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TBG 2013

GENETICS

NAME : SITI SARAH BT MOHD SAIFUDDIN D20091034843

AMEERA BT YAHYA D20091034814

NURUL HUSNA BT ALIAS D20091034858

PRACTICAL : 2 (SIMPLE MENDELIAN GENETICS IN DROSOPHILA MELANOGASTER)

DATE : 2 AUGUST 2010

LECTURERS NAME: EN. HAMZAH B ABDUL AZIZ


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Title:SIMPLE MENDELIAN GENETICS IN DROSOPHILA MELANOGASTER

Objectives:

1. To introduce normal "wild type" and various mutant phenotypes.

2. To conduct a genetics experiment this spans of generation.

3. To introduce the use of the Chi square statistic to test hypotheses concerning

expected and observed ratios.

4. To compare predicted result with actual result.

5. To determine the ratio of monohybrid cross, dihybrid cross and sex linkage cross of

Drosophila melanogaster.

6. To design genetic cross to illustrate segregation, independent assortment and sex

linkage.

7. To discuss the life cycle of Drosophila melanogaster.

8. To differentiate between male and female of Drosophila melanogaster.

9. To determine the progeny from the cross between wild type and vestigial

(monohybrid cross), wild type and vestigial, sepia eye (dihybrid cross) and wild type

and white eyes (sex linkage cross).

Introduction:

Gregor Mendel was born in 1822, is now known as Father of Genetics. He initially studies

inheritance of just one pair of contrasting traits. Mendel begins his experiment with garden

pea plant. Mendel recognized two principles that were later call Principle of Mendelian

Inheritance:

Principle of Mendelian Inheritance

1. Law of Segregation (The "First Law")

The Law of Segregation states that when any individual produces gametes, the

copies of a gene separate so that each gamete receives only one copy. A gamete

will receive one allele or the other. The direct proof of this was later found when the

process of meiosis came to be known. In meiosis, the paternal and maternal

chromosomes are separated and the alleles with the traits of a character are

segregated into two different gametes.


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2. Law of Independent Assortment (The "Second Law")

The Law of Independent Assortment, also known as "Inheritance Law", states that

alleles of different genes assort independently of one another during gamete

formation. While Mendel's experiments with mixing one trait always resulted in a 3:1

ratio between dominant and recessive phenotypes, his experiments with mixing two

traits (dihybrid cross) showed 9:3:3:1 ratios. Mendel concluded that different traits are

inherited independently of each other, so that there is no relation, for example,

between a cat's colour and tail length. This is actually only true for genes that are

notlinked to each other.

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 its 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, 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. In this experiment Drosophila are use because they are small and easily

handled, Drosophila are sexually dimorphic (males and females are different), making it is

quite easy to differentiate the sexes, and flies have a short generation time (10-12 days) and

do well at room temperature.

Life cycle of Drosop hi la melanogaster

D. melanogaster exhibits complete metamorphism, meaning the life cycle includes an egg,

larval (worm-like) form, pupa and finally emergence (eclosure) as a flying adult. This is thesame as the well-known metamorphosis of butterflies and many other insects. The larval

stage has three instars, or molts.

Life cycle by day

Day 0: Female lays eggs

Day 1: Eggs hatch

Day 2: First instar (one day in length)
http://en.wikipedia.org/wiki/Genetic_linkagehttp://en.wikipedia.org/wiki/Genetic_linkage


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Day 3: Second instar (one day in length)

Day 5: Third and final instar (two days in length)

Day 7: Larvae begin roaming stage. Pupariation (pupal formation) occurs 120 hours after

egg laying

Day 11-12: Eclosion (adults emerge from the pupa case). Females become sexually mature

8-10 hours after eclosion

The time from egg to adult is temperature- dependent. The above cycle is for a temperature

range of 21-23 degrees C. The higher the temperature, the faster the generation time,

whereas a lower (to 18 degrees C) temperature causes a longer generation time. Females

can lay up to 100 eggs/day. Virgin females are able to lay eggs; however, they will be sterile

and few in number. After the eggs hatch, small larvae should be visible in the growing

medium. If your media is white, look for the black area at the head of the larvae. Some dried

premixed media is blue to help identify larvae however this is not a necessity and with a little

patience and practice, larvae are easily seen. In addition, as the larvae feed they disrupt the

smooth surface of the media and so by looking only at the surface one can tell if larvae are

present. However, it is always a good idea to double check using a stereomicroscope. After

the third instar, larvae will begin to migrate up the culture vial in order to pupate.


