalyzed

? Basic reasoning should lead to the hypothesis that a protein catalyst is likely involved, and that this enzyme either creates new bonds that are energetically more favorable (exergonic) than the bonds it breaks AND/OR it can couple the use of an external energy source to help dissociate the strands.

It turns out that the details of this process and the proteins involved differ depending on the specific organism in question, and many of the molecular level details

are not completely understood

. There are, however, some common features in the replication of eukaryotes, bacteria, and archaea, and one of these features is that the process involves multiple different

types of

proteins in replicating DNA. First, proteins called "initiators" can bind DNA at or very near origins of replication. The interaction of the initiator proteins with the DNA helps to destabilize the double helix and also help to recruit other proteins, including an enzyme called a DNAhelicase to the DNA. Here the energy required to destabilize the DNA double helix seems to come from the formation of new associations between DNA and the initiator proteins and the proteins themselves. The DNA helicase is a multi-subunit protein important in the process of replication because it couples the exergonic hydrolysis of ATP to the unwinding of the DNA double helix. Additional proteins must

be recruited

to the initiation complex (the collection of proteins involved in

initiating

transcription). These include, but

are not limited

to, additional enzymes called primaseand DNA polymerase. While the initiators depart soon after the initiation of replication, the rest of the proteins work in concert to execute the process of DNA replication. This complex of enzymes function at Y-shaped structures in the DNA called replication forks (Figure 4). For any replication event, two replication forks can form at each origin of replication, extending in both directions.

Multiple origins of replication can be found

on eukaryotic chromosomes and some archaea, while the genome of the bacterium, E. coli, seems to encode one origin of replication.

Figure 4. At the origin of replication, a replication bubble forms.The replication bubble is composedof two replication forks, each traveling in opposite directions along the DNA.It is understoodthat the replication forks include all the enzymes required for replication to occurthey are just not drawn explicitly in the figure to provide room to illustrate the relationships between the template and new DNA strands.
Attribution: BIS2A team original image

Elongation of replication

The melting open of the DNA double helix and the assembling the DNA replication complex is just the first step in the process of replication. Nowthe process ofcreating a new strand actually needs to get started. Here, we encounter additional challenges. The first obvious issue is that of determining which of the two strands should get copied at any replication fork (i.e., Which strand will serve as a template for semi-conservative synthesis? Are both strands equally viable alternatives?). There is also the problem of getting the process of the new strand synthesis started. Can the DNA polymerase start the new strand on its own? We will discuss later the answer to the latter question and some rationale and consequences. The key idea to note at this point is that it hasbeen experimentally determinedthat DNA polymerase can NOT start strand synthesis on its own. Rather, DNA polymerase requires a short stretch of double-stranded structure followed by a single-stranded template. The enzyme primase creates a short oligonucleotide polymer of RNA (not DNA) called a primer (theseare depictedby short green lines in the figures above and below). DNA polymerase uses the primer to nucleate and grow a new strand.

During the process of strand elongation, the DNA polymerase polymerizes a new covalently linked strand of DNA nucleotides (in bacteria this specific enzyme may

be called

DNA polymerase III; in eukaryotes, polymerase nomenclature is more complex and the roles of several polymerase proteins

are not completely understood

). It turns out that one strand

is favored

over the other to serve as a template. DNA polymerase will "read" the template strand from 3' to 5' and synthesize a new strand in the 5' to 3' direction. Hypotheses to explain this universal observation usually center on the energetics associated with the addition of a new nucleotide and arguments associated with DNA repair that we will describe shortly. Let us, therefore, briefly consider the reaction involving the addition of a single nucleotide. The primer provides an important 3' hydroxyl on which to begin synthesis. The next deoxyribonucleotide triphosphate enters the binding site of the DNA polymerase and, as shown in Figure 5 below,

is oriented

by the polymerase such that a hydrolysis of the 5' triphosphate can occur. This reaction releases pyrophosphate and couples the exergonic hydrolysis of the phosphoanhydride to the synthesis of a phosphodiester bond between the 5' phosphate of the incoming nucleotide and the 3' hydroxyl group of the primer. This process repeats until deoxyribonucleotide triphosphates run out or the replication complex falls off of the DNA. In effect, DNA polymerase adds the phosphate group (5') from the incoming nucleotide to the existing hydroxyl group (3') of the previously added nucleotide.

