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**Laws of Probability and Chi Square**

Probability & Genetics

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**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?

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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

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**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”

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

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**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

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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

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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.

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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

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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

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**Step 1: Example Table should look like this: Expected**

Observed (Actual) Tongue rollers 75 90 Nonrollers 25 10

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**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

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Step 2: Calculating 2

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**Step 2: Example Using the data from before: Tongue rollers**

(90 – 75)2 / = 3 Nonrollers (10 – 25)2 / = 9 2 = = 12

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**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

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

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**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.

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

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**Bozeman Chi-squared test video**

Fill in video questions

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**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/ =

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**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.

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**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%

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