1. CHI-SQUARE
To:
Sir Shahid Mahmood
By:
Abuzar Tabassum
M.Sc Zoology 3rd semester
10040814-017
University Of Gujrat

2. Chi-Square Test
• A fundamental problem in genetics is determining
whether the experimentally determined data fits
the results expected from theory (i.e. Mendel’s
laws as expressed in the Punnett square).
• A statistical method used to determine
GOODNESS OF FIT
– Goodness of fit refers to how close the
observed data are to those predicted from a
hypothesis

3. Goodness of Fit
• Mendel has no way of solving this problem. Shortly after the
rediscovery of his work in 1900, Karl Pearson and R.A. Fisher
developed the “chi-square” test for this purpose.
• The chi-square test is a “goodness of fit” test: it answers the
question of how well do experimental data fit expectations.
• We start with a theory for how the offspring will be
distributed: the “null hypothesis”. We will discuss the
offspring of a self-pollination of a heterozygote. The null
hypothesis is that the offspring will appear in a ratio of 3/4
dominant to 1/4 recessive.

4. Formula
• To calculate the chi-square statistic following formula is used.
2 (obs exp) 2
exp
• The “Χ” is the Greek letter chi; the “∑” is a sigma; it means to sum
the following terms for all phenotypes.
• “obs” is the number of individuals of the given phenotype observed;
• “exp” is the number of that phenotype expected from the null
hypothesis.

5. • How can you tell if an observed set of
offspring counts is legitimately the result of a
given underlying simple ratio?
• For example, you do a cross and see 290
purple flowers and 110 white flowers in the
offspring. This is pretty close to a 3/4 : 1/4
ratio, but how do you formally define "pretty
close"? What about 250:150?

7. Example
• As an example, you count F2 offspring, and get 290 purple and 110
white flowers. This is a total of 400 (290 + 110) offspring.
• We expect a 3/4 : 1/4 ratio. We need to calculate the expected
numbers, this is done by multiplying the total offspring by the expected
proportions. This we expect 400 * 3/4 = 300 purple, and 400 * 1/4 =
100 white.
• Thus, for purple, obs = 290 and exp = 300. For white, obs = 110 and
exp = 100.
• Now it's just a matter of plugging into the formula:
2 = (290 - 300)2 / 300 + (110 - 100)2 / 100
= (-10)2 / 300 + (10)2 / 100
= 100 / 300 + 100 / 100
= 0.333 + 1.000
= 1.333.
• This is our chi-square value: now we need to see what it means and
how to use it.

8. The Critical Question
• Using the example here, how can you tell if your 290: 110 offspring
ratio really fits a 3/4 : 1/4 ratio (as expected from selfing a
heterozygote).You can’t be certain, but you can at least determine
whether your result is reasonable.

9. Reasonable
• What is a “reasonable” result is subjective and arbitrary.
• For most work a result is said to not differ significantly from
expectations if it could happen at least 1 time in 20. That is, if
the difference between the observed results and the expected
results is small enough that it would be seen at least 1 time in
20 over thousands of experiments, we “fail to reject” the null
hypothesis.
• For technical reasons, we use “fail to reject” instead of
“accept”.
• “1 time in 20” can be written as a probability value p =
0.05, because 1/20 = 0.05.
• Another way of putting this. If your experimental results are
worse than 95% of all similar results, they get rejected because
you may have used an incorrect null hypothesis.

10. • The test statistic is compared to a
theoretical probability distribution
• In order to use this distribution properly
you need to determine the degrees of
freedom
• If the level of significance read from the
table is greater than .05 or 5% then your
hypothesis is accepted and the data is
useful
• The hypothesis is termed the null
hypothesis which states that there is no
substantial statistical deviation between
observed and expected data.

11. Degrees of Freedom
• A critical factor in using the chi-square test is
the “degrees of freedom”.
• Degrees of freedom is the number of
phenotypic possibilities in your cross minus
one.
Or
• Degrees of freedom is simply the number of
classes of offspring minus 1.
• For our example, there are 2 classes of
offspring: purple and white. Thus, degrees of
freedom (d.f.) = 2 -1 = 1.

12. Critical Chi-Square
• Critical values for chi-square are found on
tables, sorted by degrees of freedom and probability
levels. Be sure to use p = 0.05.
• If your calculated chi-square value is greater than the
critical value from the table, you “reject the null
hypothesis”.
• If your chi-square value is less than the critical
value, you “fail to reject” the null hypothesis (that
is, you accept that your genetic theory about the
expected ratio is correct).

