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How Much Is .1% Difference In Our Dna?


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#121 deadlock

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Posted 29 June 2008 - 03:13 PM

Since you correctly said in your recent post that each mutation had to occur in a separate spot And all of your mutation/dart/card/coin examples had each mutation/dart/ace/heads happening in different positions.
And you have admitted that the probability with each mutation/dart/ace/head happening in different positions is (#mutations)/(#spots) which is the answer I have been giving for your mutation/dart/card/coin examples.I'd appreciate it if you could admit i've been giving the correct answer for your examples and perhaps apologize for the various disparaging remarks you've made about me, my intelligence, education, and mathematical ability and acknowledge that I do in-fact know what I'm doing when dealing with probability.


First, I apologize for my remarks.You really have good skills about probability, But all the problem was that we were using different assumptions and I thought we were using the same.So, you seemed a very stupid person to me because I Thought you were calculating conditional probability in a different way.If mutation is counted in the way You said, then I was wrong.

Second, In the examples I used I was not assuming the assumption that each position could be target once.We were only using different assumptions.If my assumption that each position could receive more than one mutation was right then I would be right.

Third, despite of that If we assume that 120 mutations happening in the coding region will lead to error catastrophe and that only one mutation can happen per person in the coding region then the probability 1/3*10^9 would be more realistic.I dont think human being can tolerate more than 1 mutation per person in the coding region.

If we are calculating the probability of a protein evolve we must consider the probability of harmful mutations between the 120 mutations.If a lucky mutation happens in a point but 119 harmful mutations kill the person it could not be considered in my opinion.

But I´ll use your number to do what I wanted since the beginning.

If we have 4 bases and the probability of a mutation happens in a position is 10^-8
then the probability of a mutation put a specific base in that position must be 1/3*10^8.

if we want to calculate the probability of a protein with 100 aminoacids evolving, we can do the following calculation :

We need 300 nucleic bases.200 bases from that number really code the protein.

p = 1/(3*10^8)^200.

Agree ?

#122 numbers

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Posted 29 June 2008 - 07:08 PM

First, I apologize for my remarks.You really have good skills about probability, But all the problem was that we were using different assumptions and I thought we were using the same.So, you seemed a very stupid person to me because I Thought you were calculating conditional probability in a different way.If mutation is counted in the way You said, then I was wrong.

Second, In the examples I used I was not assuming the assumption that each position could be target once.We were only using different assumptions.If my assumption that each position could receive more than one mutation was right then I would be right.

Third, despite of that If we assume that 120 mutations happening in the coding region will lead to error catastrophe and that only one mutation can happen per person in the coding region then the probability 1/3*10^9 would be more realistic.I dont think human being can tolerate more than 1 mutation per person in the coding region.

If we are calculating the probability of a protein evolve we must consider the probability of harmful mutations between the 120 mutations.If a lucky mutation happens in a point but 119 harmful mutations kill the person it could not be considered in my opinion.

But I´ll use your number to do what I wanted since the beginning.

If we have 4 bases and the probability of a mutation happens in a position is 10^-8
then the probability of a mutation put a specific base in that position must be 1/3*10^8.

if we want to calculate the probability of a protein with 100 aminoacids evolving, we can do the following calculation :

We need 300 nucleic bases.200 bases from that number really code the protein.

p =  1/(3*10^8)^200.

Agree ?

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Thank you, apology accepted.

You might have been thinking there could be more than one mutation per location but none of your examples worked that way without running each example multiple times and carrying the results of each forward.

Throwing or not throwing a single dart at a target can never yield more than one dart per target unless you are doing the experiment more than once and keeping the results of each as the start point of the next
Drawing one card from a deck that only contains 1 ace of spades can never yield more than one ace of spades drawn from that deck unless we are doing the experiment more than once and keeping the results of each as the start point of the next.
We could be carrying the results of each experiment/replication forward but the only thing we really care about is the final number of positions different after how ever many runs we do.

I only jumped in to this thread to confirm that jamesf was correct in saying the probability of a mutation occuring at a specific spot was #mutations/#spots. It took a lot longer to convince you of that than expected, and turned into a math class along the way, but as long as you do now, I really have nothing more to add to this topic.

Your prob for a specific 200 base long protein is fine but you also have to remember that evolution doesn't have a goal. Whatever works is propagated, whether it's the specific 200 base long protein your calculating for or any other 200 base protein that does something.

I'm going to let this thread die now if you don't mind, math lessons are over for now :rolleyes:

#123 deadlock

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Posted 30 June 2008 - 01:13 AM

Thank you, apology accepted.

You might have been thinking there could be more than one mutation per location but none of your examples worked that way without running each example multiple times and carrying the results of each forward.

Throwing or not throwing a single dart at a target can never yield more than one dart per target unless you are doing the experiment more than once and keeping the results of each as the start point of the next
Drawing one card from a deck that only contains 1 ace of spades can never yield more than one ace of spades drawn from that deck unless we are doing the experiment more than once and keeping the results of each as the start point of the next.
We could be carrying the results of each experiment/replication forward but the only thing we really care about is the final number of positions different after how ever many runs we do.

