• 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)?

• DNA is transcribed into RNAcharacteristics of RNA
• RNA is translated into protein
• Advantages
• Exceptions

LE 17-4
DNA molecule
Gene 1
Gene 2
Gene 3
DNA strand
(template)
3

TRANSCRIPTION mRNA 5
TRANSLATION
Protein

Codon
Amino acid
5

3


• 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 17-5
Second mRNA base

Figure 17-06

• 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

• 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 Transcription unit
5

3

Start point
RNA polymerase
DNA
Initiation
3

5

5

3

Unwound
DNA
RNA transcript
Template strand of DNA
3

5


LE 17-7a-3
Promoter Transcription unit
5

3

Start point
RNA polymerase
DNA
Initiation
3

5
5

3

Unwound
DNA
RNA transcript
Template strand of DNA
Elongation
Rewound
DNA
5

3

5

RNA transcript
3

3

5
3

5

LE 17-7a-4
Promoter Transcription unit
5
3

Start point
RNA polymerase
DNA
Initiation
3
5

5

3

Unwound
DNA
RNA transcript
Template strand of DNA
Elongation
5

3

Rewound
DNA
5

RNA transcript
3

Termination
3

5

3

5

5

3

5

Completed RNA transcript
3

3

5


LE 17-7b
Elongation
RNA polymerase
Non-template strand of DNA
RNA nucleotides
3

5

3
 end
5

Newly made
RNA
Direction of transcription
(“downstream”)
Template strand of DNA

LE 17-8
5

3

5

3

Promoter
Eukaryotic promoters
TATA box
Start point Template
DNA strand
Several transcription factors
3

5

Transcription factors
3

5

Additional transcription factors
5

3

RNA polymerase II
Transcription factors
5

RNA transcript
3

5

Transcription initiation complex

• 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).

• 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.

• 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-tRNA binding site)
E site
(Exit site)
A site (AminoacyltRNA binding site)
E P A mRNA binding site
Schematic model showing binding sites
Large subunit
Small subunit

LE 17-16a tRNA molecules
Growing polypeptide
Exit tunnel
Large subunit
E P
A
5
 mRNA
3

Computer model of functioning ribosome
Small subunit

• 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 acid attachment site
3

5

Hydrogen bonds
Anticodon
Two-dimensional structure
5

Amino acid attachment site
3

Hydrogen bonds
Anticodon
Three-dimensional structure
3

5

Anticodon
Symbol used in this book

• “charging” enzyme-amino acyl t-rna synthetase (20 different enzymes)
• Requires ATP

LE 17-15
Amino acid
Aminoacyl-tRNA synthetase (enzyme)
Pyrophosphate
Phosphates tRNA
AMP
Aminoacyl tRNA
(an “activated amino acid”)

• 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
Amino acids
Ribosome tRNA with amino acid attached
5
 mRNA
Codons tRNA
Anticodon
3


• 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
E
Growing polypeptide
Next amino acid to be added to polypeptide chain tRNA
3

5
 Codons
Schematic model with mRNA and tRNA

• 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
Large ribosomal subunit
P site
Initiator tRNA
5
 mRNA
Start codon mRNA binding site
3

Small ribosomal subunit
GTP GDP
E A
5

Translation initiation complex
3


• 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
Amino end of polypeptide
Ribosome ready for next aminoacyl tRNA mRNA
5

E
P site
A site
3

2 GTP
2 GDP
E
P A
E
P A
GDP
GTP
E
P A

• 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
Release factor
Free polypeptide
5

3

3

5

Stop codon
(UAG, UAA, or UGA)
When a ribosome reaches a stop codon on mRNA, the A site of the ribosome accepts a protein called a release factor instead of tRNA.
5

3

The release factor hydrolyzes the bond between the tRNA in the
P site and the last amino acid of the polypeptide chain. The polypeptide is thus freed from the ribosome.
The two ribosomal subunits and the other components of the assembly dissociate.

• 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)

• 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
Growing polypeptides
Completed polypeptide
Incoming ribosomal subunits
Start of mRNA
(5
 end)
End of mRNA
(3
 end)
An mRNA molecule is generally translated simultaneously by several ribosomes in clusters called polyribosomes.

LE 17-20b
Ribosomes mRNA
0.1 m m
This micrograph shows a large polyribosome in a prokaryotic cell (TEM).

• Eukaryotic M-rna is modified extensively after transcription (while its still in the nucleus)
• These modifications include
A.Polyadenylationadded to 3’ end of m-rna
B. 5’ cap
C. Intron removal

• 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

• 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
Protein-coding segment
5

5

Cap 5

UTR
Start codon Stop codon
Polyadenylation signal
3

UTR
Poly-A tail

• 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
5

Exon 1
RNA transcript (pre-mRNA)
Intron
Protein snRNA
Exon 2
Other proteins snRNPs
Spliceosome

LE 17-11-2
5

Spliceosome
Spliceosome components
5

Exon 1 mRNA
Exon 2
Cut-out intron

• Why would chromosomes carry around extra DNA that isn’t used in the final mrna?
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

• 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
Exons
1 2 3 4
Troponin T gene
5
Primary
RNA transcript
1 2 3 4 5 mRNA 1 2 3 5
RNA splicing or 1 2 4 5

Summary of Transcription and Translation

• 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

• Spontaneous mutations
• Induced mutations
• Consequences of mutations-
• A. no effect-”silent mutations”
• B. harmful mutations-(may be lethal)
• C. beneficial mutations (rare)

• 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
5

Protein
Amino end
Stop
Carboxyl end
3


LE 17-24b
Base-pair substitution
No effect on amino acid sequence
U instead of C
Missense
Nonsense
U instead of A
A instead of G
Stop
Stop
Stop

• 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
3

Wild-type hemoglobin DNA
5

3

Mutant hemoglobin DNA
5

5
 mRNA
Normal hemoglobin
3

5
 mRNA
Sickle-cell hemoglobin
3


• 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
Wild type mRNA
Protein
5

3

Stop
Carboxyl end Amino end
Base-pair insertion or deletion
Frameshift causing immediate nonsense
Extra U
Stop
Frameshift causing extensive missense Missing
Insertion or deletion of 3 nucleotides: no frameshift but extra or missing amino acid
Missing
Stop

• 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.

• Plasmids
• Restriction Endonucleases
• Agarose Gel Electrophoresis

• 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

• 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)

• 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

• Clone the H gene (use a bacteria to make copies of the gene for us)

• E.coli (lacZ(-), amp sensitve)

• 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

• 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

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)