Download 1 CHAPTER 3- DNA FUNCTION – THE EXPRESSION OF GENETIC

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CHAPTER 3- DNA FUNCTION – THE EXPRESSION OF GENETIC
INFORMATION
Questions to be addressed:
1. How is information in nucleus (DNA) transmitted to the cytoplasm (site of protein
synthesis)
2. How is mRNA made from DNA
3. How does linear DNA sequence specify the linear amino acid sequence?
4. How is the nucleic acid “code” translated into a protein sequence?
5. How is protein made?
6. How may we relate genes to biological processes?
Terminology (see also Glossary pages 655-681 and web site
http://helios.bto.ed.ac.uk/bto/glossary/)
Informational RNA provide a template for protein synthesis (mRNA)
Functional RNA function as an RNA molecule (e.g. tRNA, rRNA, snRNA)
Transcription – production of RNA from a DNA template
RNA polymerase – the enzyme which transcribes DNA into RNA
Promoter – a set of DNA sequences to which RNA polymerase binds
Repressor – a protein that binds to a DNA element and prevents transcription
Activator – a protein that binds to a DNA element and activates transcription
Codon – 3 nucleotides in mRNA that encodes an amino acid
Anticodon – 3 nucleotides in tRNA that from complementary base pairs with the codon
Wobble – the ability of the 3rd nucleotide of the anticodon to pair imprecisely, so that
the anticodon can align with several codons
Active site – the part of the protein that is required for protein function
Wild type – the standard form of a gene or characteristic
Mutation –DNA or chromosome that is different from wild type
allele: a different form of a gene caused by alteration to the DNA sequence
mutant – an organism or cell containing a mutation
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frameshift – the insertion or deletion of nucleotides results in altered translational
reading frame
genotype: the alleles contained by an individual
phenotype: the form taken by a character in a specific individual
e.g. characteristic = flower colour; phenotype = purple
dominant: an allele whose phenotype is expressed when heterozygous with another
allele
recessive: an allele whose phenotype is not expressed when heterozygous with another
allele
haplo-sufficiency – one copy of the normal gene is able to confer a normal phenotype
haplo-insufficiency – one copy of the normal gene is unable to confer the normal
phenotype
prototrophic – an organism that will survive on minimal medium (carbon source,
inorganic salts, water)
auxotrophic – an organism that will not survive on minimal medium, but whose
growth depends on supplementation of medium with a specific substance
1: How is information in nucleus (DNA) transmitted to the cytoplasm (site of protein
synthesis)
observations: nucleic acid RNA is synthesized in the nucleus, and moves into the
cytoplasm
note: RNA is different from DNA, since it:
1) contains ribose instead of deoxyribose
2) is single-stranded, but can form duplexes through complementary bonds with RNA
or DNA
3) contains uracil in place of thymine (bases are A, C, G and U)
hypothesis: RNA = intermediate between DNA and protein
hypothesis 1: RNA will move from nucleus to cytoplasm.
experiment 1) labeled uracil with radioactive 32P isotope
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expected result if hypothesis is correct?:
conclusion:
hypothesis 2: increase in a specific RNA will precede an increase in a specific protein
experiment 2: infected a bacterial cell with a virus
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virus injects genetic information (DNA) into the cell and uses bacterial biosynthetic
machinery to make protein
expected result if hypothesis is correct:
conclusion: RNA is the precursor to protein
Overall conclusion: DNA acts as a template for the production of RNA, which then
moves into the cytoplasm and acts as a template for protein synthesis. This RNA is
molecule is called messenger RNA = mRNA.
Classes of RNA (page 59-60)
1) Informational RNAs – provide information template for protein synthesis
Messenger RNAs
2) Functional RNAs = RNA that is functional as an RNA molecule and is
not translated into a polypeptide.
