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142 yyR_
yyR_
124
142 yyrr
146
142 (1
2
(
( 1 5 4 −1 4 2 ) 2 4 4 −1 4 2 ) ( 1 2 4 −1 4 2 ) 2 1 4 6 −1 4 2 ) 2
+ 142
+
+ 142
142
142 χ = 1 .0 1 + 0 .0 2 8 + 2 .2 8 + 0 .1 1 3 = 3 .4 3 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 by chance (Table 5.1 in text)
chance
Remember: Χ 2 = 3.43 and df = 3
Remember:
3.43
p value ≈ 0.3 null hypothesis (no linkage) is accepted
value
is If observed values deviate from values predicted by our hypothesis, is this
significant (reflecting a real difference between observation + theory) or insignificant
insignificant
(reflecting random sampling error or chance)? P value lets us know this…. 6. Evaluate the significance of the p value. The
6. convention is that a 0.05 p value is the boundary
convention
between accepting and rejecting the null hypothesis. P value – the probability that a deviation from the predicted numbers at least
value
as large as that observed in the experiment has occurred due solely to chance
Very small p values indicate a high degree of significant difference
significant
between observed and expected data i.e.
between
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 and so is a REAL difference)
REAL
If very low, deviation of observed from expected results becomes significant,
very
significant,
unlikely hypothesis being tested explains data (i.e. if p value = 0.01, only 1 in 100
times will that observation be due to chance alone) If high, likely hypothesis being tested is correct, deviation from expected is
high
insignificant (i.e. p value = 0.90 90 out of 100 times the observation would
occur by chance if the null hypothesis is correct)
occur Are genes A and B linked in testcross AB/ab x ab/ab ?
Are
AB/ab ab/ab
ab
ab
ab
ab χ 2= (1812.5)2 + (1512.5)2
12.5
12.5 χ 2= 2.42 + 0.5 + 2.42 + 0.5
χ 2= 5.84
Df =...
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 Spring '10
 christineDupont
 Genetics

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