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CHAPTER 16 THE MOLECULE BASIS OF INHERITANCE 1. The search for genetic material lead to DNA History • 1940’s- T.H. Morgan’s group showed that genes are located on chromosomes • But…. Are _________or DNA the genetic material?? • Most believed it to be protein • However, this was not consistent with experiments with microorganisms, like bacteria and viruses. • 1928- Frederick _________- He studied Streptococcus pneumoniae, a bacterium that causes pneumonia in mammals. • One strain, the R strain, was harmless. • The other strain, the S strain, was ______________. • Experiment- Griffith mixed ______________ S strain with live R strain bacteria and injected this into a mouse. • Result- Mouse died, but the S strain was recovered from the mouse’s blood. S R Killed S R + Killed S • Griffith called this phenomenon,______ ______________, a change in genotype and phenotype due to the assimilation of a foreign substance (now known to be DNA) by a cell. Injection Dies Lives Lives Fig. 16.2 Dies Live S found • What was the transforming substance in Griffith’s experiments?? • Finally in 1944, Oswald_________, Maclyn McCarty and Colin MacLeod announced that the transforming substance was _________. • Still, many biologists were skeptical. • In part, this reflected a belief that the genes of bacteria could not be similar in composition and function to those of more complex organisms. • In 1952, Alfred Hershey and Martha Chase showed that DNA was the genetic material of the phage T2 Fig. 16.3 • Phage T2 is a __________that infects bacteria What is a virus?? • Viruses consist of a DNA (sometimes______) enclosed by a protective coat of protein. • To replicate, a virus infects a host cell and takes over the cell’s metabolic_______________. • The T2 phage, consisting almost entirely of DNA and protein, attacks Escherichia coli (E. coli), a common _____________bacteria of mammals. • This phage can quickly turn an E. coli cell into a ________________factory that releases phages when the cell ruptures. Fig. 16.3 • The Hershey and Chase Experiment-Designed to determine the source of genetic material in the phage • Procedure: Two parallel experiments #1 Grow T2 phage in the presence of _____________sulfur, marking the ___________ but not DNA •Recall that sulfur is a component of some_____________. ATP #2 Grow T2 phage in the presence of _____________ phosphorus , marking the _______ but not proteins •Recall that __________is a component of DNA •Allow each batch to infect separate E. coli cultures. •Shortly after infection, use blender to “shake loose” any remaining phage components Fig. 5.26 Hershey-Chase experiment (cont.) • Now centrifuge – bacteria pellet and phage do not. • Tested the pellet and ________________ of the separate experiments for the presence of radioactivity. Fig. 16.4 Hershey-Chase experiment (cont.) • Results of Hershey-Chase experiment• If used radioactive sulfur (proteins labeled)- most of the radioactivity was in the______________, not in the pellet. • If used radioactive phosphorous(DNA labeled)most of the ______________ was in the pellet with the bacteria. •Conclusion- injected _______ of the phage provides the _________ information for new viruses. • _________________evidence that DNA is the genetic material in eukaryotes 1. Cells double the amount of DNA in a cell prior to _____________ and then distribute the DNA equally to each daughter 2. Diploid sets of chromosomes have twice as much DNA as the __________ sets in gametes of the same organism. • 1947- Erwin __________ had developed a series of rules based on a survey of DNA _______________ in organisms. • Known that DNA was a polymer of nucleotides consisting of a _____________ base, deoxyribose, and a phosphate group. • Known that there were four _________ -adenine (A), thymine (T), guanine (G), or cytosine (C). • Chargaff noted that the four bases are found in characteristic, but not necessarily equal, ratios • He also found a peculiar regularity in the ratios of nucleotide bases which are known as________________. Chargaff’s rules• 1. The number of adenines was approximately equal to the number of ____________ (%T = %A). • 2. The number of guanines was approximately equal to the number of ___________ (%G = %C). • _________ DNA is 30.9% adenine, 29.4% thymine, 19.9% guanine and 19.8% cytosine. 