What is a gene?
• A sequence of DNA nucleotides that specifies the primary structure of a polypeptide chain (tells the cell how to make it)
• Genes-made of nucleotides• Proteins-made of amino acids• How does a nucleotide code (in the
nucleus) specify an amino acid sequence (in the cytoplasm)?
The Central Dogma
• DNA is transcribed into RNA-characteristics of RNA
• RNA is translated into protein
• Advantages
• Exceptions
LE 17-4
DNAmolecule
Gene 1
Gene 2
Gene 3
DNA strand(template)
3
TRANSCRIPTION
Codon
mRNA
TRANSLATION
Protein
Amino acid
35
5
The Genetic code-characteristics
• Triplet (3 nucleotides=codon=info for a specific amino acid);64 different codons (3 are stop codons)
• Universal• Redundant (61 codons-20 amino acids)-
variability in third nucleotide of codon. Advantages of a redundant code?
• Non-overlapping • Exceptions (ciliates; mito/chloroplasts)
LE 17-5Second mRNA base
Fir
st
mR
NA
ba
se
(5 e
nd
)
Th
ird
mR
NA
ba
se
(3 e
nd
)
Figure 17-06
Gene Expression
• If a gene is transcribed and the m-rna is translated (the gene is expressed); a protein is made. This often changes the phenotype of the cell that produces the protein.
• Differential gene expression is involved in embryonic development and cell specialization.
• Totipotency-each cell has the genetic information for an entire organism.
• Differential gene expression results in cell specialization (differentiation)
• Hormones often play a role in gene expression
Transcription
• The first step in gene expression• Takes place in the nucleus• Requirements• A. RNA nucleotides• B. DNA template (gene)• C. Enzymes (RNA polymerase)• Only one of the two DNA strands is copied
(template strand)
LE 17-7a-2
Promoter
53
35
35
53
Transcription unit
DNA
Initiation
Start pointRNA polymerase
UnwoundDNA
RNAtran-script
Template strandof DNA
LE 17-7a-3
Promoter
53
Transcription unit
35DNA
Start pointRNA polymerase
Initiation
35
53
UnwoundDNA
RNAtran-script
Template strandof DNA
Elongation
RewoundDNA
35
53 3
5
RNAtranscript
LE 17-7a-4Promoter
35
Transcription unit
DNA
InitiationRNA polymerase
Start point
Template strandof DNA
RNAtran-script
UnwoundDNA
Elongation
3
3
53
5
5
3 5
RewoundDNA
5 3
35 35
RNAtranscript Termination
35
5 3Completed RNA transcript
LE 17-7b
ElongationNon-templatestrand of DNA
RNApolymerase
RNA nucleotides
3 end3
5
5
Newly madeRNA
Templatestrand of DNA
Direction of transcription(“downstream”)
LE 17-8
Promoter
53
35
TATA box Start point
Transcriptionfactors
53
35
Several transcriptionfactors
Additional transcriptionfactors
RNA polymerase IITranscription factors
RNA transcript
53
355
Transcription initiation complex
Eukaryotic promoters
TemplateDNA strand
Transcription-some important details
• Rate-30-60 nucleotides/second• RNA polymerase (Many forms in eucaryotes, 3
basic types in bacteria: type I transcribes r-rna, type II-mrna, types III-trna)
• Promotors-(approximately 100 nucleotides)-strong and weak promotors
• Eukaryotes-transcription factors needed to help RNA polymerase to bind to TATA box (region of promotor 25 nucleotides upstream from initiation site).
RNA products of transcription
• m-rna
• t-rna
• r-rna
• sn-RNA (small nuclear)
• mi-Rna (micro)
• Si-rna (small interfering)
Not all genes code for proteins (m-rna)-Rrna and t-rna are obvious examples
• Actually, recent discoveries indicate that a large part of the eukaryotic genome is non-coding RNA-Introns
• Small rna (micro rna and small interfering rna)-play a crucial role in the regulation of gene expression involving both transcription and translation. Rna interference (Rnai)
• We’ll talk about regulation of gene expression in Chapter 15.
