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Transcript
Chapter 25
Nucleic Acids and Protein Synthesis
 Introduction
Deoxyribonucleic acid (DNA) and ribonucleic acid (RNA) are the
molecules that carry genetic information in the cell


DNA is the molecular archive for protein synthesis
RNA molecules transcribe and translate the information from DNA so it can be
used to direct protein synthesis
DNA is comprised of two polymer strands held together by
hydrogen bonds



Its overall structure is that of a twisted ladder
The sides of the ladder are alternating sugar and phosphate units
The rungs of the ladder are hydrogen-bonded pairs of heterocyclic amine bases
Chapter 25
2
DNA polymers are very long molecules

DNA is supercoiled and bundled into 23 chromosomes for packaging in the cell
nucleus
The sequence of heterocyclic amine bases in DNA encodes the
genetic information required to synthesize proteins




Only four different bases are used for the code in DNA
A section of DNA that encodes for a specific protein is called a gene
The set of all genetic information coded by the DNA in an organism is its genome
The set of all proteins encoded in the genome of an organism and expressed at
any given time is its proteome
The sequence of the human genome is providing valuable
information related to human health

Example: A schematic map of genes on chromosome 19 that are related to
disease
Chapter 25
3
 Nucleotides and Nucleosides
Mild degradation of nucleic acids yields monomer units called
nucleotides
Further hydrolysis of a nucleotide yields:



A heterocyclic amine base
D-ribose (from RNA) or 2-deoxy-D-ribose (from DNA); both are C5
monosaccharides
A phosphate ion
The heterocylic base is bonded by a b N-glycosidic linkage to C1’
of the monosaccharide

Examples: A general structure of an RNA nucleotide (a) and adenylic acid (b)
A nucleoside is a nucleotide without the phosphate group

A nucleoside of DNA contains 2-deoxy-D-ribose and one of the following four
bases
Chapter 25
4
A nucleoside of RNA contain the sugar D-ribose and one of the
four bases adenine, guanine, cytosine or uracil
Chapter 25
5
Nucleosides that can be obtained from DNA
Chapter 25
6
Nucleosides that can be obtained from RNA
Chapter 25
7
Nucleotides can be named in several ways


Adenylic acid is usually called AMP (adenosine monophosphate)
It can also be called adenosine 5’-monophosphate or 5’-adenylic acid
Adenosine triphosphate (ATP) is an important energy storage
molecule
The molecule 3’,5’-cyclic adenylic acid (cyclic AMP) is an
important regulator of hormone activity

This molecule is biosynthesized from ATP by the enzyme adenylate cyclase
Chapter 25
8
 Laboratory Synthesis of Nucleosides and
Nucleotides
Silyl-Hilbert-Johnson Nucleosidation



An N-benzoyl protected base reacts with a benzoyl protected sugar in the
presence of tin chloride and BSA (a trimethylsilylating agent)
The trimethylsilyl protecting groups are removed with aqueous acid in the 2nd
step
The benzoyl groups can be removed with base
Chapter 25
9
Unnatural nucleotide derivatives can be synthesized from
nucleosides bearing a substitutable group on the heterocyclic ring
Chapter 25
10
Dibenzyl phosphochloridate is a phosphorylating agent for
converting nucleosides to nucleotides

The 5’-OH is phosphorylated selectively if the 2’- and 3’-OH groups are protected
Chapter 25
11
 Deoxyribonucleic Acid: DNA
 Primary Structure
The monomer units of nucleic acids are nucleotides

Nucleotides are connected by phosphate ester linkages
The backbone of nucleic acids consists of alternating phosphate
and sugar units
Heterocyclic bases are bonded to the backbone at each sugar unit
The base sequence contains the encoded genetic information
The base sequence is always specified from the 5’ end of the
nucleic acid
Chapter 25
12
 Secondary Structure
The secondary structure of DNA was proposed by Watson and
Crick in 1953
E. Chargaff noted that in DNA the percentage of pyrimidine bases
was approximately equal to the percentage of purine bases


