Unit 3 Notes

October 26, 2009
Bacterial Genetics
1. The discovery of DNA and the Griffith Experiment

What is the moral of the Griffith experiment? Does it show that DNA is the “molecule of inheritance”?
2. Properties of DNA
a. Supercoiled
i. Coiled upon itself several times
ii. Clustered up in a bunch
b. Double-stranded with antiparallel strands
i. Strands are pointed in opposite directions
ii. DNA: sugar-phosphate backbone attached to bases
1. Is what is holding the bases in the order they are in
2. They are facing in an opposite direction (picture in notebook)

c. Complementary base-pairing semiconservative replication
i. For every round of DNA replication one strand is brand new and the other one is recycled from the template strand
ii. One strand is always recycled (picture in notebook)

3. Steps of DNA replication (DRAW PICTURES!) bacteria replication
a. oriC: Where DNA replication beigns on the bacterial chromosome
i. bacteria chromosome is circular and before division is supercoiled on itself
ii. first bacteria must be unwound/uncoiled and unzipped somewhere on the DNA but cannot uncoil the entire chromosome; only a small part of it
iii. this unwinding and unzipping on part of the DNA is at the oriC site
iv. oriC  origin of chromosome replication; where the replication of the chromosome is going to start
1. has a lot of As & Ts little Gs & Cs at it because this is the sequence of bases that is the easiest to unzip because they only have a double hydrogen bond
2. Enzymes bind to the DNA here
a. DNA A  not made out of DNA it is a protein enzyme that binds to oriC site on DNA
i. Brings in another protein DNA B
b. DNA B helicase  enzyme that uncoils the DNA supercoils
b. DNA B Helicase uncoils supercoils around ori C
i. unzips DNA double strands
ii. Primase  enzyme that synthesizes a small swatch of RNA that is complementary to part of the oriC sequence/DNA sequence
1. Primease attaches to unzipped DNA, synthesizes RNA primer
2. Primase synthesizes a little primer swatch of RNA complementary to DNA at oriC (picture in notes)

October 28, 2009
Quarum Sensing – the ability of certain kinds of bacteria to change their gene expression or their phenotype at different population densities
Penotype – the genes that something expresses; set of genes that are expressed; can change your phenotype for example, tanning; set of ch. that you actually see in an organism
Genotype – the set of traits that they have DNA/genes and genetic information for
* Many bacteria have genes that are not always on/expressed under some conditions
Ex) Some bacteria commit suicide at high population densities = “quarum sensing”
S. pneumonia = genes for signal, receptor are both always on; at high population densities the signaling molecules attach to the receptor on the bacteria itself and the bacterial gene expression for bacteria cell lyses from the inside is expressed (Lyt A) gene
When all receptors have signal bound to them, Lyt A gene is expressed and kills the bacterium from the inside out
Pseudomonas aeruginosa = at high population densities makes a biofilm made out of ramnolipids, pyocyanin; made at high population densities
Genes for signal and receptor are always on; when all receptors have signal (Homoserine Lactone) molecule bound to them; genes that are expressed are rhamnolipids and pyocyanin genes for biofilm
Genes for HSL signal molecule and receptor are always on
Signaling molecule sticking to receptor causes expression of the ramnolipid gene and pyocyanin gene
This is a common soil and water bacterium; catheters can introduce pseudomonas aeruginosa and its biofilm to the human body
Vibrio fischer = gm -; expresses a gene called luciferase at high population densities  blue green light; uses same signaling molecule as pseudomona aeruginosa (HSL) gene for signal molecule and receptor are always on
When all signaling molecules are attached to receptors on bacteria then luciferase gene is expressed; makes bacteria glow somewhat blue/green
Has a symbiotic relationship with a squid called Eurympmna Scolopes
The squid is nocturnal and a predator to small animals and also is prey for larger animals; off the coast at Hawaii; the squid emitting this kind of light has the same glow as the moon does on the ocean; is able to glow and cancel its own shadow because it has an organ in its head called the “light organ” that is specifically colonized with these bacteria (vibrio fischer)
“Light organ” in the head is volonized w/ Vibrio Fischeri allows high population density of Vibrio Fischeri to express liciferase
Lecture Review
A) OriC site – bacteria have circular chromosomes; at OriC site this is where DNA replication begins on the chromosome; has a whole bunch of A and T because they only have 2 hydrogen bonds between them and not 3 like G and C
DNA has to be unwound before it can be replicated; at OriC site, locally the DNA has to be unwound; only small portion is unwound and unzipped
DNA A binds to the OriC site; and recruits DNA helicase which unwinds the DNA at the OriC site
B) Primase – kind of RNA polymerase is going to synthesize swatch of RNA complementary to bit of DNA – this will allow DNA polymerase enzyme to attach to the strand of DNA, and start copying it
To keep DNA strands separate, single-strand binding protein attaches to unzipped strands to prevent re-annealing
c. DNA polymerase attaches to the RNA primer, synthesizes DNA
i. DNA polymerase synthesizes new DNA complementary to the template strand moving in 5’  3’ direction
ii. DNA polymerase enzyme is going to bring in bases that are complementary to the bases on the single strand of DNA (template strand) and match them up and hook the bases together using a sugar phosphate backbone
iii. This enzyme can only move/travel in a 5’ to a 3’ direction
1. This is an issue since the DNA strands are antiparalell
d. One strand synthesized continuously, but opposite strand has to be synthesized in pieces – it is not synthesized continuously
i. Okazaki fragments on lagging strand
ii. The fragments have to be patched together by DNA Ligase who’s only job is to patch the copied of the lagging strand fragments together
e. DNA replication enzyme sets are moving in opposite directions around the circular chromosome meet at the terminus site, DNA replication stops here
f. Interlocked circles that we got from our complete round of DNA replication have to be uninterlocked/detached
i. Done by 2 enzymes:
1. DNA gyrase
2. Topoisomerase IV
4. Repair to errors in DNA replication
a. Spontaneous mutation rate about 1 mutant base pair per 10 billion base pairs copied
i. Size of bacteria genomes
b. A lot of bases are being copied so it is easy for mistakes/mutations to be made
i. Mutation rate 1:10 ^ 10 bases copied is different from original base in the original template strand
1. Bacterial genome sizes
a. E. Coli - ~ 5 million (5x 10^6) base pairs (bps)
i. Can live in nature w/o living in us
b. S. Aureus ~3 million bps
c. Chlamydia trachomatis ~1 million bps
i. Needs to live in a host; cannot survive out of a host
d. Mitoplasmid genitalium ~500,000 bps
i. Does not have a cell wall and cannot live outside of a host
e. Microbes that can survive independently in nature tend to have larger genomes than microbes that can only survive in a host (thus larger genomes w/ more bps) than microbes that need to live in a host
i. Viruses are an extreme example of this; they have much more minimal genomes than any bacteria do
a) They have very small genomes b/c have few genes
f. Bacteria have different sizes of genomes and different numbers of base pairs
i. Relationship between what bacteria are able to do for themselves cellularly and how many base pairs there are in their genome
c. How many mutant bacteria are in a colony of E. Coli?
i. 1x10^9 E.coli x 5 x 10^6 bps
1. = 5 x 10^15 bps in colony / 1 x 10^10 mutation rate
2. = 5 x 10^5 or 500,000
ii. Suspect that an isolated colony is al clones of one individual bacterium but there were actually 500,000 mutations that occurred
d. Mismatch repair system
i. Find mismatched bases
ii. After the DNA strands have been copied
1. Bases that are not matched correctly are identified by enzymes that find things that are not matched
2. Identified by proofreading enzymes
3. Needs to figure out which side of the DNA replication is wrong
4. How do the enzymes know which strand is the correct strand and which one we want to remove the strand from?
5. Keep older one and delete the newer one
iii. Methylated DNA is older
1. Keep this strand
2. Need to ID the methylated strand
iv. Endonuclease
1. Cuts the single strand of DNA on the non-methylated strand of DNA on either side of the kinky part on either side degrades stretch between the cuts
2. DNA polymerase comes in and does not need to synthesize a primer because there are already bases there; it synethesizes the new strand to fill in the gap
3. Tension is released now
How many mutant bacteria are there in a colony of 1 billion E. coli?
1. Protection of bacterial chromosome from degredation
a. Methylation of adenines and cytosines
i. Adding methyl residues (CH3) to As and Cs on bacterial chromosome as last DNA replication step
ii. Consequences
1. Bases miscopied in DNA replication the fact that the old DNA is methylated and the new DNA is not helps the mismatched repaired system to know which strand is wrong and which is right
iii. Advantages:
1. Allows mismatched repair system to identify old vs. new (i.e. right vs. wrong) DNA strand
2. Allows identification of potential viral DNA
b. Restriction enzymes
i. Enzymes cut DNA into little fragments when it comes across particular sequences of DNA
ii. Enzymes that recognize or bind to particular sequences of DNA bases in a DNA strand and break DNA strand there
iii. Ex) particular enzyme that recognizes the sequence 5’ ATCCCGCA3’
1. Only binds to that sequence in a chain of DNA bases
iv. Bacteria make lots of different restriction enzymes that bind to/break DNA at lots of different sequences
v. Defense against viral infection
1. Viral DNA not methylated

