BIO 370 Lecture Notes - Lecture 18: Nonsynonymous Substitution, Neutral Mutation, Alcohol Dehydrogenase

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17 May 2018
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How to test for selection?
We expect synonymous changes to be unavailable to and therefore sites at which
synonymous changes may occur will be unaffected by selection.
We expect nonsynonymous changes to be available to selection and so nonsynonymous
substitution at these sites MAY occur more or less rapidly at nonsynonymous sites
depending on the selection regime.
Ka and Ks
Compare protein encoding genes site for site.
Identify synonymous sites (e.g., 3rd codon positions for which changes in sequence will
not change amino acid).
Identify nonsynonymous sites.
Ka is the number of nonsynonymous substitutions
Ks is the number of synonymous substitutions
Ka / Ks meaning
IF the changes are, in fact, selectively neutral, Ka/Ks is expected to be 1 because
nonsynonymous substitutions will occur through drift, the same as synonymous
substitutions.
If selection is maintaining the currently common allele against occasional deleterious
mutation, Ka/Ks is expected to be < 1 (mutations will fixed more rapidly at synonymous
sites [because they are neutral]).
If selection is driving changes across a population away from one phenotype to another,
Ka/Ks is expected to be > 1.
Ka/Ks ~ 1, (near) neutrality
Ka/Ks < 1, purifying selection
keepig the allele pue
Ka/Ks > 1, positive selection
eatig positie hage
Problems?
One part of a protein can undergo purifying selection, while another is undergoing
positive selection, effectively cancelling each other out.
Solution: compare within population and between population Ka/Ks values.
We will explore this further under applications of evolutionary theory when we look at
aspects of evolution of ALDH (alcohol dehydrogenase) in humans.
What is the oleula lok ad hy a’t it keep good tie?
Another way to visualize selection vs. drift at different population sizes
Drift and selection as vectors on a graph.
Drift for a neutral allele is a vector pointing equally strong up and down.
Beneficial selection only points up.
Neutrality
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Document Summary

Ka and ks: compare protein encoding genes site for site. Identify synonymous sites (e. g. , 3rd codon positions for which changes in sequence will not change amino acid). Identify nonsynonymous sites: ka is the number of nonsynonymous substitutions, ks is the number of synonymous substitutions. If the changes are, in fact, selectively neutral, ka/ks is expected to be 1 because nonsynonymous substitutions will occur through drift, the same as synonymous substitutions. If selection is maintaining the currently common allele against occasional deleterious mutation, ka/ks is expected to be < 1 (mutations will fixed more rapidly at synonymous sites [because they are neutral]). If selection is driving changes across a population away from one phenotype to another, Ka/ks is expected to be > 1: ka/ks ~ 1, (near) neutrality, ka/ks < 1, purifying selection (cid:894)(cid:862)keepi(cid:374)g the allele pu(cid:396)e(cid:863)(cid:895, ka/ks > 1, positive selection (cid:894)(cid:862)(cid:272)(cid:396)eati(cid:374)g positi(cid:448)e (cid:272)ha(cid:374)ge(cid:863)(cid:895) If n=5, then loss of one copy of one allele causes a 10% shift in p.

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