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Sex difference

Several criteria may be used to distinguish male and female Drosophila melanogaster

Body sizefemale is generally larger than male.

Abdomen shapethe female abdomen curve to a point, the male abdomen is round

and much shorter. Figure below show Male (left) and Female (right) wild-type

Drosophila.

Mark on their abdomen - Alternating dark and light bands can be seen on the entire

rear portion of the female; the last few segments of the male are fused.


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Sex comb - On males there is a tiny tuft of hairs on the basal tarsal segment of the

front leg. This is the only sure method of distinguishing young male and female flies

(less than 2 hours old), since the other adult traits are not always immediately

recognizable. Sexing via the sex comb can also be done successfully in the pupal

stage (Hadden and Cunningham, 1970).

Sex organ at abdomen - The genitalia are the easiest and most reliable character to

use in determining sex (right; ventral view, posterior is up). Note the circle of darkly

pigmented parts in the male. In contrast, the tip of the female's abdomen is lightly

colored and pointed.


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Hereditary Traits

Before one observes their mutants, one needs to be familiar with the appearance of the wild

type Drosophila, the type found most often in natural populations of the organisms. Although

thousands of mutations in Drosophila are known, only those which are relevant to theseexercises are listed.

1. Eyes

Wild type: red, oval in shape and many-faceted

Mutants: white, black, apricot, scarlet red, pink, or brown; changes in shape and number of

facets

2. Wings

Wild type: smooth edges, uniform venation, extend beyond the abdomen

Mutants: changes in size and shape; absence of specific veins; changes in position in which

wings are held when at rest

3. Bristles

Wild type: fairly long and smooth (note distribution on head and thorax)

Mutants: shortened, thickened, or deformed bristles changes in patterns of distribution

4. Body colour

Wild type: basically gray, with pattern of light and dark areas

Mutants: black (in varying degrees), yellow, in doubtful cases, color can often be determined

most clearly on wing veins and legs

Mutantstraits can be assumed as recessive to the wild type.


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Material and apparatus

Drosophila melanogaster (male and female)

Cornmeal medium

Ether/ Flynap

Vial tube with sponge cover

Filter paper

Petri dish and its cover

Soft paint brush

Label

Procedures

Anesthetizing system

1. Ether or Fly nap is dropped on the cotton which placed under the etherizer cap and

closed the bottle for a few seconds until the ether gas fulfill the entire bottle.

2. Then, the base of the bottle is strike lightly on the palm of the hand so that the flieswill drop to the bottom.

3. Next, the bottle cap is removed, quickly replaced it with mouth of etherizer, the bottle

is inverted over the etherizer and shaked the flies into the etherizer. Dont invert the

bottle over the etherizer because the ether is heavier than air and it could flow to the

culture tube and kill the larvae and pupa. Both etherizer and culture tube are inverted

and strike slowly until the adult Drosophila drop down in to the bottom.

4. Quickly, the bottle is separated from its cover.

5. The flies are then subjected to the ether for a minute or until they ceased moving.

6. After that, the etherized flies are transferred on the filter paper.

7. If the flies revived before we finished examining them, a few drops of ether is added

to the cotton and put in the petri dish and covered it.

8. The etherized flies are examined with a dissecting microscope.

9. A soft brush is used for moving the flies about on stage of the microscope.

10. Finally, after finishing our experiment, the Drosophila is discarded in a soup water or

mineral oil, except the ones we need for the further crosses which we have to put on

dry surface in the culture bottle before they come into contact with the moist medium.


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Procedure in the experiment

The wild type and mutant flies are identified. Their morphology are examined

thoroughly before we do the crosses. We will use dissecting microscope. Here are

some traits for the experiments:

Mendel Law 1

- Ebony body

- curved wings

- Sepia eyes

- Scarlet eyes

Mendel Law 2

- Ebony body, yellow body

X-linked

- white eyes

- yellow body

- echinus eyes

- bar eyes

Procedures for monohybrid crosses

1. In monohybrid crosses, red eyes drosophila (male) and scarlet eyes (female )

drosophila was used for mating.

2. 10 male Drosophila and 10 female Drosophila are shifted into the bottle which

contains new medium/ substrate and the bottle is closed with the cotton. The

rest is killed and the traits are observed.