Transfer RNAs – tRNAs – transport amino acids to RNA during protein synthesis
Ribosomal RNAs – rRNAs – component of the ribosome
Small nuclear RNAs – snRNAs – involved in RNA processing in eukaryotes
Small cytoplasmic RNAs – scRNAs – protein trafficking in eukaryotes
2. How is RNA made from DNA?
TRANSCRIPTION = Production of mRNA from DNA template (pg. 60-64)
-in each gene, one DNA strand is used as a template for complementary base pairing
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-other strand = non-template stand
-in each gene, template strand may be different
-transcribed RNA is complementary to the template
-labeling of gene is based on direction of RNA transcript
-
(figures 3-4, 3-5, 3-7, 3-8)
enzymes which drive transcription = RNA polymerase
RNA polymerase I (pol I) – transcribes rRNA genes
RNA polymerase II (pol II) – transcribes protein coding genes
RNA polymerase III (pol III) – transcribes tRNA, snRNA, scRNA
3 steps in transcription =
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initiation
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elongation
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termination
-we will concentrate on what is known about E. coli (prokaryotic) transcription, similar
mechanisms are used in eukaryotes
RNA polymerase in E. coli:
-consists of several different subunits which have different roles during the
transcription process
initiation - holoenzyme consists of 4 subunits: α, β, β‘, σ
elongation and termination - core enzyme consists of only 3 subunits: α, β, β‘
1) initiation (figure 3-9):
-determines where a mRNA molecule will start
a set of DNA sequences (= promoter) are required to initiate transcription (note - =
upstream, + = downstream)
consensus sequences:
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-35 region = TTGACAT - recognized by σ-factor
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-10 region = TATAAT - helix opens (note A-T joined by only 2 H-bonds)
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+1 often A in CAT - beginning of transcription
2) elongation
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within a few bases of the initiation site, σ-factor dissociates, leaving the core enzyme
to continue polymerization of RNA
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reads the DNA template to create an mRNA molecule
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identity of bases added is based on complementarity to the DNA
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mRNA is complementary to the DNA from which it is synthesized
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new bases are added to the 3’ OH, thus the RNA is synthesized 5’ ---> 3’
3) termination
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determines where the mRNA molecule will end (3’ end)
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2 mechanisms are used:
a) intrinsic (figure 3-10)
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in DNA template- GC rich region followed by As
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in mRNA - hairpin loop followed by Us
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hairpin loop is a structure which causes the RNA polymerase to pause
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U-A have a weak association, and as the polymerase pauses, the RNA-DNA
heteroduplex dissociates
b) rho dependent
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requires rho protein
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rho binds to RNA at rut site, and moves along RNA towards RNA polymerase
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RNA polymerase pauses at a site within the RNA
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rho catches up, and causes RNA polymerase to dissociate from the RNA
Control of transcription
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not all genes are transcribed in all cells at all times
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numerous points at which gene expression can be controlled
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most common is at transcription initiation
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activators help RNA polymerase to initiate (positive regulation), repressors keep
RNA polymerase from initiating (negatie regulation)
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in E. coli, -35, -10 and +1 sequences are consensus sequences
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present in many but not all promoters
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a promoter which does not have the ideal consensus sequence needs help from
other proteins = activators
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access to promoter may be prevented by other proteins = repressors
RNA processing = post transcriptional modification (Figures 3-11, 3-12, 3-16)
-occurs in eukaryotes, before pre-mRNA enters the cytoplasm
1) cap of 7-methylguanosine to 5’ end by Guanyltransferase
2) endonuclease cuts 3’ end, Poly A polymerase adds poly A tail to 3’ end
3) removal of introns (intervening sequences) and splicing together of exons
(expressed sequences, those used to make protein)
RNA splicing (Figures 3-11, 3-12, 3-13, 3-14, 3-15)
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conserved sequences at exon-intron boundary are recognized and aligned by
snRNPs (protein + snRNAs) to form a splicesome
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OH group in intron attacks the 5’ splice site
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5’ exon is cleaved, and intron forms a lariat
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3’ OH cleaves the 3’ splice site, the lariat is released, and the two exons ligated
Following RNA processing, mature RNA is transported to the cytoplasm
Protein Structure
protein = a polymer of amino acids joined by peptide bonds = polypeptide
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20 amino acids, each has an amino (NH3) and a carboxy (COOH) group, differ in R
group
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peptide bond forms between amino and carboxy groups (Figure 3-17)
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polypeptide has an amino end (NH2) and a carboxyl end (COOH)
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Levels of organization (Figure 3.18)
primary structure - amino acid sequence linked by peptide bonds
secondary structure - weak bonds (e.g. e.g. H-bonds) between neighbouring amino
acids
tertiary structure - bonds (e.g. sulfide bonds) between distant amino acids
quaternary structure - interactions between different polypeptides to form a multimeric
protein
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now known that an enzyme may contain multiple polypeptides (subunits),
each encoded by a different gene
3. How is the linear nucleic acid molecule, DNA, used to generate the linear
polypetide molecules?
Hypothesis: a particular DNA sequence serves as a template for each polypeptide
expectations based on hypothesis:
1) each protein has a unique amino acid sequence
2) mutation in gene causes change in amino acid sequence
3) sequence of nucleotides relates to sequence of amino acids
Question 1: does each protein have a unique sequence?
Experiment: amino acid sequences of different polypeptides deduced through Sanger
sequencing
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Proteolytic enzymes cut between specific aa to produce specific fragments
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Movement through electorphoretic field depends on size and charge of
fragments, allowing separation of different fragments
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Movement in different buffer allows further separation
Result: each polypeptide produces a unique fragment profile (fingerprint)
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Question 2: Do mutations alter protein sequence?
Experiment: compare sequence of amino acids in protein from patients with Sickle
Cell Anemia to the sequence from normal individuals
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disease results from a mutation in a single gene
result: Sanger sequencing shows that hemoglobin fingerprint of normal people is
different from that of sickle cell people
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one amino acid difference in B subunit
conclusion: mutation ----> changes amino acid sequence
Question 3: does the sequence of nucleotides relate to sequence of amino acids –
Charles Yanofsky, Box 3-1
TRPA - gene required for tryptophan biosynthesis in E.coli
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a number of different mutations in the TRPA gene are known to result from single
nucleotide changes (these are called different alleles of the TRPA gene)
Experiment: compare position of changed nucleotide (mutation) with position of
changed amino acid
nucleotide sequence -------X--------Y----------Z-------W-------wild type protein
------tyr------leu-------thr-----gly-------
mutant x protein
------cys------leu-------thr-----gly-------
mutant y protein
------tyr-------arg-------thr-----gly-------
result:
conclusion:
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4. How do genes relate to biological processes?
Hypothesis: biological processes (biosynthetic, degradation, developmental) can be
thought of as pathways, where each pathway consists of a series of steps
Beadle and Tatum, 1942 (Box 3-1)
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choose to work on a very simple amino acid biosynthetic pathway in the fungus
Neurospora
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three mutant strains each carrying a mutant allele of 3 different genes (arg-1, arg-2,
arg-3) were available
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these failed to make the final amino acid product
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the wild type fungus is a prototroph (self-feeder, that can survive on minimal
medium containing only inorganic salts, water and a carbon source), while the
mutants are auxotrophic mutants, since they require an outside source of amino
acid
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by a series of simple and elegant experiments, Beadle and Tatum formulated the
“one gene - one enzyme theory”, which has 3 fundamental components:
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biosynthetic pathways = steps
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steps controlled by enzymes
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one gene produces one enzyme
Hypothesis 1: biosynthetic pathways = series of steps
consider a pathway by which A is converted to D through 2 intermediates, B and C:
A
B
(upstream)
C
D
(downstream)
- expect that if a defect occurs in one step, subsequent steps will not occur
Experimental observations: Mutations in 3 different Neurospora genes unable to make
arginine
= auxotrophic mutants (able to grow only if supplied with a particular substance,
i.e. arginine)
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conclusion:
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pathway is a series of 3 steps
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one gene controls step
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mutation in a gene = a defective step
How might mutant strains be rescued?
Hypothesis 2: If the stepwise pathway is correct…
Results:
Table 1: 3 auxotrophic arginine mutants were tested for growth on medium containing
several compounds. Growth (+) or no growth (-) was scored
mutant
arg-1
+
+
+
arg-2
-
+
+
arg-3
-
-
+
conclusion: compounds can rescue mutants, indicating that 1) compounds =
intermediates in pathway and 2) mutants = defects in steps
Based on the data presented in Table 1, what is the pathway for arginine biosynthesis?
Alternative Experiment:
similar information can be gained from cross-feeding experiments, in which pairs of
mutants are grown side-by-side on medium containing enough arginine to support
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weak growth. Mutant strains are scored for their ability to increase growth of (“feed”)
the neighbouring strain.
according to our pathway:
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mutant arg-3 - accumulates
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mutant arg-2 -accumulates
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mutant arg-1 - accumulates ?
question: which mutant should be able to feed another mutant?
Results:
arg-1
arg-2
arg-3
Conclusions:
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each mutant strain has a unique ability to feed other mutant strains
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each mutant represents a defect in a specific step in the biosynthetic pathway
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each mutant strain carries a mutation in a different gene
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therefore, each gene controls a different step in the biosynthetic pathway
RULE - downstream mutants feed upstream mutants
5. How is the nucleotide code translated into amino acids?
= The Genetic Code (Figure 3-20)
Question: How many nucleotides must be in each “word”?
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20 amino acids, therefore must be at least 20 words
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42 = 16
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43 = 64
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must be at lest 3 nucleotides in each word
Question: Which word = which amino acid?
experiment: synthesize an mRNA containing only Us
-add mRNA to an in vitro translation mix to produce a protein
result: protein contains only Phenylalanine
conclusion:
experiment: synthesize an mRNA containing 1/4 G and 3/4 U
p(GGG) = 1/4 x 1/4 x 1/4 = 1/64
p of any of (GUG, GGU or UGG) = 1/4 x 1/4 x 3/4 = 3/64
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compare frequency of amino acids to known frequency of codons
this sort of experiment generated the Codon Table (figure 3-20)
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now known that codon usage is universal, with the same code being recognized in
all organisms
6. How is Protein Synthesized?
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Translation = the process in which amino acid polymer of specific sequence is made
using mRNA as an information template
steps:
1) initiation (start)
2) elongation (polymerization)
3) termination (end)
components:
1) mRNA - template for protein synthesis
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2) tRNA - carries the amino acids
3) synthetase - attaches a specific amino acid to the tRNA
4) ribosome - provides machinery
-consists of rRNA + protein
5) Protein factors (IF) – factors necessary for initiation (Initiation Factors), elongation
(Elongation Factors) and termination (Release Factors)
tRNA (figures 3-21, 3-22)
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matches a codon in the mRNA to an amino acid
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each tRNA is specific to a particular amino acid
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some amino acids can be carried by tRNAs having different anticodons (accounts for
some of the degeneracy in the code)
tRNA consists of:
a) anticodon loop
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anticodon loop contains 3 bases (anticodon) which are complementary to a codon
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anticodon of tRNA binds to codon through complementary base pairing
wobble position (figure 3-22, Tables 3-2 and 3-3)
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flexibility in 3rd nucleotide (5’) position of anticodon, allows this nucleotide to pair
with nucleotides other than its complementary nucleotide
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anticodon pairs with more than 1 codon
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accounts for degeneracy of code
b) amino acid acceptance site
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3’ OH becomes linked to a specific amino acid, through the action of synthetase
Charging the tRNA
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each tRNA is recognized by a specific synthetase enzyme, which joins the
appropriate amino acid to the tRNA, to produce a charged tRNA = amino-acyl
tRNA
synthetase1
aa1 + tRNA1 -------------------> aa1-tRNA1
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+ ATP
+ AMP + Pi
uncharged tRNA
= amino-acyl tRNA
4. Ribosome
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structure which positions the mRNA, charged tRNA and appropriate factors so that
peptide bonds can be made between sequential amino acids
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consists of both RNA and protein (Figure 3-23)
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RNA in ribosome interacts with mRNA through complementary base pairing
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in prokaryotes, 2 subunits, 50S, 30S = 70S
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in eukaryotes, 2 subunits, 60S, 40S = 80S
within the ribosome, the following reactions polymerize the polypeptide (figure 324):
1) formation of peptide bond between first two amino acids
aa1-tRNA1 + aa2-tRNA2 -------> aa1-aa2-tRNA2 + tRNA1
amino-acyl tRNA
peptidyl tRNA + uncharged tRNA
2) addition of sequential amino acids
aa1-aa2-tRNA2 + aa3-tRNA3 -------------> aa1-aa2-aa3-tRNA3
peptidyl tRNA
amino-acyl tRNA
peptidyl tRNA
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A site = holds amino-acyl tRNA
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P site = holds peptidyl tRNA during elongation, holds an amino-acyl tRNA only
during initiation
Initiation of protein synthesis:
1) -binding of ribosome to mRNA
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2) -entry of first tRNA
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30S (small) subunit of ribosome + IF- translocates along mRNA
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binds to ribosome binding site (GGAGG) = Shine-Dalgarno sequence
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positions the ribosome so that the first codon of the mRNA (initiation codon = AUG)
is in P site of ribosome
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AUG in P accepts a specific amino-acyl tRNA
