Transcription,translation and genetic code(cell biology)by welfredo yu


Cells are governed by a cellular chain of
command
 DNA RNA protein
Transcription
 Is the synthesis of RNA under the direction of
DNA
 Produces messenger RNA (mRNA)
Translation
 Is the actual synthesis of a polypeptide, which
occurs under the direction of mRNA
 Occurs on ribosomes
The Genetic Code
 a non-overlapping sequence, with each amino
acid plus polypeptide initiation and termination
specified by RNA codons composed of three
nucleotides.


 Genes can be expressed at
different efficiencies
 • Gene A is transcribed
much more efficiently than
gene B
 • This allows the amount of
protein A in the cell to be
 greater than protein B
 • The lower expression of
gene B is a reason behind
 incomplete dominance
 BIOL211

 RNA is the bridge between
genes and the proteins for
which they code.
what is RNA?


Monomers of proteins are amino
acids
what are proteins made
of?


Codons
• A codon is three nucleotides in a row on
an RNA
molecule that codes for a single amino acid
• A specific three-nucleotide sequence
encodes for each
amino acid


Template Strand
• During transcription, one of the two DNA
strands, called the template strand, provides
a template for ordering the sequence of
complementary nucleotides in an RNA
transcript
– The template strand is always the same strand for
a given gene
– However, different genes may be on opposite
strands


• The genetic code is nearly
universal, shared by the simplest
bacteria to the most complex
animals
– Some species prefer certain
codons (codon bias)
• Genes can be transcribed and
translated after being
transplanted from one species
to another
EVOLUTION OF THE CODE


History: linking genes
and proteins
 1900’s Archibald Garrod
 Inborn errors of metabolism: inherited human
metabolic diseases (more information)
 Genes are the inherited factors
 Enzymes are the biological molecules that drive
metabolic reactions
 Enzymes are proteins
 Question:
 How do the inherited factors, the genes, control the
structure and activity of enzymes (proteins)?


and proteins
 Beadle and Tatum (1941) PNAS USA 27, 499–506.
 Hypothesis:
 If genes control structure and activity of metabolic enzymes,
then mutations in genes should disrupt production of
required nutrients, and that disruption should be heritable.
 Method:
 Isolated ~2,000 strains from single irradiate spores
(Neurospora) that grew on rich but not minimal medium.
Examples: defects in B1, B6 synthesis.
 Conclusion:
 Genes govern the ability to synthesize amino acids, purines
and vitamins.


and proteins
 1950s: sickle-cell anemia
 Glu to Val change in hemoglobin
 Sequence of nucleotides in gene determines sequence
of amino acids in protein
 Single amino acid change can alter the function of the
protein
 Tryptophan synthase gene in E. coli
 Mutations resulted in single amino acid change
 Order of mutations in gene same as order of affected
amino acids


Ribosomal structure
E
P A
Large subunit
Peptidyl-tRNA binding site
Aminoacyl-tRNA binding site
mRNA
5’
Exit
site
Small subunit
3’


From gene to protein:
transcription
 Gene sequence (DNA) recopied or transcribed to
RNA sequence
 Product of transcription is a messenger molecule that
delivers the genetic instructions to the protein
synthesis machinery: messenger RNA (mRNA)


Transcription: evidence
for mRNA
 Brenner, S., Jacob, F. and Meselson, M. (1961) Nature
190, 576–81.
 Question: How do genes work?
 Does each one encode a different type of ribosome
which in turn synthesizes a different protein, OR
 Are all ribosomes alike, receiving the genetic
information to create each different protein via some
kind of messenger molecule?


for mRNA
 E. coli cells switch from making bacterial proteins to
phage proteins when infected with bacteriophage T4.
 Grow bacteria on medium containing “heavy”
nitrogen (15N) and carbon (13C).
 Infect with phage T4.
 Immediately transfer to “light” medium containing
radioactive uracil.


