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186 Cards in this Set

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Differences between prokaryotes and eukaryotes and viruses
Bacteria are smaller than eukaryotes, move via flagella, have a circular dbl stranded DNA chrmsm

Eukaryotes have mitochondrian and/or chloroplasts, a nuclear membrane, a cytoskeleton, and a nucleus.

Viruses are acellular, need cell machinery to make proteins and get energy. Virus genetic material can be ssRNA, dsRNA, ssDNA, or dsDNA. A bacteriophage is a head and tail virus. One phage = one plaque. Viral life cycles can by lysogenic or lytic. A lysogenic life cycle is one in which the virus incorporated its genome into the bacterial genome where it dvides passively. A lytic life cycle consists production of more viral DNA, construction and cell lysis.
What kind of genetic material can viruses have?
Virus genetic material can be ssRNA, dsRNA, ssDNA, or dsDNA
What is a bacteriophage?
A bacteriophage is a head and tail virus. One phage = one plaque.
diff life cycles of virus?
Viral life cycles can by lysogenic or lytic. A lysogenic life cycle is one in which the virus incorporated its genome into the bacterial genome where it dvides passively. A lytic life cycle consists production of more viral DNA, construction and cell lysis.
Why are bacteria and bacteriophage good experimental organisms?
They are simple, fast and easy to grow, easy to radioactively lable, and WT is prototrophic (can grow on simple medium)
Describe what happends at the bacterial replication fork
Autoradiography shows that bacterial circular chromosomes form theta structures when replicating. Viewing a bacterial chromosome requires growing bacteria dor several generations on radioactive thiamine. The gorwing it for ½ generation on regular thiamine. The bacteria is put in liquid, gently lysed and allowed to settle on filter paper. The filter paper is exposed to X-ray film and the radioactivity os detected.
Replication starts at the ORI and is bi-directional forming a theta structure.
How do plasmids replicate?
Follows a rolling circle, or sigma form.
A break occurs in one strand and it begins to “roll off.” A New strand is synthesized behind the old ones.
In eukaryotic replcation where does it start?
There are many ORI.
What direction is DNA synth?
5’--> 3’
WHat is necessary for DNA synthesis?
needs primer- short length of nuclotides

needs template

DNA polymerase
Difference btw DNA and RNA
DNA has thiamine and dideoxyribose as a sugar. (Dideoxyribose makes the DNA more stable than RNA) DNA does NOT hydrolyze in alkali.
RNA has a uracyl instead of a thiamine and ribose as a sugar. RNA has a tendency to undergo oxidative deaminization (a change form a C to a U). The cell can detect this change and fix the RNA.
Characteristics of the ORI in prokaryotes
The origin of replication in Prokaryotes begins from a fixed origin called oriC and proceeds in both directions. There is a dnaA box to which DnaA binds. This box is repeated 5 times in the oriC and in respons to binding AT rich regions begin to unwind. More DnaA proteins coat the oriC as it is opened. Then units of then helicase DnaB bind and slide in a 5’ à 3’ direction to unzip the helix at the replication fork. Primase and DNA poly III are recruited and DNA synthesis begins.
Characteristics of the ORI in eukaryotes
In Eukaryotes there are many oriC and replication proceeds in both directions from these points. ORC first binds to the oriC and then recruits Cdc6 and Cdt1. These proteins ensure that DNA replication will take place only in S phase. They in turn recruit the MCM complext which licences the ori and allows for replication.
what are some diff types of mutations
Suppression: restores reading frame and function
Synonymous or same sense mutation: maintatins same amino acid sequence. (because of the degeneracy of the code)
Nonsense mutiation: A nonsense mutation is always profound. It results in an early stop codon
Missense mutation: Can be profound or cannot. It changes one amino acid.
Tautameric shift- all 4 bases can undergo a tautameric shift. In a tautameric shift one base pair takes on a different form and may be able to pair with a nonmatching base. This may cause a mustation if it is not fixed with a proofreading enzyme
Suppression mutation
restores reading frame and function
Synonymous or same sense mutation
maintatins same amino acid sequence. (because of the degeneracy of the code)
Nonsense mutiation
A nonsense mutation is always profound. It results in an early stop codon
Missense mutation
Can be profound or cannot. It changes one amino acid.
Tautameric shift
all 4 bases can undergo a tautameric shift. In a tautameric shift one base pair takes on a different form and may be able to pair with a nonmatching base. This may cause a mustation if it is not fixed with a proofreading enzyme
What is the start codon?
5’ AUG 3’
What happends when AUG is in a ribosome and a shine delgorno sequence preceeded it?
This is a start codon:
The start codon does not cause the insertion of a methyanine by tRNAMet but it is recognized by a special tRNA called an initiator. In bacteria, a formyl group is added to the met while it is attatched to the tRNA forming N-formylmethionine.
Griffin experiment: what is it? what does it prove?
Discovered transformation in bacteria: “transforming principle”
S live strain à mouse dies
S heat killed à mouse lives
R strain à mouse lives
R + heat killed S à mouse dies *
Avery Experiment: What is it and what does it porve?
Same experiment as Griffin. Heat killed virulent bacteria + R strain but with different things disabled(protease, DNAse, etc...)
Mouse lives only when DNA is destroyed

