Drosophila Genetics by Michael Socolich May, 2003
|
|
- William Pierce
- 8 years ago
- Views:
Transcription
1 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 ( ). 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 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 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
2 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
3 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
4 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
5 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) = = 60 progeny! log(1-.05) log(.95) 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
6 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
7 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
8 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
9 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 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
10 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: 2. The Genome of Drosophila melanogaster. Dan L. Lindsley and Georgianna G. Zimm. Academic Press, Mathers, K. The measurement of linkage in heredity, 2nd edition. Methuen, 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, Drosophila, A laboratory handbook. Michael Ashburner. Cold Spring Harbor Press,
Mendelian Genetics in Drosophila
Mendelian Genetics in Drosophila Lab objectives: 1) To familiarize you with an important research model organism,! Drosophila melanogaster. 2) Introduce you to normal "wild type" and various mutant phenotypes.
More informationThe Genetics of Drosophila melanogaster
The Genetics of Drosophila melanogaster Thomas Hunt Morgan, a geneticist who worked in the early part of the twentieth century, pioneered the use of the common fruit fly as a model organism for genetic
More informationTrasposable elements: P elements
Trasposable elements: P elements In 1938 Marcus Rhodes provided the first genetic description of an unstable mutation, an allele of a gene required for the production of pigment in maize. This instability
More informationChapter 9 Patterns of Inheritance
Bio 100 Patterns of Inheritance 1 Chapter 9 Patterns of Inheritance Modern genetics began with Gregor Mendel s quantitative experiments with pea plants History of Heredity Blending theory of heredity -
More informationLAB 11 Drosophila Genetics
LAB 11 Drosophila Genetics Introduction: Drosophila melanogaster, the fruit fly, is an excellent organism for genetics studies because it has simple food requirements, occupies little space, is hardy,
More informationName: Class: Date: ID: A
Name: Class: _ Date: _ Meiosis Quiz 1. (1 point) A kidney cell is an example of which type of cell? a. sex cell b. germ cell c. somatic cell d. haploid cell 2. (1 point) How many chromosomes are in a human
More informationChromosomes, Mapping, and the Meiosis Inheritance Connection
Chromosomes, Mapping, and the Meiosis Inheritance Connection Carl Correns 1900 Chapter 13 First suggests central role for chromosomes Rediscovery of Mendel s work Walter Sutton 1902 Chromosomal theory
More informationCCR Biology - Chapter 7 Practice Test - Summer 2012
Name: Class: Date: CCR Biology - Chapter 7 Practice Test - Summer 2012 Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A person who has a disorder caused
More informationThe correct answer is c A. Answer a is incorrect. The white-eye gene must be recessive since heterozygous females have red eyes.
1. Why is the white-eye phenotype always observed in males carrying the white-eye allele? a. Because the trait is dominant b. Because the trait is recessive c. Because the allele is located on the X chromosome
More informationHeredity - Patterns of Inheritance
Heredity - Patterns of Inheritance Genes and Alleles A. Genes 1. A sequence of nucleotides that codes for a special functional product a. Transfer RNA b. Enzyme c. Structural protein d. Pigments 2. Genes
More informationA rough guide to Drosophila mating schemes (light version 2.1) 1
A. Prokop - A rough guide to Drosophila mating schemes 1 A rough guide to Drosophila mating schemes (light version 2.1) 1 1. Why work with the fruitfly Drosophila melanogaster? More than a century ago
More information5 GENETIC LINKAGE AND MAPPING
5 GENETIC LINKAGE AND MAPPING 5.1 Genetic Linkage So far, we have considered traits that are affected by one or two genes, and if there are two genes, we have assumed that they assort independently. However,
More informationMCB41: Second Midterm Spring 2009
MCB41: Second Midterm Spring 2009 Before you start, print your name and student identification number (S.I.D) at the top of each page. There are 7 pages including this page. You will have 50 minutes for
More informationCHROMOSOMES AND INHERITANCE
SECTION 12-1 REVIEW CHROMOSOMES AND INHERITANCE VOCABULARY REVIEW Distinguish between the terms in each of the following pairs of terms. 1. sex chromosome, autosome 2. germ-cell mutation, somatic-cell
More informationLAB : PAPER PET GENETICS. male (hat) female (hair bow) Skin color green or orange Eyes round or square Nose triangle or oval Teeth pointed or square
Period Date LAB : PAPER PET GENETICS 1. Given the list of characteristics below, you will create an imaginary pet and then breed it to review the concepts of genetics. Your pet will have the following
More informationGenetics Module B, Anchor 3
Genetics Module B, Anchor 3 Key Concepts: - An individual s characteristics are determines by factors that are passed from one parental generation to the next. - During gamete formation, the alleles for
More informationA trait is a variation of a particular character (e.g. color, height). Traits are passed from parents to offspring through genes.
