1. 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
LECTURER’S NAME: EN. HAMZAH B ABDUL AZIZ
2. 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.
3. 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
not linked 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 Drosophila 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 the
same 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
4. Day 1: Eggs hatch
Day 2: First instar (one day in length)
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.
5. Sex difference
Several criteria may be used to distinguish male and female Drosophila melanogaster
• Body size – female is generally larger than male.
• Abdomen shape – the 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.
6. • 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.
8. 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 these
exercises 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
Mutants’ traits can be assumed as recessive to the wild type.
9. 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 flies
will 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. Don’t 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.
10. 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.
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
11. 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.
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.
12. 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.
13. 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
st+ st
Gamete
14. F1 st+st
All wild-type
F1 × F1 st+st × st+st
Gamete st+ st st+ st
F2 st+st+ st+st st+st stst
Ratio 3: 1
wild-type : scarlet-eye
To test Mendel’s Law of Segregation, we examined the inheritance of eye color by
crossing two pure breeding strains of Drosophila melanogaster that is wild type and
scarlet eyes. We determined which allele is dominant by setting up the cross st+st+
males × stst females as described above. The phenotypes of the progeny are shown
below.
Phenotypes Number of progeny
Males Females Total
Wild type 13 23 36
Scarlet eyes 0 0 0
15. The further whether eye color was inherited according to Mendelian laws, we crossed
the F1 progeny and examined the phenotypes of the resulting F2 flies.
Phenotypes Number of progeny
Males Females Total
Wild type 27 38 65
Scarlet eyes 7 12 19
84
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 65 63 ( 65-63)2 4/ 63
=4 =0.06
Scarlet eyes 19 21 ( 19-21)2 4/21
=4 =0.19
Totals 84 84 X2 = 0.25
The degree of freedom, df = n-1 (n= total number of categories)
= 2-1
=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
st+ st
Gamete
16. F1 st+st
All wild-type
F1 × F1 st+st × st+st
Gamete st+ st st+ st
F2 st+st+ st+st st+st stst
Ratio 3: 1
wild-type : scarlet-eyed
To test Mendel’s Law of Segregation, we examined the inheritance of eye color by
crossing two pure breeding strains of Drosophila melanogaster that is wild type and
scarlet eyes. We determined which allele is dominant by setting up the cross st+st+
males × stst females as described above. The phenotypes of the progeny are shown
below.
17. Phenotypes Number of progeny
Males Females Total
Wild type 13 23 36
Scarlet eyes 0 0 0
The further whether eye color was inherited according to Mendelian laws, we crossed the F1
progeny and examined the phenotypes of the resulting F2 flies.
Phenotypes Number of progeny
Males Females Total
Wild type 21 32 53
Scarlet eyes 6 9 15
68
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
2
Wild type 53 51 (53 – 51) 4/51
=4 = 0.08
Scarlet eyes 15 17 (15 – 17)2 4/17
=4 = 0.24
Totals 68 68 X2 = 0.32
The degree of freedom, df = n-1 (n= total number of categories)
= 2-1
=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.
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
18. 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 Mendel’s 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.
Dihybrid Crosses
Dihybrid crosses involve manipulation and analysis of two traits controlled by pairs of alleles
at different loci.
In this experiment, we cross wild type with vestigial wing, sepia eyes
e is recessive allele for sepia eyes
e+ is dominant allele for wild-type eyes
vg is recessive allele for vestigial wing shape
vg+ is dominant allele for wild-type wing shape
19. Vestigial, sepia female Vestigial sepia male Vestigial male vestigial female
The cross between wild type male and vestigial, sepia female
Wild type male vestigial, sepia female
Parent e+e+ vg+vg+ × ee vgvg
Gamete
e+ e vg
vg+
F1 e+e vg+vg
All wild-type
F1 × F1 e+e vg+vg × e+e vg+vg
e+vg+ e+vg evg+ ev e+vg+ e+vg evg+ ev
g g
F2 is 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
20. expected genotypic and phenotypic ratios from an analysis of these 16 possibilities can be
calculated.
To test Mendel’s 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
The further whether eyes colour and wing shape was inherited according to Mendelian laws,
we crossed the F1 progeny and examined the phenotypes of the resulting F2 flies.
Phenotypes Number of progeny
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 (57 - 50)2 49/50
= 50 = 49 = 0.98
sepia 14 3/16 × 88 (14- 16)2 4/16
= 16 =4 = 0.25
vestigial 11 3/16 × 88 (11 – 16)2 25/16
= 16 = 25 = 1.56
sepia, vestigial 6 1/16 × 88 (9 – 6)2 9/6
=6 =9 = 1.50
Totals 88 88 X2 = 4.29
21. The degree of freedom, df = n-1 (n= total number of categories)
= 4-1
=3
From the chi square table, p = 0.1
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 cross between wild type female and vestigial, sepia male
Wild type female vestigial, sepia male
Parent e+e+ vg+vg+ × ee vgvg
Gamete
e+ e vg
vg+
F1 e+e vg+vg
All wild-type
F1 × F1 e+e vg+vg × e+e vg+vg
e+vg+ e+vg evg+ ev e+vg+ e+vg evg+ ev
g g
F2 is 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
22. 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 Mendel’s 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 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 F1 progeny and examined the phenotypes of the resulting F2 flies.
