TITLE : Drosophila melanogaster

OBJECTIVE :

i. To calculate the offspring characteristics that given same to the Mendel’s 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 Drosophila 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. We had bred the homozygote parents of wild type and homozygote mutant.

iii. 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 F2 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.

Parents generation

Mendel Law 1 Vestigial Wing x Wild Type

Mendel Law 1 Ebony x Wild Type

Mendel Law 2 Sephia, Ebony x Wild Type

X-linked White Eye (M) x Wild Type (F)

X-linked Wild Type (M) x White Eye (F)

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RESULT & DISCUSSION:

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

values)

E (Expected values) (O-E)2

Vestigial 21 1/4 x 130 = 32 121

Wild Type 109 ¾ x 130 = 98 121

CHI-SQUARE TEST:

Degrees of freedom = 2-1

= 1

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

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X2 = ∑ (21-32)2/ 33 + (109-98)2/ 98

= 4.9

= 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.

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Mendel Law 1

Legend:E- Dominant allele for wild typee- Recessive allele for ebony body

Parent : Ebony x Wild type ee EE

G1 :

F1 : All wild typeEe

F1xF1 : 5 wild type x 5 Wild type Ee Ee

G2 :

F2 :

Chi square test: x2 = ∑(O-E)2/ E

Drosophilla types Observed data (O) Experiments’ hypothesis (E)

(O-E)2/ E

Wild type 119 ¾ x 157 = 117.75 (119-117.75) / 117.75 = 0.013

Ebony 38 ¼ x 157 = 39.25 (38-39.25) /39.25 = 0.040

Total 157 x2 = 0.053

Discussion:

M/F E eE Ee EeE Ee ee

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38 : 119Ebony Wild type

e E

E e E e

x2 = ∑(O-E)2/ E

x2 = Chi square value O = Observed data in each

categoryE = Data in each category based

on the experiments hypothesis

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.

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 Drosophilla 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 F1 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

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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 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 Red ebony Wildtype

110 50 49 12

Total 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)

Observed data based on experimenter’s hypothesis ( E)

x² = (O- E)² E

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 = 140.286

Total 221 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.

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X- linked

Parent phenotype white eye x wild type (Female) (Male)

F 2 phenotype Male Female

Red eye 25 45White eye 50 41

Characteristics Observed data (O) Observed data based on

experimenter’s hypothesis ( E)

x² = (O- E) ² E

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 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.

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X – Linked (Parent: White – eyed male to wild – type female)

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 F1 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 F2 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

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

Female (F=38) Male (M=33)Red eye White eye Red eye White eye

36 2 12 21

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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 Drosophila. 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

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White eye Vestigial wing

wildtype Red eye

Red ebony Sephia ebony

White eye