October 30, 2009
1. Ground Zero
a. Chapter 1: Doctrix Robson’s Checkered Past
i. Herpes Simplex 1 from guy named Jeff
b. Chapter 2: Viral Replication
i. Viruses need a host cell to replicate
ii. b/c they need a host cell to copy genome, not quite alive
iii. 1) Virus attaches to a host cell because of specific protein receptors on the host cell
1. Interaction btw. Specific protein on host cell surface & viral protein allows this
2. Bacteriophage: capsid stays on outside host cell, DNA injected into cell
a. Capsid – protein shell surrounding viral genome
3. Viruses that infect eukaryotic cells:
a. That virus will be engulfed by the host cell and taken in by the host cell; enzyme sin our cells break down the capsid surrounding its genome
b. Capsid broken down by our enzymes; which frees the genetic material of that virus (some viruses have DNA genomes and some have RNA genomes)
iv. 2) Viral genes expressed (i.e., genetic material of virus transcribed & translated into protein) by host cell ribosomes (machinery), etc.
1. & viral genetic material copied by host cell enzymes (DNA polymerase, etc.)
v. 3) New viral genetic material (copied) packaged into new capsid proteins  resulting in new virus particles!
1. some viruses: DNA or genetic material in general of virus alone, in cell  new functional viruses
vi. 4) viruses released from host cell
1. Eukaryotes: viruses often “bud” off of host cell  not necessarily fatal to host cell but viral infection does keep cell from doing its job  can  death (ex. HIV)
c. Chapter 3: Viral Affinity for Host Cells
i. Viruses that we get can only infect humans and may only be able to infect certain cell types in humans
ii. Because there has to be a specific interaction between host cell and viral cell it limits the cells that the virus can infect
iii. Viral interactions with host cell proteins very specific, so specific viruses can only infect specific cells
1. Herpes Simplex 1: can only infect epithelial cells and neurons; cold sore virus
a.  infects epithelial cells and neurons because these are the only cells that have the right proteins on their surface for the herpes simplex virus to attach to replicate and so forth
b. Fled from the epithelial cells and to neurons in the spinal cord that feeds the lip since they were attached at the lip because the immune system was attaching them there
2. The Curious Case of Dr. Wimmer
a. Chapter 1: Host cells, host species, and the potential for disease eradication
i. Eradication – to eliminate something utterly from planet earth; to destroy it
1. Destroy a disease so that it can never infect anyone ever again
2. Destroy viruses access to new host cells so they will all die
a. Hard if a viral disease has multiple host species
3. Good candidates for eradication:
a. Viral diseases that only infect humans and have reliable vaccine against
4. All had to do was break mode of transmission everywhere so that polio could not move from one host to another host it would be wiped off the face of the earth
b. Chapter 2: Polio
i. Polio virus: RNA genome; can only infect intestine epithelia cells and neurons
1. In nature can only do this in humans
ii. Causes low level gut symptoms that go away quickly
iii. If it is acquired when the child gets older causes paralysis of the neurons of the lower legs
iv. Eradication – to eliminate something utterly from planet earth; to destroy it
1. Destroy a disease so that it can never infect anyone ever again
c. Chapter3: Viral synthesis
3. Death Rides a Pied Horse
a. Chapter 1: The pox!
i. Large virus with double stranded DNA genome
ii. Only infects human epithelial cells
iii. Mortality Rates:
1. 30% in experienced population
2. 90% in naive population
b. Chapter 2: Hope springs eternal
c. Chapter 3: Victory
i. Wars stopped so that vaccine workers could come on and resume killing each other
ii. Small pox cases dropped dramatically after the small pox vaccination campaign
d. Chapter 4: Tragedy
i. Soviet’s kept a vial and the united states kept a vial to use as warfare
ii. The soviet’s vial was kept under lock and key until the soviet union fell
e. Chapter 5: The end?

November 4, 2009
a) OriC site is where DNA replications starts and has a lot of As and Ts which makes it easier to unwind and unzip
b) DNA A attaches to OriC site to unwind the DNA; this recruits enzyme helicase which then unwinds the DNA locally

November 6, 2009
DNA is the set of instructions that established why a particular organism has the set of characteristics that it has; molecule of inheritance
Organisms are mainly made out of proteins however
a. DNA  instructions (blueprint)
b. Protein building blocks of an organism
c. DNA has to be made into RNA instructions in order to provide the template for the proteins that we are made of
d. This process is true of all living things including bacteria
e. Gene ~ DNA instructions for a particular protein
i. Cook book has a lot of recipes but you don’t cook them all at the same time you choose which ones you want to cook
ii. Some genes are expressed or “turned on/off” at particular times
iii. Example) biofilms at high population densities of pseudomonas auerogonosas
1. High level of signal molecule triggers a change in the genes that are expressed
1. Transcription: DNA  RNA
a. Initiation
i. The start of transcription
ii. Determining which bit of DNA will be copied to RNA
iii. Promoter sequences:
1. Where transcription initiation occurs: Where DNA is unwound and unzipped so RNA copy can be made
2. A lot of As and Ts because there is only two hydrogen bonds between them which will facilitate the unzipping of the genes
3. TATA box TATAAT like sequences at almost all promoter sites
4. While have common TATA box sequence, also have variability in surrounding bases
5. Specific sequence of bases on DNA that RNA polymerase attaches to
a. RNA polymerase enzyme that synthesizes the RNA strand copy of DNA strand
iv. RNA polymerase: sigma factors
1. Enzyme that copies DNA into RNA copy
2. Only one strand of the DNA is being copied
3. RNA copy is single stranded, complementary to ONE DNA strand
4. RNA polymerase binds to promoter sequences on DNA
a. Sigma factor (subunit) of RNA polymerase that binds to DNA promoter sequences
b. Most responsible for enzyme attachment to DNA promoter sequence
c. Different sigma factors attach to different promoter sequences
d. Different sigma factors expressed under different conditions
e. Different sigma factors bind to different promoters regions on the DNA
f. Different sigma factors produced in response to different environmental conditions
v. DNA must be unwound, and unzipped before it can be transcribed
1. This is b/c single DNA strand serves as a template for RNA copy
vi. Promoter sequence on DNA
vii. Rifampin binds bacterial, but not eukaryotic or archaeal, RNA polymerase

