Drosophila Genetics by Michael Socolich May, 2003

Drosophila Genetics by Michael Socolich May, 2003 I. General Information and Fly Husbandry II. Nomenclature III. Genetic Tools Available to the Fly Geneticists IV. Example Crosses V. P-element Transformation VI. References and Figures Preface This review is intended to be a primer of Drosophila genetics for the beginning graduate or post-doctoral student, and any interested researcher. Those interested in a more in depth discussion on the topic are referred to the sources in the reference section. I. General Information and Fly Husbandry The fruit fly Drosophila melanogaster has 3 pairs of autosomal chromosomes and an X and Y chromosome. Each autosome has two arms that are simply referred to as left (L) and right (R). Each chromosome arm is numbered as follows: X (1-20), 2L (21-40), 2R (41-60), 3L (61-80), 3R (81-100), and chromosome 4 (101-102). Each chromosome arm is also numbered by recombination units, thus allowing one to know the expected recombination frequency between two genes located on the same chromosome arm. The chromosomal locations of individual genes are identified either by numerical location or by recombination units. Sex determination in Drosophila is based on the ratio of X chromosome to autosomal sets. Therefore, females which have two X chromosomes have a ratio of 1 whereas; males which have only 1 X chromosome have a ratio of 0.5. Fortunately, for the fly geneticists recombination does not occur in male flies. This quality can be exploited in certain genetic crosses as will be explained later. The life cycle of Drosophila is quite simple: eggs are layed which develop into larvae, which develop into pupae, which develop into adult flies. Flies do not "hatch" from pupal cases they "eclose". The above life cycle is dependent on the temperature at which the flies are grown. The generation time for flies grown at 25 o C is 10 days, at room temperature (21 o -22 o C) it is 12-13 days, and at 18 o C the generation time is 19 days. Female fruit flies can mate with more than one male and store sperm from multiple matings. This forces the geneticists to use virgins when conducting a genetic cross. Using non-virgins results in progeny that do not have the expected genotypes or phenotypes predicted by simple Mendellian genetics. When collecting virgins it is useful to remember that newly eclosed virgins will remain virgins for 6 hours at 25 o C, 12 hours at 21-22 o C, and 18 hours at 18 o C. An important source of information for the fly researcher is FlyBase (1). FlyBase is a website that allows one to browse for information regarding individual genes, 1

chromosomal rearrangements, fly anatomy, fly stocks, cytologic maps, and a host of other information useful to the fly researcher. Proper care and attention are necessary to ensure that one's fly stocks remain healthy. The following points need to be remembered and attended to at all time to maintain healthy stocks: 1. Ensure that fly food remains moist at all times. Dry food will inhibit larval growth and result in few flies eclosing. 2. Do not expose fly stocks to extremes in temperature for extended periods of time, that is less than 18 o C or greater than 25 o C. 3. Always transfer stocks every 3 weeks to prevent and minimize the chance of mite infestation. For the researcher working with fruit flies the pest one must always be vigilant against is mites! Mites come in many different varieties: mites that eat eggs, mites that eat fly food, and mites that feed off of adult flies (see the attached figures of various mites). Mites can decimate fly stocks! The threat of mites can be minimized by transferring stocks frequently, inspecting vials and bottles regularly, and quarantining any stock(s) that are received from another lab. Proper quarantining of a new stock involves the following procedure: 1. Transfer flies to a fresh vial upon receipt of new stock. Keep original vial. 2. Continue to transfer flies everyday for 5 days. Throw out vials from days 1 to 4. On the sixth day dump flies and keep the day 5 vial. 3. When flies eclose from day 5 vial transfer to new vial and repeat steps 1 and 2 again. 4. Inspect the original vile for the presence of any mites. Also, inspect the day 5 vial for mites. If after two generations no mites are found in any of the above vials it is safe to introduce this stock into one's flyroom. However, if mites are found then the above procedure must be continued until two successive generations of mite free vials are generated! II. Nomenclature The nomenclature used in Drosophila genetics is fairly straightforward yet to the uninitiated can be daunting. It is important that one follows the standard rules of nomenclature to properly and clearly describe the complete genotype of a fly stock. Chromosomes are written in order, as follows, with a semi-colon separating each chromosome. X/Y; 2; 3; 4 Genotypes are listed only when a mutation is present and are italicized. Recessive mutations are written in lower case (e.g. w for white gene), dominant mutations are capitalized (e.g. Roi for rough eye). If a particular allele is present that allele is superscripted (e.g. norpa P41 ). Anything not listed is assumed to be wild type. 2

