Analyzing Data - Western Washington University
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Transcript Analyzing Data - Western Washington University
Analyzing Data
c2 Test….”Chi” Square
Forked-Line Method, F2
UuDd x UuDd
1/4 UU
1/2 Uu
1/4 uu
1/4 DD
1/4 x 1/4 = 1/16 UUDD
1/2 Dd
1/4 x 1/2 = 1/8 UUDd
1/4 dd
1/4 x 1/4 = 1/16 UUdd
1/4 DD
1/2 x 1/4 = 1/8 UuDD
1/2 Dd
1/2 x 1/2 = 1/4 UuDd
1/4 dd
1/2 x 1/4 = 1/8 Uudd
1/4 DD
1/4 x 1/4 = 1/16 uuDD
1/2 Dd
1/4 x 1/2 = 1/8 uuDd
1/4 dd
1/4 x 1/4 = 1/16 uudd
Genotypes U--D-If you pick wt F2 worms, what will be the proportion of each class in the F3?
Work this out in your notebooks, for Thursday.
1/4 UU
1/2 Uu
1/4 DD
1/4 x 1/4 = 1/16 UUDD
1/2 Dd
1/4 x 1/2 = 1/8 UUDd
1/4 DD
1/2 x 1/4 = 1/8 UuDD
1/2 Dd
1/2 x 1/2 = 1/4 UuDd
Chance
Deviation
Experimental Data
Does it fit Expected
Values?
The outcomes of
segregation, independent
assortment, and
fertilization are subject to
random fluctuations from
predicted occurrences…
…as a result of chance.
As the sample size
increases, the average
deviation from the
expected fraction or ratio
is expected to decrease.
Mendel
Goodness of Fit
Mendel had no way of solving this problem,
Karl Pearson and R.A. Fisher developed the c2 “chisquare” test for this type of application.
The chi-square test is a “goodness of fit” test…
…it answers the question of how well do
experimental data fit expectations.
Null Hypothesis
null hypothesis: there is no statistical difference
between the observed value and the predicted value.
The apparent difference can be attributed to chance
alternative hypothesis: there is a statistical difference
between the observed value and the predicted value. The
apparent difference can not be attributed to chance alone.
Important: accepting or rejecting the null hypothesis
does not prove anything, it only provides a statistical
means of measuring the difference between the
observed and expected results.
Answer these
questions…
290 purple; 110 white flowers
Offspring n = 400
Classes n = 2 (purple, white)
Observed:
290 purple
110 white
Expected:
300 purple
100 white
what is the total number of
offspring?
offspring n = ______
how many different classes
(phenotypes or genotypes)
of offspring did you
observe?
classes n = _______
what are the classes, and
how many were observed
for each class?
based on genotype classes,
how many of each
phenotype do you expect in
each class in each...
Null
Hypothesis
290 purple; 110 white flowers
Offspring n = 400
Classes n = 2 (purple, white)
Observed:
290 purple
110 white
Expected:
300 purple
100 white
The observed 290 purple
flowers and 110 white
flowers are statistically no
different than the expected
values of 300 purple flowers
and 100 white flowers.
The apparent difference is
due to chance.
The Formula
O: observed
E: expected
Observed:
290 purple
110 white
purple
white
290
110
Expected:
300 purple
100 white
300
100
10
10
100
100
0.33
1
1.33
Degree of Freedom
A critical factor in using the chi-square test is the
“degrees of freedom”, which is essentially the
number of independent random variables involved,
For this analysis (c2 ), the degrees of freedom is
simply the number of classes, minus 1.
For our example, there are 2 classes of offspring:
purple and white. Thus, degrees of freedom…
(d.f.) = 2 -1 = 1.
What does the c2 value mean?
Critical values for chi-square are found on tables,
sorted by degrees of freedom and probability levels...
c2 = 1.33
Null Hypothesis
We will use p = 0.05 as the critical value,
if p < 0.05, you “reject the null hypothesis”
if p > 0.05, you “fail to reject the null hypothesis”
…that is, you accept that your genetic hypothesis about
the expected ratio is correct*.
* actually, not not correct...does this make it “TRUE”?
What does the c2 value mean?
One way to think of
this, is that if you did
this experiment 100
p = ~0.2
times, you’d get similar
results, do to chance,
roughly 20 times.
c2 = 1.33
Null Hypothesis: There is no statistical difference between
the measured values, and the expected values (with details).
Chi Square Analysis: “What is the probability that the
null hypothesis is correct?”
Reject Null Hypothesis: p = < 0.05
Wednesday
1. Pick L4, cross progeny,
P UUDD x uudd
F1
UuDd (L4)
2. Bioinformatics #2,
1. WormBase
TO DO before Friday!
Look up phenotype info.
Friday
Wormbase
Entrez
Determine phenotype(s)
for each mutant
(assignment, due at start of
class Friday), and then ID
worms in class (on Friday).
?