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Compare and contrast DNA polymerase I from DNA pol III (at least 2 similarities and 2 differences)Are these in prokaryotic or eukaryotic cells? How are the DNA polymerases of the other type of cell denoted?
a)Both are found in prokaryotic DNA replication, have 3’ to 5’ exonuclease activity. DNA pol I is used for replacing RNA primers with DNA while DNA pol III is used for actual polymerization. DNA pol I also has 5’ to 3’ exonuclease activity, while DNA pol III does not. b)Prokaryotic... Eukaryotic DNA polymerases would be denoted with greek letters (ex. DNA pol α, DNA pol δ, etc)
Describe the function of the 8proteins (dnaA, helicase, SSBP, gyrase, primase, DNA pol I, DNA pol III, ligase)
dnaA- binds oriC to initiate replicationHelicase-unwinds DNASSBP-prevents unwound strands from reannealingGyrase-creates nicks in DNA to prevent supercoiling Primase-addsRNA primers for DNA pol III to add toDNApol I-removes RNA primers, replaces with DNADNA pol III-main synthesis polymeraseLigase-seal okazaki fragments together
DNA pol 1 exonuclease activity
DNA pol 3 exonuclease activity3' to 5' AND 5' to 3' 3' to 5' (no backing up)
3' to 5'exonuclease activity =
backspace or proofreading.
DNA replication: four stages:
- Unwinding (followed by priming)
- Initiation: Origin of replication, bidirectional startpoint
- Unwinding: after initiator proteins, helicase unwinds (binding to the Lagging strand template), SSBP's keep DNA from re-annealing, and gyrase reduces supercoil ahead of the fork.
- polymerases in prokaryotes: catalyze phosphodiester bond between 3'OH and 5'PO4. Direction is always 5'to3', cannot initiate de novo. can only add to existing 3' end.
- Primase synthesizes short stretches of RNA because they can start de novo.
Describe how eukaryotes ensure that each replicon replicates only once per cell cycle.
- factors are produced in G1, which bind to origins to license. Only licensed
- origins can have initiator proteins bind. Once initiation is complete and the
- replication fork moves, the licensing factor falls off and is degraded so that
- it can’t be used again.
A gene with 200 base pairs contains 60 guanine nucleotides. How many thymine
nucleotides are there?
You determined this using conclusions from which scientist?
140 thymine nucleotides
This is based on Chargaff’s law
Describe the rho-independent termination process. What sort of things are needed for
this form of intrinsic termination to occur?
- the DNA there’s an inverted repeat in the
- sequence in the 3’ end followed by a A-rich region. When transcribed the
- inverted repeat forms a stem-loop which slows down RNA pol and destabilizes the
- DNA-RNA hybrid. This pulls the weak U-A bonds apart.
Genes A and B are linked, with a
recombination frequency of 22%. If you test crossed an individual with AB/ab, which of the following is a correct
B) 11% ab
C) 39% AB
E) 39% aB
F) 78% Ab
39% because out of 100, 22 of your progeny are recombinants, so you would get 78% parentals, each at 39%
This shows the flow of genetic info in a prokaryote
1)Which number represents the coding strand?
2)What is 6? Is it found in the mRNA?
3)What sequence would you find at 4? What binds there?
1 is the coding strand
6 is the stop codon. It will be in the mRNA
4 is the -10 consensus sequence, TATAAT. RNA pol will bind there
needs Rho protein
2 structural features: hairpin and A/U rich region at end
Come back to Transcription Review
Consensus sequence letters in a gene promoter region...
- R - any purine
- Y - any pyrimidine
- N - any nucleotide
Three primary regions of mature mRNA are the...
5' untranslated region, the protein-coding region, and the 3' untranslated protein
Describe the rho-dependent termination process. What things will you see (i.e.
sequences, proteins, structures, etc.)
