Simple Mendelian Genetics in Drosophila

Lab objectives:

1) To familiarize you with an important research organism, the fruit fly, Drosophila melanogaster.

2) Introduce you to normal "wild type" and various mutant phenotypes.

3) To help you understand the connection between presence or absence of a genetic trait in an individual and the ratio of that trait in a population.

4) To introduce the use of the Chi square statistic to test hypotheses concerning expected and observed ratios.

Drosophila biology: The common fruit fly is a model organism for genetic studies. The reason it is so widely used is because it is easily cultured in the lab, has a short generation time, and can produce many offspring. Before you begin the lab today, you should learn a little about the life cycle of fruit flies. The life cycle (from egg to adult) takes about 10 days at room temperature. Eggs are laid and hatch into first instar larvae. These larvae feed voraciously on the culture medium provided. You can observe this by looking at a culture bottle - you should see many tunnels in the medium made by small white larvae (or maggots). These first instar larvae go through several instar stages and eventually the third instar larvae crawl up the sides of the bottle away from the culture medium. There they stop and their larval cuticle hardens forming a dark brown pupa. Metamorphosis takes place during the pupal stage. Larvae tissues degenerate and reorganize forming an adult fly inside the pupal case. When metamorphosis is complete, the adult fly emerges from the pupal case. After the fly emerges, the wings expand and dry, the abdomen becomes more rotund, and the color of the body darkens.

Sexing flies: Male and female fruit flies can be distinguished from each other in three ways:

1) Only males have a sex comb, a fringe of black bristles on the forelegs.

2) The tip of the abdomen is elongate and somewhat pointed in females and more rounded in males.

3) The abdomen of the female has seven segments, whereas that of the male has only five segments.

The first task of the lab is to become proficient at sexing the flies. Practice sexing several flies and check out your diagnosis with your instructor.

Genetic notation:

In fruit fly genetics, the normal fly is called a "wild type" and any fly exhibiting a phenotypic mutation is called a "mutant". Mutant flies are given names that generally denote the type of mutation the fly exhibits. For example, the mutant "ebony" has a much darker body than the wild type fly. Each mutation is also given a letter code. Thus, in the case of ebony, the code is a lower case e. The wild type fly is denoted by a superscript + over the mutant letter code. For example, e+ denotes a wild type fly for the ebony body trait - meaning it has normal body color (not ebony). The above description is for a gene located on an autosome (a non-sex chromosome). Of course, fruit flies also have sex chromosomes and they contain a subset of genes as well. If the gene is located on a sex chromosome, we use a slightly different notation. Under normal diploid conditions a female fruit fly has two X chromosomes, a male has an X and a Y chromosome. Sex-linked genes are located on one of the sex chromosomes (usually the X chromosome). Thus, the genotypic notation for a mutant gene for white eye color on the X chromosome would look like:

Xw Xw = white-eyed female

Xw+Xw = wild type heterozygote female

Xw Y = white-eyed male

Xw+ Y = wild type male

It may take a while to get used to this notation, but it’s well worth the effort as latter classes will undoubtedly use this notation.


To get use to the idea of phenotypic mutations, you will be given several strains of mutant flies. In your lab notebook, describe their morphology paying particular attention to eye color, body color, and wing shape. You should begin with a wild type fly so that you will have some basis for comparison. You may want to draw a picture.

Determining the Type of Inheritance in a Cross

So now you know what a wild type fly and several mutant flies look like phenotypically, but that really doesn’t tell us what their genotypes are (Why not?). In order to determine the genotypes, we must look at the ratios of phenotypes of the offspring. You will be given several bottles of fruit flies. Each bottle represents the F1 (first filial) generation of a cross set up two weeks ago. Recall that there must be a P (parental) generation before you can have a F1 generation. The idea here is to work back to the parental genotypes from the ratios of the F1 phenotypes you are given today. This requires a bit of detective work, since there are several possible parental crosses.

In the first cross you are given there is a single mutation. Anesthetize the flies using ether (your instructor will demonstrate). Observe them carefully under the dissecting microscope and record your observations in your lab notebook. The first step is to identify the sex and type of fly — mutant or wild type. Once you have determined the type of mutation, you should write out all possible crosses that could have produced these offspring. Remember, a single trait can be inherited on an autosome or on a sex chromosome. Is your mutation on an autosome or sex chromosome? How would you tell?

Assume you identify 100 flies and record the following data for the offspring of an unknown cross involving a single trait.

  Male Female Total
Wild type 40 30 70
mutant 15 15 30


What are the possible crosses that could have produced this pattern of offspring? If the mutation was inherited as a simple autosomal recessive, then we might suspect the parental generation was either:

  1. w w X w+w+
  2. w+w X w+w
  3. w+w X w w
  4. w+w X w+w+

Remember a w represents the mutant trait and a w+ represents the wild type.

Which of these 4 possibilities would give you the same approximate pattern of offspring you actually observed? Which of these 4 possibilities can be ruled out? Take parental cross 1 above for example. In a cross between a homozygous wild type and a homozygous mutant type you would expect to get all heterozygous wild type offspring. Since heterozygous offspring all appear phenotypically to be wild type flies, you could now rule this cross our because you actually observed 30 mutant flies in you F1 population.

