Ch16 - The Molecular Basis of Inheritance Inheritance...

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Unformatted text preview: The Molecular Basis of Inheritance Inheritance Chapter 16 What is DNA??? What Frederick Griffith Experiment Experiment --causes Studying Streptococcus pneumoniae Studying Streptococcus pneumoniae --causes pneumonia in mammals pneumonia There are two strains: There A pathogenic (disease-causing) strain: S strain (for a smooth cell wall) (for A nonpathogenic (harmless) strain: R strain (for a nonpathogenic rough cell wall) rough On the following slide, study his experiment and On come up with your own conclusions come Living S (control) cells Mouse dies Living R (control) cells Mouse healthy Mixture of heatHeat-killed killed S cells (control) S cells and living R cells Mouse healthy Mouse dies Living S cells are found in blood sample. Important Conclusions? Conclusions? ________________________________ ________________________________ ________________________________ ________________________________ Transformation: Hershey and Chase Experiment Experiment Worked with viruses Worked that infect bacterial cells--Bacteriophages cells--Bacteriophages Set-up Set-up Batch 1: Phages were grown with radioactive sulfur (35S), which was incorporated into phage protein (pink). Batch 2: Phages were grown with radioactive phosphorus (32P), which was incorporated into phage DNA (blue). Make Some Predictions Predictions What will you see if What the protein coat is causing infection in the bacteria? the What will you see if What the DNA inside is causing the infection in bacteria? in How Infection Occurs Phage head Tail Tail fiber 100 nm DNA Figure 16.3 Bacterial cell Important Conclusions? ________________________________ ________________________________ ________________________________ ________________________________ So what does DNA look like? look What we knew prior to the 1950’s: DNA is made of deoxyribose, phosphate, and DNA 4 nitrogenous bases--see model kits nitrogenous Wilkins and Franklin determined that DNA Wilkins was made of a backbone of deoxyribose and phosphate phosphate Wilkins and Franklin also determined that the Wilkins nitrogenous bases were in the interior of the molecule molecule Let’s build what we know so far… Nitrogenous bases Sugar-phosphate backbone 5′ end O– O P 5′ CH2 O 4′ O– CH3 H H O H O 1′ 3′ H 2′ N N H H H O Thymine (T) O O CH2 O P O– H H N O H H H N N H N N H H H Adenine (A) O CH2 O P O – H H N H P N O Cytosine (C) 5′ O 4′ O– Phosphate CH2 H H 3′ OH Figure 16.5 H H H O N O H O H H O Sugar (deoxyribose) 3′ end H O N O 1′ N H 2′ H H N N N H H H Guanine (G) DNA nucleotide What we knew, cont’d What Chargaff made two key Chargaff observations observations DNA composition varied between DNA species species Human DNA is approx. 30.3% A; Human e.coli DNA is approx. 26.0% A e.coli What conclusion can you draw from What this? this? _______________________________ __ Chargaff, cont’d Chargaff, There is an equal percentage of A’s as There T’s T’s There is an equal percentage of G’s as There C’s C’s Can you decipher more of the Can structure now? structure Let’s take into account, also Let’s Franklin determined that it was doublestranded, and that the two strands were stranded, anti-parallel anti-parallel Can you figure out the structure? Figure 16.6 a, b (a) Rosalind Franklin (b) Franklin’s X-ray diffraction Photograph of DNA 5′ end O OH Hydrogen bond P – O 3′ end OH O O A T O P – O O O CH2 O O H2C O G O C O O CH2 P – O O O O O C O G O O O CH2 P O O– P H2 C – P O O O– O O O– P O O A O T O CH2 OH 3′ end O O– P O Figure 16.7b (b) Partial chemical structure O 5′ end Watson & Crick Watson Put it all together… Knowing that A and T make 2 bonds; G and C Knowing make 3 bonds make H N N CH3 O H G C A T T N N Sugar H N 1 nm N O Sugar Thymine (T) H O N H N C A C Sugar N T G T A T A A T T N H N Figure 16.7a, c O Sugar N N Figure 16.8 H H Guanine (G) 3.4 nm G A N C G N Adenine (A) A Cytosine (C) G A C 0.34 nm T (a) Key features of DNA structure (c) Space-filling model Lots of Controversy Lots Figure 16.1 Figure 16.6 a, b (a) Rosalind Franklin (b) Franklin’s X-ray diffraction Photograph of DNA Practice Exam Question Question How is DNA replicated??? replicated??? Hypotheses? _______________________________ _______________________________ _______________________________ _______________________________ Using the following experimental technique, can you make some predictions for your hypotheses? Bacteria cultured in medium containing 15 N Bacteria transferred to medium containing 14 N LL LH HH First Second replication replication Parent cell Conservative model Semiconservative model Dispersive model Predictions? P 1st 2nd The Actual Results The DNA sample centrifuged after 20 min (after first replication) LL LH LH DNA sample centrifuged after 40 min (after second replication) Conclusions? Conclusions? __________________________________________ __________________________________________ __________________________________________ _____________________ Semi-conservative Replication Replication T A T A T A C G C G C T A T A T A A T A T A T G C G C G A T A T A T C G C G C G T A T A T A T A T A T C G C G C C G A G (a) The parent molecule has two complementary strands of DNA. Each base is paired by hydrogen bonding with its specific partner, A with T and G with C. Figure 16.9 a–d (b) The first step in replication is separation of the two DNA strands. (c) Each parental strand now serves as a template that determines the order of nucleotides along a new, complementary strand. (d) The nucleotides are connected to form the sugar-phosphate backbones of the new strands. Each “daughter” DNA molecule consists of one parental strand and one new strand. The Origin of Replication