Phenotypes Number of progeny
Males Females Total
wild 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.
Class Observed Expected (O-E)2 (O-E)2/ Expected
wild type 56 9/16 × 93 (56 - 53)2 9/53
= 53 =9 = 0.17
sepia 15 3/16 × 93 (14- 17)2 9/17
23. = 17 =9 = 0.53
Vestigial 16 3/16 × 93 (16 – 17)2 1/17
= 17 =1 = 0.06
sepia, vestigial 6 1/16 × 93 (6 – 6)2 0/6
=6 =0 =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 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 ( 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 F1 progeny. Calculations from the F2 data 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 F1 progeny 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 Mendel’s 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
24. vestigial, sepia male because vestigial wing and sepia eyes does not located on sex
chromosome.
Sex linked Traits
In sex-linked inheritance, alleles on sex chromosomes are inherited in predictable patterns.
In Drosophila the 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
25. generation all had red eyes. So we can conclude that red eyes are dominant over white.
Then, F1 generations are crossed each other to give F2 progenies.
From the result of the experiment, only male Drosophila shows 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
XB XB XB Y
Gamete: + +
F1: XB+XB XB+Y XB+XB XB+Y
(all red eyes)
F1 X F1: XB+XB X XB+Y
XB XB 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
26. . 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: XB XB XB Y
+
F1: XB+XB XBY XB+XB XBY
Phenotypic ratio: 2 red eyes type female : 2 white-eyes male
F1 X F1: XB+XB X XBY
XB XB XB Y
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
27. RESULT
The female with red eyes will be crossed with white-eyed male to produce F1 progeny. 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:
Phenotypes Number of progeny
Males Females Total
Wild type 8 9 17
White eyes 0 0 0
The phenotype of F2 progeny is:
Phenotypes Number of progeny
Males Females Total
Wild type 14 33 47
White eyes 16 0 16
Class Observed Expected (O-E)2 (O-E)2/ Expected
wild type male 14 1/4 × 63 (14-16)2 4/16
= 16 =4 = 0.25
Wild type 33 2/4 × 63 (33-31)2 4/31
female = 31 =4 = 0.13
White eyes 16 1/4 × 63 (16-16)2 0/16
male = 16 =0 =0
White eyes 0 0/4 × 62 (0 – 6)2 36/6
female =6 = 36 =6
Totals 63 63 X2 = 6.38
Interpretation with the chi square value
28. 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 F1 progeny. 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:
Phenotypes Number of progeny
Males Females Total
Wild type 0 13 13
White eyes 15 0 15
The phenotype of F2 progeny is:
Phenotypes Number of progeny
Males Females Total
Wild type 14 15 29
White eyes 13 14 27
Class Observed Expected (O-E)2 (O-E)2/ Expected
wild type male 14 1/4 × 56 (14-14)2 0/14
= 14 =0 =0
Wild type 15 1/4 × 56 (15- 14)2 1/14
female = 14 =1 = 0.07
White eyes 13 1/4 × 56 (13-14)2 1/14
male = 14 =1 = 0.07
White eyes 14 1/4 × 56 (14-14)2 0/14
29. female = 14 =0 =0
Totals 29 29 X2 = 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 Drosophila shows 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 F 2 offspring. 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 F2 offsprings. This is because all possible combinations
of white eye and sex are possible and the white eye trait can be carried over to female
when F1 females are crossed with white eye male.
Conclusion
From this experiment, we learn on how to conduct a genetic experiment which spans of
generation. We also learn on how to design genetic crosses to illustrate segregation,
independent assortment and sex-linkage. There are four stages of Drosophila melanogaster
life cycles that are egg, larva, pupa, and adult. From study of its life cycle, we are able to
perform this experiment. We can differentiate the male and female of Drosophila
melanogaster based on several characteristic such as size of adult, shape of abdomen,
30. markings on abdomen, appearance of sex comb external genitilia on abdomen and sex
organ during larval stage. This making easier for us to differentiate them especially in the
experiment about sex-linkage. What is more important is we are able to recognize wild-type
flies and those with classic mutations. To test the result that we obtained, we use the Chi
square statistic to test hypotheses concerning expected and observed ratios. If the p value is
less than 0.005, so the hypothesis can be rejected.
References
Paul Arnold (2009). Human Genetics and the Fruit Fly Drosophila Melanogaster.
Retrieved March 29, 2010, from http://www.biol.org/DrosPics.htm#Misc
Christin E. Arnini (1996). Using Drosophila to Teach Genetics. Retrieved March 29,
2010,http://www.google.com.my/search?
hlen&qdrosophila+melanogaster+phenotypes&revid.
Basic Genetics: Thomas Hunt Morgan and Sex-linked Traits. Retrieved April 6, 2010
from http://library.thinkquest.org/20465/sexlinked.html
Miko, I. (2008) Thomas Hunt Morgan and sex linkage. Nature Education 1(1). Retrieved April
6, 2010 from http://www.nature.com/scitable/topicpage/Thomas-Hunt-Morgan-
and-Sex-Linkage-452#TB_inline?height=300&width=400&inlineId=trOutLine
Retrieve on 8 April 2010 at
http://www.mun.ca/biology/dinnes/B2250/DrosophilaGenetics.PDF
Retrieved on 8 April 2010 at http://www.dreamessays.com/customessays/Science
%20Research%20Papers/11452.htm
