1. Generations and Mendel – Part 4
PBG 430

2. Hypothesis Testing: Determining the “Goodness-of-Fit”
Observed ratio vs. expected ratio
Chi-square test = a statistical analysis for testing the ‘goodness-of-fit’ between an observed and expected ratio
Based on a probability value (p-value), do you accept or reject the null hypothesis?
Expected ratios in cross progeny will be a function of the following:
degree of homozygosity of the parents
generation studied
degree of dominance
degree of interaction between genes
number of genes determining the trait

3. Chi-Square statistic test
The Chi-Square statistic tests the "goodness of fit” (i.e. how closely the observed and predicted results agree).
The degrees of freedom (df) that are used for the test are a function of the number of classes.
A probability value (p-value) is associated with the calculated chi-square value by consulting an established table or an online p-value calculator.
This is a test of a null hypothesis.
The null hypothesis will always be that “the observed ratio and expected ratio are not different.”

4. The Chi-Square concept: Summary
As deviations from hypothesized ratios get smaller, the chi square value approaches 0; therefore, there is a good fit.
As deviations from hypothesized ratios get larger, the chi square value gets larger; therefore, there is a poor fit.
What determines a good vs. poor fit?
The probability of observing a deviation as large, or larger, due to chance alone.
P-values below 0.05 (i.e , 0.01, .005) lie in the area of null hypothesis rejection.

5. The Chi-Square Formula
The chi-square formula consists of several components and symbols.
X = chi
X2 = chi-square
Σ = sigma = “sum of”
O = number of individuals observed
E = number of individuals expected
X2 = [(O1 - E1)2/E1] + [(O2 – E2)2/E2] [(On - En)2/En]
O1 = number of observed members of the first class
E1 = number of expected members of the first class
O2 = number of observed members of the second class
E2 = number of expected members of the second class
On = number of observed members of the nth class
En = number of expected members of the nth class

6. Chi-Square Example Computation #1
Let’s complete a chi-square computation for a monohybrid doubled haploid (DH) ratio from the Oregon Wolfe Barley (OWB) population
The data:
Population: 82 OWB DH
Number of kernel rows in barley (Hordeum vulgare) is represented by Vrs1 and vrs1
Vrs1 is dominant, vrs1 is recessive
For simplicity, vrs1 is abbreviated as "v" in the table
Hypothesis: If we randomly sample 82 DH from the OWB population, 41 individuals will be 2-row and 41 individuals will be 6-row (i.e. a 1:1 phenotypic ratio).
Gametes from an F1 individual V v
DH genotypes VV vv
DH phenotypes
Two-row
Six-row
DH - observed number of individuals 35 47

7. Chi-Square Example Computation #1 – Cont.
We know that there are two possible phenotypes from the Vrs1 genes  a two-row barley and a six-row barley.
This gives us two phenotype classes to analyze in the chi-square equation.
Remember to sum all the results from each class to determine the Chi-Square value.
Class Number
Phenotype (Genotype)
Number Observed
Number Expected
(O – E)
(O – E)2
(O – E)2 / E 1
2-row (VV) 35 41 -6 36
0.88 2
6 row (vv) 47 6
TOTAL
N/A 82 72
Chi-square = 1.76

8. Example #1 – Degrees of Freedom & P-value
Calculate the degrees of freedom
df = number of classes - 1
Where does the calculated chi- square value fall in the table?
For this example, we calculated a chi-square value of 1.76
The p-value needed to reject the null hypothesis is 0.05 (or less). So, according to where the chi- square value fell, do you reject the null hypothesis or do you fail to reject it?
You fail to reject it! P df
0.995
0.975
0.9
0.5
0.1
0.05
0.025
0.01
0.005 1
0.000
0.016
0.455
2.706
3.841
5.024
6.635
7.879 2
0.010
0.051
0.211
1.386
4.605
5.991
7.378
9.210
10.597 3
0.072
0.216
0.584
2.366
6.251
7.815
9.348
11.345
12.838 4
0.207
0.484
1.064
3.357
7.779
9.488
11.143
13.277
14.860 5
0.412
0.831
1.610
4.351
9.236
11.070
12.832
15.086
16.750 6
0.676
1.237
2.204
5.348
10.645
12.592
14.449
16.812
18.548 7
0.989
1.690
2.833
6.346
12.017
14.067
16.013
18.475
20.278 8
1.344
2.180
3.490
7.344
13.362
15.507
17.535
20.090
21.955 9
1.735
2.700
4.168
8.343
14.684
16.919
19.023
21.666
23.589 10
2.156
3.247
4.865
9.342
15.987
18.307
20.483
23.209
25.188

9. Chi-Square Example Computation #2
Let’s complete a chi-square computation for a monohybrid F2 ratio from the OWB population
The data:
Population: 100 Oregon Wolfe Barley F2s
Hypothesis: If we randomly sample 100 F2s from the OWB population, 75 individuals will be 2-row and 25 individuals will be 6-row (i.e. a 3:1 phenotypic ratio).
Class Number
Phenotype (Genotype)
Number Observed
Number Expected
(O – E)
(O – E)2
(O – E)2 / E 1
2-row (VV or Vv) 78 75 3 9
0.12 2
6 row (vv) 22 25 -3
0.36
TOTAL
N/A
100 18
Chi-square = 0.48