I only jumped in to this thread to confirm that jamesf was correct in saying the probability of a mutation occuring at a specific spot was #mutations/#spots.  It took a lot longer to convince you of that than expected, and turned into a math class along the way, but as long as you do now, I really have nothing more to add to this topic.

Your prob for a specific 200 base long protein is fine but you also have to remember that evolution doesn't have a goal.  Whatever works is propagated, whether it's the specific 200 base long protein your calculating for or any other 200 base protein that does something.

I'm going to let this thread die now if you don't mind, math lessons are over for now :rolleyes:

View Post


First, Your concept is wrong, proteins are very specific.A protein which works on brain does not make the same thing of a protein which works on muscle.So, a protein which makes something in the wrong place does not work at all.

Second, If you want to avoid facing the impossibility of evolution, it´s your right. In reality it was expected.

#124 deadlock

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Posted 01 July 2008 - 09:10 AM

Teacher I know the class is already over, but I have made some research about mutation rates and I think I have discovered some useful information for you.

The first site shows how mutation rates are calculated :

Mutation Rates

Calculation of Mutation Rates
Instead of the mutation rate, often the "mutant frequency" is reported. The mutant frequency is simply the ratio of mutants divided by the total number of bacteria in the population. The mutant frequency is easy to measure, but is less reliable than the mutation rate because it may show large fluctuations depending upon when the first mutation appeared in the population.
The mutation rate is the number of mutations per cell division. Because the cell population is so large, the number of cell divisions is approximately equal to the number of cells in the population (N).


Posted Image

In the example shown above, if h was determined from a fluctuation test with 10^7 cells per tube, then the mutation rate would be:

Posted Image

So, You can notice that the mutation rates is measured per cell division or per gene per generation. It´s not calculated per base copied.

To confirm What I´m saying we can see another site from San Diego State University: The Zen of Mutagenesis

What is the probability of finding a mutation in one particular gene on the chromosome?

Spontaneous mutations in a particular gene are typically very rare. Treatment with mutagens can increase the probability of getting a mutation in the gene, but it also increases the probability of getting other mutations in the chromosome as well. Consider a typical gene in an average sized bacterial chromosome:

Posted Image


If there is an equal, random chance of a "hit" (mutation) in any bp on the chromosome, and (ii) there is an average of one hit per cell, then the probability that a mutation will occur in any particular gene is:

Posted Image

So teacher, you can notice that he divided the size of the gene by the size of the chromosome.If your teaching material was correct the probability would be the size of gene * probability of mutation.Then, the probability of a mutation hit a specific position in the human genome is 1/3*10^9.

Please, dont forget to correct your teaching material for the next class.

#125 deadlock

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Posted 01 July 2008 - 06:34 PM

You might have been thinking there could be more than one mutation per location but none of your examples worked that way without running each example multiple times and carrying the results of each forward.

Throwing or not throwing a single dart at a target can never yield more than one dart per target unless you are doing the experiment more than once and keeping the results of each as the start point of the next
Drawing one card from a deck that only contains 1 ace of spades can never yield more than one ace of spades drawn from that deck unless we are doing the experiment more than once and keeping the results of each as the start point of the next.
We could be carrying the results of each experiment/replication forward but the only thing we really care about is the final number of positions different after how ever many runs we do.


IT´S ABSOLUTELY FALSE !

All my examples were based on probability a posteriori. These are some of my posts :

Excuse me numbers, but it seems that your knowledge about statistics is lower than jamesf, I spent a lot of time making him understand, he´s already convinced of that, I wont waste my time trying to convince you too.But I´ll give you a list of things that if you understand maybe you realise why you are wrong.

1 - Difference between number of possibilities and number of tries.
2 - Difference between probability 'a priori' and probability 'a posteriori'
3 - Bayes Theorem.

Good Luck


I didnt say that position matters to mutation probability, I said that the probability of a mutation happens is different from the probability that mutation happens in a specific location.

1 - We have a mutation 'a priori' = 10^-8
2 - We have a mutatioin 'a posteriori' = 3 * 10^-9 ( positions of genome )

Imagine you have 10 persons with 10 deck of cards, if three aces of spades were drawn, what is the probability of those cards were drawn by person #1, #5 and #8 ?


Numbers, what you are saying is : Knowing that if a mutation happens it must happen in the position 10, then what is the probability of a mutation happens in position 10 ?

Of course your reasoning is completely flawed.The correct question is :

If a mutation happens what is the probability of it happens in position 10 ?


What happens first ? the mutation or the position ? It´s impossible to know the position before it happens. So, the probability is 'given a mutation happened' what is the probability of it has happened in position x.I can´t understand how you cannot figure out so obvious reasoning


I never used probability 'a priori' in my examples.




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