3. After a few days, the Drosophila will mate and finally the female Drosophila

will lay eggs which then will hatch. At this moment, all the parental Drosophila

has to be discarded to prevent mixed-up with the F2 generation.4. The experiment is repeated by using red eyes (female) and scarlet eyes (

male).

ANALYSIS OF F2 GENERATION

1. By using the ether/ Flynap, the F2 Drosophila is killed and is put on the filter

paper.

2. The total number of every F2 phenotype is counted. From the monohybrid

crosses, there are TWO phenotypes only. The distribution of the F2

phenotypes is tested by using X2.


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4. By using the appropriate symbols a diagram which shows the genotypes in

each crosses, from the parental stage to the F2 stage is draw.

Procedures for dihybrid crosses

1. In dihybrid crosses, wild type drosophila (male) and vestigial, sepia eyes

(female) drosophila was used for mating.

2. 10 male Drosophila and 10 female Drosophila are shifted into the bottle which

contains new medium/ substrate and the bottle is closed with the cotton. The

rest is killed and the traits are observed.

3. After a few days, the Drosophila will mate and finally the female Drosophila

will lay eggs which then will hatch. At this moment, all the parental Drosophila

has to be discarded to prevent mixed-up with the F2 generation.

4. The experiment is repeated by using wild type (female) and vestigial, sepia

eyes (male).

ANALYSIS OF F2 GENERATION

1. By using the ether/ Flynap, the F2 Drosophila is killed and is put on the filter

paper.

2. The total number of every F2 phenotype is counted. From the dihybrid

crosses, there are FOUR phenotypes.The distribution of the F2 phenotypes is

tested by using X2.

4. By using the appropriate symbols a diagram which shows the genotypes in

each cross, from the parental stage to the F2 stage is draw.

Procedures for X-linked

1. To determine sex-linked traits, backcross protocol can be used.2. Wild type female will be crossed with white eye male and wild type male will

be crossed with white eye female to produce F1 progeny.

3. Then F1 progeny is crossed each other to produce F2.

4. The F2 has been analysed and X2 test has been conducted.


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Results

Monohybrid Crosses

Figure 1 : Scarlet Drosophilla melanogaster Figure 2: Red eye Drosophila melanogaster

The crosses between wild type (male) scarlet eyes (female)

St+ is dominant allele for wild type

st is recessive allele for scarlet eyes

male normal eye (wild type) female Scarlet eye

Parent st+st+ stst

Gamete

F1 st+st

All wild-type

F1 F1 st+st st+st

Gamete

st+ st

st+ st stst+


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= 1

From the Chi square table, p = 0.5

Since the p value is greater than 0.05, it can be concluded that it is not possible to

reject the null hypothesis on the basis of this experiment.

The crosses between wild type (female) scarlet eyes (male)

St+is dominant allele for wild type

st is recessive allele for scarlet eyes

female normal eye (wild-type) male scarlet eye

Parent st+st+ stst

Gamete

F1 st+st

All wild-type

F1 F1 st+st st+st

Gamete

F2 st+st+ st+st st+st stst

Ratio 3: 1

wild-type : scarlet-eyed

st+ st

st+ st stst+


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Since the p value is greater than 0.05, it can be concluded that it is not possible to reject the

null hypothesis on the basis of this experiment.

DISCUSSION

The results of the parental cross ( st+st+ males stst females) demonstrate that the wild

type allele (st+) is dominant allele for scarlet eyes (st) as no scarlet-eyed progeny were seen

in the F1 progeny.

Calculations from the F2 data show that the ratio of normal eye (wild-type) to scarlet eyed

flies is 3.42:1. Although this ratio is very close to the expected 3:1 ratio for a monohybrid

cross, the Chi square test performed to determine whether this experimental data differed

significantly from the 3:1 ratio expected for simple monohybrid cross. The results of the Chi

square test suggest that the experimental data do not differ significantly from the expected

3:1 ratio. Specifically, there is between 50% and 90% probability that differences seen due to

chance. In this case, the differences seen are probably due to the small sample size scored

from the cross. The result of the parental cross ( st+st+ females x stst male ) show the same

result parental cross ( st+st+ males stst females)

CONCLUSION

In conclusion, the phenotype of the F1 progeny confirmed that the allele for wild type scarlet

eyes, st+ is dominant to the allele for scarlet eyes, st. The ratio of normal-eyed (wild-type) to

scarlet-eyed flies of 3.42:1 seen in the F2 is very near that of the expected 3:1 ratio for a

monohybrid cross, and the Chi square test verifies that it is within statistical limits. Therefore,

the results of this experiment confirm Mendels Law of Segregation. The result of the

parental cross ( st+st+ females x stst male ) show the same result parental cross ( st+st+

males stst females) because scarlet eye does not located on sex chromosome.


