Genetics Lecture 18, 10/10 - Bacteriophages, Genetic Experiments

How bacteriaphages accomplish genetic recombination

• Lederberg-Zinder experiment showed bacteriaphage can accomplish recombination.
• Two strains both of which are autotrophs for two genes.
  
• between the two strains place a filter: media passes through but not cells
• Plate cells from both sides of the tube on minimal media
• Plating cells from the LA-2 side there is no growth.
• From the LA-22 side you get prototrophs.
• (4:35) How do you get prototrophs from the LA 22 side? ANS: Possible presence of a filterable agent which facilitated this result. Made 3 observation about this filterable agent:
• 1.) The agent was only produced when LA-2 cells were grown in concert with LA22 cells. If they grew LA2 and then placed it with LA-22 there was no cell recombination
• 2.) DNAse: enzyme which digests DNA did not make the filterable agent inoperative.
  
• 3.) The filterable agent was dependent on the size of the pores.
• Concluded that the filterable agent was a phage (P22 phage). It was associated with the LA22 cells. Phage starts with LA22 and crosses the filter and picks up DNA from the LA 2 cells. (10:30) It then passes back across the filter and infects the LA22 cells. When this happens it donates the DNA from the LA2 cell to the LA 22 cell.
  
• (11:45) Lysogenic cell - cell thats infected with phage but not a large enought level to kill the cell.
  The prototrophs from the LA22 side where lysogenic cells with the DNA from the LA2 cell to make the cell prototrophic.

Chapter 10 - DNA structure and analysis.

• (19:35) What is the genetic material?
• Four qualities of a genetic material
  
• This molecule must have the ability to be replicated
  
• It needs to be able to serve as a repository of information
• Needs to have a mechanism to allow it to be expressed. (it needs to be converted to protein to cause a phenotype)
• It needs to be able to be changed. Mutation facilitates diversity.
• DNA has all of the above characteristics.
• (30:10) Two proposed molecules for the genetic material.
• DNA or protein (protein was originally the favored choice). It came about this way due to an incorrect hypothesis --> Tetranucleotide hypothesis: The largest DNA molecule you could make was 4 nucleotides long. This structure would contain each of the four nitrogenous bases.
• (34:10) 1927 Fredrick Griffith worked with two strains of diplococcus pneumoniae. III S strain (virulent) and the smooth strain. When he injected the IIIS strain into a mouse it died.
• IIR strain - rough strain - avirulent strain. When he injected into the mouse it lived
• Smooth strain (IIIS) - Heat killed it. He found that when he injected it lived.
  
• GRIFFITH'S CRITICAL EXP.: mixed heat killed IIS with living IIR cells and injected it into a mouse and got a dead mouse. (two avirulent strains became virulent). When analyzing the tissue of the dead mouse he recovered living IIIS cells.
  
• proposed the presence of transforming principle: a molecule which survived heat killing and able to convert the IIR cells to IIIS cells.
• Transforming principle is a logical candidate for the genetic material.
  
• (42:10) Avery, Macleod, McCarty, 1944. Wanted to figure out what the transforming principle was. They heat killed with IIIS cells and extracted lipids, some proteins and carbohydrates (cell membrane). Took this filtrate and mixed with IIR cells. Transformation occured IIS cells
• (44:50) Treat the filtrate with protease (degrades proteing). Mix with IIR cells and transformation occurs. Therefore protein is NOT the genetic material.
  
• TReat the filtrate with ribonuclease (degrades RNA). Mix with IIR cells and transformation occurs. RNA is not the genetic material.
  
• Treat the filtrate with DNAse (degrade DNA). Mix with IIR cells and transformation DOES NOT OCCUR. DNA is THE genetic material.
• (50:00) Hershey - Chase experiment: looking at the T2 virus of e. coli.
  
• Virus is a protein coat surrounding DNA
  
• took e. coli cells and grew them in two media: p32 label nucleic acid AND s35 label protein. Put phage into each culture. Expect the phage to pick up your label.
  
