Hayflick Limit Calculation

Hayflick Limit Calculation .
Andre Willers
19 Feb 2011

Synopsis :
The Hayflick Limit calculation from first principles is looked at in more detail (on request)

Discussion :

See http://andreswhy.blogspot.com "Ageing" Feb 2011

The bit below seems to need some amplification .

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" Suppose there is a p probability of damage at each mitosis event .
Then the cumulative damage ratio would be Cd= p* n(n+1)/2 , where n is the number of cell divisions . A linear summation .

From http://andreswhy.blogspot.com "NewTools" Reserves and error arguments we know that for an old system like this 1/3 error ratio would probably render it non-viable .

Thus , we can say 1/3 = p* n(n+1)/2 would solve for n at the Hayflick limit ."

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Detailed Explanation :

1, Error Probability p
TTAGGG is the switch for activating a telomere cap .
There are 4 bases , which gives 4^6=4096 possibilities .
p = 1/(4^6) = 1/4096 = 0.000244414 as discussed .

2, Reserve :
There exists a Probability Reserve to counter errors.
This term includes the various repair mechanisms , fail-safes , etc to mitigate damage
The organism can only afford to allocate a specific percentage of it's resources to this . It has been calculated (in old systems subject to evolutionary pressures) to be an average of 1/3 of unity probability .
See http://andreswhy.blogspot.com "NewTools" Nov 2008 . The subsections under "Infinite Probes" and Reserves . I am not going to repeat these here .

3.At each Cell division :
Division 1 : Error Prob = p Deduct p from Probability Reserve .
Division 2 : Error Prob = p+p (old error plus new error) Deduct 2p from Probability Reserve .
Division 3 : Error Prob = 2p+p (old errors plus new error) Deduct 3p from Probability Reserve .
……….
……….
………..
Division n : Error Prob = (n-1)p+p (old errors plus new error) Deduct n*p from Probability Reserve .

The summation 1+2+3+…+n = n(n+1)/2

The Probability Reserve is exhausted when the deductions have reduced it to zero .
Then n has reached a limit when
Probability Reserve = p*n(n+1)/2

Probability Reserve= 1/3 as discussed .

Then solving for n when
1/3 = p*n(n+1)/2 for p=1/(4^6)
gives n= 51.5866 or -52.58666

This is the Hayflick Limit .

What does it mean ?

Positive solution : The Individual Hayflick Limit .
The positive solution of n= + 51 seems fairly direct . The cells in an organism have a ceiling on the number of divisions . Cells will apoptose following a more-or-less quadratic curve . Standard old age without scientific intervention .

Negative solution : The Species Hayflick Limit .
The negative solution of n= - 52 is more speculative .
It could mean that species go extinct after 52 speciation events .
Ie , they "spontaneously" go extinct . Like an individual "spontaneously" goes extinct due to old age .
Sharks and crocs have nearly no speciation events .
The problem is where to put the zero-point .
(A new telomere switch ?)
This needs more work .

Boundaries of this argument :
The repair mechanisms postulated in Probability Reserve does not include conscious interventions like Science (ie where Beth(>1) )

A roll in the hay used to be more fun .

Andre .

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