the art and science of pcr jos. j. schall department of biology university of vermont

The Art and Science of PCR

Jos. J. SchallDepartment of BiologyUniversity of Vermont

3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’

You have extracted DNA from a tissue sample. There were thousands of cells in that tissue, soeach chromosome is represented in the extracted DNA thousands of times. To follow what happensduring the PCR process, we will follow only one segment of the DNA -- one double strand. It isshown here as a short segment, but of course the real DNA would be in very long strands of thousands of bases (if you did a nice job with your extraction). Notice that the 5’ and 3’ ends of the strands are indicated. Remember that DNA elongates during its replication from the 5’ to 3’ end ofthe strand that is elongating. Again, always keep in mind that we are following what happens to asingle copy of the DNA molecule of interest, but there will be thousands present.


By the way….suppose this DNA segment is from the mitochondrion’s genome. Most cells containthousands of mitochondria, so this segment would be present thousands of times for EACH cell. Compare this to any DNA segment from the nuclear genome that is represented once (for haploidcell such as Plasmodium) or twice (for cells of diploid organisms). Do you see why the Mitochondrial DNA will often give better results in PCR?


You wish to amplify (a billion-fold) the segment of DNA shown in RED on the strand. Why thatparticular segment? It may be the section of a gene that you wish to sequence, or some segmentthat includes a repeat region (a microsatellite), or a segment that you use to find a parasite ina tissue sample, or a hundred other reasons. The beauty of PCR is that it will amplify a specificsegment of your choice, and only that segment.Note that the segment shown in red is far shorter than the normal segment you would want to amplify (normally from about 150 to 1000), but to see what is happening, we have to showonly a short segment (to fit on the screen!).


Your first job is to design a pair of primers at the ends of the target strand of DNA. Primers are typicallyaround 20 bases long. Because the strand shown above is so short, for purposes of illustrationhere, the primers will be shown shorter than that. In practice, such short primers would not beused, because they would not be very stable when they anneal to the target DNA, and may alsomatch regions on non-target parts of the genome, so you would get a mixed product at the end.


The primer sites are shown above in BLUE. Note that the primer sites that are shown inblue are part of the segment that you want to amplify.By “primer site” is meant the location on the DNA molecule where the primer will anneal.“Anneal” means to attach. The terms “anneal”, “stick”, and “sit down” all can be used.Remember that strands of DNA with complementary bases can anneal. So, a strand with CGTA will anneal with another strand that is GCAT. So, theprimer will be the complement of the segment where you want it to sit down.

3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’

If the DNA is heated to about 95° C, the two strands will separate. We are denaturing the DNA.Sometimes this is called the “melting” of the DNA.

5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’

95° C -- strands separate


If you cool the DNA back down again…the strands will come back together….the complementarystrands will re-anneal. Because they are so long, it may take a long time for this to happen, andthere could be lots of errors when these long strand try to re-anneal.


Cool down, and strands anneal


OK, let’s do a PCR reaction. Start by a melt temperature of 95° C (some people prefer 94°). A melt of about 1 minute is fine. The polymerase that is part of the reaction is damaged slightlyby the high temperature, and is damaged more over longer melt times. So, the best melt is atemperature which allows all the DNA to denature, but not too high to cause much damage tothe polymerase.


95° C for 1 minute

Cycle 1


In the mixture are the primers. The two primers are shown in PURPLE. Remember that the sitewhere the primers will sit down are indicated in blue. Note that one primer is the complement ofone of the sites and the other primer is the complement of the second site.

The primers are present in very high concentration….a great surplus compared to the target DNA.This is important because we want the PRIMERS to anneal at the target sites rather than theother long strand of DNA at its complementary site.


5’CCGATTACGT3’3’GATTCCGACG5’

Cycle 1


The primers should not be complements to each other (just to the part of the strands where youwant them to sit down. If they are self-complements, they could anneal with each other andreduce the amount of free primers available to sit down at the desired sites on the DNA molecule.


5’CCGATTACGT3’3’GATTCCGACG5’

Cycle 1


The temperature is lowered to the “Annealing Temperature” that varies with the primers, butis typically about 50° C. Longer primers have a higher anneal temperature, and so tend to be morespecific to the desired target sites on the DNA molecule. (You need to calculate the correct annealing temperature for your specific primers (or use the information that came with the primers.)The primers find their way by kinetic action to the target segment on the DNA….the complement of the primers. Since the primers are much shorter than the genomic DNA fragments, and are there in such high concentrations, they will find their way to the target sites much more efficiently than the longer complementary strands of genomic DNA. The time needed for annealingcan be 15 seconds to a few minutes.