10. Example #2 – Degrees of Freedom & P-value
We calculated a chi-square value of 0.48…do we fail to reject or do we reject the null hypothesis?
We fail to reject it!
Therefore, we conclude that, at an acceptable level of probability, there is a 3:1 ratio of barley phenotypes in the F2 OWB population. P df
0.995
0.975
0.9
0.5
0.1
0.05
0.025
0.01
0.005 1
0.000
0.016
0.455
2.706
3.841
5.024
6.635
7.879 2
0.010
0.051
0.211
1.386
4.605
5.991
7.378
9.210
10.597 3
0.072
0.216
0.584
2.366
6.251
7.815
9.348
11.345
12.838 4
0.207
0.484
1.064
3.357
7.779
9.488
11.143
13.277
14.860 5
0.412
0.831
1.610
4.351
9.236
11.070
12.832
15.086
16.750 6
0.676
1.237
2.204
5.348
10.645
12.592
14.449
16.812
18.548 7
0.989
1.690
2.833
6.346
12.017
14.067
16.013
18.475
20.278 8
1.344
2.180
3.490
7.344
13.362
15.507
17.535
20.090
21.955 9
1.735
2.700
4.168
8.343
14.684
16.919
19.023
21.666
23.589 10
2.156
3.247
4.865
9.342
15.987
18.307
20.483
23.209
25.188

11. Chi-Square Example Computation #3
Let’s complete a chi-square computation for a dihybrid ratio with a doubled haploid (DH) from the OWB population
The data:
Population: 82 Oregon Wolfe Barley DH
Loci being considered:
Number of kernel rows is represented by V and v
Seedling type in barley is represented by W and w
The expected frequencies of gametes used to produce haploid plants are:
25% for VW, 25% for Vw, 25% for vW, and 25% for vw
After chromosome doubling, this would give the following ratios:
Genotypic  1 VVWW: 1 VVww: 1 vvWW: 1 vvww
Phenotypic  1 two-row normal seedling: 1 two-row white stripe seedling: 1 six-row normal seedling: 1 six-row white stripe seedling
Hypothesis: If we randomly sample 82 DHs from the OWB population, they will segregate into a 1:1:1:1 phenotypic and genotypic ratio.

12. Chi-Square Example Computation #3 – Cont.
We know that there are four possible phenotypes from the V and W genes.
With a population of 82 DH individuals and the hypothesized 1:1:1:1 ratio, we can calculate the expected number of individuals for each class to be 20.5
Class Number
DH Genotype
Number Observed
Number Expected
(O – E)
(O – E)2
(O – E)2 / E 1
VV WW 20
20.5
-0.5
0.25
0.012 2
VV ww 15
-5.5
30.25
1.476 3
vv WW 25
4.5
20.25
0.988 4
vv ww 22
1.5
2.25
0.110
TOTAL
N/A 82 53
Chi-square = 2.586

13. Example #3 – Degrees of Freedom & P-value
Because there were four different classes evaluated, we now have 3 degrees of freedom.
We calculated a chi-square value of 2.586…do we fail to reject or do we reject the null hypothesis?
We fail to reject it.
Therefore, we conclude that, in this dihybrid analysis of V and W genes, at an acceptable level of probabaility, there is a 1:1:1:1 ratio of barley phenotypes in the DH OWB population. P df
0.995
0.975
0.9
0.5
0.1
0.05
0.025
0.01
0.005 1
0.000
0.016
0.455
2.706
3.841
5.024
6.635
7.879 2
0.010
0.051
0.211
1.386
4.605
5.991
7.378
9.210
10.597 3
0.072
0.216
0.584
2.366
6.251
7.815
9.348
11.345
12.838 4
0.207
0.484
1.064
3.357
7.779
9.488
11.143
13.277
14.860 5
0.412
0.831
1.610
4.351
9.236
11.070
12.832
15.086
16.750 6
0.676
1.237
2.204
5.348
10.645
12.592
14.449
16.812
18.548 7
0.989
1.690
2.833
6.346
12.017
14.067
16.013
18.475
20.278 8
1.344
2.180
3.490
7.344
13.362
15.507
17.535
20.090
21.955 9
1.735
2.700
4.168
8.343
14.684
16.919
19.023
21.666
23.589 10
2.156
3.247
4.865
9.342
15.987
18.307
20.483
23.209
25.188

14. By now you should be able to…
Perform a chi-square test to determine the “goodness of fit” of an observed ratio to an expected ratio.
You will be given observed values for each class and the generation of the progeny studied. Then, you should:
Formulate the expected ratio based on the generation of the progeny and your knowledge on Mendelian inheritance (remember the Punnett Square!).
Calculate the expected values for each class based on the expected ratio.
Calculate the chi-square value (the formula will be given).
Determine the p-value for the calculated chi-square using the table (range of p-values) or an online calculator (exact p-value).
Based on a cut-off p-value, you will then fail to reject or reject the null- hypothesis.
Interpret the results of the chi-square test in terms of number of genes determining the trait, and the degree of dominance.