Analysis of Plasmid pBr322

[messi19]

having a problem with a few questions regarding the analysis of the plasmid. i have got the electrophoresis image and from there i have worked out the distance migrated for each band. it is from question 2 that i am unsure of what to do and how the graph should turn out. i would really appreciate any help i get.

 

 

1) From the photograph determine the migration distances (in mm) of the molecular weight standard DNA bands and of the bands of pBR322 cut with the different enzymes and construct a table with this data.

i have done this question. and got the folowing results

Fragment size (bp) Migration distance (mm) log of fragment size

3500 56 3.54

3200 59 3.51

4000 55 3.6

4300 53 3.63

4300 53 3.63

4300 53 3.63

 

2) Construct a standard curve (graph) of migration distance versus log molecular weight (in base pairs - bp) for the molecular weight standard bands plotting log10 marker fragment size (bp) against distance of fragment migration. Use the standard curve to determine the size of your pBR322 fragment bands from all the different combinations of enzymes - in Lanes 2-4 – and confirm the size of pBR322 for the single enzyme digests Lanes 5-7. Present your data in tabular form appropriately labelled.

 

3) What is the size of pBR322? How did you work this out from your results?

 

4) Using the data you have generated. Draw a representation of the plasmid showing the relative positions of the three restriction endonuclease sites.

 

thank you

Edited July 26, 2012 by messi19

[ecoli]

2) Construct a standard curve (graph) of migration distance versus log molecular weight (in base pairs - bp) for the molecular weight standard bands plotting log10 marker fragment size (bp) against distance of fragment migration. Use the standard curve to determine the size of your pBR322 fragment bands from all the different combinations of enzymes - in Lanes 2-4 – and confirm the size of pBR322 for the single enzyme digests Lanes 5-7. Present your data in tabular form appropriately labelled.

 

You already have the data to do this. Just graph the log of molecular weight vs migration distance and interpolate the point in question.

 

3) What is the size of pBR322? How did you work this out from your results?

If you digest the whole plasmid into fragments, the whole will simply be the sum of its parts.

 

4) Using the data you have generated. Draw a representation of the plasmid showing the relative positions of the three restriction endonuclease sites.

 

This is given by the fragment sizes.

[messi19]

thanks for the help. however when i start the graph with both points starting from 0 then i cant make a curve as the points are all located in one small area.

 

so i decdied to plot a graph with migration distance going from 50 -60mm and log of fragment size going from 3.5 to 3.64. This was much better and did give me a curve. the next problem came when i had to Use the standard curve to determine the size of your pBR322 fragment bands from all the different combinations of enzymes - in Lanes 2-4 – and confirm the size of pBR322 for the single enzyme digests Lanes 5-7. the first 3 lanes contain 2 bands and the bands migration points are 104, 109 and 118mm. and these just cant be plotted on the graph as i dont have enough space or enough points to make the curve from. any idea on where im going wrong. just to be sure the migration distance goes on the x axis with the log on the y axis?

 

thanx

 

 

 

Edited July 26, 2012 by messi19

[ecoli]

thanks for the help. however when i start the graph with both points starting from 0 then i cant make a curve as the points are all located in one small area.

 

so i decdied to plot a graph with migration distance going from 50 -60mm and log of fragment size going from 3.5 to 3.64. This was much better and did give me a curve. the next problem came when i had to Use the standard curve to determine the size of your pBR322 fragment bands from all the different combinations of enzymes - in Lanes 2-4 – and confirm the size of pBR322 for the single enzyme digests Lanes 5-7. the first 3 lanes contain 2 bands and the bands migration points are 104, 109 and 118mm. and these just cant be plotted on the graph as i dont have enough space or enough points to make the curve from. any idea on where im going wrong. just to be sure the migration distance goes on the x axis with the log on the y axis?

 

thanx

 

 

 

So you're saying the problem is that the migration distance of your test bands are farther than your ladder fragments? If so, you need to use a different ladder. You can't extrapolate band migration, since migration is nonlinear.

 

edit: actually could you put the picture of the gel up? I have a feeling you're not graphing the ladder fragments.

[messi19]

http://imageshack.us/photo/my-images/52/scan0002fm.jpg/

 

thats the pic of the gel and the distances

[ecoli]

Ah.. thought this might be your problem. A standard curve must be made with your standard to calculate the mol weights of your test fragments. The question says as much: "Construct a standard curve (graph) of migration distance versus log molecular weight (in base pairs - bp) for the molecular weight standard bands" [emphasis mine].

 

You make the curve with the fragments from you ladder (lane 1) and interpolate the weights of the bands in lanes 2-7

[messi19]

oh right. ok well ill do the graph and the points now and let u know how i get on..

 

thanks a lot!

