C HI S QUARED (X²) By: Christina Patellos & Bismah Warraich

W HAT IS CHI SQUARE ? Chi Square is a statistical test that allows us to determine differences between our observed and expected measurements. The initial purpose is to establish whether or not the two variables in the experiment have a significant relationship or differences in measurements are due to chance or sample errors. This statistical test is usually used when we are dealing with distinct data.

W HAT IS C HI S QUARE USED FOR ? Chi square is used to test the validity of a null hypothesis. A null hypothesis is the hypothesis that states there is no significant difference between observed and expected values. More so, any differences that occur would be because of sampling or experimental error.

H OW DO WE CALCULATE CHI SQUARE ? There is a formula! X² = Σ (o-e)² e o stands for observed values e stands for expected values Σ stands for the ‘sum of’

D EGREE OF FREEDOM Degree of freedom (df) is used to accept or deny the null hypothesis. Df is equivalent to the number of categories minus one. Here’s an example of a degree of freedom chart:

L ET ’ S TRY TO APPLY IT TO AN EXAMPLE … In fern plants purple leaves (R) are dominant to yellow leaves (r). Plants from the offspring of a cross between a purple plant and yellow plant were used in a lab. A student counts 329 purple and 299 yellow leaves on one fern. Calculate the chi-squared value for the null hypothesis that the purple parent was heterozygous for purple leaves. Give your answer to the nearest tenth.

D ON ’ T FORGET THE PUNNET SQUARE ! A cross between a heterozygous purple leaved fern plant (Rr) and a yellow leaved fern plant (rr) would yield offspring that display a 1:1 ratio between purple and yellow leaves. R r r Rr rr rRr rr

L ET ’ S MAKE A TABLE ! Of the 628 leaves,it would be expected that 314 would be purple and 314 would be yellow (Expected values). Phenotype Observed Value Expected Value Obs - Exp (Obs-Exp)²(Obs-Exp)² / Exp Purple leaves Yellow leaves Σx² = 1.44

N OW TRY A PROBLEM OUT ON YOUR OWN ! RULES: Get into teams of four (turn to the table behind you) Work out the problem AS A TEAM Whichever team correctly finishes the problem first gets a special prize

P RACTICE P ROBLEM Watermelons have genes for bitter taste (Su) and explosive rind (e). Explosive rinds are recessive. Non-bitter watermelons are also recessive and have the genotype (susu). A farmer wants to determine if the genes assort independently. She performs a testcross between a bitter/nonexplosive hybrid and a plant homozygous recessive for both traits. The following offspring are produced: bitter/non-explosive – 88 bitter/explosive – 68 non-bitter/non-explosive – 62 non-bitter/explosive – 81 Hint: A cross between Susu/Ee x susu/ee watermelon plants is expected to produce offspring with a 1:1:1:1 phenotypic ratio.