Fruit Fly BasicsDrosophila melanogaster

Wild Type Phenotype

• Red eyes• Tan Body• Black Rings on abdomen• Normal Wings

Males vs. Females

• Males = smaller, black patch on abdomen, more rounded abdomen

• Females = larger, more pointed abdomen

Why Fruit Fly Genetics?

• Small• Short Life Cycle• Easily observable characteristics• 3 pairs of autosomes and 1 pair of sex

chromosomes

MutationsPhenotype Description

Eye Shape Normal Wild typeEyeless Eyes reducedEye Color Red Wild typeWhite WhiteSepia Brown to black with

ageWings Normal Wild TypeVestigial Wings ReducedApterous WinglessDumpy Wings Truncated

Counting Fruit Flies

• Make a table that records all the different phenotypes

• Count males and females separately

Phenotype Males Females TotalWild type 17 43 60White Eyes 20 0 20

Phenotype Males Females TotalWild type 35 40 75Vestigial Wings 14 11 25

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

Phenotype Males Females TotalWild type 35 40 75Vestigial Wings 14 11 25

Phenotype Males Females TotalWild type 17 43 60White Eyes 20 0 20

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)

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.

• 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.