3) IF + GTP + fMet-tRNA -----> enters P site
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IFs released, 50S (large) subunit joins
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at the completion of initiation, both subunits of ribosome are attached to the mRNA,
with the initiation codon and fMet-tRNA in the P site
Elongation
1. entry of aa2-tRNA2 into A site
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requires EF + GTP
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aa2-tRNA2 in A site and f-met-tRNA in P-site
2) formation of peptide bond, translocation
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peptide bond forms between aa2 and f-Met
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releases f-Met from tRNA1
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fMet-aa2-tRNA2 (peptidyl tRNA) in A site, empty tRNA in P-site
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empty tRNA is released
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ribosome moves to position peptidyl-tRNA2 in P site
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translocation requires EF
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aa3-tRNA3 enters A site (requires EF + GTP)
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many cell toxic compounds inhibit polypeptide elongation
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antibiotic fusidic acid - mimics EF, allowing it to bind to ribosome, but cannot
translocate ribosome
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diphtheria toxin - inhibits EF, therefore stops translocation
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termination
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ribosome encounters a STOP codon (UAA, UGA, or UAG)
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3 codons UAG, UAA, UGA are not recognized by any tRNA = STOP codons
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these codons are recognized by RFs, which enter A site instead of a charged tRNA
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entry of RF causes release of polypeptide from P site
Protein Function
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two major classes of protein, structural (e.g. myosin, keratin) and active (e.g.
enzymes such as PEP carboxylase, alcohol dehydrogenase)
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specificity of protein action is due to specificity of physical interaction
e.g. enzyme and substrate interaction (Fig 3-23)
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active site is a region of the protein critical for the interaction and therefore for
protein function (Fig 3-24)
7. Relating mutations to protein function and phenotypic changes
Types of mutations (see Chapters 10 and 11 for more detail):
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the normal DNA sequence is called “wild type” and is designated either by
UPPERCASE letters, or by a “+” superscript (nomenclature depends on the
organism)
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a mutant allele that changes the DNA sequence is designated by either lowercase
letters, or by the absence of the “+” superscript
Examples of simple mutations
1. single nucleotide change
2. single nucleotide addition or deletion (= frameshift)
Phenotype = the outward manifestation of the genotype, i.e. the form an individual
takes as the result of the genes present in that organism
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the normal organism, containing a wild type set of genes, will display a wild type
phenotype
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alterations to DNA sequence (mutations) within the gene may alter the time, place
or level of expression (if mutation is in the regulatory region), or the efficiency of
protein function (if mutation is in the coding region) (Figure 3-27)
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if mutation alters expression or efficiency of protein, a change in phenotype may
arise
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e.g. gene involved in short term memory in flies
question: based on the mutant phenotype, what does the wild type DUNCE+ allele do?
NOTE: Basic principle of genetic analysis – a gene and its function are identified
based on what happens to the organism when the gene is malfunctioning (usually
through mutation)
1. Geneticists identify genes important to a process through mutation
2. Geneticists identify region of a gene important to its function through mutation
(Figure 3-27, Figure 3-29)
Dominant and Recessive alleles
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dominance and recessive relationships are determined based on the phenotype of
an individual who is heterozygous (two different alleles are present) at the gene
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can only be defined in diploid individuals
Dominant: an allele that expresses its phenotype in the heterozygote
Recessive: an allele whose phenotype is not expressed in the heterozygote
haplo-sufficiency or haplo-insufficiency of a gene determines whether loss-of –
function mutant alleles are dominant or recessive (Figure 3-30)
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Haplo-sufficiency: If one copy (“haplo”) of the normal gene is able to confer the
normal phenotype (i.e. the heterozygote is normal)
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Haplo-insufficiency: If one copy of the normal gene is unable to confer the normal
phenotype (i.e. the heterozygote is mutant)
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e.g.
haplo-sufficient gene
haplo-insufficient gene
genotype
phenotype
genotype
phenotype
a+a+
normal
B+B+
normal
a+a
normal
B+B
defective
aa
defective
BB
defective
(i.e. a+ is dominant to a)
(i.e. B is dominant to B+)