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Laboratory: Inferring Lateral gene transfer
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  1. Laboratory:Inferring Lateral gene transfer

  2. PREVIOUS LABS • You inferred gene transfer by phylogenetic analysis of a gene TODAYS LAB • You will infer gene transfer by • G+C content • GC3 bias at the third codon position • Codon Bias

  3. How lateral gene transfer can occur? • Plasmids • Transposons • Phages • DNA uptake from the environment

  4. Why study gene transfer? • origin of genes • predict expression levels • organisms can acquire new metabolic capacity (resistance to drugs, ability to degrade compounds that are persistent…)

  5. What we are going to study? • The gene bphA3 in a Transposon in the bacteria called Ralstonia Oxalatica. • Why study this microorganism? • Gram negative bacteria • degrade several chloroaromatic compounds

  6. Chromossome Transposon Where is this bphA3 gene? Bacterial cell

  7. What is this gene bph? Responsible for the production of enzymes that breakes a compound called Chlorobiphenyl, which is toxic, carcinogenic and persistent in the environment. When broken, can be used as carbon source also.

  8. Count Data = # of times codon is used out of total codons analyzed (i.e. 280 out of 6003).

  9. RSCU values = frequency with which a codon is used compared to other synonymous codons i.e. Phe has 2 codons. If each were used equally, RSCU for UUU = 1 and RSCU for UUC = 1.

  10. Start codons, Stop codons, Tryptophan codons should not be used in the analysis. Start codons and Tryptophan have only one codon and therefore no bias is possible. Stop codons occur only once per protein and will skew results if included.

  11. Run Chi-Square test to see if %Codon usage for entire genome/transposon is significantly different than your gene bphA3 • Chi Square: (Observed – Expected)2/ Expected ∑((O - E)2/E)

  12. Expected Values (Average of all ORF’s from entire genome or transposon) For Each Codon, Chi-Square will subtract observed from expected, square that, then divide by expected. It will then sum all of these together. This final number is the Chi-Square Statistic Observed Values (Codon usage of single gene-bphA)

  13. Chi-Square Statistic Incorrect values, do not use!

  14. There are only 41 degrees of freedom There are 6 Leu codons. Once 5 of these are known, the 6th can be determined = 5 degrees of freedom. If done for all AA’s, = 41 degrees of freedom.

  15. If your Chi-Square statistic is greater than the value in the table for any given Degree of Freedom (df), than your results are statistically significant e. g. Chi-Square Stat = 17.8, df= 8. Results are statistically significant, P>0.01, P<0.05.