Genetics WorkShop Number Three The Chi-Square Worksheet - COMPLETED by Dr Jamie Love 2002 - 2010

Mendel's data from one experiment was ...
P = smooth seeds crossed with wrinkled seeds
F₁ = all smooth seeds (so smooth is dominant and wrinkled is recessive)
F₂ = 5,474 smooth seeds and 1,850 wrinkled seeds

1. What ratio did he observe?
5474 / 1850 = 2.9589189 : 1 = 2.96 : 1

2. What ratio did he expect?
3 : 1

You should understand that the chi-square compares the NUMBER (not ratio) observed to the NUMBER (not ratio) expected.
You already know the number observed.
Smooth = 5474
Wrinkled = 1850

3. What is the total number of seeds?
7324

4. What number of wrinkled is expected?
7324 / 4 = 183

5. What number of smooth is expected?
1831 X 3 = 5493 or 7324 X 3/4 = 5493

6. What is the difference between observed and expected smooth?
5474 - 5493 = -19

7. What is the difference between observed and expected wrinkled?
1850 - 1831 = 19

8. What is the square of the difference between the observed and expected smooth?
-19² = 361 or -19 X -19 = 361

9. What is the square of the difference between the observed and expected wrinkled?
19² = 361 or 19 X 19 = 361

10. What is the square of the difference between the observed and expected smooth, divided by the expected number of smooth?
361 / 5493 = 0.06572 = 0.066

11. What is the square of the difference between the observed and expected wrinkled, divided by the expected number of wrinkled?
361 / 1831 = 0.19716 = 0.197

12. What is the sum of the "squared differences per expected"?
0.066 + 0.197 = 0.263 the ² = 0.263
Therefore, the chi-square for this experiment is ² = 0.263.

A. Significance Level
Any chi-square larger than the value from the 5% Table indicates an experiment in which the ratios observed are so far off the ratios expected that we have to conclude that the ratios expected are wrong!
B. Degrees of Freedom
The "degrees of freedom" are one less than the number of classes.

13. Name all the different classes in the experiment (earlier).
Smooth and Wrinkled

14. How many degrees of freedom were in that experiment?
2 - 1 = 1

Here's a portion of the Chi Square Significance Table. 15. Is the chi-square you calculated within the boundary of "the possible"? Yes! We calculated a 2 = 0.263. With one degree of freedom we could have a chi-square up to 3.84 before we would become suspicious that the observed data was in a ratio too far removed from the ratio we tested.

Degrees of Freedom 5 % Significance Levels 1 3.84 2 5.99 3 7.81 4 9.49

When doing a Chi-square it helps to set it up as a table and (perhaps helps) to understand that all we have been doing is represented by the equation ² = [(O - E)²/E]

Consider these results among the F₂ = 4,400 yellow seeds and 1,624 green seeds.
Use the space below to work out this chi-square using a table (as explained in the workshop).
First, set up a table.
Second, enter the data. Remember, data is what is observed. So data goes in the "observed" (O) column.
Next fill in the "expected" (E) column.
Finally, fill in the rest of the table.

Phenotypes	O	E	O-E	(O-E)2	(O-E)2 E
Yellow	4400	6024 X 3/4 4518	4400 - 4518 -118	-1182 13,924	13924 / 4518 3.08
Green	1624	6024 X 1/4 1506	1624 -1506 -118	-1182 13,924	13924 /1506 9.24
Total	6024	6024			12.32

Is the chi-square you calculated here within the boundary of "the possible"? (Use the Table from earlier.)
NO! ² = 12.32 but, with one degree of freedom we cannot accept any ratio that gives us a chi-square larger than 3.84.

Do we accept that these results are within acceptable range of a 3 : 1 ratio?
No! We must reject the 3 : 1 ratio. This data is far off the 3 : 1 ratio.

Consider these results from a dihybrid cross
30 red tall
65 white tall
83 red short
206 white short

Based upon these numbers, which phenotypes are dominant and recessive for the two loci?
The biggest group is the white shorts so they must be the doubly dominant class. - white shorts can be assigned the genotype W-S-.
The least represented group, would be the doubly recessive so the red talls are the "1" in the 9 : 3 : 3 :1 ratio and have the genotype wwss.
The other two classes make up the "3" in the ratio. The white talls have the genotype W-ss and the red shorts are wwS-.

In the space below, arrange your computation table starting with the "observed".
Then determine the "expecteds".
Finally, fill in the rest of the table and calculate the chi-square.

Phenotypes	O	E	O-E	(O-E)2	(O-E)2 E
White and short (W-S-)	206	24 X 9 216	206 - 216 10	102 100	100 / 216 0.463
Red and short (wwS-)	83	24 X 3 72	83 - 72 11	112 121	121 / 72 1.681
White and tall (W-ss)	65	24 X 3 72	65 - 72 -7	-72 49	49 / 72 0.681
Red and tall (wwss)	30	24 X 1 24	30 - 24 6	62 36	36 / 24 1.500
Total	384	384			4.325

How many "classes" (categories, groups) are in this experiment?
Four (Red and tall, White and tall, Red and short, White and short)

Now, how many degrees of freedom are in this experiment?
Three (4 -1 )

Does the 9 : 3 : 3 : 1 ratio fit the data? Yes! With three degrees of freedom you can have a chi-square as large as 7.81 before we would be beyond our 5% significance.

Degrees of Freedom 5 % Significance Levels 1 3.84 2 5.99 3 7.81 4 9.49

What is the expected ratio of boys to girls?
1 : 1

What is the degrees of freedom in that example?
There are two categories (classes) so there is one degree of freedom.

If a particular IVF clinic can, indeed, increase the odds, would you expect the chi-square to be above or below the value of 3.84 (which I got from the table above)?
If the IVF clinic can change the ratio from the expected 1 : 1 then the chi-square, calculated on the number of daughters (or sons) born, would be greater than 3.84.

Store A made $1,000,000
Store B made $3,000,000
Store C brought in $5,000,000.
How would you use the chi-square to test the idea that these stores are different - beyond luck? (Don't do the chi-square - just tell me how you would set it up.)
You would "expect" a 1 : 1 : 1 ratio in the revenues if they were all the same. You would expect
Store A = $3,000,000
Store B = $3,000,000
Store C = $3,000,000
You could now find, for each store, the difference between expected and observed revenues, square the difference, divide that by the expected and then add all three together to get a chi-square value.

Store A serves a population that is only a quarter the size of the communities served by stores B and C. Can you redo the chi-square? How?
The information about the populations tells you that there are four times as many likely customers for stores B and C as A. You can express that as a ratio of 1 : 4 : 4.

And finally, what is the degree of freedom for this three-store problem?
There are three categories (Stores, A, B and C) so there are two degrees of freedom.