Ribosomal RNA and ribosomes
• R-rna; one of two important components of ribosomes (other is protein-some of the proteins are enzymes). 60% r-rna; 40% protein.
• Ribosomes consist of 2 subunits • Ribosomes needed to translate proteins• “workbench of protein synthesis”• Position t-rna (which is attached to a specific
amino acid) on the codon of a m-rna• Result is the synthesis of a protein (whose
amino acid sequence is specified by the m-rna; which is transcribed from a gene)
LE 17-16b
P site (Peptidyl-tRNAbinding site)
E site (Exit site)
mRNAbinding site
A site (Aminoacyl-tRNA binding site)
Largesubunit
Smallsubunit
Schematic model showing binding sites
E P A
LE 17-16a
tRNAmolecules
Exit tunnelGrowingpolypeptide
Largesubunit
mRNA 3
Computer model of functioning ribosome
Smallsubunit
5
E P A
Ribosomal –rna processing
T-rna
• Single polynucleotide chain folded into a complex 3-D shape (inter-chain H bonding). 75-80 nucleotides in length
• Binds a specific amino acid (involvement of amino-acyl-trna-synthetase
• Attaches to a specific m-rna codon via its anticodon
• How many different t-rna’s are there? 61? Actually only 45 (wobble)
LE 17-14a
Amino acidattachment site
Hydrogenbonds
3
5
Two-dimensional structureAnticodon
Amino acidattachment site
35
Hydrogenbonds
Anticodon Anticodon
Symbol used in this bookThree-dimensional structure
3 5
“Charging” t-rna with its specific amino acid
• “charging” enzyme-amino acyl t-rna synthetase (20 different enzymes)
• Requires ATP
LE 17-15Amino acid Aminoacyl-tRNA
synthetase (enzyme)
Pyrophosphate
Phosphates
tRNA
AMP
Aminoacyl tRNA(an “activatedamino acid”)
Messenger Rna (m-rna)
• Contains the information for the primary sequence of a polypeptide chain
• Consists of codons
• Binds to ribosomes
• T-rna binds to m-rna (codon/anticodon)
LE 17-13
Polypeptide
tRNA withamino acidattached
Ribosome
tRNA
Anticodon
35
mRNA
Aminoacids
Codons
Translation
• Codons (m-rna) read by ribosomes/t-rna• Polypeptide chain produced• 3 steps in translation-• A. initiation• B. elongation• C. termination• Translation is a process that consumes a
tremendous amount of energy (ATP and GTP)
LE 17-16c
Amino end
mRNA
5
3
Growing polypeptide
Next amino acidto be added topolypeptide chain
tRNA
Codons
Schematic model with mRNA and tRNA
E
Translation-Initiation
• Initiation codon is AUG
• T-rna that bonds to AUG has an anticodon UAC-this carries the amino acid methionine
• Requires a GTP molecule
• Requires proteins called initiation factors.
LE 17-17
Met
GTPInitiator tRNA
mRNA
53
mRNA binding site
Smallribosomalsubunit
Start codon
P site
5 3
Translation initiation complex
E A
Largeribosomalsubunit
GDP
Met
Translation-Elongation
• The elongation cycle takes about 60 milliseconds
• During elongation, one m-rna codon is read and then the ribosomes moves down the message to the next codon.
• Binding of incoming t-rna to the A site of the ribosome requires a GTP
• Translocation-requires a GTP
LE 17-18
Ribosome ready fornext aminoacyl tRNA
mRNA
5
Amino endof polypeptide
E
Psite
Asite
3
2
2 GDP
E
P A
GTP
GTP
GDP
E
P A
E
P A
Translation-Termination
• When the ribosome reaches a termination codon, it causes the m-rna/ribosome complex to separate
• No t-rna binds to the termination codon.
• Release factors
• Newly made polypeptide chain is released (folds into its characteristic 3-D shape)
LE 17-19
Releasefactor
Stop codon(UAG, UAA, or UGA)
5
3
5
3
5
Freepolypeptide
3
When a ribosome reaches a stopcodon on mRNA, the A site of theribosome accepts a protein calleda release factor instead of tRNA.