November 9, 2009
b. Elongation
i. Process by which RNA strand is actively synthesized by RNA polymerase
ii. RNA strand synthesized by pairing complementary RNA bases w/ DNA template strand & hooking those bases together w/ sugar-phosphate backbone
1. Sugar is ribose in the sugar-phosphate backbone
2. RNA sugar = ribose
3. RNA uses Uracil instead of Thymine to pair up with Adenine
iii. No matter where the promoter sequence is it has a lot of Ts and As
c. Termination
i. RNA polymerase must detach from DNA strand
ii. DNA strand is folded back on itself “hairpin loop”
iii. Since RNA polymerase can only move on a single strand of DNA the RNA polymerase falls off when it gets to this hairpin loop
iv. RNA polymerase reaches hairpin loop, can’t get past it b/c RNA polymerase can only move along ss DNA, so RNA polymerase falls off DNA strand, releasing new ss RNA
2. Kinds of RNA
a. mRNA
i. messenger RNA
ii. single stranded RNA that was transcribed off of the DNA and is going to be translated into protein strand
iii. single stranded RNA copy of DNA gene, will be translated to amino acid sequence
iv. tend to be very short lived—only exist for a few seconds if they aren’t actively being transcribed or translated
b. tRNA
i. transfer RNA
ii. involved a lot in the translation of messenger RNA into an amino acid sequence
iii. transfer RNA has a clover leaf structure
iv. also a single stranded RNA but is base-paired on itself much like the “hairpin loop”
v. Clover leaf shaped structure
vi. on the one side of tRNA there is an amino acid attached to it
vii. on the other side there is a swatch of just a few bases that can base pair with a sequence on the messenger RNA
1. this is called an Anti-Codon
viii. tend to be very long lived—exist for days or even weeks
c. rRNA
i. ribosomal RNA
ii. main thing that ribosomes are made of
iii. ribosome are the site in the cell that actually do the translation of mRNA into amino acid sequences
iv. long lived – exist for days or even weeks

November 11, 2009
1. Overview of translation:
a. DNA gene mRNA transcriptamino acid chain
b. Codon: 3 mRNA nucleotide sequence that codes for specific amino acid
i. 3-base long sequence on mRNA that corresponds to particular amino acid
1. Total of 64 possible different codons
2. Only 20 different amino acids that correspond to those codons
3. Have to have more than one possible codon for each amino acid
ii. Redundant codons—Why?
1. Idea that we have more than one codon per amino acid
iii. Know how to use table of codons, like the one attached
c. Ribosome binds to mRNA
i. Single stranded mRNA
ii. Ribosome attaches to mRNA at the Shine-Dalgarno sequence
1. Specific series of bases on the mRNA that are complementary to a part of the RNA sequence in the ribosomal RNA
2. Complementary to a sequence of bases in rRNA
d. AUG= start codon; also, methionine
i. Ribosome moves 1 base at a time down mRNA until it reaches AUG
1. This defines “Reading Frame”
ii. Every three bases after the AUG start codon now is a codon
iii. Each codon corresponds to a particular amino acid
iv. tRNAs bearing amino acids brought in by ribosome, matched to codons
e. amino acids linked into a chain by ribosome
i. tRNA are clover leaf shaped
1. Clover leaf-shaped w/ anticodon on one side, amino acid on other
2. rRNA brings in tRNAs and matches them with codons on mRNA
3. tRNA has the amino acid on the other end of it opposite the anti-codon that is paired up with the codon on mRNA
f. at stop codon (UAA, UAG, or UGA), ribosome releases mRNA and amino acid chain
i. rRNA forms bond btwn amino acids that are adjacent to each other—tRNA detached from amino acid
ii. continues down the line until the stop codon is reached
iii. there is no tRNA that corresponds to the stop codon (no anti-codon)
1. there is no tRNA that has an anti-codon that corresponds to the stop codon
iv. b/c no tRNA w/ anticodon complememntaray to stop codons , the rRNA pauses at stop codon-
1. proteins called “Release Factors” interact (bind) with the stop codon instead
a. they also are enzymes that catalyze the reaction detaching the amino acid chain from ribosome & ribosome from mRNA j
b. A+ stop codon, Release Factors bind w/ ribosome, mRNA stop codon, RF detach ribosome from mRNA, amino acid chain from ribosome
g. Polyribosomes
i. One mRNA w/ lots ribosomes attached to it, in different stages of translating mRNA
Problems on Exam
Given a DNA sequences that will include in it a promoter region and one it is transcribed will contain a shine dalgarno sequences
Transcribe DNA  RNA and RNA amino acid sequence
Have to identify the promoter region/sequence (TATATATA) but shine delgarno sequence will be given
2. Kinds of mutations
a. Any change in the DNA sequence of an organism; NOT a change in gene expression
b. Point mutations
i. Mutation to DNA that changes just one base
ii. Missense
1. Mutation that changes the sequence of the DNA and changes only ONE base in the DNA sequences but this is enough to change the amino acid sequence expressed
2. Changes to DNA sequences  change to amino acid sequence of protein
3. Changes the mRNA transcribed and the amino acid sequence translated
iii. Nonsense
1. DNA changed so that the mRNA has a stop codon earlier than it did before the mutation
2. Change to DNA sequence so there’s now a early stop codon in mRNA, so short protein results
iv. Silent
1. Way more codons than amino acids
2. Many amino acids that have a codon that codes for them
3. Changes in DNA sequence but the resulting in no change to the amino acid sequence
c. Frameshift mutations
i. Add or delete 1-2 bases from our DNA sequence
ii. Shifts the reading frame 1-2 bases shifting the mRNA sequence as well as the amino acid chain sequence
New Handout
A story about gene expression and Armistice Day
1. Clostridium perfringens
a. Gm+ bacteria that live in the soil
b. Soil bacterium common in U.S. and Europe
c. Strict anaerobes!
i. Cant grow in the presence of any oxygen which is why they are found in the soil
d. Spore-forming
i. Spores
1. Very hardy/sturdy non-metabolic state that some gm+ bacteria can form into/enter
2. When they are spores they are not using any energy/dividing/transcribing/translating
3. They hang out in this spore state
ii. Spore-forming sigma factors
1. Specific sigma factors expressed in the presence of oxygen
2. Spore genes are not on all the time; only when the bacteria are among too much oxygen
3. Spore-gene sigma factors in high [O2], low [H2O]
4. Turned on when in the presence of oxygen
e. When actively growing it expresses a virulence factor
f. If environment is dry, too much 02  sigma factor specific to genes needed for spore formation transcription of spore genes  translation of those spore proteins
i. C. perfringens has what it needs to make itself into a dense coat (spore) and not be metabolically active but still live
g. If spores find themselves in a moist environment that is warm, with very low (o2) concentrations, then a different set of signma factors is going to be produced
i.  vegetative sigma factors; sigma factor associated w/ the bacteria actively growing
ii. These sigma factors  transcription of growth-related genes (including phospholipase) translation of those proteins or growth related proteins (including phospholipase)
2. Virulence Factors
a. Proteins that express a particular symptom in a host
b. Phospholipase
i. Enzyme that breaks down phospholipids which breaks down our cells
ii. Produced by C. perfringens when in a warm, moist, O2 low environment
c. Soldiers boots became an anaerobic environment with soil in them containing this bacteria
i. Tissue is killed by phospholipase
d. Exotoxins
i. Phospolipase
ii. Regulation of phospholipase gene expression
e. Acinteobacter baumanii (“Iraqibacter”)