1 Biology Chapter 10 Study Guide Trait A trait is a variation of a particular character (e.g. color, height). Traits are passed from parents to offspring through genes. Genes Genes are located on chromosomes
More informationName: 4. A typical phenotypic ratio for a dihybrid cross is a) 9:1 b) 3:4 c) 9:3:3:1 d) 1:2:1:2:1 e) 6:3:3:6
Name: Multiple-choice section Choose the answer which best completes each of the following statements or answers the following questions and so make your tutor happy! 1. Which of the following conclusions
More informationI. Genes found on the same chromosome = linked genes
Genetic recombination in Eukaryotes: crossing over, part 1 I. Genes found on the same chromosome = linked genes II. III. Linkage and crossing over Crossing over & chromosome mapping I. Genes found on the
More information(1-p) 2. p(1-p) From the table, frequency of DpyUnc = ¼ (p^2) = #DpyUnc = p^2 = 0.0004 ¼(1-p)^2 + ½(1-p)p + ¼(p^2) #Dpy + #DpyUnc
Advanced genetics Kornfeld problem set_key 1A (5 points) Brenner employed 2-factor and 3-factor crosses with the mutants isolated from his screen, and visually assayed for recombination events between
More informationAn efficient mutagenesis screen to generate duplications of polytene section 8 on the X chromosome of Drosophila melanogaster.
Lilly, Brenda, and Juan Botas. 1999. An efficient mutagenesis screen to generate duplications of polytene section 8 on the X chromosome of Drosophila melanogaster. Dros. Inf. Serv. 82: 105-112. An efficient
More informationAP: LAB 8: THE CHI-SQUARE TEST. Probability, Random Chance, and Genetics
Ms. Foglia Date AP: LAB 8: THE CHI-SQUARE TEST Probability, Random Chance, and Genetics Why do we study random chance and probability at the beginning of a unit on genetics? Genetics is the study of inheritance,
More informationCHAPTER 15 THE CHROMOSOMAL BASIS OF INHERITANCE. Section B: Sex Chromosomes
CHAPTER 15 THE CHROMOSOMAL BASIS OF INHERITANCE Section B: Sex Chromosomes 1. The chromosomal basis of sex varies with the organism 2. Sex-linked genes have unique patterns of inheritance 1. The chromosomal
More informationHeredity. Sarah crosses a homozygous white flower and a homozygous purple flower. The cross results in all purple flowers.
