Title: Identification of Wild Type Characters and Sex in Drosophila melnogaster
Introduction: This lab was done using the fruits of Drosophila melanogaster flies. Drosophila melanogaster is a useful tool in genetic studies of diploid organisms because it is small, short life cycle. it is easily and cheaply erected and handled, and produces many from one copulation.
Question: What are the results when certain types of drosophila melanogaster (cuttleeye x cuttlefish species, white eyed female x wild cuttlefish species, and feather traces x cuttlefish eye) are crossed respectively? Are these results consistent with Mendelian genetic laws?
Hypothesis:
• In one number of crosses, if the alleles for sepia eyes are homozygous recessive and not sex-linked, the F1 generation of the first cross will show no offspring with the trait, and the F2 generation will show only about 25% of that generation. cuttlefish 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, all males of the F1 generation will inherit the trait from homozygous recessive mothers who display the trait and all female offspring. the carriers will be heterozygous.
• In the number three cross, if the alleles for the traces of the wings and eyes are sepia, both are homozygous recessives, not conjugation of the sexes, nor conjunction of the genes, the ratio 9:3:3:1, showing 9/16 red and normal eyes. sepia wings, 3/16 eyes and normal wings, 3/16 red eyes and wing traces, and 1/16 wings showing sepia eyes and eyes.
Materials: 1) Microscopic dissection
2) filter paper
3) Petri dish in the lid
4) male bowl, female of both sexes (wild eyes, cuttlefish, white eyes, wing traces) fruit flies
5) a bucket of ice
6) Freezer Pack
The procedure:
•Brown flies are placed in a bucket of ice. After a few minutes, the flies are anesthetized by the cold temperature in the ice bucket. Then Peter’s dish was placed in ice packs or a 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 placed in a Petri dish covered with a lid. The fruit flies are then examined under a microscope.
Data: 1) Cuttlefish eye x wild type
2) White-eyed female x wild male species
3) Traces of wings x sepia eyes
Cuttlefish eyes x wild type
* Examination of F1 and F2 hybrids of 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″
Cuttlefish eyes, normal wings 5 3 8
White-eyed female x wild male species
* Examination of F1 hybrids and F2 generation #2
F1 generation # males # females Total
Red eyes, normal horns 0 21 21
White eyes, normal horns 19 0 19
F2 generation # male # female
Red eyes, normal wings 9 11 20
White eyes, normal wings 12 8 20
Traces of wings x sepia eyes
*Examination of F1 hybrids and F2 generation #3
F1 # males # females Total
Red eyes, normal wings 22 18 40
F2 # males # females
Red eyes, normal wings 12 3 25
Red Eyes, Footprints, Wings 3 3 6
Cuttlefish eyes, normal wings 2 3 5
Cuttlefish eyes, Wing traces 2 2 4
Results:
F2 Phenotype # Observed # Expected (o-e)^2/e
Cuttlefish eyes 32 30 0.13
Red eyes 8 X 0.4
Total Chi square 0.53
• The degree of freedom is 1 because only two species are considered for sepia eyes and red eyes. According to the chi-square chart, there is a 51% chance that this testcross will light up. When the chi-square 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
Total Chi square 2.31
• The degree of freedom in this experiment is 1, because again only two types are used: red eyes and white eyes. According to the chi-square chart, there is about a 0% chance of a testcross based on 2, thus proving the hypothesis of this testcross to be strong.
F2 Phenotype # Observed # Expected (o-e)^2/e
Red eyes, normal wings 25 22.5 .28
Red eyes, traces of wings 6 7.5 .3
Cuttlefish eyes, normal wings 5 7.5 .83
Cuttlefish eyes, wing tracks 4 2.5 .9
Chi square sum 2.13
There are 3 degrees of freedom in this test, because there are four classes: red eyes + normal wings, red eyes + footprints, cuttlefish eyes + normal wings, cuttlefish eyes + wing traces. According to the chi-square chart, there is about a 55% chance that this testcross will light up. However, when the chi-square value is greater than 0.05, this hypothesis is valid.
conclusion:
In this experiment, chi-square analyzes or Mendelian genetic laws were used to determine the wild-type characteristics and sex of Drosophila Melnogastr.
Chi-square tests were used to analyze the results of the experiments and observe how much they deviated from the expected results in each hypothesis. The chi-square analysis in this experiment proved that any of the previously stated hypotheses were valid and true.
In one cross number, my hypothesis was correct. All progeny of the F1 generation had no phenotype that showed a homozygous recessive trait. Because cuttlefish males do not possess eyes, it can be deduced from females of the P generation that the allele for cuttlefish eyes is not sex-linked. Furthermore, in the F1 generation cross, the F2 generation showed an expected phenotype ratio of approximately 25% of the offspring exhibiting sepia eyes.
In cross number 2, my hypothesis was correct. The allele for white eyes is linked to the sex chromosome and is homozygous recessive. This is clear because in the F1 generation all the males have inherited the recessive allele from the female of the P generation, thus all the males of the F1 generation have white eyes. The F2 generation also showed consistent Mendelian genetic results as there were 50% males showing the trait and 100% females showing the trait.
In cross number 3, my hypothesis was correct. Alleles for cuttlefish eyes and wing traces are not linked, homozygous recessive and not sex-linked.
Question:
What are the results when certain types of melanogaster Drosophila (cuttleeye x cuttlefish species, white eyed female x wild male species, and feather traces x cuttlefish eye) are respectively crossed? Are these results consistent with Mendelian genetic laws?
Results:
A chi-square test was used to analyze the results and observe how much they deviated from the expected result 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 eye and red eye, the degree of freedom (df) is 1 . Chi square (x2) = .13333 + .4 = .53333. According to the chi-square chart, this value was based on a testcross probability of about 51%. When the chi-square value is greater than 0.05, the hypothesis stated above holds true and it can be concluded that the cross-test 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 there are two races working: white-eyed and red-eyed, the degree of freedom (df) is 1. Chi-square (x2) = 0 + 0 = 0. According to the chi-square chart, this value is about a. 0% chance of testcross being bid. There is absolutely no error; The test cross should not shine and be fair, so the thesis for testcross 2 is 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. Wing tracks 67.5-1.52.25.3
Cuttlefish eyes, normal wings57.5-2.55.25.7
Cuttlefish eyes, Wing traces 42.51.52.25.9
We work with four categories: eye & amp; normal winged, red-eyed & winged tracks, sepia eyes & normal winged, and eye-catching & winged trace, the degree of freedom (df) is 3. Chi-square (x2) = .23333 + .3 + .7 + .9 = 2.13333. According to the chi-square chart, this value shines with a 55% chance of a testcross. When the chi-square value is greater than 0.05, although the hypothesis previously stated holds true and this test cross is considered fair and Female horse.