The release factor hydrolyzes thebond between the tRNA in theP site and the last amino acid of thepolypeptide chain. The polypeptideis thus freed from the ribosome.
The two ribosomal subunitsand the other componentsof the assembly dissociate.
Summary of energy demands for protein synthesis
• A rough estimate is that for every amino acid incorporated into a polypeptide chain, 3 ATP/GTP are consumed
A. Charging the amino acid (1 ATP)B. Binding of incoming t-rna into the A site (1
GTP)C. Translocation (1 GTP)D. So a small protein (120 amino acids in length)
would cost the cell 360 ATP/GTP to make (the equivalent of 12 glucose molecules going through aerobic cell respiration)
Polyribosomes
• A single ribosome can translate an average-sized polypeptide in about 1 minute
• Several ribosomes can translate the same message one after the other.
• Increases the efficiency of protein production
LE 17-20a
Incomingribosomalsubunits
Growingpolypeptides
Completedpolypeptide
Start ofmRNA(5 end)
End ofmRNA(3 end)
Polyribosome
An mRNA molecule is generally translated simultaneouslyby several ribosomes in clusters called polyribosomes.
LE 17-20b
Ribosomes
mRNA
This micrograph shows a large polyribosome in a prokaryotic cell (TEM).m0.1
M-rna modifications
• Eukaryotic M-rna is modified extensively after transcription (while its still in the nucleus)
• These modifications include
A.Polyadenylation-added to 3’ end of m-rna
B. 5’ cap
C. Intron removal
M-RNA modifications
• Poly A tail
• A. added to the 3’ end of the m-rna
• B.30-200 Adenine nucleotides
• C. roles-regulation of transport of m-rna out of the nucleus; regulation of degradation of m-rna in the cytoplasm; helps m-rna attach to small ribosomal subunit
M-RNA modifications (continued)
• 5’ cap
• A. Modified guanine nucleotide stuck onto 5’ end of m-rna
• B. Roles- positioning of m-rna on small ribosomal subunit in initiation; protects m-rna from degradation
LE 17-9
5Protein-coding segment
5 Start codon Stop codon Poly-A tail
Polyadenylation signal
5 3Cap UTR UTR
Introns
• Many eukaryotic genes have nucleotide sequences that don’t code for amino acids (Introns)
• Introns separate coding sequences (exons). Split genes
• Introns must be removed from the m-rna before it is translated (introns have nucleotide sequences that indicate splicing sites)
• Splicesomes are molecular machines that remove introns from m-rna
LE 17-11-1
Exon 15
Intron
Other proteins
Protein
snRNA
snRNPs
RNA transcript (pre-mRNA)
Spliceosome
Exon 2
LE 17-11-2
5
Spliceosome
Cut-outintron
mRNA
Exon 1 Exon 25
Spliceosomecomponents
Significance of introns
• Why would chromosomes carry around extra DNA that isn’t used in the final m-rna?
A. Expensive to maintain (energy). B. Splicing out introns is a risky business
(what if it’s done incorrectly)C. With these disadvantages, there must be
an advantage or natural selection would not favor this arrangement
Benefits of Introns
• Evolution of protein diversity
• One gene can be alternatively spliced in a number of different ways to form several different types of m-rna (alternative splicing)
• Human antibody genes-about 500 genes can code for billions of different antibody molecules because of alternative splicing.
Figure 15.12
DNA
PrimaryRNAtranscript
mRNA or
Exons
Troponin T gene
RNA splicing
1 2 3 4 5
1 2 3 5 1 2 4 5
1 2 3 4 5
Summary of Transcription and Translation
Mutation
• An alteration in the nucleotide sequence of a DNA molecule (chromosome)
• Chromosomal mutations (duplications; deletions; inversions)
• Point mutations-alterations of one or a few nucleotides in a gene
Point mutations
• Spontaneous mutations
• Induced mutations
• Consequences of mutations-
• A. no effect-”silent mutations”
• B. harmful mutations-(may be lethal)
• C. beneficial mutations (rare)
Spontaneous mutations
• Base pairing errors; why aren’t they corrected by DNA repair enzymes?
