MENDELIAN GENETICS CHI SQUARE ANALYSIS

AP BIOLOGY

What is Chi-Squared?l In genetics, you can predict genotypes

based on probability (expected results)l Chi-squared is a form of statistical analysis

used to compare the actual results (observed) with the expected results

l NOTE: C2 is the name of the whole variable – you will never take the square root of it or solve for C

CHI SQUARE FORMULA:

NULL HYPOTHESIS:l The hypothesis is termed the null

hypothesis which states l That there is NO substantial

statistical deviation (difference) between observed values and the expected values.

l In other words, the results or differences that do exist between observed and expected are totally random and occurred by chance alone.

l If the null hypothesis is supported by data• The assumption is that mating is random and

normal gene segregation and independent assortment occurred.

• Note: this is the assumption in all genetic crosses! This is normal meiosis occurring and we would expect random segregation and independent assortment.

l If the null hypothesis is not supported by data• The deviation (difference) between what was

observed and what the expected values were is very far apart…something non-random must be occurring….

CHI SQUARE VALUE:

l In order to determine the probability using a chi square chart you need to determine the degrees of freedom (DF)

l DEGREES OF FREEDOM: is the number of phenotypic possibilities in your cross minus one. l DF = # of groups (phenotype classes) – 1

l Using the DF value, determine the probability or distribution using the Chi Square table

l If the level of significance read from the table is greater than 0.05 or 5% then the null hypothesis is accepted and the results are due to chance alone and are unbiased.

DF VALUE:

Step 1: Calculating C2

l First, determine what your expected and observed values are.

l Observed (Actual) values: That should be something you get from data that you measure

l Expected values: based on probabilityl Suggestion: make a table with the expected

and actual values

Step 1: Examplel Observed (actual) values: Suppose you

have 90 tongue rollers and 10 nonrollers(recessive)

l Expected: Suppose the parent genotypes were both Rr à using a punnett square, you would expect 75% tongue rollers, 25% nonrollers

l This translates to 75 tongue rollers, 25 nonrollers (since the population you are dealing with is 100 individuals)

Step 1: Examplel Table should look like this:

Expected Observed (Actual)

Tongue rollers 75 90

Nonrollers 25 10

Step 2: Calculating C2

l Use the formula to calculated C2

l For each different category (genotype or phenotype calculate (observed – expected)2 / expected

l Add up all of these values to determine C2

Step 2: Examplel Using the data from before:l Tongue rollers

(90 – 75)2 / 75 = 3l Nonrollers

(10 – 25)2 / 25 = 9l C2 = 3 + 9 = 12

Step 3: Determining Degrees of Freedom

l Degrees of freedom = # of categories – 1l Ex. For the example problem, there were two

categories (tongue rollers and nonrollers) àdegrees of freedom = 2 – 1

l Degrees of freedom = 1

Step 4: Critical Value l Using the degrees of freedom, determine

the critical value using the provided tablel Df = 1 à Critical value = 3.84

If X2 is greater than the critical value then reject the null hypothesis.

Step 5: Conclusionl If C2 > critical value…reject null…

there is a statistically significant difference between the actual and expected values.

l If C2 < critical value…accept null…there is a NOT statistically significant difference between the actual and expected values.

Step 5: Examplel C2 = 12 > 3.84à There is a statistically significant difference

between the observed and expected population

Tiger Examplel Two tigers at the zoo have 4 cubs.l Two of the four cubs are albino tigers.l Albinism is recessive, what are the genotypes

of their orange parents?l What is the expected number of albino tigers?

Apply Chi-Squared

Accept or reject null?

Chi-squared and Hardy Weinberg

l Review: If the observed (actual) and expected genotype frequencies are the same then a population is in Hardy Weinberg equilibrium

l But how close is close enough?l Use Chi-squared to figure it out!l If there isn’t a statistically significant difference

between the expected and actual frequencies, then it is in equilibrium

Ferret Example

Ferrets Expected Observed (Actual)

BB 74 78

Bb 72 65

bb 18 21

Example

l C2 CalculationBB: (78 – 74)2 / 74 = 0.21Bb: (72 – 65)2 / 72 = 0.68bb: (18 – 21)2 / 18 = 0.5C2 = 0.21 + 0.68 + 0.5 = 1.39

l Degrees of Freedom = 3 – 1 = 2l Critical value = 5.99l C2 < 5.99 à there is not a statistically

significant difference between expected and actual values à population DOES SEEM TO BE in Hardy Weinberg Equilibrium

EXAMPLE:DIHYBRID FRUIT FLY CROSS

x

Black body -eyeless

Wild type

F1: all wild type

F1 CROSS PRODUCED THE FOLLOWING OFFSPRING

5610 1881

1896 622

Wild type (normal body

and eyes)

Normal body -eyeless

Black body -eyeless

Black body –w/ eyes

ANALYSIS OF THE RESULTS:

l Once the total number of offspring in each class is counted, you have to determine the expected value for this dihybrid cross.

l What are the expected outcomes of this typical dihybrid cross? (9:3:3:1)Ø 9/16 should be wild typeØ 3/16 should be normal body eyelessØ 3/16 should be black body wild eyesØ 1/16 should be black body eyeless.

NOW CONDUCT THE ANALYSIS:Phenotype Observed Hypothesis

Wild 5610

Eyeless 1881

Black body 1896

Eyeless, black body 622

Total 10009

To compute the expected value multiply the expected 9/16:3/16:3/16:1/16 ratios by 10,009

CALCULATING EXPECTED VALUES:

l To calculate the expected value:l Multiply the total number of offspring times the

expected fraction for each phenotype classl TOTAL = 10,009

l Wild-type expected value: 9/16 x 10,009 = 5634l Eyeless expected value: 3/16 x 10,009 = 1878l Black body expected value: 3/16 x 10,009 = 1878l Black body & Eyeless expected value: 1/16 x

10,009 = 626

NOW CONDUCT THE ANALYSIS:

Phenotype Observed Hypothesis

Wild 5610 5634

Eyeless 1881 1878

Black body 1896 1878

Eyeless, black body 622 626

Total 10009

l Using the chi square formula compute the chi square total for this cross:Ø (5610 - 5630)2 / 5630 = 0.07Ø (1881 - 1877)2 / 1877 = 0.01Ø (1896 - 1877)2 / 877 = 0.20Ø (622 - 626)2 / 626 = 0.02Ø x2 = 0.30

l How many degrees of freedom? Ø 4 phenotype classes – 1 = 3 Ø Accept or reject null?

CALCULATING X2: