Staredit Network > Forums > Serious Discussion > Topic: The Origin of Life - Theories
The Origin of Life - Theories
Feb 19 2010, 6:32 pm
By: RetPallylol
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Mar 17 2010, 3:05 am Vi3t-X Post #41



Quote from MasterJohnny
I do not think exogenesis is very possible because from what I know of astronomy. Space has extreme temperatures and radiation. Cells would be screwed in space.
If I recall correctly, life on Earth is speculated to have begun from a collision with the earth, where many of the early elements which were, at the time, abundant on the Earth pressured together to form proteins. My memory is a bit hazy, but I believe they were Oxygen, Hydrogen, Carbon and Nitrogen.

EDIT: Just noticed a lot of people posted already. :P



None.

Mar 17 2010, 3:33 am CecilSunkure Post #42



Quote from Vrael
P(B|A) = P(AB)/P(A) = P(B)/P(A)
P(C|AB) = P(ABC)/P(AB) = P(C)/P(B)
P(D|ABC) = P(ABCD)/P(ABC) = P(D)/P(C)
P(E|ABCD) = P(ABCDE)/P(ABCD) = P(E)/P(D)

Then with some algebra skills, and the knowledge we want P(ABCDE)
P(ABCDE) = P(E)*P(ABCD)/P(D)
P(ABCD) = P(D)*P(ABC)/P(C)
P(ABC) = P(C)*P(AB)/P(B)
P(AB) = P(B)/P(A)

When we substitute into each expression, we end up with
P(ABCDE) = P(E)*P(D)*P(C)*P(B) / P(D)*P(C)*P(B)*P(A)

P(ABCDE) = P(E)/P(A), which is much larger than P(A)*P(B)*P(C)*P(D)*P(E)
Right I agree. Here is what I meant earlier:

We were representing different things with the same notation, so I defined what each of my notations were.

P(A) is the probability of A happening.
P(BA) is the probability of B happening, assuming all previous steps are completed.
P(B) represents the probability that B would happen, with the assumption that A did not occur.
P(CBA) is the probability of C happening, assuming all previous steps are completed.
P(C) represents the probability that C would happen, with the assumption that A and B did not occur.

So, the probability that all three steps successfully occur is P(CBA) * P(BA) * P(A), which is what I've been trying to say this entire time. This is different from P(A) * P(B) * P(C), which is what you pointed out, and which is also not what I meant in my previous post with the picture.

Post has been edited 1 time(s), last time on Mar 17 2010, 3:39 am by CecilSunkure.



None.

Mar 17 2010, 4:21 am Vrael Post #43



Quote from CecilSunkure
So, the probability that all three steps successfully occur is P(CBA) * P(BA) * P(A), which is what I've been trying to say this entire time.
This is still false. If step C is concentric with B and concentric with A (this is in terms of a graphical representation, like if you draw a big box, commonly referred to the "sample space", which is all the possible events that may occur, then step "A" would be some subset of the sample space, and B would be a subset of A, and C a subset of B, and so on),
then P(CBA)*P(BA)*P(A) = P(C)*P(B)*P(A), since P(CBA) = P(C), and P(BA) = P(B), and P(A) = P(A).

The point is, you can't simply multiply the individual probabilities together to get the probability of all of them occuring, unless they are independant events. Even if the steps are not concentric as I have suggested they are, you still can't just multiply them together since they're still not independant.

The probability that 3 of the steps occur is going to be P(C)/P(A), assuming A has occured. If A hasn't occured, the probability is just P(C) since step C occurs with some probability P(C).



None.