- A rho protein binds to the rut sequence and moves towards the 3’ end. A secondary
- structure may be present to slow down the RNA pol so that rho catches up. Rho
- has helicase activity so when it catches up to the RNA pol, it separates the
- H-bonds of the DNA-RNA hybrid
3 scientific experiments:
- Salmonella pneumonia, IIR and
- IIIS strains
- IIR alone=mouse lives
- IIIS alone=mouse dies
- Heated IIIS=mouse lives
- Heated IIIS+live IIR=mouse dies
Avery, Macleod, McCarthy
- Some “transforming factor” can be
- passed on to other cells
- Same as Griffiths
- Isolated RNA, DNA, protein from
- heat-killed IIIS.
- RNA+ ribonuclease+ IIR= mouse dies (transformation
- protein + protease +IIR=
- dies (transformation occurred)
- IIIS DNA +nuclease + IIR=
- Mouse lives (no transformation)
DNA is the transforming factor
Hershey and Chase
T2 bacteriophage, E. coli
- Batch 1: grew phage in 35S, inflected bacteria. Blended up the
- bacteria, centrifuged. Found radioactivity in the liquid (protein shells were
- 2: Repeated with phage in 32P. Found radioactivity in pellet (cells)
- Protein doesn’t get transferred to host
- cell, DNA does. DNA is hereditary material.
In eukaryotes, transcription begins when ____binds to the ____at -25. After many other TFII proteins, including _____ and ____ (both of which act to stabilize the complex) bind, RNA pol binds. Lastly, _____binds, which acts as a _____enzyme to unwind DNA. Much further upstream, there is a regulatory promoter where _____ proteins bind to the enhancer to increase
- In eukaryotes, transcription begins when _TFIID___binds to the _TATA Box___at -25. After many other TFII proteins, including _TFIIA____ and _TFIIB___ (both of which act to stabilize the complex) bind, RNA pol binds. Lastly, _TFIIH____binds, which acts as a _helicase_enzyme to unwind DNA. Much further upstream, there is a regulatory promoter where _activator___ proteins bind to the enhancer to increase
Describe two ways in which DNA
replication is terminated.
1)Replication forks meet and replication is complete
- 2)A termination protein Tus binds to a specific sequence Ter.
- (in E. coli)
If RNA were the universal genetic
material instead of DNA, how would it have affected the Avery experiment and
the Hershey-Chase experiment?
Explain these experiments in your answer.
- In the Avery, MacLeod and McCarty
- experiment, they isolated the transforming factor from Griffith’s experiments
- and in separate test tubes, added RNase, DNase,
- and Protease. When they inserted the treated transforming factors back into the
- mice, those treated with RNase and Protease died while the DNase ones lived, meaning DNA is the
- transforming factor. If RNA were the genetic material, the RNase treated transforming factor would’ve
- In Hershey-Chase, they took T2
- bacteriophages and used 32P and 35S radioactive isotopes. They found the 32P in
- the cells, so DNA is the genetic material. Their results would still be the
- same if RNA were the genetic material because it still has phosphates.
Telomerase is used to compensate
for what issue?
What are the two structural
components of telomerase?
How does telomerase work?
- During transcription of eukaryotic chromosomes, the ends shorten with each replication
- where the primer is removed but DNA is not added. Telomerase is used to extend
- the ends to protect it (like shoelace aglets).
It is made of protein and an RNA template
- The RNA acts as a template to build upon the 3’ end of the chromosome. This
- elongation lets another Okazaki fragment to be formed where the shortened end
What are three post-transcriptional modifications that eukaryotic mRNA undergo? What is the relevance of each (i.e. why do they occur, what are their roles)?
- -5’ G-cap: a methylguanine added in reverse orientation, used
- for efficient translation (it is recognized by the small ribosomal unit for
- binding to occur).
- 3’ polyA
- tail- 50-250 adenines added after a cleavage site on the 3’ end. Used to orient
- the mRNA properly on the ribosome.
- Splicing- Introns are taken out by spliceosomes and the exons are joined together.
- Alternative splicing allows for more proteins to be coded from the same
- transcript, so less DNA is needed.
