FCH 532 Lecture 4
Chapter 5: DNA
The Watson and Crick double helix model for DNA
Forces in DNA Double Helix
• DNA Double helix is stabilized by two types of
forces:
1. H-bonds between complementary bases on opposite
strands:
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2 H-bonds in A-T pair
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3 H-bonds in G-C pair
2. Van der Waals forces and hydrophobic interactions
between “stacked bases”
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Aromatic bases have p-electrons that interact via
attractive Van der Waals forces.
Structure of DNA determines
heredity
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Watson-Crick bp structure will allow any sequence on one
polynucleotide strand as long as the opposite strand has
complementary sequence.
Each polynucleotide strand can act as the template for its
complementary strand.
In order to replicate, the parental strands must separate so that a
complementary daughter strand can be synthesized on each parent
strand.
Results in duplex (double-stranded) DNA consisting of one
polynucleotide parental strand from the parental molecule and
another from the newly synthesized daughter strand.
This is called semi-conservative replication.
Shown by the Meselson-Stahl experiment in 1958.
Increased density of DNA by labeling with 15N and monitored the
overall DNA density as a function of growth using equilibrium density
gradient centrifugation.
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Denaturation and renaturation
• Duplex DNA can be heated above a certain
temperature to separate the complementary strands
into a random coil conformation.
• Denaturation is followed by a change in the physical
properties of DNA.
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Figure 5-14 Schematic representation of the strand
separation in duplex DNA resulting from its heat
denaturation.
Denaturation is cooperative
• DNA can be monitored by UV absorbance.
• When DNA denatures, UV abs is due to aromatic
bases and increases compared to the double
stranded DNA
• Results from disruptions of electronic interactions
among nearby bases.
• This is called the hyperchromic effect.
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Figure 5-15 UV absorbance spectra of native and
heat-denatured E. coli DNA.
Denaturation is cooperative
• The hyperchromic effect takes place over a narrow
temperature range.
• Indicates that collapse of one part of the DNA duplex will
destabilize the rest of the structure (cooperative
process).
• Melting curves are used to demonstrate the stability of
the DNA double helix and determine the melting
temperature (Tm) which is the midpoint of a melting
curve.
• Tm is dependent on the
– solvent
– concentrations and types of ions
– pH
– Mole fraction of GC base pairs
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Figure 5-16 Example of a DNA melting curve.
Denaturation is cooperative
• The hyperchromic effect takes place over a narrow
temperature range.
• Indicates that collapse of one part of the DNA duplex will
destabilize the rest of the structure (cooperative
process).
• Melting curves are used to demonstrate the stability of
the DNA double helix and determine the melting
temperature (Tm) which is the midpoint of a melting
curve.
• Tm is dependent on the
– solvent
– concentrations and types of ions
– pH
– Mole fraction of GC base pairs
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Figure 5-17 Variation of the melting temperatures, Tm,
of various DNAs with their G + C content.
Denatured DNA can be renatured
• If a solution of DNA is rapidly cooled below the Tm, the
resulting DNA is only partially base paired.
• However, if the temperature is maintained at 25 ºC
belowe the Tm, the base paired regions will rearrange
until DNA completely renatures.
• These are called annealing conditions and are
important for hybridization of complementary strands of
DNA or RNA-DNA hybrid double helices.
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Figure 5-18 Partially renatured DNA.
Size of DNA molecules
• DNA molecules are very large.
• Mass can be determined by
– ultracentrifugation
– length measurements by electron microscopy
– Autoradiography-technique in which the position of a
radioactive substance in a sample is recorded by exposure
to film.
• Contour lengths - end to end lengths of stretched out
native molecules of DNA.
• Genome - complement of genetic information
• kb - kilobase pair = 1000 bp
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Figure 5-19 Electron micrograph of a T2
bacteriophage and its DNA.
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Figure 5-20 Autoradiograph
of Drosophila melanogaster
DNA.
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Table 5-2
Sizes of Some DNA Molecules.
Size of DNA molecules
• DNA is highly susceptible to mechanical damage outside
of the cell.
• Shearing forces generated by ordinary lab techniques
can result in shearing of the DNA into small pieces.