Mar 17 2010, 4:41 am CecilSunkure Post #44



Quote from Vrael
Quote from CecilSunkure
So, the probability that all three steps successfully occur is P(CBA) * P(BA) * P(A), which is what I've been trying to say this entire time.
This is still false. If step C is concentric with B and concentric with A (this is in terms of a graphical representation, like if you draw a big box, commonly referred to the "sample space", which is all the possible events that may occur, then step "A" would be some subset of the sample space, and B would be a subset of A, and C a subset of B, and so on),
then P(CBA)*P(BA)*P(A) = P(C)*P(B)*P(A), since P(CBA) = P(C), and P(BA) = P(B), and P(A) = P(A).
But I wasn't saying that. I was saying that P(CBA) is the probability of just C happening, assuming that the other two outcomes occurred successfully (A and B), thus isolating the occurrence of C. So no matter if you say the probability of C occurring in sequence from A --> C, or you saying that the probability of each sequence is separate, both outcomes are the same. I am not saying that P(C) = P(CBA) * P(BA) * P(A).

Say there is a 1/5 chance of each step occurring, assuming all previous steps before each step happened. Then P(A) = 1/5. Now, assuming P(A) was successful, P(BA) now has a 1/5 chance of being completed. In order for both of these steps to be completed, in the correct sequence, you would just multiply 1/5 by 1/5, similarly to how you would try to land on a certain side of a dice twice in a row.

Now, the probability of B happening without assuming A to already happen, could be completely different, since B requires A, just as you can't land on two of the same die face without rolling twice. So, P(B) could be 1/100, assuming A has not been successfully completed.



None.

Mar 17 2010, 5:00 am rayNimagi Post #45



Quote from CecilSunkure
Say there is a 1/5 chance of each step occurring, assuming all previous steps before each step happened. Then P(A) = 1/5. Now, assuming P(A) was successful, P(BA) now has a 1/5 chance of being completed. In order for both of these steps to be completed, in the correct sequence, you would just multiply 1/5 by 1/5, similarly to how you would try to land on a certain side of a dice twice in a row.
There would be many molecules of P(A) available to create B. Anyone who has taken high school Chemistry knows that elements react to form more than just one molecule at a time. In fact, there are so many atoms that react at once that the standard unit for chemical reactions is the mole, equivalent to (6.02 * 1023) (where 1 mole of Carbon-12 equals 12 grams). Thus even if the probability of any reaction happening by random chance is increased tremendously for every gram of the substances present. Surely there was a couple thousand kilograms of carbon and nitrogen on Earth available to form compounds. All it takes is one single hit on the right formula, and eventually that self-replicating polymer has years, decades, even millions of years to populate the Earth without disturbance.

And as CAFG said:
Quote from ClansAreForGays
1) They calculate the probability of the formation of a "modern" protein, or even a complete bacterium with all "modern" proteins, by random events. This is not the abiogenesis theory at all.

2) They assume that there is a fixed number of proteins, with fixed sequences for each protein, that are required for life.
There is not just one way of creating life, there are possibly millions of ways to create it. Many proteins created in modern cells have multiple DNA codes for the same protein.



Win by luck, lose by skill.

Mar 17 2010, 5:05 am Syphon Post #46



Quote from CecilSunkure
Quote from Vrael
Quote from CecilSunkure
So, the probability that all three steps successfully occur is P(CBA) * P(BA) * P(A), which is what I've been trying to say this entire time.
This is still false. If step C is concentric with B and concentric with A (this is in terms of a graphical representation, like if you draw a big box, commonly referred to the "sample space", which is all the possible events that may occur, then step "A" would be some subset of the sample space, and B would be a subset of A, and C a subset of B, and so on),
then P(CBA)*P(BA)*P(A) = P(C)*P(B)*P(A), since P(CBA) = P(C), and P(BA) = P(B), and P(A) = P(A).
But I wasn't saying that. I was saying that P(CBA) is the probability of just C happening, assuming that the other two outcomes occurred successfully (A and B), thus isolating the occurrence of C. So no matter if you say the probability of C occurring in sequence from A --> C, or you saying that the probability of each sequence is separate, both outcomes are the same. I am not saying that P(C) = P(CBA) * P(BA) * P(A).

Say there is a 1/5 chance of each step occurring, assuming all previous steps before each step happened. Then P(A) = 1/5. Now, assuming P(A) was successful, P(BA) now has a 1/5 chance of being completed. In order for both of these steps to be completed, in the correct sequence, you would just multiply 1/5 by 1/5, similarly to how you would try to land on a certain side of a dice twice in a row.

