Yeast Lab: Reassortment

Question 1

Goal

Answer 1

• Crossing yeast mutant strains to illustrate
  
  • complementation
  • allelism
  • independent assortment of chromosomes
  • recombination during meiosis

Question 2

Complementation

Answer 2

• Restoration of wild type function
• Diploids formed between mutant haploids can have wild type phenotype
• Indicates mutations in different genes

Question 3

To recombine several mutant genes in one strain you can …

Answer 3

• haploid cells can be mated to form a diploid cell
• diploid can be induced to undergo meiosis, resulting in haploid progeny with new combinations of mutations

Question 4

Two mating types: MATa and MATα

Answer 4

• Haploid yeast exist as different mating types: a and α
• Each type secretes a specific factor (hormone) that binds only to receptors on the other type of cell.
• When exposed to each other’s mating factors, a pair of cells of opposite type change shape, grow toward each other, and fuse (mate)
• cell cycle is paused, cell fusion and nuclear fusion occurs
• The new diploid a/α contains all the genes of both original cells, a combination of genetic resources that provides advantages to the cell’s descendants
• Haploid cells of opposite mating type can form diploids while cells of the same mating type cannot

Question 5

Diploid selection

Answer 5

• Can select for diploid cells resulting from the mating of two different haploid strains by using a medium that will not allow either haploid to grow, but on which the diploid can grow

Question 6

Inducing meiosis (sporulation) in diploid

Answer 6

• induced by plating cells on a medium lacking nitrogen
• cells undergo sporulation and enter meiosis
  
  • when a diploid cell divides twice to produce four haploid cells
  • produces 4 haploid ascospores encased in a protective outer layer called the ascus
  • segregation of alleles and independent assortment of chromosomes
• haploid meiospores are extremely resistant to adverse environmental conditions
  
  • can be germinated with nutrient‑rich medium
  • recessive mutant alleles will be revealed
  • mutant genes can be obtained through crossing and subsequent analysis of the haploid progeny

Question 7

drop‑out media

Answer 7

• Media lacking specific nutrients for selection of transformants or identification of mutants
• yeast that are unable to synthesize the nutrient will fail to grow
• example: leu drop‑out medium
  
  • media lacking leucine
  • allows identification of yeast that are unable to synthesize leucine or yeast that have gained the ability to synthesize leucine

Question 8

adenine mutant yeast strains

Answer 8

• in yeast, adenine is synthesized as adenosine monophosphate (AMP)
  
  • 12 steps in pathway
  • a mutation in any one of the genes in the pathway require adenine in the medium to survive (auxotrophic)
• ADE1 and ADE2 mutants have a visible phenotype – they accumulate a red pigment
• color morphology
  
  • results from accumulation of p-ribosylaminoimidazole (AIR), an intermediate compound
  • mutation in 6th step
    
    • red colonies
    • strain ade2
    • cannot convert AIR to CAIR
    • AIR is not red but is oxidized to form a red pigment
  • mutations in 7th step
    
    • pink colonies
    • strain ade1
    • convert AIR to CAIR
    • enough AIR accumulates resulting in pink colonies
  • Mutations in steps 1-5, and 8-12
    
    • white colonies
    • AIR does not accumulate.
• using yeast strains with mutations in ADE1 and ADE2
  
  • # denotes order of their identification, not order in pathway

Question 9

genotypes of the haploid progeny

of diploid cells are the result of

Answer 9

• independent assortment of chromosomes
• recombination during meiosis

Question 10

Meiospore (haploid spores) enrichment

Answer 10

• spore produced by meiosis
• Sporulation is never 100% efficient
• There will still be diploid cells on the sporulation plate which will need to be released and enriched
• Zymolase
  
  • enzyme preparation from the bacterium Arthrobacter luteus
  • contains an endo-1,3-β-glucanase
  • lyses the vegetative diploid cells and loosens the ascus walls, which will result in enrichment in the meiospores in the samples

Question 11

Segregation of alleles

Answer 11

• each haploid cell inherits one or the other of the alleles from the diploid cell

