Exercise 6B: Codon
Usage
STA
4953 (Spring
2001)
Due 3/8/2001
Find out what the GCG program "Codonfrequency" does. Run the
program on the hemoglobin sequences retrieved in Exercise #5.
When several codons code for the same amino acid, they are called
synonomous codons. In gerneral, do you think synonomous codons are used
equally? Give some evidence to support your answer.
Solution:
Observing the results of Exercise 6A, it does appear as if the synonomous
codons are not used equally. But, to have a statistical certainty that the
conclusions drawn from observing the data are correct, a statistical
goodness of fit test must be performed, such as a X2
test.
Choose one of the amino acids to run the test on. The test on the
amino acid, Leucine, from the Delta chain would be as follows:
H0: All of the codons coding for Leucine in the Delta chain are
used equally: pUUA = pUUG = pCUA =
pCUC = pCUG = pCUU = 3 /
18 = 1 / 6.
H1: pX does not equal 1 / 6
for some x = {UUA, UUG, CUA, CUC, CUG, CUU}.
You may choose to run a (i) Pearson's goodness of fit test, or (ii)
Liklihood ratio test.
(i) Pearson's goodness of fit test:
X2 = sum(i={UUA, UUG, CUA, CUC, CUG, CUU}) [(Oi -
Ei)2/
Ei]
= [(0-3)2/3] +
[(0-3)2/3] + [(0-3)2/3] +
[(3-3)2/3] +
[(15-3)2/3] +
[(0-3)2/3]
= 60
Using a Chi-square (right-tailed) table look up the value for a
Chi-square random variable with 5 degrees of freedom and a p-value of 0.05
to get 11.07. Any value of the test statistic (calculated above) that is
larger than 11.07, will lead to a rejection of the null hypothesis,
Ho. Since 60 > 11.07, we reject Ho.
(ii). Liklihood ratio test
G2 = 2 * {sum(i={UUA, UUG, CUA, CUC, CUG,
CUU}) [Oi *
log(Oi /Ei)]}
= 2 * [(0) * log(0/3) +
(0) *
log(0/3) + (0) *
log(0/3) + (3) *
log(3/3) + (15) *
log(15/3) + (0) *
log(0/3) ]
= 48.2831
Since 48.2831 > 11.07, we reject Ho.
Since both the Pearson and Liklihood ratio tests led to a rejection of
the null hypothesis, we would conclude that there is evidence to
suggest that synonomous codons are not used equally.
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