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Unformatted text preview: hought of as a “null hypothesis” – null hypothesis , or H0, can be any hypothesis – here, H0 is the hypothesis that the trait is being determined by a single Mendelian factor with dominance in a monohybrid cross • the alterna5ve hypothesis, or H1, is that there is not a Mendelian basis for the phenotypic ra7o that we are measuring Chi square test of goodness of ﬁt • Chi square distribu7on with k degrees of freedom = distribu7on of the sum of squared devia7ons of k independent data points (standard normal random variables) • how do we determine k, the number of degrees of freedom? • here we have two types of progeny: tall and short, and 86 total progeny tall short total 66 20 86
for a fixed # of observations
(fixed marginal value of a
1 row x 2 column data table
(in red)
once we fill one of the red cells => we have 1
degree of freedom the other cell is automatically
determined (i.e. is not
independent of the first cell) in general • for a list of categories (only 1 column or row) – # degrees of freedom k = # categories – 1 – “one way chi square test” • for a table with N rows and C columns – # degrees of freedom k = – (N 1) x (C 1) – “two way chi square test” calcula7ng chi square tall short total observed 66 20 86 expected =(86)(3/4)= 64.5 =86(1/4) = 21.5  observed expected 1.5 1.5  (obs exp)^2 2.25 2.25  (obs exp)^2/exp 0.035 0.105 0.14 chisquare = 0.14 with 1 degree of freedom
in MS Excel type the following into a cell
=chidist(0.14,1)
returns
0.708, which is the P value what is the P value? • it is the chance of genng the data that you got, by chance, if the expected ra7o were actually the value in the popula7on you are sampling – scoring phenotypic/genotypic ra7os of oﬀspring can be thought of as sampling the meio7c output of the parents what does the P value mean? • conven7on: if P<0.05 • which means, there is less than a 1 in 20 chance of genng the data by chance if the expected ra7o is the true popula7on ra7o • if P<0.05, then this is a “sta7s7cally signiﬁcant” result; i.e. something other than the expected ra7o is the actual ra7o in the popula7on – i.e. something other than a Mendelian mechanism is determining the cross outcome P value in hypothesis tes7ng • in general, if P < 0.05, we reject the null hypothesis (H0) and accept the alterna7ve hypothesis (H1) coming to a conclusion from our cross data •
•
•
• 66 tall, 20 short (ra7o = 3.3:1) P = 0.708 => we cannot reject the null hypothesis these data are consistent with a single, Mendelian locus with dominance in a monohybrid cross in class exercise 3 • let’s do a chi square goodness of ﬁt for a single locus cross with genotype data • make up some data for – #DD – #Dd – #dd...
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This note was uploaded on 10/01/2013 for the course BIO 321 taught by Professor John during the Spring '12 term at SUNY Stony Brook.
 Spring '12
 john
 DNA, RNA

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