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).
?