Heredity 1. Sarah is doing an experiment on pea plants. She is studying the color of the pea plants. Sarah has noticed that many pea plants have purple flowers and many have white flowers. Sarah crosses
More informationBio EOC Topics for Cell Reproduction: Bio EOC Questions for Cell Reproduction:
Bio EOC Topics for Cell Reproduction: Asexual vs. sexual reproduction Mitosis steps, diagrams, purpose o Interphase, Prophase, Metaphase, Anaphase, Telophase, Cytokinesis Meiosis steps, diagrams, purpose
More informationGenetics Lecture Notes 7.03 2005. Lectures 1 2
Genetics Lecture Notes 7.03 2005 Lectures 1 2 Lecture 1 We will begin this course with the question: What is a gene? This question will take us four lectures to answer because there are actually several
More information2 GENETIC DATA ANALYSIS
2.1 Strategies for learning genetics 2 GENETIC DATA ANALYSIS We will begin this lecture by discussing some strategies for learning genetics. Genetics is different from most other biology courses you have
More informationAnswer Key Problem Set 5
7.03 Fall 2003 1 of 6 1. a) Genetic properties of gln2- and gln 3-: Answer Key Problem Set 5 Both are uninducible, as they give decreased glutamine synthetase (GS) activity. Both are recessive, as mating
More informationGenetics for the Novice
Genetics for the Novice by Carol Barbee Wait! Don't leave yet. I know that for many breeders any article with the word genetics in the title causes an immediate negative reaction. Either they quickly turn
More informationBasics of Marker Assisted Selection
asics of Marker ssisted Selection Chapter 15 asics of Marker ssisted Selection Julius van der Werf, Department of nimal Science rian Kinghorn, Twynam Chair of nimal reeding Technologies University of New
More informationPRACTICE PROBLEMS - PEDIGREES AND PROBABILITIES
PRACTICE PROBLEMS - PEDIGREES AND PROBABILITIES 1. Margaret has just learned that she has adult polycystic kidney disease. Her mother also has the disease, as did her maternal grandfather and his younger
More information7A The Origin of Modern Genetics
Life Science Chapter 7 Genetics of Organisms 7A The Origin of Modern Genetics Genetics the study of inheritance (the study of how traits are inherited through the interactions of alleles) Heredity: the
More informationLAB : THE CHI-SQUARE TEST. Probability, Random Chance, and Genetics
Period Date LAB : THE CHI-SQUARE TEST Probability, Random Chance, and Genetics Why do we study random chance and probability at the beginning of a unit on genetics? Genetics is the study of inheritance,
More informationInfluence of Sex on Genetics. Chapter Six
Influence of Sex on Genetics Chapter Six Humans 23 Autosomes Chromosomal abnormalities very severe Often fatal All have at least one X Deletion of X chromosome is fatal Males = heterogametic sex XY Females
More informationA and B are not absolutely linked. They could be far enough apart on the chromosome that they assort independently.
Name Section 7.014 Problem Set 5 Please print out this problem set and record your answers on the printed copy. Answers to this problem set are to be turned in to the box outside 68-120 by 5:00pm on Friday
More informationBiological Sciences Initiative. Human Genome
Biological Sciences Initiative HHMI Human Genome Introduction In 2000, researchers from around the world published a draft sequence of the entire genome. 20 labs from 6 countries worked on the sequence.
More informationPractice Problems 4. (a) 19. (b) 36. (c) 17
Chapter 10 Practice Problems Practice Problems 4 1. The diploid chromosome number in a variety of chrysanthemum is 18. What would you call varieties with the following chromosome numbers? (a) 19 (b) 36
More informationTerms: The following terms are presented in this lesson (shown in bold italics and on PowerPoint Slides 2 and 3):
Unit B: Understanding Animal Reproduction Lesson 4: Understanding Genetics Student Learning Objectives: Instruction in this lesson should result in students achieving the following objectives: 1. Explain
More informationGENETIC CROSSES. Monohybrid Crosses
GENETIC CROSSES Monohybrid Crosses Objectives Explain the difference between genotype and phenotype Explain the difference between homozygous and heterozygous Explain how probability is used to predict
More informationBiology 1406 Exam 4 Notes Cell Division and Genetics Ch. 8, 9
Biology 1406 Exam 4 Notes Cell Division and Genetics Ch. 8, 9 Ch. 8 Cell Division Cells divide to produce new cells must pass genetic information to new cells - What process of DNA allows this? Two types
More informationChapter 13: Meiosis and Sexual Life Cycles
Name Period Chapter 13: Meiosis and Sexual Life Cycles Concept 13.1 Offspring acquire genes from parents by inheriting chromosomes 1. Let s begin with a review of several terms that you may already know.