Now, the probability of B happening without assuming A to already happen, could be completely different, since B requires A, just as you can't land on two of the same die face without rolling twice. So, P(B) could be 1/100, assuming A has not been successfully completed.

Use correct notation please. This is hurting my eyes. P(C|(B∩A)) is what you're looking for. Probability of A given B and C have occurred. The probability of this is P(C∩B∩A)/P(B∩A)

P(A|B) for P(A) = .2, and P(B) = .2 is .4 only if A and B are independent. As you later state this is not the case, you're wrong about that. It's actually...

P(A) = .2
P(B) = .2

P(A|B) = P(A∩B)/P(B)
P(A|B) = .04/.2
P(A|B) = .2

The dice analogy is not apt, as those events are independent.



None.

Mar 17 2010, 5:59 am Vrael Post #47



Quote from Syphon
Use correct notation please. This is hurting my eyes. P(C|(B∩A)) is what you're looking for. Probability of A given B and C have occurred. The probability of this is P(C∩B∩A)/P(B∩A)
I did use the correct notation. I even drug out my probability book to make sure, and I was correct the first time. P(AB) is the probability of the intersection of A and B, the upside down intersect U sign that I don't know how to type isn't necessary. I'll type your way for your convenience now, though, even though I find it easier to read my way.

Quote from Syphon
P(A|B) for P(A) = .2, and P(B) = .2 is .4 only if A and B are independent. As you later state this is not the case, you're wrong about that. It's actually...

P(A) = .2
P(B) = .2

P(A|B) = P(A∩B)/P(B)
P(A|B) = .04/.2
P(A|B) = .2

The dice analogy is not apt, as those events are independent.
I think you meant ".04" in that top line there, just a typo probably.

However, if A and B are dependant, then P(A∩B) does not equal .04.
P(A|B) = P(A∩B)/P(B)

P(A∩B) = P(B)*P(A|B) = P(B∩A) = P(A)*P(B|A)
The individual probabilities tell us nothing about the conditional probability if we know nothing about the condition.

Quote from CecilSunkure
Say there is a 1/5 chance of each step occurring, assuming all previous steps before each step happened.
What you are saying is this:
P(A) = 1/5
P(B|A) = 1/5
P(C|A∩B) = 1/5
P(D|A∩B∩C) = 1/5
P(E|A∩B∩C∩D) = 1/5

The probability that all events occur, or P(A∩B∩C∩D∩E), in this particular case, will be (1/5)^5.

However, this is different from saying P(A) = .2, P(B) = .2, ect ect. In which case we would need to know some information about the conditional statements in order to figure out the probability of all the steps occuring.

Post has been edited 3 time(s), last time on Mar 17 2010, 6:44 am by Vrael.



None.

Mar 17 2010, 7:05 am Syphon Post #48



Quote from Vrael
Quote from Syphon
Use correct notation please. This is hurting my eyes. P(C|(B∩A)) is what you're looking for. Probability of A given B and C have occurred. The probability of this is P(C∩B∩A)/P(B∩A)
I did use the correct notation. I even drug out my probability book to make sure, and I was correct the first time. P(AB) is the probability of the intersection of A and B, the upside down intersect U sign that I don't know how to type isn't necessary. I'll type your way for your convenience now, though, even though I find it easier to read my way.

Quote from Syphon
P(A|B) for P(A) = .2, and P(B) = .2 is .4 only if A and B are independent. As you later state this is not the case, you're wrong about that. It's actually...

P(A) = .2
P(B) = .2

P(A|B) = P(A∩B)/P(B)
P(A|B) = .04/.2
P(A|B) = .2

The dice analogy is not apt, as those events are independent.
I think you meant ".04" in that top line there, just a typo probably.

However, if A and B are dependant, then P(A∩B) does not equal .04.
P(A|B) = P(A∩B)/P(B)

P(A∩B) = P(B)*P(A|B) = P(B∩A) = P(A)*P(B|A)
The individual probabilities tell us nothing about the conditional probability if we know nothing about the condition.

