Lecture 6 - BIS101-001 Lecture 6 Bacterial genetics and...

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Unformatted text preview: BIS101-001 Lecture 6. 1/26/09 Bacterial genetics and mapping Announcements- Pick up review quiz after Wednesday in boxes outside MCB ofFce.- Midterm 1 is this Wednesday 1/28 in class. 100 points, covers material from lectures 1-5 Show your work! No calculators Must be in pen for regrade request- Practice midterms on course website (“Resources” section) LECTURE 6: OUTLINE Statistical signifcance oF linkage UseFulness oF bacterial genetics Experimental methods Gene transFer in bacteria: 1) Conjugation: bacterial sex- Bacterial gene mapping by conjugation- Bacterial gene mapping by recombination 2) TransFormation (DNA uptake) 3) Transduction (bacteriophages) Chi-square analysis Is there statistical evidence for linkage? For Ab / aB x ab / ab testcross, Ab /ab 280 aB /ab 270 AB /ab 220 ab /ab 230 Calculated map distance = 220 + 230 / 1000 = 45cM Two possibilities 1) Linked genes 45cM apart 2) Random deviation from expected 1:1:1:1 ratio Chi-square analysis For Ab / aB x ab / ab testcross, Ab /ab 280 aB /ab 270 AB /ab 220 ab /ab 230 Null hypothesis = simplest and most straightforward explanation. In this case, assume genes are unlinked and random ¡uctuation caused observed progeny Chi-square analysis Chi-squared formula: Chi-squared = total of (observed - expected) 2 observed for all classes Chi-squared = (280-250 ) 2 + (270-250 ) 2 + (220-250 ) 2 + (230-250 ) 2 250 250 250 250 3.6 + 1.6 + 3.6 + 1.6 = 10.4 Consult a chi-squared table Degrees of freedom (df) = # classes - 1 In our case, 4 gamete types - 1 = 3 degrees of freedom Consult a chi-squared table...
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Lecture 6 - BIS101-001 Lecture 6 Bacterial genetics and...

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