Study Guides (238,278)
Canada (115,052)
Biology (180)
BIOL 2104 (5)

Genetics - lab exam review

24 Pages
Unlock Document

Carleton University
BIOL 2104
Amanda Helleman

Ioana D. 2010 Genetics BIO 2133 – lab exam review Lab #1 - Introduction to Flylab - It’s a simulation program that helps us investigate various aspects of Mendelian genetics - We can study the mode of transmission of various mutations in Drosophila melanogaster(fruit fly) - All traits in Fly lab are either dominant or recessive - Incomplete dominance CANNOT be studied using this simulation - We can cause phenotypic changes in the fruit fly: bristles, body color, antennal form, eye color, etc. - When selecting a mutation the mutation you select will always be homozygous unless the trait is lethal in the homozygous condition. In this case, FlyLab generates a fly that is heterozygous four that mutation - Only 1 allele per trait can be chosen *Design* - allows selection of the phenotypes of males and females - after selection of flies we must choose ’10 000 flies’ for the # of offspring we want to produce - the total # of offspring may not be exactly 10 000 b/c of FlyLab built-in experimental error - chi square analysis helps us test hypothesis: e.g if the total chi square is less than the critical chi square provided to us ( i.e 0.05, n-1 degrees of freedom), we cannot reject the hypothesis **Understand genetic notation A) Monohybrid cross: RI - + (that means female is RIRI and male is RI+RI+) P generation : RI RI x RI+RI+ F1 generation  RI RI RI+ RI RI+ RI RI+ RI+ RI RI+ RI RI+  All offspring are heterozygous Ioana D. 2010 F2 generation: RI+RI x RI+ RI RI+ RI RI+ RI+RI+ RI+RI RI RI+RI RI RI Phenotypic ratio: 1:2:1 B) Dihybrid cross SV –RI (that means the female is SV SV RI+RI+ and the male is SV+SV+ RI RI ) *each parent can only give only one type of gamete in this case. So we only have 1 possible offspring phenotype P generation: SV SV RI+RI+ x SV+SV+ RI RI F1 generation: SV+RI SV RI+ SV+SV RI+RI  All offspring are heterozygous F2 generation: SV+SV RI+RI x SV+SV RI+RI SV+ RI+ SV+ RI SV RI+ SV RI SV+ RI+ SV+SV+ RI+RI+ SV+SV+ RI+RI SV+SV RI+RI+ SV+SV RI+RI SV+ RI SV+SV+ RI+RI SV+SV+ RIRI SV+ SV RI+RI SV+SV RI RI SV RI+ SV+SV RI+RI+ SV+SV RI+RI SV RI+ SV SV RI+RI SV RI SV+SV RI+RI SV+SV RI RI SV SV RI+RI SV SV RI RI Phenotypic Ratio: 9:3:3:1 Ioana D. 2010 Trihybrid cross: + - SV, SS, EY This means, in P generation Females: SV+SV+ SS+ SS+ EY+EY+ x males: SV SV SS SS EY EY In the F1 generation, just like before, all offspring will be heterozygous: i.e SV+SV SS+ SS EY+ EY In the F2 generation, we cross SV+SV SS+ SS EY+ EY x SV+SV SS+ SS EY+ EY SV+SS+EY+ SV+SS+EY SV+SS EY SV+ SS EY+ SV SS+ EY+ SV SS+ EY SV SS EY+ SV SS EY SV+SS+EY+ SV+SV+SS+SS+EY+EY+ Etc….. SV+SS+EY SV+SV+SS+SS+EY+EY SV+SS EY SV+SV+SS+SS EY+EY SV+ SS EY+ SV+SV+ SS+SS EY+EY+ SV SS+ EY+ SV+SV SS+SS EY+EY+ SV SS+ EY SV+SV SS+SS+ EY+ EY SV SS EY+ SV+SV SS+SS EY+EY+ SV SS EY SV+SV SS+SS EY+EY 3 We have 8 possible gametes, b/c we have 3 genes found in heterozygous state: 2 = 8 Ratio: 27:9:9:9:3:3:3:1 (this should all add up to 64) Lab 2a – Flylab -testcrosses - An organism that displays a dominant phenotype may be either a homozygous or heterozygous genotype. - To conduct a testcross, we need to take a fly of unknown genotype and cross it with a fly that is homozygous recessive for that trait - Female fly has one mutation, male has a different mutation Select an F1 female fly and cross it with a male fly that has both mutations AP – SE Female : APAP SE+SE+ x Male: AP+AP+ SE SE F1 cross offspring: AP+AP SE+SE Ioana D. 2010 Now we cross a female: AP+AP SE+SE x male: AP AP SE SE (double mutant)  this is our test- cross AP SE AP+SE+ AP+AP SE+SE AP+ SE AP+AP SE SE AP SE+ AP AP SE+ SE AP SE AP AP SE SE Phenotypic ratio: 1:1:1:1 Lab #2b – Introduction to the genetics of Drosophila melanogaster - Drosophila melanogaster is one of the most useful eukaryotic organisms used in the study of genetics Some convenient characteristics: - Easy and economical to culture in the lab - Small size : easy to handle and store - Prolific breeding  female can lay 100’s of eggs - Short generation times: 9-11 days at 25 degrees, 14 days at 23 degrees - Large # of external characteristics - Giant chromosomes in salivary glands - The culturing medium