Fruit Fly Mendelian Genetics

Title: Identification of Wild Type Characters and Sex in Drosophila melnogaster
Introduction: This lab was performed using Drosophila melanogaster fruit flies. Drosophila melanogaster is a useful tool in genetic studies of diploid organisms because it is small, has a short life cycle, is easily and inexpensively raised and handled, and produces many progeny from a single mating.

Question: What are the results when certain types of Drosophila melanogaster (sepia eyes x wild type eyes, white eyed females x wild type males, and vestigial wings x sepia eyes) are crossed respectively? Do these results cohere to the laws of Mendelian genetics?

Hypothesis:
�In cross number one, if the allele for sepia eyes is homozygous recessive and not sex-linked, then the F1 generation of the first cross will display no offspring with the trait, and the F2 generation will display only about 25% of that generation with sepia eyes and the other 75% will be wild type.
�In cross number two, if the allele for white eyes is homozygous recessive and sex-linked on the X chromosome, then all the males of the F1 generation will inherit the trait from the homozygous recessive mothers that show the trait and all the female offspring will be heterozygous carriers.
�In cross number three, if the alleles for vestigial wings and sepia eyes are both homozygous recessive, not sex-linked, and not linked genes, then the ratio of 9:3:3:1, showing 9/16 red eyes and normal wings, 3/16 sepia eyes and normal wings, 3/16 red eyes and vestigial wings, and 1/16 showing both vestigial wings and sepia eyes.

Materials: 1) Dissection Microscope
2) Filter paper
3) Petri dish with lid
4) Vials of male, female of respective types (wild, sepia eyes, white eyes, vestigial wings) fruit flies
5) Ice bucket
6) Freezer pack

Procedure:

�Fruit flies were put into the ice bucket. After few minutes, flies were anesthetized by the cold temperature in the ice bucket. Then the Petri dish was put to the ice pack or freezer pack. The entire surface of the Petri dish was in contact with the freezer pack. The cap was unplugged and the fruit flies were put into the Petri dish covered with lid. Then the fruit flies were put underneath the microscope to be examined.

Data: 1) Sepia eyes x wild type
2) White eyed females x wild type males
3) Vestigial wings x sepia eyes

Sepia eyes x wild type

* Examination of the F1 hybrids and the F2 generation #1

F1 generation # male # female Total
Red eyes, normal wings (wild type) 19 21 40

F2 Generation #male #female Total
Red eyes, normal wings (wild type) 15 17 32
Sepia eyes, normal wings 5 3 8

White eyed females x wild type males

* Examination of the F1 hybrids and the F2 generation #2

F1 generation # males # females Total
Red eyes, normal wing 0 21 21
White eyes, normal wing 19 0 19

F2 generation # males # females Total
Red eyes, Normal wings 9 11 20
White eyes, normal wings 12 8 20

Vestigial wings x sepia eyes

* Examination of the F1 hybrids and the F2 generation #3

F1 # males # females Total
Red eyes, normal wings 22 18 40

F2 # males # females Total
Red eyes, normal wings 12 3 25
Red eyes, Vestigial wings 3 3 6
Sepia eyes, Normal wings 2 3 5
Sepia eyes, Vestigial wings 2 2 4

Results:

F2 Phenotype #Observed # Expected (o-e)^2/e
Sepia eyes 32 30 0.13
Red eyes 8 10 0.4
Sum of Chi squared 0.53

�The degree of freedom is 1 because there are only two classes accounted for: sepia eyes and red eyes. According to a chi-square chart there is a 51% chance that this testcross is biased. However, since the chi-squared value is greater than 0.05, the hypothesis is valid.

F2 Phenotype #Observed # Expected (o-e)^2/e
Red eyes 20 20 0
White eyes 20 20 0
Sum of Chi squared 2.31

�The degree of freedom in this test is 1 because again there are only two classes being used: red eyes and white eyes. According to a chi-square chart there is about 0% chance of testcross 2 being biased, thus proving the hypothesis for this testcross valid.