If more than one mutation is present on a chromosome, the mutations are listed from left to right corresponding to the left and right arms of each chromosome. arr1 cn bw If a mutation is homozygous then the mutation is written only once as follows: cn bw if heterozygous then written as follows: cn bw or cn bw/+ + or cn bw/+ + + Deficiencies: Df(2L)VA= Deficiency of the left (L) arm of chromosome 2 that includes the gene Venea abnormeis (VA). Transpositions: Tp(1;3)HF308= Transposition involving the X and 3rd chromosome. Inversions: In(2LR)SMC8= Inversion of the left (L) and right (R) arm of the 2nd chromosome. Translocations: T(1;3)Th1= Translocation between the 1st (X) chromosome and 3rd chromosome. Commas follow rearrangements and indicate mutations present: e.g. In(2LR)SM1, al 2 Cy cn 2 sp 2 = Inversion involving the left (L) and right (R) arms of chromosome 2 with the following mutations present: aristaless (al), Curly (Cy), cinnabar (cn), and speck (sp). The above information can be obtained either through FlyBase (1) or The Genome of Drosophila melanogaster (2). III. Genetic Tools Available to the Fly Geneticists There are three tools that one can utilize when performing genetic crosses: (1) balancers, (2) phenotypic markers, and (3) non-recombination in males. Balancer chromosomes have the following properties that make them powerful: multiple inversions within the same chromosome, one or more dominant markers, usually 2-4 recessive markers, and lethality as homozygotes. Each chromosome in Drosophila has its own set of balancers. These balancers are numbered numerically and each has its own dominant and recessive markers, though some balancers are "upgrades" of earlier balancers and often share the same markers. The table below is a partial list of the major balancers for each chromosome and the dominant marker associated with each balancer. Due to its small size chromosome 4 has no balancers since homologous recombination does not occur or is infrequent. Chromosome Balancer Dominant marker X FM3 B=Bar eye FM4 " FM6 " FM7a, b, c " 3

2 CyO Cy=Curly wings SM1 " SM2 " SM5 " SM6a, b a=curly wings, b=curly wings and rough eye (Roi) 3 TM1 Me=Moiré TM2 Ubx=Ultrabithorax (bristles on the halteres) TM3 Sb=Stubble (short bristles) TM6 Ubx=Ultrabithorax (bristles on the halteres) TM6B Hu=Humeral, Tb=Tubby TM6C Tb=Tubby TM8 DTS=Dominant temp. sensitive, Sb=Stubble TM9 DTS=Dominant temp. sensitive, Sb=Stubble An example of the multiple inversions present in a balancer is exemplified by TM6B (In(3LR)TM6B, Hu Tb e) that has the following chromosomal numbering 61A 87B- 86C 84F-86C 84B-84F 84B-75C 94A-100F 92D-87B 61A-63B 72E-63B 72E- 75C 94A-92E 100F-100F. The presence of these inversions within the same chromosome prevents homologous recombination. Since most recessive mutations have no phenotype, balancers allow the geneticists to indirectly follow the recessive mutation (by scoring dominant markers) without losing the mutation (due to inhibited recombination). Balancers are also used to maintain chromosomal deficiencies that would otherwise be lethal due to the deficiency. The number of phenotypic markers available to the geneticists is one factor that makes genetics in Drosophila so unique and easy compared to other organisms. Drosophila genetics is replete with both recessive and dominant markers that allow one to select flies by eye phenotype, body phenotype, bristle phenotype, larval phenotype, wing phenotype, etc. Phenotypic markers are especially valuable when using balancers. By using Mendel's laws and following the dominant phenotypic markers associated with a balancer one is able to confidently and successfully produce a homozygous recessive (non-phenotypic) stock following a multi-generation cross. The last genetic tool is the previously mentioned lack of recombination in male fruit flies. This phenomenon can be used in certain genetic crosses without worrying about losing the gene of interest by recombination while in the unbalanced state. More importantly, since most large scale mutagenesis screens start with potentially mutagenized male flies one need not worry about losing the mutation during meiosis. IV. Example Crosses Before presenting any examples of mating crosses, the issue of recombination between two genetic loci needs to be addressed. When trying to either separate or recombine two genetic loci on the same chromosome arm it is necessary to determine beforehand the expected recombination frequency between the two loci and the number of progeny one must screen to find the fly with the desired genotype. The expected recombination frequency between two loci can be ascertained by simply determining the 4

