translation and transcription Flashcards
Griffith’s experiment
step- by- step
he used 4 different treatments on mice and saw how they reacted to each
worked with two strains of bacteria
one pathogenic(S cells) and one harmless(R cells)
when he gave a mouse living S pathogenic cells it dies
when he gave a mouse living R, nonpathogenic cells, it lived
mice given heat killed S cells, survived
lastly, he gave mice a mixture of heat-killed S cells and living R cells
mice died
the heat killed remains of the pathogenic strain made some of the living non pathogenic cells become pathogenic
called this transformation
hershey-chase experiment
Radioactive sulfur was used to determine if proteins carried genetic material and radioactive phosphorus was used to determine if DNA used genetic material
experiment tested which macromolecule was going to enter into the bacteria and act like a blueprint for more phage(which carries the genetic material to facilitate this)
result: dna entered and the protein stayed outside the bacteria so dna has to be the genetic material
hershey-chase experiment
step by step
two different batches of experiments
one batch where phage were labeled with radioactive protein and another with radioactive DNA
the phage were allowed to interact with bacteria cells and centrifuged to be detached once done
then the material found in bacterial cells were determined (whether protein or dna entered)
If we conducted the Hershey-Chase experiment again using radioactive N15, where would you expect to find the radioactive signal (and what part of the macromolecule is being labelled?
both inside the bacteria and outside the bacteria since both protein and dna macromolecules contain N
why do cells divide
growth
repair
to reproduce
conservative model of dna rep
idea that both of the parental strands remained after replication
the parental dna strands acetic as a template for daughter strands and then reassembled into original parent double helix
after second replication = three daughter double helix’s and one parental double helix
semiconservative model
the true model of dna rep
the two parental strands separate and each individual strand acts as a template for the synthesis of a new complementary strand
after first rep, two double helix’s each with one parental strand and one daughter strand
after second rep, two double helix’s with one parental one daughter strand, and then two double helix’s with just daughter strands
dispersive model
replication results in strands of dna with a mixture of both daughter and parental strand sections
sections of daughter and parental strands after replication
meselson-stahl experiment
bacteria cultured in medium with 15N(heavy isotope) was transferred into a medium with 14N(lighter isotope)
DNA sample was centrifuged after the first replication and there was one weighted band of DNA
DNA sample centrifuged after the second replication revealed two bands, one heavier weighted band and one lighter weighted band
semi conservative model is the only model that would have one band in the first replication and two in the second
Looking at the experimental result after the first replication of the DNA, where there is one uniform weighted band of DNA, what model of DNA replication can be ruled out?
conservative
where does DNA replication start
at origins of replication
- short section of DNA with specific sequence of DNA
proteins that recognize these sequence attach to the DNA to separate the two strands, opening a replication bubble
replication fork
replication of DNA proceeds in both directions after the two dna strands are opened and create a replication bubble
each end of the replication bubble has a replication fork where DNA is being unwound to enable replication
what does the extension of the polymer of nucleic acids require
a dehydration reaction per each monomer added
the dehydration reaction between each nucleotide monomer results in the lengthening of DNA strand, and a phospho-dyester bond between each monomer
DNA polymerases
responsible for catalyzing the formation of the phosphodiester bond
add nucleotides to chain of dna strands
can only read/move from 3’ to 5’ direction (adds from 3’ to 5’)
can only add a new nucleotide at the 3’ end
requires a primer to start adding nucleotides
how does dna replication begin
helical unzips/unwinds DNA and creates replication fork
topoisomerase releases tension from the helicase unzipping DNA so that the DNA strands don’t break apart
single-strand binding proteins act as anchors to hold DNA open
primase attaches to parental strand to write the RNA primer that DNA polymerase can attach to
the leading strand
dna polymerase III attaches to the RNA primer and moves from 3’ to 5’ on the parental strand to add nucleotides to primer
creates a continuous 5’ to 3’ daughter strand towards the replication fork
the lagging strand
dna polymerase moves from 3’ to 5’ on the parental strand so for the lagging strand its moving away from the replication fork
makes okazaki fragments
dna polymerase iii attaches to rna primers and detaches once it reaches another rna primer ahead of it on the parental strand
dna polymerase i replaces the rna primer fragments with dna
dna ligament forms bonds between the dna fragments
purpose of telomerase
every time you replicate DNA you lose a chunk of DNA at the end
