How to calculate occurrence frequency of restriction enzyme sites on DNA fragment?
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- Опубліковано 11 вер 2024
- Probability is the branch of mathematics concerning numerical descriptions of how likely an event is to occur, or how likely it is that a proposition is true. The probability of an event is a number between 0 and 1, where, roughly speaking, 0 indicates impossibility of the event and 1 indicates certainty.
thanks very much for this beautiful explanation. it is helping me a lot to understand the problems our lecturer use to give us.
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Thanks for explaining the rare problems
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You are welcome! Thanks for watching.
Such a relaxing explanation.... Thank u
When we wrote 1/4 for first G, then how could we write 1/4 for other G
as because then the number of G has reduced by 1.
How are these probabilities independent of each other because we want that particular sequence together.
Everywhere it is showing like this.
Could you please explain.
1. There is infinite supply of bases so this improbability prime with replacement.
2. The multiplication rule is used when calculating the probability of A and B. The two probabilities are multiplied together. To find probability A and B and C all probabilities multiplied.
Hello! I know I'm late but I have to ask: I translate 1/4096 as that there are 4096 possible sequences of 6 bases. Having that on my mind i say that every 4096×6 bases I will find the sequence ,where is my thought process lacking?
Your mistake that you think that sequence jumps by 6 bases - but it moves one base a time. so 12 bases produce 12 variants not 2.
@@GeneticsLessons Thank you so much I was so locked in that thought process!!!!
A very big thank you sir! you are my hero!
The restriction site for Van91I is 5′ CCANNNN↓NTGG 3′. How many times can Van91I cut a 4 Mb
bacterial genome with 50% GC content?
I just saw this. Sorry for even replying. But I was searching for something and saw this comment. So the answer would be like-
The sequence for the enzyme is given (12 bases total), In the question it is given that 50% is GC content therefore, 25% would be G and 25% C. Same goes for A and T. Now we know that N in the sequence represent any base (A/T/G/C). In the sequence G = 2 C = 2 A= 1 T= 1. So if you want to make 25:25:25:25 ratio. You would need to substitute 1 N=G, 1N = C and 2 (N) = A and 2(N) = T. Now we have 50 percent G and 50 percent C in the sequence. Now the probability of finding an A/T or a G/ C is 1/2 . So when you write probability for each base as I said. It would be 1/2^12 which is 1/4096.
So your sequence would be present at 1 in every 4096 places. The genome size given is 4mb= 4000000 bases. So 4000000/4096 ~ 976 times or ~ 1000 times
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Sir that means when they are telling us to calculate average size of the fragment produced is 1 bp in every 4096 bp ?
No, they are asking how frequently we can expect such frequency to happen and this will also give us a length of the fragment.
@@GeneticsLessons thnku sir
Hey that was really good! keep it up
Thanks, will do!
how the calculation will be affected if seqence is RGCGCY (where R is A or G and Y is C or T )
Could you post the full problem?
Is the answer 16384....See all the possibilities(there are only four possibilities in total) and add them all !
I assume equal probability for all ( i.e 25%)
Дорогой Николай, ваши уроки великолепны, хотела спросить, нет ли субтитров на русском?
Субтитры можно включить нажав на кнопку "СС" но не думаю что они точные.
much better explanation than my dogshit prof
The restriction endonuclease Hind III recognizes the sequence AAGCTT.If the genomic DNA sequence is cleaved with Hind III,What will be the average size of the fragment produced?
Here is a solution: ua-cam.com/video/EAPxcCr37rA/v-deo.html
Why we take 5 sites?? And from where 1/5 comes..plz tell
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Why we are doing the reciprocal at 9:02
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thank you
What is this calculation called?
No special name "restriction site probability frequency" something like that....
very nice
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