13. Chi-Square Table

14. Using the Table
• In our example of 290 purple to 110 white, we
calculated a chi-square value of 1.333, with 1 degree
of freedom.
• Looking at the table, 1 d.f. is the first row, and p =
0.05 is the sixth column. Here we find the critical
chi-square value, 3.841.
• Since our calculated chi-square, 1.333, is less than
the critical value, 3.841, we “fail to reject” the null
hypothesis. Thus, an observed ratio of 290 purple to
110 white is a good fit to a 3/4 to 1/4 ratio.

15. Chi-square with more than one
degree of freedom

16. 9:3:3:1

17. phenotype observed expected expected
proportion number
round 315 9/16 312.75
yellow
round 101 3/16 104.25
green
wrinkled 108 3/16 104.25
yellow
wrinkled 32 1/16 34.75
green
total 556 556

18. Finding the Expected Numbers
• You are given the observed numbers, and you determine the
expected proportions from a Punnett square.
• To get the expected numbers of offspring, first add up the
observed offspring to get the total number of offspring. In
this case, 315 + 101 + 108 + 32 = 556.
• Then multiply total offspring by the expected proportion:
--expected round yellow = 9/16 * 556 = 312.75
--expected round green = 3/16 * 556 = 104.25
--expected wrinkled yellow = 3/16 * 556 = 104.25
--expected wrinkled green = 1/16 * 556 = 34.75
• Note that these add up to 556, the observed total offspring.

19. Calculating the Chi-Square Value
• Use the formula.
• X2 = (315 - 312.75)2 / 312.75
+ (101 - 104.25)2 / 104.25
+ (108 - 104.25)2 / 104.25
+ (32 - 34.75)2 / 34.75
= 0.016 + 0.101 + 0.135 + 0.218
= 0.470.
2 (obs exp) 2
exp

20. D.F. and Critical Value
• Degrees of freedom is 1 less than the number of
classes of offspring. Here, 4 - 1 = 3 d.f.
• For 3 d.f. and p = 0.05, the critical chi-square value is
7.815.
• Since the observed chi-square (0.470) is less than the
critical value, we fail to reject the null hypothesis.
We accept Mendel’s conclusion that the observed
results for a 9/16 : 3/16 : 3/16 : 1/16 ratio.
• It should be mentioned that all of Mendel’s numbers
are unreasonably accurate.

21. Chi-Square Table

22. Consider Another example:
For Drosophila melanogaster
• Gene affecting wing shape • Gene affecting body color
– c+ = Normal wing – e+ = Normal (gray)
– c = Curved wing – e = ebony
• Note:
– The wild-type allele is designated with a + sign
– Recessive mutant alleles are designated with lowercase
letters
• The Cross:
– A cross is made between two true-breeding flies (c+c+e+e+ and
ccee). The flies of the F1 generation are then allowed to mate
with each other to produce an F2 generation.

23. • The outcome
– F1 generation
• All offspring have straight wings and gray bodies
– F2 generation
• 193 straight wings, gray bodies
• 69 straight wings, ebony bodies
• 64 curved wings, gray bodies
• 26 curved wings, ebony bodies
• 352 total flies

24. Applying the chi square test
Step 1: Propose a null hypothesis that allows
us to calculate the expected values based on
Mendel’s laws
• The two traits are independently assorting

25.  Step 2: Calculate the expected values of the four phenotypes,
based on the hypothesis
• According to our hypothesis, there should be a 9:3:3:1 ratio
on the F2 generation
Phenotype Expected Expected Observed number
probability number
straight wings, 9/16 9/16 X 352 = 198 193
gray bodies
straight wings, 3/16 3/16 X 352 = 66 64
ebony bodies
curved wings, 3/16 3/16 X 352 = 66 62
gray bodies
curved wings, 1/16 1/16 X 352 = 22 24
ebony bodies

26.  Step 3: Apply the chi square formula
(O1 – E1)2 (O2 – E2)2 (O3 – E3)2 (O4 – E4)2
+ + +
E1 E2 E3 E4
(193 – 198)2 (69 – 66)2 (64 – 66)2 (26 – 22)2
+ + +
198 66 66 22
0.13 + 0.14 + 0.06 + 0.73 Expected Observed
number number
198 193
1.06
66 64
66 62
22 24