F2 Phenotype #Observed # Expected (o-e)^2/e
Red eyes, normal wings 25 22.5 .28
Red eyes, vestigial wings 6 7.5 .3
Sepia eyes, normal wings 5 7.5 .83
Sepia eyes, vestigial wings 4 2.5 .9
Sum of Chi squared 2.13

�The degree of freedom in this test is 3 because there are 4 classes being delt with: red eyes + normal wings, red eyes + vestigial wings, sepia eyes + normal wings, sepia eyes + vestigial wings. According to a chi-squared chart there is about a 55% chance that this testcross is biased. However, since the chi-squared value is greater than 0.05 the hypothesis is valid.

Conclusion:

In this experiment chi-square analysis or the laws of Mendelian genetics were used to determine to determine the wild type characteristics and sex of Drosophila melnogaster.

The chi-squared tests were used to analyze the experiment results and observe how much they deviated from the expected outcome as outlined in each respective hypothesis. The chi-squared analysis in this experiment proved each earlier stated hypothesis valid and true.

In cross number one, my hypothesis was correct. All the offspring of the F1 generation had no phenotype that displayed the homozygous recessive trait. Because males did not inherit sepia eyes from the females of the P generation it can be deduced that the allele for sepia eyes is not sex-linked. Furthermore in the cross of the F1 generation, the F2 generation showed an expected ratio of offspring phenotypes of approximately 25% of the offspring exhibiting sepia eyes.

In cross number 2, my hypothesis was correct. The allele for white eyes is sex-linked to the X chromosome and is homozygous recessive. This is apparent because in the F1 generation all males inherited the recessive allele from the female of the P generation, thus making all the males of the F1 generation exhibit white eyes. The F2 generation also showed expected results cohering to Mendelian genetics as there were 50% males showing the trait and 50% females showing the trait.

In cross number 3, my hypothesis was correct. The allele for sepia eyes and vestigial wings are non linked, homozygous recessive, and not sex linked.

Question:
What are the results when certain types of Drosophila melanogaster (sepia eyes x wild type eyes, white eyed females x wild type males, and vestigial wings x sepia eyes) are crossed respectively? Do these results cohere to the laws of Mendelian genetics?

Results:
The chi-squared test was used to analyze the results and observe how much they deviated from the expected outcome as outlined in the hypothesis.

Key:
O = Observed
E = Expected

For testcross 1 (F2 generation):

Table 7: Chi-square analysis for testcross 1
PhenotypeOEO-E(O-E)2(O-E)2/E
Sepia-eyed323024.13333
Red-eyed810-24.4

Since we are working with two classes: sepia eyes and red eyes, the degree of freedom (df) is 1. Chi-squared (x2) = .13333 + .4 = .53333. According to a chi-squared chart, this value with about a 51% chance of the testcross being biased. Since the chi-squared value is greater than 0.05, the hypothesis stated earlier holds true and it can be concluded that the testcross is fair.

For testcross 2 (F2 generation):

Table 8: Chi-square analysis for testcross 2
PhenotypeOEO-E(O-E)2(O-E)2/E
Red-eyed2020000
White-eyed2020000

Since we are working with two classes: white eyes and red eyes, the degree of freedom (df) is 1. Chi-squared (x2) = 0 + 0 = 0. According to a chi-squared chart, this value with about a 0% chance of the testcross being biased. There is no error at all; the testcross is not biased and is fair, thus, the hypothesis for testcross 2 holds true.

For testcross 3 (F2 generation):

Table 9: Chi-square analysis for testcross 3
PhenotypeOEO-E(O-E)2(O-E)2/E
Red-eyes, Normal wings2522.52.55.25.23333
Red-eyes. Vestigial wings67.5-1.52.25.3
Sepia eyes, Normal wings57.5-2.55.25.7
Sepia eyes, Vestigial wings42.51.52.25.9

Since we are working with four classes: red-eyed & normal winged, red-eyed & vestigial winged, sepia-eyed & normal winged, and sepia-eyed & vestigial winged, the degree of freedom (df) is 3. Chi-squared (x2) = .23333 + .3 + .7 + .9 = 2.13333. According to a chi-squared chart, this value with about a 55% chance of the testcross being biased. Since the chi-squared value is greater than 0.05 though, the hypothesis stated earlier holds true and this testcross is considered fair and unbiased.

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