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Figure 5-21 The central dogma of molecular biology.
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Figure 5-22 Gene expression.
Transcription
• Catalyzed by RNA polymerase.
• Couples NTPs (ATP, CTP, GTP, UTP) to make RNA
• (RNA)n residues + NTP  (RNA)n+1 residues + P2O74• 5’  3’ nucleotides are added to the free 3’-OH group
• Nucleotides must meet Watson-Crick base pairing
requirements with the template strand
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Figure 5-23 Action of RNA polymerases.
Transcription
• Transcribes only one template DNA strand at a time.
• RNA polymerase will move along the duplex DNA it is
transcribing and creates a transcription bubble
• This forms a short DNA-RNA hybrid with newly
synthesized RNA.
• DNA template strand is read 3’  5’
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Figure 5-24 Function of the transcription bubble.
Transcription
• DNA template contains control sites consisting of specific base
sequences that specify where the RNA polymerse initiates
transcription and the rate of transcription.
• activators and repressors control the sites in prokaryotes.
• Transcription factors bind to these sites in eukaryotes.
• messenger RNA (mRNA) - RNAs that encode proteins
• Rates at which cells synthesize a protein are determined
by the rate at which mRNA synthesis is initiated.
• Promoter-in prokaryotes-a sequence that precedes the
transcriptional initiation site.
Transcription
• Prokaryotes can control transcriptional initiation in complex
manners.
• Example E. coli lac operon.
• Has 3 consecutive genes (Z, Y, and A) that are necessary to
metabolize lactose.
• In the absence of lactose, the lac repressor protein binds a
control site in the lac operon called an operator.
• This prevents the RNA polymerase from initiating transcription.
• If lactose is present, some of the lactose is converted to
allolactose which binds to the lac repressor causing it to fall of
the operator sequence.
• This allows RNA polymerase to initiate transcription of the
genes.
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Figure 5-25 Control of transcription of the lac operon.
Eukaryotic RNA undergoes posttranscriptional modification
• In order for mRNAs in eukaryotes to become functional, they
must undergo modifications.
• 7-methylguanosine-containing “cap” is added to the 5’ end.
•  250 nucleotide polyadenylic acid [poly(A)] tail is added to
the 3’ end.
• Undergo gene splicing in which RNA segments called introns
are excised from the RNA and the remaining exons are
rejoined to form the mature mRNA.
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Figure 5-26 Post-transcriptional processing of
eukaryotic mRNAs.
mRNA
• In prokaryotes, transcription and translation both
take place in the cytosol.
• Prokaryotic mRNAs have a short lifetime (avg. 1-3
min). They are degraded by nucleases.
• Rapid turnover in prokaryotes allows the prokaryote
to respond quickly to the environment.
• In eukaryotic cells, RNAs are transcribed and posttranslationally modified in the nucleus, then sent to
cytosol.
• Eukaryotic mRNAs have lifetimes of several days.
Translation: Protein synthesis
• Polypeptides are synthesized from mRNA by ribosomes.
• Ribosomes are 2/3 rRNA (ribosomal RNA) and 1/3 protein.
• Prokaryote ribosomes approx. 2500 kD, eukaryotes 4300 kD
• Transfer RNAs (tRNAs) deliver amino acids to the ribosome.
• mRNA sequences can be broken down to codons-consecutive
3-nucleotide segments that specify a particular amino acid.
• Once the mRNA binds to the ribosome, they specifically bind
to the tRNA that is covalently linked to an amino acid.
Figure 5-27 Transfer RNA (tRNA) drawn in its
“cloverleaf” form.
tRNA has 76 nucleotides
Has an anticodoncomplementary sequence
to the mRNA sequence
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Amino acid is linked to the
3’ end of the tRNA to form
aminoacyl-tRNA.
tRNAs are “charged” with
amino acids by specific
enzymes (aminoacyltRNA synthetases or
aaRSs)
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Figure 5-28 Schematic diagram of translation.
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Figure 5-29 The ribosomal reaction forming a peptide
bond.
Genetic code
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Correspondence between the sequence of bases in a codon and the
amino acid residue it specifies.
