Posted on 07/03/2006 12:32:51 PM PDT by Al Simmons
*LOL*
BTW, are you in Europe or do you have the same insomnia I've currently got????
It's a shame that you chose to become personal so quickly . How long ago I signed up does not affect whether I am right or wrong.
AA is 1 base pair out of 1953 in the bacteria gene under debate.
A does not pair with A. A pairs with T. And when we list a genetic sequence, we don't usually give the complementary bases, because anyone who knows the rudiments can figure out what the complementary strand is. So AA pairs with TT.
You evolutionists need to bone up on simple math.
If you don't understand the structure of DNA, you are going to make mistakes even before you get to doing any math.
By the way, because pairs such as AG & GA could be seen as duplicates, the formula that I gave in the earlier post for 3^1953 was used instead of 6^1953.
DNA strands have a direction. They are by convention given from 5' end to 3' end. So AG is not the same as GA, and they can't be 'seen as duplicates', unless you don't understand the chemical structure of DNA.
I hope you don't think it's impolite for me to suggest that you should learn some elementary molecular biology before you try to argue about it.
Let's say we have 2 bases, and the original sequence is AA. Then the list of sequences accessible by a single point mutation is (AT, AG, AC, TA, GA, CA). That's six, which is 3*2, not 3^2.
If we have three bases, starting from AAA, we get (AAT, AAG, AAC, ATA, AGA, ACA, TAA, GAA, CAA). That's 3*3 = 9, not 3^3 = 27. And so on. You do know * is a multiplication sign, right?
617 posted on 07/07/2006 9:43:05 PM CDT by DanDenDarA does not pair with A. A pairs with T. And when we list a genetic sequence, we don't usually give the complementary bases, because anyone who knows the rudiments can figure out what the complementary strand is. So AA pairs with TT.
Oh, brother.
Dan, Dan, Dan...
"AAA" doesn't equal 3 base pairs. "AA" does not equal 2 base pairs.
So you got the biology wrong. Then after listing "AA" you contracted yourself with "A does not pair with A."
Then you got the math wrong: it's not 3*1953, it's 3^1953.
Congrats. You're 0 for 3.
There are n = 3*1953 = about 6000 single-point mutations. There are about n^2 two-point mutations. 6000^2 = 36,000,000 is close enuff to 20,000,000. So it looks like every mutation that differed at one or two locations occurrred.
No, js1138's claim was that every possible point mutation occurred and was observed in the experiment. See post #577.
You are trying to re-state the initial claim as if having a mutation at each point was the same as having every possible point mutation at every point.
Disingenuous.
A longer sequence makes it easier to see.
From the initial sequence:
agtcctgagtctacgtatcgata
We get the following single nuceotide changes:
agtcctaagtctacgtatcgata
agtccttagtctacgtatcgata
agtcctcagtctacgtatcgata
agtcctgggtctacgtatcgata
agtcctgtgtctacgtatcgata
agtcctgcgtctacgtatcgata
agtcctgaatctacgtatcgata
agtcctgattctacgtatcgata
agtcctgactctacgtatcgata
3*3=9 single nucleotide mutations
From his we would get, in addition to all of the above:
agtcctaggtctacgtatcgata
agtcctatgtctacgtatcgata
agtcctacgtctacgtatcgata
agtcctaaatctacgtatcgata
agtcctaattctacgtatcgata
agtcctaactctacgtatcgata
agtcctagatctacgtatcgata
agtcctatatctacgtatcgata
agtcctacatctacgtatcgata
agtcctaggtctacgtatcgata
agtcctagttctacgtatcgata
agtcctagctctacgtatcgata
.
.
.
.
And so on...
until will get all permutations of those three nucleotides (minus duplicates) many of which are *not* single nucleotide mutations.
33=27 multiple nucleotide mutations
Troll elsewhere, dude. You're not even amusing.
Tell the truth, dude. The alternative, well, you know what that makes you?
You don't even know the point under debate...
There have been very recent experiments in heat resistant bacteria in which every possible point mutation in the relevant genes was observed. Therefore there are at least some instances in which the source of variation -- random or not -- is irrelevant. ...
Or just try to be an man in this one, and admit you were wrong.
"every possible point mutation"
I'm sorry, dude.
It's OK. Most evolutionists botch the biology and the math, just like you did. I seldom expect more.
...the others tend to have more clever, more original insults than you, though.
Dude? What, are you 12?
What are you, 98?
This restriction to a single point deviation from the original string is what restricts the math to the number of 'other' possible bases at a given 'point' times the number of possible 'points' that can change.
In the sequence CTAG:
If we desire to create a single point mutation we can change the C, the T, the A or the G. If we change more than one of the four, such as the CT, the TA, CG or even CA, then we have gone beyond the initial condition of only changing a single point.
Your math includes all single point changes, all two point changes, all three point changes and all four point changes less duplications.
In light of the above, the argument appears to revolve around whether or not multiple nucleotide changes are to be included in the calculation or, looked another way, whether the 'initial' string for each iteration can contain mutations from a previous instance.
How big are Blue Whale eyes?
El Toro is a vegetarian, but the Matador is very wary of him!
Details....
"we KNOW they got to be there; somewhere..."
-- EvoDude
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