More informationGene Mapping Techniques
Gene Mapping Techniques OBJECTIVES By the end of this session the student should be able to: Define genetic linkage and recombinant frequency State how genetic distance may be estimated State how restriction
More informationGenetics 1. Defective enzyme that does not make melanin. Very pale skin and hair color (albino)
Genetics 1 We all know that children tend to resemble their parents. Parents and their children tend to have similar appearance because children inherit genes from their parents and these genes influence
More informationFact Sheet 14 EPIGENETICS
This fact sheet describes epigenetics which refers to factors that can influence the way our genes are expressed in the cells of our body. In summary Epigenetics is a phenomenon that affects the way cells
More informationDRAGON GENETICS LAB -- Principles of Mendelian Genetics
DragonGeneticsProtocol Mendelian Genetics lab Student.doc DRAGON GENETICS LAB -- Principles of Mendelian Genetics Dr. Pamela Esprivalo Harrell, University of North Texas, developed an earlier version of
More informationBiology Behind the Crime Scene Week 4: Lab #4 Genetics Exercise (Meiosis) and RFLP Analysis of DNA
Page 1 of 5 Biology Behind the Crime Scene Week 4: Lab #4 Genetics Exercise (Meiosis) and RFLP Analysis of DNA Genetics Exercise: Understanding how meiosis affects genetic inheritance and DNA patterns
More informationMendelian and Non-Mendelian Heredity Grade Ten
Ohio Standards Connection: Life Sciences Benchmark C Explain the genetic mechanisms and molecular basis of inheritance. Indicator 6 Explain that a unit of hereditary information is called a gene, and genes
More information5. The cells of a multicellular organism, other than gametes and the germ cells from which it develops, are known as
1. True or false? The chi square statistical test is used to determine how well the observed genetic data agree with the expectations derived from a hypothesis. True 2. True or false? Chromosomes in prokaryotic
More information17. A testcross A.is used to determine if an organism that is displaying a recessive trait is heterozygous or homozygous for that trait. B.
ch04 Student: 1. Which of the following does not inactivate an X chromosome? A. Mammals B. Drosophila C. C. elegans D. Humans 2. Who originally identified a highly condensed structure in the interphase
More informationX CHROMOSOME OF DROSOPHILA MELANOGASTER.
FITNESS EFFECTS OF EMS-INDUCED MUTATIONS ON THE X CHROMOSOME OF DROSOPHILA MELANOGASTER. I. VIABILITY EFFECTS AND HETEROZYGOUS FITNESS EFFECTS JOYCE A. MITCHELL2 Laboratory of Genetics, University of Wisconsin,
More informationGene Therapy and Genetic Counseling. Chapter 20
Gene Therapy and Genetic Counseling Chapter 20 What is Gene Therapy? Treating a disease by replacing, manipulating or supplementing a gene The act of changing an individual s DNA sequence to fix a non-functional
More informationGenetic Mutations. Indicator 4.8: Compare the consequences of mutations in body cells with those in gametes.
Genetic Mutations Indicator 4.8: Compare the consequences of mutations in body cells with those in gametes. Agenda Warm UP: What is a mutation? Body cell? Gamete? Notes on Mutations Karyotype Web Activity
More informationHuman Blood Types: Codominance and Multiple Alleles. Codominance: both alleles in the heterozygous genotype express themselves fully
Human Blood Types: Codominance and Multiple Alleles Codominance: both alleles in the heterozygous genotype express themselves fully Multiple alleles: three or more alleles for a trait are found in the
More informationStocker & Gallant 1 Fly Husbandry. 8093 Zurich, Switzerland, phone: +41-44-6333679, fax: +41-44-6331051, email:
Stocker & Gallant 1 Fly Husbandry Getting started: An overview on raising and handling Drosophila Overview chapter: non-standard format Hugo Stocker 1 & Peter Gallant 2 1 Institute for Molecular Systems
More informationHardy-Weinberg Equilibrium Problems
Hardy-Weinberg Equilibrium Problems 1. The frequency of two alleles in a gene pool is 0.19 (A) and 0.81(a). Assume that the population is in Hardy-Weinberg equilibrium. (a) Calculate the percentage of
More informationIMSR File Format. Author: Mark Airey Last Modified: September 02, 2010 01:47. Table 1: New format of IMSR Data File
1 IMSR Provider Data-Files IMSR File Format Author: Mark Airey Last Modified: September 02, 2010 01:47 Institutions that contribute to the IMSR will provide their data on a regular schedule as tabdelimited
More informationsomatic cell egg genotype gamete polar body phenotype homologous chromosome trait dominant autosome genetics recessive
CHAPTER 6 MEIOSIS AND MENDEL Vocabulary Practice somatic cell egg genotype gamete polar body phenotype homologous chromosome trait dominant autosome genetics recessive CHAPTER 6 Meiosis and Mendel sex
More informationBioBoot Camp Genetics
BioBoot Camp Genetics BIO.B.1.2.1 Describe how the process of DNA replication results in the transmission and/or conservation of genetic information DNA Replication is the process of DNA being copied before
More informationTwo copies of each autosomal gene affect phenotype.