when a DNA strand gets too short, it signals the cell to die —> apoptosis
some cells need to keep being replicated to keep us functioning
avoid damage to such DNA and DNA shortening by making sure the strands don’t get too short
Telomerase is the enzyme responsible for replacing the DNA that is lost to replication
telomeres —> arms of DNA that telomerase is adding DNA to
what are the general steps of getting from DNA to mRNA
transcription of DNA into pre-mRNA
RNA processing of pre-mRNA into mRNA
occurs in the nucleus
transcription of DNA to mRNA
one DNA strand is used as a template strand and read by RNA Polymerase (3’ to 5’)
the other strand is called non-template strand/coding strand (5’ to 3’)
nucleotide sequence is the same as RNA strand but with U instead of T
how do you identify a gene
you use the coding strand to identify the gene
find promoter/TATA box
find start codon (ATG)
find stop codon (TAG, TAA, TGA)
transcription
the formation of a specific rna sequence from a specific dna sequence
what is required for transcription
a dna templtate
the four ribonucleoside triphosphates
a rna polymerase
salts and pH buffers if done in a test tube
rna polymerase
reads template dna strand from 3’ to 5’
catalyzes addition of nucleotides in a 5’ to 3’ direction
doesn’t require primers or helicase
gene structure that tells rna polymerase how to transcribe
promoter with initiation site that tells the promoter which nucleotide to start transcribing
sigma and transcription factors
terminator that tells rna polymerase to fall off
three steps of transcription
three steps of transcription
why is the termination sequence necessary in rna transcription
rna polymerase ignores start and stop codons in transcription and transcribes parts of DNA that won’t be transcribed if it followed the start and stop codon
the terminator sequence then causes unstable binding of the polymerase which causes it to fall off
processing mRNA after transcription
directly after transcription you get pre-mRNA
you process pre-mRNA by modifying both ends of the strand and removing introns
how are the ends of pre-mRNA modified
a modified guanine nucleotide is added to the 5’ end (5’ cap)
poly A tail is added to the 3’ end
splicing
introns are spliced out and exons are brought together
exons are the parts of transcript that are actually read by ribosomes and made into proteins
spliceosomes selectively remove introns
snRNps bind to the end of each intron and direct spliceosomes to introns
splicing results in diversity of proteins because
any part of the sequence can be deemed an intron or exon and thus the resulting protein will differ
translation occurs where
occurs in cytoplasm
3 main components of translation
mature, processed mRNA template
tRNA with attached amino acids
ribosome
large and small subunits
codons
triplets of nucleic acids that correspond to amino acids
tRNA
links the information contained in mRNA codons with specific amino acids
how does tRNA link the information contained in mRNA codons with specific amino acids
anticodon portion connects to transcribed mRNA sequence
amino acid attachment site has a sequence where a specific amino acid that corresponds to the specific amino acid, attaches
through tRNA synthetase
tRNA synthetase
attachment of tRNA to the appropriate amino acid is called charging
tRNA synthetaae is the protein that charges tRNA
tRNA and its respective amino acid enter active site of protein
ATP synthetase catalyzes covalent bonding of amino acid to tRNA
ribosomal structure in translation
large and small subunits come together in initiation
small subunit has an mRNA binding site
large subunit has an A site, P site, and E site
A site is where tRNA enters, P site is where polypeptide chain grows
E site is where the tRNA’s are released after they add amino acids to the chain
three steps of translation
initiation
elongation
termination
initiation step of translation
mRNA from transcription binds to mRNA binding site of small ribosomal subunit
first tRNA has anticodon of start codon AUG and attaches to psite of small ribosomal unit
large ribosomal unit completes the initiation complex by binding to the small subunit
elongation step of translation
charged tRNA enters A site and anticodon binds to codon of mRNA
tRNA in patients donates amino acid chain to incoming amino acid on the new tRNA
large subunit catalyzes peptide bond in P site between amino acid and existing polypeptide chain
first tRNA moves to E site and exits ribosomal complex and ribosome moves down the mRNA strand from 5’ to 3’ direction (n to c)
termination step of translation
when the A site of the complex reaches a stop codon
UAA, UAG, or UGA
release factor enters
hydrolysis between tRNA and amino acid/polypeptide chain
polypeptide is released from the complex
why is polyribosomes important
several ribosomes can simultaneously translate a single mRNA molecule
mass production of polypeptides is possible
this is important because mRNA is short-lived in the cytoplasm
why is mRNA short-lived in the cytoplasm
ribosomes are competing against a molecule called exonucleases
its job is to destroy mRNA and it cuts a couple nucleotides at a time
proteins function in specific compartments
sometimes proteins only function in specific cellular locations
this information is placed in the protein primary structure as a target peptide
no sequence means the peptide goes to cytosol 9