Similarly, if this mutation was inherited as a sex-linked trait, you might predict one of the following parental crosses:

  1. XwXw x XwY
  2. Xw+Xw+ x Xw+Y
  3. Xw+Xw x XwY
  4. Xw+Xw x Xw+Y

Again you can do each cross and quickly rule out the ones that do not fit with your observed pattern of offspring. Assume you have ruled out all crosses but a parental cross between two heterozygotes for a trait located on an autosome (i.e. a parental cross of w+w x w+w). In this cross you would expect to see a phenotypic ratio of 3 wild type for every 1 mutant type regardless of the sex of the fly. Does this expected outcome fit with your observed data? It sure looks like it, but how can you be sure?

To test your hypothesis that the observed ratio of 70:30 is the same as the expected ratio of 3:1, we can use a statistic called the Chi square statistic.

Chi square statistic:

A chi square (X2) statistic is used to investigate whether two distributions of categorical variables differ from each other. In statistical testing we always use a null hypothesis that there is no difference between the distributions. (note: Chi square tests can only be used on actual numbers and not on percentages, proportions, means, etc.)

In our case, we can use the actual observed number of flies of each type as our observed values. We can find the expected number of flies of each type for a 3:1 ratio by using the same number of flies (100 flies) and dividing by 4 to give us expected values of 75:25 for a 3:1 ratio of 100 flies.

Calculate the chi square statistic x2 by completing the following steps:

  1. For each observed number in the table subtract the corresponding expected number (O — E).
  2. Square the difference [ (O —E)2 ].
  3. Divide the squares obtained for each cell in the table by the expected number for that cell [ (O - E)2 / E ].
  4. Sum all the values for (O - E)2 / E. This is the chi square statistic.

For our example, the calculation would be:

  Observed Expected (O — E) (O — E)2 (O — E)2/ E
Wild type 70 75 5 25 0.31
Mutant type 30 25 5 25 1.00
totals 100 100      

Having now obtained our chi square statistic x2 = 1.31, we look up in a table of the Chi Square X2 distribution the probability attached to it. Before we can do this, however, we need to know the degrees of freedom. When a comparison is made between one sample and another, a simple rule is that the degrees of freedom equal (number of columns minus one) x (number of rows minus one) not counting the totals for rows or columns. For our data this gives (2-1) x (2-1) = 1. Entering the Chi square distribution table with 1 degree of freedom and reading along the row we find our value of x2 (1.31) lies between 0.455 and 2.706. The corresponding probability is 0.5<P<0.1. This is well below the conventionally accepted significance level of 0.05 or 5%, so the null hypothesis that the two distributions are the same is verified. In other words, when the computed x2 statistic exceeds the critical value in the table for a 0.05 probability level, then we can reject the null hypothesis of equal distributions. Since our x2 statistic (1.31) did not exceed the critical value for 0.05 probability level (3.841) we can accept the null hypothesis that a ratio of 70:30 is the same as a 75:25 ratio (within 5% error).

Chi Square distribution table

Df 0.5 0.10 0.05 0.02 0.01 0.001
1 0.455 2.706 3.841 5.412 6.635 10.827
2 1.386 4.605 5.991 7.824 9.210 13.815
3 2.366 6.251 7.815 9.837 11.345 16.268
4 3.357 7.779 9.488 11.668 13.277 18.465
5 4.351 9.236 11.070 13.388 15.086 20.517



To put this into context, it means that we do have a 3:1 ratio of wild type to mutant offspring and that these offspring could have come from a cross between two heterozygotes.

To make the chi square calculations a bit easier, plug your observed and expected values into the following applet.



Pre-Lab Questions:

Complete the answers to these questions on a separate sheet of paper and be prepared to hand it in at the very beginning of lab.

1) You are given a monohybrid cross between a wild type fruit fly and a mutant fly called "apterous" (ap) which lacks wings. If both flies are homozygous, how would you denote the cross described? In other words, write out the parental genotypes in the form used for fruit fly genetics.

2) In the cross described in question #1, how would you write the cross if both parents were heterozygous?


3) What would the expected phenotypic ratio of offspring from the cross is question #1? Show the cross and the results.


4) What would be the expected phenotypic ratio of offspring in question #2? Show the cross.


5) If the trait you crossed was the result of a gene located on the X sex chromosome (not an autosome), how would you denote the cross in question #1? Complete the resulting cross and give the expected ratio of offspring of each genotype and phenotype.


6) If the trait you crossed was the result of a gene located on the X sex chromosome (not an autosome), how would you denote the cross in question #2? Complete the resulting cross and give the expected ratio of offspring of each genotype and phenotype.

7) Suppose you had a cross involving two traits (dihybrid) located on autosomes. If you crossed a homozygous ebony body male with a homozygous apterous wing female, what would the F2 genotypic and phenotypic ratios be assuming you produced the F2 by crossing two of the F1 flies together? Show all your work for full credit.