Lecture 9 By way of review, let's consider the general question of how genetic mapping studies can be used to locate a gene that has been identified by an allele with an interesting phenotype. For example, the CLOCK mutation in mouse was identified as a semi-dominant mutation that disrupts the normal circadian rhythm this mutation was isolated after mutagenesis of animals with ethyl-nitorosourea and then screening their offspring for abnormal activity at a time when normal mice would be sleeping Lets assume that we have at our disposal a large number of genetic markers spaced evenly along all of the mouse chromosomes (later in the course we will discuss DNA-based markers which can be found in an almost unlimited number in the mammalian genome) and that we have the capability to screen about 1000 mice for recombinants in a mapping experiment. The question of how precisely we can locate the CLoCK gene can be considered to be a question of the resolving power of a mapping experiment. Two points on a genetic map can only be resolved if a recombination event that separates them can be found. The smallest interval that can be resolved on average if 1000 progeny screened is 1 recombinant in 1000 which corresponds to a map distance of w 0.1 cM on either side of the CLoCK mutation. Thus we would be able to map the CLOCK mutation to an interval of o2 cm In mice, and in humans, a simple rule of thumb is that 1 cM corresponds to a physical distance of about 1 Mbp. Therefore our mapping experiment would locate the CLOCK gene to a region of dna about 200 kbp in size. Such an interval may contain two to ten gene sequences so additional work using recombinant DNA methods would be needed to precisely identify the CLOCK gene. However, there is still no substitute for recombinational mapping studies as a way to initially narrow down the approximate location of a gene
Lecture 9 Lecture 9Lecture 9 By way of review, let’s consider the general question of how genetic mapping studies can be used to locate a gene that has been identified by an allele with an interesting phenotype. For example, the CLOCK mutation in mouse was identified as a semi-dominant mutation that disrupts the normal circadian rhythm. This mutation was isolated after mutagenesis of animals with ethyl-nitorosourea and then screening their offspring for abnormal activity at a time when normal mice would be sleeping. Let’s assume that we have at our disposal a large number of genetic markers spaced evenly along all of the mouse chromosomes (later in the course we will discuss DNA-based markers which can be found in an almost unlimited number in the mammalian genome) and that we have the capability to screen about 1000 mice for recombinants in a mapping experiment. The question of how precisely we can locate the CLOCK gene can be considered to be a question of the resolving power of a mapping experiment. Two points on a genetic map can only be resolved if a recombination event that separates them can be found. The smallest interval that can be resolved on average if 1000 progeny are screened is 1 recombinant in 1000 which corresponds to a map distance of ~ 0.1 cM on either side of the CLOCK mutation. Thus we would be able to map the CLOCK mutation to an interval of 0.2 cM. In mice, and in humans, a simple rule of thumb is that 1 cM corresponds to a physical distance of about 1 Mbp. Therefore our mapping experiment would locate the CLOCK gene to a region of DNA about 200 kbp in size. Such an interval may contain two to ten gene sequences so additional work using recombinant DNA methods would be needed to precisely identify the CLOCK gene. However, there is still no substitute for recombinational mapping studies as a way to initially narrow down the approximate location of a gene
To review how we can categorize tetrads consider the following cross Trp1- Leu2+ x Trpl- Leu2 Trpl-Leu2+, Trpl*Leu2-, Trpl*Leu2*, Trpl-Leu2 This is a Tetratype Trpl-Leu2+ Trpl-Leu2*, Trpl*Leu2-, Trpl*Leu2 This is a PD Trpl*Leu2+, Trpl*Leu2*, Trpl-Leu2-, Trpl-Leu2 This is an NPD Now consider a cross that involves mutations that have the same phenotype Leu1-Leu2+ x Leu1+Leu2 Notice that even though we can't always unambiguously assign genotypes to each spore clone we can distinguish the three tetrad types Tetratype genotype ph phenotype genotype ph phenotype genotype phenotype 1+2 Leu Leut Le 12 1+2- Le 1+2- Leu 12 Lc 12+ Le 1+2 12- Lc Here are the steps used in tetrad analysis cOunt tetrads of each type to get four numbers: PD, T, NPD, total ii) If the classes are in the following ratio: PD: T: NPD the genes are unlinked and you are finished IO) If PD >> NPD, then the genes are linked and the distance between them is given by T+6NPD Distance in CM 100
To review how we can categorize tetrads consider the following cross: Trp1– Leu2+ x Trp1+ Leu2– Trp1–Leu2+, Trp1+Leu2–, Trp1+Leu2+, Trp1–Leu2– This is a Tetratype TetratypeTetratype Trp1–Leu2+, Trp1–Leu2+, Trp1+Leu2–, Trp1+Leu2– This is a PD Trp1+Leu2+, Trp1+Leu2+, Trp1–Leu2–, Trp1–Leu2– This is an NPD Now consider a cross that involves mutations that have the same phenotype. Leu1–Leu2+ x Leu1+Leu2– Notice that even though we can’t always unambiguously assign genotypes to each spore clone we can distinguish the three tetrad types. Tetratype TetratypeTetratype PD NPD genotype phenotype genotype phenotype genotype phenotype 1+2+ Leu+ 1–2+ Leu– 1+2+ Leu+ 1–2– Leu– 1–2+ Leu– 1+2+ Leu+ 1+2– Leu– 1+2– Leu– 1–2– Leu– 1–2+ Leu– 1+2– Leu– 1–2– Leu– Here are the steps used in tetrad analysis: i) Count tetrads of each type to get four numbers: PD, T, NPD, total ii)If the classes are in the following ratio: PD : T : NPD 1 : 4 : 1 the genes are unlinked and you are finished. iii)If PD > > NPD, then the genes are linked and the distance between them is given by: T + 6NPD Distance in cM = x 100 2 Σ
