Transcript By controlling Protein Synthesis

From Gene to
Protein
Question?
How
does DNA control a
cell?
By controlling Protein
Synthesis.
Proteins are the link between
genotype and phenotype.
Central Dogma
DNA
Transcription
RNA
Translation
Polypeptide
Explanation
DNA
- the Genetic code or
genotype.
RNA - the message or
instructions.
Polypeptide - the product for
the phenotype.
Genetic Code
Sequence
of DNA bases that
describe which Amino Acid
to place in what order in a
polypeptide.
The genetic code gives the
primary protein structure.
Genetic Code
Is
based on triplets of bases.
Has redundancy; some AA's
have more than 1 code.
20 amino acids: 64 codons
Code Redundancy
Third
base in a codon shows
"wobble”.
First two bases are the most
important in reading the code
and giving the correct AA.
The third base often doesn’t
matter.
Code Evolution
The
genetic code is nearly
universal.
Ex: CCG = proline (all life)
Reason - The code must have
evolved very early. Life on
earth must share a common
ancestor.
Reading Frame and
Frame Shift
The
“reading” of the code is every
three bases (Reading Frame)
Ex: the red cat ate the rat
Frame shift – improper groupings
of the bases
Ex: thr edc ata tat her at
The “words” only make sense if
“read” in this grouping of three.
Transcription
Process
of making RNA from
a DNA template.
Only one strand is used as a
template.
Where in the cell does
transcription take place?
Eukaryotes:
nucleus
Prokaryotes: cytoplasm
Transcription Steps
1.
2.
3.
4.
RNA Polymerase Binding
Initiation
Elongation
Termination
RNA Polymerase
Enzyme
for building RNA
from RNA nucleotides.
Binding
Requires
that the enzyme
find the “proper” place on the
DNA to attach and start
transcription.
Binding
Is
a complicated process
Uses Promoter Regions (start
region) on the DNA (upstream
from the information for the
protein)
Initiation
Actual
unwinding of DNA to
start RNA synthesis.
Elongation
RNA
Polymerase untwists
DNA 1 turn at a time.
Exposes 10 DNA bases for
pairing with RNA nucleotides.
Elongation
Enzyme
moves 5’
3’.
Rate is about 60 nucleotides
per second.
Comment
Each
gene can be read by
sequential RNA Polymerases
giving several copies of RNA.
Result - several copies of the
protein can be made.
Termination
DNA
sequence that tells RNA
Polymerase to stop.
Ex: AATAAA
RNA Polymerase detaches
from DNA after closing the
helix.
Final Product
Pre-mRNA
This
is a “raw” RNA that will
need processing.
Modifications of RNA
1. 5’ Cap
2. Poly-A Tail
3. Splicing
5' Cap
Modified
Guanine nucleotide
added to the 5' end.
Protects mRNA from
digestive enzymes.
Recognition sign for
ribosome attachment.
Poly-A Tail
150-200
Adenine nucleotides
added to the 3' tail
Protects mRNA from
digestive enzymes.
Aids in mRNA transport from
nucleus.
RNA Splicing
Removal
of non-protein
coding regions of RNA.
Coding regions are then
spliced back together.
Introns
Intervening
sequences.
Removed from RNA.
Exons
Expressed
sequences of
RNA.
Translated into AAs.
Result
Introns - Function
Left-over
DNA (?)
Way to lengthen genetic
message.
Old virus inserts (?)
Way to create new proteins.
Translation
Process
by which a cell
interprets a genetic message
and builds a polypeptide.
Where in the cell does
translation take place?
Eukaryotes
and prokaryotes:
cytoplasm because that’s
where the ribosomes are
located.
Materials Required
tRNA
Ribosomes
mRNA
Transfer RNA = tRNA
Made
by transcription.
About 80 nucleotides long.
 Carries AA for polypeptide
synthesis.
Structure of tRNA
Has
double stranded regions
and 3 loops.
AA attachment site at the 3'
end.
1 loop serves as the
Anticodon.
Anticodon
Region
of tRNA that base
pairs to mRNA codon.
Usually is a compliment to
the mRNA bases, so reads
the same as the DNA codon.
Example
DNA
- GAC
mRNA - CUG
tRNA anticodon - GAC
Ribosomes
Two
subunits made in the
nucleolus.
Made of rRNA (60%)and
protein (40%).
rRNA is the most abundant
type of RNA in a cell.
Large subunit
Proteins
rRNA
Both sununits
Large Subunit
Has
3 sites for tRNA.
P site: Peptidyl-tRNA site carries the growing polypeptide
chain.
 A site: Aminoacyl-tRNA site holds the tRNA carrying the next
AA to be added.
E site: Exit site
Translation Steps
1. Initiation
2. Elongation
3. Termination
Initiation
Brings
together:
mRNA
A
tRNA carrying the 1st AA
2 subunits of the ribosome
Elongation Steps:
1. Codon Recognition
2. Peptide Bond Formation
3. Translocation
Codon Recognition
tRNA
anticodon matched to
mRNA codon in the A site.
Peptide Bond
Formation
A
peptide bond is formed
between the new AA and the
polypeptide chain in the
P-site.
After bond formation
The
polypeptide is now
transferred from the tRNA in
the P-site to the tRNA in the
A-site.
Translocation
tRNA
in P-site is released.
Ribosome advances 1 codon
tRNA in A-site is now in the
P-site.
Process repeats with the next
codon.
Termination
Triggered
by stop codons.
Release factor binds in the
A-site instead of a tRNA.
 H2O is added instead of AA,
freeing the polypeptide.
Ribosome separates.
Prokaryotes
Comment
Polypeptide
usually needs to
be modified before it
becomes functional.
Examples
Sugars,
lipids, phosphate
groups added.
Some AAs removed.
Protein may be cleaved.
Join polypeptides together
(Quaternary Structure).