AP Biology Laws of Probability and Chi Square Probability & Genetics.

AP Biology

Laws of Probability and Chi Square

Probability & Genetics

AP Biology

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

AP Biology

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)

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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 / 75 = 3 Nonrollers

(10 – 25)2 / 25 = 9 2 = 3 + 9 = 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 > 3.84There is a statistically significant

difference between the observed and expected population

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Bozeman Chi-squared test video http://www.youtube.com/watch?v=WXP

BoFDqNVk&edufilter=vBrBKiVMlaMnBrdX3oXR-Q&safe=active

Fill in video questions

AP Biology

Animal Behavior Chi SquareWet Dry

Observed value 8.9 1.1

Expected value 5 5

Chi Square = Σ (O – E)2 / E

(8.9 – 5)2 / 5 + (1.1-5)2 / 5

15.21 / 5 + 15.21 / 5

30.42/ 5 = 6.084

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2 variable – wet and dry so1 degree of freedom

6.084 is higher than 3.841 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: 19.5%, Orange: 20.2%, Red: 20%, Purple: 20.6%. 

Color distribution for M&Ms

Brown15% Yellow12% Orange20% Red 13% Green 16% Blue 24%