Also the mole percentage of adenine Is nearly equal to that of thymine
The mole percentage of guanine is nearly equal to cytosine
Chargaff also noted that the ratio of A and T versus G and C varies
by species but the ratio is the same for different tissues in the
same organism
Chapter 25
13
X-ray crystallographic data showed the bond lengths and angles
of purine and pyrimidine bases

X-ray data also showed DNA had a long repeat distance (34 Å)
Based on this data, Watson and Crick proposed the double helix
model of DNA (next slide)





Two nucleic acid chains are held together by hydrogen bonding between the
bases on opposite strands
The double chain is wound into a helix
Each turn in the helix is 34Å long and involves 10 successive nucleotide pairs
Each base pair must involve a purine and a pyrimidine to achieve the proper
distance between the sugar-phosphate backbones
Base pairing can occur only between thymine and adenine, or cytosine and
guanine; no other pairing has the optimum pattern of hydrogen bonding or would
allow the distance between sugar-phosphate backbones to be regular
Chapter 25
14
Chapter 25
15
Specific pairing of bases means the two chains of DNA are
complementary

Knowing the sequence of one chain allows one to also know the sequence of the
other
Chapter 25
16
 Replication of DNA (see next slide)
The DNA strand begins to unwind just prior to cell division
Complementary strands are formed along each chain (each chain
acts as a template for a new chain)
Two new DNA molecules result; one strand goes to each daughter
cell
Chapter 25
17
Chapter 25
18
 RNA and Protein Synthesis
“The central dogma of molecular genetics”
A gene is the portion of a DNA molecule which codes for one
protein

Proteins have many critical functions, e.g., catalysis, structure, motion, cell
signaling, the immune response, etc.
DNA resides in the nucleus and protein synthesis occurs in the
cytoplasm


Transcription of DNA into messenger RNA (mRNA) occurs in the nucleus
mRNA moves to the cytoplasm and the translation into proteins occurs using two
other forms of RNA: ribosomal RNA (rRNA) and transfer RNA (tRNA)
Chapter 25
19
 Transcription: Synthesis of Messenger RNA (mRNA)
In the nucleus a DNA molecule partially unwinds to expose a
portion corresponding to at least one gene
Ribonucleotides with complementary bases assemble along the
DNA strand

Base-pairing is the same in RNA, except that in RNA uracil replaces thymine
Ribonucleotides are joined into a chain of mRNA by the enzyme
RNA polymerase
Chapter 25
20
An intron (intervening sequence) is a segment of DNA which is
transcribed into mRNA but not actually used when a protein is
expressed
An exon (expressed sequence) in the part of the DNA gene which
is expressed
Each gene usually contains a number of introns and exons

Introns are excised from mRNA after transcription
Chapter 25
21
 Ribosomes - rRNA
Protein synthesis is catalyzed in the cytoplasm by ribosomes


A ribosome consists of approximately two thirds RNA and one third protein
A ribosome is a ribozyme ( an reaction catalyst made of ribonucleic acid)
A ribosome has 2 large subunits


The 30S subunit binds the mRNA that codes for the protein to be translated
The 50S subunit catalyzes formation of the amide bond in protein synthesis
Transfer of an amino acid to the growing peptide chain is aided by
acid-base catalysis involving an adenine in the 50S subunits
See Figure 25.14, page 1238
Chapter 25
22
 Transfer RNA (tRNA)
Transfer RNAs (tRNAs), specific to each amino acid, transport
amino acids to complimentary binding sites on the mRNA bound
to the ribosome

More than one tRNA codes for each amino acid
tRNA is comprised of a relatively small number of nucleotides
whose chain is folded into a structure with several loops


