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Find number of restriction fragment size in chromosome
#1
Hi all,
I am prepare in entrance exam. I need help.
A restriction endonuclease has the recognition sequence G/AATTC, where "/" indicate the cut side. This sequence is found, on average, once every 'X' residues in a chromosome. 'X' =
Options
a. 146 base - pairs
b. 200 base - pairs
c 256 base - pairs
d. 4096 base - pairs
advance thanks reply as soon as possible
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#2
Dear Kanagasundar87,

Can you please specify the chromosome in question like its form, source etc.? Is it of human or any other animal etc?
This will help to calculate the same. Or you may share the course type on which these objective questions are being asked. This will give hint to identify the answer.

Background::
As you may know, chromosome or even plasmids form / shape contribute to number of fragments. Example whenever these are circular pieces of DNA, the fragment number of equals the number of cuts from the restriction enzymes. If plasmid or chromosome is cut at one place by one restriction enzyme then the fragment will be only one. Cutting the circle one will yields only one fragment. If the restriction cuts in two places, it will result in two fragments; with three places, three fragments and so on etc.
In case of linear chromosomes, the situation is slightly different. Cutting a linear chromosome at one place yield two fragments. Similarly cutting it at two places results in three fragments etc. So the number of fragments depends upon the type of cut as well as the original base pair of the chromosome in question (That is chromosome source and form).
A restriction map of a chromosome will indicate all of the cuts as made by the restriction enzymes. One can count the spaces or gap between cuts to find out the number of fragments that are produced.
Also partial DNAse/RE digestion of chromatin reveals its nucleosome structure. Because DNA portions of nucleosome core particles are more accessible for RE(Restriction Enzyme type I-IV) than Dnase. DNA gets digested into fragments equal to multiplicity of distance between nucleosomes (like 180, 360, 540 base pairs etc.).
Hope this will help you to find out the answer. Or please specify the RE type, its source, Chromosome in question to identify the exact answer.
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#3
hi sir,
I search answer in web and finally i got it answer. But I did not understand I have poor english knowledge so any grammatical mistakes sorry. Can you explain the answer
Answer is 4096 bp


Expected frequency of cleavage depends on the complexity of the recognition sequence.
For instance, for the common 6-base cutter, EcoRI, the frequency of its recognition sequence (GAATTC) is ¼ x ¼ x ¼ x ¼ x ¼ x ¼ = once per 4,096 base pairs.

For the more relaxed 6 base cutter, BsrFI (PuCCGGPy), this is ½ x ¼ x ¼ x ¼ x ¼ x ½ = once every 1024 base pairs.
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#4
A restriction endonuclease is an enzyme which cuts the DNA sequence at the middle instead of at the terminal like an exonuclease. Restriction endonucleases are molecular scissors that cut the sequence at a specific region known as the recognition site. Depending on the type of Restriction Endonuclease, the cut ends can be blunt (clean cut) or sticky( hanging ends) . Most RE’s recognition sites are composed of 4 to 6 nucleotides.

EcoR1 is a restriction endonuclease derived from the bacteria Escherichia coli. EcoR1’s recognition site is GAATTC , it scans the DNA sequence for this site and makes a sticky cut.

NNNNNNNNNNNNGAATTCNNNNNNNNNNNNNNNN

EcoR1

NNNNNNNNNNNNG
AATTCNNNNNNNNNNNNNNNN
( where N can be any nucleotide A,G,T or C)
These sticky ends can easily join back together and hence the name.
As per your question, you have to calculate the number of times (frequency) that the recognition site for EcoR1 will be present in a given DNA sequence. Here’s the solution:

Desired Sequence – G A A T T C
Size of sequence – 1 2 3 4 5 6
Random DNA sequence of 6 ntds- N N N N N N
Now each N can be any 1 of 4 possibilities - A,G,C or T
Hence for each N there are 4 options, and there are 6 Ns.
Therefore, the frequency of GAATTC is 4x4x4x4x4x4 ( 4 multiplied 6 times) or 46. This is equal to 4096 bp. Therefore, the recognition site for EcoR1 can occur once every 4096 bp.
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Find number of restriction fragment size in chromosome00