LAB REPORT DROSOPHILA MELANOGASTER

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LAB REPORT DROSOPHILA MELANOGASTER

  1. 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. 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. 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. 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. 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. 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.
  7. 7. Dorsal view Ventral view of abdomen
  8. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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

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