Gene Fine Structure So far we have been considering the phenotypic consequences of mutations in different genes and the relative positions of genes along chromosomes. Now we are going to look at the internal structure of genes themselves and to consider the different kinds of mutations that are found in genes. For the next eleven lectures we will examine the genes of bacteria and their viruses, known as phage(some special features of eukaryotic genes will come later). First, lets see how the life cycle of phage can be exploited to carry out classical genetic manipulations E coli cell DNA synthesis Phage protein expression head assembly Lysis DNA packaging pl lawn of bacteria In order to do genetics with phage it is necessary to have recognizable variants of phage The easiest mutants to work with have obvious distinctive plaque morphologies make small plaques because fewer particles are released n phage normally infects cells to make more phage. But about 1%o of the time phage becomes quiescent in host cell and prevents other phage from infecting that cell. This property of phage n is known as immunity. The quiescent pl Is known as a lysogen er mutants make clear plaques because they can't form lysogens lysogens wild-type unbid plaque clear plaque
Gene Fine Structure So far we have been considering the phenotypic consequences of mutations in different genes and the relative positions of genes along chromosomes. Now we are going to look at the internal structure of genes themselves and to consider the different kinds of mutations that are found in genes. For the next eleven lectures we will examine the genes of bacteria and their viruses, known as phage (some special features of eukaryotic genes will come later). First, let’s see how the life cycle of phage can be exploited to carry out classical genetic manipulations. Phage cross In order to do genetics with phage it is necessary to have recognizable variants of phage. The easiest mutants to work with have obvious distinctive plaque morphologies λ phage mi–: make small plaques because fewer particles are released λ phage normally infects cells to make more phage. But about 1% of the time phage becomes quiescent in host cell and prevents other phage from infecting that cell. This property of phage λ is known as immunity. The quiescent phage is known as a lysogen. cI– mutants make clear plaques because they can’t form lysogens
Phage Cross: Infect bacteria with 10 of each phage per cell to insure that every cell has both kinds of phage replicating inside. Allow one round of growth then plate out the morphology of the plaques Plaque type Frequency lg clear (parent 0.4 sm turbid (paren 04 lg turbid (recomb. 0.1 sm clear (recomb.) 0.1 For this cross the measured distance between mi and cr is number of recombinants 100×100 0 m u(map units) We can also use this system to find the distance between two mutations in the same gene. For example, lets say that we have isolated two different cr mutants that give clear plaques. After infection of E. coli with w10 phage per cell of each mutant, the resulting phage are plated and their plaque morphology is examined. Most of the phage will still be ef, but recombination between mutations will produce wild type phage with turbid plaques. If 2 out of 1,000 phage make turbid plaques then the measured map distance between the two mutations is 0. 4 m u. Note that in this example we are only counting wild type recombinants and there should be an equal number of double-mutant recombinants- both classes of recombinants must be accounted for in calculating the map distance -1 cI-2
Phage Cross: Infect bacteria with ~10 of each phage per cell to insure that every cell has both kinds of phage replicating inside. Allow one round of growth then plate out the phage that are produced to look at the morphology of the plaques. Plaque type Frequency lg clear (parent) 0.4 sm turbid (parent) 0.4 lg turbid (recomb.) 0.1 sm clear (recomb.) 0.1 For this cross the measured distance between mi– and cI– is: number of recombinants 20 x 100 = x 100 = 20 m.u. (map units) total 100 mi cI 20 mu We can also use this system to find the distance between two mutations in the same gene. For example, let’s say that we have isolated two different cI– mutants that give clear plaques. After infection of E. coli with ~10 phage per cell of each mutant, the resulting phage are plated and their plaque morphology is examined. Most of the phage will still be cI–, but recombination between mutations will produce wild type phage with turbid plaques. If 2 out of 1,000 phage make turbid plaques then the measured map distance between the two mutations is 0.4 m.u. Note that in this example we are only counting wild type recombinants and there should be an equal number of double-mutant recombinants — both classes of recombinants must be accounted for in calculating the map distance. cI-1 cI-2 0.4 mu
