Linkage Mapping in Drosophila Melanogaster Genetics: Fall 2012 Joshua Hanau
Introduction: An experiment was performed in order to determine the presence and degree of gene linkage in Drosophila Melanogaster. Gene linkage describes whether or not two or more genes are located on the same chromosome in a eukaryotic individual. The degree of gene linkage can describe the relative distance between two linked genes on a single chromosome. Drosophila Melanogaster, which is also known by the common name Fruit Fly, is an ideal organism to use for genetic studies. Drosophila proliferates rapidly, are small, are cheap to maintain, and their phenotypes are easily observable. Drosophila males and females display numerous gender dimorphisms, which enable gender segregation, and ultimately, controlled mating. Females have a striped black posterior, on the dorsal side of their abdomen, while males have a solid black posterior, on the dorsal side of their abdomen. Females have no external genitalia, on their ventral posterior end, while males do have external genitalia, on their ventral posterior end. Females have a larger elongated abdomen, when compared to males. Males have sex combs on their anterior appendages, while females do not. These dimorphisms (Figure 1) enable females to be separated from males before they are ready to mate, thus enabling controlled mating.
Figure 1: Ventral Gender Dimorphisms in Drosophila [I] A three- factor cross was the method used to determine whether or not three drosophila genes were linked, and if so, to what degree. The three genes that were analyzed displayed different phenotypes when they were in mutant and wild type forms. The first of these genes corresponded to the fly s eye- color. While wild type flies have dull red colored eyes, the mutants displayed brighter orange colored eyes. The second gene analyzed corresponded to the bristles located on the dorsal side of the fly s thorax. Wild type flies had long straight bristles, while the mutants had short and curled bristled. The third gene analyzed corresponded to the body color being displayed on the dorsal side of the fly s abdominal section. The wild type flies
displayed a regular pattern of alternating black and tan sections, which differed in males and females (Figure 2). The mutants, displayed the same basic pattern, only the black sections were faded, and less pronounced than they were in the wild type strain. Figure 2: Body Color Pattern in Wild Type Drosophila [II] After performing the three- factor cross, the ratio of the F2 progeny (Filial generation #2) phenotypes could be used to determine whether or not the three genes were linked, and if they were, to determine the degree of the linkage. This data could then be compared to a map of known drosophila genes, and the identity of the genes could potentially be inferred from the literature.
Methods: At the start of the experiment, two strains of flies were crossed. The first strain was a true- breeding wild type stock, which proves that it was homozygous for all three wild type alleles, or +/+ ; +/+ ; +/+, where + indicates the wild type allele for any given gene. The second strain was a true breeding triple mutant, which proves it was homozygous for all three mutant alleles, or e/e ; br/br ; bc/bc, where e indicates the unknown mutant eye color allele, br indicates the unknown mutant bristle allele, and bc indicates the unknown mutant body color allele. To perform a cross, virgin females of one strain must be placed in a vial with males of the other strain. Virgin females can be identified by their pale coloring, their folded wings, their enlarged abdomens, and by the presence of a meconium (Figure 3). Not all four of these signs are required, but the presence of a meconium is the most important sign of recent eclosion (emergence from the pupa casing), and the best guarantee of a female fly being a virgin, since drosophila females do not mate for the first few hours (around 8 on average) after eclosion. Using virgin females in all crosses is crucial. Female drosophila can only mate with one male, and they then store that male s sperm for the rest of their adult life cycle. Therefore, in order to control mating, females must be isolated soon after they eclose from their pupa casing, while they still exhibit the features listed above, and have a guaranteed virgin status.