Dies under other conditions (DNA is intact)

futher pointed to DNA as hereditary material.
Hershey and Chase Experiment: What is it and what does it prove?
Used T2 bacteriophage
Labeled DNA with radioactive Phosphorus
Labeled Proteins with radioactive sulfur
(There is no sulfur in DNA and no phosphorus in protein)
Infected Ecoli
Sheared phage ghosts off of bacteria with kitchen blender
Separated ghosts from bacteria in centrifuge
Found: labeled phosphorus INSIDE bacterial cells
Labeled sulfur in phage ghosts
THEREFORE: transforming principle is DNA
Chargaffs Rules
1. A = T
2. G = C
3. A + G = T + C
4. A + T ≠ always G + C
When is melting point of DNA high>
Melting point is higher at higher G + C content
describe Molecular Density Gradient centrifugation
Use cesium chloride. If it is spun at high speeds for many hours ions tend to be pushed down to the bottom of the tube and an ion gradient is established with hisgest ion density at bottom. DNA centrifuges with the cesium chloride forms a band at the position identical with its density in the gradient. DNA of different densities form bands at different places.

If asked about what percentage if heavy/light, do a punnet sq
where do you find Phosperous atoms?
In DNA not protein.
Meselson-Stahl experiment: what is it and what did it prove?
replication method of DNA is semiconservative

cells are grown in 15N (heavy) medium as a label for several generations and then transferred into a normal light 14N form of medium for one or two cell divisions. Molecules were separated by density with cesium chloride gradient centrifugation.

