Unformatted text preview: btained from a breeding experiment
to determine the total number of offspring, the
number of different classes of offspring and the
number of offspring observed in each class
2. Calculate the number of offspring that would
be expected for each class if the null hypothesis
were correct (percentage predicted x total
number of offspring)
30 The Chi Square Test
3. χ2 = Σ (Number observedNumber expected)2
Number expected 4. Determine the degrees of freedom (df) :
number of classes 1 5. Use the chisquare value and number of
degrees of freedom to determine a p value :
the probability that a deviation from the
predicted numbers at least as large as
that observed in the experiment will occur
31
by chance (Table 5.1 in text) The Chi Square Test
6. Evaluate the significance of the p value.
The convention is that a 0.05 p value is
the boundary between accepting and
rejecting the null hypothesis. Very small p
values indicate a high degree of significant difference 0.05 = 5% chance that the observation
is due to chance alone
(in other words, a 95% probability
that the observation is not due
to chance. It is a REAL difference)
32 chromatids from Ps
F1
Are genes A and B linked in testcross AB/ab x ab/ab ? F2 progeny
AaBb
aabb
Aabb
aaBb observed
17
14
8
11
50 expected no linkage Null hypothesis: 2 gene pairs are assorting independently
Hypothesis predicts # of parental phenotypes = # of
recombinant phenotypes (1/2 parental class, 1/2 recombinant)
Expected values would be 25 for each class phenotype! 33 parentals Are genes A and B linked in testcross AB/ab x ab/ab ?
progeny
AaBb
aabb
Aabb
aaBb observed
17
14
8
11
50 expected …if unlinked, then independent assortment:
Ab AB ab aB F1 AaBb
test cross
aabb ab Ab
ab
R AB
ab
P ab
ab
P aB
ab
R 34 Are genes A and B linked in testcross AB/ab x ab/ab ?
progeny
AaBb
parental class
aabb
Aabb
recombinant
aaBb
class observed expected
17
25
31
14
8
25
19
11...
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This document was uploaded on 02/25/2014.
 Winter '14
 Genetics

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