Question 12

recombination

Answer 12

• Physical exchange between nonsister chromatids of homologous chromosomes
• Crossing-over
• Breakage and rejoining of DNA
• New combination of alleles on a chromosome

Question 13

linkage

Answer 13

• Genes close together on a chromosome more likely remain together during meiosis
• the greater the distance between genes the more likely a recombination event will occur
• Alleles of different genes on the same chromosome are considered linked if they are inherited together more frequently than expected if they were on separate chromosomes
• If the recombination frequency between genes is 50%, genes assort independently
• if the genes are on separate chromosomes, they will have segregation of alleles, and independent assortment of chormosomes
• if the genes are on the same chromosomes, possibility of recombination or linkage

Question 14

Week 10: Mate Yeast

create crosses between 4 different strains of yeast cells

Answer 14

• Purpose: mate yeast
• create crosses between 4 different strains of yeast cells
  
  • each strain differentiated by their mating types, and mutations
• crosses were plated on minimal medium
  
  • minimal medium supplemented with leucine, uracil, tryptophan and adenine
  • selects for diploid cells ONLY generated from the mating
  • haploids are not expected to grow cuz each strain has a mutation in a gene synthesizing a required product not provided by the medium
• See mating genotypes and results
  
  • Copy and paste the following link:
  • https://drive.google.com/file/d/1TOdIaiBK_WtynKti9XVwMmoVXd0Sdvdv/view?usp=sharing
• Complementation brings about selection of diploids
  
  • all are white

Question 15

Week 11: Induce sporulation

Answer 15

Week 11:

• Purpose: Prepare yeast for sporulation
• Plate on YPD to prepare for sporulation
  
  • rich medium can improve sporulation

Week 11 – Follow up

• Purpose: Induce sporulation
• Plate to sporulation medium
  
  • Induce sporulation by plating on nutrient poor medium
    
    • low glucose/nitrogen
    • high concentration of potassium acetate, which kills vegetative cells
  • Medium induces meiosis and production of haploid ascospores contained in the remnants of the cell, the ascus
• Example
  
  • Diploid ADE2/ade2 placed on a nutrient poor medium
  • Medium induces meiosis
    
    • segregation of alleles
    • Independent assortment of chromosomes
  • After meiosis we expect
    
    • 50% of haploids with wild type ADE2 allele
    • 50% of haploids with mutant ade2 allele

Question 16

Week 12

Meiospore enrichment

(spores produced by meiosis)

Answer 16

Week 12

• Purpose: Lyses of diploid cells and meiospore enrichment
• Sporulation is not 100% efficient and diploid cells will remain on the plate. To release and enrich them, zymolase is used
  
  • enzyme from Arthrobacter luteus
  • contains an endo-1,3-β-glucanase
  • lyses vegetative diploid cells, loosens ascus walls, resulting in enrichment of meiospores

Week 12 - Follow up

• Purpose: Meiospore enrichment and germinate meiospores
• Dilutions of 1:25, 1:50, 1:75 and 1:100 sporulated diploids were prepared
  
  • Different dilutions used in order to have well separated colonies
  • Copy and paste link
  • https://drive.google.com/file/d/1frR5XNHq2wQyFfIwsrp_3nLQd43WZ-F-/view?usp=sharing
• Plate meiospores on YPD and adenine medium
  
  • YPD is used as rich medium to encourage growth
  • Diploid cells have undergone meiosis, and have 4 haploids per cell
  • Adenine was added to the medium
    
    • for haploids that have mutations in an ADE1 or ADE2 gene
      
      • S30: MATα - met3, leu2
      • X29: MATα - met14, leu1, ura3, his2, ade1, trp1, gal1
      • YP4: MATa - leu2, ura3, his3, ade2, trp1, lys2
    • allows us to differentiate the mutants, by looking for pink/red color
      
      • mutations in ade1 (pink) or ade2 (red)
      • see “adenine mutant yeast strains” brainscape card

Reasoning

• Mutant phenotypes on diploids are more difficult to identify than on haploids.
  
