TITLE
: Drosophila melanogaster
OBJECTIVE :
i.
To calculate the offspring characteristics that given same to the Mendel\u2019s law of inheritance.
ii.
To make students creative to do the experimental design the crosses of flies.
INTRODUCTION:
Drosophila melanogaster is a small, common fly found near unripe and rotted fruit. It has
been in use for over a century to study genetics and lends itself well to behavioral studies. Thomas Hunt Morgan was the preeminent biologist studying Drosophila early in the 1900's. Morgan was the first to discover sex-linkage and genetic recombination, which placed the small fly in the forefront of genetic research. Due to it's small size, ease of culture and short generation time, geneticists have been using Dr os ophila ever since. It is one of the few organisms whose entire genome is known and many genes have been identified. Fruit flies are easily obtained from the wild and most biological science companies carry a variety of different mutations. In addition these companies sell any equipment needed to culture the flies. Costs are relatively low and most equipment can be used year after year. There are a variety of laboratory exercises one could purchase, although the necessity to do so is questionable. PROCEDURE:
There are a few methods in fainting the flies. In this practical, we are given chloroform of anesthetization kit for fainting the flies. The lab assistants show to us the technique. The newly fainted flies should be examined immediately. Always ready with cotton or cotton buds with the anesthetizer before hands as to avoid the sudden awaken of the flies.
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Part I
i.
Firstly we identify the wild type and mutant flies. We had examined their morphology thoroughly before we do the crosses. We use the dissecting microscope in lab to examine the flies.
ii.
We had design our experiment properly in order to yield expected results.
iii.
We choose the traits that can breed through based on autosomal and sex-linked.
iv.
Before we hands the flies, we show our experimental design to our Tutor for approval. Then we inform the lab assistant the flies that we want to use.
Part II
i.
We have chosen five pairs of traits that follow: Mendel Law 1, Mendel Law 2 and X-linked.
ii. iii.
We had bred the homozygote parents of wild type and homozygote mutant. When we see the larvas have emerged from the medium, we took out the parents and kill them immediately. The Tutor and lab assistants had shown us how to kill the flies. We also make sure that no one of flies flying in the lab.
iv.
We let the larvas grow. We placed the virgin females in new tubes with their male siblings. When we see another generation of larvas (F2) emerge, we killed the F1.
v.
Every day we go to lab to see if we get the F 2 generation or not. If we have that flies, we anesthetize the adult F2 and we count.
vi.
We had analysed the results using Chi-square.
Mendel Law 1 Mendel Law 1 Mendel Law 2 X-linked X-linked
Parents generation Vestigial Wing x Wild Type Ebony x Wild Type Sephia, Ebony x Wild Type White Eye (M) x Wild Type (F) Wild Type (M) x White Eye (F)
RESULT & DISCUSSION:
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Mendel Law 1
Law 1
: Vestigial Wing x Wild Type
Legend
: S- Dominant allele control the wild type wing trait s- Recessive allele control the vestigial wing trait
Cross
: VV x vv
Result (f2 generation): 3:1
Characteristic
O (observation
Vestigial Wild Type
21 109
E (Expected values)
values)
1/4 x 130 = 32 \u00be x 130 = 98
(O-E)2 121 121
CHI-SQUARE TEST: Degrees of freedom = 2-1 =1 X2 = \u2211 (O-E)2 / E Where, O = Observed data in each category E = Observed data in each category based on the experimenter\u2019s hypothesis S = Sum of the calculation for each category
X2 = \u2211 (21-32)2/ 33 + (109-98)2/ 98 = 4.9
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= 1%< X2 <5% = 0.01< X2 < 0.05 Discussion:
From the result, the observation value is 21 for vestigial wing and 109 for wild type wing. So, wild type is dominant to vestigial wing. Based on the observations, the hypothesis is ‘Wild type (V) is dominant to the Vestigial (v). From the chi-square test, the chi-square value is lower than 0.05. So, that means the null hypothesis is rejected for the experiment. From the chi-square test the data and hypothesis do not have a good fit. During the experiment, the data are collected not accurately due to the lost of the Drosophila melanogaster during the researcher change the place of the Drosophila melanogaster and the another factor is due to the fatal problem.
Mendel Law 1
Legend: E- Dominant allele for wild type e- Recessive allele for ebony body 4
Parent :
G1
:
F1
:
F1xF1 : G2
Ebony ee
Wild type EE
E
e All wild type Ee 5 wild type Ee
x
Ee
5 Wild type
: E
F2
x
:
e
E
38 : 119 Ebony M/F EWild typee E Ee Ee E Ee ee
Chi square test: x2 = ∑(O-E)2/ E Drosophilla types
Observed data (O)
Wild type
119
Ebony
38
Total
157
e
x2 = ∑(O-E)2/ E x2 = Chi square value O = Observed data in each category E = Data in each category based on the experiments h othesis
Experiments’ (O-E)2/ E hypothesis (E) ¾ x 157 = 117.75 (119-117.75) / 117.75 = 0.013 ¼ x 157 = 39.25 (38-39.25) / 39.25 = 0.040 x2 = 0.053
Discussion:
Mendel Law 1, or Law of Segregation states that, the two copies of a gene segregate from each other during transmission from parent to offspring. This law predicts that the phenotypes of the F2 generation will be in ratio of 3:1 of dominant to recessive traits.
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Therefore, for this experiment, the hypothesis of Mendels Law 1 is said to be 3:1 ratio of wild type to ebony type of Drosophilla melanogaster.
