Lecture 2 Using Mutants to study Biological processes

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. 1993. Mutant analysis, a key tool for the study of metabolism and development. Plant Journal 3: 303 -308. • Westhoff et. al. Molecular Plant Development: from gene to plant. Chapter 3: 39 -65.

Arabidopsis Genome Annotated Gene Function Year 2018 - 33, 602 genes Annotation = assignment of gene function based on sequence alone. # of genes investigated in detail (phenotype, biochemistry, cell biology, 3 molecular biology) is only a few thousand.

Why use mutants? Pick the best answer! ***i. Clicker! A mutant phenotype is valuable because: A. it can be used to identify genes involved in specific biological processes. B. it can be used to associate a gene sequence with a specific biological process. C. it can be used to find the DNA sequence of a gene defined by a phenotype. D. it may be useful commercially. E. All of the above.

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 -use phenotype to find associated DNA sequence The identification of a gene by DNA sequence = reverse genetics -use sequence to find a mutant with a defect in that gene

Forward genetics is a powerful approach to understand gene function? Researchers need both phenotypic and biochemical (protein function) information about their gene to understand its role. Forward genetics—isolation of mutants affecting a specific aspect of biology The mutant phenotype provides: -- 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 (M 1 generation)

Mutagenesis in Arabidopsis Mutagenized cells are heterozygous in diploids Westhoff Fig. 3. 1

Mutagenesis: generation for screening **i. Clicker!! A mutagen induces multiple mutations in the genome of the cells exposed (In Arabidopsis = M 1 generation). Arabidopsis is a diploid organism that reproduces by selffertilization. What generation would you normally screen to identify homozygous mutants? A. B. C. D. E. M 0 M 1 M 2 M 3 M 4

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 (M 1 generation) The M 1 plants are not typically screened for mutant phenotypes-Why?

Mutagenesis in Arabidopsis Westhoff Fig. 3. 1

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 (M 1 generation; mutations are heterozygous in diploids). --Those mutations in germline cells are passed on to the next generation (M 2 generation). --In plant species that self fertilize (eg. Arabidopsis) the M 2 population will include some plants homozygous for mutations. Therefore: The M 2 generation is typically screened for mutant phenotypes

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 (abbreviation) = = GEORGE (uppercase, italics) 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 (M 2) 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 What are the competing hypotheses? How do you test your hypothesis?

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 F 1 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 F 1 plants and determine the number and type of mutant phenotypes among the F 2 progeny.

Genetic analysis **i. Clicker!! You screened a large mutagenized (M 2) population and found three plants with curled leaves, sepals and petal (Crl plants). One mutant you cross to wild type. The F 1 progeny appears wild type. In the F 2 you find 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 You can conclude that the phenotype is: A. B. C. D. E. due to a recessive mutation in a single nuclear gene. due to a dominant mutation in a single nuclear gene. due to a recessive mutation in each of two nuclear genes. due to a dominant mutation in each of two nuclear genes. due to a dominant mutation in one nuclear gene and a recessive mutation in a second nuclear gene.

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 F 1 plants and determine the number and type of mutant phenotypes among the F 2 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.

Make a prediction based on the data ***i. Clicker! You have a crl phenotype that segregates like a single recessive nuclear mutation. In the crl mutant you discover a homozygous missense mutation in the DNA sequence of a gene (X) and want to know if the missense mutation causes the phenotype. You cross the crl mutant to wild type and self the F 1 and select four crl mutants from the F 2 progeny. You sequence gene X in each of the mutants and find that two of the mutants are homozygous for the mutation and two are heterozygous for the mutation? You can conclude that the missense mutation in gene X: A. is the cause of the crl phenotype. B. is not the cause of the crl phenotype. C. may be the cause of the phenotype but there is uncertainty.

Basic Genetic Analysis of Mutants (Co)Segregation analysis: If two aspects of a phenotype (eg. Curly leaves, curly petals) segregate together (if all plants with curly leaves also have curly petals and vice versa) an F 2 population 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 F 2 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

Make a prediction based on the data ***i. Clicker! If you cross two mutants with the same phenotype that are homozygous for different recessive mutations in the same crl gene, what phenotype do you expect? A. Wild type B. crl C. Phenotype intermediate between wild type and crl D. Impossible to tell E. I forgot my iclicker today

Make a prediction based on the data ***i. Clicker! If you cross two crl mutants with the same phenotype that are homozygous for recessive mutations in different genes what phenotype do you expect? A. Wild type B. crl C. Phenotype intermediate between wild type and crl D. Impossible to tell E. I forgot my iclicker today

Complementation test Phenotype Mutant 1 crl Mutant 2 crl Mutant 3 crl All mutants are recessive to wild type. All segregate as single nuclear genes. Cross every mutant to every other mutant. The following table shows the phenotypes of the F 1 progeny of crossing two of the three unc mutants. Parent 2 Parent 1 cross crl#1 crl#2 crl#3 crl#1 crl wild type crl#2 crl#3 crl How many new genes have been identified? Possibilities: 1, 2, 3

Make a prediction based on the data ***i. Clicker! How many crl genes have been identified by the complementation test data in the previous slide? A. 0 B. 1 C. 2 D. 3

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 1 Crl Mutant 2 Crl Mutant 3 Crl All mutants are recessive to wild type. All segregate as single nuclear genes. How many genes have been identified? Possibilities: 1, 2, 3 Test Mutant 1 Crl x Mutant 2 Crl Deduction Mutant 1 crl 1 -1/ crl 1 -1 x Mutant 2 crl 1 -2/ crl 1 -2 Result F 1 Crl F 1 crl 1 -1/ crl 1 -2 Conclusion: Mutants 1 and 2 fail to complement and must be homozygous for mutations in the same gene

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