for mRNA
 If genes encode different ribosomes, the newly
synthesized phage ribosomes will be “light”.
 If genes direct new RNA synthesis, the RNA will contain
radiolabeled uracil.
 Results:
 Ribosomes from phage-infected cells were “heavy”,
banding at the same density on a CsCl gradient as the
original ribosomes.
 Newly synthesized RNA was associated with the heavy
ribosomes.
 New RNA hybridized with viral ssDNA, not bacterial
ssDNA.


for mRNA
 Conclusion
 Expression of phage DNA results in new phage-
specific RNA molecules (mRNA)
 These mRNA molecules are temporarily associated
with ribosomes
 Ribosomes do not themselves contain the genetic
directions for assembling individual proteins


Transcription: overview
 Transcription requires:
 ribonucleoside 5´ triphosphates:
 ATP, GTP, CTP and UTP
 bases are adenine, guanine, cytosine and uracil
 sugar is ribose (not deoxyribose)
 DNA-dependent RNA polymerase
 Template (sense) DNA strand
 Animation of transcription


 Features of transcription:
 RNA polymerase catalyzes sugar-phosphate bond
between 3´-OH of ribose and the 5´-PO4.
 Order of bases in DNA template strand determines
order of bases in transcript.
 Nucleotides are added to the 3´-OH of the growing
chain.
 RNA synthesis does not require a primer.


 In prokaryotes transcription and translation are
coupled. Proteins are synthesized directly from the
primary transcript as it is made.
 In eukaryotes transcription and translation are
separated. Transcription occurs in the nucleus, and
translation occurs in the cytoplasm on ribosomes.
 Figure comparing eukaryotic and prokaryotic
transcription and translation.


Transcription: RNA
Polymerase
 DNA-dependent
 DNA template, ribonucleoside 5´ triphosphates, and Mg2+
 Synthesizes RNA in 5´ to 3´ direction
 E. coli RNA polymerase consists of 5 subunits
 Eukaryotes have three RNA polymerases
 RNA polymerase II is responsible for transcription of
protein-coding genes and some snRNA molecules
 RNA polymerase II has 12 subunits
 Requires accessory proteins (transcription factors)
 Does not require a primer


Transcription: promoter
recognition
 Transcription factors bind to promoter sequences and
recruit RNA polymerase.
 DNA is bound first in a closed complex. Then, RNA
polymerase denatures a 12–15 bp segment of the DNA
(open complex).
 The site where the first base is incorporated into the
transcription is numbered “+1” and is called the
transcription start site.
 Transcription factors that are required at every promoter
site for RNA polymerase interaction are called basal
transcription factors.


Promoter recognition: promoter sequences
 Promoter sequences vary considerably.
 RNA polymerase binds to different promoters with
different strengths; binding strength relates to the
level of gene expression
 There are some common consensus sequences for
promoters:
 Example: E. coli –35 sequence (found 35 bases 5´ to the
start of transcription)
 Example: E. coli TATA box (found 10 bases 5´ to the
start of transcription)


Promoter recognition:
enhancers
 Eukaryotic genes may also have enhancers.
 Enhancers can be located at great distances from the
gene they regulate, either 5´ or 3´ of the transcription
start, in introns or even on the noncoding strand.
 One of the most common ways to identify promoters
and enhancers is to use a reporter gene.


Promoter recognition:
other players
 Many proteins can regulate gene expression by
modulating the strength of interaction between the
promoter and RNA polymerase.
 Some proteins can activate transcription (upregulate
gene expression).
 Some proteins can inhibit transcription by blocking
polymerase activity.
 Some proteins can act both as repressors and
activators of transcription.


