Tuesday, 6 October 2009

Finding scientific references given in talks or lectures

I have been asked how you can track down papers online when a lecturer gives a paper reference as, for example:

Clin Med. (2003) 3, 333-7

The easiest way to track this down is to use 'Single Citation Matcher'.

Go to pubmed and the link can be found in the left hand menu. Alternatively, follow this direct link:
http://www.ncbi.nlm.nih.gov/corehtml/query/static/citmatch.html

On the page enter the required information, so using the above reference, Clin Med. (2003) 3, 333-7, you should end up with:

Single citation manager

Click the 'Go' button and you should be taken to the paper (and any link to the full paper if available).

Abstract1

The abstract for the paper - don't forget there may be a link to the full paper on the far righthand side of the page.

Monday, 5 October 2009

Protein sequence to DNA - Degeneracy calculation

I have had a some questions about reverse translation of protein to DNA, and degeneracy....

The protein sequence is: THERIGHTREADINGFRAME

It is 20 amino acids, and therefore you will need 60 bases to encode it. So....

Protein Seq:  T  H  E  R  I  G  H  T  R  E  A  D  I  N  G  F  R  A  M  E 
DNA Seq:     ACNCAYGARMGNATHGGNCAYACNMGNGARGCNGAYATHAAYGGNTTYMGNGCNATGGAR    
Full DNA:    ACACACGAACGAATAGGACACACACGAGAAGCAGACATAAACGGATTCCGAGCAATGGAA
               T  T  G  T  C  T  T  T  T  G  T  T  C  T  T  T  T  T     G
               G        G  T  C     G  G     C     T     C     G  C
               C        C     G     C  C     G           G     C  G
                      AGG            AGG                     AGG
                        A              A                       A
Number codons: 4  2  2  6  3  4  2  4  6  2  4  2  3  2  4  2  6  4  1  2
So, 4 x 2 x 2 x 6 x 3 x 4 x 2 x 4 x 6 x 2 x 4 x 2 x 3 x 2 x 4 x 2 x 6 x 4 x 1 x 2 = 2,038,431,744 or 2 x 109 possible DNA sequences would encode the protein sequence.

This is a big number, however, compared to the total number of possible DNA sequences you could have for a 60 base sequence, it is small.

The total number of DNA sequences you could have for a 60 base sequence is 4 x 4 x 4.... sixty times, or 460, which is equal to 1.3 x 1036 possible sequences. Of those 1.3 x 1036 sequences only 2,038,431,744 would encode THERIGHTREADINGFRAME. Or in percentage terms, (2,038,431,744 / 1.3 x 1036) x 100 = 0.0000000000000000000000002% (2 x 10-25%) of all the possible sequences would encode THERIGHTREADINGFRAME.

Sunday, 4 October 2009

Calculating Dilutions

Blog Post Bonus: A free eBook on dilutions is available on Gumroad
I have received a number of emails about dilution calculations.

The type of question is:

You have been asked to dilute a 10 mg/ml solution to give 5 ml of a 1.5 mg/ml solution.

How many ml of water would you use?

How many ml of the stock protein solution would you use?

The way I would work this out is by saying that 5 ml of 1.5 mg/ml contains 5 x 1.5 mg = 7.5 mg.

If the stock is 10 mg/ml then I need to take a volume that would give 7.5 mg. So, the number of ml of the stock I need is 7.5 / 10 = 0.75 ml (that is, 0.75 ml of a 10 mg/ml solution contains 7.5 mg).

Therefore, I have 0.75 ml of the stock, and the final volume is 5 ml, so we need to add 5 - 0.75 = 4.25 ml of water.

The answer:

How many ml of water would you use? 4.25 ml

How many ml of the stock protein solution would you use? 0.75 ml

Although I prefer to work it out as above you can use a formula:

M1V1 = M2V2

Where:
M1 = Concentration 1
V1 = Volume 1
M2 = Concentration 2
V2 = Volume 2
Therefore:
M1 = Concentration 1 = 10 mg/ml
V1 = Volume 1 = Unknown X
M2 = Concentration 2 = 1.5 mg/ml
V2 = Volume 2 = 5 ml

or:
M1V1 = M2V2 10 mg/ml x Unknown X = 1.5 mg/ml x 5 ml

Rearranging to solve for Unknown X:

Unknown X = (1.5 mg/ml x 5 ml) / 10 mg/ml Therefore, Unknown X = 0.75 ml

Using the formula is the same as the first solution, but in the first solution, I don't have to remember the formula....

Blog Post Bonus: A free eBook on dilutions is available on Gumroad
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