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:

• 2 H-bonds in A-T pair

• 3 H-bonds in G-C pair

2. Van der Waals forces and hydrophobic interactions between “stacked bases”

• Aromatic bases have -electrons that interact via attractive Van der Waals forces.

Structure of DNA determines heredity

• 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.

Figure 5-14 Schematic representation of the strand separation in duplex DNA resulting from its heat

denaturation.

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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.

Figure 5-15 UV absorbance spectra of native and heat-denatured E. coli DNA.

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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

Figure 5-16 Example of a DNA melting curve.

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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

Figure 5-17 Variation of the melting temperatures, Tm, of various DNAs with their G + C content.

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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.

Figure 5-18 Partially renatured DNA.

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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

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.P

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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.

Figure 5-21 The central dogma of molecular biology.

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Figure 5-22 Gene expression.

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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

Figure 5-23 Action of RNA polymerases.

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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’

Figure 5-24 Function of the transcription bubble.

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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.

Figure 5-25 Control of transcription of the lac operon.

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Eukaryotic RNA undergoes post-transcriptional 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.

Figure 5-26 Post-transcriptional processing of eukaryotic mRNAs.

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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 post-translationally 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.

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tRNA has 76 nucleotides

Has an anticodon-complementary sequence to the mRNA sequence

Amino acid is linked to the 3’ end of the tRNA to form aminoacyl-tRNA.

tRNAs are “charged” with amino acids by specific enzymes (aminoacyl-tRNA synthetases or aaRSs)

Figure 5-28 Schematic diagram of translation.

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Figure 5-29 The ribosomal reaction forming a peptide bond.

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Genetic code• 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.

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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.

Figure 5-31 Action of DNA polymerases.

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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.

Figure 5-32aReplication of duplex DNA in E. coli.

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Figure 5-32bReplication of duplex DNA in E. coli.

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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.

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.

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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.

Figure 5-35 Function of DNA ligase.

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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.

Figure 5-36 The 35 exonuclease function of DNA polymerase I and DNA polymerase III.

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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.