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Experiment 4: Tetrad analysis in yeast As part of life cycle of Saccharomyces cerevisiae, the a/diploid cells can be induced to sporulate (undergo meiosis to produce haploid spores). They do so when any one of a number of essential elements becomes growth rate limiting (carbon source, for example). In their natural environment, hard times will be met seasonally when the nutrients needed to support vegetative growth become unavailable. Under these conditions, yeast, like many fungi, will form spores arrested for growth and protected by a tough cell wall from the harsh conditions. During sporulation diploid yeast cells undergo meiosis (Figure 1). Each chromosome replicates to form a bivalent chromosome. Homologous chromosomes then pair, undergo recombination, and line up on the spindle apparatus. Each homologue then segregates and goes to an opposite pole in the first meiotic division (reduction division). In the second division each bivalent associates with the spindle apparatus and then each chromatid composing the bivalent goes to an opposite pole. This process, like meiosis in higher eukaryotes, generates 4 meiotic products (ascospores) or haploid yeast cells. The 4 ascospores derived from a single meiotic event are contained in an ascus or sac. This sac protects the four cells from perishing during the starvation conditions. Unlike gametogenesis in many higher eukaryotic organisms, these 4 haploid meiotic products (gametes) can be collected as a unit (the ascus) and all 4 meiotic products can be grown vegetatively as haploid cells.

Figure 1: Life cycle of yeast Once growth conditions improve, the four cells will then germinate growing vegetatively as haploid yeast cells. Each of these haploid meiotic products are either of the a or mating type (2 each). The a and


mating type cells can then mate to form an a/ diploid cell that can then be induced to undergo meiosis. Hence, the life cycle continues. The ability to grow all 4 haploid products of yeast meiotic events facilitates our ability to do genetic analyses with this organism. The ascal wall can be partially digested by treatment with an enzyme (glusulase: a snail gut enzyme). This treatment releases the 4 meiotic products that are derived from a single diploid cell that has undergone an independent meiotic event. The 4 spores from a single meiosis tend to be sticky and stay together as a tetrad. A tetrad placed on a YEPD plate can be viewed under a specially adapted microscope used for dissecting tetrads. This dissecting microscope includes a micromanipulator. The micromanipulator is essentially a thinly drawn-out glass needle that can be visualized under the microscope. This needle can be moved in an x, y, and z coordinate system. The thin tip of the needle can touch the agar surface and pick up a tetrad. It can then be used to separate the 4 meiotic products as well as pick up each individual ascospore from the agar surface and place it back down on the agar surface. This enables you to place each of the four meiotic products of a single meiosis on an agar petri plate equidistant from each other and in a contiguous array. Each spore will then germinate on the YEPD agar plate and grow vegetatively into a colony. The end result is that the 4 spores grow into 4 separate colonies. Each colony represents a pure population of haploid yeast cells that are genetically identical to the single haploid meiotic product originally placed down at that position on the agar surface. These colonies can then be replica plated to other types of media to analyze their growth requirements. If the original diploid strain contained an auxotrophic mutation one can then analyze the genetic nature of the mutation that confers the auxotrophy and study its relationship to mutations in other genes. Mutations in yeast chromosome are inherited in accordance to Mendelian rules. One advantage of doing genetic analyses with yeast is that it is simpler and faster than analyses in other eukaryotic organisms. You are able to grow all 4 haploid gametes of a meiotic event and therefore analyze a given trait or phenotype independent of further breeding or the test crosses required for genetic analyses in obligate diploid organisms like Drosophila melanogaster. The first order of business in performing tetrad analysis is to be sure that the phenotype that you are observing is a result of a mutation in a single nuclear gene. We can be sure of this if the phenotype is inherited during meiosis in a predictable fashion that is consistent with Mendel's 1st principle (Alleles Segregate). Let's take an example that you would be familiar with, namely, a mutation in one of the histidine biosynthetic genes. If you crossed a His1+ yeast strain of the a mating type to a his1- mutant of the mating type and sporulated the heterozygous His1+/His1- diploid, you would generate 4 meiotic products (Figure 2). Each of the 4 colonies would grow on the enriched media contained in a YEPD plate. Upon replica plating them to SD-histidine medium you would find that 2 of the colonies (and therefore the original gametes) are His1+ and the other 2 colonies are his1-. Two of the meiotic products would have the same phenotype as one of the parents (His1+) and the other two meiotic products would have the same phenotype as the other parent strain (his1-). Crossing over between the HIS1 gene and its centromere would not affect this ratio and the same ratio should be observed in each set of tetrads derived from any diploid that resulted from this mating. In practice, we would dissect at least 20 tetrads from 20 independent heterozygous diploids (20 independent meiotic events) to be sure that the segregation