From the cross between F1 generation, they produce two types of Dr os ophilla offsprings which are ebony type and wild type. Thus n= 2. As degree of freedom, df= n-1, therefore df value is 1. With the df=1, and the Chi Square value equal to 0.053, we have obtained results that near to 0.0642 value under probability of 0.80. Therefore from the table of Chi Square values and Probability, the obtained chi square value from experiment which is 0.053, are expected to occur 80% of the time based on random chance alone when hypothesis are correct. This low value of chi square indicates that the observed value that we obtained from the experiments is actually fit the hypothesis. Thus, our result can be well accecpted. The results obtained from the experiments fits the hypothesis value of Mendels law 1. The gene separate from each other during transmission from parent to offspring and cause the cross of F 1 to F1 producing F2 offspring that obey Mendels law 1, with 3:1 ratio of wild type to ebony type of Drosophilla melanogaster.
Mendel Law 2
Mendel stated for the Law 2, two different genes will randomly assort their alleles during the formation of haploid cells. The expected ratio is usually 9:3:3:1 Parent phenotype:
Sephia, ebony x
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wild type
(Female)
(Male)
From the cross above, the predicted outcome is sephia ebony, sephia wild type, red ebony and wildtype. Sephia ebony
Sephia wildtype
110
50
Red ebony 49
Total
Wildtype 12
221
The ratio get from this data is 9:3:3:1 To test the validity: Using the Chi-Square Test Degrees of freedom = 4-1 =3 X2 = ∑ (O-E)2 / E Where, O = Observed data in each category E = Observed data in each category based on the experimenter’s hypothesis S = Sum of the calculation for each category
Through Chi- Square Test:
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Characteristics
Observed data (O)
x² = Observed data based on experimenter’s hypothesis ( E) E)²
(O-
Sephia ebony
110
9/16 x 221 = 124.3
1.645
Sephia wildtype
50
3/16 x 221 = 41.4
1.786
Red ebony
49
3/16 x 221 = 41.4
1.395
Wildtype
12
1/16 x 221 = 14
Total
221
E
0.286 5.112
Discussion: At first, we have known that df = 3. Thus, from the chi square the value obtained is 5.112. From the table of chi square this value is greater than 4.642 and less than 7.815. So from that we know that P value is located between 0.05 and 0.20 or 5% to 20%. Since the value is between 5% and 20%, and not exceed 0.05, this value is accepted and the data and the hypothesis have a good fit. From the cross, the number of sephia ebony is higher than others and the wildtype has the lower number.
X- linked
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Parent phenotype
white eye (Female)
x
F 2 phenotype
Male 25 50
Red eye White eye Characteristics
wild type (Male)
Observed data (O)
Observed data based on experimenter’s hypothesis ( E)
Female 45 41 x² =
(O-
Red eye male
30
¼ x 166 = 41.5
3.187
Red eye female
45
¼ x 166 = 41.5
0.295
White eye male
50
¼ x 166 = 41.5
1.741
White eye female
41
¼ x 166 = 41.5
0.006
Total
166
E) ²
E
5.229
Discussion:
With df = 3, the chi square value of 5.229 is slightly greater than 4.642 and less than 7.815 which give a P value between 0.05 and 0.20 or 5% to 20%. Since the value is between 5% and 20%, based on random chance alone, thus the hypothesis is accepted. The data and the hypothesis have good fit even though the P value is quite low.
X – Linked (Parent: White – eyed male to wild – type female)
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Female (F=38) Red eye White eye 36
Male (M=33) Red eye White eye
2
12
21
Discussion:
Based on our result in the table above, the experiment is done successfully. To get this result, we have crossed the white – eyed male to a wild – type or true – breeding red – eyed female. The result for the offspring or the F1 generation has been recorded. All of the F1 offspring had red eyes due to red is dominant compare to white. Then, F1 offspring were mated to each other to obtain an F2 generation. After we crossed the F1 offspring with each other, the result is recorded just like in the table above. We obtained 38 female and 33 male. We got 36 red – eyed female and 2 white – eyed female while 12 red – eyed male and 21 white – eyed male. If we compare to the experiment that have been done by Thomas Hunt Morgan who is confirmed about the chromosome theory of inheritance, there is different is some part of the same experiment. He also obtained that all of the F 1 offspring had red eyes following X-link cross ratio of F1 which all progeny should be red. It same goes to us. But the different is F 2 generation. He most notably that no white – eyed female offspring were observed in the F2 generation. According to expected theory we should never get the female F2 progeny with white eye. So, in this experiment, we might be late in observing the F2 progeny as we conclude that the 2 white eye female of F2 progeny is actually not F2 generation but the F3 generation. We also conclude that, even if the 2 white eye female are from F2 generation, the dominant allele of red eye in the X chromosome 10
undergo mutation, so, it did not show the characteristic of dominant allele and the recessive allele on the other X chromosome becomes more dominant than the mutated allele. Hence, both Drosophila melanogaster shows recessive appearance. It suggested that the pattern of transmission from parent to offspring depend on the sex of the offspring and on the alleles that they carry. That’s why he concluded that the eye color alleles are located on the X chromosome because the prediction was confirmed experimentally. For our result, it doesn’t follow the Mendel’s Law or Morgan’s proposal because we obtained the white – eyed female. But our experimental just undergo one time compared to Morgan’s experiments. So, might be there is error that occur during we were mating the fruit flies or also known as D r os ophila. But, the number of the white – eyed female is only 2. So, it very closes to the theory.
CONCLUSION:
As the conclusion, the objective of this experiment is accepted where the students have knowledge and creative to design the experiment of crosses the flies. We also can identify the characteristics of Drosophila melanogaster using the dissecting microscope.
REFERENCE:
Robert J. Broker ( 2005 ). Genetics analysis & Principles. McGraw Hill International Edition.
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APPENDIX:
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Pictures of Drosophila melanogaster
White eye
Vestigial wing
wildtype
Red eye
Red ebony
Sephia ebony
White eye 13