found:
parental: one heavy band
fisrt gen: band in middle (1/2 heavy, 1/2 normal)
second gen: two bands one all light, one 1/2 heavy 1/2 light
Replication fork
zone where the dbl helix is unwinding
Leading strand
New strand synthesized towards the replication fork 5’--> 3’
Lagging strand
New strand synthesized away from replication fork with Okazaki fragments.
Okazaki fragment
short (1000-2000) nt stretched of newley synthesized DNA on the lagging strand
DNA polymerase I
Removes RNA primer and fills in with DNA. Has 3’ à 5’ exonuclease activity and 5’ à 3’ exonuclease activity (the Kornberg enzyme) A distributive enzyme.
DNA polymerase II
back up role in DNA repair
DNA polymerase III
The main DNA polymerse in bacteria. Uses RNA as a primer. Has 3’ à 5’ exonuclease activity as a proofreading function. A processive enzyme with an associated sliding clamp.
DNA polymerase III holoenzyme
The main prokaryotic DNA polymerase. It has10 subunits, there are 10 polIII holoenzymes per cell, it is highly processive (synthesizes 1000 nucleotides per second) and has NO 5’ to 3’ exonuclease activity. The most important individual subunits are:
Alpha- This is the actual polymerase. There are 2 alpha subunits per holoenzyme, one for the leading and one for the lagging strand.
Epsilon: the 3’ to 5’ exonuclease (proofreading or editing activity)
Beta: sliding clamp or processicity factor. Looks like a doughnut and there is one per each alpha subunit.
Gamma: clamp loader, requires and recognizes a 3’ hydroxyl end (the 3’ end of a primer) to use ATP to load the clamp.
alpha subunit of DNA polIII holoenzyme
This is the actual polymerase. There are 2 alpha subunits per holoenzyme, one for the leading and one for the lagging strand.
Epsilon subunit of DNA polIII holoenzyme
: the 3’ to 5’ exonuclease (proofreading or editing activity)
Beta subunit of DNA polIII holoenzyme
sliding clamp or processicity factor. Looks like a doughnut and there is one per each alpha subunit.
Gamma subunit of DNA polIII holoenzyme
clamp loader, requires and recognizes a 3’ hydroxyl end (the 3’ end of a primer) to use ATP to load the clamp.
How is eukaryotic replication diff from bacteria?
Eukaryotic replication is not highly processive because there is no sliding clamp
How is euk replication similar to bacteria?
One continuous, one discontinuous strand; helicase; topoisomerase
Integrase
A site specific recombining enzyme in specialized transduction
Telomerase
An enzyme important in synthesizing the telomeres of a lagging strand of linear DNA. Telomerase adds a noncoding sequence to the 3’ end. It carries a small RNA molecule that acts as a template for the polymerization of the telemetric repeat unit
Telomere
- the ends of linear DNA
What is the difference between a telomere and a broken chromosome
A telomere has end is stable, a broken chromosome (usually sticky ends)? and can be translocated across the chromosome.
How do telomeres stabilize?
Because telomeres have a repeating sequence with a 3’ overhang, the overhang can loop arount in a D-loop-T-loop to form a stable structure.
Base stacking interactions
Interactions between electrons of bases stacked above each other. They stabilize the double helix. A decrease in base stacking interactions results in an increase of UV absorption (hyperchromicity)
PCNA
proliferating cell nuclear antigen acts like a sliding clamp in eukaryotes?
DNA ligase
Joins knicks by reforming a phoshodiester bond. (Fills in Gap left by poly 1 and poly 3 on lagging strand)
DNA topoisomerase I
relaxes supercoiling of DNA due to unwinding by breaking a single strand of the DNA and then rejoining them.
DNA topoisomerase II
Relieves catenation by breaking dbl strand of DNA. Catenation happened at the end of the replication of a circular chromosome.
DNA helicase
part of the replisome that breaks DNA hydrogen bonds. It unzips the dbl helix
DnaA protein
first step in assembly of the replisome. DnaA binds to a specific 13-bp sequence (called a DnaA box) that is repeated 5 times in the OriC. In reponse the origin encircles the DnaA proteins and is unwound at a cluster of AT rich region. DnaA then reqruits DnaB.
DnaB
a helicase that binds after DnaA and slides in a 5’ à 3’ direction to unzip fork.
Primase
an RNA polymerase made up of the central protein primosome that synthesizes a short stretch of complementary RNA to a DNA template. No exonuclease activity.
Single stranded binding protein
prevents single stranded DNA from reforming a dbl helix by stabilizing the single-stranded form
Nucleosome
basic unit of chromatin which consists of DNA wrapped around histone proteins.
Chromatin Assembly Factor I (CAF-I
protein that binds histones and targets them to the replication fork, where they can be assembled with new DNA.
Proliferating cell nuclear antigen (PCNA)
eukaryotic version of the clamp protein.
RNA polymerase holeoenzyme
The bacterial RNA polymerase that scans DNA for a promoter sequence. It is made up of four subunits of the core enzyme plus the sigma factor. It needs ATP, UTP, CTP, GTP, a DNA template. It does not need a primer in vitro.
Sigma factor (σ)-
In bacteria the part of the RNA polymerase holeoenzyme that binds to the –35 and –10 regions of the promoter and postitions the holeoenzyme to initiate transcription at the correct start site. It also helps melt the DNA at the –10 region. Once RNA poly is bound sigma disassociates from the holeoenzyme. Different sigma factors recognize different promoter regions.
Transcription bubble
Region of single stranded DNA where the template strand is exposed.
RNA polymerase I
In eukaryotes, transcribes rRNA genes, 18S and 28S (the larger subunits (except 5S rRNA)
RNA polymerase II
In eukaryotes, transcribes all protein-coding genes (mRNA) and some snRNAs
RNA polymerase III
In eukaryotes, transcribes small functional RNA genes (tRNA, some snRNA, and 5S rRNA)
Reverse transcriptase
RNA dependant DNA polymerase Not all RNA viruses have reverse transcriptase
Ribozyme
an RNA that acts as a catalyst
Types of RNA
mRNA- intermediate btw DNA and protein.
Functional RNA- RNA that is not intended for translation and serves its own purpose

such as:
Transfer RNA (tRNA)- molecules responsible for bringing the correct amino acid to the mRNA in translation

