7 replies to this topic

[silentears]

G1= chromosomal DNA from Vibrio fischeri
G2= chromosomal DNA from Vibrio fischeri cut with Sal I restriction enzyme to obtain the lux operon
P1= plasmid DNA from Escherichia coli pUC18
P2= plasmid DNA from Escherichia coli pUC18 cut with Sal I restriction enzyme

why some is lower some is higher?
i know smaller size of dna can travel further and uncut DNA is travel further than the DNA cut with Sal I, but the results doesnt match

G1 suppose to be lower then G2 as it is un cut..... any body can help me?

i still got 2 more questions... What is the range of restriction enzyme fragment sizes gernerated from Sal I digestion of Vibrio fischeri DNA.. is the question asking that they want to know the Sal I enzyme cut which part of the DNA? or the size of the DNA after cutting with Sal I.

2nd) As Sal I recognize a 6 base sequence, what is the theireutcak size distribution you would expect.? is the question asking  how big ish the size of the enzyme?

lastly thanks in advance   (sorry for the bad photo)


Thanks in advance to those who can help

Edited by silentears, 16 May 2009 - 05:57 PM.

[bob1]

Looks like homework:

What are your answers to the questions?

Why should uncut DNA travel faster than cut DNA?  - cut DNA should be smaller in theory!  There are some special cases though, which yous should be able to look up in a text book - look for coiling of DNA.

For your other questions:
1) Size of DNA after digestion.
2) see #1 questions appear to be the same.

[silentears]

thanks bob1 for ur info.. erm does anybody know where can i get the genomic sequence for Vibrio fischeri so that i can see the range of restriction fragment sizes that are generated from the Sal I digestion of Vibrio fischeri

[bob1]

Here:http://www.ncbi.nlm....vibrio fishceri

[silentears]

erm i found the genomic sequence
for example as below the sequence which are recognise by Sal I  G TCGA C


5701 ctgattgagc ttgtaatttg actataattt cttggatgtt tattgtcgac ttttgtgttt
42661 gtttgtcaga tattattaat aagtcgacat ttttttctat tctacctggt actcatagta
49801 tgtaaaggcg gcaagagaat atcagaacaa aggtgtcgac tctctaatgc tattgccacc
77821 ataaaattac gtcgacctca tccaatggca tgtcacagat tatggtgagc atgaaaatgg

what should i do now in order to find the range of the restriction fragment sizes?
  need serious help pls
thanks bob1 again fore helping




Sal I recgonisation is G TCGA C

in order to find the range..... wad should i do?

Edited by silentears, 29 May 2009 - 06:09 AM.

[phage434]

An easy way to address this is to look at it probablistically.  Each genome has a specific GC vs. AT ratio.  Let g be the probability of GC in the genome.  Let a be the probability of AT.  a = 1 - g.

The recognition site for SalI is gtcgac.  The probability of this sequence is (.5 * g)(.5 * a)(.5  * g)(.5 * g)(.5 * a)(.5 * g), or 2^-6 * a**2 times g**4.  For g = .50, this is 2^-12.  This means that the length of DNA fragments between these cut sites will be on average 2^12, or 4096 bp.

Now, for Vibrio fischeri, the gc content is 38.3%.  g=.383, a=.617.

p = 2**-6 * (.383)**4 * (.617)**2 =0.0001276

1/p = 7837 bp

So, the average restriction enzyme fragment length would be around 7.8 Kb.

This analysis ignores the (sometimes important) effects of di- tri- and tetra- nucleotide statistics, and the effects of local G vs. C bias present in many genomes, but it is a pretty good start.

[silentears]

phage434, on May 30 2009, 04:34 AM, said:

An easy way to address this is to look at it probablistically.  Each genome has a specific GC vs. AT ratio.  Let g be the probability of GC in the genome.  Let a be the probability of AT.  a = 1 - g.

The recognition site for SalI is gtcgac.  The probability of this sequence is (.5 * g)(.5 * a)(.5  * g)(.5 * g)(.5 * a)(.5 * g), or 2^-6 * a**2 times g**4.  For g = .50, this is 2^-12.  This means that the length of DNA fragments between these cut sites will be on average 2^12, or 4096 bp.

Now, for Vibrio fischeri, the gc content is 38.3%.  g=.383, a=.617.

p = 2**-6 * (.383)**4 * (.617)**2 =0.0001276

1/p = 7837 bp

So, the average restriction enzyme fragment length would be around 7.8 Kb.

This analysis ignores the (sometimes important) effects of di- tri- and tetra- nucleotide statistics, and the effects of local G vs. C bias present in many genomes, but it is a pretty good start.

erm wad i meant is are we able to find the range of restriction fragment sizes?
eg wad is the smallest range size and the largest range size this is wad i am asking

[bob1]

Find an enzyme cutting program such as enzyme X (Mac only) or ApE (A plasmid Editor) and load it, then use that is a good way.  Other wise, given that you have the genomic DNA and the number of restriction sites, you should be able to calculate the fragment sizes by simple arithmetic.