November 16, 2009
New Handout
Took bacterial samples out of the paper and found that many of the bacteria were resistant to a lot of antibiotics
Figure 2: Letters that are circled correspond to the mutations that occurred; not limited to As and Ts and Cs and Gs; The letters are one letter long abbreviations for amino acids; MIC mcg/mL means Minimum Inhibitory Concentration; concentration of Cipro that is needed to inhibit the growth of certain bacteria; Cipro attacks DNA gyrase so DNA cannot unwind
MIC = minimum amount you have to have before the bacteria are killed by the abx; figure has minimum concentration of cipro needed to inhibit the growth of certain bacteria
Mutation is not silent, frameshift, could be missence mutation because it changes one DNA that changes the amino acid sequence of the resulting protein

1. A story about H1N1 and World War I
a. 1918 flu was eradicated
b. 50 million people died but for the people who didn’t die of the 1918 flu, virtually every single human being got the flu and became resistant to it so the strain changed because everyone was resistant to it
2. Classification of viruses
a. Viruses have narrow host ranges
i. Viruses tend to infect particular host cells or a few species at most
1. Most viruses have only 1-2 host species
2. There are some viruses that can infect multiple hosts however
ii. Implications for disease eradication
1. What traits must a disease have for it to be potentially eradicable
2. The disease would ideally have to only infect humans; if there was a vaccine it is easier to get all humans vaccinated
3. For a disease to become eradicable
a. Pathogen has to require a host for it to survive
b. That host that it needs to survive has to be us or humans only
c. Need to have a solid vaccine or preventive measure to prevent the spread from host to host and break that chain of transmission
d. New strain of flu every season because if someone gets it and survives they become resistant to it