recombination frequency difference between two loci. Recombination frequencies can be obtained from either The Genome of Drosophila melanogaster (2) or FlyBase (1). Deciding how many progeny to screen can be calculated using Mather's (3) formula: N = -log(1-p) log(1-f) where N = number of progeny one needs to screen p = probability of obtaining a recombinant f = expected recombination rate between two loci The more closely two loci are located the greater the number of progeny that must be screened before a recombinant is found. Conversely, if two loci are distant from each other, then one need screen only a small number of progeny. Two genes located on different arms of the same chromosome will sort independently from each other as follows from Mendel's law of independent assortment. Below are several examples of crosses or mating schemes. Brief explanations will be given at each step explaining what is being done and why. Also, the phenotype of the expected flies will be given. Only the genotypes of the desired classes are listed for each mating scheme. The reader is encouraged to determine all theoretical classes that can be produced in each example below by making Punnett squares of each mating. Cross 1: During the course of conducting some electrophysiological studies it is necessary to test a y w norpa P41 fly. This fly can be created from the following two stocks: y w (yellow body color and white eyes) and norpa P41 (red eyes and mutation in the norpa gene). To generate the desired fly, y w norpa P41, one must find a recombinant between the white (w) and norpa loci. All three genes are located on the X chromosome in the following order (starting from the tip of the X chromosome): y w norpa. The recombination value associated with each gene is as follows: y (0.0%), w (1.5%), and norpa (6.5%). Therefore the expected recombination rate, between white and norpa, is 5% (6.5% - 1.5%). The number of recombinants that one must screen to have 95% confidence of finding the desired fly is: N = -log(1-.95) = -log(.05) = -1.301 = 60 progeny! log(1-.05) log(.95) -0.022 The mating scheme is: F1 Red eyed norpa norpa P41 X y w White eye, yellow body color male. Y y w virgin female. 5

F2 X chromosome balancer FM7A X norpa P41 Red eyed female. male. Y y w Recombination occurs here during meiosis. F3 White eyed, yellow body y w norpa P41? X FM7A X chromosome virgin female male. Potential norpa re- Y FM7A balancer. FM7A is not combinant. Set up 60 single homozygous lethal. pair matings. F4 X chromosome balancer FM7A X y w norpa P41? Yellow body, white eye male sibling generated Y FM7A potential norpa female from F3 cross. recombinant. Cross to FM7A male sibling. F5 Potential male recombinant. y w norpa P41? X y w norpa P41? Potential recombinant sibling Y FM7A Virgin female. y w norpa P41? X y w norpa P41? Potential homozygous Y y w norpa P41? recombinant stock. Test for norpa by Western or electrophysiology. In the above cross one can select against recombination between the yellow and white loci by selecting only those males having white eyes and a yellow body at the F3 stage. FM7A unlike other balancers is not homozygous lethal. Cross 2: Given the following two stocks (w; arr1 cn bw; arr2 and w;; ninae e) produce the following fly: w; arr1 cn bw; arr2 ninae e. Again one is trying to find a recombinant between arr2 and ninae, however, since they are on different arms of the same chromosome they will sort independently from each other and there will simply be a 50/50 chance that a particular progeny will contain both mutations. The mating scheme is: F1 w; SM6B; TM6B e X w; arr1 cn bw; arr2 w; SM1; TM2 X w; +; ninae e Y Sco MKRS w arr1 cn bw; arr2 Y Sco MKRS w; +; ninae e F2 arr1, arr2 over balancers. w; arr1 cn bw; arr2 X w; SM1; TM2 ninae virgin over White eyes, curly wings, Y SM6B TM6B e w + ninae e double balancer. humeral, tubby, rough White eyes, curly eyes. wings, Ubx. 6

