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Biology 1001A
Tom Haffie

WESTERN UNIVERSITY B I O L O G Y 1001A October 27, 2012 Time: 2 Hours Student No. _________________ Test Room ___________ Row _____ I NSTRUCTIONS - F OLLOW THE CHECK LIST ! (√ ) On your Scantron sheet Fill the bubbles completely Print name Print clearly • Use HB pencil only Signature Do your best • No stray marks or doodles Instructor Haffie/MacDougall-Shakleton • Make all erasures complete Course Bio 1001A Student number Print clearly/ Bubble neatly Yes - Calculators are permitted Exam Code 111 – Very important (non-programmable only) Section Leave it blank Answer Sheet Leave it blank No – Borrowing is not allowed (√ ) On this Test Book Student number No names please Do not write your name on the cover Test Room The room you’re in right now Your Scantron answers will be emailed to Row number We will tell you this your UWO email account within a week. Review copies of this test will posted Indicate your answers in the test book. Leave no questions blank. on OWL for study purposes. 1) Please place your ID prominently on your desk and sign the) Indicate your answers in both the test paper and on the attendance sheet when it comes to you. Scantron. 2) There are 42 questions in this test. Check your paper 5) It is your responsibility to transfer all answers from the test carefully. book to the Scantron sheet within the 2 hour test period. 3) There is only one fully correct option for each question.6) When finished, please stay seated and raise your hand. Part marks may be awarded. Answer all questions. We will collect both your test book and your Scantron sheet. Test books will be returned to you in tutorial. Warning The Scantron marking program has a cheating analysis feature that compares answer patterns for all papers. It alerts us to similarities. We then check seating arrangements. Do not sit near your study partners or write the same test codes. Keep your work directly in front of you. Please help us avoid any and all misunderstandings during these tests. Clarifications :Please defend your arguments on this page only. Comments will not be accepted after the test. Q # Ans Clarification Alt Ans 50 D If you mean Tom and Beth shop together, then my answer changes B October 2012 Biology 1001A Page 1 of 12 Circle the best single letter choice for each of the following questions before transferring your answers to your computer sheet. Note, for “multiple -multiple” style questions, more than one option may be correct (e.g. 1, 2 & 3 only). Part marks may be available for choosing some of the correct answers but choosing any incorrect answer earns a grade of “0”. 1. Which of the following statements about retroviruses is correct? 1. Drugs that treat retroviruses generally have more severe side effects than drugs that treat DNA - based viruses. 2. Retroviruses earned their name because they evolve ‘backward’; their genomes have become smaller and simpler over evolutionary time. 3. Immediately after a retroviral particle (virion) fuses with a host cell, the retroviral RNA is translated into proteins to assemble new virions. 4. Retroviruses tend to have higher mutation rat es than DNA-based viruses. A. 1, 2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1, 2, 3 and 4 are correct. 2. The use of antiviral drugs suffers from a paradox: the more we use them, the less effective they become in controlling vira l infections. Why? A. Antiviral drugs put a selection pressure on the virus, favouring the spread of drug -resistant strains. B. Antiviral drugs stimulate viral populations to grow faster. C. Antiviral drugs increase viral mutation rates; some of the resulting mutat ions make viruses more drug-resistant. D. Antiviral drugs promote horizontal gene transfer (HGT) of drug -resistant alleles into drug-sensitive strains. 3. The CCR5-Δ32 allele, which provides protection against HIV, is more common in some human populations than others. However, the geographic distribution of this allele probably cannot be explained by selection imposed by HIV. Why not? 1. The human populations with the highest frequency of CCR5 -Δ32 are not the ones that have been most heavily affected by HIV . 2. HIV is not truly “alive”, so it cannot influence the evolution of living organisms. 3. HIV has only been infecting humans for a few (human) generations, so there has not been time for much of an evolutionary response. 4. The CCR5-Δ32 mutation may have first occurred long before HIV began infecting humans . A. 1, 2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1, 2, 3 and 4 are correct. Student Number __________________________ Code 111 Code 111 Code 111 October 2012 Biology 1001A Page 2 of 12 4. What is the main advantage of treating a viral infection with a “cocktail” of multiple drugs at the same time, instead of with a single drug? A. Drug cocktails reduce the viral mutation rate, making it unlikely that the virus wilexperience a mutation that improves its resistance to any of the drugs used . B. Drug cocktails allow each drug to be taken a t a lower dose, so the risk of side effects is lower compared with taking high doses of a single drug. C. Drug cocktails trains the patient’s immune system to recognize a variety of viral coat proteins, so they are better able to deal with the virus than if the patient took a single drug. D. Drug cocktails are more likely than a single drug to remain effective, because a given virion is unlikely to experience a mutation that improves its resistance simultaneously to all the drugs in the cocktail. 5. Which of the following would be a consequence of reductive evolution? 1. Smaller genome size. 2. Increased amount of junk DNA. 3. Decrease in time taken to complete the cell cycle. 4. Inability to adapt to extreme environments. A. 1,2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1,2,3 and 4 are correct 6. Which of the following statements about the Last Universal Common Ancestor (LUCA) is correct? 1. It had only ribozymes to catalyze reactions. 2. It was the first form of life, develop ing about 4 billion years ago. 3 It probably had an RNA genome. 