Lecture_8

# Lecture_8 - EEMB 129 LECTURE 8 2010 I Linkage(Chapter 5 X2...

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EEMB 129 - LECTURE 8 2010 1 I. Linkage (Chapter 5) X2 Test Tetrad Analysis X 2 Test When testing for linkage, how different from 50% do the recombinant classes need to be before we conclude that two genes are linked? X 2 = sum of the (Obs - Exp) 2 Exp As the Obs approaches the value of the expected, then the X 2 value is small If the Obs is fairly different than the expected, then the X 2 value will be larger AaBb x aabb Expected if independently OBS assorting O - E (O - E) 2 (O-E) 2 /E AaBb (AB) 18 12.5 5.5 30.25 2.42 Aabb (Ab) 7 12.5 -5.5 30.25 2.42 aaBb (aB) 10 12.5 -2.5 6.25 0.5 aabb (ab) 15 12.5 2.5 6.25 0.5 50 50 5.84 degrees of freedom: # of freely varying parameters. If you examine 50 individuals, any three classes can vary freely, but once you specify how many individuals are in three classes, the last class is fixed. for 3 degrees of freedom, we can then look at a X 2 table Now look at a Chi-square table: Find the row for the degrees of freedom Look across to see where your X 2 value lies Look up to the probability to see between which values your value lies X 2 , df = 3 2.37 < 5.84 < 6.25 corresponding P 0.50 p 0.10 Therefore IF the genes truly were independent (the null hypothesis), THEN you would have between 10-50% chance of obtaining a X 2 value of 5.84

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EEMB 129 - LECTURE 8 2010 2 Tetrad Analysis: There are a number of questions about meiosis that cannot be answered by traditional genetic methods (i.e., with diploid organisms such as fruit flies).
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## This note was uploaded on 10/16/2010 for the course EEMB 129 taught by Professor Ajnarivera during the Spring '09 term at UCSB.

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Lecture_8 - EEMB 129 LECTURE 8 2010 I Linkage(Chapter 5 X2...

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