5’CCGATTACGT3’

3’GATTCCGACG5’

50° C Annealing Temperature

Primer (purple) anneals at targetsite on the DNA strand

Cycle 1


Take another minute to be sure that the primers are the complementary sequence to thetarget on the DNA.


5’CCGATTACGT3’

3’GATTCCGACG5’


Cycle 1


It is possible that the primers could anneal at the wrong location on the DNA. First, there couldbe another place in the genome where approximately that particular 20+ base sequence occurs(the primers are a few bases “off”). This is called misprimingand can occur when the anneal temperature is too low. If the temperature is low, then the annealedprimers do not have to be very stable….they won’t pop off. At higher temperatures, the connectionbetween primer and DNA has to be stable, and therefore has to match very well. This is why ahigher annealing temperature assures a more specific match between primer and the target DNA.Can you think of reasons why we might want to have a LOW annealing temperature to increaseMispriming?? Perhaps we had poor information to use to design primers, so we try to “push”the reaction and get band(s).


5’CCGATTACGT3’

3’GATTCCGACG5’


Cycle 1



5’CCGATTACGT3’

3’GATTCCGACG5’

We now want to encourage the strand of DNA to elongate…that is, the short strand (thePrimer) must now elongate to produce a complete complementary strand. That strand will elongate fromits 5’ to its 3’ end. To elongate, it needs bases to add. These are the ATCG bases, andare called the dNTP’s. Let’s add equal amounts of A,T,C and G dNTP’s to the mix.

Cycle 1



5’CCGATTACGT3’

3’GATTCCGACG5’

GA

T

C

Shown here are only one A, one T, one C, and one G, but there are vast numbers added.Where do we get the primers and dNTP’s? We don’t have to make them ourselves,but instead we buy them from one of many vendors. But, what makes the short strand (only the primers so far) begin to attach the right dNTP’s?We need an enzyme that will encourage this process. This is a “DNA polymerase.”

Cycle 1



5’CCGATTACGT3’

3’GATTCCGACG5’

GA

T

C

There are many kinds of DNA polymerase available. One vital feature is that the enzyme mustnot be damaged very much by the high temperature of the melt stage (the 95° C). The mostcommon enzyme used is “taq” named for the species of hot springs bacteria from which it washarvested. There are others, such as “Vent” that are used. Vent is more stable, and does someProof-reading to double check for errors when the strand elongates. Vent is more costly, andalso does not produce product that is very good for cloning into bacteria. So, taq is the mostpopular.We will show the enzyme as a V here.

Cycle 1



5’CCGATTACGT3’

3’GATTCCGACG5’

GA

T

C

The enzyme sits at the elongation site and encourages the process (and sometimes will “proofread” to be sure the right base is added depending on the enzyme used).

The strand will elongate from its 5’ to 3’ end…so let’s watch this process.

V

V

Cycle 1



5’CCGATTACGTCCTA3’

3’GATCGATTCCGACG5’

Go to next slides to watch the strand elongate.

The elongation temperature is typically 72° C…the optimal temperature for taq enzyme.

V

V72° C -- Elongation

Cycle 1



5’CCGATTACGTCCTA3’

3’GATCGATTCCGACG5’

At 72° the primers would pop off (the annealing temperature used is usually the highest possiblewithout causing the primers to pop off. Why do they not pop off when the temperature is raisedfor elongation? The increase in temperature is fairly slow….so bases are already being added, albeit slowly, but just enough are being added to keep the strand attached. By the time theTemperature reaches 72° the elongating strand is long enough to stay put! If you are ampinga fragment that has a high AT content, such as mitochondrial DNA, you should use a lowerextension temperature, perhaps 65 or 68.

V

V72° C -- Elongation

Cycle 1



5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAAT3’

Go to next slides to watch the strand elongate.

V

V

72° C -- Elongation

3’ATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’

Cycle 1



5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’

There, the process has been completed.

Notice that we have replicated the target sequence (the one shown in red and blue), but weHave also replicated part of the nontarget sequence. Do you see where? Go to next slideTo see the replicated non-target part of the DNA molecule.

.

72° C -- Elongation

3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’




The bars show the part of the DNA molecule that was replicated that was not target.





That was the first cycle of our PCR reaction. Let’s look again at the strands. Our two originalstrands are shown at top and bottom of the four. The red is the target DNA we wish to amplify.The blue is the site where the primer will anneal. The two middle strands are the ones thatwere just produced by the taq polymerase. We have the primers that were added (shown in purple) and the new strands are shown in black.


End of Cycle 1




.


