Question

First off thanks for helping me!
Please consider the 45 base pair:

5’-CGCACCTGTGTTGATCACCTAGCCGATCCACGGTGGATCCAAGGC-3’
|||||||||||||||||||||||||||||||||||||||||||||
3’-GCGTGGACACAACTAGTGGATCGGCTAGGTGCCACCTAGGTTCCG-5’

Tip: If you want to see a review of PCR, we recommend this amazing animation and virtual lab made by the University of Utah.

You also have a collection of short DNA primers:

Primer 1: 5’-CGTGGA-3’
Primer 2: 5’-TGTGTT-3’
Primer 3: 5’-ATCCAA-3’
Primer 4: 5’-CCTTGG-3’
You put together several reaction mixtures, run them in the thermocycler under conditions that should produce a product if the reaction mixture is correct.

All mixtures contain:

5’-CGCACCTGTGTTGATCACCTAGCCGATCCACGGTG
GATCCAAGGC-3’

The 45 bp DNA fragment
DNA polymerase that can function at high temperature
The necessary buffer conditions
For each mixture, indicate whether or not the PCR reaction would produce a product. If a product would be produced, give the length of that product in base-pairs (bp); if no product would be produced, give zero “0” for the length.



1
Added: dATP, dGTP, dCTP, Primer 2, Primer 4
- unanswered

2
Added: dATP, dGTP, dCTP, dTTP, Primer 2
- unanswered
3
Added: dATP, dGTP, dCTP, dTTP, Primer 2, Primer 4
- unanswered
4
Added: dATP, dGTP, dCTP, dTTP, Primer 1, Primer 3
- unanswered
5
Added: dATP, dGTP, dCTP, dTTP, Primer 2, Primer 3

Answer 1

So you need see how far the strand will be copied.

For example:

1. Added: dATP, dGTP, dCTP Primer 2, Primer 4

There is no dTTP. Meaning the sequence will be copied UNTIL we encounter a T.

We need to consider both primers AT BOTH ENDS.

Primer 2:

5’-TGTGTT-3’

5’-CGCACCTGTGTTGATCACCTAGCCGATCCACGGTGGATCCAAGGC-3’

no match

5’-TGTGTT-3’
3’-GCGTGGACACAACTAGTGGATCGGCTAGGTGCCACCTAGGTTCCG-5’


Fragment: 5'- TGTGTTC-3'

Primer 4: 5’-CCTTGG-3’

5’-CGCACCTGTGTTGATCACCTAGCCGATCCACGGTGGATCCAAGGC-3’
3’-GGTTCC-5'

Fragment: 5’-CCTTGG-3’ (no extension because T is next)

3’-GCGTGGACACAACTAGTGGATCGGCTAGGTGCCACCTAGGTTCCG-5’

no match

Answer 2

I'm sorry, so when you say no match… what exactly does that mean? I don't want to sound rude, I just don't fully understand that part.

Answer 3

By no match, i mean that there is no exact complementary sequence on the strand.

Answer 4

thank you for clearing that up. so does that mean that there is no possible product stead or target sequence that can b created?

Answer 5

For that one strand, no. As you can see though, on the other (complementary) strand, there is a match.

Answer 6

oh gosh. i am so so sorry. i must seem so dumb because you clearly know what you're doing! but um didn't you actually find matching strands for both?

" 5’-TGTGTT-3’
3’-GCGTGGACACAACTAGTGGATCGGCTAGGTGCCACCTAGGTTCCG-5’

and

5’-CGCACCTGTGTTGATCACCTAGCCGATCCACGGTGGATCCAAGGC-3’
3’-GGTTCC-5'

please let me know if i misunderstood. and i agree that these are the primers, but how is the target sequence deduced from the primers location? once again, sorry for bothering you, but you seem very knowledgeable in the subject

Answer 7

There was only 1 strand that matched each primer Note that I did one primer at a time for clarity. The top strand has primer 2 and the bottom has primer 4.

So yes, in part 1 each primer matched one of the strands.

The sequence produced (extended) is deduced by going towards 3' (because Pol synthesizes the strand 5' to 3'). So just follow the 3' end until you hit a "T" (for part 1.)

Answer 8

last two questions. (p.s. I'm giving you a great review. you've had more patience with me than my tutor has had so far!)

1. from which primer do i go till i hit a t. bc if its from primer two the strand would be 7 or 2 nucleotides long <- are either of those values correct (do you count the primer's nucleotides when you're counting the # of bases?) but if i start at primer 4 it would be 6 or 0 (again this depends on if i count the primer's nucleotides)

2. and how do i know when to stop when dATP, dTTP, dGTP, and dCTP are added?

The best part about you is that you didn't give me the answers, you are teaching me the process so I can answers the others and gain the knowledge. Thank you again so much aaronq!!!!!! I wish more teachers and tutors could be like you!

Answer 9

Thanks for the kind words! I do try to not give answers because i feel it actually hinders your education. Anyways, for 2, the strand would be extended to completion (till the end) because all the deoxyribonucleotides are present.

For 1. you need to look at the strand direction. The extension is always 5' to 3'.

5’-TGTGTT-3’ ----> extension
3’-GCGTGGACACAACTAGTGGATCGGCTAGGTGCCACCTAGGTTCCG-5’

I screwed up a little bit, the first fragment should be (this is probably why you got confused)

Fragment: 5'- TGTGTTCGA-3'


5’-CGCACCTGTGTTGATCACCTAGCCGATCCACGGTGGATCCAAGGC-3’
extension <--- 3’-GGTTCC-5'

Fragment: 5’-CCTTGGA-3’

Answer 10

Correction*** first fragment is: 5'- TGTGTTGA-3'

Answer 11

of course! You really deserve it!
1. just to make sure i fully understand, you stop when you reach a T or when you reach an A that will be transcribed to a T?

2. and when you say for strand 2 it will be extended to the end, does that mean till it reaches the next primer? or literally the end of the sequence?

3. and finally, from your work am i to understand that the primer's nucleotides ARE included when counting the length (in bp) of the final product?

Answer 12

1. yes, you stop when you have to insert a T, meaning that you reached an A on the strand being copied.

2. Not for strand 2, for question 2. When you have all dNTPs present.

For example,

5’-TGTGTT-3’ ----> extension
3’-GCGTGGACACAACTAGTGGATCGGCTAGGTGCCACCTAGGTTCCG-5’

you'll get the strand until the end, 5’-TGTGTT...AAGGC-3’

3. I assume you would, I dont see why the primers wouldn't be counted. If were to electrophorese this on a gel, it would be characterized as the whole size, and would not exclude the primer.

Answer 13

Thank you so much again! I'm going to go ahead an right that review right away so I remember how helpful you are!

Answer 14

no problem! if you have other questions dont hesitate to ask!