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Old 07-30-2012, 01:59 PM   #1
owenman
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Default bowtie2 minimum score threshold

I am using bowtie2 version 2.0.0-beta7

A careful reading of the bowtie2 manual section for local alignments makes it sounds like the best score you can possibly get is 2 times the read length:

"A base that matches receives a bonus of +2 by default."

"The best possible score in local mode equals the match bonus times the length of the read."

But the very next paragraph says:

"In local alignment mdoe, the default minimum score threshold is 20 + 8.0 * ln(L), where L is the read length."

This seems contradictory: If a perfect match receives a score of 2 times the read length, how can you ever get a score that is 20 plus 8 times the length?

Am I missing something?
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Old 07-30-2012, 02:10 PM   #2
nickloman
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My reading is that it's 20 + 8 * the natural log of the read length.
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Old 07-30-2012, 03:44 PM   #3
owenman
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Ah, right. I guess my "careful reading" wasn't careful enough.

I wasn't expecting a natural log - I was reading ln(L) as "the length of read L".

It's clear now that you point it out. And I also missed the section on writing formulas farther down in the manual: "The available function types are constant (C), linear (L), square-root (S), and natural log (G)." How embarrassing!

Thank you, nickloman, for your reply.
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Old 08-01-2012, 04:58 PM   #4
cam.jack
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The first paragraph you read is for end-to-end alignment and the 2nd is for local alignment. They are different methods with different scoring systems.
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Old 08-31-2012, 03:59 PM   #5
joseph
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Default --score-min bowitie2

Hi
I am using bowtie2.0.0-beta7 in --local mode to align SE 50 bases reads; I would like to change the parameter --score-min (default in --local mode is G,20,8) to increase the alignment score. Any suggestions?
Thank you,
Joseph Dhahbi
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