Figure 3: Virgin Female Drosophila Image [III] The general pattern of a cross was as follows (Figure 4). Once 4-5 virgin females were isolated, they were placed in a vial with 8-10 males from the other cross strain. The flies would mate, and the females began laying eggs shortly thereafter. After the eggs were laid, the larva, or 1 st instars, would emerge within a day or so. The larva would continue to grow, and pass through the 2 nd and 3 rd instar phases. During the 3 rd instar phase, the larva would climb up the wall of the vial and form a pupa case. Around four days later, and around 10 days after the cross was started, a new generation of adult flies would emerge. The parental generation was
removed from the vial by this point, to ensure that the two generations were kept separate. Figure 4: Drosophila Life Cycle [IV] Assuming that none of the three genes are sex linked, the progeny of this initial cross would all display wild type phenotypes, since the wild type is presumably dominant, and all the offspring would be heterozygous for all three genes (+/e ; +/br ; +/bc). However, if one or more genes are sex linked, then there is a possibility that the male offspring, who are hemizygous for x linked genes, would display the mutant phenotype. To test for sex linkage, reciprocal crosses were
performed. This means that there were two crosses performed between the two strains, with the genders of each strain alternated in the two crosses. One cross was between wild type males and female mutants, while the other was between male mutants and wild type females. If one or more genes were sex linked, then the male F1 generation (Filial generation #1) offspring of the reciprocal crosses would always have the same phenotype as their mothers for that one or more genes, instead of simply having the wild type phenotype that would be expected for all F1 flies. The second cross was a test cross. A test cross is when heterozygous individuals (F1 virgin females) are crossed with double mutant individuals (Parental mutant males). This cross is performed in order to determine whether or not the genes assort independently. If the genes assorted independently, then all 8 potential phenotype combinations (2 phenotypes for 3 genes. 2 3 = 8) would ideally show up with the same frequency, 12.5%, in the F2 generation. If the parental phenotypes are displayed the majority of the time, or far greater than the expected 12.5% (i.e. no need for chi squared analysis), then it can be assumed that the three genes are linked. Since linked genes are on the same chromosome, it makes sense that the genes are inherited together the majority of the time, and it explains the lack of independent assortment. Additionally, any phenotypes that are not parental, in the linkage scenario, must be the result of crossing over during meiosis, which is also known as meiotic recombination. Thomas Hunt Morgan s research showed that the frequency of recombination between two genes is directly proportional to the physical distance between them. Thus, using recombinant frequencies, an RF map of
the chromosome, which is similar to the physical map (but not identical), can be constructed using recombinant frequencies. The progression of crosses performed is shown below (Figure 5). Figure 5: Progression of Crosses Performed P: +/+ ; +/+ ; +/+ x e/e ; br/br ; bc/bc + reciprocal cross F1: e/e ; br/br; bc/bc x +/e ; +/br ; +/bc F2: Eight possible phenotypes (See results section) Results: The results of the reciprocal crosses suggest that all three genes are sex linked, since the male offspring of the cross with the mutant phenotype females were mutant for all three genes. By extension, all three genes are linked, since there is only one sex chromosome. Therefore, the ratio of F2 phenotypes were expected to, and in fact did, deviate from the 1:1:1:1:1:1:1:1 ratio that would be expected in a case of independent assortment. The number and percentage of each phenotype is displayed below (Table 1). In addition to there being eight possible phenotypes, there are eight possible genotypes that can result from this type of test cross. Each genotype corresponds to a phenotype, where the wild type allele (+) is always
expressed over the mutant type. Eight testcrosses were performed, resulting in a total of 515 progeny. Table 1: F2 Progeny Ratios Genotype/Phenotype Number of F2 Progeny % of Total Progeny 1: +/e ; +/br ; +/bc 181 35.1% 2: e/e ; br/br ; bc/bc 186 36.1% 3: +/e ; +/br ; bc/bc 12 2.3% 4: e/e ; br/br ; +/bc 21 4.1% 5: +/e ; br/br ; +/bc 48 9.3% 6: e/e ; +/br ; bc/bc 63 12.2% 7: e/e ; +/br ; +/bc 2 0.4% 8: +/e ; br/br ; bc/bc 2 0.4% Using these ratios a recombination map of the X chromosome (because they are sex linked) can be constructed. For example, in the parental phenotypes (# s 1+2) the gene for eye color and bristles are the same type of allele (both wild, or both mutant). Thus, any offspring where they display different types of alleles on the maternal homologue must be a case where recombination took place between the two genes. Thus, recombination took place between these two genes in progeny numbers 5,6,7, and 8. By adding up the percentages of recombinants, the total recombination frequency between the two genes can be calculated to be 22.3%. Therefore, the genes for eye color and bristles should be located around 22.3 map units (map units) apart on an RF map. The distance between the other sets of genes