Ribosomal RNA (rRNA)- the major components of ribosomes

Small Nuclear RNAs (snRNAs) in eukaryotes, a part of a system that further processes RNA transcripts. Some snRNAs guide the modification of rRNA’s others unite with other proteins to form the spicosome that removes introns from eukaryotic mRNA.
mRNA
intermediate btw DNA and protein.
Functional RNA
RNA that is not intended for translation and serves its own purpose
Transfer RNA (tRNA)-
functional RNA
molecules responsible for bringing the correct amino acid to the mRNA in translation
Ribosomal RNA (rRNA
Functional RNA
the major components of ribosomes
Small Nuclear RNAs (snRNAs)
Functional RNA
in eukaryotes, a part of a system that further processes RNA transcripts. Some snRNAs guide the modification of rRNA’s others unite with other proteins to form the spicosome that removes introns from eukaryotic mRNA.
Spicosome
A functional splicing unit made up of small nuclear ribonucleoproteins (snRNPs)
Coding strand
strand that is NOT used by RNA polymerase as a template but that resembles the mRNA transcript in content. When giving the DNA sequence for a gene give this strand.
Template strand
The strand used as a template to synthesize the mRNA transcript but is opposite as far as content.
Stages of transcription:
Initiation-
Prokar: Binding of RNA polymerase to the promoter with the help of sigma factors
Euk: Binding of RNA poly to promoter with help of GTFs. First stem is the binding TBP to the TATA box. Then the Preinitiation Complex (PIC) is assembled
Elongation- synthesis of mRNA
Prok: RNA holoenzyme synthesizes mRNA off of the template strand
Euk: RNA poly II becomes disassociated form GTF and synthesizes the primary transcript off the template strand. GTF stay at the promoter and recruit another RNA poly II for multiple transcripts. Elongation id ended when an enzyme recognized a conserved AAUAAA or AUUAAA sequence and cuts of the transcript 20 bases further.
Termination- end of transcription.
Prok: can be intrinsic or rho dependant.
EUK: an enzyme recognizes AAUAAA or AUUAAA and stops transcription 20 bases down.
Initiation
Prokar: Binding of RNA polymerase to the promoter with the help of sigma factors
Euk: Binding of RNA poly to promoter with help of GTFs. First stem is the binding TBP to the TATA box. Then the Preinitiation Complex (PIC) is assembled
intiation of transcription in prokaryotes
Binding of RNA polymerase to the promoter with the help of sigma factors
initiation of transcription in eukaryotes
Euk: Binding of RNA poly to promoter with help of GTFs. First stem is the binding TBP to the TATA box. Then the Preinitiation Complex (PIC) is assembled
Elongation of transcription
synthesis of mRNA
Prok: RNA holoenzyme synthesizes mRNA off of the template strand
Euk: RNA poly II becomes disassociated form GTF and synthesizes the primary transcript off the template strand. GTF stay at the promoter and recruit another RNA poly II for multiple transcripts. Elongation id ended when an enzyme recognized a conserved AAUAAA or AUUAAA sequence and cuts of the transcript 20 bases further.
Elongation of transcription in prok
RNA holoenzyme synthesizes mRNA off of the template strand
Elongation of transcription in euk
RNA poly II becomes disassociated form GTF and synthesizes the primary transcript off the template strand. GTF stay at the promoter and recruit another RNA poly II for multiple transcripts. Elongation id ended when an enzyme recognized a conserved AAUAAA or AUUAAA sequence and cuts of the transcript 20 bases further.
termination of transcription
end of transcription.
Prok: can be intrinsic or rho dependant.
EUK: an enzyme recognizes AAUAAA or AUUAAA and stops transcription 20 bases down.
Why is transcription more complicated in eukaryotes?
1. more genes; more non-coding DNA makes initiation more difficult;
2. Eukaryotes have a nucleus where RNA is synthesized. RNA processing must take place before it is exported outside of the nucleus.
3. chromatin in Eukaryotes
Intrinsic termination
termination where terminator sequence contains about 40 bp ending in a GC rich region followed by a run of six or more A’s. The G and C’s interact with each other and form a hairpin loop in the mRNA transcript. As the RNA poly synthesized the weak bonding U’s from the A rich region it backtracks, comes across the hairpin loop, and falls off.
Rho dependant termination
A hexamer called rho binds to the rut (rho utilization site), RNA polymerase pauses and rho mediates the disassociation of RNA poly.
General transcription factors (GTF’s
enzymes that aid in transcription
RNA processing
modification of RNA in eukaryotes before it is transported outsider the nucleus. These processes include: the addition of a cap at the 5’ end, the addition of a 3’ poly A tail, and splicing to eliminate introns. It is usually contratranscriptional.
Primary transcript (pre-mRNA):
mRNA transcripts without any processing or modifications
Mature RNA (mRNA):
mRNA with modifications and processing
Promoter
a specialized DNA sequence located upstream (5’ of the gene) to the start of transcribes region where RNA polymerase binds. Prokaryotic promoters have –35 and –10 regions. Eukaryotic promoters
PIC
preinitiation complex that consists of 6 GTF and RNA polymerase II core in eukaryotes
Consensus sequence
a general promoter template for an organism
5’ UTR
5’ untranslated region. The segment of mRNA between the start of transcription and the first translation codon.
3’ UTR
3’ untranslated region. The segment of DNA that is transcribed into mRNA but does not code for protein.
