MENDELIAN GENETICS CHI SQUARE ANALYSIS Ms. Day Honors Genetics

CHI SQUARE ANALYSIS The chi square analysis allows you to use statistics to determine if your data is “good” or not. allows us to test for deviations of observed frequencies from expected frequencies. In our Corn lab, we are using laws of probability to determine possible outcomes for genetic crosses (ex: Punnett Squares)

The following formula is used You need 2 different hypotheses: 1. NULL Hypothesis Data are occurring by chance and it is all RANDOM! 2. Alternative Hypothesis Data are occurring by some outside force. It is NOT by chance and it is NOT RANDOM!

Two Types of Hypotheses: 1. NULL HYPOTHESIS states that there is no substantial statistical deviation between observed and expected data. a hypothesis of no difference (or no effect) is called a null hypothesis symbolized H0 In other words, the results are totally random and occurred by chance alone. The null hypothesis states that the two variables are independent, or that there is NO relationship to one another.

Two Types of Hypotheses: 2. ALTERNATIVE HYPOTHESIS states that there IS a substantial statistical deviation between observed and expected data. a hypothesis of difference (or effect) is called a alternative hypothesis symbolized H1 In other words, the results are affected by an outside force and are NOT random and di NOT occur by chance alone. Ex of outside “forces”: Gene linkage, favoritism, miscounting, etc.

2 Types of Chi Square Problems Non-genetic Null Hypothesis: Data is due to chance and is completely random. There is no preference between the groups/categories. Alternative Hypothesis Data is NOT due to chance and there IS a preference between the groups/categories. Data is not random.

2 Types of Chi Square Problems 2. Genetic Null Hypothesis: Data is due to chance and is random. There is no gene linkage affecting independent assortment & segregation IF there are 2+ genes involved. Punnett Square probabilities are EXPECTED!!! Alternative Hypothesis Data is due NOT to chance and is NOT random. There COULD BE gene linkage affecting independent assortment & segregation IF there are 2+ genes involved. Punnett Square probabilities are NOT expected !!!

Let’s look at a fruit fly cross and their phenotypes x Black body, eyeless (bbee) Wild type (BBEE) F1: all wild type (BbEe)

Black body, eyeless (R, R) Black body, Wild type (R, D) F1 x F1 (BbEe x BbEe) 5610 1881 Wild type (D, D) Eyeless, Wild type (D, R) 622 1896 Black body, eyeless (R, R) Black body, Wild type (R, D)

Analysis of the results Once the data numbers are in (collected), you have to determine the expected value of this cross using Punnett Square outcomes. This is your hypothesis called the null hypothesis (no gene linkage is occuring). What are the expected outcomes of this cross? F1 Cross: BbEe x BbEe 9/16 should be wild type 3/16 should be normal body eyeless 3/16 should be black body wild eyes 1/16 should be black body eyeless.

The following formula is used If your alternative hypothesis is supported by data you are claiming that mating is random as well as segregation and independent assortment. If your alternative hypothesis is not supported by data you are seeing that the deviation (difference) between observed and expected is very far apart  something non- random must be occurring…could be GENE LINKAGE, MISCOUNTING, PREFERENCES, etc!!!

Now Conduct the Analysis: To compute the null hypothesis value take 10009/16 = 626 (a.k.a- 1/16 of total offspring)

Now Conduct the Analysis: To compute the hypothesis value take 10009/16 = 626

Calculate (o-e)2/ e for EACH phenotype Using the chi square formula compute the chi square total for this cross: Calculate (o-e)2/ e for EACH phenotype (5610 - 5630)2/ 5630 = .07 (1881 - 1877)2/ 1877 = .01 (1896 - 1877 )2/ 1877 = .20 (622 - 626) 2/ 626 = .02 Sum all numbers to get your chi square total  2= .30 Determine how many degrees of freedom are in your experiment 4 (phenotype) groups– 1 = 3

I Have my Chi Square Total (X2)…. What next? Figure out which hypothesis is accepted: your NULL hypothesis is a 9:3:3:1 ratio will be seen due to genetics (independent assortment/ segregation) The alternative hypothesis = any change from the expected 9:3:3:1 is due to SOME OUTSIDE FORCE! DATA IS NOT RANDOM! To figure which hypothesis is accepted, you need to use the CHI SQUARE TABLE, which list CRITICAL VALUES (Chi Square values)!

HOW DO YOU USE THIS TABLE PROPERLY? Using this Table… You need to determine your probability (p) value using the Chi Square distribution table HOW DO YOU USE THIS TABLE PROPERLY? you need to determine the degrees of freedom Degrees of freedom is the number of groups (categories) in your data minus one (1) If the level of significance read from the table is greater than .05 or 5% then your hypothesis is accepted and the data is useful

Remember…our chi square value was X2 = 0.30 This value is useful b/c we can obtain the probability that the data occurs (and the probability that the data are an error)

CHI SQUARE TABLE RANDOM DATA Accept Null Hypothesis CHI-SQUARE DISTRIBUTION TABLE RANDOM DATA NOT RANDOM DATA Accept Null Hypothesis Reject Null Hypothesis Probability (p) Degrees of Freedom 0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001 1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83 2 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82 3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27 4 1.06 1.65 2.20 3.36 4.88 7.78 9.49 13.38 18.47 5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52 6 1.63 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46 7 2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32 8 2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12 9 3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88 10 3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59 In biological applications, a probability 5% is usually adopted as the standard conventional criteria for probability to have statistical significance is 0.001-0.05

We therefore accept our null hypothesis. Looking statistical values up on the chi square distribution table tells us the following: the PROBABILITY (P) value read off the table places our chi square number of .30 closer to .95 or 95% (~94%) This value means that there is a 6% chance that our results are biased and due to gene linkage. In other words, the probability of getting our results is 94%. 94% of the time when our observed data is this close to our expected data, this deviation is due to random chance. We therefore accept our null hypothesis.

First, we will do a “m & m” Chi Square Analysis Lab” Plain M & M’s 30% brown, 20% each of yellow and red and 10% each of orange, green and blue. Peanut M & M’s 20% each of brown, yellow, red & blue 10% each of green and orange