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Tuesday, January 30, 2007

QUESTIONS THAT CAME UP

1) HOW MUCH INFO FOR FULL MARKS?
It depends. If it’s just a table to fill in with « + » and « -« , then you don’t need to give much info at all.
For restriciton mapping, you will want to show your work (bunches of circles or lines all scribbled with one enzyme or the other…), but make sure that YOU SHOW THE FINAL ANSWER CLEARLY, neatly, and with all the symbols that you invented properly explained/defined.
For experimental procedures, you’ll have to give alll the key info without unimportant details. To identify key info think about things like would the strain of bacteria that i use matter? Does it have to be a specific genotype? If so, indicate it, it’s key info! Would I need to plate on a specific medium? Where would I get mutants from? What is the source of the DNA that i am proposing to clone? All key info! What would I be looking for in my screen?

Some technical info on the various types of blot, on PCR, etc, can be found under the link "techniques ch11 and 12 stuff".




2) HOW TO DO RESTRICTION MAPPING?
First, it’s important to realize that restriction enzymes simply recognize specific sequences and cut the DNA within those sequences (they are like cissors).
Restriction mapping is an ANALYTICAL TECHNIQUE. You always do complete digestions for restriction mapping (unless the enzyme does not work properly and the reaction ends up being incomplete).
In order to determine a restriction map, you need to know the sizes of the restriction fragment produced by your enzyme(s) of interest from your initial DNA molecule. You do this by running the products of the restriction digestion on a gel. You also need to know if the initial DNA molecule is circular or linear.

Short explanation : for each restriction digest, draw a molecule (circle or line) and place restriction sites for the enzyme used. Then look at double digests and try to superimpose the maps for the two corresponding single digests. And so on. It’s a lot of trial and error.

Long explanation :
To analyze the results of a set of restriction digests you start by drawing out your DNA molecule (a line or a circle, depending on the situation) and look at the products of one single restriction digest, let’s say with enzyme RA. If you see one band on your gel, and it is 12kb, and your DNA is circular, then it means that enzyme RA likely has only one site that it recognizes. You can draw it anywhere on your circle.
Then look at the results with enzyme RB, let’s say you see 2 bands, one that is 4kb and one that is 8 kb. It means that RB cuts at 2 sites that are 4kb away from each other. You can draw that on a new circle.
Now you’ll probably find the results of a double digest with RA and RB. Let’s say it gives you 3 bands that are 5kb, 4kb and 3kb, respectively. Look for a band size that is in common with either the single RA digest or the single RB digest…here we have the 4kb one. This suggests that the bands of 3 and 5kb represent a ‘chopped up’ version of the 8kb band that you had in the RB digest. And what could be chopping it up? Probably enzyme RA cuts somewhere within that 8kb region, at a site that is 5kb away from one RB site and 3kb away from the other.


23 Comments:

Anonymous Anonymous said...

Hi Pam!
I was wondering if it's possible to put more than 1 restriction enzyme sites on the same location on the chromosome.

Thanks! :)

3:00 PM  
Anonymous Anonymous said...

hi Pam...I understand how to do this type of restriction mapping, but how do you go about attempting to do a question like number 3 in the practice midterm 2006 that's posted on the course website? There are multiple differnt double digests...I'm not sure how to go about mapping the DNA. Maybe you can walk us through it in a step by step fashion like you did for the restriction mapping. THANKS A LOT!!!

3:09 PM  
Anonymous Anonymous said...

I'm looking on the practice midterm (july 30, 2003)in your package.

How come the undigested DNA is shorter than the others?.. I think it's supposed to be 10kb in length, but how come the undigested DNA is 7? Is it because something's inserted into it?

Thanks :)

~Teresa

3:30 PM  
Anonymous Anonymous said...

Undigested DNA: if it is circular-I don't have the question with me, is it circular?-, it does not run like the linear DNA (it 'looks like' it's shorter because it is more compacted than linear DNA, so it runs 'faster').

Cheers

Pam

4:17 PM  
Anonymous Anonymous said...

Double digests only:
this requires trial and error, but it's similar to the 'regular' mapping.
Try to start with a map derived from the 1st double digest, without specifying which sites are for EcoRI and which ones are for the other enzyme.
Go through each digest this way...at some point you'll find some 'overlaps' that will allow you to locate the EcoRI sites. Then, go from there....
If it seems like it's not working, just try to put sites in at the appropriate distance, and keep going until you run into a contradiction. When you do, you know you'll have to change something...

Cheers

Pam

4:22 PM  
Anonymous Anonymous said...

what is meant by overlaps?

9:04 PM  
Anonymous Anonymous said...

even if you do all the digests, how do you know if you are doing it correctly?

9:06 PM  
Anonymous Anonymous said...

I think for overlaps, it means that one specific size shows up in more than one band. Correct me if I am wrong.

9:09 PM  
Anonymous Anonymous said...

Hi Pam and others!

I am stuck on this little question: It's from the July 2005 question 1 with the table and you fill in the +/-.

for (i): I-OcZ+Y-/I+O+Z-Y+

I know that Oc is not trans complement to O+ but the I,Z and Y are trans to each other. So for this question, does it mean that even if an inducer is present or absent, all the proteins would be made? Would any of the proteins be constitutively made?

I'm not quite sure if I'm right, so plese help me out. I find it a little difficult to get my logic straight sometimes when doing these problems.

Thank you

9:17 PM  
Anonymous Anonymous said...

Hi,
Yeah I get a little confused on those problems as decribed above as well. Would you mind just give a brief description as to what we should do at the above problem, Pam?

thank you so much for your help.

J

2:24 PM  
Anonymous Anonymous said...

exactly!!! the steps that Pam listed above seem to make sense, but then when it comes down to actually trying to attempt the question, you just don't know where to begin. these multiple double digest questions are so confusing!!!

2:43 PM  
Anonymous Anonymous said...

Hi there confused Lac-operoner

Well, I think you have the first part correct, in the presence or absece of an inducer the Z gene will be expressed since the operator is Oc, meaning that the repressor protien cannot bind to it, but sine our Y gene on that molecule is Y-, it won't be expressed. But looking at the plasmid strand the only way the Y gene will be expressed is when an inducer is present since the operator, O, is normal and the repessor protien made by the I gene is still able to bind to this location. Again, on the plasmid, the Z gene is mutated and will not be expressed, but we don;t need to care about that since it will be expressed on the chromosomal DNA. So, to summerize, the Z gene will be expressed regardless of an inducer or not where as the Y gene will only be expressed when an inducer is present.
Hope that helps!!!

2:50 PM  
Anonymous Anonymous said...

can anyone help with this multiple double digest business?

8:51 PM  
Anonymous Anonymous said...

APOLOGIES!
I am looking at the 2006 question with the northern blot and the ORFs and am realizing that told some of you something completely wrong. I am so sorry!
The top band in the blot on the right can only be one thing: an ORF that MUST hybridize to at least part of the 700bp fragment given, BUT WE DON'T WANT IT TO HYBRIDIZE WITH THE Puc ORF! (If it did, it should also show up in the blot on the left as well). You'll find that there's only one ORF that fits both criteria.

Double digests questions: see new post!

Pam

9:21 PM  
Anonymous Anonymous said...

Thank you, kind lac-operon helper!

1:05 AM  
Anonymous Anonymous said...

Sorry Pam, but I still dont get what you are saying for the blot question

1:10 AM  
Anonymous Anonymous said...

The blot question:
Get your diagram of the various reading frames and starts and stops in front of you...now:

part a) of the question had a 700bp fragment and we were told that it contains the whole coding sequence for the Puc protein, which is 120 aa. We can easily find it as that fragment contains only one ORF that is long enough (i.e. must be 360 bp=120X3, between a start and a stop).

In part B) we take two identical RNA samples and we run them on 2 separate gels, and do a northern blot of each gel. For one (the one the left) we use the Puc coding sequence as a probe. So, if Puc is expressed in the cells where we got our RNA from, we should see a band showing up in our blot-in fact we see it, and it's just under 400bp, which matches with the 360 bp calculated above. Nothing else lights up, which tells us that our samples do not contain any RNAs that have a part of the sequence in common with the Puc sequence.

In the second blot we use the whole 700bp illustrated in part a) as a probe. therefore, any RNAs that have part of their sequence in common with our 700bp fragment will show up. One of them will obviously be the 360bp Puc ORF.
Now, we also see a 1kb band, so there must be, in our map, a start codon that is not followed by stop for at least 1000bp, and the stretch of DNA between this start and stop (there are three such start codons), but our ORF that gives the 1kb band must also have no parts in common with the Puc ORF. There is only one such start codon.
Is it any better?

9:49 AM  
Anonymous Anonymous said...

why can't in the search for the 1kb band ORF have anything in common with the Puc ORF?

11:33 PM  
Anonymous Anonymous said...

hi pam, in response to your 9:49 comment for a)

i found ~360bp start to stop sequences in RF +1, -2, -3. which one is the correct one?

i must be missing something to understand these diagrams...u said we could find it easily...

thanks for ur help in advance

4:27 PM  
Anonymous Anonymous said...

Well...you need an ORF that goes from start to stop without interruptions, and it also needs to be all within the 700bp fragment given to you.
The longest UNINTERRUPTED ORF in RF+1 is less than 200bp. In RF -2 the longest one is under 200bp too, and in RF-3 it is about 230bp.

You need to find one where the distance between the start and the closest STOP is 360bp!

Good luck!

4:34 PM  
Anonymous Anonymous said...

thank you 4:34!
so +2

for b) i went with finding RF that ends with a start but no stop visble. I came up with +1, +3, -1. or are the sequence available for hybridizing in +1,-1 too short?

4:42 PM  
Anonymous Anonymous said...

For b) you are right, those 3 could work. EXCEPT...it can't be +1 or +3 because if it was, then it would also show up in the first blot (it would also hybridize to the 'Puc coding sequence probe"). So must be -1

4:51 PM  
Anonymous Anonymous said...

thanks, i completely understand now! wow u got this mt aced =P

4:58 PM  

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