pseudostatistically calculating DNA restriction

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lucilius
lucilius's picture
pseudostatistically calculating DNA restriction

Hallo,

I have the following question that I can not solve: Some BamHI restriction sites in a particual DNA fragment are no longer cleaved when the DNA is pretreated with M.MspI . What percentage of possible BamHI sites is no longer sensitive for BamHI restriction (pseudostatistically)?
With: M.MspI C*CGG and BamHI G/GATCC

I have no clue how to solve this.

I know that the changes are 1/16 that you have a BamHI restriction site that has 2 GG after the 2 CC at the end.

But then what?

I would think that you have 1/16 change that there are 2 GG's behind the 2 C's and that the change that 1 or 2 of those C are methylated is 3 out of 4 (either first C methylated or second C methylated or both methylated or none of the 2 methylated and this last option is the only one that will still be cut by the BamHI, so in 3 out of 4 it will not cut).

So I would end up saying 1/16 that you have 2 G's behind the 2 C's and then 3/4 that it doesnt cut, making it 3/64 and then I would multiply it by 2 because it doenst matter whetter its the 5'-3' or 3'-5' DNA fragment making it 3/32 in the end.
But is this correct?

Anyone an idea?

Sami Tuomivaara
Sami Tuomivaara's picture
lucilius,

lucilius,

I had hard time following your train of thought, so I just put down my thoughts. The recognition site is destroyed if either of the following methylation scenarios occur. The odds of each scenario is in parentheses (assuming independent occurrences of all nucleotides at all green positions at probability 1/4) after each sequence. The resulting methylated Cs are marked with black background:

5'-CCGGGATCCNNN-3'
3'-GGCCCTAGGNNN-5' (1/4 * 1/4 * 1/4 = 1/64)

or

5'-NCCGGATCCNNN-3'
3'-NGGCCTAGGNNN-5' (1 * 1/4 * 1/4 = 1/16)

These two scenarios are mutually exclusive, hence we add their probabilities, the total probability is
P = 1/64 + 1/16 = 5/64.

Identical scenarios can be drawn for the blue sequences, and their probability sum is also 5/64. Since the green and blue scenarios are independent and only one (at least one) of them needs to occur, the joint probability P for this is complementary to "neither green or blue end is modified", (of which probability is (59/64)^2). Hence,

P = 1 - (59/64)^2 = appr. 0.15 or 15%.

I don't know whether M.MspI methylates already methylated sequence, but it doesn't matter since only one methylation is required to demolish the recognition site.

Other people should comment on this as well, since I could have made a mistake...

Cheers,

lucilius
lucilius's picture
Dear Suola,

Dear Suola,

you are right, I didnt see it like that.
5'-CCGGGATCCNNN-3'
3'-GGCCCTAGGNNN-5' (1/4 * 1/4 * 1/4 = 1/64)

or

5'-NCCGGATCCNNN-3'
3'-NGGCCTAGGNNN-5' (1 * 1/4 * 1/4 = 1/16)

Is indeed correct.

I can follow your argumentation, however I would simply say that changes are: 1/16 and 1/64 and this on both sides (acctually you simple "flip" the fragment upside down and start from 3'-5' then ), meaning that you would get: (1/16 + 1/64 ) x 2 = 5/32 = 0.156= 15.6% .

However in this scenario you accept that the M. MspI would methylate always (100%) and I am not sure if this is true?

I do have to say I made an error in my previous post in the M. MspI: I said that it would methylate either the first, second or both C , however I can only methylate the first C , so only C (methylated or not) C (not meth.) GG can occur.
However question remains: will M. MspI always methylate, will it work for 100% ?

Is fo then I would think 15.6% is ok, or your way of calculating P is also ok.
(I have not used your way of calculating because for this excersice you do not need to know the statistical formules I think.

any other insight would indeed be welcoma as Suola allready stated.

Sami Tuomivaara
Sami Tuomivaara's picture
lucilius,

lucilius,

I have to disagree with you about how to treat the green and blue ends. Only if probabilistic events are mutually exclusive, you can directly sum the probabilities. The methylation of green and blue ends are independent (assuming the sequences are random and don't depend on the sequence on the other end, in other words the palindromicity doesn't extend beyond the BamHI sequence) events and at least one of them has to occur (green methylated, blue methylated, or both methylated), hence you have to calculate the probability as shown.

The calculation was indeed made assuming that the methylation efficiency is 100%. Whether this is the case in test tube or in vitro, I don't know.

Cheers,