for Drosophila was yeast  served as a source of food for the developing larvae - Drosophila should be cultured at temps between 18 and 25 degrees - Development is very slow at low temps - Above 26 degrees there is high risk of abnormal development - At 23 degrees. The time from egg laying to emergence at pupae is 14 days - Because of their small size it is necessary to 1) Use a microscope for examination 2) Anaesthetize the flies How do we anaesthetize them? - Carbon dioxide  non-toxic gas, it anaesthetizes flies for long periods of time without damage  When CO2 is removed they are revived instantly  CO2 is heavy and will thus stay in a vial or container Ioana D. 2010 - Ice  fruit flies are heterothermic (cold blooded). Thus can be immobilized by cold without killing them (We made a cold plate, by filling a petri dish with ice ) - Ether  dangerous fire hazard and can be ignited explosively by a spark Drosophila life cycle - 4 stages: egg, larva, pupa, adult - The length of the cycle depends on the temperature at which flies are kept - At 25 degrees the cycle will be complete in about 10 days - At 20 degrees, about 15 days - We must NOT expose Drosophila cultures to high T, because it can result in sterilization or death - Low temperature (e.g 10 degrees) can result in prolonged life cycle and reduce viability - Optimum T: 25 degrees 1. Egg: - Female starts laying eggs on the second day after emergence from pupa - 0.5 mm length egg, ovoid in shape, white in color - Embryonic development takes about 1 day (25 degrees); a larva emerges from the egg case 2. Larva: - White, segmented, worm like Black mouth parts (jaw hooks) - No eyes (completely blind) - Lacks appendages - Breathes by trachea (has a pair of spiracles i.e air pores a the anterior and posterior ends of the body) - Oxygen moves through the spiracle, trachea, and oxygen exchange occurs b/w tracheole walls and muscle cells via extracellular fluid - Rapid eating and growing - There are 3 subdivisions of this stage  instars st nd  1 and 2 instars end in moults i.e complete shedding of the skin and mouth parts  3 instar ends by pupation - Larval stage takes about 4 days for completion (3 instar is 4.5 mm long) 3. Pupa: - Reorganizational stage of the cycle Ioana D. 2010 - All larval structures are destroyed and adult structures are developed from embryonic tissues called anlagen, or imaginal discs (these have been lying formant in the animal) - Pupal stage takes about 4 days at 25 degrees to be complete 4. Adult: - This is the reproductive stage of the lifecycle - Animal emerges from the puparium - The wings are initially unexpanded, but they expand within an hour - The body is initially elongated but soon assumes a more rotund form - At first, adults are relatively light in color, but in a few hours they darken and assume characteristic adult colors - They mate about 6 hrs after they have emerged from the puparium - Fertilization is internal - Sperm is stored in the spematheca and ventral receptacles of the female, and then released gradually in the oviduct as the eggs are passed through the oviduct and into the vagina - Female begins to deposit eggs about 2 days after it has emerged - 50-75 eggs/ day in the first few days - Egg production decreases with time - Average life span: 37 days at 25 degrees Fly Anatomy: - 3 parts: head, thorax, abdomen Head: - 6 fused segments: - Antennae - Aristae - Proboscis - tongue - Compound eyes - Ocelli – simple eyes - Bristles Thorax: - 3 fused segments - Prothorax – the sex comb in males is present here - Mesothorax - Metathorax - Thoracic spiracles - 2 of them - Bristles Abdomen: Ioana D. 2010 -7-8 segments in females, 5-6 in males - tergites -sternites -abdominal spiracles -genital region **Check the diagrams in the lab manual. Unfortunately I wasn’t able to find good diagrams on google. How can we distinguish Males from Females? 