Let’s now change the color of the primers that were added to show they are no longer“primers” but part of the new DNA molecule.




.


The entire new strand is shown in black. On our two original strands we still show the targetsequence in red and the primer sites (part of the target sequence) in blue.




.


It will be easier from this point to convert the DNA strands we have here from ATCG formatto lines showing the segments of the DNA of interest in different colors.So, the target sequence will continue to be shown in RED, and the rest of the molecule willbe shown in BLACK. Just keep in mind that the primer sites are part of the target sequenceand there are two different primers present. Let’s see what that will look like...

3’5’

.This may make it easier to follow the strands from this point. Notice that each molecule hasan overhang of unwanted (nontarget) DNA at one end and a double strand of unwanted DNAat the other.

Let’s call this kind of molecule an “A” molecule….one with unwanted DNA at each end, oneend double stranded and the other a single strand overhang.

5’3’

3’3’5’

5’

Results from Cycle 1

3’5’

.We now do the whole thing over.

There is still lots of the the dNTP’s left, lots of primers left, and lots of taq. The taq does not get used up (it’s an enzyme!), but the heating and cooling does start to wear it out. after 35 cycles, about only half of the taq will still function, but still there is plenty.Start the process over by raising the temperature back up to 95° C.

5’3’

3’3’5’

5’

3’

5’

.The two DNA molecules denature and we have four strands.

5’

3’

3’

3’5’

5’

95° Melt

Cycle 2 begins

3’

5’

.Lower the temperature to the annealing temp and the primers will attach.

5’

3’

3’

3’5’

5’

50° Annealing temperature

3’

5’

.Raise the temperature, and the strands elongate.

5’

3’

3’

3’5’

5’

72° Elongation

Cycle 2

3’

5’

.Here is the final result…we now have four DNA molecules. In two cycles we went fromone to four, so we are doubling the DNA with each cycle. This is why it is called thechain reaction, because each cycle builds on the previous cycle, a kind of chain reaction. Thus, the name of Polymerase Chain Reaction, or PCR.

5’

3’

3’

3’5’

5’

72° Elongation

Result of Cycle 2

3’

5’

.Notice that you still have an extra overhang of DNA on each segment that you haveamplified. Let’s look at the 4 DNA molecules. We have 2 of the A types, so we have not increased theirnumber (we had 2 A’s at the end of Cycle 1), and we now have two molecules with unwantedDNA only at one end, and it is a single stranded piece. Let’s call that the “B” type of DNAmolecule. Notice that we don’t have any of our pure target DNA (no pure red strands).

5’

3’

3’

3’5’

5’

72° Elongation

Result of Cycle 2

3’

5’

.Melt...

5’

3’

3’

3’5’

5’

Cycle 3 begins

3’

5’

.Anneal and elongate to get this result….We now have 8 DNA ds molecules. Because of spaceLimitations, sometimes a double stranded piece of DNA will show up as a wide line (sorry).

Let’s look at the kind of strands present. We have 2 A type (no increase), 4 B’s (so they havedoubled), and a new kind of segment, one that is double stranded pure target. That is ourgoal! Call these two DNA molecules the Targets.

5’

3’

3’

3’5’

5’

Results of Cycle 3

3’

5’

.The two targets will replicate each cycle, so we have 2 here at the end of Cycle 3,and will have 4 at the end of Cycle 4, 8 at the end of cycle 5, and so on. These two strandswill be the ancestor of millions of strands after 35 or so cycles. And, all of those strands willbe the exact sequence you wanted to amplify. But, what happens to the other 6 strands here?Let’s put our two “target” fragments off-screen (but whatever is off-screen will double eachCycle) to make it easier to follow the future results.

5’

3’

3’

3’5’

5’

Results of Cycle 3

3’

5’

.At the end of Cycle 3 we have gone from 1 to 8 DNA molecules:Again, let’s keep track….2 are Target2 are A’s with nontarget DNA at both ends4 are B’s with nontarget DNA at one end

5’

3’

3’

3’5’

5’

Results of Cycle 3

3’

.

5’

Results of Cycle 3

5’5’ 3’

We will keep track of only one of each of these kinds of segment, so the others will be putoff-screen along with the Target sequences we just put away. But, whatever happens to oneof the A’s or B’s happens to the others as well.

Run another cycle….

2 A type DNA segments

4 B type DNA segments

3’

.

5’

Results of Cycle 4

5’ 3’

Each A type molecule produces one A and one B.Each B type molecule produces one B and one Target.

Our total after 4 cycles is 16 molecules and these are:8 of the target DNA segments2 A’s6 B’s

3’

.