F3 Independent assortment w; arr1 cn bw; ninae e X w; SM6B; TM6B e Double balancer occurs here during w SM1 arr2 Y Sco MKRS male. meiosis. White eyes, curly wings. F4 Potential arr2, ninae w; arr1 cn bw; arr2? ninae e X w; SM6B; TM6B e Backcross males recombinant. Ebony Y SM6B TM6B e w; Sco MKRS to double balancer body color, white eyes, virgin females. curly wings, humeral, tubby, rough eyes. Set up individual crosses using these males. F5 Sibling mating. White w; arr1 cn bw; arr2? ninae e X w; arr1 cn bw; arr2? ninae e eyes, curly wings, rough Y SM6B TM6B e w SM6B TM6B e eyes, tubby, humeral, ebony body color. Homozygous stock. w; arr1 cn bw; arr2? ninae e In the above cross the following trick was used: ninae and e are closely linked genetically and by selecting an ebony F4 potential recombinant male ones knows that ninae is present. Therefore, one need only screen for the absence of arr2 in the final stock instead of having to screen for the absence of both arr2 and ninae. Since arr2 and ninae have sorted independently of each other one need only screen a few lines to find a recombinant (remember each line has a 50/50 chance of being a double mutant). Cross 3: One wishes to make a w; inad; trp fly. The available stocks are w; inad; + and w; +; trp. Because both mutations are recessive and therefore do not have a phenotype to follow it is necessary to outcross each stock to different double balancer lines. This allows one to indirectly follow the mutations via the dominant markers for each balancer. The mating scheme is: F1 w; inad; + X w; SM6B; TM6B w; +; trp X w; SM1; TM2 Y inad; + w Sco MKRS Y + trp w Sco MKRS F2 White eyes, curly wings, w; SM6B; TM6B X w; SM1; TM2 White eyes, curly rough eyes, tubby, Y inad + w +; trp wings, Ubx. humeral. F3 White eyes, curly wings, w; SM1; TM6B X w; SM1; TM6B tubby, humeral. Sibling Y inad trp w; inad trp cross. 7

w; inad; trp Double mutant stock. inad; trp As stated previously, each line was outcrossed to different double balancer stocks so that the individual mutants could be followed by selecting for the dominant markers from each balancer. To produce the double mutant stock one switches the balancers for both the second and third chromosome as was done when selecting progeny from the F2 cross. V. P-element Transformation A powerful technique available to the researcher is the use of P-elements to transform flies. Using P-elements one can reintroduce a wild type copy of a mutated gene into a null or mutated background. One can also introduce genes that have been altered (e.g. destroying a phosphorylation site) or tagged (e.g. GFP fusions). P-elements are transposable pieces of DNA that randomly insert themselves into genomic DNA. P-elements that are used to generate transgenic animals are engineered without the transposase gene that catalyzes DNA insertion. A separate piece of DNA ( 2-3) that contains the transposase gene must be co-injected with the P-element in order for the P-element to insert itself within the genomic DNA. Transformants are selected using a variety of phenotypic markers such as eye color, body color, resistance to antibiotics, etc. Genes expression, from the incorporated P-element, is driven by a variety of different promoters depending on the experimental design. P-element DNA and 2-3 DNA are injected into the posterior pole of the developing embryo where the germline cells are located. The intention is to transform these germline cells so that a stable, transmissable transformant is created. Embryos that survive the injection procedure are allowed to develop into flies that represent unique P-element insertion events. Individual flies are backcrossed to the injection stock to select for germline transformants. Outcrossing to the injection stock is necessary because not every insertion is into the germline cells but rather into somatic cells. Backcrossing to the injection stock is performed twice to ensure the selection of stable germline transformants. Isolation of germline transformants allows one to proceed with generating a homozygous stock. A generic homozygous producing mating scheme is shown below for a P-element that is marked with the white gene (this will produce a fly with red eyes) and was injected into w 1118 (a white eyed stock) flies. F1 Red eyed male w 1118 X w; SM6B; TM6B Double balancer transformant. Location Y w; Sco MKRS virgins of P-element unknown. F2 Red eyed male over w; SM6B; TM6B X w; SM6B;TM6B Red eyed sibling double balancers Y + + w + + female virgin over double balancers. F3 Red eyed homozygous stock. w; +; + Location of P-element unknown. + + 8

