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ISU BBMB 405 - Exam 3 Study Guide
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BBMB 405 1st Edition Exam 3 Study Guide Lectures 23 33 Chapter 28 Lecture 23 28 4 DNA replication does not take place in the absence of the ribonucleotides ATP CTP GTP and UTP Propose an explanation A The construction of new DNA begins with DNA primase which produces RNA primer onto which DNA polymerase extends the DNA strand RNA primer requires ribonucleotides Production of the new DNA strand requires this RNA primer so DNA replication cannot occur without ribonucleotides 6 a How fast does template DNA spin expressed in revolutions per second at an E coli replication fork b What is the velocity of movement in micrometers per second of DNA polymerase III holoenzyme relative to the template A a At the E coli replication fork the speed of the spin of the template DNA can be calculated in this way the DNA template usually spins at the rate of 1000 nucleotides per second A B DNA has 10 4 nucleotides per turn therefore 1000 10 4 96 15 96 2 revolutions per second b The velocity of the movement of DNA polymerase II holoenzyme relative to the template is the DNA template usually spins at the rate of 1000 nucleotides per second the axial distance between each nucleotide is 3 4 angstroms for 1000 nucleotides it is 3400 angstroms sec which can also be written as 0 34 micrometers sec 7 Why would replication come to a halt in the absence of topoisomerase II A In the process of replication topoisomerase II induces super coils in DNA If the enzyme is absent the DNA becomes tightly wound that the movement of the replication complex becomes energetically impossible 8 One form of a plasmid shows a twist of Tw 48 and a writhe of Wr 3 What is the linking number What would the value of writhe be for a form with twist Tw 50 if the linking number is the same as that for the preceding form A The linking number Ln can be calculated using twist Tw and writhe Wr numbers in the following equation Ln Tw Wr Using the given values the linking number can be calculated Ln 48 3 51 If the linking number is the same using the twist to calculate writhe Wr Ln Tw 51 50 1 9 Telomerase is not active in most human cells Some cancer biologists have suggested that activation of the telomerase gene would be a requirement for a cell to become cancerous Explain why it might be the case A The telomeres at the end of the chromosomes will shorten progressively when the telomerase is inactive in the cell This leads to abnormalities of the proteins in the cell and ultimately cell death Cells with active telomerase the life of the cell is prolonged because telomeres are resynthesized Cancer cells tend to keep dividing Many cancer cells have activated telomerase which help the cell keep dividing and replicating Hence the activation of telomerase genes might cause cancerous cells 11 Suppose that you wish to make a sample of DNA duplex highly radioactive to use as a DNA probe You have a DNA endonuclease that cleaves the DNA internally to generate 3 OH and 5 phosphoryl groups intact DNA polymerase I and radioactive dNTPs Suggest a means for making the DNA radioactive A To make DNA molecule radioactive radioactive nucleotides need to be incorporated into unlabeled DNA molecule 1 2 3 4 Cleave the DNA at several points by brief exposure to DNA endonuclease Occasional nicks would be generated each with free 3 OH and 5 phosphoryl group Add DNA polymerase I and radioactive dNTPs The nicked DNA strand would be degraded by DNA polymerase I using its 5 to 3 exonuclease and radioactive nucleotides incorporated by its polymerase activity The nicks however would not be sealed because of the absence of ligase activity The DNA formed after this process would be highly radioactive 12 Suppose that replication is initiated in a medium containing moderately radioactive tritiated thymine After a few minutes of incubation the bacteria are transferred to a medium containing highly radioactive tritiated thymidine Sketch the autoradiographic pattern that would be seen for a undirectional replication and b bidirectional replication each from a single origin Answer could not be found 14 DNA photolyases convert the energy of light in the near ultraviolet or visible region 300 500 nm into chemical energy to break the cyclobutane ring of pyrimidine dimers In the absence of substrate these photoreactivating enzymes do not absorb light of wavelengths longer than 300 nm Why is the substrate induced absorption band advantageous A The enzymes DNA photolyases convert energy of light in near UV or visible region 300500nm to break cyclobutane ring of pyrimidines In the absence of a pyrimidine dimer these enzymes do not absorb wavelengths longer than 300nm This feature of these enzymes is an advantage because deleterious side reactions are prevented These enzymes might be damaged by light of wavelength exceeding 300nm in the absence of pyrimidine dimers 16 With the assumption that the energy required to break an average base pair in DNA is 10 kJ mol 1 2 4 kcal mol 1 estimate the maximum number of base pairs that could be broken per ATP hydrolyzed by a helicase operating under standard conditions A The hydrolysis of ATP to form ADP releases energy in the amount of 30 5 kJ mol Divide this amount by amount of energy needed to break base pair by DNA helicase 30 5 kJ mol 10 kJ mol 3 05 base pairs Therefore one ATP molecule can break about three base pairs with DNA helicase 20 The illustration below shows four petri plates used for the Ames test A piece of filter paper white circle in the center of each plate was soaked in one of four preparations and then placed on a petri plate The four preparations contained A purified water control B a known mutagen C a chemical whose mutagenicity is under investigation and D the same chemical after treatment with liver homogenate The number of revertants visible as colonies on the petri plates was determined in each case a What was the purpose of the control plate which was exposed only to water b Why was it wise to use a known mutagen in the experimental system c How would you interpret the results obtained with the experimental compound d What liver components would you think are responsible for the effects observed in preparation D A The Ames test is used to determine mutagenicity of chemical compounds that act as potential mutagens The concept behind the test is reversion of modified Salmonella stain that lacks the ability to grow in the absence of histidine The reversion is tested by the number of colonies growing on histidine


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ISU BBMB 405 - Exam 3 Study Guide

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