1. Fruit Fly Basics Drosophila melanogaster

2. Wild Type Phenotype Red eyes Tan Body Black Rings on abdomen Normal Wings

3. Males vs. Females Males = smaller, black patch on abdomen, more rounded abdomen Females = larger, more pointed abdomen

4. Why Fruit Fly Genetics? Small Short Life Cycle Easily observable characteristics 3 pairs of autosomes and 1 pair of sex chromosomes

5. Mutations PhenotypeDescription Eye Shape NormalWild type EyelessEyes reduced Eye Color RedWild type White Sepia Brown to black with age Wings NormalWild Type VestigialWings Reduced ApterousWingless DumpyWings Truncated

6. Counting Fruit Flies Make a table that records all the different phenotypes Count males and females separately PhenotypeMalesFemalesTotal Wild type 174360 White Eyes200 PhenotypeMalesFemalesTotal Wild type 354075 Vestigial Wings141125

7. Interpreting the Results Look for trends that will allow you to predict which trait is dominant and which trait is recessive. Look for trends that will allow you to predict if traits are linked. Analyze trends to determine the Parental Cross PhenotypeMalesFemalesTotal Wild type 354075 Vestigial Wings141125 PhenotypeMalesFemalesTotal Wild type 174360 White Eyes200

8. Testing your Hypotheses Chi Squared (χ 2 )test Statistical test used to compare observed data to the expected data according to a specific hypothesis Tests the “null hypothesis” (states that there is no significant difference between the expected and observed result)

9. Calculating Chi- Squared State the hypothesis being tested and the predicted results. Gather the data by conducting the proper experiment (or, if working genetics problems, use the data provided in the problem). Determine the expected numbers for each observational class. Calculate χ 2 using the formula.

10. Use the chi-square distribution table to determine significance of the value. Determine degrees of freedom (number of categories minus 1) and locate the value in the appropriate column. Locate the value closest to your calculated χ 2 on that degrees of freedom row. Move up the column to determine the p value. (probability that the deviation of the observed from that expected is due to chance alone) Biology standard is p > 0.05. This means you would expect any deviation to be due to chance alone 5% of the time or less. State your conclusion in terms of your hypothesis. If the p value for the calculated χ 2 is p > 0.05, accept your hypothesis. 'The deviation is small enough that chance alone accounts for it. If the p value for the calculated χ 2 is p < 0.05, reject your hypothesis, and conclude that some factor other than chance is operating for the deviation to be so great.