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GeneticsMidterm1-2008Makeup.pdf

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Department
Biology
Course
Biology 2581B
Professor
Jim Karagiannis
Semester
Winter

Description
SURNAME / INITIALS ____________________ (Please Print) Biology 281b FIRST MIDTERM TEST MAKEUP Feb 14, 2008 7pm -10pm Version 111 25% of Final Mark 3 hours INSTRUCTIONS This Test paper is composed of 40 equally weighted questions covering 16 sides (excl. cover). Please ensure that it is complete. Although there are no deliberate tricks on this paper there is ample opportunity to make careless oversights; READ THE QUESTIONS CAREFULLY. Answers will be subject to computerized matching analysis to detect cheating. Since it is impossible to provide equal access to instructors during the Test, proctors will NOT answer questions of interpretation. All answers must be recorded on the Scantron sheets within the 3 hour time limit. 1 . Fill in your NAME at the top of the scantron sheet and the exam paper 2. STUDENT NUMBER - Print the digits of your student number in the squares provided. Mark the corresponding bubble in the column below each printed digit. 1 3. SECTION AND ANSWER SHEET - Leave these areas blank. 4. In the EXAM CODE section enter 111 4 Mark one choice from the alternatives provided for each question. 5. Wrong answers will not cancel correct answers. Part marks may be available. 6. Submit your test paper and the Scantron sheet SEPARATELY. 1. During your summer job at Virotech, you isolate a previously unknown virus. Analysis of its genome reveals that it is composed of a double stranded DNA molecule containing 28% A (adenine). Based on this information, what would you predict the % C (cytosine) to be? a. 22% b. 28% c. 36% d. 72% e. Cannot be determined from the information given. 2. Which of the following has the capacity to store the most information. (PLEASE NOTE: i) Twenty different amino acids can be used to construct polypeptide chains ii) the English alphabet has 26 letters, iii) a binary system has only 2 characters). a. A DNA sequence composed of 22 nucleotides. b. A polypeptide chain composed of 17 amino acids. c. A 16 character message written using the English alphabet. d. A message of 20 characters written in binary. e. All of the above have equal capacities to store information. 2 3. In the Hershey and Chase experiment designed to determine the molecule of heredity, what was radiolabeled with 32P? a. protein b. DNA c. mRNA d. rRNA e. None of the choices are correct. 4. The chromosome can be considered a structure important for the _______________ of DNA. a. expression b. segregation c. packaging d. None of the above choices are correct. e. a), b) and c) are correct. 5. A diploid nucleus of the roundworm C. elegans contains about 4x10 nucleotide pairs. Assume that all the nuclear DNA is packaged into nucleosomes. How many molecules of histone H1 would be required to package the DNA. a. 1x10 6 b. 2x10 6 c. 4x10 5 8 d. 2x10 e. 1x10 8 6. In 1952, Robert Corey, and Nobel laureate Linus Pauling proposed a triple-helix model for the structure of DNA. This model was later shown to be incorrect. Which of the following experimental results are not consistent with the triple-helix model. 3 a. Griffiths demonstration that mixtures of heat killed S strains and live R strains could cause lethal infections in mice. b. Chargaffs rule. c. Hershey and Chases demonstration that DNA, and not protein, was injected into host bacterial cells when infected by bacteriophage. d. Both A and B are inconsistent. e. Both B and C are inconsistent. 7. Which of the following accurately describes the results of Avery's experiments that defined "the active principle" (derived from heat killed S strains) that was capable of transforming living R cells into virulent S cells. a. the active principle was present in a purified sample that contained almost nothing but DNA b. protein digesting enzymes destroyed the ability of the active principle to transform living R cells to virulent S cells. c. the extraction of lipids by ultracentrifugation destroyed the ability of the active principle to transform living R cells to virulent S cells. d. DNA-digesting enzyme, DNase, had no affect on the ability of the active principle to transform living R cells to virulent S cells. e. all the above are accurate. 8. Which of the following molecules would be in the least energetically favourable state to allow for the initiation of transcription. a. a negatively supercoiled, circular dsDNA molecule b. an overwound, linear ds DNA molecule c. a bacterial plasmid exhibiting (-) superhelicity d. a circular ds DNA molecule with neither (+) or (-) superhelicity e. All of the above would be in an equally favourable state. 9. In a certain breed of plants, dark green is determined by the dominant allele "G" and light green is determined by the recessive allele "g". The heterozygote shows 60% penetrance of the dark green phenotype. In a cross of Gg Gg individuals what phenotypes would be expected in 400 progeny plants? a. 220 dark green; 180 light green b. 180 dark green; 220 light green c. 400 dark green d. 175 dark green; 225 light green e. none of the above are correct 4
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