Novemeber 18, 2009
Programmed Ribosomal Frameshift: Ribosome shifts forward or backward on the mRNA,
PRF B gene (programmed ribosomal framshift) – codes for the protein release factor 2 makes the ribosome fall off at a specific stop codon (UGA and UAA). RF2 recognizes UGA and UAA.
PRFB gene codes for realese factor 2 which’s job is to recognize these two stop codons and detach the RNA from the mRNA
What will happen to translation if mRNA if:
1) There’s a lot of RF2 protein available? The A.A. sequence would not stop and it would shift forward 1 because it is a +1 shift. So the ribosome shifts forward one and gives it a new reading frame. This is NOT a framshift mutation. The ribosome moves just one base forward.
RF2 production is regulated because if there is a lot of RF2 present then we don’t need ot waste energy making RF2 so there is very early stop codon. If RF2 is not present then the AA chain goes on until it codes for a RF2.
2) There’s very little/no RF2 protein available? RF1 will still cause it to fall off
1. Virus structure
a. Nucleic acid (NOTE: some viruses have RNA genomes, not DNA ones.)
i. Virus genomes come in a variety of sizes
1. Can be DNA OR RNA
2. Variety of viral genome sizes
3. A large virus such as small pox
a. Has a DNA genome that is about 300,000 base pairs in length
4. Smallpox will have 300-1500 proteins coded in its genome
5. Polio
a. RNA genome that is only about 3,000 base pairs in length
b. If there is no RFS can only get 3-15 different proteins that it has in its genome
6. Generally, 1 gene or protein ~ 1,000 base pairs long
7. 3 bases in a codon so you can shift over one base or two bases; but if you shift over 3 bases in the same reading frame again
ii. Viroid: Self-replicating double-stranded DNA circle
1. Independently replicating circle of DNA
b. Protein coat
i. Capsid
1. Outer protein layer
2. Made of capsomeres
ii. Capsomeres
1. Small repeating protein units that make up capsid
2. Genes for capsomeres are viral genes not genes that they get from the host cell; from VIRAL genome
iii. Virus shapes: Viruses come in a limited range of shapes. Why?
1. There are virtually only 2 shapes for a virus cell to be
a. Viruses only have so many genes t o work w/
b. Virus does not have a lot of space to work with a lot of capsomere genes to change the shape
c. This is because the viruses have few genes and as a result can code for few proteins and as a result only have a few capsomere protein possibilities which code for the same shape
2. Polyhedral
a. Shaped like a soccer ball
b. Has a capsid which is the same outer protein layer surrounding nucleic acid
c. Can infect eukaryotes
3. Helical
a. Shape of viruses that infect eukaryotic cells
b. Like a telephone cord
c. Has outer protein layer (capsid) with amino acids within the tube
d. Can infect eukaryotes
4. Both helical and polyhedral are able to infect eukaryotic cells
5. Complex (only bacteriophages)
a. Mashe up of polyhedral shapes and helical shaped viruses
b. Shape of T4 phages
c. ONLY bacteriophages are this shape; this shape NEVER infects human cells only infects BACTERIA
c. Enveloped viruses also have a membrane, derived from host cell membrane (NOTE: enveloped viruses only infect eukaryotic cells)
i. Enveloped virus
1. Virus that has a cell membrane outside of the capsid
2. Mainly infect eukaryotes
3. Many though not all eukaryote-infecting viruses have a phospholipid bilayer membrane
4. Envelope is made of phospholipids and is called a phospholipid membrane
5. This membrane is taken from the host cell and wrapped around the outside of the capsid
6. This membrane is attached by another kind of protein (glycoproteins)
d. Glycoproteins
i. Proteins responsible for attaching a virus to a host cell
ii. Attach membrane to capsid in envelope
iii. Used to identify subcategories of viruses
iv. Actually attaches the virus to the host cell