Correct base pairing, or selection of correct nucleotide to add at each step,

is accomplished

by structural constraints felt by the DNA polymerase and the energetically favorable hydrogen bonds formed between complementary nucleotides. The process

is energetically driven

by the hydrolysis of the incoming 5' triphosphate and the energetically favorable interactions formed by the inter-nucleotide interactions in the growing double helix (base stacking and complementary base pairing hydrogen bonds). Note that the energetics of nucleotide addition do not technically prevent a strand growing in the 3' to 5' direction. The key difference in this “backwards” synthesis scheme is that the energy “source” for synthesis would need to come from a nucleotide already incorporated into the growing strand rather than the new incoming nucleotide (which this might be an important selective disadvantage

is discussed

briefly). After elongation has started a different DNA polymerase (in bacteria we usually call this enzyme DNA Polymerase I) comes in to remove the RNA primer and to synthesize the remaining bit of missing DNA.

As discussed in more detail in class, the movement of the replication fork induces winding of the DNA in both directions of replication. Another ATP consuming enzyme called topoisomerase helps to relieve this stress.

Figure 5. DNA polymerase catalyzes the addition of the 5' phosphate group from an incoming nucleotide to the 3' hydroxyl group of the previous nucleotide. This process creates a phosphodiester bond between the nucleotides while hydrolyzing the phosphoanhydride bond in the nucleotide.
Source: http://bio1151.nicerweb.com/Locked/m...h16/elong.html

Leading and lagging strand

The discussion above about strand elongation describes the process ofnewstrand synthesis if that strandis synthesizedin the same direction as the replication fork is or appears to be moving along the DNA. This strand canbe synthesizedcontinuously andis calledthe leading strand. However, both strands of the original DNA double helix mustbe copied. Since the DNA polymerase can only synthesize DNA in a 5' to 3' direction, the polymerization of the strand opposite of the leading strand must occur in the opposite direction that helicase, or front of the replication fork, is traveling. This strandis calledthe lagging strand, anddue togeometric constraints, must be synthesized through a series of RNA priming and DNA synthesis events into short segments called Okazaki fragments. As noted, the initiation of synthesis of each Okazaki fragment requires a primase to synthesize an RNA primer, and each of these RNA primers mustbe removedand replaced with DNA nucleotides by a different DNA polymerase. The covalent bonds between each of the Okazaki fragment can notbe madeby the DNA polymerase and must thereforebe formedby yet another enzyme called DNAligase. The geometry of lagging strand synthesis is difficult to visualize and willbe coveredin class.

Figure 6. The lagging strand is createdin multiple segments. A replication fork shows the leading and lagging strand. A replication bubble shows the leading and lagging strands.
BIS2A Team original image