SECTION 7.1 CHROMOSOMES AND PHENOTYPE Study Guide KEY CONCEPT The chromosomes on which genes are located can affect the expression of traits. VOCABULARY carrier sex-linked gene X chromosome inactivation
More informationChapter 4 The role of mutation in evolution
Chapter 4 The role of mutation in evolution Objective Darwin pointed out the importance of variation in evolution. Without variation, there would be nothing for natural selection to act upon. Any change
More informationInstructor s Key for GloFish Protocol
Instructor s Key for GloFish Protocol Summary of Supplemental Materials Supplemental Material 1 Animal training homework Supplemental Material 2 GloFish Instructor s Key Supplemental Material 3 Powerpoint
More informationBiology 1406 - Notes for exam 5 - Population genetics Ch 13, 14, 15
Biology 1406 - Notes for exam 5 - Population genetics Ch 13, 14, 15 Species - group of individuals that are capable of interbreeding and producing fertile offspring; genetically similar 13.7, 14.2 Population
More informationBioSci 2200 General Genetics Problem Set 1 Answer Key Introduction and Mitosis/ Meiosis
BioSci 2200 General Genetics Problem Set 1 Answer Key Introduction and Mitosis/ Meiosis Introduction - Fields of Genetics To answer the following question, review the three traditional subdivisions of
More informationDNA Determines Your Appearance!
DNA Determines Your Appearance! Summary DNA contains all the information needed to build your body. Did you know that your DNA determines things such as your eye color, hair color, height, and even the
More informationThe Making of the Fittest: Natural Selection in Humans
OVERVIEW MENDELIN GENETIC, PROBBILITY, PEDIGREE, ND CHI-QURE TTITIC This classroom lesson uses the information presented in the short film The Making of the Fittest: Natural election in Humans (http://www.hhmi.org/biointeractive/making-fittest-natural-selection-humans)
More informationChromosomal Basis of Inheritance. Ch. 3
Chromosomal Basis of Inheritance Ch. 3 THE CHROMOSOME THEORY OF INHERITANCE AND SEX CHROMOSOMES! The chromosome theory of inheritance describes how the transmission of chromosomes account for the Mendelian
More informationGenetics 301 Sample Final Examination Spring 2003
Genetics 301 Sample Final Examination Spring 2003 50 Multiple Choice Questions-(Choose the best answer) 1. A cross between two true breeding lines one with dark blue flowers and one with bright white flowers
More informationMendelian inheritance and the
Mendelian inheritance and the most common genetic diseases Cornelia Schubert, MD, University of Goettingen, Dept. Human Genetics EUPRIM-Net course Genetics, Immunology and Breeding Mangement German Primate
More informationBiology Final Exam Study Guide: Semester 2
Biology Final Exam Study Guide: Semester 2 Questions 1. Scientific method: What does each of these entail? Investigation and Experimentation Problem Hypothesis Methods Results/Data Discussion/Conclusion
More informationChapter 13: Meiosis and Sexual Life Cycles
Name Period Concept 13.1 Offspring acquire genes from parents by inheriting chromosomes 1. Let s begin with a review of several terms that you may already know. Define: gene locus gamete male gamete female
More informationCystic Fibrosis Webquest Sarah Follenweider, The English High School 2009 Summer Research Internship Program
Cystic Fibrosis Webquest Sarah Follenweider, The English High School 2009 Summer Research Internship Program Introduction: Cystic fibrosis (CF) is an inherited chronic disease that affects the lungs and
More informationAbout The Causes of Hearing Loss
About 1 in 500 infants is born with or develops hearing loss during early childhood. Hearing loss has many causes: some are genetic (that is, caused by a baby s genes) or non-genetic (such as certain infections
More informationPaternity Testing. Chapter 23
Paternity Testing Chapter 23 Kinship and Paternity DNA analysis can also be used for: Kinship testing determining whether individuals are related Paternity testing determining the father of a child Missing
More informationMeiosis is a special form of cell division.
