Biol200_december1996 - Mm Wu BIOLOGY 200 Section 103 Dr Kasinsky Practice final exam(this years final will be 100 points 2V2 hours 250 marks

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Unformatted text preview: Mm _:/_, Wu, BIOLOGY 200 Section 103, Dr. Kasinsky Practice final exam (this years final will be 100 points) 2V2 hours. 250 marks QUESTION 1 (45 points) WW9 For each of the cellular molecules listed in alphabetical order in column A, find the corresponding ‘ molecule listed in alphabetical order in column B that is most closely associated with it. For each of these pairs of molecules indicate the following: (1 point) 1. The correct pair or molecules (1 point) 2. The common cellular function in which these two molecules participate. (Be specific) ‘ (I point) 3. Is the common cell type in which these molecules function a prokaryotic of a eukaryotic cell? . (2 points) 4. One major difference between each of the molecules of the pair. COLUMN A COLUMN B 1) actin AUG 2) deoxyribonucleic acid polymerase III dynein ~3) N—formyl methionine primase 4) signal peptide signal recognition particle receptor 5) sodium-glucose co-transporter - sodium-potassium adenosine triphosphatase 6) tubulin 7;. - ' (“WWW V NOTE: You can only choose a molecule ONCE from column A and once from Column B in this question. Please be brief. - QUESTION 2 (48 points) Mills») For each of the cellular structures listed in alphabetical order in column A, find the corresponding structure listed in alphabetical order in column B that is most closely associated with it. For each of these pairs of molecules indicate the following: (1 point) 1. The correct pair of structures. ( 1 point) 2. The common cellular function in which the two structures participate. (Be specific). (4 points) 3‘ HOW do each of the structures in the pair participate in that common function COLUMN A COLUMN B l) basal body axoneme 2) coated vesicle cis Golgi compartment 3) desmosome endosome 4) kinetochore interpolar microtubule 5) thylakoid stroma 6) trans Golgi network tonofilament NOTE: You can only choose a molecule ONCE from column A and once from Column B in this question. Please be brief. QUESTION 3 (42 points ) ((34le For each of the experimental proeedures listed below, describe how they have increased our knowledge of cell structure and’function. ' - - yup: - a) homopolymer mRNA b) DNA: mRNA hybrid c) tRNA cys - tRNA ala d) pulse/chase autoradiography QUESTION 4 (95 points) (gale) ' Use the genetic code on page 106 (Alberts, et al.) to answer the following questions: a) (15 points) Indicate the sequence of the transcription and translation products of the following piece of DNA in the nucleus. ASSume there are no introns present 3' TACCAATI‘GGA'I'I‘C’ITACTATI’ITATAATC 5' (template strand) ' 5' AAGAATGATAAAATA’ITAG 3' b) (15 points) If an identical piece of DNA were also present in the mitochondrion of this mammalian cell, would it have identical transcription and translation productsTExplain in detail. 0) (5 points) Name the exact compartments in the cell in which the i} nuclear and ‘ 3 mitochondrial transcription and translation products are made. (1) Now, assume that the nuclear DNA does have introns as indicated below: 3' TACCAA(intr0n)TTGGATTCI‘(intron)TACTATITI‘ATAATC 5' (template) 5‘ ATGG’IT(intr0n)AACCTAAGA(intron)ATGATAAAATATTAG 3' e) (10 points) Outline fully the processing of the HnRNA transeribed from this DNA. How would it differ for the mitochondrial DNA? 0 (10 points) What is the currently accepted evolutionary origin of mitochondrial DNA? Explain. g) (30 points) Assume that subunit A of a protein found in the inner membrane of the mammalian mitochondriOn is coded by a nuclear gene and that subunit B of the same protein is coded by a mitochondrial gene. Starting with the mature messenger RNA in the nucleus. indicate in detail how subunit A will be synthesized and incorporated into its final destination. h) ( 10 points) Starting with mRNA in the mitochondrion, indicate how subunit B is likely to be synthesized and incorporated into its final destination. QUESTION 5 (20points total) { "i Indicated below are 4 sets of experimental findings that we have not previ0usly encountered in ' Biology 200 lectures. Based on your present understanding of cell structure and function, do your best to provide a plausible explanation for each one of them: a) (5 points) DNA fragments can be generated by treating intact chromatin with a nuclease enzyme. Analysis by gel electrophoresis shows that the DNA fragments obtained are all multiples of 200 base pairs (bp) in length ie 200 bps, 400 bps. 600 bps etc. Treatment of naked DNA with nuclease does not generate any such pattern of fragments. Why? Explain briefly. b) (5 points) N ucleOplasmin is a large protein with quaternary structure, consisting of five heads' and five .tails .Vlt is made in the cytoplasm and travels to the nucleus via the nuclear pores. How it does so is mysterious. However, recent experiments involving cleavage of nucleoplasmin’ heads. from and pulse—chase autoradiography reveals that radioactive heads show silver grains in the cytoplasm whereas radioactive tails show silver grains in the nucleus. Intact nucleoplasmin that is made radioactive also shows grains in the nucleus. Why? Explain briefly. c) (5 points)-In the disease known as familial hypercholesterolemia, individuals are particularly susceptible to heart attacks from deposition of cholesterol in the arteries. Measurement of low density lipoprotein-cholesterol in the blood plasma of such individuals reveals very high levels. Why? Explain briefly. -- . ' ' d) (5 points) During the early development of the frog embryo. a broad central region of the ectoderm, the outemiost cell layer, thickens, pinches off from the rest of the cell sheet and forms a neural tube. Formation of this tube is due in part to elongation of cells in a direction perpendicular to the sheet. This elongation is disrupted by the drug colchicine. Why? Explain briefly. *****THE END****** ...
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This note was uploaded on 04/13/2008 for the course BIOL 200 taught by Professor Unknown during the Spring '97 term at The University of British Columbia.

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Biol200_december1996 - Mm Wu BIOLOGY 200 Section 103 Dr Kasinsky Practice final exam(this years final will be 100 points 2V2 hours 250 marks

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