Unformatted text preview: 7.03 Genetics – Lectures 33-36 TA: Yun Song 912-484-2511 | [email protected] Main Concepts Metazoan Genetic Analysis - Mutant Screens o Dominant / Recessive Mutations o X-Linked Mutations o C. elegans – hermaphrodites: simplification of screen for recessive mutations o Markers / balancer chromosomes Characteristics of balancers • Suppress recombination between homologs during meiosis • Dominant visible phenotype • Recessive lethality OR recessive visible phenotype P0: +/+ x BAL1/BAL2 F1: m1/BAL1 x BAL1/BAL2 F2: m1/BAL1 x m1/BAL1 F3: P’types: - ¼ m1/m1 no dom BAL p’type ½ m1/BAL1 dom BAL p’type ¼ BAL1/BAL1 dead OR Characterize Mutation o Mode of inheritance: dominant/recessive test o Nature of the allele Mode of Inheritance Recessive Dominant Nature of Allele Partial Null Hypermorph Neomorph Dominant-negative Haploinsufficient Loss/Gain of Function Loss Loss Gain Gain Loss Loss 7.03 Genetics – Lectures 33-36 TA: Yun Song 912-484-2511 | [email protected] Dosage Experiments: Recessive allele m1/m1 x Df/BAL Degree of Mutant P’type m1/Df > m1/m1 m1/Df = m1/m1 m1/Df Dominant allele m1/+ x Dp/BAL Degree of Mutant P’type m1/Dp > m1/+ m1/Dp < m1/+ m1/Dp Nature of allele Hypermorph Dom-neg Haploinsufficient Nature of allele Partial Null Determine Gene Number o Recessive Mutations: complementation test o Dominant & Recessive Mutations: cis/trans test Determine Gene Function EX: Mutant phenotype = tall Mode of Inheritance Nature of Allele Normal Gene Function Recessive Dominant Partial lof / null Hypermorph Neomorph Dominant-negative Haploinsufficient Inhibit growth Promote growth ?? Inhibit growth Inhibit growth - - Determine Gene Location o Linkage studies: determine linkage to other known mutations or polymorphisms o Rescue experiment: determine if introduction of exogenous copy (e.g. via transgenesis) of gene restores wild-type phenotype Determine Expression: reporter genes Alternate Perturbation Techniques (w/o generating mutants) o RNAi: dsRNA used to knock down gene expression o Introduction of dominant-negative allele of protein via transgenesis Developmental Genetics - Maternal effects: mother’s genotype determines progeny phenotype because gene product is required during early embryonic development - Mosaic: animals with cells of different genotypes, but genotypes derived from same source o How are mosaics generated? Mitotic recombination o Use for mosaics: study location (which tissue) gene product is required 7.03 Genetics – Lectures 33-36 TA: Yun Song 912-484-2511 | [email protected] - Cell Autonomy o cell autonomous: gene product required in cell in which gene is transcribed o cell nonautonomous: gene product required in neighboring cell WT Gene Function - Fix damaged DNA - Promote apoptosis - Regulate cell cycle - Promote cell growth Gain/Loss of F’n Loss of function – Recessive Gain of function – Dominant Region Mutation Occurred Coding region, cisregulatory region Cis-regulatory region Cancer Type of Gene Tumor Suppressor Oncogene Questions Based on Lecture 1. Mutant Screens You are working with Drosophila melanogaster and would like to isolate recessive mutations on chromosome 2. melanogaster nd following isolate recessive mutations on 6) You are working with Drosophila You performathewould like tomutagenesis screen (see schematic below).
chromosome 2. You perform the following mutagenesis screen (see schematic below). You mutagenize male flies mutagenize male flies gene, which confers a cinnabargene, which confers a cinnabar eye. You You carrying the recessive cn carrying the recessive cn eye. You mate the mutagenized males to females heterozygous for SM6, females heterozygous carrying the Cy mate the mutagenized males to a chromosome 2 balancer for SM6, a chromosome 2 balancer marker, which confers a dominant curly wing which confers a as a mutation in cinnabarphenotype as well as a mutation in carrying the Cy marker, phenotype as well dominant curly wing (cn), which recessively confers cinnabar eyes. Certain F1 flies are mated to SM6 heterozygotes to cinnabar (cn), which recessively confers cinnabar eyes. Certain F1 flies are mated to SM6 obtain an F2 generation. F2 siblings are crossed to obtain F3 animals that can be screened for heterozygotes to recessive mutations induced in P0. obtain an F2 generation. F2 siblings are crossed to obtain F3 animals that can be screened for recessive mutations induced in P0.
