Study Guide for Exam 3Objectives for Lecture 19: Structure of Nucleic Acids• Be able to describe and interpret the experiment on the replication of the bacterial virus in which it was demonstrated that DNA is the genetic material. Know the roles of radioactive sulfur and phosphorus in this experiment.- Transformation: a change in genotype and phenotype due to the assimilation of external DNA by a cell- Griffith & Avery: A harmless strain of bacterium was turned into disease causing when its DNA was mixed with the disease causing DNA, concluding that the transforming agent was DNA.- Bacteriophages, phages: (bacteria eaters) virus’ that infect bacteria - Hershey & Chase- showed that DNA is the genetic material by devising experiment showing that only one of the two components of T2 actually enters the E. coli cell during infection In this experiment, they used a radioactive isotope of sulfur to tag protein in one batch of T2 and a radioactive isotope of phosphorus to tag DNA in a second batch These were tested to see which one would enter the E. coli and therefore be capable of reprogramming them• Know the components of a nucleotide and how nucleotides are joined together to make nucleic acids (phosphodiester bonds are formed between the 5’ carbon on sugar of one nucleotide with 3’ carbon on the sugar of the next nucleotide) , know how the 5’ and 3’ ends of a nucleic acid differ.-Nucleotides are composed of: a nitrogenous base [Adenine, Guanine, Thymine, or Cysteine], a pentose sugar called deoxyribose, and a phosphate group-double helix- two strands of DNA-Nitrogenous base is connected to the 3’ carbon of sugar- nucleoside-phosphate group is added to 5’ carbon of sugar- nucleotide-Any linear chain of nucleotides has a free 5’ phosphate on one end, and a free 3’ hydroxide on the other. A chain of DNA thus has polarity.-different DNA molecules differ only in their sequence of nitrogenous bases• DNA- sugar is deoxyribose-nitrogenous bases are Adenine, Guanine, Cysteine, and Thymine-almost always exists as a double helix-strands are held together by hydrogen bonding between nitrogenous bases-A with T and G with C-only way pairings will work is if strands have opposite polarity 5’ to 3’ “Antiparallel: oriented in opposite directions to each other”• RNA- sugar is ribose-nitrogenous bases are Adenine, Guanine, Cysteine, and Uracil• Be able to describe the Watson and Crick model for the structure of DNA (double stranded, antiparallel helix, held together by complementary base pairing), and know which bases pair with each other.-Adenine forms two hydrogen bonds with Thymine-Guanine forms three hydrogen bonds with CytosineObjectives for lecture 20: DNA Replication• Conservative model- The two parental strands reassociate after acting as templates for new strands, thus restoring the parental double helix. • Semiconservative model- The two strands of the parental molecule separate, and each functions as a template for synthesis of a new, complimentary strand. Each of the two daughter molecules will have one old strand from the parent and one newly made strand.• Dispersive model- Each strand of both daughter molecules contains a mixture of old and newly synthesized DNA.• Meselson Stahl Experiment- E. coli was cultured for several generations in 15NH4. Then the bacteria were transferred to a medium with only 14NH4, a lighter isotope. -DNA could be distinguished by different densities by centrifuging DNA extracted from the bacteria-the first replication (after 20 mins) produced a band of hybrid DNA, eliminating the conservative model-the second replication produced both light and hybrid DNA, which refuted the dispersive model and supported the semiconservative model• That DNA polymerase requires a DNA template, a primer, synthesizes 5' to 3', and uses deoxynucleotide triphosphates, understand what 5’-3’ synthesis means (the polymerase adds a nucleotide to the hydroxyl on the 3’ end of the growing nucleotide chain).-each nucleotide added to a growing DNA strand comes from a nucleoside triphosphate which is a nucleoside (a sugar and a base) with three phosphate groups. EX: ATP• Know that DNA replication begins at sites called origins, and proceeds bidirectionally away from the origin. This produces two replication forks that move away from the origin and both strands of DNA are replicated simultaneously at each fork. Know what the leading strand and the lagging strand are.-leading stand- DNA polymerase III can synthesize a complementary strand continuously by elongating the new DNA in the mandatory 5’3’ direction. DNA pol III simply nestles in the replication fork on that template strand and continuously adds nucleotides to the complimentary strand as the fork progresses. Only one primer is required. -lagging strand- DNA pol III must work along the other template strand in the direction away from the replication fork. In contrast to the leading strand, the lagging strand is synthesized discontinuously, as a series of segments. These segments are called Okazaki fragments. Each fragment must be primed separately. Requires many primers• Replication fork- a Y shaped region where the parental strands of DNA are being unwound• Helicase- enzymes that untwist the double helix at the replication forks, separating the two parental strands and making them available as template strands• Primase- and enzyme that synthesizes the primer. It starts an RNA chain from a single RNA nucleotide, adding RNA nucleotides one at a time, using the parental DNA strand as a template. The completed primer is generally 5 to 10 nucleotides long and is base-paired to the template strand. The new DNA strand will start from the 3’ end of the RNA primer.• DNA polymerase III- adds a DNA nucleotide to the RNA primer and then continues adding DNA nucleotides, complementary to the parental DNA template strand, to the growing end of the new DNA strand. This is 500 nucleotides per second in E. coli and 50/second in humans• DNA polymerase I- replaces the RNA nucleotides of the primers with DNA versions, adding them one by one onto the 3’ end of the adjacent Okazaki fragment.• DNA ligase- joins the final nucleotides of this replacement DNA to the first DNA nucleotide of the Okazaki fragment whose primer was just replaced. It joins the sugar-phosphate backbones of all the Okazaki fragments into a continuous DNA strand• Know the importance of DNA repair
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