Download Chi Square with Fruit Flies

Fruit Fly Lab Write-up
The Fly Family Tree
• P Generation- Set up by the supply
company-removed before shipping.
• F1 Generation- The flies you started your
culture with.
• F2 Generation- The flies you counted.
These are the offspring of the F1.
Specific Write-up reminders:
• Lab is a formal write-up.
• You must include your raw data.
• You need a null hypothesis (Ho) and an
alternative hypothesis (Ha).
The Hypothesis
• Ho: There is no significant difference
between the expected and observed
results, following the phenotypic ratio of
1:1:1:1 (enter your ratio here!).
• Ha: There is a significant difference
between the expected and observed
results following the phenotypic ratio of
1:1:1:1.
Step 1
• Decide on what type of cross you had.
There were three possibilities.
– Monohybrid
– Dihybrid
– Sex linked
• Ask:
– Do I have any white eyes
– Does only one trait vary?
– Do two traits vary?
Total each category
• Sex-linked (4 categories)
– Red eyed ♀
– White eyed♀
-Red eyed ♂
-White eyed ♂
• Monohybrid (2 categories) for instance:
– Apterous vs. wild
– Sepia vs. wild
• Dihybrid (4 categories) for instance:
– Wild/wild
– Apterous/sepia
-sepia/wild
-apterous/wild
Step 2
• Apply the Chi-square statistic.
=Σ
(O-E)2
E
•
X2
•
•
•
•
O=observed number in a phenotypic category
E=expected number in a phenotypic category
X2= “chi-square”
Σ = sigma “sum of”
Example #1 Monohybrid
• Data: 69 wild (red), 31 sepia=100 total flies counted
• A monohybrid cross is a 3:1 ratio according to the
punnett square so you expect:
• 3 out of 4 to be dominant or ¾
• 1 out of 4 to be recessive or ¼
• Multiply the total number counted by the ratio to get the
expected number:
• 100 x ¾ = 75
100 x ¼ = 25
• Solve: X2 = (69-75)2 + (31-25)2
75
25
.48
• X2= 1.92
+
1.44
=1.92
Interpreting the Chi Square
• Decide how many degrees of freedom you
have in this problem.
• D of F = The number of phenotypic
categories – 1.
• There are two categories in the sample
problem: 2-1=1 degree of freedom.
• Look your Chi Square value up in the chart
in your lab book.
• The X2 (Chi-Square) Distribution
Chart
Probability Values
df
0.05
0.025
0.01
0.001
•
1
3.841
5.024
6.635 10.828
2
5.991
7.378
9.210 13.816
3
7.815
9.348 11.345 16.266
4
9.488 11.143 13.277 18.467
5
11.070 12.833 15.086 20.515
6
12.592 14.449 16.812 22.458
Use the .05 probability column or “significance level”
If your chi-square value is less than this value you “fail to reject” your null
hypothesis.
This means you are 95% confident any variation between your population and
the one you thought you had (in the null hypothesis) is due to chance alone (
5% of the true null hypotheses will be rejected).
If your chi-square value is greater you reject your null hypothesis (there is little
chance they are the same population).
Our problem: X2= 1.92
Critical value= 3.842
Fail to reject
Example #2 Dihybrid Cross
Observed data
Eye
Wing
Expected Values:
25
Wild
Wild
47 x 9/16 = 26.4
9
Wild
Apterous
47 x 3/16 = 8.8
8
Sepia
Wild
47 x 3/16 = 8.8
5
Sepia
Apterous
47 x 1/16 = 2.9
47 total
Calculations:
X2 = Σ (O-E)2
E
X2 = Σ (25-26.4)2 + (9-8.8)2 + (8-8.8)2 + (5-2.9)2
26.4
8.8
8.8
2.9
.07 +
.0045 +
.07 +
D of F = 4-1= 3 Critical value= 7.815
1.5 = 1.67
X2 is less
1.67< 7.815
Fail to reject
Example #3 Sex-Linked
• P1= Xr Xr x XRY
Xr
Xr
XR
XR Xr
XR Xr
Y
Xr Y
Xr Y
XR
Xr
Xr
XR Xr
Xr Xr
Y
XR Y
Xr Y
• F1 = XR Xr x Xr Y
• F2 = 1:1:1:1
Expected: Total x ¼
Observed:
♀ red
78
♀ red
71
♀ white
69
♀ white
71
♂ red
62
♂ red
71
♂ white
73
♂ white
71
Calculations:
X 2 = Σ (O-E) 2
E
X 2 = Σ (78-71) 2 + (69-71) 2 + (62-71) 2 + (73-71) 2
71
71
71
71
D of F = 4-1= 3
X 2 is less than critical value
= 1.95
Critical value= 7.815
1.95>7.815
Fail to reject
Why do we “fail to reject” instead of accept?
•
In the words of Gerry Rau, Lincoln American School in Taiwan on the AP
Biology Teachers List:
•
A chi-square less than the critical value leads to the conclusion that the null
hypothesis cannot be rejected, which is slightly different from being
accepted. It may be that it cannot be rejected (proven false) due to small
sample size, large variation in the data, etc. The distinction is that it may be
possible to prove something false, but impossible to prove something true.
Example: If someone says ‘I always tell the truth’, you only need to show
one example of a lie to prove it false, but a thousand examples of telling the
truth, while providing substantial evidence, cannot prove it true.