1. Laws of Probability and Chi Square
Probability & Genetics

2. Probability & Genetics
Calculating probability of making a specific gamete is just like calculating the probability in flipping a coin
probability of tossing heads? 50%
probability making a P gamete…
Outcome of 1 toss has no impact on the outcome of the next toss
probability of tossing heads each time? 50%
probability making a P gamete each time?

3. Rule of Addition
Chance that an event can occur 2 or more different ways
SUM of the separate probabilities
Use for heterozygous possibilities
Two ways to be heterozygous: Pp or pP
Key word is “or”.
Ex: Probability of getting 2 or a 6 on the roll of a die. 1/6 + 1/6 = 2/6 = 1/3
Ex: Probability of having offspring with dominant phenotype? PP or Pp or pP ¼+ ¼ + ¼ = ¾

4. Rule of multiplication
Chance that 2 or more independent events will occur together
probability that 2 coins tossed at the same time will land heads up
probability of pp or PP offspring
Ex: Probability of getting a head and a tail with two different coins.
½ x ½ = 1/4
Key word is “and”

5. Calculating
Probability of
Pp x Pp
½ x ½ = ¼ = PP
½ x ½ = ¼ = pp
What about Pp?

6. Calculating Dihybrid Probability
Rule of multiplication application with Dihybrid crosses:
heterozygous parents — YyRr
probability of producing yyrr?
probability of producing y gamete = 1/2
probability of producing r gamete = 1/2
probability of producing yr gamete
= 1/2 x 1/2 = 1/4
probability of producing a yyrr offspring
= 1/4 x 1/4 = 1/16

7. What is Chi-Squared?
In genetics, you can predict genotypes based on probability (expected results)
Chi-squared is a form of statistical analysis used to compare the actual results (observed) with the expected results
NOTE: 2 is the name of the whole variable – you will never take the square root of it or solve for 

8. Chi-squared
If the expected and observed (actual) values are the same then the 2 = 0
If the 2 value is 0 or is small then the data fits your hypothesis (the expected values) well.
By calculating the 2 value you determine if there is a statistically significant difference between the expected and actual values.

9. Step 1: Calculating 2
First, determine what your expected and observed values are.
Observed (Actual) values: That should be something you get from data– usually no calculations 
Expected values: based on probability
Suggestion: make a table with the expected and actual values

10. Step 1: Example
Observed (actual) values: Suppose you have 90 tongue rollers and 10 nonrollers
Expected: Suppose the parent genotypes were both Rr  using a punnett square, you would expect 75% tongue rollers, 25% nonrollers
This translates to 75 tongue rollers, 25 nonrollers (since the population you are dealing with is 100 individuals)
Tongue rolling is a dominant trait

11. Step 1: Example Table should look like this: Expected
Observed (Actual)
Tongue rollers 75 90
Nonrollers 25 10

12. Step 2: Calculating 2 Use the formula to calculated 2
For each different category (genotype or phenotype calculate
(observed – expected)2 / expected
Add up all of these values to determine 2

13. Step 2: Calculating 2

14. Step 2: Example Using the data from before: Tongue rollers
(90 – 75)2 / = 3
Nonrollers
(10 – 25)2 / = 9
2 = = 12

15. Step 3: Determining Degrees of Freedom
Degrees of freedom = # of categories – 1
Ex. For the example problem, there were two categories (tongue rollers and nonrollers)  degrees of freedom = 2 – 1
Degrees of freedom = 1

16. Step 4: Critical Value
Using the degrees of freedom, determine the critical value using the provided table
Df = 1  Critical value = 3.84

17. Step 5: Conclusion If 2 > critical value…
there is a statistically significant difference between the actual and expected values.
If 2 < critical value…
there is a NOT statistically significant difference between the actual and expected values.

18. Step 5: Example
2 = 12 > 3.84
There is a statistically significant difference between the observed and expected population

19. Bozeman Chi-squared test video
Fill in video questions

20. Animal Behavior Chi Square
Wet
Dry
Observed value
8.9
1.1
Expected value 5
Chi Square = Σ (O – E)2 / E
(8.9 – 5)2 / (1.1-5)2 / 5
15.21 / / 5
30.42/ =

21. 2 variable – wet and dry so 1 degree of freedom
6.084 is higher than so must reject null hypothesis.
Something influenced the Pill bugs.

22. In 2002 The distribution for Skittles is: Green: 19. 7%, Yellow: 19
In 2002 The distribution for Skittles is: Green: 19.7%, Yellow: 19.5%, Orange: 20.2%, Red: 20%, Purple: 20.6%. 
Color distribution for M&Ms
Brown15% Yellow12% Orange20% Red 13% Green 16% Blue 24%