550 likes | 837 Views
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: H-bonds between complementary bases on opposite strands: 2 H-bonds in A-T pair 3 H-bonds in G-C pair
E N D
FCH 532 Lecture 4 Chapter 5: DNA
Forces in DNA Double Helix • DNA Double helix is stabilized by two types of forces: • H-bonds between complementary bases on opposite strands: • 2 H-bonds in A-T pair • 3 H-bonds in G-C pair • Van der Waals forces and hydrophobic interactions between “stacked bases” • Aromatic bases have p-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.
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. Page 90
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. Page 90
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
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. Page 91
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.
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. Page 91
Figure 5-20 Autoradiograph of Drosophila melanogaster DNA. Page 92
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 5-22 Gene expression. Page 93
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
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’
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. Page 95
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. Page 95
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. 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) Page 95
Figure 5-29 The ribosomal reaction forming a peptide bond. Page 96
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.
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.
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.
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. Page 100
Figure 5-34 Replacement of RNA primers by DNA in lagging strand synthesis. Page 101
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. Page 101
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.