Exonuclease
chews in and destroys DNA or RNA form ends
Endonuclease
cuts from within
Replisome
The large complex consisting of DNA polymerase that coordinates the activity at the replication fork
Poly III holoenzyme
consistis of two catalytic cores (one for leading, one for lagging strand) and many accessory proteins. The accessory proteins form bridges that connect the catalytic cores, coordinating the synthesis of the leading and lagging strands.
Sliding clamp
A Poly III holoenzyme accessory protein which encircles DNA like a doughnut and keeps Poly III attatched to the DNA.
A distributive enzyme
adds only a few nt before falling off the template. (primase
A processive enzyme
stays attatched for a long time and adds many NTs. (poly III)
ORC
Origin recognition complex. A protein requires to assemble a replisome in Eukaryotes. It binds to sequences in yeast origins and reqruits other proteins (Cdc6 and Cdt1) Then this combo of proteins recruit the replicative helicase, called the MCM complex.
Cdc6 and Cdt1
proteins that help initiate replication in eurkaryotes. They are synthesized during late mitosis and gap 1 and destroys by proteolysis after synthesis have begun. They ensure that the replisome is assembled only before S phase.
MCM complex
Eukaryotic helicase that is said to licence the origin
Kozak sequence
like the shine delgarno sequence but in vertebrates. It is purine rich (A or G)
Shine-Delgarno Sequence
: In prokaryotes a special sequence in the 5’ untranslated region that preceeds a start codon. It pairs with the 3’ end of an rRNA called the 16S rRNA, in the 30S subunit. The pairing correctly positions the initiator codon in the P site.
Initiation factors
Proteins that help initate translation:
prok:
IF1: initiation factor helps initiator tRNA enter P site
IF2: initiation factor helps initiator tRNA enter P site
IF3: initiation factor keeps 30 Subunit dissacociated from 50 S.
IF1 (prok)
initiation factor helps initiator tRNA enter P site
IF2 (prok)
initiation factor helps initiator tRNA enter P site
IF3 (prok)
initiation factor keeps 30 Subunit dissacociated from 50 S.
Formyl-methianine
the changed form of methianine that is attatched to the initiator tRNA. It is later removed.
TATA box
place in the Eukaryotic promoter (–30) where the first transcription event takes place. This is where the TATA binding protein (TBP), one part of one of the 6 GTF that recruit RNA poly II binds. It attracts other GTF.
TBP
TATA binding protein, one part of one of the 6 GTF that recruit RNA poly II binds. It attracts other GTF.
Carboxyl Tail domain (CTD):
): A part of RNA poly II that helps RNA poly II disassociate with GTF at the promoter. Initiation of Eukaryotes ends and elongation begins when CTD is phosphorylated by one of the GTFs. Important in coordination transcriptional modifications.
5’ mRNA capping
process coordinated by the CTD that helps protect mRNA from degradation and helps initiate translation. A cap consists of a 7-methylguanosine residue linked to the transcript by three phosphate groups.
Poly A tail
150- 200 A’s added to the 3’ end of eukaryotic mRNA after an enzyme recognized a polyadenylation signal.
Polyadenylation signal
: In euk the conserved AAUAAA or AUUAA region that signals the end of elongation and beginning of termination and where the poly A tail is added.
Alternative splicing
different arrangements of exons from same initial mRNA usually in diff cell types or at different stages in development.
GU- AG rule
Introns usually have a GU at the 5’ end and AG at the 3’ end. This is where they will be cut.
How does splicing occur?
snRNP’s recognize conserved sequences within the pre mRNA introns such as the GU-AG regions and the A residue that is 15 to 45 nucleotides upstream from the 3’ splice site. The snRNP’s make up a sliceosome. Components of the sliceosome interact with the CTD and attach to intron and exon sequences. The sliceosome moves the mRNA in such a way that it performs two cuts, one between the GU (5’) region and the A residue and one btw the GU and AG regions.
What is the RNA world theory?
The discovery was made that RNA can sometimes self-splice introns. This suggested that because RNA can act as both genetic material and a reaction catalyst it could have been the genetic material in the first cells.
Pribnow box
the –10 sequence in prokaryotic promoter
CAAT box
Is an unsteam promoter element in eukaryotes that is –40 to –110 from the transcription site and controls levels of transcription.
Hayflick limit
the limit of the number of divisions a cell can undergo due to the deactivation of telomerase and the loss of end chromosome genetic material. Cancer cells have no Hayflick limit and most likely have a reactivation of telomerase.
What did Yanofsky prove? How?
Yanofsky proved the colinearity og gene and protein by indicing mutations in the gene for triptophan synthetase alpha subunit, making it inactive. He then, by recombination, mapped the mutant sites on the DNA and biochemically identified changes in the amino acid structure of the protein. He then showed that the mutations in the DNA occurred in the same order as the changes in the protein and the distance between them corresponded.
Why must the genetic code of DNA code for amino acids in triplets?
If 1 base = 1 amino acid and there are 4 bases and 20 amino acids
41= 4 possible amino acids
If 2 bases = 1 amino acid
42 = 16 possible amino acids
If 3 bases = 1 amino acid
43 = 64 possible amino acids (which is more than enough