2. Watson and Crick discovered the double helix by building models to conform to Xray data • By the beginnings of the 1950’s, the race was on to move from the structure of a single DNA strand to the three-dimensional structure of DNA. • Among the scientists working on the problem were Linus____________, in California, and Maurice Wilkins and Rosalind __________ , in London. The basic structure of DNA was known by 1952 • The ___________ group of one nucleotide is attached to the _______ of the next nucleotide in line. • The result is a “_____________ ” of alternating phosphates and sugars, from which the bases project. Fig. 16.3 T Bases A C G Fig. 16.5 How was the structure of DNA resolved?? • Answer: By ________ crystallography – (done by Maurice Wilkins and Rosalind Franklin)………. • In this technique, X-rays are diffracted as they passed through aligned fibers of purified DNA. • The diffraction pattern can be used to deduce the three-dimensional shape of molecules. • ….AND by _________ • James Watson learned from their research that DNA was _________ in shape and he deduced the width of the helix and the spacing of bases. Franklin Fig. 16.6 • Watson and his colleague Francis Crick began to work on a model of DNA with two strands, the________________ . • Using molecular models made of wire, they first tried to place the sugar-phosphate chains on the________ . • However, this did not ______ the X-ray measurements and other information on the chemistry of DNA. • The key _______________ came when Watson put the sugar-phosphate chain on the outside and the nitrogen bases on the inside of the double helix. The double helix- a _________ homology Pairs of nitrogen bases, one from each strand, form rungs. The ladder forms a twist every _____ bases Fig. 16.7 The _______________ chains of each strand are like the side ropes of a rope ladder. • Pairing like nucleotides did not fit the uniform diameter indicated by the X-ray data. •A __________________pair would be too wide and a pyrimidine-pyrimidine pairing would be too short. •Only a ____________________ pairing would produce the _____ diameter indicated by the X-ray data. • In addition, Watson and Crick determined that chemical side groups off the nitrogen bases would form hydrogen bonds, connecting the two strands. • Adenine could form two _________bonds only with thymine • Guanine could form ______ hydrogen bonds only with cytosine. • This finding __________ Chargaff’s rules. Fig. 16.8 • The linear sequence of the four bases can be varied in countless ways. • Each gene has a unique order of nitrogen bases. • In April 1953, Watson and Crick published a succinct, one-page paper in Nature reporting their double helix model of DNA. The Nobel Prize in Chemistry was awarded came in later years 3. Base pairing is at the heart of DNA replication Base pairing enables existing DNA strands to serve as templates for new complimentary strands • In a second paper, Watson and Crick published their ______________ for how DNA replicates. When a cell copies a DNA molecule, each strand serves as a __________ for ordering nucleotides into a new ____________________ strand. Fig. 16.9 Doublestranded DNA 1.Separate strands 2.Add complemen tary bases 3.Connect bases But how was the DNA copied?? Three models proposed • #1-Watson and Crick’s model -___________________ replication (i.e. each of the daughter strands will have one old strand and one newly made strand. • #2- The _______________model- (an all new copy of DNA is made) • #3- The _______________model- daughter stands contain both old and new DNA Parent DNA 1st replication 2nd replication Fig. 16.10 Conservative SemiConservative Dispersive Which model is correct?? • Late 1950s - Matthew Meselson and Franklin_____ provided evidence for the _______________model. Experiment 1. Label bacterial DNA with a heavy ________ of Nitrogen (15N), then __________ in 14N medium 2. _______replicated strands by density in a centrifuge. Prediction: 20 min 40 min Fig. 16.11 Which model is correct?? Results After 1st replication- only _______(15N-14N) DNA detected Thus, conservative model is____________ After 2nd replication- both light and hybrid (15N-14N) DNA detected Thus, the ________________ model is incorrect Why? Because no light DNA should be present here Thus, semiconservative model is correct. 4. DNA replication requires a large number of enzymes and other proteins A. DNA replication is fast and efficient Time to copy the entire genome and divide into daughter cells: • Bacterium E. coli (5 million base pairs- )- 40 minutes • Human cells- (6 billion base pairs)- 6 hours ___________ • Less than one error per billion nucleotides. Machinery • More than a dozen enzymes and other proteins required B. Where does replication start? – Answer- The___________________________ Bacteria- one site in DNA recognized by specific proteins Origin of replication (cont.) • Eukaryotes- ________________ of origin sites per chromosome. 