Transcription: chain
initiation
 Chain initiation:
 RNA polymerase locally denatures the DNA.
 The first base of the new RNA strand is placed
complementary to the +1 site.
 RNA polymerase does not require a primer.
 The first 8 or 9 bases of the transcript are linked.
Transcription factors are released, and the
polymerase leaves the promoter region.
 Figure of bacterial transcription initiation.


elongation
 Chain elongation:
 RNA polymerase moves along the transcribed or
template DNA strand.
 The new RNA molecule (primary transcript) forms a
short RNA-DNA hybrid molecule with the DNA
template.


termination
 Most known about bacterial chain termination
 Termination is signaled by a sequence that can form
a hairpin loop.
 The polymerase and the new RNA molecule are
released upon formation of the loop.
 Review the transcription animation.


Transcription: mRNA
synthesis/processing
 Prokaryotes: mRNA transcribed directly from DNA
template and used immediately in protein synthesis
 Eukaryotes: primary transcript must be processed to
produce the mRNA
 Noncoding sequences (introns) are removed
 Coding sequences (exons) spliced together
 5´-methylguanosine cap added
 3´-polyadenosine tail added


Transcription: mRNA
 Removal of introns and splicing of exons can occur
several ways
 For introns within a nuclear transcript, a spliceosome is
required.
 Splicesomes protein and small nuclear RNA (snRNA)
 Specificity of splicing comes from the snRNA, some of
which contain sequences complementary to the splice
junctions between introns and exons
 Alternative splicing can produce different forms of a protein
from the same gene
 Mutations at the splice sites can cause disease
 Thalassemia • Breast cancer (BRCA 1)


Transcription: mRNA
 RNA splicing inside the nucleus on particles called
spliceosomes.
 Splicesomes are composed of proteins and small
RNA molecules (100–200 bp; snRNA).
 Both proteins and RNA are required, but some
suggesting that RNA can catalyze the splicing
reaction.
 Self-splicing in Tetrahymena: the RNA catalyzes its
own splicing
 Catalytic RNA: ribozymes


genetic code
 Central Dogma
 Information travels from DNA to RNA to Protein
 Is there a one-to-one correspondence between DNA,
RNA and Protein?
 DNA and RNA each have four nucleotides that can form
them; so yes, there is a one-to-one correspondence between
DNA and RNA.
 Proteins can be composed of a potential 20 amino acids;
only four RNA nucleotides: no one-to-one correspondence.
 How then does RNA direct the order and number of amino
acids in a protein?


genetic code
 How many bases are required for each amino acid?
 (4 bases)2bases/aa = 16 amino acids—not enough
 (4 bases)3bases/aa = 64 amino acid possibilities
 Minimum of 3 bases/aa required
 What is the nature of the code?
 Does it have punctuation? Is it overlapping?
 Crick, F.H. et al. (1961) Nature 192, 1227–32.
(http://profiles.nlm.nih.gov/SC/B/C/B/J/ )
 3-base, nonoverlapping code that is read from a fixed
point.


genetic code
 Nirenberg and Matthaei: in vitro protein translation
 Found that adding rRNA prolonged cell-free protein
synthesis
 Adding artificial RNA synthesized by polynucleotide
phosphorylase (no template, UUUUUUUUU)
stimulated protein synthesis more
 The protein that came out of this reaction was
polyphenylalanine (UUU = Phe)
 Other artificial RNAs: AAA = Lys; CCC =Pro


genetic code
 Nirenberg:
 Triplet binding assay: add triplet RNA, ribosomes, binding
factors, GTP, and radiolabeled charged tRNA (figure)
 UUU trinucleotide binds to Phe-tRNA
 UGU trinucleotide binds to CYS-tRNA
 By fits and starts the triplet genetic code was worked out.
 Each three-letter “word” (codon) specifies an amino acid or
directions to stop translation.
 The code is redundant or degenerate: more than one way to
encode an amino acid


Translation
 Components required for translation:
 mRNA
 Ribosomes
 tRNA
 Aminoacyl tRNA synthetases
 Initiation, elongation and termination factors
 Animation of translation


Translation: initiation
 Ribosome small subunit binds to mRNA
 Charged tRNA anticodon forms base pairs with the
mRNA codon
 Small subunit interacts with initiation factors and
special initiator tRNA that is charged with
methionine
 mRNA-small subunit-tRNA complex recruits the
large subunit
 Eukaryotic and prokaryotic initiation differ slightly


Translation: initiation
 The large subunit of the ribosome contains three
binding sites
 Amino acyl (A site)
 Peptidyl (P site)
 Exit (E site)
 At initiation,
 The tRNAfMet occupies the P site
 A second, charged tRNA complementary to the next
codon binds the A site.