27.  Step 4: Interpret the chi square value
– The calculated chi square value can be used to obtain
probabilities, or P values, from a chi square table
• These probabilities allow us to determine the likelihood that the
observed deviations are due to random chance alone
– Low chi square values indicate a high probability that the
observed deviations could be due to random chance alone
– High chi square values indicate a low probability that the
observed deviations are due to random chance alone
– If the chi square value results in a probability that is less than
0.05 (ie: less than 5%) it is considered statistically significant
• The hypothesis is rejected

28.  Step 4: Interpret the chi square value
– Before we can use the chi square table, we have to determine
the degrees of freedom (df)
• The df is a measure of the number of categories that are
independent of each other
• If you know the 3 of the 4 categories you can deduce the
• df = n – 1
– where n = total number of categories
• In our experiment, there are four phenotypes/categories
– Therefore, df = 4 – 1 = 3
– Refer to Table

29. 1.06

30.  Step 4: Interpret the chi square value
– With df = 3, the chi square value of 1.06 is slightly greater than
1.005 (which corresponds to P-value = 0.80)
– P-value = 0.80 means that Chi-square values equal to or
greater than 1.005 are expected to occur 80% of the time due
to random chance alone; that is, when the null hypothesis is
true.
– Therefore, it is quite probable that the deviations between the
observed and expected values in this experiment can be
explained by random sampling error and the null hypothesis is
not rejected. What was the null hypothesis?

31. If your hypothesis is supported by data
•you are claiming that mating is random and so is
segregation and independent assortment.
If your hypothesis is not supported by data
•you are seeing that the deviation between observed
and expected is very far apart…something non-random
must be occurring….

32. Let’s look at an other fruit fly cross
x
Black body, eyeless wild
F1: all wild

33. F1 x F1
5610
1881
1896 622

34. Analysis of the results
• Once the numbers are in, you have to
determine the cross that you were using.
• What is the expected outcome of this cross?
• 9/16 wild type: 3/16 normal body eyeless:
3/16 black body wild eyes: 1/16 black body
eyeless.

35. Now Conduct the Analysis:
Phenotype Observed Hypothesis
Wild 5610
Eyeless 1881
Black body 1896
Eyeless, black body 622
Total 10009
To compute the hypothesis value take 10009/16
= 626

36. Now Conduct the Analysis:
Phenotype Observed Hypothesis
Wild 5610 5634
Eyeless 1881 1878
Black body 1896 1878
Eyeless, black body 622 626
Total 10009
To compute the hypothesis value take 10009/16
= 626

37. 2 (obs exp) 2
exp
• Using the chi square formula compute the chi
square total for this cross:
• (5610 - 5630)2/ 5630 = .07
• (1881 - 1877)2/ 1877 = .01
• (1896 - 1877 )2/ 1877 = .20
• (622 - 626) 2/ 626 = .02
• 2= .30
• How many degrees of freedom?

38. 2 (obs exp) 2
exp
• Using the chi square formula compute the chi
square total for this cross:
• (5610 - 5630)2/ 5630 = .07
• (1881 - 1877)2/ 1877 = .01
• (1896 - 1877 )2/ 1877 = .20
• (622 - 626) 2/ 626 = .02
• 2= .30
• How many degrees of freedom? 3

39. CHI-SQUARE DISTRIBUTION TABLE
Accept Hypothesis Reject
Hypothesis
Probability (p)
Degrees of
0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001
Freedom
1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83
2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82
3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27
4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.38 18.47
5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52
6 1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46
7 2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32
8 2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12
9 3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88
10 3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59

40. • When reporting chi square data use the
following formula sentence….
With degrees of freedom, my chi square
value is , which gives me a p value
between % and %, I therefore
my null hypothesis.
• This sentence would go in the “reults” section
of your formal lab.
• Your explanation of the significance of this
data would go in the “discussion” section of
the formal lab.

41. • Looking this statistic up on the chi square
distribution table tells us the following:
• the P value read off the table places our chi
square number of .30 close to .95 or 95%
• This means that 95% of the time when our
observed data is this close to our expected
data, this deviation is due to random
chance.
• We therefore accept our null hypothesis.

42. • What is the critical value at which we
would reject the null hypothesis?
• For three degrees of freedom this value for
our chi square is > 7.815
• What if our chi square value was 8.0 with 4
degrees of freedom, do we accept or reject
the null hypothesis?
• Accept, since the critical value is >9.48 with
4 degrees of freedom.