  • Diploids display mutations if they are homozygotes to the mutant gene or heterozygotes for a dominant mutant gene
  • Haploids can display a mutant phenotype just by carrying the mutant gene because they lack a wild-type copy to mask the mutant copy; making haploids the preferred choice to analyze mutations
• only selecting pigmented colonies
  
  • coloration confirms they have a mutation
  • All other cells will have a white color
    
    • not feasible to visually detect if these white cells are mutants or diploids, since no further experiments are scheduled to differentiate them
  • Selecting for pigmented colonies only, ensures we have selected only mutants
  • This restricts our selection to only those mutants with ade2 or ade1 mutations
  • Any mutated white haploid will be missed and not included in any further analysis

Question 17

Week 13

Examination of plated meiospores

Answer 17

Week 13

• Examination of plated meiospores
• HIS3 is on the same chromosome (XV) as ade2
  
  • If far apart, there’s a high probability of a recombination event
  • If very close together
    
    • little or no recombination will occur
    • likelihood of the parental alleles staying together would be high
    • co-inheritance of the his3 and ade allele in the progeny would be higher than predicted (based on independent assortment of chromosomes)
• Most genes in this class are on different chromosomes
  
  • principle of independent assortment of chromosomes can be applied to predict frequency of any particular progeny haploid genotype from a cross of known genotypes
  • If more parental genotypes than expected were recovered, then that would indicated the genes are close together on the same chromosome
• S30-YP4
  
  • S30: MATα - met3, leu2
  • YP4: MATa - leu2, ura3, his3, ade2, trp1, lys2
  • No mutants for ADE1: ADE1/ ADE1
  • Expect 50% red, 50% white
    
    • ADE2/ADE1
    • ade2/ADE1
  • Will see an excess of white colonies, because of haploids with wild type ADE2 allele and any remaining diploids
  • Haploids replica plated to histidine dropout and leucine dropout plates.
  • What % of the haploids do you expect to grow on the leucine dropout plates?
    
    • leu2/leu2: 0%
  • HIS3 is on the same chromosome (XV) as is ade2. If these 2 genes are far apart on the chromosome what would you expect in the haploid progeny when plated on histidine dropout plates?
    
    • I would expect recombination
    • HIS3/ade2
      
      • recombination event
      • neither original haploid had this genotype
    • HIS3/ADE2
      
      • missed
      • we are selecting for ade1 or ade2
    • his3/ADE2
      
      • recombination event
      • neither original haploid had this genotype
      • missed. we are selecting for ade1 or ade2
    • his3/ade2
    • 50% will survive on the histidine dropout
    • 50% had recombination event
• X29-YP4
  
  • X29: MATα - met14, leu1, ura3, his2, ade1, trp1, gal1
  • YP4: MATa - leu2, ura3, his3, ade2, trp1, lys2
  • Expect 75% pink/red, 25% red (ade2), 50% pink (ade1)
    
    • ADE1/ADE2
      
      • recombination event, neither original haploid had this genotype
      • this one will be missed, since we are selecting for ade1 or ade2
    • ADE1/ade2
    • ade1/ADE2
    • ade1/ade2
      
      • recombination event, neither original haploid had this genotype
  • Haploids replica plated onto uracil dropout, leucine dropout, and lysine dropout media
  • What % of haploids do you expect to require leucine, uracil, and lysine?
    
    • Leucine: LEU1/leu1, LEU2/leu2
      
      • LEU1/LEU2
      • LEU1/leu2
      • leu1/LEU2
      • leu1/leu2
      • 75% mutants
    • Lysine
      
      • LYS2/lys2
      • 50% mutants
    • Uracil
      
      • ura3/ura3
      • 100%
  • Predict coinheritance of lysine and leucine wild type or requiring phenotypes
    
    • Wild type
      
      • LYS2 = 0.5
      • LEU1 and LEU2 = 0.25
      • Lysine and leucine wt: (0.5)(0.25) = 0.125
    • Mutants
      
      • lys2 = 0.5
      • leu1, leu2 = 0.75
      • Lysine and leucine mutants: (0.5)(0.75) = 0.375

Question 18

Calculations

Answer 18

https://drive.google.com/file/d/1frR5XNHq2wQyFfIwsrp_3nLQd43WZ-F-/view?usp=sharing