Presentation on theme: "Laws of Probability and Chi Square"— Presentation transcript:
1Laws of Probability and Chi Square Probability & Genetics
2Probability & Genetics Calculating probability of making a specific gamete is just like calculating the probability in flipping a coinprobability of tossing heads? 50%probability making a P gamete…Outcome of 1 toss has no impact on the outcome of the next tossprobability of tossing heads each time? 50%probability making a P gamete each time?
3Rule of AdditionChance that an event can occur 2 or more different waysSUM of the separate probabilitiesUse for heterozygous possibilitiesTwo ways to be heterozygous: Pp or pPKey word is “or”.Ex: Probability of getting 2 or a 6 on the roll of a die. 1/6 + 1/6 = 2/6 = 1/3Ex: Probability of having offspring with dominant phenotype? PP or Pp or pP ¼+ ¼ + ¼ = ¾
4Rule of multiplication Chance that 2 or more independent events will occur togetherprobability that 2 coins tossed at the same time will land heads upprobability of pp or PP offspringEx: Probability of getting a head and a tail with two different coins.½ x ½ = 1/4Key word is “and”
5CalculatingProbability ofPp x Pp½ x ½ = ¼ = PP½ x ½ = ¼ = ppWhat about Pp?
6Calculating Dihybrid Probability Rule of multiplication application with Dihybrid crosses:heterozygous parents — YyRrprobability of producing yyrr?probability of producing y gamete = 1/2probability of producing r gamete = 1/2probability of producing yr gamete= 1/2 x 1/2 = 1/4probability of producing a yyrr offspring= 1/4 x 1/4 = 1/16
7What 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 resultsNOTE: 2 is the name of the whole variable – you will never take the square root of it or solve for
8Chi-squaredIf the expected and observed (actual) values are the same then the 2 = 0If 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.
9Step 1: Calculating 2First, 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 probabilitySuggestion: make a table with the expected and actual values
10Step 1: ExampleObserved (actual) values: Suppose you have 90 tongue rollers and 10 nonrollersExpected: Suppose the parent genotypes were both Rr using a punnett square, you would expect 75% tongue rollers, 25% nonrollersThis translates to 75 tongue rollers, 25 nonrollers (since the population you are dealing with is 100 individuals)Tongue rolling is a dominant trait
11Step 1: Example Table should look like this: Expected Observed (Actual)Tongue rollers7590Nonrollers2510
12Step 2: Calculating 2 Use the formula to calculated 2 For each different category (genotype or phenotype calculate(observed – expected)2 / expectedAdd up all of these values to determine 2
14Step 2: Example Using the data from before: Tongue rollers (90 – 75)2 / = 3Nonrollers(10 – 25)2 / = 92 = = 12
15Step 3: Determining Degrees of Freedom Degrees of freedom = # of categories – 1Ex. For the example problem, there were two categories (tongue rollers and nonrollers) degrees of freedom = 2 – 1Degrees of freedom = 1
16Step 4: Critical ValueUsing the degrees of freedom, determine the critical value using the provided tableDf = 1 Critical value = 3.84
17Step 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.
18Step 5: Example2 = 12 > 3.84There is a statistically significant difference between the observed and expected population
19Bozeman Chi-squared test video Fill in video questions
212 variable – wet and dry so 1 degree of freedom 6.084 is higher than so must reject null hypothesis.Something influenced the Pill bugs.
22In 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&MsBrown15% Yellow12% Orange20% Red 13% Green 16% Blue 24%