Translation: elongation
 Elongation
 Ribosome translocates by three bases after peptide
bond formed
 New charged tRNA aligns in the A site
 Peptide bond between amino acids in A and P sites is
formed
 Ribosome translocates by three more bases
 The uncharged tRNA in the A site is moved to the E
site.


Translation: elongation
 EF-Tu recruits charged tRNA to A site. Requires
hydrolysis of GTP
 Peptidyl transferase catalyzes peptide bond
formation (bond between aa and tRNA in the P
site converted to peptide bond between the two
amino acids)
 Peptide bond formation requires RNA and may
be a ribozyme-catalyzed reaction


Translation: termination
 Termination
 Elongation proceeds until STOP codon reached
 UAA, UAG, UGA
 No tRNA normally exists that can form base pairing
with a STOP codon; recognized by a release factor
 tRNA charged with last amino acid will remain at P
site
 Release factors cleave the amino acid from the tRNA
 Ribosome subunits dissociate from each other
 Review the animation of translation

48
Genetic code:
Def. Genetic code is the nucleotide base sequence on DNA ( and
subsequently on mRNA by transcription) which will be translated into a
sequence of amino acids of the protein to be synthesized.
The code is composed of codons
Codon is composed of 3 bases ( e.g. ACG or UAG). Each codon is
translated into one amino acid.
The 4 nucleotide bases (A,G,C and U) in mRNA are used to produce the
three base codons. There are therefore, 64 codons code for the 20 amino
acids, and since each codon code for only one amino acids this means
that, there are more than one cone for the same amino acid.
How to translate a codon (see table):
This table or dictionary can be used to translate any codon sequence.
Each triplet is read from 5′ → 3′ direction so the first base is 5′ base,
followed by the middle base then the last base which is 3′ base.

49
Examples: 5′- A UG- 3′ codes for methionine
5′- UCU- 3′ codes for serine
5′ - CCA- 3′ codes for proline
Termination (stop or nonsense) codons:
Three of the 64 codons; UAA, UAG, UGA do not code for any amino
acid. They are termination codes which when one of them appear in
mRNA sequence, it indicates finishing of protein synthesis.
Characters of the genetic code:
1- Specificity: the genetic code is specific, that is a specific codon
always code for the same amino acid.
2- Universality: the genetic code is universal, that is, the same codon is
used in all living organisms, procaryotics and eucaryotics.
3- Degeneracy: the genetic code is degenerate i.e. although each codon
corresponds to a single amino acid,one amino acid may have more than
one codons. e.g arginine has 6 different codons (give more examples
from the table).

51
Gene mutation (altering the nucleotide sequence):
1- Point mutation: changing in a single nucleotide base on the mRNA
can lead to any of the following 3 results:
i- Silent mutation: i.e. the codon containg the changed base may code
for the same amino acid. For example, in serine codon UCA, if A is
changed to U giving the codon UCU, it still code for serine. See table.
ii- Missense mutation: the codon containing the changed base may code
for a different amino acid. For example, if the serine codon UCA is
changed to be CCA ( U is replaced by C), it will code for proline not
serine leading to insertion of incorrect amino acid into polypeptide chain.
iii- Non sense mutation: the codon containing the changed base may
become a termination codon. For example, serine codon UCA becomes
UAA if C is changed to A. UAA is a stop codon leading to termination
of translation at that point.