4. It possessed peptidyl-transferase activity. A. 1,2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1,2,3 and 4 are correct 7. The Miller-Urey experiment was a huge breakthrough in our understanding of the evolution of life. However, creationists argue that the biochemistry of life is distinctly different from that of the Miller-Urey experiment. Which of the following statements supports the creationist's argument ? A. Life requires energy while the Miller -Urey experiment did not. B. Water is required for all life but was not required for the Miller -Urey experiment. C. The amino acids produced by the Miller -Urey experiment are not required for life. D. The Miller-Urey experiment produced two forms of each sugar while all life produces only one. Student Number __________________________ Code 111 Code 111 Code 111 October 2012 Biology 1001A Page 3 of 12 8. Both DNA and RNA are nucleic acids that contain genetic information. Which of the following statements accurately compares DNA and RNA? A. Tautomeric shifts occu r only in DNA, not RNA. B. Phosphodiester bonding is identical in DNA and RNA. C. DNA contains a 6 carbon sugar while RNA contains a 5 carbon sugar. D. Hydrogen bonding is important for maintaining the structure of RNA, but not DNA. 9. If you are registered in this course, you are a human. From which of the following groups are you descended? A. LUCA B. Gorillas C. Both A and B are correct. D. Neither A nor B is correct. 10. To which of the following groups are you (a human) related? 1. Bacteria 2. Marsupials (e.g. kangaroo) 3. Plants 4. Chimpanzees A. 1, 2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1, 2, 3 and 4 are correct. 11. Until fairly recently, living things were classified as belonging to either the Plant Kingdom or the Animal Kingdom. What is the main problem with this system of classification? A. It does not reflect evolutionary relationships: some animals are more closely related to plants than they are to other animals. B. It does not reflect the diversity of life on earth: some living things are neither plants nor animals. C. It does not account for convergent evolution: some plants are carnivorous, and some animals use photosynthesis. D. The term “kingdom” is both sexist and un-democratic. 12. DNA sequence data suggest that modern humans last shared a comm on ancestor with Old World monkeys about 25 million years ago. In contrast, we last shared a com mon ancestor with New World monkeys about 40 million years ago. What can you conclude about evolutionary relationships among these groups? A. Humans are more closely related to Old World monkeys than to New World monkeys. B. Old World monkeys are more closely related to New World monkeys than to humans . C. Both A and B are correct. D. Neither A nor B is correct. Student Number __________________________ Code 111 Code 111 Code 111 October 2012 Biology 1001A Page 4 of 12 13.Arabidopsis is a model diploid plant species. The value of C for Arabidopsis is 110 Megabases (Mb). (1Mb = 1 million base pairs) How much DNA would there be in the nucleus of an Arabidopsis gametophyte cell during prophase of mitosis? A. 55 mB B. 110 mB C. 220 mB D. 440 mB 14.Birds have a relatively low value of C relative to related organisms such as turtles. Small genome size is believed to be the result of the high metabolic demands of flig- favouring smaller cells that must therefore have smaller nuclei. Which of the following would you predict to be dramatically less common, or fewer, in the genome of birds relative to turtles? A. genes B. chromosomes C. transposable elements D. single nucleotide polymorphisms 15.Which of the following stateme nts most accurately summarizes a relationship between genome size and organismal complexity ? A. On average, humans have the largest genome size of all animals . B. On average, animals have larger genome sizes than plants. C. On average, multi-cellular eukaryotes have larger genome sizes than uni -cellular eukaryotes. D. On average, eukaryotic organisms have larger genome sizes than prokaryotic organisms. 16.Which of the following sequences in the human genome is classified as “junk” DNA? 1. retrotransposons 2. telomeres 3. proviruses 4. centromeres A. 1,2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1,2,3 & 4 are correct. Student Number __________________________ Code 111 Code 111 Code 111 October 2012 Biology 1001A Page 5 of 12 17.This screen shot from an animation shown in class shows the structure of a replisome . The template DNA is approaching from the bottom left; the leading strand is leaving toward the right and the lagging strand is at the top left. The arrows identify a protein complex called the “sliding clamp”. What is the function of the sliding clamp? A. The sliding clamp prevents unwound single -stranded regions on the template strand from binding with themselves before the polymerase can replicate them. B. The sliding clamp triggers proof -reading if DNA polymerase creates a mismatch by mistake. C. The sliding clamp holds newly synthesized Okazaki fragments in place until ligase can seal any remaining “nicks”. D. The sliding clamp helps DNA polymerase III stay attached to the template strand. 18.Meselson and Stahl used bacterial cells for their famous experiment involving the incorporation of heavy nitrogen into DNA during replication followed by separation of different DNA into distinct bands in cesium chloride gradient tubes. What difference would you expect in the banding pattern in the tubes if you repeated the Meselson and Stahl experiment using eukaryotic cells instead of bacteria? A. Since eukaryotes have the same fundamental mechanism of DNA replication as prokaryotic cells, there would be no difference in the banding patterns in the tubes . B. Since eukaryotes have larger genomes, all the bands would travel further down the tubes. C. Since eukaryotes have more chromosomes, there would be more bands in the tubes. D. Since eukaryotes have faster polymerases, more DNA would be produced and t he bands would be wider. 19.All biology students learn that cells require nucleotides for DNA replication during S phase of the cell cycle. But what if you discovered cells taking up nucleotides outside of S phase? What process might require nucleot ides to be taken up outside of normal S phase in somatic cells? 1. Replacing primers. 2. Movement of mobile elements. 3. Adding telomeres. 4. Excision repair of thymine dimers. A. 1,2 & 3 only B. 1 & 3 only C. 2 & 4 only D. 4 only E. All of 1,2,3 & 4 are correct. Student Number __________________________ Code 111 Code 111 Code 111 October 2012 Biology 1001A Page 6 of 12 20.Bakers yeast, Saccharomyces cerevisiae , is a single-celled eukaryote with linear chromosomes. The sequence of the telomere repeat on the end of yeast chromosomes is 5’TGTGGGTGTGGTG TGTGGGTGTGGTG TGTGGGTGTGGTG 3’ 3’ACACCCACACCAC ACACCCACACCAC ACACCCACACCAC 5’ What must be the sequence of the RNA template carried by
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