zero ds target molecules 4 the 1st 2 cycles in PCR - WHY???? (Jul/24/2007 )
i have read the following :-
PCR amplification of DNA fragments.
cycle no. no. of double stranded target molecules
1 0
2 0
3 4
4 8
my qs is,why in the 1st two cycles i have no ds target molecules????
can i find any explanation??? 
i have looked for this in many books but no one mentions s.th like this...!!!!
Try to imagine what happens in the first two cycles of PCR:
First cycle - you have your source DNA (ds) and no product yet. The source DNA gets denatured, so you have the forward and the backward strand. Next step - annealing of the primers - one primer binds to each strand (F-Primer to reverse strand, R-Primer to forward strand). After the elongation, you end up with the original source strands that have a small double-stranded part - the primer binding site + the newly synthetized fragment that is normally longer than your target!
Second cycle - denaturation - you have one long and one short forward strand and one long and one short reverse strand. Primer annealing to each site and elongation - you end up with: 2 source strands, again with a short double stranded part (as after cycle 1) and two shorter strands with a double stranded part that has the exact length of your target. But still, no double-stranded target molecule, as all four DNA molecules have different F and R strand lengt. During the third cycle, you have for the first time primers annealing to your target moelcule (ss) and an elongation just until the end of the target molecule, which leaves you with ... tatatata ... your ds target molecule! If you count closely what kind of molecules entered the third round, you will easily see that you should leave the third round with: 4 ds target DNA molecules + the results from cycle 2.
Does this answer your question? If not, try to make a drawing of it ...
Krümel
And btw - I move this post to the Homework Questions ... it fits better!
thanks alot krümelmonster 
i really don't know how to thank you 
correct me if am wrong,my PCR device starts counting dsDNA amplicons that are totally new i.e no parental strand is present.
am i right??
i have a q.,and forgive me if it seems silly...
u've said that i will end up having a SMALL ds part,then u said that the newly synthesized fragment is LONGER than my target...
do you mean that no matter how long the newly synthesized fragment is,it remains as a small portion regarding to the whole DNA sample i have ???
i added wrong values in the table
cycle no. no. of ds target molecules
1 0
2 0
3 2
4 4
5 8
let me say that the PCR considers cycle no. 3 as the first cycle and then starts exponentially (2^n)
in my drowing i got 4 ds target DNA molecules at the end of cycle 3,so why in the table isn't like this ???
thanks a bunch
am i right??
in my drowing i got 4 ds target DNA molecules at the end of cycle 3,so why in the table isn't like this ???
thanks a bunch
I'attach a drawing I made. You will find exatly 2 dsTarget molecules.
thanks alot krümelmonster
am so sorry for bothering you...
am so grateful
first : i would like to explain my opinion about the counting issue.. ( kindly find the attached figure )
and plz correct me,
PCR starts counting primer-defined amplicons,as written under the figure.is this right??
what i mean that PCR considers cycle 3 as cycle 1,is regarding to the formation of the dsDNA target molecules
because in the 1st 2 cycles the count is zero...am not refering to any scientific concept,and am not ignoring the role of the first 2 cycles...what i mean is;
cycle 1 0
cycle 2 0
cycle 3 2 DNA target strands
cycle 4 4 DNA target molecules
cycle 5 8 DNA target molecules
now,in order to apply the equation,
no. of target DNA molecules= 2^n,n= no. of cycle
i consider cycle 3 as no. 1 .... to make this equation applicable
my questions are,
what is the importance of having anchored DNA ( or semi-bounded DNA ) during my PCR reaction???
as i understand the PCR counts the ds green lines ( primer defined amplicons) ONLY,so why do they ( anchored) exist??
and why they are semi-bounded ???
kindly accept my apology for causing any troubles
thanks..
[attachment=3311:Figure_7.doc]
Nightingale, the PCR counts every newly synthetized DNA, i.e. also in the first two round you have doubling of your target sequence! The numbers in your table are not realted to what the PCR counts but to the amount of double-stranded target molecules synthetized. And ho sould the bound or semi-bound DNA vanish? It needs to be there if you want the reaction to start?
The quotation 2^n is only valid when you accept the fact that also the bound or semi-bound DNA counts for the quantification.
Sorry, I am not able to make this more clear, I think you have mixed two different things up: the amount of ds amplicon (green) is what you find in your table. The amount of synthetized dsDNA is what quantifies the PCR.
exactly...
am really so sorry,i feel that i made you mad...
am really so grateful to you...
finally,i understood..
THANK YOU SOOOO MUCH.
Hi Nightingale, no you didn't make me mad. It is only sometimes hard to cope with the fact that I am not able to find the correct words for what I want to tell and than it is more confusing. I am glad, that I could help you 