!1!HW#5 Chpts 15 & 16 !1. Which of the following is/are accurate statement(s) as to why a short RNA segment (‘primer’) has to be synthesized in order for DNA synthesis to ensue? A. Because DNA polymerase requires a free 3’ –OH group upon which to build dNTPs, and the short RNA segment has one. B. Because unlike DNA polymerase, primase does not require primer to begin synthesis of RNA polymers, so cells are able to build the short RNA segments to ‘prime’ the start of DNA synthesis without a free 3’ –OH group already in place. C. Because Okazaki fragments use short RNA segments to synthesize lagging-strand DNA fragments. D. Because DNA polymerase I moves in the 5’ to 3’ direction, removing RNA primer ahead of it and replacing the ribonucleotides with the appropriate deoxyribonucleotides. E. A & B (select E if both A and B are accurate statements explaining why primer is needed for DNA synthesis). 2. Which!of!the!following!is!NOT!true!about!RNA!primer?!A)!It!supplies!the!free!3’!hydroxyl!group!necessary!for!the!synthesis!of!mRNA.!B)!!It!is!a!few!nucleotides!long!and!when!no!longer!needed,!is!removed!by!DNA!polymerase!I.!C)!!It!serves!as!a!primer!for!DNA!synthesis.!D)!It!uses!a!DNA!sequence!template.!E)!It!supplies!the!free!3’!hydroxyl!group!necessary!for!the!formation!of!phosphod iester!bon ds.! 3. Which of the following incorrectly describes the function of the enzyme described? A. DNA polymerase I proofreads DNA base pairs while DNA synthesis is occurring. B. DNA ligase catalyzes the formation of a phosphodiester bond between adjacent Okazaki fragments. C. DNA helicase is involved in unwinding the DNA helix by breaking hydrogen bonds. D. Primase is capable of catalyzing polymerization of ribonucleotides into RNA without requiring a primer. E. Telomerase catalyzes the synthesis of DNA from an RNA template that the telomerase contains. 4. Which of the following provides the BEST explanation for why, in DNA synthesis, the lagging strand is discontinuous in nature, whereas the leading strand is continuous? A) Because the leading strand leads into the replication fork. B) Because DNA polymerase only works in the 5’ to 3’ direction. C) Because single-strand DNA binding proteins (SSBPs) attach to the separate strands of the DNA helix and prevent them from reforming only in the leading strand. D) Because the origin of replication only allows replication to proceed in one direction. E) Because in the lagging strand, primase continues to open the replication fork, exposing new single-stranded DNA to be replicated. 5. Hutchinson-Gilford progeria is a syndrome that results in the accelerated aging of infants and children. Balding, circulatory trouble, and fragile bone structure typically begin during infancy, and the children typically do not live through their teenage years. When scientists looked at the cells of these individuals, they found that the A. cells had higher amounts of telomerase than amounts in healthy individuals. B. cells had lower amounts of DNA polymerase I than amounts in healthy individuals. C. DNA had shorter telomeres than those of healthy individuals. D. DNA had shorter genes than those of healthy individuals. E. both A and C (select E if both A and C are accurate). 6. Examine the sketch below of a replication fork for DNA synthesis (see next page). The arrows indicate direction of DNA synthesis. Note that in this example, there aren’t any Okazaki fragments formed. How might this replication fork be possible? A. A DNA polymerase has been engineered to build DNA without a primer. B. A DNA polymerase has been engineered to synthesize DNA without using a template. C. A DNA polymerase has been engineered to proofread in the 5’ to 3’ direction. D. A DNA polymerase has been engineered to be able to build DNA also in the 3’ to 5’ direction. E. None of the above (select E if none of the above are explanations for how it would be possible).!2! Question #7 Sketch: 7. Examine the sketch above of a replication bubble for DNA synthesis. The origin of replication is labeled for you. The arrows indicate direction of DNA synthesis. Note that in this sketch, Okazaki fragments aren’t indicated as broken fragments in the diagram (in other words, a continuous arrow is shown in region containing Okazaki fragments, just for the purpose of the question). Based on the information provided, identify the 3’ and the 5’ ends on the DNA strands. A. W= 3’ , X= 5’ , Y= 3’ , Z= 5’ D. W= 5’ , X= 3’ , Y= 5’ , Z= 3’ B. W= 3’ , X= 5’ , Y= 5’ , Z= 3’ E. W= 5’ , X= 3’ , Y= 5’ , Z= 5’ C. W= 5’ , X= 3’ , Y= 3’ , Z= 5’ Examine the below figure showing the relationship between telomere length of chromosomes, and number of cell divisions, then answer the next two questions.!3!8. Which of the following BEST describes why the Germ cells have the longest telomere length even after multiple cell divisions, whereas telomere length declines in somatic (‘normal’) cells? A. Telomerase levels are high in germ cells, to protect the length of the cells’ chromosomes, whereas the levels decline in normal cells, resulting in a limited number of cell divisions possible based in part on the declining length of the telomeres. B. Telomerase levels are low in germ cells, to protect the length of the cells’ chromosomes, whereas the levels increase in normal cells, resulting in a limited number of cell divisions possible based in part on the length of the telomeres. C. Telomerase levels are the same for all cell types; germ cells have the longest telomere length because they are able to continue dividing throughout life. D. Telomerase levels are the same for all cell types; germs cells continue to divide because they have the longest telomeres. E. None of the above (select E if none of the above correctly explain these relationships). 9. Which of the following accurately describe(s) what happens at the point labeled ‘telomerase activation’ (see figure previous page)? A. Telomerase enzyme is activated in normal cells when they turn into cancerous cells, resulting in continuous cell divisions of the cancerous cells. B. Telomerase enzyme is de-activated in normal cells when they turn into cancerous cells, resulting in continuous cell divisions of the cancerous cells. C. Telomere length of cancerous cells remain constant as cell divisions continue to occur, whereas those of normal cells continue