Figure 2: Yeast meiosis. A diploid heterozygous for his auxotrophy is followed. Note the 2+:2- ratio of tetrads. This indicated that the his- phenotype is due to a single gene. ratio was 2+:2- and to statistically eliminate the possibility that the histidine auxotrophy could have been a result of two mutations, one in each of two unlinked genes. This test to determine that a mutation is due to a single gene should have been performed with the histidine auxotrophs you isolated in Experiment 1 prior to performing the complementation tests (Why?). This test of whether or not a mutation is due to a single gene would have been done in a research situation, but due to time constraints in a lab course we had to skip this very important test. Once you know that the phenotype you are observing is due to a mutation in a single nuclear gene and that yeast obeys the first Mendelian law for this trait, we can study the genetic relationship (linkage, epistasis) of this gene to other known and characterized genetic loci in the yeast genome (other auxotrophies for other biosynthetic pathways; e.g. leucine, uracil and inositol mutants). Linkage can be determined by analyzing the tetrads (tetrad analysis) derived from crosses of haploid yeast strains. There are three kinds of genetic linkage relationships which can be established by tetrad analysis and which are the main points of our experiment: 1. Are the two genes being examined genetically linked and, thus, on the same chromosome? 2. Are two genes unlinked to each other and, thus, either far apart from each other on the same chromosome or located on different chromosomes? 3. Are two genes unlinked to each other and located on different chromosomes but jointly linked to the centromeres of their respective chromosomes? Note that there are two separate linkage questions being asked: 1. Are the genes genetically linked to each other?" 2. If not linked to each other, are the two genes jointly linked to their respective centromeres? Tightly Linked Genes Let's assume a hypothetical situation whereby you have an a mating type strain that has a mutation, x-, in a nuclear gene which is recessive to its wild type allele, X+. You then mate this a strain to an strain that has a mutation, z-, in a nuclear gene which is recessive to its wild-type allele, Z+. Furthermore, assume both of these genes are tightly linked (close) to each other such that no recombination can occur between them. We can observe this situation by tetrad analysis after the diploid is induced to undergo meiosis


(Figure 3). The first thing to note is that when each phenotype is examined individually, it will segregate 2+:2- in the tetrad. Looking at gene X, 2 meiotic products will be X+, the other 2 will be x-. Looking at gene Z, 2 will be Z+ while 2 will be z-. This is because chromosome homologs segregate in the first division of meiosis.

Figure 3: The production of yeast tetrads which exhibit the parental ditypes. If there had been two recombination events between X and Z involving two strands, you would see the same end result ­ parental ditype. How these phenotypes segregate relative to each other in the 4 different meiotic products tells us about the linkage relationship between the X and the Z genes. In our case because there is no recombination occurring, 2 meiotic products will be xZ and the other two Xz. These are the parental genotypes and we refer to this tetrad type as a parental ditype. There are only two types of haploid meiotic products and both are parental types. Given the tight linkage of the two genes, if we dissected more and more tetrads from independent meiotic events we would find that nearly all tetrads would be of the parental ditype. Thus, the two mutations, x and z, behave almost as if they were alleles of each other. When two mutations segregate as if they were alleles at the same locus, this tells us that the two mutations are tightly linked on the same chromosome. Less Tightly Linked Genes If the Z gene were further away from the X gene on the same chromosome then the chances of Z behaving as an allele of X decreases. This is because as the physical distance between the two genes increases the chance of a recombination event between the two genes will also increase (Mendel's 2nd Law-Genes Assort). For obvious reasons the first type of recombination event that will occur as we begin to increase the distance between X and Z is a single cross-over event between X and Z (Figure 4). This type of event will generate recombinant genotypes in two of the four meiotic products. One spore will be like one parent, xZ, and a second like the other parent, Xz. The third and fourth products represent the recombinant genotypes xz and XZ. Thus, a single cross-over event between two genes generates 4 meiotic products each of which has a different genotype (and phenotype). This type of tetrad is referred to as a tetratype. There are four different types of meiotic products. Note that recombination can occur between any two strands, one from each homologue, of the four chromatids.


There are 4 possible combinations of strands that can be involved in a recombination event to arrive at the tetratype ascus. Also, remember that "strand" refers here to one of the two chromatids that make up the bivalent. This "strand" is the base-paired, double helix DNA that is exchanged during recombination. Don't confuse the usage of the word strand used here for recombination events with the more recent usage of the word strand as in single or double-stranded DNA.

Figure 4: Examples of the production of tetratypes in yeast meiosis. Tetra types can also be produced by two cross over events involving 3 strands If the Z gene were even further from the X gene on the same chromosome then the chances of multiple cross-over events occurring between x and z will also increase. Double cross-over events involving the same two strands will regenerate parental ditype tetrads. Double cross-over events involving three strands of the homologs will generate tetratype tetrads. However, double cross-over events involving all 4 strands of the homologues produce a third type of tetrad called a non-parental ditype (Figure 5). There are only two types of meiotic products and both show a recombinant relationship of the two genes. In non-parental ditype tetrads produced from our strain, two of the meiotic products will have the genotype xz and the other two products will have the genotype XZ.