Futhermore, Crick and Brenner showed that the reading frame can be fixed with the addition or deletion of 3 bases.
Describe the Crick Brenner Experiment
Induced a single deletion or addition of a base with the chemical proflavin in phage T4 so that they would not grow on K strain bacteria. Then they found revertants but after analysis found that they were not identical to the wild type strain. So a second suppression mutation must have occurred. The suppression mutation was either a deletion or an addition of a single base that restored the reading frame. The did more exact experiments to prove that 3 deletions would be “fixed” with 3 additions.
What do you need to make mRNA in vitro? Why would you do this?
This was a way to figure out the amino acid mRNA code. Went on to prove code degeneracy.
Need mRNA (U, G, C, A) nucleotides, a polynucleotide prosphorylase (the protein factor), ribosomes, and charged tRNAs. The amino acids are assembled at random.
Why is the genetic code degenerate?
1. To protect against mutation
Wobble
loose pairing of the third base (5’) in the anticodon with mRNA
Isoaccepting tRNA
tRNA that accepts the same amino acid but are transcribed from different tRNA genes.
What are the stop codons?
UAA, UAG, UGA, UGG
What direction is mRNA read? TRNA?
mRNA is read 5’ --> 3’
tRNA is read 3’ --> 5’
aminoacytl-tRNA synthetase
attatched amino acids to tRNA. There are at least 20 kinds of this enzyme each specific to an amino acid/tRNA combo.
Inosine
a rare base in tRNA that can pair with U, C, or A.
charged tRNA
one with an attatched amino acid
nonsense suppressor
: a mutated tRNA that had an anitcodon that recognizes a termination codon. Supresses a nonsense mutation. Usually these tRNA mutations happen in the minor tRNA (low codon preference in organism) and bacteria usually have two stop codons. These features protect against the mutation being lethal.
What are the sizes of prokaryotic and eukaryotic ribosomes?
Prok 30S and 50S. Together 70S
Euk 40S and 60S. Together 80S
What are the positions where tRNA fits into a ribosome and what happends there?
EPA 3’
The A site is where tRNA enters first. This is called the aminoacyl site. The A site binds an incoming aminoacyl tRNA whose anticodon matched the codon in the A site of the 30S subunit.
The P site is called the peptidyl site of the 30 S subunit. The tRNA in the P site contains the growing amino acid chain. The amnio acid chain exits through a tunnel in to 50 S subunit.
The E site contains a deacylated tRNA that is ready to be released from the ribosome.
The A site
where tRNA enters first. This is called the aminoacyl site. The A site binds an incoming aminoacyl tRNA whose anticodon matched the codon in the A site of the 30S subunit.
The P site
called the peptidyl site of the 30 S subunit. The tRNA in the P site contains the growing amino acid chain. The amnio acid chain exits through a tunnel in to 50 S subunit.
The E site
contains a deacylated tRNA that is ready to be released from the ribosome.
Decoding center
area in 30 S subunit ribosome that ensures that only tRNAs carrying anticodons that match the codon will be accepted into the A site
Peptidyl transferase center
area in the 50 S subunit of the ribosome where cognate tRNAs associate and a peptide bond is catalyzed
How is translation initiated in Prokaryotes? Eukaryotes?
In prokaryotes Initiation factors in the 30 S subunit help move mRNA into the ribosome and lead the initator tRNA bind to the P site. Then initiation factors attatch the large subunit to the small subunit. This can happen as transcription is taking place.
In eukaryotes, Eukaryotic initiation factor E4 (elF4E) binds to the 5’ cap of the mRNA. Then elF4G and elF4A bind to elF4E forming the complex elF4F. The small subunit the initiator tRNA and elF2 combo recognize elF4F and the mRNA moves into the small subunit, until the AUG codon is in. Then elF2 releases and the large subunit binds.
How is translation initiated in Prokaryotes?
In prokaryotes Initiation factors in the 30 S subunit help move mRNA into the ribosome and lead the initator tRNA bind to the P site. Then initiation factors attatch the large subunit to the small subunit. This can happen as transcription is taking place.
How is translation initiated in Eukaryotes?
In eukaryotes, Eukaryotic initiation factor E4 (elF4E) binds to the 5’ cap of the mRNA. Then elF4G and elF4A bind to elF4E forming the complex elF4F. The small subunit the initiator tRNA and elF2 combo recognize elF4F and the mRNA moves into the small subunit, until the AUG codon is in. Then elF2 releases and the large subunit binds.
EF-Tu
Elongation factor Tu is a protein factor that attaches to an aminoacyl-tRNA to form a ternary complex. It is removed once the tRNA enters the A site. It can then form a peptide bond.
EF-G
Elongation factor G is a protein factor that is GTP driven. It moves into the A site and moves the tRNA down into the P and E sites. It then leaves and the A site is open again.
What is the role of GTP in protein synthesis?
GTP can bind to protein initatiation factors and change their state so that they can no longer bind to the ribosome. Specifically IF2.
Polyribosome
In eukaryotes, many ribosomes translate one sequence of mRNA. Interactions between the poly A tail and the 5’ end of the mRNA form a circle that makes it easy for ribosomes to attatch again to the mRNA once they are done.
Release factors
proteins that recognize stop codons by use of tripeptides not anticodons. The do not participate in forming a peptide bond.
RF1: recognizes UAA or UAG
RF2: regognizes UAA or UGA
RF3: aids RF1 and RF2
Native conformation
a correctly folded protein
Nascent
newly synthesized protein
Chaperones
: a class of proteins that help other proteins fold
What are some common port-translational modifications?