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F2is cross using Punnett square

e+vg+ e+vg e vg+ e vg

e+vg+ e+e+ vg+vg+ e+e+vg+vg e+e vg+vg+ e+e vg+vg

e+vg e+e+vg+vg e+e+vgvg e+e vg+vg e+e vgvg

e vg+ e+e vg+vg+ e+e vg+vg ee vg+vg+ ee vg+vg

e vg e+e vg+vg e+e vgvg ee vg+vg ee vgvg

F2 Phenotype ratio: 9 wild-type: 3 sepia: 3 vestigial: 1 sepia vestigial

In a dihybrid cross, each of the F1 parents can produce four different gamete types, so there

are 16 (= 4 x 4) possible offspring combinations. Because the two traits show complete

dominance and separate independently of each other (Law of Independent Assortment), the

expected genotypic and phenotypic ratios from an analysis of these 16 possibilities can be

calculated.To test Mendels Law of Independent Assortment, we examined the inheritance of eyes

colour and wing shape by crossing two pure breeding strains of Drosophila melanogaster

that is wild-type and vestigial, sepia. We determined which allele is dominant by setting up

the cross e+e+ vg+vg+ males ee vgvg females as described above. The phenotypes of the

progeny are shown below.

Phenotypes Number of progeny

Males Females Total

Wild type 9 13 22

Vestigial, sepia 0 0 0

e+vg

+ e

+vg evg

+ evg e

+vg

+ e

+vg evg

+ evg


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The further whether eyes colour and wing shape was inherited according to Mendelian laws,

we crossed the F1progeny and examined the phenotypes of the resulting F2 flies.


Males Females Total

wild type 23 34 57

sepia 5 9 14

vestigial 4 7 11

sepia, vestigial 2 4 6

88

To determine the statistical relevance of the data, we performed the Chi square test on our

F2 data.

Class Observed Expected (O-E)2 (O-E)2/

Expected

wild type 57 9/16 88

= 50

(57 - 50)2

= 49

49/50

= 0.98

sepia 14 3/16 88

= 16

(14- 16)2

= 4

4/16

= 0.25

vestigial 11 3/16 88

= 16

(1116)2

= 25

25/16

= 1.56

sepia, vestigial 6 1/16 88

= 6

(96)2

= 9

9/6

= 1.50

Totals 88 88 X = 4.29

The degree of freedom, df = n-1 (n= total number of categories)

= 4-1= 3

From the chi square table,p= 0.1

Since thepvalue is greater than 0.05, it can be concluded that it is not possible to reject the



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The cross between wild type female and vestigial, sepia male

Wild type female vestigial, sepia male

Parent e+e+ vg+vg+ ee vgvg

Gamete

F1 e+e vg+vg

All wild-type

F1 F1 e+e vg+vg e+e vg+vg

F2is cross using Punnett square

e+vg+ e+vg e vg+ e vg

e+vg+ e+e+ vg+vg+ e+e+vg+vg e+e vg+vg+ e+e vg+vg

e+vg e+e+vg+vg e+e+vgvg e+e vg+vg e+e vgvg

e vg+

e+

e vg+

vg+

e+

e vg+

vg ee vg+

vg+

ee vg+

vg

e vg e+e vg+vg e+e vgvg ee vg+vg ee vgvg

F2 Phenotype ratio: 9 wild-type: 3 sepia: 3 vestigial: 1 sepia vestigial

e+vg+ e vg

e+vg+ e+vg evg+ evg e+vg+

e+vg evg+ evg


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In a dihybrid cross, each of the F1 parents can produce four different gamete types, so there

are 16 (= 4 x 4) possible offspring combinations. Because the two traits show complete

dominance and separate independently of each other (Law of Independent Assortment), the

expected genotypic and phenotypic ratios from an analysis of these 16 possibilities can be

calculated.

To test Mendels Law of Independent Assortment, we examined the inheritance of eyes

colour and wing shape by crossing two pure breeding strains of Drosophila melanogaster

that is wild-type and vestigial, sepia. We determined which allele is dominant by setting up

the cross e+e+ vg+vg+ males ee vgvg females as described above. The phenotypes of the

progeny are shown below.


