Drosophila Melanogaster Genetics - Lab Report Example

Drosophila Study: Ebony Body and Bar Eye Mutations Abstract: Drosophila Melanogaster genetics was studied using the ebony body color mutation and the bar eye mutation. After creating four crosses, phenotype numbers were counted and it was determined from the first cross data that the bar eye mutation was dominant and the ebony body color mutation was recessive, since ebony body color was not present in the first cross. From the data, the bar eye mutation appeared to be a heterozygous double bar eye mutation, in which there is a double duplication of a chromosome segment. This was due to the fact that no difference in degree of bar-eyed mutation was observed, in contrast to the other bar-eye mutation possibilities. The bar eye mutation was consistent with being sex-linked, occurring on the X chromosome, since different normal/mutant ratios between females and males were observed within some of the crosses; normal/mutant ratios were found to be the same for females and males with the ebony mutation, showing that it is not sex-linked. No intergenic interactions were observed, since the normal/mutant ratios of one mutation did not appear to have any effect on the ratios of the other mutation. This is in line with the two mutations occurring on different chromosomes. Introduction and Review of Literature This genetics experiment was a Drosophila “dew lover” Melanogaster “dark gut” (fruit fly) study, or “Study of the Dark gutted dew lover”. The purpose of this study was to determine the genetics of Drosophila from the phenotype counts. Drosophila is one of the most useful organisms for genetic analysis. The Drosophila melanogaster species is linked to atleast 11 other species of the Drosophila genus, and was one of the first organisms to be used in genetic analysis, so its genome is known well. Drosophila have been used in sex-linkage and genetic recombination studies. The genome of D. melanogaster is studied in order to unravel “the molecular mechanisms underlying development, behavior, aging, and many other processes common to metazoans for which Drosophila is such an excellent model” (Adams 2195). D. melanogaster have been used to study, for instance, deleterious mutation accumulations by insertions of a transposable element, copia. “About half of the major morphological mutations in flies are caused by TE insertions” (Houle 17). Drosophila continue to be a good source for studying mutation genetics. The usefulness of studying D. melanogaster can be seen from the way it has been used to further studies of human disease. D. melanogaster has been used to study human cancer, neurological disease, and other disease conditions. Drosophila genetics has led to “understanding of signaling pathways involved in oncogenesis” (Botas 589), pathways such as Ras, which when activated cooperates with genes in tumor formation. The fast pace of Drosophila research can provide a large amount of data for further mammalian research. In this study, two mutations were studied simultaneously. The original female fly had a bar-eye mutation. The original male fly had an ebony color mutation. Subsequent generations were studied in order to determine the properties of the two mutations, and to discover if there were interactions between the two mutations when they occurred simultaneously. The bar mutation results in a reduced number of facets in the eye, making the eye smaller and bar shaped instead of oval. Bar phenotype results from an X-linked duplication (Pierce 241). The bar phenotype is an incompletely dominant, X-linked trait. This mutation is caused when a segment on the X-chromosome is duplicated. The possible phenotypes are: 1) Wild-type female: B+B+ (1 bar region on each chromosome pair). The flies have normal eye size. 2) Heterozygous bar female: B+B. The eyes are a little smaller. There is 1 extra bar region on one chromosome of the chromosome pair. 3) Homozygous bar female is BB. The eyes are much smaller. This is the same as the hemizygous male, which is denoted BY or B/Y. There is 1 extra bar region on both chromosomes of the chromosome pair. 4) Heterozygous double bar female is B+B0. The eyes are very much smaller. There are two extra bar regions on one chromosome (3 bar regions altogether) and just 1 bar region on the other chromosome of the chromosome pair. The capitalized B means that the mutation is dominant. A “+” sign means that no mutation is present on that chromosome. A “0” sign after the B means that there is a double mutation on that chromosome. So a B+B fly has one chromosome (the B+) that has no mutation, and one chromosome (the B) that has a mutation. These are called the individual’s two homologs. During meiosis each pair of homologs separate, and 1 of each pair is transmitted to a given progeny. The dominant marker Bar (B) is carried by the X-chromosome balancer FM7a (Greenspan 13). Drosophila has four chromosome pairs, marked as: X, 2, 3, 4. The B mutation is found at map position 1-57.0 (Graf, van Shaik & Wurgler 26), which is on the first chromosome pair, the X chromosome, and 57 units from the left end of that chromosome. There were approximately equal numbers of females and males in the phenotype counts for ebony body color mutations, showing that the ebony body color mutation is not sex-linked. This is confirmed by the literature, where ebony body color is found to be on the 3rd chromosome (an autosome and not an X or Y sex chromosome): map position 3-70.7 (Graf, van Shaik & Wurgler 26). Furthermore, since the ebony mutation occurs on a different chromosome from the bar mutation, the two mutations should show no signs of there being intergenic interactions (epistatis). Materials and Methods This experiment used a pure breeding mutant P1 made of ebony body mutation males and pure breeding second mutant P2 made of bar eye mutation females. These two groups were crossed, resulting in P1 x P2 = F1generation for subsequent crosses and to generate data for analyzing allelic interactions. Three test crosses were formed: the F1 generation was crossed again with P1 for 100 flies; F1 was crossed with P2 for 100 flies; and finally, F1 was crossed with F1 to form 300 flies. After forming these test crosses, the number of each phenotype, or observable property, for each cross was counted. From the data, we were able to determine nearly all the allelic interactions: dominant/recessive, partial dominance, codominance, etc. and/or intergenic interactions (epistatis) and/or sex linked genes. The methods of conducting the crosses and all of the laboratory procedures were performed according to Flagg’s manual. Results The phenotype numbers that were counted for each of the three crosses are illustrated in table 1. The normal/mutant ratios appear to be about the same for females and males, except for in the bar mutations of P2 x F1 and F1 x F1 crosses. Table 1- Phenotype counts for crosses. P1 x F1 P2 x F1 F1 x F1 Normal eyed males 38 25 112 Mutant eyed males 12 25 38 Normal eyed females 37 38 56 Mutant eyed females 13 12 94 Normal color males 25 50 112 Mutant ebony males 25 0 38 Normal color females 24 50 113 Mutant ebony females 26 0 37 Discussion The original male fly (P1) had an ebony body mutation: this is symbolized by ee. The lower case symbol “e” means that this ebony body mutation is recessive. The original female fly (P2) had a normal body color, which is symbolized by e+e+. When they were crossed, this produced the F1 generation, which were all normal body colored. This is what would be expected, since when this first generation is crossed, the resulting cross (F1) gets one gene from the male and one gene from the female. The only combination is ee+. Since ebony body color is a recessive gene, this ee+ combination appears as the physically normal body color phenotype. When the F1 generation is crossed with itself, producing the 300 F1 X F1, approximately ¼ were ebony bodied, and approximately ¾ were normal body color. This is because when we cross ee+ with ee+, the combinations (taking just one gene from each ee+) are: ee, ee+, e+e, e+e+. Since this is a recessive gene, only the ee genotype results in an ebony color phenotype. For the 100 flies in the P1 X F1 cross, this is crossing ee X ee+. The combinations are: ee, ee+, ee, ee+, so that there are approximately ½ ebony colored mutants and ½ normal body colored flies. And again the males and females should be approximately distributed in these two groups, showing that the ebony mutation is not sex-linked. Finally, for the 100 flies in the P2 X F1 cross, this is crossing e+e+ X ee+, so that the combinations are: e+e, e+e+, e+e, e+e+, which means all the flies in this group are normal. This holds 100% because there is no other possibility of inheritance. To analyze the crosses in terms of the bar mutation, there are three possibilities to consider. We know that the original male fly (P1) had normal eyes, so he is a B+Y. We only know that the original female (P2) had a bar mutation, but as we have seen , there are three different ways the mutation can be expressed: heterozygous, homozygous, and heterozygous double. Our data showed that there were some normal eyed flies in the first generation, F1. This ruled out a P2 female being BB, since a B+Y x BB results in BY, BY, B+B, B+B; all the progeny are mutants. In the first cross, there were no males that had even more seriously bar-eyed males than the original mutant female. There were also no females in the first cross that had bar eyes that looked any different from the original mutant female, P2. If this original mutant female P2 was a B+B, then we would have a B+Y x B+B cross, resulting in the combinations