Nearly universal.
4 possible bases (U[T], C, A, and G) can occupy three positions of
codon, therefore 43 = 64 possible codons.
61 codons specify amino acids, and three UAA, UAG, and UGA are
stop codons (cause ribosome to end polypeptide synthesis and
release the transcript).
All but two amino acids (Met, Trp) are specified by more than one
codon.
Three (Leu, Ser, Arg) are specified by six codons.
Synonyms-multiple codons can code the same amino acid.
tRNA may recognize up to 3 synonymous codons because the 5’
base of a codon and 3’ base of the anticodon can interact in ways
other than via Watson-Crick base pairs.
Translation is initiated at the AUG codon (Met) but this tRNA differs
from the tRNA for internal amino acid the Met codon.
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Figure 5-30 Nucleotide reading frames.
DNA replication
• DNA is replicated similar to RNA with some differences:
• 1. Deoxynucleotide triphosphates (dNTPs) are used
instead of NTPs
• 2. Enzyme is the DNA polymerase
• Other differences:
• RNA polymerase can link together two nucleotides on
DNA template, but DNA polymerase can only extend (in
the 5’ to 3’) direction an existing polynucleotide that is
base paired to the template strand.
• DNA polymerase needs an oligonucleotide primer to
initiate synthesis.
• Primers are RNA.
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Figure 5-31 Action of DNA polymerases.
DNA strands replicated in different
ways
• DNA strands are simultaneously replicated.
• Takes place at replication fork - junction where the two
parental DNA are pried apart and where the two
daughter strands are synthesized.
• Leading strand is continuously copied from the 3’ to 5’
parental template in the 5’ to 3’ direction
• Lagging strand is discontinuously replicated in pieces
from the 5’ to 3’ parental strands.
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Figure 5-32a Replication of duplex DNA in E. coli.
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Figure 5-32bReplication of duplex DNA in E. coli.
DNA strands replicated in different
ways
• DNA strands are simultaneously replicated.
• Takes place at replication fork - junction where the two
parental DNA are pried apart and where the two
daughter strands are synthesized.
• Leading strand is continuously copied from the 3’ to 5’
parental template in the 5’ to 3’ direction
• Lagging strand is discontinuously replicated in pieces
from the 5’ to 3’ parental strands.
• E. coli has 2 DNA polymerases necessary for survival.
DNA polymerase III (Pol III) synthesizes the leading
strand and most of the lagging strand.
• DNA polymerase I (Pol I) removes RNA primers and
replaces them with DNA. This enzymes also has a 5’ to
3’ exonuclease activity.
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Figure 5-33 The 5¢ ® 3¢ exonuclease function of DNA
polymerase I.
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Figure 5-34 Replacement of RNA primers by DNA in
lagging strand synthesis.
Lagging strand synthesis
• Synthesis of the leading strand of DNA is completed by
the replacement of the RNA primer by DNA.
• Lagging strand is completed after nicks between multiple
disconinuously synthesized segments are sealed by
DNA ligase.
• Catalyzes the links of 3’-OH to 5’-phosphate groups.
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Figure 5-35 Function of DNA ligase.
Errors in DNA sequence can be
corrected
• RNA polymerase has an error rate of 1 in 104 base pairs
in E. coli.
• Pol I and Pol III have 3’  5’ exonuclease activities.
• This activity degrades the newly synthesized 3’ end of a
daughter strand one nucleotide at a time to edit out
mistakes that are sometimes incorporated.
• Other enzymes are present that detect and correct errors
in DNA damage that occurs from UV radiation and
mutagens (chemical substances that damage DNA) and
hydrolysis.
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Figure 5-36 The 3¢ ® 5¢ exonuclease function of DNA
polymerase I and DNA polymerase III.
Molecular cloning
• Clone-a collection of identical organisms that are derived
from a single ancestor.
• Molecular cloning techniques - genetic engineering,
recombinant DNA technology.
• Main idea is to insert a DNA segment of interest into an
automously replicating cloning vector so that the DNA
segment is replicated with the vector.
• Cloning into a chimeric vector in a suitable host
organism results in large amounts of the inserted DNA
segment.