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The Art and Science of PCR Jos. J. Schall Department of Biology University of Vermont

The Art and Science of PCR Jos. J. Schall Department of Biology University of Vermont. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’.

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The Art and Science of PCR Jos. J. Schall Department of Biology University of Vermont

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  1. The Art and Science of PCR Jos. J. Schall Department of Biology University of Vermont

  2. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ You have extracted DNA from a tissue sample. There were thousands of cells in that tissue, so each chromosome is represented in the extracted DNA thousands of times. To follow what happens during the PCR process, we will follow only one segment of the DNA -- one double strand. It is shown 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 of the strand that is elongating. Again, always keep in mind that we are following what happens to a single copy of the DNA molecule of interest, but there will be thousands present.

  3. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ By the way….suppose this DNA segment is from the mitochondrion’s genome. Most cells contain thousands 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 haploid cell such as Plasmodium) or twice (for cells of diploid organisms). Do you see why the Mitochondrial DNA will often give better results in PCR?

  4. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ You wish to amplify (a billion-fold) the segment of DNA shown in RED on the strand. Why that particular segment? It may be the section of a gene that you wish to sequence, or some segment that includes a repeat region (a microsatellite), or a segment that you use to find a parasite in a tissue sample, or a hundred other reasons. The beauty of PCR is that it will amplify a specific segment 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 show only a short segment (to fit on the screen!).

  5. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ Your first job is to design a pair of primers at the ends of the target strand of DNA. Primers are typically around 20 bases long. Because the strand shown above is so short, for purposes of illustration here, the primers will be shown shorter than that. In practice, such short primers would not be used, because they would not be very stable when they anneal to the target DNA, and may also match regions on non-target parts of the genome, so you would get a mixed product at the end.

  6. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ The primer sites are shown above in BLUE. Note that the primer sites that are shown in blue 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, the primer will be the complement of the segment where you want it to sit down.

  7. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 95° C -- strands separate 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ 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.

  8. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ Cool down, and strands anneal 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ If you cool the DNA back down again…the strands will come back together….the complementary strands will re-anneal. Because they are so long, it may take a long time for this to happen, and there could be lots of errors when these long strand try to re-anneal.

  9. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 95° C for 1 minute 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ 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 slightly by the high temperature, and is damaged more over longer melt times. So, the best melt is a temperature which allows all the DNA to denature, but not too high to cause much damage to the polymerase.

  10. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ In the mixture are the primers. The two primers are shown in PURPLE. Remember that the site where the primers will sit down are indicated in blue. Note that one primer is the complement of one 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 the other long strand of DNA at its complementary site.

  11. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ The primers should not be complements to each other (just to the part of the strands where you want them to sit down. If they are self-complements, they could anneal with each other and reduce the amount of free primers available to sit down at the desired sites on the DNA molecule.

  12. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ 50° C Annealing Temperature 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ Primer (purple) anneals at target site on the DNA strand The temperature is lowered to the “Annealing Temperature” that varies with the primers, but is typically about 50° C. Longer primers have a higher anneal temperature, and so tend to be more specific 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 annealing can be 15 seconds to a few minutes.

  13. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ 50° C Annealing Temperature 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ Take another minute to be sure that the primers are the complementary sequence to the target on the DNA.

  14. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ 50° C Annealing Temperature 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ It is possible that the primers could anneal at the wrong location on the DNA. First, there could be another place in the genome where approximately that particular 20+ base sequence occurs (the primers are a few bases “off”). This is called mispriming and can occur when the anneal temperature is too low. If the temperature is low, then the annealed primers do not have to be very stable….they won’t pop off. At higher temperatures, the connection between primer and DNA has to be stable, and therefore has to match very well. This is why a higher 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 increase Mispriming?? Perhaps we had poor information to use to design primers, so we try to “push” the reaction and get band(s).