B+B+, B+B, B+Y, BY. The B+B+ and B+Y are normal. The BY is the bar-eyed hemizygous male, with eyes that are much smaller, analogous to the BB female. These eyes would appear smaller than the original B+B female. Since our data didn’t show a difference in the size of the eye mutations across these two generations, we can rule out this possibility. This leaves the only other possibility, that the original female P2 was a B+B0, or heterozygous double bar female. This would result in a B+Y x B+B0 cross, with combinations: B+B+, B+B0, B+Y, B0Y for the F1 generation. The B+B+ and B+Y are normal females and normal males, respectively. The B0Y is a mutant male, and the B+B0 is the same as the original female P2. This first cross therefore has approximately ½ normal males, ½ mutant males, and the same for the females. When we cross P1 x F1, we are crossing P1, which is B+Y male, with the two F1 female possibilities: B+B+ and B+B0. For the B+Y x B+B+ cross, with combinations B+B+, B+B+, B+Y, B+Y, all flies are normal; crossing B+Y x B+B0 results in the same cross as F1 itself. This results in ¾ normal males, ¼ mutant males, and the same ratios for females. The P2 x F1 cross means we are crossing P2, which is B+B0 female, with the two F1 male possibilities: B+Y and B0Y. The B+B0 x B+Y again gives the F1 results. The B+B0 x B0Y cross has combinations B+B0, B0B0, B+Y, B0Y. These are all mutant except the B+Y. In this case we have approximately ½ normal males, ½ mutant males, but ¾ mutant females and ¼ normal females. The F1 x F1 cross includes four combinations: a B+Y x B+B+ cross, which is the same as the first part of the P1 x F1, all normal flies; B0Y x B+B+; B+Y x B+B0, with F1 results, approximately ½ males normal, ½ males mutant and the same ratios for females; and B0Y x B+B0 cross, which is the same as the last part of the P2 x F1 cross. The B0Y x B+B+ has combinations B+B0, B+B0, B+Y, B+Y, or all normal males and all mutant females. Adding up these combinations shows that approximately ¾ of the males were normal, ¼ males mutant; but for females 3/8 were normal, 5/8 mutant. In the 300 F1 x F1 flies, there were 150 males and 150 females. Approximately ¾ the males should be normal eyed, or 112.5. The rest, 37.5, should be mutant eyed. Approximately 3/8 of the females would be normal eyed, or 56.25. The rest, 93.75, would be mutant eyed. We would also expect approximately 75 ebony bodied mutants, and 225 normal bodied flies, with males and females approximately equally distributed in these two groups. In the 100 P1 x F1 flies, for 50 males and 50 females, there should be approximately 37.5 normal eyed males, 12.5 mutant eyed males, and the same numbers for females. We would also expect approximately equal numbers of ebony mutations and normal body colored flies. In the 100 P2 x F1 flies, there should be approximately 25 normal eyed males and 25 mutant eyed males. There are approximately 37.5 mutant eyed females and 12.5 normal eyed females. This cross should contain only normal body colored flies. There does not appear to be any intergenic interactions between the two mutations. This is because the data does not appear to show a correlation in the number of phenotype counts of one mutation when the other mutation’s phenotype counts increase/decrease. In conclusion, our results have shown that the bar eye mutation is dominant, and the ebony body color mutation is recessive. Our results show that the bar eye mutation was a heterozygous double bar eye mutation. The bar mutation is sex-linked and carried on the X chromosome. The fact that this mutation is sex-linked can be seen in the way the phenotype count ratios of normal flies to mutant flies were different for females than for males. This is in contrast to the ebony mutation, which is not sex-linked: in this case, the phenotype counts are roughly the same for both males and females. There were no intergenic interactions since the two mutations did not affect each other, being carried on different chromosomes. References Adams, M.D., S.E. Celnike, R.A. Holt et al (2000). “The genome sequence of Drosophila melanogaster.” Science 287 (5461): 2185–95. Botas, Juan. “Drosophila researchers focus on human disease.” Nature Genetics, v39, n5, p589, May 2007. Flagg, Raymond. Carolina Drosophila Lab Manual. Graf, Ulrich, Nancy van Shaik & Friedrich E. Wurgler. Drosophila Genetics: A Practical Course. New York: Springer-Verlag, 1992. Greenspan, Ralph J. Fly Pushing: The Theory and Practice of Drosophila Genetics. Plainview, NY: Cold Spring Harbor Laboratory Press, 1997. Houle, David and Sergey V. Nuzhdin. “Mutation accumulation and the effect of copia insertations in Drosophila melanogaster.” Genetical Research. Feb 2004. Vol. 83, Iss. 1; p. 7 Pierce, Benjamin. Transmission and Population Genetics: A Short Course. W.H.Freeman, 2006. Read More