Lecture 2: Using Mutants to study Biological processes Objectives: 1. Why use mutants? 2.How are mutants isolated? 3. What important genetic analyses must.

Slides:



Advertisements
Similar presentations
Multiple Choice Questions
Advertisements

Single Gene Inheritance
Lecture 2 Section 2 Exceptions to Mendelian Genetics Today 1.Administrative issues 2.Mopping up concepts and terminology 3.Problems solving 4.Comments.
Review Mendel’s “rules of the game”
Genetics notes For makeup. A gene is a piece of DNA that directs a cell to make a certain protein. –Homozygous describes two alleles that are the same.
Segregation, Assortment, and Dominance Relationships A.Genes and alleles B.Random segregation C.Independent assortment D.Assortment vs. Linkage E.Dominance.
Heredity and Evolution
David Sadava H. Craig Heller Gordon H. Orians William K. Purves David M. Hillis Biologia.blu B – Le basi molecolari della vita e dell’evoluzione Genetics:
Key Area : Genetic Control of Metabolism in Micro-organisms Unit 2: Metabolism and Survival.
Patterns of inheritance
Genetics can be used to characterize biological pathways Epistasis tells which gene products are involved in common pathways and which act earlier or later.
Forward Genetics What is forward genetics?
© 2005 Prentice Hall Inc. / A Pearson Education Company / Upper Saddle River, New Jersey Genomics Book (20 chapters)  Chapter 7: High throughput.
Competency Goal # 3 The learner will develop an understanding of the continuity of life and the changes of organisms over time. (Weight 28%) Part 3.
Bioinformatics Lecture 2: molecular biology. Essential concepts of evolution – The Basic Tenets of evolution: Adaptability and stability in an environment.
1.Generate mutants by mutagenesis of seeds Use a genetic background with lots of known polymorphisms compared to other genotypes. Availability of polymorphic.
Mendel wondered if genes that determine different traits affect one another. He did an experiment to find out. Mendel found that the gene for seed shape.
Rules: Cell phones off Computers only for class-related work No food or drink in lab room Text Book: Hartwell et al Genetics from Genes to Genomes, third.
Genetics Chapters 9 and 12.
2 March, 2005 Chapter 12 Mutational dissection Normal gene Altered gene with altered phenotype mutagenesis.
Microbial Genetics (Micr340) Lecture 6 Genetic Analysis.
Rules: Cell phones off Computers only for class related work No food or drink in lab room Text Book: Hartwell et al., 2nd Edition 2004 Genetics from Genes.
Plant Thermotolerance - hot1-3 (HSP101) HORT 301 9/30/2008.
Chapter 5 Heredity.
Genetics Chapter 11.
Chapter 11 Introduction to Genetics. Chromosomes and Cells Two general types of cells –Somatic cells-body cells that make up the tissues and organs –Gametes-sex.
Unit 4 Vocabulary Review. Nucleic Acids Organic molecules that serve as the blueprint for proteins and, through the action of proteins, for all cellular.
Using mutants to clone genes Objectives 1. What is positional cloning? 2.What is insertional tagging? 3.How can one confirm that the gene cloned is the.
Chapter 7: Genes and Inheritance Family resemblance: how traits are inherited Lectures by Mark Manteuffel, St. Louis Community College.
 Explain how genes are responsible for certain traits that we acquire from our parents?  In complete sentences, explain meiosis and explain its role.
Forward genetics and reverse genetics
Introduction to C. elegans and RNA interference Why study model organisms?
Fig Chapter 12: Alternative approaches to mutational dissection.
Genome-Scale Mutagenesis Introduction Model systems –Yeast –Mouse Implications for science.
Module II Mendelian Genetics & Probability Theory.
Classical Genetics The Legacy of Gregor Mendel Or The Monk with the Missing “Peas”
Pea plants have several advantages for genetics.
INTRODUCTION TO GENETICS
FOOTHILL HIGH SCHOOL SCIENCE DEPARTMENT Chapter 11 Introduction To Genetics Section 11-3 Exploring Mendelian Genetics.
Ch. 11 Introduction to Genetics
Mendelian Genetics Genetics Lecture III. Biology Standards Covered 2c ~ students know how random chromosome segregation explains the probability that.
The study of patterns of inheritance and variations in organisms.
Mendel and Meiosis Chapter 11. Mendel’s Laws of Heredity Heredity – passing on traits from parents to offspring Gametes – sex cells; they have a haploid.
Mendelian Genetics AP Biology Unit 3 Mendel’s Experiments Crossbred Pea Plants P, F1, F2 generations.
BIOL 433 Plant Genetics Term 2, Instructors: Dr. George Haughn Dr. Ljerka Kunst BioSciences 2239BioSciences Tel
Exam II Lectures and Text Pages I. Cell Cycles – Mitosis (218 – 228) – Meiosis (238 – 249) II. Mendelian Genetics (251 – 270) III. Chromosomal Genetics.
Mutations to Aid in Gene Study By: Yvette Medina Cell Phys
6.3 Mendel and Heredity KEY CONCEPT Mendel’s research showed that traits are inherited as discrete units.
Molecular Characterization of an Unknown P-Element Insertion in Drosophila melanogaster.
Chapters 13 & 14 GENETIC ENGINEERING & THE HUMAN GENOME.
General Genetics Chapter 14 Mendel and the Gene Idea.
 Structural genes: genes that contain the information to make a protein.  Regulatory genes: guide the expression of structural genes, without coding.
Unit 3 Lesson 4 Heredity Copyright © Houghton Mifflin Harcourt Publishing Company.
Finding a gene based on phenotype Model organisms ’s of DNA markers mapped onto each chromosome – high density linkage map. 2. identify markers linked.
Chapter 2: Single Gene Inheritance Jan 10 th 2014 Office Hours: Thurs & Fri 9:00 – 10:00 am.
Genetics Mendelian Genetics Genetic Engineering. Gregor Mendel Used pea plants to experiment on genetic traits Pea plants can self-pollinate, producing.
BIOL 433 Plant Genetics Term 2,
IB syllabus 3.4.U2 Gametes are haploid so contain only one allele of each gene. 3.4.U3 The two alleles of each gene separate into different haploid daughter.
Sequence recognition P(M|s) = P(M)* P(s|M) / P(s) (Bayes' theorem)
Map-based cloning of interesting genes
Mendelian Genetics Section 6.3, 6.4. and 6.5.
Relationship between Genotype and Phenotype
Relationship between Genotype and Phenotype
INTRODUCTION TO MOLECULAR GENETICS
Lecture 2: Using Mutants to study Biological processes
BIOL 433 Plant Genetics Term 2,
The Basis of Heredity.
Using mutants to clone genes
INTRODUCTION TO MOLECULAR GENETICS
Presentation transcript:

Lecture 2: Using Mutants to study Biological processes Objectives: 1. Why use mutants? 2.How are mutants isolated? 3. What important genetic analyses must be done immediately after a genetic screen for mutants?

Reading References: Cove Mutant analysis, a key tool for the study of metabolism and development. Plant Journal 3: Westhoff et. al. Molecular Plant Development: from gene to plant. Chapter 3:

Information obtained from sequencing the Arabidopsis genome Genome size: 115 Mbp sequenced + 10 Mbp highly repetitive: (rDNA, Centromere) = 125 Mbp Sequence is 99.99% accurate

Annotation Identification of genes: How? Use known sequence features of genes to predict: Open reading frames, Splice junctions, promoter elements, base composition, translation initiation sites. Refine with cDNA sequence. Predict: Arabidopsis 27,000 (estimates) C. elegans19,000 D. melanogaster14,000

Gene function Define gene function on the basis of Biochemical role Based on sequence similarity to known proteins Biological process? Cellular function? Plant Functional Genomics Chris Somerville* and Shauna Somerville SCIENCE VOL non-coding

Scope of the problem All genes in Arabidopsis have been identified through genome sequencing so cloned genes are readily accessible. But of 27,000 genes:# genes Known biochemical/ cellular/biological function 1,000 Predicted biochemical function from annotation16,000 Unknown10,000

Why use mutants? Researchers need both phenotypic and biochemical (protein function) information about their gene to understand its role. The identification of a gene by mutant phenotype = forward genetics. Using a cloned gene to find a mutant phenotype = reverse genetics

Forward genetics Goal: Identify genes with a role in a specific biological process by finding mutants defective in that process. Mutant phenotypes provide -- information concerning the role of the gene in vivo. -- method by which to clone the gene that is mutated.

Mutagenesis 1. Mutants are generated by exposing a population of organisms to a mutagen and allowing the individuals in the population to reproduce. mutagens = irradiation (UV, Xray, fast neutron, etc.), chemicals (ethyl methane sulfonate, nitrosoguanidine etc.), insertional elements (transposons, TDNA) 2. The mutagen induces multiple mutations in the genome of the cells exposed (M1 generation)

Mutagenesis in Arabidopsis Westhoff Fig. 3.1 Mutagenized cells are heterozygous in diploids

Mutagenesis 1. Mutants are generated by exposing a population of organisms to a mutagen and allowing the individuals in the population to reproduce. mutagens = irradiation (UV, Xray, fast neutron, etc.), chemicals (ethyl methane sulfonate, nitrosoguanidine etc.), insertional elements (transposons, TDNA) 2. The mutagen induces multiple mutations in the genome of the cells exposed (M1 generation) The M1 plants are not typically screened for mutant phenotypes-Why?