• Effects:• A. no effect-silent mutation (redundancy of
genetic code; alteration of a non-critical amino acid)
• B. Positive effect-rare• C. negative effect-missense mutations;
nonsense mutations
LE 17-24a
Wild-type
mRNA
Protein
Amino end Carboxyl end
Stop
35
LE 17-24bBase-pair substitution
No effect on amino acid sequenceU instead of C
Stop
MissenseA instead of G
Stop
NonsenseU instead of A
Stop
Sickle cell anemia
• Results of a spontaneous missense mutation
• Result-altered hemoglobin molecule
• Effect-Depends on the environmental conditions and number of copies of the defective gene you inherited.
LE 17-23
Wild-type hemoglobin DNA
mRNA
3 5 53
5 3 35
Mutant hemoglobin DNA
mRNA
Normal hemoglobin Sickle-cell hemoglobin
Induced mutations
• Caused by environmental damage• Radiation (UV)- T-T dimers; excision
repair enzymes; xerdoerma pigmentosa • Chemicals-Common result-base pair
addition or deletion• Result of addition or deletion (frame shift
mutation)-missense or nonsense • Worst scenario-addition/deletion of 1 or 2
nucleotides at the beginning of a gene
LE 17-25
Base-pair insertion or deletion
Frameshift causing immediate nonsense
Extra U
MissingFrameshift causingextensive missense
Insertion or deletion of 3 nucleotides:no frameshift but extra or missing amino acid
Missing
Stop
Stop
Amino end Carboxyl end
Stop
Wild type
mRNA
Protein5 3
Mutations and Cancer
• Many mutations make cells cancerous
• 90% of known carcinogens are mutagens
• Ames test-screens potential chemicals for being carcinogens by seeing if they are mutagens
• Bacteria are the test subjects in the Ames test.
Employee Resource Manual
• Plasmids
• Restriction Endonucleases
• Agarose Gel Electrophoresis
Plasmids
• Small extrachromosomal pieces of DNA found in some bacterial species
• May carry additional genes (such as antibiotic resistance)
• Can be genetically modified and used as vectors for genetic engineering
PUC 18-Plasmid
Restriction Endonucleases
• Produced by some bacteria as a defense against virus infection
• Cleave DNA at specific bases sequences (different recognition site for each different enzyme)
• Can be used to join DNA from 2 different sources (plasmid DNA and genomic DNA)
ECOR1
Agarose Gel Electrophoresis
• Separates DNA based upon size differences
• DNA is pulled through a gel by an electric current
• (-) charged DNA is pulled to the positive pole of the apparatus.
• Smaller pieces of DNA migrate through the gel faster than larger pieces of DNA
Agarose Gel Electrophoresis
Procell
Procell in Action
What is your first job assignment?
• Clone the H gene (use a bacteria to make copies of the gene for us)
What kind of bacteria do we use to clone the H gene?
• E.coli (lacZ(-), amp sensitve)
Where is the H gene located?
• Lambda virus
How do you get the cloned gene into the bacteria so the bacteria can copy it?
• Transform E.coli (lacZ(-), amp sensitve) with PUC 18-lambda plasmid (heat shock and osmotic shock)
Lambda virus genes have
been inserted into the plasmids here
How do you get lambda genes into PUC 18 plasmid?
• Incubate PUC-18 and lambda with EcoRI, ligate products
How many different plasmids do you get when you mix PUC 18 and lambda, both of which have been
ECOR1 and then ligated?
7
Would all 7 of the plasmids be recombinant (have lambda DNA)?
No!
How do you tell if bacteria have been transformed successfully with PUC-18
plasmid?
• They will grow on amp agar.
How can you distinguish whether plasmids that transformed bacteria were recombinant (lambda and PUC-18) or nonrecombinant (pUC-18 only)?
Plate the transformed cells on Xgal-amp agar
E.Coli transformed With nonrecombinant Plasmids (PUC-18)
E.Coli transformed With recombinant Plasmids (PUC-18/lambda)