  15. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ We now want to encourage the strand of DNA to elongate…that is, the short strand (the Primer) must now elongate to produce a complete complementary strand. That strand will elongate from its 5’ to its 3’ end. To elongate, it needs bases to add. These are the ATCG bases, and are called the dNTP’s. Let’s add equal amounts of A,T,C and G dNTP’s to the mix.

  16. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ C A G T 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ 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.”

  17. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ C A G T 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ There are many kinds of DNA polymerase available. One vital feature is that the enzyme must not be damaged very much by the high temperature of the melt stage (the 95° C). The most common enzyme used is “taq” named for the species of hot springs bacteria from which it was harvested. There are others, such as “Vent” that are used. Vent is more stable, and does some Proof-reading to double check for errors when the strand elongates. Vent is more costly, and also does not produce product that is very good for cloning into bacteria. So, taq is the most popular. We will show the enzyme as a V here.

  18. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGT3’ V C A G T V 3’GATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ 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.

  19. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTA3’ V 72° C -- Elongation V 3’GATCGATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ Go to next slides to watch the strand elongate. The elongation temperature is typically 72° C…the optimal temperature for taq enzyme.

  20. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTA3’ V 72° C -- Elongation V 3’GATCGATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ At 72° the primers would pop off (the annealing temperature used is usually the highest possible without causing the primers to pop off. Why do they not pop off when the temperature is raised for 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 the Temperature reaches 72° the elongating strand is long enough to stay put! If you are amping a fragment that has a high AT content, such as mitochondrial DNA, you should use a lower extension temperature, perhaps 65 or 68.

  21. Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAAT3’ V V 3’ATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ 72° C -- Elongation Go to next slides to watch the strand elongate.

  22. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’ 3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ 72° C -- Elongation There, the process has been completed. Notice that we have replicated the target sequence (the one shown in red and blue), but we Have also replicated part of the nontarget sequence. Do you see where? Go to next slide To see the replicated non-target part of the DNA molecule. .

  23. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’ 3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ The bars show the part of the DNA molecule that was replicated that was not target.

  24. End of Cycle 1 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’ 3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ That was the first cycle of our PCR reaction. Let’s look again at the strands. Our two original strands 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 that were just produced by the taq polymerase. We have the primers that were added (shown in purple) and the new strands are shown in black.

  25. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’ 3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ . 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.

  26. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’ 3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ . The entire new strand is shown in black. On our two original strands we still show the target sequence in red and the primer sites (part of the target sequence) in blue.

  27. 3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’3’TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCGCACGTCAGATCG5’ 5’CCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGCGTGCAGTCTAGC3’ 3’ TTAGCCTAGGCTAATGCAGGATCCGCTTGCTACTAGCTATCCTTTAGCTCCTCCCTAGCATTCCGACG5’ 5’AATCGGATCCGATTACGTCCTAGGCGAACGATGATCGATAGGAAATCGAGGAGGGATCGTAAGGCTGCAGTCTAGC3’ . It will be easier from this point to convert the DNA strands we have here from ATCG format to 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 will be shown in BLACK. Just keep in mind that the primer sites are part of the target sequence and there are two different primers present. Let’s see what that will look like...

  28. 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ Results from Cycle 1 . This may make it easier to follow the strands from this point. Notice that each molecule has an overhang of unwanted (nontarget) DNA at one end and a double strand of unwanted DNA at the other. Let’s call this kind of molecule an “A” molecule….one with unwanted DNA at each end, one end double stranded and the other a single strand overhang.

  29. 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ . 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.

  30. Cycle 2 begins 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ 95° Melt . The two DNA molecules denature and we have four strands.

  31. 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ 50° Annealing temperature . Lower the temperature to the annealing temp and the primers will attach.

  32. Cycle 2 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ 72° Elongation . Raise the temperature, and the strands elongate.