Males Females Total

Wild type 13 21 34

Vestigial, sepia 0 0 0

The further whether eyes colour and wing shape was inherited according to Mendelian laws,

we crossed the F1progeny and examined the phenotypes of the resulting F2 flies.


Males Females Totalwild type 23 33 56

sepia 6 9 15

vestigial 6 10 16

sepia, vestigial 2 4 6

93

To determine the statistical relevance of the data, we performed the Chi square test on our

F2 data.


Expected

wild type 56 9/16 93

= 53

(56 - 53)

= 9

9/53

= 0.17

sepia 15 3/16 93

= 17

(14- 17)2

= 9

9/17

= 0.53

Vestigial 16 3/16 93 (1617)2

1/17


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= 17 = 1 = 0.06

sepia, vestigial 6 1/16 93

= 6

(66)2

= 0

0/6

= 0

Totals 93 93 X2= 0.76

The degree of freedom, df = n-1 (n= total number of categories)

= 4-1

= 3

From the chi square table,p= 0.95

Since thepvalue is greater than 0.05, it can be concluded that it is not possible to reject the


DISCUSSION

The results of the parental cross ( e+e+vg+vg+males ee vgvg females) demonstrate that

the wild type allele (e+e+vg+vg+) is dominant allele for sepia, vestigial alleles (ee vgvg) as

only wild-type progeny were seen in the F1progeny. Calculations from the F2data show that

the ratio of wild-type to sepia to vestigial and to sepia, vestigial flies is 9.5:2.3:1.8:1. Although

this ratio is close to the expected 9:3:3:1 ratio for a dihybrid cross, the Chi square test

performed to determine whether this experimental data differed significantly from the 9:3:3:1

ratio expected for simple dihybrid cross. The results of the Chi square test suggest that the

experimental data do not differ significantly from the expected 9:3:3:1 ratio. Specifically,

there is between 10% and 50% probability that differences seen due to chance. In this case,

the differences seen are probably due to the small sample size scored from the cross.

CONCLUSION

In conclusion, the phenotype of the F1progeny confirmed that the allele for wild type, e+e+

vg+vg+is dominant to the allele for sepia, vestigial, ee vgvg. The ratio of wild type to sepia to

vestigial and to sepia, vestigial of 9.5:2.3:1.8:1 seen in the F2 is very near that of the

expected 9:3:3:1 ratio for a dihybrid cross, and the Chi square test verifies that it is within

statistical limits. Therefore, the results of this experiment confirm Mendels Law of

Independent Assortment. The result of the parental cross between wild type male and

vestigial, sepia female show the same result parental cross between wild type female and

vestigial, sepia male because vestigial wing and sepia eyes does not located on sex

chromosome.


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Sex linked Traits

In sex-linked inheritance, alleles on sex chromosomes are inherited in predictable patterns.

In Drosophilathe locus for eye color is located on the X chromosome. The allele for red eye

color, which is normal in wild flies, is dominant to the mutant allele for white eyes.

In the left hand example, homozygous red eyed females (RR) mate with homozygous white

eyed males (w-). In the offspring, all the daughters are red eyed heterozygotes (Rr)and all

sons are red eyed homozygotes (R-). In the right hand, homozygous white eyed females (rr)

mate with homozygous red eyed males (R-). In the offspring, all the daughters are red eyed

heterozygotes (Rr) and all sons are white eyed homozygotes (r-).

Females have two chromosomes X (with a locus for eye color), they might be homozygous

or heterozygous for either allele. Males, carry only one X chromosome, are always

hemizygous. They carry only the one X chromosome inherited from their mother, and it

determines their eye color.

The first cross is between normal female with red eyes and the mutant male with white eyes.

The first cross is to determine whether the white or red eyes were dominant. The F1

generation all had red eyes. So we can conclude that red eyes are dominant over white.

Then, F1generations are crossed each other to give F2 progenies.


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From the result of the experiment, only male Drosophilashows white eyes. X-linked inherited

diseases occur far more frequently in males because they only have one X chromosome.

Females must receive a copy of the gene from both parents to have such a recessive

disease. However, they will still be carriers if they receive one copy of the gene.