• take each of the phages and place them in new cultures without label. The genetic material will be injected into the cell. Expect that whichever molecule was serving as the genetic material, the progeny phages will be labeled.
• p32 progeny phages were labeled
  
• s35 progeny phages were not labeled
• this supports the idea that DNA is the genetic material for VIRUSES. Check out wikipedia for a good explanation
  

Genetics Lecture 17, 10/8 - Plasmids, Transformation, Bacteriophage

• • F Factor is a plasmid. Plasmid is a small circular, extrachromosomal, piece of DNA. Plasmids are replicated when the host cell goes through DNA replication.
• • Types of Plasmids
• o F plasmid: contains genes necessary for a fertile cell. Allows the formation of the sexpilus. During the transfer F plasmid is transmitted to the recipient cell.
• o R plasmid: two important regions, transfer region → allows the plasmid to be passed between cells, AND R-determinant → codes for resistance to an antibiotic.
• o Colplasmid: commonly found in e. coli. It produces a toxin called colicin. A cell that lacks the colplasmid will be killed by the toxin. A cell WITH the plasmid will be immune to the toxin that it produces.
• • Transformation (6:25) – small pieces of DNA that are transferred from one cell to the next. Two STEP Process: entry (DNA enters into the cell) AND recombination (DNA enters the host cells genome).
• o Entry: requires a bacterial cell to be in a state of competence. A competent cell is one that has the ability to take up exogenous DNA. DNA enters and one of the two strands degrades.
• o DNA strand looks for regions of homology, places where it is similar to the host cells DNA. It will then Bind to the region of homology.
• o Recombination: host genomes DNA is removed and the DNA that is entered becomes part of the host genome.
• o Following replication you have one cell with two copies of your newly replicated DNA.
• • Transformation can be used for Mapping (14:45)
• o 1954 – first demonstration that bacterial genes can be linked by transformation (in pnuemococcus).
• o When transformation occurs 19,000 to 20,000 base pairs of DNA are brought it. This is a small gragment in comparision to a large genome.
• o If two genes are co-transformed (transformed on the same piece of DNA) they are relatively close to each other.
• o EX. Mutation on B. subtilis
• • Have a mutation that leaves a cell that requires galactose to grow.
• • Take a number of pieces of DNA from the B. subtilis genome and transform them into the cell. (20:40)
• • We will transform the mutant strain and determine what percentage of cells NO LONGER require galactose. We will use the strains from 4 separate quadrants along the genomes. Whatever piece it is nearest will have a greater percentage of cells that will be able to grow without galactose (the mutant trait). This in turn will allow you to have a general idea of where the mutation is.
• • The higher the percentage of co-transformation between two genes, the more closely linked in the genome they are.
• • Bacteriophage (38:20) – virus of a bacterial cell
• o Transduction mapping – bacteriaphage is used to transmit DNA from one cell to the next. The ADVANTAGE to transduction mapping is that we can take up more DNA.
• o T4 phage : You have DNA surrounded by the phage head. The phage head is an Icosohedral (20 sided) protein coat. Head is connected to the tail. In the Tail there is a contractile sheath. The contractile sheath functions to punch open holes in the host cell membrane. Tail fibers recognize and bind the host cell specifically
• • Life cycle of the bacteriophage (44:30)
• o Bind to the host cell, mediated by the tail fibers.
• o The contractile sheath opens a hole in the membrane of the cell and the DNA is injected.
• o Viral DNA hijacks the replication and transcription and translation machinery of the cell. It then starts to replicate itself
• o Start to assemble mature phage.
• o You eventually reach a critical mass. When 200 phages have been produced the phages produce a compound called lysozyme. This functions to lyse the host cell and release the new phages.

