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d_kennetz
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Sure, so you start out with what is called a bitscore, which is a normalized to the score calculated from the alignment between the 2 seqs which depends on the following equation. It is independent of database size:

(lambda * S - ln(k))/(ln)2

Then, the p-value of a local blast is just:

1/2^bitscore

So if your bitscore is 15, you need 1/32768 alignments before you will get a score as good or better (highly similar sequences) BY CHANCE ALONE. This addresses your "BY CHANCE" question.

Earlier I said the bitscore was independent of database size. The E value is just the p-value above normalized to the database size (so it is dependent on the db size) by the following equation:

(query length * database length * p-value)

which simplifies a bit to:

E = (query length* (db length/2^bitscore)

I hope this helps!

Sure, so you start out with what is called a bitscore, which is a normalized to the score calculated from the alignment between the 2 seqs which depends on the following equation. It is independent of database size:

(lambda * S - ln(k))/(ln)2

Then, the p-value of a local blast is just:

1/2^bitscore

So if your bitscore is 15, you need 1/32768 alignments before you will get a score as good or better (highly similar sequences).

Earlier I said the bitscore was independent of database size. The E value is just the p-value above normalized to the database size (so it is dependent on the db size) by the following equation:

(query length * database length * p-value)

which simplifies a bit to:

E = (query length* (db length/2^bitscore)

I hope this helps!

Sure, so you start out with what is called a bitscore, which is a normalized to the score calculated from the alignment between the 2 seqs which depends on the following equation. It is independent of database size:

(lambda * S - ln(k))/(ln)2

Then, the p-value of a local blast is just:

1/2^bitscore

So if your bitscore is 15, you need 1/32768 alignments before you will get a score as good or better (highly similar sequences) BY CHANCE ALONE. This addresses your "BY CHANCE" question.

Earlier I said the bitscore was independent of database size. The E value is just the p-value above normalized to the database size (so it is dependent on the db size) by the following equation:

(query length * database length * p-value)

which simplifies a bit to:

E = (query length* (db length/2^bitscore)

I hope this helps!

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d_kennetz
  • 581
  • 3
  • 16

Sure, so you start out with what is called a bitscore, which is a normalized to the score calculated from the alignment between the 2 seqs which depends on the following equation. It is independent of database size:

(lambda * S - ln(k))/(ln)2

Then, the p-value of a local blast is just:

1/2^bitscore

So if your bitscore is 15, you need 1/32768 alignments before you will get a score as good or better (highly similar sequences).

Earlier I said the bitscore was independent of database size. The E value is just the p-value above normalized to the database size (so it is dependent on the db size) by the following equation:

(query length * database length * p-value)

which simplifies a bit to:

E = (query length* (db length/2^bitscore)

I hope this helps!