Figure 5: Example of the production of nonparental ditypes in yeast meiosis. Therefore if X and Z genes are linked (but not very tightly) and not completely free to recombine, and we analyze many tetrads from independent meiotic events from the cross, xZ to Xz, we will observe some tetrads which are parental ditype (PD), some which are non-parental ditype (NPD) and some tetratype (TT) tetrads. For two genes on the same chromosome that are genetically linked, you would always see significantly more PDs than NPDs. NPDs will always be the smallest category because, for linked genes, the mechanism that produces NPDs is a fraction of the double crossovers that occur. Completely Unlinked Genes For two genes that are very far away from each other on the same chromosome we would expect to see a ratio of PD: NPD: TT of 1:1:4. The same types of tetrads (and same ratios) are also seen if two genes are on different chromosomes and at least one of those genes is free to recombine with its centromere. Thus, if you conclude that two genes are completly unlinked, that is all you can say about these two genes. You will be unable to say whether they are on the same or different chromosomes. (Why? Draw it.) Centromere Linked Genes Tetrad analysis (and only tetrad analysis) can also be used to detect and analyze the relationship of a pair of genes with their centromeres. It is important to stress that centromere linked genes are NOT linked to each other. As mentioned above, if the X and Z gene are located on different chromosomes and at least one of the genes was unlinked to its centromere we would detect a 1 PD: 1 NPD: 4 TT ratio. However, let's assume that gene X is very close to and restricted in its ability to recombine with its centromere and that the Z gene is also close to the centromere of its chromosome (Figure 6). Only a few single cross-over events can occur between Z and its centromere. For genes on different chromosomes, only crossovers between a gene and its centromere can give rise to the TT tetrad. Indeed, for genes on different chromosomes, the TT class is a measure of the sum of the genetic distance between each gene and its respective centromere. If the Z gene is so close to its centromere that no single cross-over events can occur (and, as proposed, the X gene also cannot recombine with its centromere) then, in the cross x to z, we would eliminate the TT recombinant class. In this case the number of PD = NPD, but TT = 0. This is an extreme case. Not every centromere-linked gene is so close to its centromere that it will show no recombination. However, we can still detect centromere linkage by noting in the tetrad analysis of a cross that PD ~ NPD, and TT << 2(PD+NP). Note that PD always ~ NPD because the two genes are unlinked (they are on different chromosomes). Unlike the case where the two genes are on the same chromosome, NPDs do not arise because of double cross-overs when the two genes are on different chromosomes. Rather NPDs arise because of the alignment of the two independent chromosomes on the mitotic spindle in the first meiotic division and the alignment that gives rise to NPDs is just as likely as the one that gives rise to PDs. [If you could not see all four products of meiosis in a tetrad, but only saw a population of gametes (this is, of course, the case for most organisms) derived from these tetrads, what you would see is 50% recombinant gametes and 50% parental.]


Figure 6: Unlinked yeast genes, with each gene linked to its centromere. Because the chromosome arrangement at meiosis is just as likely to be that in the A as B, the parental ditype (A) and nonparental ditype (B) are equally likely to occur. Determining Linkage Distances You should first distinguish whether two genes in a cross are linked or are unlinked (either by being completely unlinked or by being unlinked to each other but centromere linked). The best operational test of linkage asks the question "Are these two genes unlinked?" because all unlinked genes (whether completely unlinked or centromere linked) show a specific expected ratio for certain tetrad types. Only after answering the above question do we ask of the unlinked genes "Are these two genes completely unlinked or are they centromere linked (centromere linked genes are not linked to each other)?". Again, answer the question by using an expected ratio and analyzing whether your observations fit that expectation. The best operational test of centromere linkage asks the question "Are these two genes completely unlinked?" because completely unlinked genes show a specific expected ratio of tetrad types.


At the end of this analysis, you should have grouped gene pairs into three categories: linked, completely unlinked, and centromere linked. At this point you are ready to determine linkage distances for all distances that are measureable (unlinked genes, of course, do not have linkage distances). If the two genes are linked, we can calculate the map distance between them by using a formula that takes double cross-overs into account. Basically, this formula uses the number of NPDs (which represents a fraction of double crossovers) to estimate the total number of double crossovers. The formula for calculating map distances for pairs of linked yeast genes is:

1% recombination is usually expressed as the genetic map unit centiMorgan (cM). Don't confuse this with centimeter (cm). For centromere linked genes, you can calculate the sum of the distances between each gene and its respective centromere. If one of the genes is so tightly linked to its centromere that it almost never recombines, then we can use this gene as a marker to calculate the map distance of the other mutation relative to its centromere using the formula:

This formula does not and can not use NPDs as a measure of double crossovers because NPDs are not due to double crossovers in centromere linked genes.