Genetics Lecture 16, 10/6 - Infectious Heredity, Bacteria, Mapping

Chapter 9 overheads available here.http://docs.google.com/Presentation?id=dhqwrndc_393g3pmdwfm

Chapter 6 overheads available here. http://docs.google.com/Presentation?id=dhqwrndc_384d93nzm8c


Infectious heredity: a trait that is passed to a eukaryotic cell by living in symbiosis with a microorganism

• • Example of infectious heredity in Paramecium urelia
• o Killer phenotype → these cells secrete a compound which is toxic to any cell not having the kller phenotype
• • Two things necessary for killer phenotype
• o Kappa particle a microorganism in the cytoplasm
• o Dominant k allele
• o Paramecium can undergo conjugation (Sexual reproduction) → Genetic exchange occurs during conjugation. Also during conjugation cytoplasmic exchange sometimes occurs.
• • Cytoplasmic exchange is necessary to pass on the killer phenotype
• • For a Heterozygote that undergoes autogamy (asexual reproduction) it will produce two offspring → One homozygous dominant and one homozygous recessive.
• • Maternal Effect – (Non mendelian heredity) (7:50)
• o Gene products in the egg which influence the phenotype. This gene product influences the early development of your offspring.
• o EX. Shell coiling in the snail (linnea peregra)
• • Two types of coils: dextral - right handed coil (dominant trait) and sinistral: left handed coil (recessive trait)
• • Snails are hermaphrodites. They can undergo sexual reproduction or self fertilization.
• • The offspring’s phenotype is directly based on the genotype of the mother. Not based on the offspring’s genotype. An offspring from a mom with a dominant D will be dextral. A mom with a recessive d will be sinistral.
• • Maternal genotype to offspring phenotype.
• • Chapter 6 bacteria and bacteriophages (19:35)
• • Bacterial growth – two types of growth media
• o Complete media – media which contains all of the necessary components for growth. Also referred to as rich media.
• o Minimal media – contains basic chemical requirements for growth.
• o Auxotrophic mutant – any strain which fails to grow on minimal media. Typically has a mutation in a gene necessary for making some component.
• • Bacterial growth curve:
• o Lag phase: immediately after entereing the culture, period of little or not growth. Adjustment time
• o Log phase: cells begin to grow rapidly. Doubling of the population every 30 to 60 minutes
• o Stationary phase: begins when some component of your media becomes limiting. Populations level off and there is little to no new growth
• • How they exchange information. (32:00)
• o In 1946 the first report was issued that bacterial cells can undergo conjugation: one bacterial cell sending information to a second.
• o Experiments which showed conjugation could occur (34:20)
• • Start with Two strains in complete media, both auxotrophs (see pic). They are auxotrophic for different genes.
• • Plated each strain on minimal media as a control. Neither strain grow on minimal media. In a different culture the two strains were mixed the two strains together and incubated overnight.
• • Took the mix and plated on minimal media – produced some prototrophs. This occurred at a rate of 1 in every 10^7 colonies.
• • This indicates that some exchange of information has occurred.
• • Bacteria have the ability to undergo unidirectional transfer.
• • F+ cell is a cell that can donate information. F- cell receives information. Following the transfer the F+ cell becomes F- and vice versa.
• o Does transfer require physical contact between the cells? DAVIS U-TUBE experiment (40:45)
• • Take auxotrophic strains and place them in a u tube. Inbetween the strains place a glass filter - Allows media to be exchanged BUT Cells cannot pass across the filter
• • Plate media from both sides of the filter on to minimal media – found no growth from either side.
• • Indicates that- physical contact is required for gene transfer. This becomes the basis for mapping bacterial genes.
• • 1950 – scientist working with an F+ strain treating it with nitrogen mustard. Found a strain which underwent recombination more frequently . . . 1 in every 10^4 cells. Hfr strains – high frequency recombination strains.
• o Basic mapping experiment to find how Hfr gets to F-. (47:35)
• • Hfr strain with four mutations
• • F- strain without theses mutations
• • 1.) Mix the Hfr strain and the F- strain then (2.)Give them 10 min for conjugation. (3) Pull out some cells, blend your cells. Why? Conjugation is facilitated through a sexpilus long projection from one cell to another through which DNA can pass.
• • plate these cells to look for the passage of the mutation
• • repeat at 15 minutes steps 2 and 3
• • at 20 minutes repeat steps 2 and 3
• • at 30 minutes repeat steps 2 and 3
• • based on your data (results of Hfr mapping)
• azir tons leu+ gal+

Genetics Lecture 15, 10/3 - TEST 2 Review

Test 2!

No lecture.