One arm of the tRNA always terminates in the sequence cytosine-cytosineadenine, and it is here the amino acid is attached
On another arm is a sequence of three bases called the anticodon, which binds
with the complementary codon on mRNA
The mRNA genetic code is shown on the next page
The structure of a tRNA molecule is shown in Figure 25.15, page
1240
Chapter 25
23
Chapter 25
24
 The Genetic Code
The genetic code is based on three-base sequences in mRNA
Each three-base sequence corresponds to a particular amino acid



The fact that three bases are used to code for each amino acid provides
redundancy in the overall code and in the start and stop signals
N-formyl methionine (fMet) is the first amino acid incorporated into bacterial
protein and appears to be the start signal
fMet is removed from the protein chain before its synthesis is complete
Chapter 25
25
 Translation
Translation is peptide synthesis by a ribosome using the code
from an mRNA
The following is an example (see figure on next page):










An mRNA binds to a ribosome
A tRNA with the anticodon for fMet associates with the fMet codon on the mRNA
A tRNA with anticodon UUU brings a lysine residue to the AAA mRNA codon
The 50S ribosome catalyzes amide bond formation between the fMET and lysine
The ribosome moves down the mRNA chain to the next codon (GUA)
A tRNA with the anticodon CAU brings a valine residue
The ribosome catalyzes amide bond formation between Lys and Val
The ribosome moves along the mRNA chain and the process continues, e.g., with
the tRNA for phenylalanine binding to the ribosome
A stop signal is reached and the ribosome separates from the mRNA
At this point the polypeptide also separates from the ribosome
The polypeptide begins to acquire its secondary and tertiary
structure as it is being synthesized
Several ribosomes can be translating the same mRNA molecule
simultaneously
Protein molecules are synthesized only when they are needed

Regulator molecules determine when and if a particular protein will be expressed
i.e. synthesized
Chapter 25
26
Chapter 25
27
 Determining the Base Sequence of DNA
The Chain-Terminating (Dideoxynucleotide) Method






DNA molecules are replicated in such a way that a family of partial copies is
generated; each DNA copy differs in length by only one base
Random chain-termination is done by ‘poisoning’ a replication reaction with a low
concentration of 2’3’-dideoxynucleotides, which are incapable of chain elongation
at their 3’ position
The 2’3’-dideoxynucleotides are labeled with covalently attached colored
fluorescent dye molecules, with each color representing a base type
The partial copies are separated according to length by capillary electrophoresis
The terminal base on each strand is detected by the color of laser-induced
fluorescence as each DNA molecule passes the detector
A four-color chromatogram is generated (see Figure 25.17, page 1246)
Automation of high-throughput ‘dideoxy’ sequencing made
possible completion of the Human Genome Project by the 50th
anniversary of Watson and Crick’s elucidation of the structure of
DNA in 2003
Chapter 25
28
 Laboratory Synthesis of DNA
Solid-phase methods for laboratory synthesis of DNA are similar
to those used for laboratory synthesis of proteins



The solid phase is often controlled-pore glass (CPG)
Protecting/blocking reagents are needed (e.g., the dimethoxytrityl and bcyanoethyl groups)
A coupling reagent (1,2,3,4-tetrazole) is used to join the protected nucleotides
Chapter 25
29
Chapter 25
30
 The Polymerase Chain Reaction (PCR)
PCR is an extraordinarily simple and effective method for
exponentially multiplying (amplifying) the number of copies of a
DNA molecule.


PCR beginning with a single molecule can lead to 100 billion copies in an
afternoon
The Nobel Prize was awarded to K. Mullis in 1993 for invention of PCR
PCR requires:





A sample of the DNA to be copied
The enzyme DNA polymerase
A short ‘primer’ sequence complimentary to the template DNA
A supply of A, C, G, and T nucleotide triphosphate monomers
A simple device for thermal cycling during the reaction sequence
The PCR process is summarized on the next 2 slides
Chapter 25
31
Chapter 25
32
Chapter 25
33