Quote from CecilSunkure
Say there is a 1/5 chance of each step occurring, assuming all previous steps before each step happened.
What you are saying is this:
P(A) = 1/5
P(B|A) = 1/5
P(C|A∩B) = 1/5
P(D|A∩B∩C) = 1/5
P(E|A∩B∩C∩D) = 1/5

The probability that all events occur, or P(A∩B∩C∩D∩E), in this particular case, will be (1/5)^5.

However, this is different from saying P(A) = .2, P(B) = .2, ect ect. In which case we would need to know some information about the conditional statements in order to figure out the probability of all the steps occuring.

I didn't mean you, I meant the guy that is calculating probabilities wrong up there. Ya it was a typo.

EDIT - P(A|B) = P(A) by definition if they're dependent.

Post has been edited 1 time(s), last time on Mar 17 2010, 8:57 pm by Syphon.



None.

Mar 17 2010, 4:30 pm CecilSunkure Post #49



Quote from Vrael
The probability that all events occur, or P(A∩B∩C∩D∩E), in this particular case, will be (1/5)^5.

However, this is different from saying P(A) = .2, P(B) = .2, ect ect. In which case we would need to know some information about the conditional statements in order to figure out the probability of all the steps occuring.
Yeah, I agree with this >.<

I don't think I ever meant different, I was just using improper notation, since I didn't know proper.

Quote from Syphon
P(A|B) for P(A) = .2, and P(B) = .2 is .4 only if A and B are independent. As you later state this is not the case, you're wrong about that. It's actually...

P(A) = .2
P(B) = .2

P(A|B) = P(A∩B)/P(B)
P(A|B) = .04/.2
P(A|B) = .2

The dice analogy is not apt, as those events are independent.
That's calculating the chance of each individual event. Although, if the second event will not land the right side of the die, since the second event will not happen if the first fails, you then multiply the two probabilities of each individual event occurring when all previous events are assumed complete, which is a 1/25 chance.

Quote from rayNimagi
And as CAFG said:
Quote from ClansAreForGays
1) They calculate the probability of the formation of a "modern" protein, or even a complete bacterium with all "modern" proteins, by random events. This is not the abiogenesis theory at all.

2) They assume that there is a fixed number of proteins, with fixed sequences for each protein, that are required for life.
There is not just one way of creating life, there are possibly millions of ways to create it. Many proteins created in modern cells have multiple DNA codes for the same protein.
Okay, fair enough.
Quote from CecilSunkure
Like I said, cells, the most basic units of life, ( http://en.wikipedia.org/wiki/Cell_theory )...
So you cannot call something alive, without having all of the necessary pieces to be considered a cell.
Quote from CecilSunkure
The point wasn't that cells are perfectly irreducibly complex, but that the complexity necessary to have functioning and living material is a very complex starting point.
Even so, for something to qualify as life, there is a high degree of complexity.

Quote from Syphon
Regardless of this, the Earth is not a closed system. The second law of thermodynamics does not apply to it. It applies to the Universe.
I didn't say earth was a closed system. If I consider the system to be the chemicals that are to generate life, then any energy or matter than becomes unusable to this system is the result of an entropic effect; the chaos of random chance acting upon a system of chemicals makes it seem unlikely, to me, that a chemical pool could last an extremely long time without being disturbed enough to end the process, long enough to generate life.

Quote from Syphon
Cells are most definitely not irreducibly complex.
Quote from CecilSunkure
The point wasn't that cells are perfectly irreducibly complex, but that the complexity necessary to have functioning and living material is a very complex starting point.


Post has been edited 4 time(s), last time on Mar 17 2010, 4:53 pm by CecilSunkure. Reason: Minor changes.



None.

Mar 19 2010, 7:09 pm Vrael Post #50



Quote from CecilSunkure
Yeah, I agree with this >.<

I don't think I ever meant different, I was just using improper notation, since I didn't know proper.
I think you may have missed the main point here, but the topic isn't really about math.

My point pertaining to the topic was, that as each successive step is reached, it makes the following step more probable in terms of abiogenesis, and not less probable, as multiplying values less than 1 would indicate.



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