Termination of replication

Telomeres and telomerase

The ends of replication in circular bacterial chromosomes pose few practical problems. However, the ends of linear eukaryotic chromosomes pose a specific problem for DNA replication. Because DNA polymerase can add nucleotides in only one direction (5' to 3'), the leading strand allows for continuous synthesis until the end of the chromosomeis reached; however, as the replication complex arrives at the end of the laggingstrandthere is no place for the primase to "land" and synthesize an RNA primer so that the synthesis of the missing lagging strand DNA fragment at the end of the chromosome canbe initiatedby the DNA polymerase. Without some mechanism to help fill this gap, this chromosomal end will remain unpaired and willbe lostto nucleases. Over time, and several rounds of replication, this wouldresult inthe ends of linear chromosomes getting progressively shorter, ultimately compromising the ability of the organism to survive. These ends of the linear chromosomesare knownas telomeres, and nearly all eukaryotic species have evolved repetitive sequences that do not code for a specific gene. As a result, these "non-coding" telomeres act as replication buffers andare shortenedwith each round of DNA replication instead of critical genes. For example, in humans, a six base-pair sequence, TTAGGG, repeats 100 to 1000 times at the end of most chromosomes. Besides acting as a potential buffer, the discovery of the enzyme telomerase helped in understanding how chromosome endsare maintained. Telomerase is an enzyme composed of protein and RNA. Telomerase attaches to the end of the chromosome by complementary base pairing between the RNA component of telomerase and the DNA template. Telomerase uses the RNA as a complementary strand for the short elongation of its complement. The cell repeats this process many times. Once the lagging strand templateis sufficiently elongatedby telomerase, primase will create a primer followed by DNA polymerase which can now add nucleotides complementary to the ends of the chromosomes. Thus, the ends of the chromosomesare replicated.

Figure 7. The ends of linear chromosomesare maintainedby the action of the telomerase enzyme.

Telomerase is not active in adult somatic cells. Adult somatic cells that undergo cell division continue to have their telomeres shortened. This means that telomere shorteningis associatedwith aging. In 2010, scientists found that telomerase can reverse some age-related conditions in mice, and this may have potential in regenerativemedicine.1 Telomerase-deficient micewere usedin these studies;thesemice have tissue atrophy, stem-cell depletion, organ system failure, and impaired tissue injury responses. Telomerase reactivation in these mice caused the extension of telomeres, reduced DNA damage, reversed neurodegeneration, and improved functioning of the testes, spleen, and intestines. Thus, telomere reactivation may have potential for treating age-related diseases in humans.


Possible NB Discussion Point

Imagine that researchers have invented a drug for humans that upregulates telomerase activity. Would you be excited or would you be skeptical about being part of a clinical trial? Do you trust this drug? Can you think of any negative consequences to activating telomerase to a level that it would not “naturally” be at? Assuming scientists proved without doubt that their drug has no negative side effects -- what are some of the ethical questions you would consider?


Differences in DNA replication rates between bacteria and eukaryotes

DNA replication hasbeen extremely well studiedin bacteria, primarily because of the small size of the genome and the large number of variants available. E. coli has 4.6 million base pairs in a single circular chromosome, and all of it gets replicated in approximately 42 minutes, starting from a single origin of replication and proceeding around the chromosome in both directions. This means that the polymerase adds approximately 1000 nucleotides per second to a growing strand. The process is much more rapid than in eukaryotes. Table 1 summarizes the differences between bacterial and eukaryotic replications.

Table1.Differences between bacterial and eukaryotic replication.

Differences between prokaryotic and eukaryotic replication
PropertyBacteriaEukaryotes
Origin of replicationSingleMultiple
Rate of polymerization per polymerase1000 nucleotides/s50 to 100 nucleotides/s
Chromosome structureCircularLinear
TelomeraseNot presentPresent

Link to external resources

Click through a tutorial on DNA replication.

Replication design challenge: proofreading

When the cell replicates its DNA, it does so in response to environmental signals that tell the cell it is time to divide. The ideal goal of DNA replication is to produce two identical copies of the double-stranded DNA template and to do it in an amount of time that does not pose an unduly high evolutionarily selective cost. This is a daunting task when you consider that there are ~6,500,000,000 base pairs in the human genome and ~4,500,000 base pairs in the genome of a typical E. coli strain and that Nature has determined that the cells must replicate within 24 hours and 20 minutes, respectively. In either case, many individual biochemical reactions need to take place.

While ideally replication would happen with perfect fidelity, DNA replication, like all other biochemical processes, is imperfect—bases may

be left

out, extra bases may

be added

, or

bases may be added

that do not properly base-pair. In many organisms, many of the mistakes that occur during DNA replication

are corrected

promptly by DNA polymerase itself via a mechanism known as proofreading. In proofreading, the DNA polymerase "reads" each newly added base via sensing the presence or absence of small structural anomalies before adding the next base to the growing strand. In doing so,

a correction can be made

.

