Question 1:

Is Log Odds ratio, which is Log (qij/eij), the same as Log Odds score? Or, Log Odds score is the value in BLOSUM score matrix, which equals 2*Log (qij/eij)?

The log-odds ratio, s(i,j) = log qij/pi pj, is indeed the log-odds score for a pair of characters (here, "log" stands for natural logarithm).  On the other hand, the entries of actual scoring matrices (e.g., BLOSUM) are typically computed using different bases for the log and various scaling factors.

For instance, in BLOSUM matrices, eij = pi pj; therefore, the log-odds score for the pair of characters i, j is s(i,j) = log qij / eij.  For BLOSUM62, however, base-2 logs are used instead of natural logs, and the result is multiplied by 2 (the scoring unit in this case is called "half-bits").  Thus, the score is s'(i,j) = 2 * log2 qij / eij, where log2 stands for log base 2.  These are the values in BLOSUM scoring matrix.  (See p. 103 of the text.)

When comparing scores, it is important to know the unit in which they are provided.  The conversion factor between these units and log-odd scores is

(essentially) the lambda value that we saw in class.  For BLOSUM62, s(i,j) =

(1/lambda) * s'(i,j), where lambda = (1/2) * log 2.  Therefore, the BLOSUM scores are just the log-odds scores multiplied by a scaling factor.

Question 2:

I am very confused about how to convert the log odds score of an alignment to odds score.

For examples, one of lab 2 question,

DEDEDEDE

DDDDDDDD

I calculated the log odds score in the fllowing way. Here, I indicated log base 2 as log as following.

The odds score of finding a D to E substitution in an alignment is qDE/eDE=0.10/0.05=2.

The log odds score for the D to E substitution in bits is log 2= 1.  The odds score of finding a D to E substitution in an alignment is qDE/eDE=0.8/0.1=8.  The corresponding log odds score in bits is log 8= 3.  The log odds score of the following alignment is 1+3+1+3+1+3+1+3=16.

To convert to odds score, I don't know which of the following ways are correct?

(1): odds score of the alignment is the anti log odds socre = 2^16=65536.

(2): odds score of the alignment equal to the sum of odds score of individual pairs= 8+2+8+2+8+2+8+2=40.

Could you please let me know which one is correct?

The log-odds score is just the log of the odds score.  The conversion factor depends on the base used for the log.  If the log-odds score was computed as S = log_b D, where b is the base and D is the odds score, then D = b^S (b raised to the S).  An odds scores is typically the product of probability ratios.  Translated to log-odds, this becomes the sum of logs of probability ratios.  Thus, option (1) below is the correct one, provided the invidual probabilities are computed correctly.

Question 3:

Average profiles were introduced in

Gribskov et al. PNAS. 84 (13): 4355 (1987).

The paper

Gribskov, M. and Veretnik, S., Methods Enzymol. 266, 198-212, 1996

discusses evolutionary profiles and compares them with average profiles.