5’

Results of Cycle 4

Again, here is what we have after 4 cycles.

Do you see the pattern emerging? After each cycle there are still a total of 2 A DNA molecules,two more B’s, and more than doubling of the Targets. As the cycles go on, the Targets willdouble (with a few more added from the B’s), the B’s will increase only by 2 each cycle, andthe A’s will stay the same.

So, continuing this..for the next cycle

2 of these

6 of these

8 of these

3’

.

5’

Results of Cycle 5

For 5 cycles we have a total of 32 molecules, 22 of the 32 are the target.

Let’s predict how many A and B DNA molecules will result from 35 cycle:A’s = Still a total of 2!B’s = 34 x2 = 68. Why multiply by 34? Because the first 2 show up only at cycle 2, then2 more are added with each cycle.Targets = Billions!!…see more below

2 of these

8 of these

22 of these

3’

.

5’

Results of Cycle 6

After 6 cycles there are 64 copies, and 52 of them will be target….

2 of these

10 of these

52 of these

3’

.

5’

Results of Cycle 7


2 of these

12 of these

114 of these

3’

.

5’

Results of Cycle 8


2 of these

14 of these

240 of these

3’

.

5’

Results of Cycle 9


2 of these

16 of these

494 of these

3’

.

5’

A = 2B = 18Target = 1,004

Target

A

B

Cycle 10

3’

.

5’

A = 2B = 20Target = 2,026

So, with future cycles, a very, very small proportion of the product will be the nontargetDNA. So, we can state the number of products by just doubling each cycle…1, 2, 4, 8 and soon for 35 cycles.

Target

A

B

Cycle 11

After 20 cycles, there will be

1.049 million copies of the target DNA segment

(Remember that each copy of that segment in the genomicDNA is being copied….from all the cells used in theoriginal extraction)


1.073 BILLION copies of that target DNA


34 BILLION copies of the target

35 cycles is a typical number in many PCR programs.

So, at the end of the reaction, there will be some genomicDNA left in the vial (which was never amplified becausethe primers were specific for one small part of the genome),and some of the segments with nontarget ends. But, thetarget DNA will be there in vast numbers….34 billion forevery single one that was in the reaction at the start.

The PCR Program (set on the thermal cycler)

Predwell of 94° for 1 or 2 minutesAll of the Genomic DNA denatures to get a good, clean

start to the reaction (not needed for 2nd PCR in nested)

35 Cycles of:

Melt of 94° for 30 secondsAnnealing at 50° for 30 to 60 secondsExtension at 72° for 30 to 90 seconds (shorter for short DNA)

Postwell of 72° for 2 minutes to 10 minutes to get nice produce, especially needed when cloning the product in bacteria

Cold hold at 4° C until put into the refrigerator or freezer

Optimizing the PCR Program

• Most important is the annealing temperature

T = 4 (G+C) + 2 (A+T)

Is a good estimate, but on-line programs can do a moresophisticated analysis and give a more precise temperature.

Start low….mispriming, but at least will usually work, thenincrease until get optimal yield (to high will give no resultsbecause the primers cannot sit down in a stable way to beginthe reaction). Finding the optimal temperature can be done on a gradient cycler in a single PCR run.

Optimizing the PCR Program

Some tricks to get better product by altering the annealingTemperature:

• Run at high temp for first few cycles to get some perfectMatches, then run at low temperature for maximum efficiency.

• Or, do a High-Stringency Bounce. If primers match perfectlyand have a high annealing temperature, and if the DNA beingamplified is short (< 800 bp), can bounce from the melt toannealing and extension very quickly with only a 15 secondpause at each step. This is a quick program and often producesA` very clean, single product.

Optimizing the PCR Mix:

• MgCl2 is added typically at 1.5 mM, but can increase thisSometimes to get better results. The Mg is a cofactor for theTaq enzyme. It is not used up, but can become inactivated byThe presence of substances in the extracted DNA buffer. TheHighest concentration that would work is about 6 mM.

• Be sure that the primers are in high concentration.

• Add some “PCR Helper” such as 0.1% to 1% BSA (BovineSerum Albumin) or Gelatin, or DMSO, or Glycerol. This Sometimes increases the efficiency by 10% and gives a 15xIncrease in the PCR product!

Genomic DNA Concentration

• Too low…no product• Too high…no product. Too much DNA is more often a problem that too little! Often need to cut the DNA by factorof 1:10 or even 1:100. Watch for DNA remaining in the wellsof the gel…a sign that there was too much present, and too much for the primers to work properly

The End!

the art and science of pcr jos. j. schall department of biology university of vermont

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