can be calculated in the same way (Table 2). The one small point to note is that in between br and bc, the percentage of double recombinants needs to be counted twice, since two crossover events took place between these two genes in those scenarios. Table 2: Recombinant Frequencies Alleles of Interest Calculation Recombinant Frequency e and br 9.3 + 12.2 + 0.4 + 0.4 22.3% (m.u.) e and bc 2.3 + 4.1 + 0.4 + 0.4 7.2% (m.u.) br and bc 2.3 + 4.1 + 9.3 + 12.2 + 29.5% (m.u.) 2(0.4) + 2(0.4) Using the tabulated data, an RF map for these three genes can be constructed (Figure 6). A useful trick for constructing a recombinant map is to look for the flipped allele in the least common phenotype, which is always the result of the double crossover. This flipped allele in the double recombinant will always be located between the other two genes [V]. Since the allele for eye color differs from the two parental strains in the double recombinant, the gene for eye color must be located between the other two genes on the recombinant map. Figure 6: RF Map of Three Genes of Interest
Conclusions: By comparing these results to a map of known genes on the drosophila X chromosome (Figure 7), the identity of the three genes can be inferred. The three genes must correspond to eye color, bristle shape, and body color. The distances between the three known genes must also be similar to the distances calculated between the three unknown genes in this lab report. Figure 7: RF Map of Drosophila X Chromosome [VI] This small segment of the Drosophila X chromosome contains the most likely candidates for our three unknown genes. The calculated distance between our bristle and body color mutations is 29.5 map units, while in this map, the distance between forked bristles and tan bodies is 29.2 maps units. Since both of these mutations sound like they could correspond with our mutant phenotypes, it is reasonable to conclude that the two genes that contain these two mutant alleles are the same as our first unknown genes. Our eye color mutation is located 7.2 maps
units away from our body color mutation, and 22.3 map units away from our bristle mutation. On this map of known genes, the mutation for Vermillion eyes is located 5.5 maps units away from the tan body mutation, and 23.7 map units away from the forked bristles mutation. These results are similar enough to conclude that our eye color mutation might be caused by a mutant vermillion allele. Thus, it is reasonable to conclude that our three unknown genes, correspond with the genes whose mutated forms result in the tan body, Forked bristles, and vermillion eyes mutations. This inference was much easier to make due to the use of a reciprocal cross. The reciprocal crosses showed that all three genes were sex linked, which led to the conclusion that all three genes were located on the X chromosome. This narrowed down the search for the identity of the unknown alleles considerably. Although the results were pretty close (assuming these are the correct genes), there was some error, which has to be explained. The most obvious cause is human error. It was not always easy to distinguish the mutant body color from the wild type. This may have resulted in incorrect phenotype tallying, and therefore, faulty recombinant frequencies. The same is true for eye color, which was not always 100% discernable. The bristle mutation was the only one that was completely obvious in all cases.
Another source of error may have been due to a failed separation of generations. In one case, flies were scored from a vial 22 days after the cross was set up. It is possible that some of the flies that were counted belonged to the F3 generation, and may have interfered with the results. To verify the identity of the genes a complementation test could be performed. Flies that are known to be heterozygous for the three genes mentioned above could be crossed with the unknown mutant parental strain. If the result were a wild type phenotype (or phenotypes), then the hypothesis of the genes identity of that gene (or genes) must be incorrect, since complementation suggests the mutations are on different genes. If mutant phenotype flies were prevalent in 50% of the offspring, for one or all three genes, then the identity of that mutant allele, or alleles, will be confirmed. Citations: [I]: http://flymove.uni- muenster.de/genetics/flies/malefemale/malefemalepage.html?http&&&flymove.u ni- muenster.de/genetics/flies/malefemale/malefemaletxt.html [II]: http://www.ableweb.org/volumes/vol- 10/15- pollock.pdf [III]: See [I] [IV]: See [II]
[V]: Griffiths, Anthony; Wessler, Susan; Carroll, Sean; Doebley, John (2012). Introduction to Genetic Analysis- 10 th Edition. New York City: W.H. Freeman and Company. Pages 134-135. [VI]: http://www.blacksage.com/apbiology/handouts/drosophilamap.htm