Page 1 of 6 KEY CONCEPT Meiosis is a special form of cell division. BEFORE, you learned Mitosis produces two genetically identical cells In sexual reproduction, offspring inherit traits from both parents
More informationWhat is Cancer? Cancer is a genetic disease: Cancer typically involves a change in gene expression/function:
Cancer is a genetic disease: Inherited cancer Sporadic cancer What is Cancer? Cancer typically involves a change in gene expression/function: Qualitative change Quantitative change Any cancer causing genetic
More informationY Chromosome Markers
Y Chromosome Markers Lineage Markers Autosomal chromosomes recombine with each meiosis Y and Mitochondrial DNA does not This means that the Y and mtdna remains constant from generation to generation Except
More informationP1 Gold X Black. 100% Black X. 99 Black and 77 Gold. Critical Values 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51
Questions for Exam I Fall 2005 1. Wild-type humbugs have no spots, have red eyes and brown bodies. You have isolated mutations in three new autosomal humbug genes. The mutation Sp gives a dominant phenotype
More informationVon Mäusen und Menschen E - 1
Von Mäusen und Menschen E - 1 Mus musculus: Genetic Portrait of the House Mouse E - 3 Outline Mouse genome Mouse life cycle Transgenic protocols Addition of genes by nuclear injection Removal of genes
More informationGenetic Technology. Name: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.
Name: Class: Date: Genetic Technology Multiple Choice Identify the choice that best completes the statement or answers the question. 1. An application of using DNA technology to help environmental scientists
More informationBio 102 Practice Problems Mendelian Genetics and Extensions
Bio 102 Practice Problems Mendelian Genetics and Extensions Short answer (show your work or thinking to get partial credit): 1. In peas, tall is dominant over dwarf. If a plant homozygous for tall is crossed
More informationB2 5 Inheritrance Genetic Crosses
B2 5 Inheritrance Genetic Crosses 65 minutes 65 marks Page of 55 Q. A woman gives birth to triplets. Two of the triplets are boys and the third is a girl. The triplets developed from two egg cells released
More informationBest Practices for Efficient Mouse Colony Management
Best Practices for Efficient Mouse Colony Management Dominique Kagele, Ph.D. Technical Information Services Overview of Today s Presentation Mouse reproduction Factors affecting breeding performance Data
More informationAnimal Behavior. Evaluation copy
Animal Behavior Computer 11 Perhaps one of the most difficult fields of biology to study is ethology, the study of animal behavior. Observation of a behavior is simple; interpreting what has been observed
More informationDrosophila Embryo Injection Order Form #2 For Multiple Samples as Injection Mix v2.0.4 Please use multiple pages if more than 3 orders
BestGene Inc http://www.thebestgene.com BestGene Inc, 2140 Grand Ave. Suite#205, Chino Hills, CA 91709, USA Tel: +1-888-821-9155 Fax: +1-888-822-8445 Drosophila Embryo Injection Order Form #2 For Multiple
More informationI somatic vigor and, at the same time, show disturbances in the gonads
STUDIES ON HYBRID STERILITY IV. TRANSPLANTED TESTES IN DROSOPHILA PSEUDOOBSCURA TH. DOBZHANSKY AND G. W. BEADLE W. G. Kerckhoff Laboratories of the Biological Sciences, Culijornia Institute of Technology,
More information!!!!!!!!!!!!!!!!!!!!!!!!!!