cn/cn (x) SM6/+ males females F1 offspring (x) SM6/+ F2 sibling crosses F3 generation A. What is the phenotype of the F1 animals that you cross to SM6/+? Explain. m1,cn / +,SM6 = cinnabar eyes, curly wings
A. What is the phenotype of the F1 animals that you cross to SM6/+? Explain. Bsiblings would you would you choose for the F2 sibling crosses B. Which . Which siblings choose for the F2 sibling crosses (both genotype and(both genotype and phenotype)? phenotype)? / +,SM6 = cinnabar eyes, curly wings m1,cn C. List all F3 genotypesgenotypes and with ratios. C. List all F3 and phenotypes, phenotypes, with ratios. What phenotype would you select to D. Describe the marker phenotypes you would use to identify the flies with your mutations acquire mutant? in the F3 generation. ¼ m1,cn / m1,cn = cinnabar eyes olegless mutant we want E. You isolate 3 mutations that recessively confer a nly phenotype. Describe with ½ (note: legless flies cannot mate) how you would crosses m1,cn / +,SM6 = cinnabar eyes, curly place these mutations into ¼ SM6 / SM6 = (show the eyes, curly complementation groupscinnabar phenotypes that allow you to identify flies of interest with corresponding genotypes). F. You find that all three mutations fail to complement. To map one sample mutation, you perform crosses to two deficiency lines (the lines are kept as heterozygotes: Df1/cn and Df2/cn; note that Df1 and Df2 delete two different regions of chromosome 2 and do not delete cn). Describe your crosses and show the phenotypes that allow you to identify flies of interest with corresponding genotypes. C. Possible Genotypes (occurring in a 1:2:1 ratio) from F2 sibling cross: m cn/m cn 7.03 Genetics – Lectures 33-36 m cn/SM6 TA: Yun Song SM6/SM6 912-484-2511 | [email protected] D.E. Youmutant mutationscn/ m cn would have cinnabar eyes but NOT curly wings. The isolate 3 flies m that recessively confer a legless phenotype. Describe with crosses
(note: legless flies cannot mate) how you would place these mutations into complementation groups (show the phenotypes that allow you to identify flies of interest with corresponding E. gSet up the following crosses: enotypes). If any non-curly, cinnabar-eyed flies do not have a legless phenotype, then those two mutants of the complement and are in eyes (but WITHOUT curly wings) have If any alleles flies with cinnabar different genes. a legless phenotype, then the two mutations m1 and m2 (or m1 and m3, or m2 and m3) fail to complement each other and are therefore in the same gene. If flies with cinnabar eyes do not have the legless phenotype, then the mutations complement each other 2. Characterizing Mutant Alleles and are on different genes. You are studying animal Z and are interested in finding the genes responsible for its leg F. mutants, all with leg defects. Set up the following crosses: m1 cn/SM6 (x) Df1/cn and m1 cn/SM6 (x) Df2/cn. You want to analyze flies lacking both cinnabar eyes and curly wings (genotypes m1 cn/Df1 Animals homozygous for m1 (genotype m1/m1) have 2 legs. andomozygous m2 animals have 16 legs. H m1 cn/Df2). If m1 and Df1 cover same gene, then those flies will have the legless phenotype. If they do not cover the same gene, then they will be phenotypically A. Given normal. the information from the following crosses, determine whether these mutations are
dominant or recessive. m1 (x) WT yields progeny with 8 legs. m2 (x) WT yields progeny with 14 legs. m1 is recessive, m2 is dominant You are interested in determining what type of mutations you have isolated and decide to perform dosage studies. (Df = deficiency, Dp = duplication, BAL=balancer; the alleles used cover the gene of interest, as determined by prior mapping.) B. You isolate all of the non-balancer progeny from m1/m1 (x) Df1/BAL crosses. These animals (genotype m1/DF1) have 2 legs. Propose the type of mutation/allele that m1 is. m1 is null allele C. All of the non-balancer progeny from m2/m2 (x) Df3/BAL crosses have 12 legs. [Df3/BAL animals have 8 legs.] All of the non-balancer progeny from m2/m2 (x) Dp3/BAL crosses have 18 legs. Propose the type of mutation/allele that m2 is. m2 is hypermorphic D Crosses between m1/m1 and m2/m2 yield heterozygous progeny that have 12 legs. What does this suggest about the relationship between m1 and m2? m1 and 2007 m2 are allelic Problem Set 1 Answer Key formation. Animal Z normally has 8 legs. You perform a mutagenesis screen and isolate two 7.22 Fall 6 7.03 Genetics – Lectures 33-36 TA: Yun Song 912-484-2511 | [email protected] 3. Cancer You are studying a family in which HNPCC, a type of colon cancer, is prevalent. Colon cancer is prevalent in this family because of the germline transmission of a mutation in the MLH1 gene, which normally encodes a protein that helps to fix DNA damage. If you analyzed a non-cancerous skin cell from one of the affected individuals: How many WT alleles of MLH1 would be present? How many mutant alleles of MLH 1 would be present? 1 1 Of these, how many would have resulted form somatic mutations? 0 If you analyzed a tumor cell from one of the affected individuals: How many WT alleles of MLH1 would be present? How many mutant alleles of MLH1 would be present? 0 2 Of these, how many would have resulted form somatic mutations? 1 ...
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