53
Types of point mutation:
U A A (termination codon) Nonsense mutation
↑
U C A → U C U Silent mutation
(codon for serine) (codon for serine)
↓
C C A ( codon for proline) Missense mutation:
Give other examples on missense mutation which leads to some Hb
disease.

54
2- Frame- shift mutation:
deletion or addition of one or two base to
message sequence, leading to change in
reading frame (reading sequence) and the
resulting amino acid seuence may become
completely different from this point.

55
Translation
Components required for protein synthesis:
1- Amino acids: all amino acids involved in the
finished protein must be present at the time of
protein synthesis.
2- Ribosomes: the site of protein synthesis. They are
large complexes of protein and rRNA. In human,
they consist of two subunits, one large (60S) and one
small (40S).
3- tRNA: at least one specific type of tRNA is required to transfer
one amino acid. There about 50 tRNA in human for the 20 amino
acids, this means some amino acids have more than one specific
tRNA. The role of tRNA in protein synthesis is discussed before.
(amino acid attachment and anticodon loop).
4- aminoacyl-tRNA synthetase: This is the enzyme that catalyzes the
attachment of amino acid with its corresponding tRNA forming
aminoacyl tRNA

56
5- mRNA: that carry code for the protein to be synthesized
6- protein factors: Initiation, elongation and termination (or release)
factors are required for peptide synthesis
7- ATP and GTP : are required as source of energy.
Steps: (movie)
1- Initiation:
Initiation (start) codon is usually AUG which is the codon of
methionine, so the initiator tRNA is methionnyl tRNA (Met. tRNA).
a- The initiation factors (IF-1, IF-2 and IF-3) binds the Met. tRNA with
small ribosomal subunit then to mRNA containing the code of the
protein to be synthesized. IFs recognizes mRNA from its 5' cap

57
b-This complex binds to large ribosomal subunit forming initiation complex in which
Met. tRNA is present in P- site of 60 ribosomal subunit.
NB:- tRNA bind with mRNA by base pairing between codon on mRNA and anticodon
on tRNA.
- mRNA is read from 5′ → 3′ direction
P-site: is the peptidyl site of the ribosome to which methionyl tRNA is placed (enter).
2- Elongation: elongation factors (EFs) stimulate the stepwise elongation of
polypeptide chain as follow:
a- The next aminoacyl tRNA (tRNA which carry the next amino acid specified by
recognition of the next codon on mRNA) will enter A site of ribosome
A site or acceptor site or aminoacyl tRNA site:
Is the site of ribosome to which each new incoming aminoacyl tRNA will enter.

b) ribosomal peptidyl transferase enzyme will transfer methionine
from methionyl tRNA into A site to form a peptide bond between
methionine and the new incoming amino acid to form dipeptidyl
tRNA.
c) Elongation factor-2 (EF-2), (called also, translocase): moves
mRNA and dipeptidyl tRNA from A site to P site leaving A site free to
allow entrance of another new aminoacyl tRNA.
The figure shows the repetitive cycle of elongation of chain. Each
cycle is consisting of
1) codon recognition and the entrance of the new aminoacyl tRNA
acid ( amino acid carried on tRNA) into A site,
2) The growing chain in P site will moved to A site with peptide
bond formation with the new amino acid
3) Translocation of growing chain to P site allowing A site free for
enterance of new amino acid an so on………………….. Resulting in
elongation of poly peptide chain.

59
repetitive cycle of elongation

60
3- Termination: occurs when one of the three stop codons (UAA, UAG or
UGA) enters A site of the ribosome. These codons are recognized by release
factors (RFs) which are RF-1, RF-2, RF-3. RFs cause the newly synthesized
protein to be released from the ribosomal complex and dissociation of
ribosomes from mRNA (i.e. cause dissolution of the complex)

Transcription,translation and genetic code(cell biology)by welfredo yu

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