Cross between red eye (wild type) female X white eye male

B+ Dominant allele codes for red eye

B Recessive allele codes for white-eye

Parental: Red eye (female) X white eye (male)

Genotype: XB+ XB+ X XB Y

Gamete:

F1: XB+XB XB+Y XB+XB XB+Y

(all red eyes)

F1X F1: XB+XB X XB+Y

Gamete:

F2: XB+XB+ XB+Y XB+XB XBY

Phenotypic ratio: 2 red eyes female : 1 red eye male : 1 white-eye male

XB+

XB+

XB Y

XB+

XB X

B+ Y


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. Cross between white-eye female x wild type male

B+ Dominant allele codes for red eye type (wild type)

B Recessive allele codes for white-eye

Parental: white-eye (female) X Red eye (male)

Genotype: XB XB X XB+ Y

Gamete:

F1: XB+XB XBY XB+XB XBY

Phenotypic ratio: 2 red eyes type female : 2 white-eyes male

F1X F1: XB+XB X XBY

Gamete:

F2: XB+XB XB+Y XBXB XBY

Phenotypic ratio: 1 red eye female : 1 red eye male : 1 white-eye female : 1 white-eye

male

X X X+ Y

XB+

XB X

B Y


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RESULT

The female with red eyes will be crossed with white-eyed male to produce F1progeny. Then,

this F1 were crossed each other

R is dominant allele for wild type eyes (red eyes)

w is recessive allele for white eyes

The phenotype of F1 progeny is:


Males Females Total

Wild type 8 9 17

White eyes 0 0 0

The phenotype of F2progeny is:


Males Females Total

Wild type 14 33 47

White eyes 16 0 16


Expected

wild type male 14 1/4 63

= 16

(14-16)2

= 4

4/16

= 0.25

Wild type

female

33 2/4 63

= 31

(33-31)2

= 4

4/31

= 0.13

White eyes

male

16 1/4 63

= 16

(16-16)2

= 0

0/16

= 0

White eyes

female

0 0/4 62

= 6

(06)2

= 36

36/6

= 6


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Totals 63 63 X2= 6.38

Interpretation with the chi square value

df = 3

With df = 3, the chi square value of 6.38 is slightly greater than 4.642 (which correspond to

P > 0.20). Therefore P = 0.20 show that expected to occur are 20% of the time. 80% error

occurred. Hypothesis accepted.

The male with red eyes will be crossed with white-eyed female to produce F1progeny. Then,

this F1 were crossed each other

R is dominant allele for wild type eyes (red eyes)

w is recessive allele for white eyes

The phenotype of F1 progeny is:


Males Females Total

Wild type 0 13 13

White eyes 15 0 15

The phenotype of F2progeny is:


Males Females Total

Wild type 14 15 29

White eyes 13 14 27


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Expected

wild type male 14 1/4 56

= 14

(14-14)2

= 0

0/14

= 0

Wild type

female

15 1/4 56

= 14

(15- 14)2

= 1

1/14

= 0.07

White eyes

male

13 1/4 56

= 14

(13-14)2

= 1

1/14

= 0.07

White eyes

female

14 1/4 56

= 14

(14-14)2

= 0

0/14

= 0

Totals 29 29 X = 0.14

Interpretation with the chi square value

df = 3

With df = 10, the chi square value of 0.14 is slightly greater than 0.115 (which correspond to

P > 0.99). Therefore P = 0.99 show that expected to occur are 99% of the time. 1% error

occurred. Hypothesis accepted.

Discussion

From the result of the experiment, in first cross only male Drosophilashows white eyes. X-

linked inherited diseases occur far more frequently in males because they only have one X

chromosome. Females must receive a copy of the gene from both parents to have such a

recessive disease. However, they will still be carriers if they receive one copy of the gene.

From the results that obtain theoretically in cross 1, it is proven according to the Morgan

(1910) that the gene for white eyes in Drosophila is located on the X chromosome and not

the Y chromosome. This can be shown from the cross where wild type female is cross with

white eye male and the result shows that 2:1:1 ratio in F2offspring. The white eye type is

inherited in male because male has only one X chromosome which means that the male

phenotype is not reflective of a dominant or recessive trait, but it reflects only the sex

chromosomes that the male fly carries. This state of male genotype which has only one X

chromosome is termed as hemizygous.

As from cross 2, female flies carry the mutant gene. When it is crossed with normal male,

ratio of 1:1:1:1 can be seen in the F2offsprings. This is because all possible combinations


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