Describe a method of
transcriptional termination in eukaryotes (there are many, but you guys only
- pre-mRNA is cleaved at a cleavage site after the polyadenylation consensus sequence. RAT1 attaches
- to the leftover bit of RNA and uses its exonuclease activity to chew up the strand
- until it reaches the RNA pol, where it and the RNA pol fall off.
three forms of DNA exist?
& Crick’s DNA model is which form?
Describe at least 4 major features
of the Watson Crick model of DNA
- three forms of DNA exist?
- & Crick’s DNA model is which form?
- Describe at least 4 major features
- of the Watson Crick model of DNA
Yikes! You are looking at different
cells and in each one, a different protein/enzyme has been mutated so that
they’re no longer functional. Describe what will happen to the DNA, and state whether initiation
will occur and whether DNA is made.
-DNA Pol I
- Pol I- The RNA primers wouldn’t be removed from the 5’ end of the leading
- strand/Okazaki fragments. Initiation does occur and DNA is made.
- vGyrase- there will be a lot of tension
- from torsional strain so unwinding will be limited. Initiation does occur and
- DNA is made.
- vdnaA- the origin of replication will
- not be unwound. Initiation will not
- occur, DNA will not be made.
- Helicase- The replication fork
- won’t proceed past the initial replication bubble. Initiation will occur, and
- DNA is made.
- Ligase- The Okazaki fragments won’t
- be joined together. Initiation does occur and DNA is made.
Some quick notes to remember
- Always include 5’ and 3’ labels for RNA and DNA, and N and C for proteins
- whenever possible.
- The C terminus comes before
- the stop codon (ex. N-Met-Ile-Leu-C Stop, NOT N-Met-Ile-Leu-Stop-C),
- otherwise it implies there’s a stop amino acid (which there is not!!).
- If you come across a stop codon, you don’t have to continue translating afterwards
- (unless explicitly specified) since translation has stopped. It won’t be on the
- protein sequence.
Follow directions! “inserts a nucleotide” is NOT an effect on the protein.
Be careful! Translation of mRNA NEEDS a start codon. No start codon=no protein.
study! one of the four big problems!
coding strand in transcription is the same term as
non-template strand. (5' to 3')
Template strand will be 3' to 5'
splicing is a two step process
requires consensus sequences at both 5' and 3' ends of introns and exons and the sequence in the intron known as the 'branch point'.
The spliceosome is 5 different RNA molecules, small nuclear RNAs associated with proteins snRNP(snurps)
In Prokaryotic transcription can happen and right away, almost simultaneously, translation can be happening too, because...
there's no editing and no nucleus to leave, etc.
5' cap helps with what ?
- -where to start the process
- -protects the ends from outside nucleases
- 1. TFIID binds to TATA box
- 2. TFIIA and TFIIB bind (stabilizes complex)
- 3. TFIIF and RNA polymerase holoenzyme bind
- 4. TFIIE binds, causing the jaws of the polymerase to open
- 5. TFIIH binds and acts as a helicase to unwind the DNA
- 6. Activator proteins bound to enhancer or regulatory promoter can interact with basal apparatus (mediator) and increase level of transcription.
"Core promoter" - basal transcription factors
proteins, that along with RNA polymerase II, are enough to intiate minimal or 'basal' levels of transcription. These are also called 'general transcription factors'
RNA Pol II Promoters, two primary parts:
core promoter, regulatory promoter
Bacterial termination of transcription
both rho dependent and independent use hairpin loops in RNA from sequence to pause the RNA polymerase
Transcription in Prokaryotes 3 stages
initiation, elongation, termination
3' to 5' is the template strand, the other is the coding strand or non template strand?
RNA polymerases, one type in prokaryotes:
(4 in Eukaryotes) RNA pol 1, 2 and 3 in animals, a 4 in plants
core enzyme + sigma =holoenzyme
core enzyme - RNA polymerase , sigma ensures that it binds in a stable manner, without sigma, it cannot distinguish between promoter sequence and other DNA sequences.