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B1002B Feb13 term test.pdf

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Department
Biology
Course
Biology 1002B
Professor
Tom Haffie
Semester
Winter

Description
B I O L O G Y 1002B February 10, 2013 Time: 2 Hours Student number Room Row Print Name _______________________________________________ ______ 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 Maxwell • Make all erasures complete Course Bio 1002B 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 your Test Book (√ ) Student number Legibly please Name Print your name Scantron answers will be emailed to you. Test Room The room you’re in right now Marks will be posted 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 sign th4) Indicate your answers in both the test paper and on the attendance sheet when it comes to you. Scantron. 2) There are 11 pages in this test. Check your paper careful5). It is your responsibility to transfer all answers from the test book to the Scantron sheet within the 2 hour test period. 3) There is only one fully correct answer for each question. Part marks may be awarded. Answer all questions. We do not 6) 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 arran gements. 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 B Biology 1002B February Term Test 2013 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 4 or 5 choices. 1. In comparing the eyespot in Chlamydomonas and the eye in humans, which of the following statements is correct? A. The two organs are homologous. B. The eyespot is an ancient organ, while the eye evolved more recently. C. The eyespot does not contain chlorophyll but can influence photosynthesis. D. Photochemical reactions only take place in the eye. 2. Lecture 1 featured a video clip that showed normal phototaxis in wild -type Chlamydomonas cells and the lack of phototaxis in a strain of mutant cells. Which of the following hypotheses is supported by the responses of the two strains to a directional light source? 1. Photosynthesis in the mutant is more sensitive to high light. 2. The mutant may not be able to transport ATP to the flagella. 3. The eyespot must be defective in the mutant strain. 4. The mutant may have higher expression of enzymes that detoxify toxic oxygen (e.g. oxygen radicals, reactive oxygen species...) 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 3. The figure at the right shows the structure of a rod cell that is found in the retina of the mammalian eye. Which of the following statements correctly explains the structure /function of a rod cell? A. The rod is a specialized cell in which no protein synthesis takes place. B. The arrow points to the plasma membrane. During phototransduction the inside of the cell becomes positively charged compared to the outside. C. The bracket identifies 13 photoreceptor disks each containing one photoreceptor molecule. D. In the dark, this cell “communicates with” the next cell through the release of glutamate. Code 111 Code 111 Code 111 Biology 1002B February Term Test 2013 Page 2 4. A number of diseases that affect vision are caused by mutation s in genes that code for proteins of the phototransduction pathway. Which of the following scenarios would disrupt normal visual processing? 1. Mutation that alters a transmembrane domain of transducin. 2. Mutation to opsin gene that prevents trans-retinal binding to opsin. 3. Mutation to the gene that codes for retinal. 4. Mutation to opsin gene that decreases transducin binding to opsin. 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 5. There are four possible fates for the energy associated with the excited state of a pigment - transfer, heat loss, fluorescence and photochemistry. Which of the following statements correctly describes one or more of these fates? 1. When the organizati on of a photosystem is disrupted, the amount of heat loss increases. 2. The fluorescence emitted during the in -class demonstration was red light but at a longer wavelength than chlorophyll absorbed. 3. For transfer to occur, the excited state of the rece iver pigment must be lower than the excited state of the donor pigment. 4. Fluorescence results in an electron decaying from a higher to a lower excited state. 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 6. Bioluminescence is the generation of light by an organism (e.g. what a firefly does). Applying what you know about how molecules absorb and utilize the energy from a photon of light - which of the following statements most likely explains bioluminescence? A. Chemical energy is used to excite an electron in a pigment molecule and light is emitted as the electron decays back down to the ground state. B. It results from the absorption of light at a particular wavelength and emission of that light at a shorter wavelength. C. Photons of light are trapped within a special region of the cell and then released in response to various stimuli. D. ATP from oxidative phosphorylation is used to convert an electron into a photon of light. Code 111 Code 111 Code 111 Biology 1002B February Term Test 2013 Page 3 7. As polymers of amino acids with different charges, proteins can fold into specific conformations based on certain bonding arrangements. Which of the following descriptions of protein folding is correct ? A. The primary structure of a protein is lost when you incubate it with urea. B. Secondary structure is driven mostly by hydrophobic interactions. C. The native structure of a protein may require covalent bonding between R groups. D. Tertiary structure of a monomer protein requires ATP to drive folding. 8. The figure at right of a Western blot shows the change 0 6 12 24 h in abundance of the FLACO protein following a heat shock response in Chlamydomonas.   Compared to the 6 h lane, w hich of the following explanations for the low abundance at 24 h is correct? A. The low abundance is because of a decrease in the rate of mRNA decay. B. The low abundance is because of a decrease in transcription in the cytosol. C. The low abundance is because of an increase in the activity of the proteosome. D. The low abundance is because of an increase in the rate of translation in the cytosol. 9. All proteins have a defined half-life and eventually need to be replaced. Which of the following statements is consistent with the above statement? 1. According to the second law of thermodynamics, proteins break down because they are highly ordered. 2. The transcription and translation associated with house -keeping proteins is constitutive. 3. If the half-life is dependent somewhat on the reactive con ditions of its surroundings then the protein that binds P680 must turnover faster than most. 4. The entropy of a system and its surroundings must always increase. 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 111 Code 111 Code 111 Biology 1002B February Term Test 2013 Page 4n i rt g e l a e o e n pl . c 10. The image at right shows a stained agarose gel - the first u step in Northern blotting. H L S E Based on your knowledge of basic biology and the Northern blotting technique, which of the following interpretations of this gel is correct? 1. The spleen lane contains thousands of different kinds of mRNA. 2. Compared to human tissue, the mRNAs of E.coli are a different size. 3. Ribosome RNA structure is identical in all human cells. 4. In heart, lung and spleen, all genes show constitutive expression. 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 image at the right shows the predicted structure of an ABC- transporter protein. (The lines represent the membrane borders.) A Which of the following statements about an ABC transporter is correct? 1. Like all proteins, alpha helices are only found in the transmembrane domain. 2. All ABC transporters share a nearly identical domain A structure. 3. The underlying mechanism of transport only B found in eukaryotic cells. 4. Domain B contains more charged amino acids than domain A. 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 111 Code 111 Code 111 Biology 1002B February Term Test 2013 Page 5 12.Many chemical reactions in the human body can proceed spontaneously. Why? A. Because body temperature is relatively high. B. Because of the low entropy of the products. C. Because there is a readily available supply of ATP. D. Because of the high potential energy of the reactants . 13. Some spontaneous reactions take 20 million years for half of the reactants to be converted into products while other reactions take 10 se conds. What is the basis for this difference in rate? A. Reactions take longer the closer their ΔG is to zero. B. The rate is dependent upon the concentration of substrate molecules. C. Reactions where substrate molecules can more rea dily reach the transition state occur faster. D. The rate of the reaction is dependent upon a s upply of energy - thus coupled reactions that utilize ATP are the fastest. 14. Hexokinase isolated from a bacterium living in An tarctica fails to catalyze the phosphorylation of glucose at room temperature. From an evolutionary perspective, why must this be true? A. Because the maintenance of enzyme flexibility at low temperature is essential. B. This maintains defined organism temperature niches and decre ases competition. C. Because at low temperatures there is less energy available to maintain protein structure. D. Because natural selection favours higher stability in low temperature habitats. 15.An uncharacterized polypeptide has been isolated from the high Arctic. Given the data shown at the
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