BBMB 405 1st EditionExam #3 Study Guide Lectures: 23 - 33Chapter 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 secondb) The velocity of the movement of DNA polymerase II holoenzyme relative to the templateis: 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/sec7. 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 becomesenergetically 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 = 19. 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. Cleave the DNA at several points by brief exposure to DNA endonuclease2. Occasional nicks would be generated, each with free 3’ OH and 5’ phosphoryl group3. Add DNA polymerase I and radioactive dNTPs4. 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 theabsence 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 (300-500nm) 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 helicase20. 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