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2014 Term Test 1

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Department
Biology
Course
Biology 1001A
Professor
Tom Haffie
Semester
Fall

Description
B I O L O G Y 1002B T ERM T EST February 8, 2014 Time: 2 Hours Student number Room Row Print Name _______________________________________________ ______ I NSTRUCTIONS - F OLLOW THE C HECK 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 Maxwell • Make all erasures complete Course Bio 1002B Student number Print clearly/ Bubble neatly Yes - Calculators are permitted Exam Code 222 – Very important (non-programmable only) Section Leave it blank Answer Sheet Leave it blank No – Borrowing is not allowed (√ ) On your Test Book Student number Legibly please Scantron answers will be emailed to you. Name Print your name Marks will be posted Test Room The room you’re in right now Row number We will tell you this to OWL after all tests are finished. Indicate your answers in the test book. Leave no questions blank. 1) Please place your ID prominently on your desk and si4) Indicate your answers in both the test paper and on the attendance sheet when it comes to you. Scantron. 2) There are 12 pages in this test. Check your paper ca5)fIt is your responsibility to transfer all answers from the test 3) There is only one fully correct answer for each question. Part the Scantron sheet within the 2 hour test period. marks may be awarded. Answer all questions. We do no6) When finished, please stay seated and raise your hand. subtract wrong from right. We will collect both your test book and Scantron sheet. 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. Warning 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 40 D If you mean Denis has a dog named Watson, then my answer changes to B Biology 1002B February Term Test 2014 Page 1 Circle the best single letter choice for each of the following questions before transferring your answers to your computer sheet. Note: Questions may have 3, 4 or 5 choices. 1. Its particular evolutionary relationship with other organisms is one reason that Chlamydomonas is such a useful experimental organism. Which of the following descriptions of Chlamydomonas evolutionary relationships is correct? A. Chlamydomonas is only distantly related to any multicellular organism. B. Chlamydomonas is more closely related to plants than to animals . C. Chlamydomonas is single-celled; therefore it is closely related to Bacteria. D. Chlamydomonas diverged from the lineage leading to fungi about 6 billion years ago. 2. To make things easier, scientists tend to use only one of the two different mating types of Chlamydomonas when doing most experiments in the laboratory. Which of the following topics could NOT be studied using this approach? A. DNA replication B. Heterozygote advantage C. Chromosome segregation during mitosis D. Phenotypic consequences of muta ted genes 3. The table below shows comparative genomic data for three organisms (Mb = million base pairs). E. coli Chlamydomonas Human Genome size 4 Mb 120 Mb 3,000 Mb Protein coding genes 4,000 15,000 20,000 Assuming that the coding region of a typical gene is roughly the same length in all organisms, which of the following genomes has the lowest proportion of non -coding DNA? A. E. coli B. Chlamydomonas C. Human 4. Pigments used to harvest light are not freely floating in the cytoplasm; they are bound to proteins. With respect to light being used as an energy sourcwhich of the following is a consequence of having pigments bound in such pigment-protein complexes? A. Efficient energy transfer from one pigment molecule to another. B. Increased ability of light absorption to be used for heat generation. C. Increased number of possible conjugated arrangements of the pigment. D. A change in shape of the protein upon light absorption by the pigment. Code 222 Code 222 Code 222 Biology 1002B February Term Test 2014 Page 2 5. The car1 mutation of Chlamydomonas results in cells unable to synthes ize the particular component of the eye-spot that contains carotenoid. Which of the following characteristics is likely shown by car1 mutant cells? A. They are unable to move. B. They are unable to perceive blue light. C. They are unable to determine the direction of a light source. D. They are unable to cause depolarization of the plasma membrane. 6. Phototransduction in the human eye and in Chlamydomonas may be homologous processes. Which of the following characteristics of phototransductionis found in both humans and Chlamydomonas? 1. A component is a seven-membrane-spanning-domain protein that binds a pigment. 2. It is triggered by photoisomerization of retinal. 3. It results in a change in electrical potential across the plasma membrane. 4. It involves the protein opsin, which serves as an ion channel. A. 1,2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1,2,3 and 4 are correct 7. Bioluminescence is widespread in nature among insects, fungi and bacteria . A recent report has shown that biofluoresence is also common, especially in fish. With respect to their ability to emit light, how are biofluorescent fish different than bioluminescent fish? A. Biofluorescent fish don't need a pigment; bioluminescent fish do . B. Biofluorescent fish need to consume ATP; bioluminescent fish do not. C. Biofluorescent fish have to be exposed to light; bioluminescent fish do not. D. Biofluorescent fish requir e electrons in an excited state; bioluminescent fish do not. 8. Methionine synthase is an enzyme that requires vitamin B 12as a cofactor. Even though this enzyme is essential, humans have to obtain vitamin B 12from their diet. Why do you suppose that we can't synthesize vitamin B 12 ourselves? A. Humans lack a key enzyme required for vitamin B 12 synthesis. B. Humans lack the gene that codes for vitamin B . 