Mutagenesis 1. Mutants are generated by exposing a population of organisms to a mutagen and allowing the individuals in the population to reproduce. mutagens = irradiation (UV, Xray, fast neutron, etc.), chemicals (ethyl methane sulfonate, nitrosoguanidine etc.), insertional elements (transposons, TDNA) 2. The mutagen induces multiple mutations in the genome of the cells exposed (M1 generation; mutations are heterozygous in diploids). --Those mutations in germline cells are passed on to the next generation (M2 generation). --In plant species that self fertilize (eg. Arabidopsis) the M2 population will include some plants homozygous for mutations. Therefore: The M2 generation is typically screened for mutant phenotypes

Mutagenesis in Arabidopsis Westhoff Fig. 3.1

Genetic Nomenclature (Arabidopsis, yeast) A gene is typically named after the mutant phenotype or the biological function with which it was identified. Mutant phenotype=George or george (italics) Gene name=GEORGE (uppercase, italics) (abbreviation)=GEO mutant alleles=geo-1(lowercase italics) geo-2 geo-3 dominance/recessiveness is not indicated protein=GEO(uppercase, no italics)

Basic Genetic Analysis of Mutants (what you should know genetically about any mutant you find) You screened a large mutagenized (M2) population and found three plants with curled leaves, sepals and petals. Hypothesis: Each plant is homozygous for a recessive allele of a single nuclear gene that is needed for the leafy organs of the plant to develop normally. Phenotype = Curly Leaf (Crl) Gene = CRL alleles = crl-1, crl-2, crl-3 How do you test your hypothesis? What are the competing hypotheses?

Basic Genetic Analysis of Mutants 1. Is the mutant phenotype heritable? Allow the plant to self fertilize. Does the phenotype show up in the next generation? Yes  heritable No  phenotype is probably not due to mutation.

Basic Genetic Analysis of Mutants 2. Is the mutant phenotype due to a recessive, codominant or dominant mutant allele? Cross the mutant plant to wild type. If the F1 progeny phenotype is: i) wild type then the mutant allele is recessive (most common) ii) mutant then the mutant allele is dominant iii) intermediate between wild type and mutant then the mutant allele is co-dominant.

Basic Genetic Analysis of Mutants 3. Is the phenotype of a each mutant due to mutation of one or more than one nuclear genes? Self fertilize the F1 plants and determine the number and type of mutant phenotypes among the F2 progeny. ¾ wild type, ¼ curly leaves,petals,sepals (mutant allele recessive) or ¾ curly leaves,petals,sepals, ¼ wild type (mutant allele dominant) =single nuclear mutation 9/16 wild type, 3/16 curly leaves(normal petals and sepals), 3/16 curly sepals and petals (normal leaves), 1/16 curly leaves, petals, sepals = two nuclear genes, one required for normal leaf development and another required for normal floral organ development.

Basic Genetic Analysis of Mutants (Co)Segregation analysis: If two aspects of a phenotype (eg. Curly and white leaves) and segregate together (if all plants with curly leaves also have white leaves and vice versa) an F2 population then the mutation(s) causing the phenotypes are closely linked and may be caused by a single mutation. If two aspects of a phenotype can be observed separately in an F2 population (plants with only curly or white leaves) then they are not caused by the same mutation and are due to mutations in at least two different genes (a single recombinant would indicate that two traits are not due to the same mutation).

Basic Genetic Analysis of Mutants Three Crl mutants were found. Do they represent mutations in three different genes or three alleles of the same gene? If the mutants are recessive to wild type and the phenotype segregate as a single nuclear gene then the question can be answered by a: Complementation test

The idea of a complementation test is that for a mutant that is homozygous for a recessive (loss-of function) mutation that results in a phenotype, the phenotype can be rescued (complemented) if at least one normal (wild type) copy of the gene is introduced. A normal copy of the gene can be introduced by crossing the mutant to a wild type plant (classical complementation) or introducing a copy by transformation.

Complementation test Mutant 1Mutant 2Mutant 3 CrlCrlCrl All mutants are recessive to wild type. All segregate as single nuclear genes. How many genes have been identified? Possibilities: 1, 2, 3 Test Deduction Mutant1 xMutant 2Mutant1 xMutant 2 CrlCrl crl1-1/ crl1-1 crl1-2/ crl1-2 ResultF1 Crl crl1-1/ crl1-2 Conclusion: Mutants 1 and 2 fail to complement and must be homozygous for mutations in the same gene

Complementation test Test Deduction Mutant 1 xMutant 3Mutant1 x Mutant 3 CrlCrl crl1-1/ crl 1-1CRL 1/ CRL 1 CRL 2/ CRL 2crl 2-1/ crl 2-1 Result F1 Normal leaves, sepals,petals crl-1/ CRL1 CRL2/ crl2-1 Conclusion: Mutants 1 and 3 are in different To have a wild type phenotype the F1 complementation groups. Mutants must be heterozygous for mutant Are homozygous for mutant alleles alleles of two different genes. of different genes. Conclusion: The three Crl mutants identify two different genes required for normal leafy organs. CRL1 and CRL2