Generation of transgenic flies is not complete without determining which chromosome the P-element inserted into. This information is necessary if one anticipates crossing transgenic animals into various mutant backgrounds. By using male transformant flies one can immediately determine if the P-element is on the X chromosome (see the following mating scheme). If the P-element is on the X chromosome than all F2 females will have red eyes and all F2 males will have white eyes. Both F2 males and F2 females are crossed against each other and the balancers selected against to generate a homozygous stock. F1 Red eyed male w 1118 P[w + ]; +; + X w; SM6B; TM6B Double balancer transformant. P-element Y + + w; Sco MKRS virgin female. is on X chromosome. F2 All males white eyed w; SM6B; TM6B and w 1118 P[w + ]; SM6B; TM6B All females red Y + + w + + eyed. If the P-element is not on the X-chromosome then both F2 males and F2 females will have red eyes. Determining whether the P-element is on the 2nd or 3rd chromosome requires outcrossing the F2 double balancer males, from the above cross, to the injection stock and following how the phenotypic marker segregates in the resulting progeny. The male transformant fly used below is generated from the outcross to the double balancer stock. Red eyed male w 1118 ; SM6B; MKRS X w 1118 ; +; + Virgin injection transformant. In Y P[w + ] P[w + ] w 1118 + + stock females. reality only one P-element is present Three classes of flies are produced from the above cross. Depending on how the red eye phenotype segregates one can determine the chromosome that the P-element is on. The three classes are: w 1118 /Y; SM6B/+; +/P[w + ] w 1118 /Y; SM6B/+; MKRS/+ w 1118 /Y; +/P[w + ]; MKRS/+ All males of this class have red eyes which means P-element on 3rd chromosome. All males of this class have white eyes. No insertion information is revealed. All males of this class have red eyes which means P-element on 2nd chromosome. One uses MKRS instead of TM6B because the dominant marker on MKRS (stubble bristles) is easier to score than the dominant marker on TM6B (humeral). MKRS is generally not used as a balancer because it does not suppress recombination very well even though it does contain inversions. For an exact location of where the P-element has inserted itself one can map the P- element by chromosomal in situs. Homozygous P-element transformants which are lethal or are sterile require the presence of a balancer for continuous propagation. Though P- elements do insert themselves randomly there are "hotspots" within the Drosophila genome that disproportionately attract P-element insertions. 9

Insertion location greatly effects the expression level of the transformed gene and hence any resulting phenotype. Accordingly, a minimum of two independent P-element lines must be evaluated to control for any variance in expression. Two independent lines must also be tested to ensure that any observed phenotype is due to the introduced gene and not to a neomorphic phenotype resulting from where the P-element inserted into the genome. VI. References and Figures 1. FlyBase: http://flybase.bio.indiana.edu:82/ 2. The Genome of Drosophila melanogaster. Dan L. Lindsley and Georgianna G. Zimm. Academic Press, 1992. 3. Mathers, K. The measurement of linkage in heredity, 2nd edition. Methuen, 1951. The following are excellent sources of fly genetics and fly biology. Fly Pushing: The theory and practice of Drosophila genetics. Ralph Greenspan. Cold Spring Harbor Laboratory Press, 1997. Drosophila, A laboratory handbook. Michael Ashburner. Cold Spring Harbor Press, 1989. 10