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On the contrary, 3^1953 is greater than 20 million. The number of single point mutations in a gene of length 1953 is 3*1953, not 3^1953.

Let's take it nice and easy.

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?

"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."

Incorrect. Nice of you to just join up to post such rubbish, though.

AA is 1 base pair out of 1953 in the bacteria gene under debate. A single point mutation (presuming canonical Crick for an ideal B formation double helix - omiting other possible pairs such as AA GG GA CC UU (or TT) CU AC AU (or AT) GC GU (or TT) pairings) would give us AT, AG, AC, TA, GA, or CA.

2 base pairs would then give us 6^2 possibilities.

E.g., instead of AA, GG for the first two base pairs, we could have any of the 6 combinations for AA mentioned above (AT, AG, AC, TA, GA, or CA) in combination with all 6 possibilities for GG (GA, GT, GC, AG, TG, CG).

That's 6 * 6 possibilities for 2 base pairs, which is represented as 6^2.

3 base pairs would give us 6 * 6 * 6 possibilities, represented as 6^3.

You evolutionists need to bone up on simple math. So far no fewer than 2 of you have failed "sequencially" on this thread.

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.

WHAT???