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DNA sequence evolution

  • Thread starter bowlbase
  • Start date
  • #1
146
2

Homework Statement


A strand of length L begins life as all A's. Assume that each letter evolves independent of all the rest until today, 1000 generations later. Within each generation there is a ##\mu## probability that the letter mutates to either C, G, T. Finally, assume that once a letter mutates that it cannot mutate again.
Calculate the number of A's as a function of ##\mu##. Then equate this expectation to ##N_A## and write down a function for ##\mu## in terms of##N_A##.

Homework Equations




The Attempt at a Solution


So, I have 1000 generations where each A has the possibility to mutate to something else with probability ##\mu##. The first generation the total number of A's is ##N_A=L##. The second generation we must multiply each A by the mutation probability. Since there is L A's we will get: ##N_A=\mu L##. The third generation occurs and we have to multiply the current number of A's by ##\mu## again. Which gives us ##N_A=\mu \mu L##. Taking this to 1000 generations we'd have ##N_A= \mu^{1000-1} L## which doesn't really seem likely at all.

Any suggestions, or is this correct?
 

Answers and Replies

  • #2
130
30
What your solution is working toward is the number of non-A's in a given generation. What you want is to apply the opposite probability, the probability of not mutating.

For example, think if the probability was 1% to mutate. After the 1st generation, you would expect .99L A genes and .01L non-A genes. if you just took NA = μL, you would effectively be saying that NA in the first generation is (.01)L which would actually be the Nnot-A

So, your concept of multiplying the probability successively is correct, but you just need to use the right probability.
 
Last edited:
  • #3
146
2
You are right. I had a feeling that I was getting the opposite result that I was meaning to get. I should have realized I had them mixed up! Thanks!
 
  • #4
130
30
No problem!

And one more thing, if the second generation is P2L, and the third is P3L. Wouldn't the 1000th be P1000L? Just wondering since you put P1000-1L
(assuming P is the corrected probability)

Edit: nevermind, the first generation is just L, haha :oops:
 

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