 

did the graph as well as the curve and the interpolated the points to get the log. question 3 like u said is the sum of the parts. but which fragment sizes am i adding up. i take it its not the log fragment size so it should be the ones based on the gel?

Edited July 26, 2012 by messi19

[ecoli]

oh right. ok well ill do the graph and the points now and let u know how i get on..

 

thanks a lot!

 

did the graph as well as the curve and the interpolated the points to get the log. question 3 like u said is the sum of the parts. but which fragment sizes am i adding up. i take it its not the log fragment size so it should be the ones based on the gel?

No, the log transform is just so it fits the linear model. Remember, the inverse of log is the exponential - log10; 10^x.

[messi19]

yeah did the inverse of the log , which gives me the fragment size of the logs. i presume these are the base pairs. for the single digests i have 3981. but i dont know what i add onto this to get the size of the plasmid cos i have 3 double digests which have smaller values such as 630,794 and 501

 

ahh . i believe i have figured it out . there is a double digest that also contains 3981 + 501. this would give me an answer of 4482

 

the other double digests are 3162 and 2511

Edited July 26, 2012 by messi19

[messi19]

Correct me if wrong but for the restriction map the sizes have to add up too the same value for each of the enzymes. But I don't believe mine do unless im using the wrong values

Edited July 27, 2012 by messi19

[ecoli]

Correct me if wrong but for the restriction map the sizes have to add up too the same value for each of the enzymes. But I don't believe mine do unless im using the wrong values

 

Yes, they should... the size doesn't change. But keep in mind for a single digest you're cutting a circular plasmid once (assuming the plasmid only has 1 restriction site), to get a 'linear' piece of DNA, so the band running after a single digest will be the size of the whole plasmid.

[messi19]

the value i would be looking at is the log value surely as these do add up to the same value. but i dont use these for the restriction mapping? i use the antilog for the base pairs and mapping. so u only add up the smaller bands of the double digests to get the fragment size. im just getting confused so much with this.

Edited July 27, 2012 by messi19

[ecoli]

the value i would be looking at is the log value surely as these do add up to the same value. but i dont use these for the restriction mapping? i use the antilog for the base pairs and mapping. so u only add up the smaller bands of the double digests to get the fragment size. im just getting confused so much with this.

 

The only reason to convert to the log base pair is so the graph of migration vs fragment size is linear. After finding the log fragment size of the bands from your standard, convert those numbers into the fragment size on only work with those numbers.

 

Working with these real numbers, the sum of the double digest should equal the band from the single digest, allowing some error for accuracy/gel resolution.

[messi19]

the log pairs converted to the standard is the antilog. my 3 double digests are roughly aaround my single digest.single digest is 3981. double digests are 3792, 3305 and 4482. so i take it the size of the plasmid is one of those from the double digest but how do u work out which one

[ecoli]

the log pairs converted to the standard is the antilog. my 3 double digests are roughly aaround my single digest.single digest is 3981. double digests are 3792, 3305 and 4482. so i take it the size of the plasmid is one of those from the double digest but how do u work out which one

 

You need to be clearer with your language. A lane representing a doubly digested plasmid (with two different restriction enzymes) should have two bands per lane - assuming the plasmid has single restriction sites for both of the REs you used. Now be clear... what are the sizes for each band in each lane, and what restriction enzymes did you use for each lane and how many of what kind of restriction sites does your plasmid have?

[messi19]

lane 2 - EcoRI & PstI digested - 3162 , 630

Lane 3 - BamHI& PstI digested - 2511,794

Lane 4 - BamHI& EcoRI digested - 3981, 501

Lane 5 BamHIdigested - 3981

Lane 6 PstIdigested - 3981

Lane 7 EcoRI digested - 3981

[ecoli]

So lanes 5, 6, 7 are the size of the full plasmid.. but something funky is obviously going on with your measurements of the double digests. These gels are fairly low res so its ok if the numbers don't work out perfectly, the the sums of each of the bands in lanes 2-4 seem way off.

 

You might improve accuracy in your standard curve by doing a simple linear regression to get the equation of the best fit line and interpolate from that (can be done easily in MS excel or other spreadsheet software).

[messi19]

So lanes 5, 6, 7 are the size of the full plasmid.. but something funky is obviously going on with your measurements of the double digests. These gels are fairly low res so its ok if the numbers don't work out perfectly, the the sums of each of the bands in lanes 2-4 seem way off.

 

You might improve accuracy in your standard curve by doing a simple linear regression to get the equation of the best fit line and interpolate from that (can be done easily in MS excel or other spreadsheet software).

 

yeah there was something going on so went through my results again starting from the log till the end and its that lane 3 which seems to be producing the odd result but ill leave that and move onto the restriction map. il give it a go and then ask if i have any problems.thank you though for all the help

Edited July 27, 2012 by messi19