If the polymerase detects that a newly added base has

been paired

correctly with the base in the template strand, it adds the next nucleotide. If, however, a wrong nucleotide

is added

to the growing polymer, the misshaped double helix will cause the DNA polymerase to stall, and it will eject the newly made

strand

from the polymerizing site.

Consequently, the

DNA strand will enter an exonuclease site. In this site, DNA polymerase can cleave off the last several nucleotides that

were added

to the polymer. Once the polymerase removes the incorrect nucleotides, the DNA strand can return to the polymerizing site and new nucleotides will

be added

again. This proofreading capability comes with some trade-offs: using an error-correcting/more accurate polymerase requires time (the trade-off is speed of replication) and energy (always an important cost to consider). The slower you go, the more accurate you can be. Going too slow, however, may keep you from replicating as fast as your competition, so figuring out the balance is key.

We call errors that

are not corrected

by proofreading mutations.

Figure 1. Proofreading by DNA polymerase corrects errors during replication.

Replication mistakes and DNA repair

Although DNA replication is typically a highly accurate process and proofreading DNA polymerases helps to keep the error rate low, mistakes still occur. Besides errors of replication, environmental damage may also occur to the DNA. Such uncorrected errors of replication or environmental DNA damage may lead to serious consequences. Therefore, Nature has evolved several mechanisms for repairing damaged or incorrectly synthesized DNA.

Mismatch repair

Errors not corrected during replication may instead get corrected after replication completes;

we

know this kind of repair as mismatch repairs. Specific enzymes recognize the incorrectly added nucleotide and excise it, replacing it with the correct base. But, how do mismatch repair enzymes recognize which of the two bases is the incorrect one?

In E. coli, after replication, the nitrogenous base adenine gains a methyl group;

this

means that directly after replication, the parental DNA strand will have methyl groups, whereas the newly synthesized strand lacks them. Thus, mismatch repair enzymes can scan the DNA and remove the wrongly incorporated bases from the newly synthesized, non-methylated strand by using the methylated strand as the "correct" template from which to incorporate a new nucleotide. In eukaryotes, the mechanism

is not as well understood

, but we believe it to involve recognition of unsealed nicks in the new strand, and a short-term, continuing association of some replication proteins with the new daughter strand after replication has completed.

Figure 2. In mismatch repair, the incorrectly added baseis detectedafter replication. The mismatch repair proteins detect this base and remove it from the newly synthesized strand by nuclease action.The gap is now filledwith the correctly paired base.

Nucleotide excision repair

Nucleotide excision repair enzymes replace incorrect bases by making a cut on both the 3' and 5' ends of the incorrect base. The entire segment of DNAis removedand replaced with correctly paired nucleotides by the action of a DNA polymerase. Once the basesare filledin, it seal the remaining gap with a phosphodiester linkagecatalyzedby the enzyme DNA ligase.This repair mechanism is often employedwhen UV exposure causes the formation of pyrimidine dimers.

Figure 3. Nucleotide excision repairs thymine dimers. When exposed to UV, thymines lyingadjacentto each other can form thymine dimers. In normal cells, theyare excisedand replaced.

Consequences of errors in replication, transcription, and translation

Something key to think about:

Cells have evolved a variety of ways to make sure DNA errors are both detected and corrected. We have already discussed several of them. But why did so many mechanisms evolve? From proofreading by the various DNA-dependent DNA polymerases, to the complex repair systems. Such mechanisms did not evolve for errors in transcription or translation. If you are familiar with the processes of transcription and/or translation, think about what the consequences would be of an error in transcription. Would such an error affect the offspring? Would it be lethal to the cell? What about errors in translation? Ask the same questions about the process of translation. What would happen ifthe wrong amino acid is accidentally putinto the growing polypeptide during translation? How do these contrast with DNA replication? If you are not familiar with transcription or translation, don't fret. We'll learn those soon and return to this question again.



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