November 20, 2009
Helical and polyhedral viruses are the only viruses that infect humans.
a. Virion = single virus particle
2. Viral replication
a. Attachment/absorption
i. Attachment
1. Virus and potential host cell
2. Virus has to attach to the potential host cell
3. Interactions between viral proteins (usually glycoproteins) & receptors on host cell
a. Receptors – is what the virus attaches to but is not its point it is used by the host cell to do other jobs
4. Shapes of viral glycoproteins and receptors on the host cell fit together
5. Virus sticks to host cell
ii. Absorption:
1. Once virus attaches to host cell the cell has an impulse to pull the virus into itself
2. Cell pulls in virion attached to surface
3. This creates a vacuole containing the virus
b. Penetration/decoating
i. Vacuole containing virion fuses w/ lysosome, which contains acids and digestive enzymes, breaks down proteins
ii. This dissolves the capsid which frees the nucleic acid of the virus
c. Replication of nucleic acids
i. Virions have nucleic acids & a few enzymes needed to start the host ‘s replication of their nucleic acids packaged in the viral capsids—these enzymes are also now free to help viral genome get replicated
ii. DNA viruses
1. Enzymes (viral) help get viral DNA into nucleus (eukaryotic viruses) and patch viral DNA into host chromosome (both eukaryotic viruses and bacteriophages)
iii. RNA viruses
1. RNA as genome
2. Need their own enzyme like RNA polymerase that instead of copying DNA to RNA it copies RNA genome to more RNA
3. No transcription is done
4. Uses host ribosomes and tRNA and amino acids to start translating their RNA into the proteins they code for
iv. Retroviruses
1. RNA genome but instead of enzymes that just copy the RNA to more RNA copies, they have another enzyme that is called reverse transcriptase
2. Copies RNA genome to dsDNA (double stranded DNA copy) copies
3. That DNA version of viral genome then patched into host cell chromosome and all the rest is transcribed and translated into proteins
4. Example: HIV
d. Synthesis of viral proteins
i. If we have a DNA virus who’s DNA has been patched into host chromosome, now we need to synthesize a bunch of new viral proteins
1. Will need a TATATA box (promoter sequence) and an AUG start codon on the RNA
ii. DNA virus proteins transcribed and translated off of viral genes in viral DNA that’s now in host chromosome
iii. Copies of the DNA genome are being made as well while the transcription and translation is taking place
iv. RNA virus
1. RNA itself is a transcript so it does not have to be transcribed it just has to be translated by the host ribosomes and amino acids and tRNAs
v. Retrovirus
1. Mix between DNA virus and RNA virus
2. RNA is coped into a double stranded DNA and patched into chromosome and does it like a DNA virus would to copy its genome
3. DNA version of viral RNA genome patched into host chromosome transcribed and translated; RNA copies of viral genome transcribed to be packaged into new virions
e. Assembly of new virions
i. When viral genes have been expressed (i.e. proteins coded for by genes have been made) viral particles spontaneously form (capsomeres aggregate around viral genetic material, glycoproteins collect on capsomeres)
f. Release of new virions from host cell
i. Lysis of host cell
1. If the host cell dies all the virus particles are then released
2. Cell destroyed, this frees viral particles to infect new cells
a. Ex) T4 phages and E. Coli ; E.coli that weren’t resistant were caused to blow up and die
b. Not common of viruses that infect eukaryotic cells
ii. Budding of host cell
1. Most common for viruses that infect eukaryotic cells
2. Virus particle twists off part of the host cell membrane and comes out leading to new enveloped virus with its membrane envelope being pulled off of the host cell membrane
3. Virion uses glycoproteins to grab on to host cell membrane, wrap self in membrane and yank itself away—most common way of releasing virions from eukaryotic host cells
4. All enveloped viruses have to bud off of the host cell they cant come out by being lysed from host cells
3. Consequences of viral infection on host cell
a. Death of cell
i. Happens but usually not right away
ii. Host cell could die if directly lysed by the virus
iii. Cell directly lysed by virus
iv. Cell isn’t directly lysed by the viral cell but dies not directly b/c of virus
1. Cell dies because our immune system recognizes that the cell is a virus cell and kills the cell b/c its virally-infected
2. Viral replication taxes cell function seriously enough that cell cannot keep up with day to day tasks and dies
3. Does not happen often
b. Disruption of cell function
i. Disruption of cell function
ii. Cell is alive but unable to do much more than make virions
iii. HIV kills because it does not kill immune cells but makes them do nothing with their day but make more HIV virions and cannot fight off other bacterial infections
c. “Transformation” of normal cell into cancerous cell—why?
i. Transformation – change of a genetic function of a cell
1. Changing a cell from having one set of genetic characteristics to having another set of characteristics
2. Rough bacteria in mouse experiment  glossy bacteria
ii. Ex) HPV type 16 and 18 cause cancer  infect epithelial cells and integrate their own DNA into host cell DNA; DNA genome of HPV integrates into host cell DNA
1. Prevents epithelial cells from committing suicide on schedule or apoptosis in infected cells  makes cell cancerous
2. If continued replication then there is more cells for HPV to infect
iii. Cancer cell is a cell that replicates and replicates and never apoptosis
4. Viral culture
a. Animals or animal cells
b. Have to have the right kind of host cell that the virus infects in order to grow the virus
c. Must grow viruses in appropriate host cell
“Immortalized” human cells : Henrietta Lacks and HeLa cells
Henrietta Lacks died of cervical cancer due to HPV 18
Henrietta lacks cells are grown in culture since 1953; her cells grow and grow because they are cancerous