1. Sex comb  only found in males. A row of 10 stout black bristles on the uppermost tarsal joint of the first pair of legs (can be identified in the pupal stage) 2. Abdomen female abdomen has 7 visible segments, has separate dark bands along the dorsal surface. A male has 5 visible segments; dark bands on the last few segments are fused 3. Genital region  female has anal plates and lightly colored ovipositor plates. Male has anal plates and darkly pigmented genital arch and penis Female Drosophila Male Drosophila Ioana D. 2010 Look for: 1) bands (fused for males, separate for females) 2) sex comb ( only in males, anterior legs) Feeding: **Check diagrams in the manual - Extensive tubular proboscis - During feeding it is lowered and salivary secretions are pumped out onto food - Dissolved food flows by capillary action along the tiny grooves on the surface of the 2 pads found on the ends of the proboscis - Usually fruit flies eat fruit or fermenting grain mixtures - Sugar: energy source - Yeast: provides protein and vitamins - Both larvae and adults feed on the same food source Notation used in Drosophila genetics - Normal fly = wild type  genes yielding the normal phenotype are designated + - + can describe the entire genome or one single gene depending on the context - Mutant alleles:  If recessive they are designated small letters ( ec)  If dominant, the first letter is capitalized (Cy)  E.g wildtype gene: ec+, dominant to ec  Cy+, dominant to Cy - If an individual is designated only by mutant symbols e.g ( dp, ec, ey), we can assume that the individual is homozygous for the traits and contain wild-type alleles at other loci - If the individual is heterozygous for 1 or more genes, we must specify the genes on both homologous chromosomes - E.g genes are on the same chromosome dp cn+ bw or dp + bw dp+ cn bw+ + cn + e.g genes on different chromosomes dp cn+ bw or dp + bw dp+ cn bw+ + cn + P1: parental generation F1 : offspring of the P1 i.e first filial generation F2: second filial generation Backcross: a cross b/w an F1 individual and a P individual. Ioana D. 2010 **any backcross in which we have a recessive parental stock is known as a testcross Exp #2 c – Polytene chromosomes - In viruses, blue-green algae and bacteria, each cell contains a single chromosome, consisting only of DNA - In protozoa, algae (except blue-green), fungi + all multicellular organisms, the DNA is associated with protein, and organized into more than a single chromosome - Drosophila virilis was used for this experiment b/c it has a larger 3 instar larva - In the cells of salivary glands, the homologous chromosomes are permanently synapsed - The cells do not divide. They enlarge while their chromosomes are duplicated regularly  this chromosome duplication without cell division is called endomitosis and the chromosomes are called polytene (many stranded) chromosomes - Puffs – enlarged nonbanded regions of the chromosomes, whose genes are involved in very active DNA transcription Method - Use the 3 instar larva and place it in a drop of insect Ringer’s solution on a slide. The dissection is done directly in this solution, under 20-30x magnification (dissecting microscope), and DARK background - Dissecting need is placed behind the black hooks and the other near the posterior end of the larva - Upon dissection, the intestine remains attached to the head, along with the thing salivary glands on either side - After blotting excess Ringer’s solution, add aceto- orcein stain (red)  this allows us to identify the chromosomes under the compound microscope -Drosophila melanogaster has a diploid # of chromosomes of 8 (1 pair of sex chromosomes, 3 pairs of autosomes) - sex chromosomes are designated #1 - each chromosome has a special constricted section called centromere (attaches the chromosomes to a spindle fiber during mitosis and meiosis ) - chromosomes 1 & 4 have the centromeres located at the ends of their chromosomes - chromosomes 2 & 3 have centromeres located in the middle - X chromosome  appearance of a straight rod - chromosomes 2 &3 have V shaped appearance ( have a right arm and a left arm) - Chromosome 4 is small and dot-like ( in males we have a J-shaped Y chromosome) Ioana D. 2010 - All chromosomes join one another in a large unbanded mass called chromocenter consisting of all centromeres of the salivary gland - If male – the Y chromosomes are located in the chromocentere (therefore, invisible) - the X chromosome is thin, difficult to discern - Chromosome 4 is also difficult to distinguish b/c of its small size Exp 3a – Using Chromatography to characterize Drosophila melanogaster mutants - Genes regulate cellular chemistry by controlling the synthesis of enzymes (one – gene- one – enzyme hypothesis) - Eye color in Drosophila is due to pigments deposited in the outer region of the compound eye - Pigments are produced in a series of steps, each catalyzed by a different enzyme - An inactive enzyme due to a gene mutation leads to an increase in substrate concentration - If substrate is pigmented, the color is added to the color of other pigments present -> resultant mixture determines the eye color of the fly 2 major eye pigments: - Brown pigments (ommochromes)  tryptophan derivatives - Red pigments (pteridines)  7 of them in wild-type drosophila - Wild-type color : red-brick - The 2 classes of pigments are produced through separate biosynthetic pathways - Brown eyed mutants: lack red pigments - Reddish eyed mutants: lack brown pigment - White eyed mutants: lack both pigments - In sepia mutants, we have large amounts of sepiapterin present in the eye Ioana D. 2010 - The flies with eye- color mutations have pteridine patterns that are different from that of wild-type flies Paper chromatography - Allows us to characterize the chemical differences b/w Drosophila’s having different genetic constitutions, in terms of eye color - Fruit flies heads are crushed along one edge of a rectangular piece of filter paper - Paper is then placed in appropriate solvent, which is drawn up the paper by capillary action - Solvent passes through the spots where fruit fly heads have been crushed, and dissolve substances soluble in it - The dissolved substances are separated from one another depending on their chemical and physical properties - the pteridines are light-sensitive and can be detected under a UV lamp ** through paper chromatography we can show that a wild-type heterozygote (with normal eye color), is not fully normal at all ** recessive mutations (e.g. X-linked white eyes) affect the pteridine profiles of the wild-type heterozygotes - the distance traveled on the filter paper is an identifying feature of a compound - Rf (ratio-to-front-value): = distance from base line to centre of pteridine spot/ distance from base line to solvent front ** sex differences exist with respect to pteridines. Therefore, the analysis must be carried out on the same sex ** the filter paper must not be touched with the fingers ** solvent in the contained must not touch the spots where the flies have been crushed, nor should the filter paper cylinder touch the sides of the contained Ioana D. 2010 Lab results: - There are no notable differences in Rf were noticed b/w male and female individuals because there are no variations in the chemical structures of pigments between genders - In wild-type Drosophila, all 8 pigments are present - In vermillon mutants ommochrome is absent, only orange pteridines being present - Ommochrome was only visible under normal light - Drosopterins and sepiapterins were visible under both UV and normal light - Isosepiapterin highest migration distance - Ommochrome  lowest migration distance Visible colors: - Ommochrome (doesn’t travel. It’s insoluble in solvent) = brown - Drosopterin = orange - Sepiapterin (travels most) = yellow Under UV, Isosepiapterin travels the most  Yellow fluorescence Experiment # 3b – Lethal mutations and epistasis - When selecting a lethal mutation, FlyLab makes the fly heterozygous for the trait - FlyLab does not reduce the # of offspring produced (when a lethal mutation is involved) - It simply rescales the probabilities among existing phenotypes i.e # of offspring remains the same Cy - Cy  Female : Cy+ Cy x Male: Cy+ Cy F1 Cy+ Cy Cy+ Cy+Cy+ Cy+Cy Cy Cy+Cy CyCy (die) Only 3 out of the 4 phenotypes survive Therefore: 2:1 phenotypic ratio F2: F1 x F1 i.e Cy+Cy x Cy+Cy (same thing as above) Ioana D. 2010 Cy+ Cy Cy+ Cy+Cy+ Cy+Cy Cy Cy+Cy CyCy (die) Phenotypic ratio : 2:1 ---------------------------------- If we cross a wild type with a mutant possessing a lethal mutation: F1 : Cy+Cy+ x Cy+Cy Cy+ Cy+ Cy+ Cy+Cy+ Cy+Cy+ Cy Cy+Cy Cy+Cy Phenotypic ratio: 1:1 F2: F1 x F1 i.e Cy+Cy x Cy+Cy Cy+ Cy Cy+ Cy+Cy+ Cy+Cy Cy Cy+Cy CyCy (die) Phenotypic ratio: 2:1 -------------------------------------- Now we mate flies that possess 2 lethal mutations: Cy/AR – Cy/AR i.e Female = Cy+Cy AR+AR x Male = Cy+Cy AR+AR Cy+AR+
More Less

Related notes for BIOL 2104

Log In


Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.