Replication Origin of replication Parental (template) strand Daughter (new) strand Bubble 0.25 µm Replication fork Two daughter DNA molecules In this micrograph, three replication bubbles are visible along the DNA of a cultured Chinese hamster cell (TEM). DNA Elongation DNA Is catalyzed by enzymes called DNA Is Polymerases, which add nucleotides to the 3’ end of a growing strand end New strand 5′ end Sugar Phosphate A T Base CG G P Template strand 5′ end 3′ end P P T OH Nucleoside triphosphate 3′ end A T C G C G C A T A OH 3′ Pyrophosphate end PP C 5′ end 2P C 5′ end How might the antiparallel structure of the double helix cause an issue? double How could we solve the problem? Leading Strand: DNA Polymerase III continuously elongates the DNA, moving toward the replication fork toward Lagging Strand: must work in the Lagging opposite direction, creates Okazaki fragments These must be joined together by DNA Ligase Ligase Another Problem Another DNA Polymerase III can only add to an DNA existing 3’ end; it cannot initiate transcription transcription How do we solve this problem??? Primase Primase Primase adds an initial DNA or RNA Primase primer to which DNA Polymerase will add nucleotides add Multiple primers are needed on the lagging Multiple strand--why?? strand--why?? DNA Polymerase I replaces the primers DNA with DNA with Ligase binds the strands together There is another problem… problem… DNA Polymerase I also can only add to DNA an existing 3’ end an This is fine for all the primers on the inside This (from the Okazaki fragments) (from But what about the primer on the very But end??? end??? End of parental DNA strands 5′ Leading strand Lagging strand 3′ Last fragment Previous fragment 5′ Lagging strand 3 Primer removed but′ cannot be replaced with DNA because no 3′ end available for DNA polymerase 3′ RNA primer Removal of primers and replacement with DNA where a 3′ end is available 5′ Second round of replication 5′ New leading strand 3′ New lagging strand 5′ 3′ Further rounds of replication Shorter and shorter daughter molecules How can we fix this??? this??? Telomeres Telomeres 1 µm What about germ cells? cells? What would happen if the chromosomes in the What germ cells (those that perform meiosis) kept getting shorter? getting _________________________________________ _________________________________________ _________________________________________ ____________ Telomerase: Other Enzymes Involved Involved Helicase unwinds the double helix and unzips Helicase the strands the Single-strand Binding Proteins: bind to singlestranded DNA to stabilize it Topoisomerase: corrects over-winding ahead Topoisomerase: of the replication fork of 1 Primase joins RNA nucleotides 3′ Template into a primer. strand 3′ 5′ 5′ RNA primer After reaching the next 3 RNA primer (not shown), DNA pol III falls off. 5′ 3′ After the second fragment is 4 primed. DNA pol III adds DNA nucleotides until it reaches the 5′ 3′ first primer and falls off. 5 DNA pol 1 replaces the RNA with DNA, adding to5′ 3 the 3′ end of fragment 2. ′ DNA ligase forms a bond 6 between the newest DNA and the adjacent DNA of 3′5′ fragment 1. 3′ 5′ 5′ 3′ 1 Okazaki fragment 3′ 5′ 1 2 1 2 1 3′ 5′ 3′ 5′ 7 The lagging strand 2 1 3′ 5′ Overall direction of replication in this region is now complete. Overall direction of replication Leading strand Origin of replication Lagging strand 1 Helicase unwinds the parental double helix. 2 Molecules of singlestrand binding protein stabilize the unwound template strands. Lagging strand Leading strand 3 The leading strand is synthesized continuously in the 5′ → 3′ direction by DNA pol III. OVERVIEW DNA pol III Leading strand 5′ Replication fork 3′ Parental DNA Primase DNA pol III Primer 4 Primase begins synthesis of RNA primer for fifth Okazaki fragment. 5 DNA pol III is completing synthesis of the fourth fragment, when it reaches the RNA primer on the third fragment, it will dissociate, move to the replication fork, and add DNA nucleotides to the 3′ end of the fifth fragment primer. Figure 16.16 4 6 DNA ligase DNA pol I Lagging strand 2 1 3 DNA pol I removes the primer from the 5′ end of the second fragment, replacing it with DNA nucleotides that it adds one by one to the 3′ end of the third fragment. The replacement of the last RNA nucleotide with DNA leaves the sugarphosphate backbone with a free 3′ end. 3′ 5′ 7 DNA ligase bonds the 3′ end of the second fragment to the 5′ end of the first fragment. Proofreading Proofreading Errors in completed DNA are about one in Errors 10 billion nucleotides; however, initial pairing errors are 100,000 times more common (1 in 100,000 bases) common DNA polymerases proofread newly made DNA DNA as each base is added DNA If they evade detection, mismatch repair is If done after the fact by enzymes that recognize mismatched DNA recognize Nucleotide excision repair repair 1 A thymine dimer distorts the DNA molecule. 2 A nuclease enzyme cuts the damaged DNA strand at two points and the damaged section is removed. Nuclease 3 Repair synthesis by a DNA polymerase fills in the missing nucleotides. DNA polymerase DNA ligase 4 DNA ligase seals the Free end of the new DNA To the old DNA, making the strand complete. Practice Exam Question Question Viruses enter your cells and take over the Viruses cell’s DNA replicating machinery to mass produce themselves. Which of the following would not need to be borrowed from the host cell in order to replicate itself (i.e., what does the virus bring with it)? what A. B. C. D. The DNA template DNA Polymerase III Primase Single-stranded binding proteins ...
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This document was uploaded on 11/01/2011 for the course BIO 181 at BYU.

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