November 30, 2009
Surprise Kitty
Host-pathogen interactions:
Size of Bacteria:
Size of Virus: 1-5 micrometers 10 ^ -6
RNA viruses use RNA replicase to copy genetic material
• Viral Culture
o Since viruses require host cell to replicate themselves, therefore you cannot grow viruses in pure culture the way you can bacteria on ager; you have to provide them a host cell. Virus that infected humans and chickens; you could culture the virus in the chickens
• Cell culture:
o If we need to grow a virus that only infected humans is to use cell culture.
o Human cancer cells are usually used because the divide indefinitely.
o Unlike normal cells that commit cellular suicide
o Have to provide cancer cells with nutrients; must be placed in flask or dish w/ nutrients
o Cells then are placed across the bottom of the flask
o Fluid placed on top of cells similar to blood plasma
• Henrietta Lax
o Most common source of human cancer cells that we use in growing viruses
o Her cells are called helicells after her
o Common cultured cancer cells = HeLa cells (name for person who died of her cervical cancer) in 1953
o HPV 18 killed her – she was infected w/ this which among other things destroys the pathway in normal epithelial cells that makes the epithelial cells commit suicide after too many generations (apoptosis). This made infected cells cancerous
o Used her cells to make more HPV-18
o Cancerous cells are misshapen; the nuclei are humongous; these are both consequences of these cells cranking out more virus particles
• Do viruses want to kill their host?
o NO; they need the host to keep replicating
Questions to Consider:
How do pathogens and hosts influence each others’ evolution?
Does ecology contribute to the ability of pathogens to cause disease?
Do pathogens inevitably evolve to become less virulent to their hosts?
Why do bacteria make “toxins”?
What is the difference between a pathogen and normal flora?
1. Terminology
a. Incidence vs. prevalence
i. Incidence of a disease – number of new cases of a disease in a population / total number of people in whole population
ii. Prevalence – total number of cases of a disease no matter if new or old cases; total number of cases in a population / total number of people in population
b. Epidemic
i. Increase in incidence of disease; increase in number of new cases of a disease
c. Reservoir
i. Where microbe (that  disease) exists in nature
1. HIV reservoir is humans or the human body
2. Clostridium perfringens reservoir is soil
3. V. Fischeri reservoir is squid, seawater
ii. Over time, most microbes become less virulent in reservoir especially if their reservoir is a living thing
d. Vector
i. Organism that transmits infectious disease from host  host
1. Ex) Black Plague – Y Pestis (plague) vector = flea
e. Virulent
i. Ability of microbe to cause disease in a host
f. Virulence factor
i. Any protein that is produced by a pathogenic microbe (kind that can cause disease) that contributes to that microbe’s ability to cause symptoms
g. Invasive disease
i. Infections disease where symptoms b/c of microbes growing in what , in a healthy host, is a sterile site
1. Heart
2. Lungs are NOT sterile
3. Blood
4. Muscle tissues
5. Brain
ii. Invasive because it is more difficult for the microbes to get to the sterile site in the host
iii. Invasion factors
1. Subcategory of virulence factors
2. Chemicals produced by pathogens that allow those pathogens to get into a normally-sterile tissue
3. Pathogen = microbe that causes a disease
iv. Disease due to microbe growing in a normally sterile environment
h. Toxigenic disease
i. Infectious disease caused by microorganisms where the symptoms are caused by a toxin produced by the microbe growing in a normally NOT sterile environment producing “toxin” that  symptoms in host
ii. Productions of toxin protein that causes symptoms in the host
1. Ex) Toxic Shock Syndrome from tampons
a. S. Aureus can produce a protein call TSST-1
iii. Exotoxin
1. Protein expressed/secreted by actively-growing bacteria leading to symptoms in a host
iv. Endotoxin – Lipid A of LPS
1. Not secreted by actively growing bacteria, it is simply just part of the gram- bacteria membrane and can only be released if the bacteria dies
2. Two examples of tonxigenic bacterial disease
a. Botulism
i. Clostridium Botulinum
1. Like other clostridia sp.
2. Strict anaerobes
3. Spore-forming
a. In response to high oxygen conditions
b. Spores
i. Very sturdy, non-metabolic ; suspended-animation state of bacteria
4. Environment is the soil (reservoir)
ii. Botulism Toxin
1. Actively growing Clostridium Botulinum expresses Botulism Toxin (exotoxin)
2. Protein that is expressed by actively growing bacteria which has the effect of causing symptoms in us when we ingest it
4. The most toxic substance known to man on planet earth
5. Deadly dose: 1ng/kg weight of person
6. Botulism toxin
a. Blocks acetylcholine receptors in our neurons (all neurons that feed muscles) which leads to suffocation
b. Blockage of these motor neuron receptors causes paralysis of diaphragm cause suffocation
7. BOTOX in face
iii. Botulims food poisoning
1. Boiling canned foods does not kill the botulism spores but kills the actively growing bacteria
2. Prepping of canned food drives all oxygen out of the jar of vegetables creating an anaerobic environment
3. This environment created by canning process allows for botulism spores to germinate and grow into actively growing bacteria of C. Botulinum spores
4. They express the botulism toxin gene creating a lot of botulism toxin in the vegetables that were canned
5. Anything that denatures proteins (heat) can kill off the botulism and it will be safe
6. Don’t have to actually ingest the botulism bacteria but ingest the toxin
iv. Infant botulism and normal flora
1. Human body has a lot of not normally sterile environments that are anaerobic
2. Guts are one of these sites
3. Human gut is a non-sterile environment, anaerobic
4. C. Botulinum spores in soil, elsewhere
5. Normal flora prevents the botulism spores from having a place to grow
6. Gut a good place for C. Botulinum to grow – except that in healthy adult, normal flora fills spots for C. Botulinum to grow
7. Infants when born are sterile and do not acquire normal flora until after they are born
a. Infants have little enormal flora to displace C. Botulinum bacteria
b. Should not feed baby honey; they do not have the normal flora to displace the C. Botulinum and as a result they make botulism toxin and they die
8. “Floppy baby syndrome” due to interactions between clostridium botulinum spores that the baby ate from being fed honey and spores germinated in the baby’s gut and there was no normal flora to displace the botulism; while growing produced toxin and traveled throughout the rest of the babies tissues
b. Cholera
i. Vibrio Cholerae
ii. Cholera toxin
iii. Cystic fibrosis and heterozygote advantage
iv. Copepods
v. Epidemic cholera and lysogenic conversion
3. Two examples of invasive bacterial diseases
a. Plague
i. Yersinia pestis
ii. Virulence factors
iii. Fleas and rats
iv. Human history
b. Scarlet fever
i. Streptococcus pyogenes
ii. Virulence factors
iii. Strep throat and flesh-eating bacteria
iv. What happened to scarlet fever?
4. Ewald’s hypothesis
What is the difference between a pathogen and a symbiont?

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