15.3 15.2 15.1 14 13.3 13.1 12 11 11.1 11.2 12 13.1 13.2 13.3 14 • At each origin site, the DNA strands separate forming a replication “_________ ” with ___________________ at each end. • The replication bubbles _____________as the DNA is replicated and eventually _________ . Fig. 16.12 15 21 22 23.1 23.3 31.1 31.2 31.3 32 33.1 33.3 34 35.1 • DNA polymerases add the _________ (nitrogen base, deoxyribose, and a triphosphate tail). • Rate of elongation • 500 nucleotides/second in bacteria • 50 per second in human cells. The raw nucleotides are nucleoside triphosphates. Two phosphates are removed upon addition of nucleotide Fig ure not in text One _________ connects nucleotides • The strands in the double helix are______________. • The sugar phosphate____________run in opposite directions. 5’ end “ _________ “ terms derived from the carbon number in the ribose sugar ring 3’ end • Each DNA strand has a 3’ end with a free _________ group attached to deoxyribose and a _______ with a free __________ group attached to deoxyribose. • The 5’ -> 3’ direction of one strand runs counter to the 3’ -> 5’ direction of the other strand. Fig. 16.13 • At the replication fork, one parental strand (3’-> 5’ into the fork), the_________________, can be used by polymerases as a template for a continuous complimentary strand. 3’ Problem- DNA polymerases can only add nucleotides to the free _______ of a growing DNA strand. •But how is the other stand copied?? SolutionThe other parental strand (5’->3’ into the fork), the__________________ , is copied away from the fork in short segments (___________ fragments) Fig. 16.14 How are Okazaki fragments generated?? 1. The enzyme _________ adds 10 nucleotides of RNA to DNA template • This RNA sequence is called a__________ 2. DNA polymerase then starts at the ______ end of the primer 3. The primer is then replaced by DNA nucleotides and joined to the previous Okazaki fragment by_____________________ . Note- Okazaki fragments are only __________ nucleotides long, thus ___________ are used during replication Fig. 16.15 Summary of DNA replication • At the replication fork • the leading stand is copied ____________________ from a single primer • the lagging strand is copied in ______________ using many primers. Fig. 16.16 5. Enzymes proofread DNA during its replication and repair damage in existing DNA • Nucleotide pairing errors occur at a rate of one error per __________ base pairs. • Yet the final error rate is only one per billion nucleotides. How do we account for this discrepency?? Answer- _______________________ a. ____________________ proofreads each new nucleotide against the template nucleotide as soon as it is added. If incorrect, it corrects it. b. Each cell continually monitors and repairs its genetic material, with over ______ repair enzymes identified in humans. • In_______________ , special enzymes fix incorrectly paired nucleotides. • A hereditary defect in one of these enzymes is associated with a form of colon cancer. • In ___________________ repair, a nuclease cuts out a segment of a damaged strand. • The gap is filled in by _______________ and ligase. Fig. 16.17 6. The ends of DNA molecules are replicated by a special mechanism Problem: The usual replication machinery provides no way to ____________ the 5’ ends of daughter DNA strands. • Repeated rounds of replication produce ___________________ DNA molecules. Solution: An enzyme called________________ Fig. 16.18 Fluorescent “dots” Fig. 16.19a are telomeres Background • The ends of eukaryotic chromosomal DNA molecules, the_____________ , have special nucleotide sequences. • In human telomeres, this sequence is typically TTAGGG, repeated between _______________________ times. • Telomeres protect genes from being eroded through multiple rounds of DNA replication. How? •Telomerase uses a short molecule of _______ as a template to extend the 3’ end of the telomere. Fig. 16.18 A few interesting facts about telomerase • Telomerase is not present in most cells of ____________ organisms. • Thus, the DNA of dividing somatic cells and cultured cells does tend to become ____________ and may limit the _______________ of cells. • Telomerase is present in • ______________ cells, ensuring that zygotes have long telomeres. • ______________ somatic cells. • This overcomes the progressive shortening that would eventually lead to self-destruction of the cancer. Big question- Is telomere shortening responsible for the life span of a human??