Polyermase Chain Reaction Question

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ferbus
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Polyermase Chain Reaction Question

Post by ferbus » Sun Feb 11, 2007 9:12 pm

I am not sure how to begin this question; can anyone give me some help?

You are interested in amplifying the target DNA sequence shown in bold font below:


5’ACATAAATCATGTATACGAATTCTGGCACGTGAGTTCAACTTGTTGATAGTTT3’


3’TGTATTTAGTACATATGCTTAAGACCGTGCACTCAAGTTGAACAACTATCAAA5’

Which set of primers would you use for this purpose?

Set a) 5’ ATACATGATTTAT 3’ and 5’ TAGTTGTTCAAC 3’

Set b) 5’ ATACATGATTTAT 3’ and 5’ ATCAACAAGTTG 3’

Set c) 5’ ATAAATCATGTAT 3’ and 5’ ATCAACAAGTTG 3’

Set d) 5’ ATAAATCATGTAT 3’ and 5’ TAGTTGTTCAAC 3’

2.) Identify the forward and reverse primers

3.) Show the sequence on the DNA of interest where the primers could be bound.

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Post by MrMistery » Mon Feb 12, 2007 11:35 am

well what you need is to determine a set of primers that will each be complementary to the 3' ends of the DNA, since the polymerase moves from 5 to 3.. not to hard...
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Post by Agronomy » Thu Feb 15, 2007 6:18 pm

I'm confused, because none of the answers seem right to me, shouldn't the forward be

3'TGCTTAAGACCG,etc 5'
and

the other one be

5'ACGAATTCT,etc 3'

so lost on this problem...

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Post by MrMistery » Thu Feb 15, 2007 6:22 pm

Yes they should if you ask me. Maybe they gave you the answer choices for another exercise :lol:
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Post by Agronomy » Thu Feb 15, 2007 7:02 pm

no they didn't give the answer choices for another exercise because this is a laboratory question, so it has to be one of those answers..any other ideas?

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Post by Agronomy » Wed Feb 21, 2007 2:52 am

please can someone help me, I really don't know how to answer this question, I"ve tried to think about it, the answer is ONE of the choices above!!

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Post by Vagabond » Wed Feb 21, 2007 2:38 pm

Hey. You can be kinda funny with this one and tell them that the sequence to amplify is so short that you would just synthesize it LOL

Ok really here ya go

You ds DNA molecule is this:

5’ACATAAATCATGTATACGAATTCTGGCACGTGAGTTCAACTTGTTGATAGTTT3’
3’TGTATTTAGTACATATGCTTAAGACCGTGCACTCAAGTTGAACAACTATCAAA5’

Taq moves 5’ to 3’ thus you can use the following:


5’ACATAAATCATGTATACGAATTCTGGCACGTGAGTTCAACTTGTTGATAGTTT3’
---------------------------------------------------------------------3' GTTGAACAACTA 5'
5’ ATAAATCATGTAT 3’
3’TGTATTTAGTACATATGCTTAAGACCGTGCACTCAAGTTGAACAACTATCAAA5’

so Set C looks to be the one you need
---------------------------------------

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Post by Agronomy » Wed Feb 21, 2007 3:37 pm

ohh, okay that makes sense! But how do we know what the differnece is between the forward and ther everse primers?

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Post by Vagabond » Wed Feb 21, 2007 3:48 pm

It is usually based off the location of the ATG start codon...
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Post by Agronomy » Wed Feb 21, 2007 3:51 pm

so which one of the sequences in set C would be the foward primer? neither one of them have that codon in them?

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Post by Vagabond » Wed Feb 21, 2007 3:58 pm

ok, since the above sequence has not given us the start codon I would say that one would call the primer 5’ ATAAATCATGTAT 3’ the forward and the other the reverse simple because I read from right to left . . .
---------------------------------------

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Coincidence? I think not. . .

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Post by Agronomy » Wed Feb 21, 2007 4:04 pm

that makes sense, and for the last question I'm just confused cuz, on the 3'-5' sequence the part where the primer is bound is the TATTTAGTACATA, but for teh reverse primer it's bound on the opposite side! So, what should I put for number 3?

I really appreciate your help, i'm starting to undersatnd how to do this...

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