Phosphatation (adding of a phosphate via phosphatase)
Removal of phosphates (kinase)
Removal of beg MET
Signal sequence
at amino-term end for secretion
Nuclear localization sequence
signal seguence for movement out of nuclear pore.
Interin
segment of a protein that is able to excise itself and rejoin the remaining portions (the exteins) with a peptide bond.
Prototrophic
can grow on defined (minimal) medium of inorganic salts and organic carbon. Can synth all amino acids, vitamins, nucleotides etc..
Auxotrophic
cannot synthesize an essencial growth factor
What are the three different classes of mutants that can be used for selection?
Auxotrophs, mutants with the inability to utilize a certain sugar source, mutants that are antibiotic resistant/sensitive, mutants that are phage resistant.
Transformation
When a cell takes up naked DNA. A cell must competent in order to take up DNA. Cells become competent just before growth stops in B subtilis Recombination must occur between the donor DNA and the recipient chromosome. For two markers to be transferred via transformation, an even number of cross-over events must occur. Not all cells become competent and not all calls can be transformed. Bacillus subtilis can be transformed and is prototrophic and is favored for experiments. Streptococcud pneumoniae can be transformed but requires complex media.
How do you determine linkage between two markers by transformation?
Two markers are said to be linked if they are close enough to be on the same transforming fragment of DNA. You would do a cross:
Donor: is protorophic
Recipient: trp- his-
After mixing DNA with recipience cells and allowing time, cells are plated onto min medium + trp. Since there is NO his in the medium only cells that are transformed his+ will grow. This selects for his+ tansformants. The resulting transformants are replica plated onto min media WITHOUT trp. Only trp+ will grow. These are double transformants (trp+ and his+) If 20% are trp+, there is a 20% linkage between trp and his. Linkage is a function of gene closeness and molecular weight of donor DNA.
How do you determine order of linked markers by transformation?
3 point cross!
Donor: a+ b+ c+
Recipient: a- b- c-
Select for transformants and find linkage numbers. Lowest numbers found indicate a quadruple crossover.
Conjugation
Transfer of chromosomal markers with the help of an F plasmid or Hfr integrated plasmid. Cell to cell contact is required. The F plasmid codes for genes that allow for sex pilli to be attatched to another cell. The other cell is drawn in close, the F plasmid nicks and one strand is transferred to the recipient cell 5’ first. Chromosomal markers are transferred also. Hrf cells are ones in which the F factor is integrated into the chromosome.
What are the differences btw F+ and Hfr cell?
F+ cells don’t tranferer chromosomal genes as well as Hfr but they do transfer themselves at a high freq while Hfr don’t transfer the F factor frequently. F+ are sensitive to acrine orange because they are outside of the chromosome while Hfr are not.
Counter selection
selection against donor cells so that it does not grow on selective medium.
How do you determine the order of chromosomal marker transfer in Hfr cells?
You need selectable markers for recipient and counter selection for donor. Determine the frequency of unselected markers in the recipient. Order follows order of transfer.
Why does Hfr seem linked to the last marker?
Transfer starts with a nick in the middle of the oriT of the F factor. It will only be complete again in the recipient cell if the whole chromosome is transferred, which is rare. Only if you select for the last maker will you see Hfr recipients.
Episome
: a genetic element that can exsist on its own or be integrated into the chromosome.
F’
An Hfr that, when it excises itself from the chromosome, has chromosomal material attatched. Is used to produce partial diploids and to test dominance of markers. Transfer can occur as a result of crossover btw F’ and chrmsm.
Generalized Transduction
The transfer of any gene from donor to recipient by a phage vector. Any chromosomal marker can be transferred. A virus head can only handle about 1% of the bacterial chromosome in size. If two bacterial genes are within 1% of the chrmsms length they will be co-tranducible.
Specialized transduction
Phage lysogenic lamda is an episome that always integrates at a specific site between gal and biotin gene. This is because of the enzyme integrase. If lamda takes parts of other genes it is defective and cant replicate
How do you determine if two makers are co-transducible?
Two and three factor crosses.
Horizontal transmission
markers passes along diff types of bacteria with self replication plasmids. NOT DNA or chrms transfer
Constitutive enzyme synthesis
enzyme is continually produced in fixed amounts
Adaptive enzyme synthesis
Inducible: An enzyme is produced (and gene turned on) in the presence of an inducer. The inducer is usually the enzyme’s subratrate. (Beta- galactosidase is iducible)
Repressible: An enzyme’s production is turned off, repressed, by the co-repressor which is usually the end product. The enzymes for argentine synthesis are repressed by argentine.
how does the Lac operon work?
Lac Operon ( an inducibe operon):
The lac Operon controls the production of three enzymes from three genes: The lac Z gene codes for the protein Beta-galactosidase which cleaves lactose. The lac Y gene codes for the protein Permease which transports lactose into the cell. Lac A codes for an enzyme that is needed in lactose metabolism. The Lac I gene which codes for the Lac repressor is upstream of the Lac promoter and operator. The operator is downstream of the promoter and is where the repressor binds. The loc Operon is induced by lactose. Lactose binds with the repressor and allows transcription.