12 C. Vitamin B 12 contains an amino acid that humans cannot synthesize. D. Vitamin B 12 requires post-translational processing that humans cannot do. Code 222 Code 222 Code 222 Biology 1002B February Term Test 2014 Page 3 9. mRNA decay (breakdown) can play an important role in controlling protein abundance. Which of the following scenarios correctly describes a relationship between mRNA decay and protein abundance? A. A decrease in transcription with an increase in the rate of mRNA decay can result in increased protein abundance. B. An increase in transcription with an increase in the rate of mRNA decay can result in no change in protein abundance. C. An increase rate of protein synthesis but failure to form an apoprotein can be explained by a decrease in mRNA decay. 10. Which of the following statements about protein folding is correct ? 1. Urea can prevent a protein from folding into the correct tertiary structure. 2. Hydrogen bonding is involved in the formation of the tertiary structure but not the secondary structure. 3. The native conformation of a protein is the conformation with the lowest energy. 4. ATP is needed for most proteins to fold correctly. A. 1,2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1,2,3 and 4 are correct 11.The images at the right show agarose gel and Northern blot analysis for the presence of transcripts of hexK, a gene that codes for hexokinase, a glycolysis enzyme, in four different species. (A) shows the agarose gel stained for DNA, and (B) shows the final Northern blot image. Based on your knowledge of glycolysis and the Northern blotting technique, what is the most likely reason that there is no band in the E. coli lane shown in Figure B? A. Being a prokaryotic organism, E. coli does not have a hexokinase gene. B. Under the experimental conditions used, the hexK was not transcribed. C. Northern blot analysis is not possible in E.coli because it lacks ribosomal RNA. D. The probe used is not similar enough to the E. coli transcript to bind to it. Code 222 Code 222 Code 222 Biology 1002B February Term Test 2014 Page 4 12. Melting ice represents a very interesting thermodynamic system. Which of the following is an accurate conclusion arising from the fact that ice melts spontaneously at room temperature? A. Exothermic reactions can be spontaneous. B. Endothermic reactions can be endergonic. C. A decrease in entropy can occur without a phase change. D. Spontaneous reactions can be driven solely by an increase in entropy. 13.Melanie and Abdullah are studying free energy and exergonic/endergonic reactions . Melanie says..."Endergonic reactions bring in free energy and exergonic reactions are those that release free energy." Abdullah says..."yes, sounds good....but don't exergonic reactions need to bring in free energy to get to the transition state ?" Melanie says...."yes....damn, now I'm really confused, thermodynamics suc ks!" Which of the following statements could you offer that would clarify this situation? A. Melanie and Abdullah are not taking into account the i ncrease in entropy that occurs during exergonic reactions. B. Thermodynamics considers only the change in the free energy of the products relative to the reactants. C. The transition state only applies to endergonic reactions, it does not apply to exergonic reactions. D. Melanie is incorrect, exergonic reactions do not release free energy. 14.Maya is in a debate with Ryan who thinks life goes against the Second Law of Thermodynamics (SLT). He tells her "the idea that life maintains low entropy goes against the SLT, which states that the entropy of a closed system must always increase....energy spontaneously spreads out". Which of the following mistakes has Ryan made that Maya can nicely point out? 1. The SLT states that the entropy of a system plus its surroundings must always increase. 2. Life is an open system; it ma intains low entropy by bringing in energy from its surroundings. 3. A living system increases the entropy of the surroundings by releasigases and heat. 4. The fact that energy spreads out is not the same thing as entropy. A. 1,2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1,2,3 and 4 are correct Code 222 Code 222 Code 222 Biology 1002B February Term Test 2014 Page 5 15.The figure at right shows the activity of the enzyme rubisco. The enzyme was isolated from three different species of cyanobacteria, each adapted to a different temperature environment. Which of the following interpretations of the figure is correct? 1. That the optimal activity progressively increases from species A to C reflects the influence of temperature on the rate of a reaction. 2. Rubisco activity in species C falls to zero at 42°C. This is probably because the enzyme becomes too rigid. 3. The difference in rubisco activity between species A and C may be due to differences in the primary sequence of the protein. 4. In general, the tertiary structure of rubisco in species B is due to more and stronger bonding arrangements compared to species C. A. 1,2 and 3 B. 1 and 3 C. 2 and 4 D. 4 only E. All of 1,2,3 and 4 are correct 16.Enzymes are the major group of biological catalysts. Which of the following statements about enzymes is correct? A. To offset the fact that some enzymes have a slow catalytic cycle, their abundance within the cell is usually high. B. To proceed quickly, all exergonic processes within cells are dependent upon enzymes. C. Enzymes increase the activation energy of a spontaneous reaction. D. Enzymes convert endergonic reactions into exergonic reactions. 17.In many lectures Maxwell kept going on and on about C-H bonds. What is so special about C -H bonds? A. They represent a high entropy form of energy. B. They are strongly electronegative and are easily reduced. C. They are the source of readily accessible electrons that are found in food. D. They are an important bonding arrangement ensuring the proper quaternary structure of proteins. Code 222 Code 222 Code 222 Biology 1002B February Term Test 2014 Page 6 18. The Peyton1 mutant of E. coli results in a desaturase always being present in the cell. How do Peyton1 cells compare to wild-type cells? A. Peyton1 cells will grow more slowly at low temperature. B. Peyton1 cells will have membranes that are less fluid at higher temperatures. C. Peyton1 cells will have greater desatura se abundance at low
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