Experimental Protocol for Experiment 4: This experiment requires you to analyze the genetic linkage relationship of several different genes in yeast. You will establish whether they are linked or unlinked and calculate map distances if they are linked. As in most mapping experiments, you will be concerned only with a single pair of genes at any one time. Experimental Organism: The meiotic events you will analyze stem from a diploid strain of yeast S. cerevisiae that was induced to undergo meiosis. The diploid strain was derived by mating an a mating type of the yeast strain, TD28, with the genotype ura3 -, ino1-, with an mating type strain, 44-1B, with the genotype leu2-, his4-. The ura3 mutation causes a defect in uracil biosynthesis. The leu2 mutation causes a defect in leucine biosynthesis. The ino1 mutation causes a defect in inositol biosynthesis. You are already familiar with a his4 mutation. In addition, we will determine the mating type of the ascospores (see Part II).

Part I: Determining Tetrad class (PD, NPD or TT) Week 1: 1. Partners will receive a YEPD plate that contains several rows of tetrads that have been micromanipulated to separate the four meiotic products (ascospores). Each meiotic product has grown into a small colony of haploid yeast. Each tetrad represents an independent meiotic event that occurred in one diploid cell derived from the mating of the haploid yeast strains TD28 and 44 1B. This dissection was performed for you. 2. Replica plate the tetrad plate to a velvet and then, without changing the velvet, make an imprint on: A. SD-histidine plate (has added leu, ura, and ino) B. SD-leucine plate (has added his, ura, and ino) C. SD-uracil plate (has added his, leu, and ino) D. SD w/o inositol plate (has added his, leu, and ura)

3. Incubate all plates overnight at 30°C for 2 days. Refrigerate the plates until the next class.

Each plate will enable you to independently score one of the auxotrophic markers segregating in this cross.

Week 2: 1. For each selective plate record the growth of each meiotic product of each tetrad. Use the score sheet to record this data for each of the 10 tetrads on your plate (see lab discussion). For each of your 10 tetrads determine whether it represents a PD, a NPD or a TT type for each of the following gene pairs: a) HIS4 and LEU2 b) HIS4 and INO1 c) HIS4 and URA3 d) URA3 and LEU2 e) URA3 and INO1 2. For each single gene, record whether a tetrad gave a 2:2 (or 3:1 or 4:0) ratio. You can only score tetrads in which all 4 spores grew on complete media. Why? 3. Write down which allele combinations are parental (use + for wild type and - for mutant) for each of the gene pairs you will score.

f) INO1 and LEU2


Part II. Determination of Mating Type in Yeast Haploid yeast cells exist as either an a or an mating type. Only a cells can mate with cells to form an a/ diploid and vice versa. a cells cannot mate with a cells and cells cannot mate with cells to form a diploid cell. Mating type in yeast is determined by a single nuclear gene referred to as the MAT locus that either contains information of the a mating type or the information of the mating type. In the cross you are analyzing, the mating type genotype of the parent strain TD28 is MATa and the mating type genotype of the other parent strain, 44-1B, is MAT. As part of tetrad analysis the mating type of each of the meiotic products is routinely determined. Mating type is determined by mating ascospores with tester strains of known mating type. As part of Experiment 4 you are to determine the mating type of each of the meiotic products from your cross of TD28 to 44-1B. You will be given a lawn of each of the mating type tester strains, 17/14 (MATa his1-) and 17/17 (MAT his1-). Week 1: 1. Label one YEPD plate 17/17 and a second YEPD plate 17/17 2. Using a new velvet, replica plate 17/14a tester lawn to velvet 3. Place the YEPD plate labeled above (17/14a) on the velvet 4. Using a new velvet, replica plate 17/17 tester lawn to velvet 5. Place the YEPD plate labeled above (17/17) on the velvet. Week 2: Replica plate each of these mating plates to a Min -Ino plate. Min -Ino medium lacks leucine, histidine, uracil, and inositol. In the following week you will score for mating type.


TETRAD ANALYSIS Plate Type: YEPD l Spore # 1 2 3 4 1 His2 3 4 +/- Ratio 1 Leu2 3 4 +/- Ratio 1 Ura2 3 4 +/- Ratio 1 Ino2 3 4 +/- Ratio 1 MAT 17/14 2 3 4 +/- Ratio 1 MAT 17/17 2 3 4 +/- Ratio 1 MAT a/ 2 3 4 Gene Pairs His-Leu His-Ura His-Ino Leu-Ura Leu-Ino Ura-Ino His-MAT Leu-MAT Ura-MAT Ino-mat 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Plate Number: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20



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