THE MOLECULAR BASIS OF INHERITANCE I. History A. Discovery of “transformation” – a change in genotype and phenotype due to the uptake of external DNA by a cell 1. Griffith 1920s did experiments with Streptococcus pneumoniae (p294 fig16. 2) a. took two strains of S. pneumoniae, one virulent, one not b. heat killed virulent strain, then mixed them with the living nonvirulent strain c. living nonvirulent strain became virulent d. nonvirulent strain took on virulent strain’s DNA ? became virulent e. see p294 fig16. 2 S strain = virulent, R = nonvirulent f. ventually Griffith’s work lead way to more studies on DNA being the carrier of genetic info. B. Proof that viral DNA and not viral protein contains genetic information to make more viral particles 1. Hershey and Chase 1950s p295 fig16. 4 a. knew that viruses could infect bacteria and make more viruses using the host cell’s replicating ‘machinery’ b. background: sulfur gets incorporated into virus’ protein/phosphorus into virus’ DNA c. took T4 (bacteriophage) and plated with a lawn of E. coli and radioactively labeled sulfur, result = T4 with radioactive labeled protein (DNA not labeled) d. took T4 and plated with E. oli and radioactively labeled phosphorus = T4 with radioactively labeled DNA (protein not labeled) e. background: when virus + bacteria is spun down, viral particles in supernatant and bacteria in pellet f. took T4 (S-labeled) infected new E. coli lawn, spun down, found S-radioactive labels in supernatant g. took T4 (P-labeled) infected new E. coli lawn, spun down, found P-radioactive labels in pellet h. result = it’s the DNA that’s injected into the host to make more virus (even plated these spun down pellet bacteria, and they lysed and released new virus C. Discovery of the structure of DNA 1.
Watson and Crick a. used an x-ray crystallography picture (p297) by Franklin to determine DNA as a double-helical structure b. review p298 – A pairs with T and G with C/ A and G are purines and C and T are pyrimidines/double hydrogen bonds between A and T, and triple between G and C II. DNA Replication A. 3 models of DNA replication p300 fig 16. 10 1. Conservative model – the parental helix splits, copies, then goes back together again to remain intact while a second entirely new copy is made 2. Semiconservative model – the parental helix splits, copies and remains a part of the two new helixes 3.
Dispersive model – the parental helix splits unevenly, copies and remains a part of the two new helixes but in pieces B. Experimental proof p300 fig16. 11 1. added radioactively labeled heavy nitrogen to replicating bacteria, then placed this culture into radioactively labeled light nitrogen (used to distinguish strands) 2. allowed bacteria to replicate again, results gave hybrid DNA strands (ruled out conservative model) (note: both hybrids half and half and totally mixed look the same, so semiconservative and dispersive models both upheld this time- see below) 3. llowed bacteria to replicate again, results gave hybrid strands and only light double strands (ruled out dispersive model since all should be mixed if this was right) C. Origins of replication p301 fig16. 12 1. origin of replication – site where DNA replication begins a. proteins recognize a specific sequence on the template DNA, open the dsDNA to make a bubble, and begin replication b. replication fork – location on DNA strand where new DNA strand is growing 1. prokaryotes plasmid (single circular dsDNA helix) have one origin of replication and replication occurs in both directions 2. ukaryotes have linear dsDNA have many origins and replication occurs in both directions D. Elongation of new DNA 1. DNA polymerase – enzyme that synthesizes the new DNA strand by adding nucleotides to the growing strand 2. DNA polymerase receives energy to do this by nucleotides being nucleoside triphosphate (CTP, GTP, ATP, TTP) since they lose Pii = exergonic reaction to supply energy E. DNA is antiparallel p302 1. carbon numbering – carbon attached to base is 1’, count clockwise, carbon attached to phosphate group is 3’, carbon attached to other phosphate group is 5’ 2. be able to find 5’ vs 3’ end . (p302 fig 16. 14) replication occurs 5’ ? 3’, so strand being made in this direction is called the leading strand and replication occurs toward the replication fork 4. lagging strand is replication that occurs 5’ ? 3’ but replication moves away from the replication fork a. lagging strand produces Okazaki fragments which must be connected with DNA ligase p303 fig 16. 15 F. Priming DNA synthesis (getting replication started) p303 fig16. 15 1. primer – existing RNA polynucleotide on the template DNA strand since DNA polymerase cannot just start adding new nucleotides on its own a. rimer is laid down by enzyme primase b. only one primer required for leading strand to begin synthesizing/new RNA primer required for each lagging strand beginning c. DNA polymerase eventually replaces RNA nucleotides with DNA ones and occurs before ligase connects any lagging DNA strands G. Other assisting proteins 1. helicase – enzyme that unwinds dsDNA at the replication form 2. single-strand binding proteins – hold apart template DNA while replication occurs **FINAL GOOD SUMMARY P304 fig 16. 16 III. DNA Proofreading and Repair
A. Mismatch repair 1. as DNA polymerase lays down nucleotides, if it notices a mismatched one to template, will remove and replace with correct one 2. Excision repair p305 fig16. 17: consists of nuclease – enzyme that can cut out damaged segments of a DNA strand, then new nucleotides are filled in based on what the other DNA strand sequence is by DNA polymerase and ligase IV. Replication of the ends of DNA strands p306 fig16. 18 A. DNA polymerase can only add nucleotides to a 3’ end (since it grows in a 5’ ? 3’ direction) B.
For lagging strand, there is no problem since it replaces RNA primer and joins DNA with ligase C. For leading strand, there is a problem, since the 3’ end of the template strand has a RNA primer, which cannot be replaced with DNA nucleotides (by DNA polymerase) since there is no 3’ end to start from (DNA polymerase cannot just add nucleotides opposite of the DNA template strand – must use a RNA primer) D. This results in successive replicated strands becoming shorter and shorter – the remedy? E. Telomeres – eukaryotic cells have short repetitive nucleotide sequences that do not code for anything 1. elomeres protect the cell from false alarms that there is DNA damage and cause the cell to die since losing these ends don’t mean anything (note that prokaryotes do not have this problem since their DNA is circular with no “end”) 2. but when telomeres are lost, are they replaced? Yes by telomerase – enzyme that works in conjunction with DNA polymerase to add length to telomeres a. p306 fig16. 19 have shortened “just made” DNA strand b. telomerase is associated with an RNA strand and DNA polymerase c. telomerase lines up the RNA strand with the 3’ DNA strand to serve as a template to have the 3’ end grow d. hen the RNA strand serves as a primer for new growth onto the 5’ strand, then the primer is removed e. result is an elongated DNA strand that was shorted during replication *telomerase is not present in most cells of multicellular organisms (like us) *DNA of older individuals tends to be shorter *telomerase is abundant in germ line cells – those that give rise to gametes *researchers find telomerase in cancer cells – makes sense since these cells replicate often and would have very short DNA (possible cancer therapy is to target their telomerase)