Figure S7 cdl-1 3'UTR gld-1 binding site (4..10) let-7 binding site (326..346) Figure Legends Figure S1. gld-1 genetically interacts with nhl-2 and vig-1 during germline development. (A) nhl-2(ok818) enhances
More informationCommonly Used STR Markers
Commonly Used STR Markers Repeats Satellites 100 to 1000 bases repeated Minisatellites VNTR variable number tandem repeat 10 to 100 bases repeated Microsatellites STR short tandem repeat 2 to 6 bases repeated
More informationX Linked Inheritance
X Linked Inheritance Information for Patients and Families 2 X linked Inheritance The following will give you information about what X linked inheritance means and how X linked conditions are inherited.
More informationThe Making of the Fittest: Evolving Switches, Evolving Bodies
OVERVIEW MODELING THE REGULATORY SWITCHES OF THE PITX1 GENE IN STICKLEBACK FISH This hands-on activity supports the short film, The Making of the Fittest:, and aims to help students understand eukaryotic
More informationConsequently agglutination occurs only when blood from different individuals. (1920) was unable to detect agglutination in mixtures of blood from 50
102 GENETICS: S. 0. BURHOE PROC. N. A. S. (1946); and Welch, A. D., Heinle, R. W., Nelson, E. M., and Nelson, H. V., Ann. N. Y. Acad. Sci., 48, 347 (1946). 6 Heinle, R. W., and Welch, A. D., Ann. N. Y.
More informationLECTURE 6 Gene Mutation (Chapter 16.1-16.2)
LECTURE 6 Gene Mutation (Chapter 16.1-16.2) 1 Mutation: A permanent change in the genetic material that can be passed from parent to offspring. Mutant (genotype): An organism whose DNA differs from the
More informationDeterministic computer simulations were performed to evaluate the effect of maternallytransmitted
Supporting Information 3. Host-parasite simulations Deterministic computer simulations were performed to evaluate the effect of maternallytransmitted parasites on the evolution of sex. Briefly, the simulations
More informationBasic Concepts Recombinant DNA Use with Chapter 13, Section 13.2
Name Date lass Master 19 Basic oncepts Recombinant DN Use with hapter, Section.2 Formation of Recombinant DN ut leavage Splicing opyright lencoe/mcraw-hill, a division of he Mcraw-Hill ompanies, Inc. Bacterial
More informationInheritance of Color And The Polled Trait Dr. R. R. Schalles, Dept. of Animal Sciences and Industry Kansas State University
Inheritance of Color And The Polled Trait Dr. R. R. Schalles, Dept. of Animal Sciences and Industry Kansas State University Introduction All functions of an animal are controlled by the enzymes (and other
More informationGENOMIC SELECTION: THE FUTURE OF MARKER ASSISTED SELECTION AND ANIMAL BREEDING
GENOMIC SELECTION: THE FUTURE OF MARKER ASSISTED SELECTION AND ANIMAL BREEDING Theo Meuwissen Institute for Animal Science and Aquaculture, Box 5025, 1432 Ås, Norway, theo.meuwissen@ihf.nlh.no Summary
More informationGenetics Test Biology I
Genetics Test Biology I Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Avery s experiments showed that bacteria are transformed by a. RNA. c. proteins.
More informationTest Two Study Guide
Test Two Study Guide 1. Describe what is happening inside a cell during the following phases (pictures may help but try to use words): Interphase: : Consists of G1 / S / G2. Growing stage, cell doubles
More informationMDM. Metabolic Drift Mutations - Attenuation Technology
MDM Metabolic Drift Mutations - Attenuation Technology Seite 2 Origin of MDM attenuation technology Prof. Dr. Klaus Linde Pioneer in R&D of human and animal vaccines University of Leipzig Germany Origin
More informationEvolution (18%) 11 Items Sample Test Prep Questions
Evolution (18%) 11 Items Sample Test Prep Questions Grade 7 (Evolution) 3.a Students know both genetic variation and environmental factors are causes of evolution and diversity of organisms. (pg. 109 Science
More informationLecture 2: Mitosis and meiosis
Lecture 2: Mitosis and meiosis 1. Chromosomes 2. Diploid life cycle 3. Cell cycle 4. Mitosis 5. Meiosis 6. Parallel behavior of genes and chromosomes Basic morphology of chromosomes telomere short arm
More information