---- I repress gene ---….----P—Operator—LacZ—LacY—LacA—

When glucose levels are low, cAMP level high, lactose level high, and cap high the repressor is turned off by lactose so it cannot bind to the operator. The cAMP made as glucose levels are low binds with CAP then binds to the promoter and increases transcription rate. CAP, catabolite activator protein, when bound with cAMP interacts with RNA poly to increase RNA poly’s affinity for the lac.promoter.
The lac repressor can work in cis, meaning that in a heterozygote with one functioning Lac I gene (and a second lac-), the Lac operon is still regulated correctly. If the operon region is mutated so that it cannot bind the repressor than the genes will be expressed even in the presence of the repressor and absence of the inducer.
If the Lac I gene is an Is gene (which is dominant to I) it produces a repressor that cannot bind with the inducer. The repressor always binds with the operator region in this case.
can the lac repressor work is cis?
The lac repressor can work in cis, meaning that in a heterozygote with one functioning Lac I gene (and a second lac-), the Lac operon is still regulated correctly. If the operon region is mutated so that it cannot bind the repressor than the genes will be expressed even in the presence of the repressor and absence of the inducer.
What is the Lac IS mutant?
If the Lac I gene is an Is gene (which is dominant to I) it produces a repressor that cannot bind with the inducer. The repressor always binds with the operator region in this case.
How does the ara operon work
In the absence of arabanose, araC bind with O2 and I1, to bend the DNA so that the RNA polymerase cannot bind.
With Arabanose, arabanose binds with araC so it now binds with I1 and I2 and not to O2. The pomotor can now bind RNA poly. If glucose levels are low CAP- cAMP can also bind and initiate transcription.

---araC---…---O2--------O1------P cap site and promotor ----I1—I2---- PROMOTOR----

No arabanose:
________________O2______
| araC
| araC
|___O1___Pcap_____I1__I2____PR_ß RNA poly can’t bind


Arabanose:
Cap Cap araCaraC RNA polyà
---araC---…---O2--------O1------P cap site and promotor ----I1—I2---- PROMOTOR----