  33. Result of Cycle 2 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ 72° Elongation . Here is the final result…we now have four DNA molecules. In two cycles we went from one to four, so we are doubling the DNA with each cycle. This is why it is called the chain reaction, because each cycle builds on the previous cycle, a kind of chain reaction. Thus, the name of Polymerase Chain Reaction, or PCR.

  34. Result of Cycle 2 3’ 5’ 5’ 3’ 3’ 5’ 5’ 3’ 72° Elongation . Notice that you still have an extra overhang of DNA on each segment that you have amplified. Let’s look at the 4 DNA molecules. We have 2 of the A types, so we have not increased their number (we had 2 A’s at the end of Cycle 1), and we now have two molecules with unwanted DNA only at one end, and it is a single stranded piece. Let’s call that the “B” type of DNA molecule. Notice that we don’t have any of our pure target DNA (no pure red strands).

  35. 5’ 3’ Cycle 3 begins 3’ 5’ 5’ 3’ 5’ 3’ . Melt...

  36. Results of Cycle 3 5’ 3’ 3’ 5’ 5’ 3’ 5’ 3’ . Anneal and elongate to get this result….We now have 8 DNA ds molecules. Because of space Limitations, 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 have doubled), and a new kind of segment, one that is double stranded pure target. That is our goal! Call these two DNA molecules the Targets.

  37. Results of Cycle 3 5’ 3’ 3’ 5’ 5’ 3’ 5’ 3’ . 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 strands will be the ancestor of millions of strands after 35 or so cycles. And, all of those strands will be 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 each Cycle) to make it easier to follow the future results.

  38. Results of Cycle 3 5’ 3’ 3’ 5’ 5’ 3’ 5’ 3’ . At the end of Cycle 3 we have gone from 1 to 8 DNA molecules: Again, let’s keep track…. 2 are Target 2 are A’s with nontarget DNA at both ends 4 are B’s with nontarget DNA at one end

  39. Results of Cycle 3 5’ 3’ 5’ 3’ 5’ 2 A type DNA segments 4 B type DNA segments . We will keep track of only one of each of these kinds of segment, so the others will be put off-screen along with the Target sequences we just put away. But, whatever happens to one of the A’s or B’s happens to the others as well. Run another cycle….

  40. Results of Cycle 4 3’ 5’ 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 segments 2 A’s 6 B’s

  41. Results of Cycle 4 3’ 5’ 2 of these 6 of these 8 of these . 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 will double (with a few more added from the B’s), the B’s will increase only by 2 each cycle, and the A’s will stay the same. So, continuing this..for the next cycle

  42. Results of Cycle 5 3’ 5’ 2 of these 8 of these 22 of these . 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, then 2 more are added with each cycle. Targets = Billions!!…see more below

  43. Results of Cycle 6 3’ 5’ 2 of these 10 of these 52 of these . After 6 cycles there are 64 copies, and 52 of them will be target….

  44. Results of Cycle 7 3’ 5’ 2 of these 12 of these 114 of these . After 7 cycles there are 128 copies, and 114 of them will be target….

  45. Results of Cycle 8 3’ 5’ 2 of these 14 of these 240 of these . After 8 cycles there are 256 copies, and 240 of them will be target….

  46. Results of Cycle 9 3’ 5’ 2 of these 16 of these 494 of these . After 9 cycles there are 512 copies, and 484 of them will be target….

  47. Cycle 10 3’ 5’ A B Target . A = 2 B = 18 Target = 1,004

  48. Cycle 11 3’ 5’ A B Target . A = 2 B = 20 Target = 2,026 So, with future cycles, a very, very small proportion of the product will be the nontarget DNA. So, we can state the number of products by just doubling each cycle…1, 2, 4, 8 and so on for 35 cycles.

  49. After 20 cycles, there will be 1.049 million copies of the target DNA segment (Remember that each copy of that segment in the genomic DNA is being copied….from all the cells used in the original extraction)

  50. After 30 cycles, there will be 1.073 BILLION copies of that target DNA

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