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Biology 2581B
Jim Karagiannis

SURNAME / INITIALS ____________________ (Please Print) Biology 281b FIRST MIDTERM TEST Feb 9, 2008 - 9am -12 noon Version 111 25% of Final Mark 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. 1 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. 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. 2 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 14% T (thymine). Based on this information, what would you predict the %C (cytosine) to be? a. 14% 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. 3 (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 36 nucleotides. b. A polypeptide chain composed of 31 amino acids. c. A 24 character message written using the English alphabet. d. A message of 164 characters written in binary. e. All of the above have equal capacities to store information. 3. In the Hershey and Chase experiment designed to determine the molecule of heredity, what was radiolabeled with 35S? 4 a. protein b. DNA c. mRNA d. rRNA e. None of the choices are correct. 4. The chromosome can be considered a dynamic organelle for the _______________ of DNA. a. expression b. segregation c. replication d. packaging e. All of the choices are correct. 5. A diploid nucleus of Drosophila melanogaster contains 8 about 2x10 nucleotide pairs. Assume that all the nuclear DNA is packaged into nucleosomes. How many molecules of histone H3 would be required to package the DNA. 6 a. 1x10 b. 2x10 6 5 5 c. 4x10 d. 2x10 8 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. a. Chargaffs rule. b. Griffith’s demonstration that mixtures of heat killed S strains and live R strains could cause lethal infections in mice. 6 c. Hershey and Chase’s demonstration that DNA, and not protein, was injected into host bacterial cells when infected by bacteriophage. d. Both B and C are inconsistent. e. Both A 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 did not affect the ability of the “active principle” to transform living “R” cells to virulent “S” cells. c. the extraction of lipids by ultracentrifugation did not affect the ability of the “active principle” to transform living “R” cells to virulent “S” cells. activity d. DNA-digesting enzyme, DNase, destroyed the ability of the “active principle” to transform living “R” cells to virulent “S” cells. e. all the above are accurate. 7 8. Which of the following molecules would be in the most 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 75% penetrance of the dark green phenotype. In a cross of Gg × Gg individuals what phenotypes would be expected in 400 progeny plants? a. 250 dark green; 150 light green b. 250 light green; 150 dark green c. 400 dark green d. 175 dark green; 225 light green e. none of the above are correct 8 10. Your aunt Bertha is a judge in a civil trial where a young man is attempting to prove that he is the illegitimate child of a very wealthy old man who has recently died. He wishes to be included in the distribution of his wealth. After considering all the testimony about how this person was conceived, the key evidence seems to come down to one fact. The wealthy man and the mother of the young man are both deaf but the young man is not. Therefore the lawyer of the family argues that this proves the wealthy man is not the father. Your aunt Bertha comes to you for advice. What do you tell her? 9 a. The lawyer is correct, it would be impossible for the young man to be the son of the wealthy man. b. The lawyer is wrong, this evidence in fact proves that the young man must indeed be the son of the wealthy man. c. The lawyer is wrong, despite this evidence it would still be possible for the young man to be the son since the son may develop deafness later in life. d. The lawyer is wrong, the mother and father simply may be homozygous recessive for mutations in different genes leading to deafness. e. The lawyer is correct, but only if the son does not develop deafness later in life. 11. In rats, the gene for the pigment (P) is dominant to no pigment (p). The gene for black (B) is dominant to the gene for cream (b). If a pigment gene (P) is absent, genes B and b are inoperative and the rats are albino in colour. Predict the genotypes and phenotypes of the F2 of a cross between a black rat homozygous for the (B) and (P) alleles, and an albino homozygous for cream. Assume that the two genes are not linked. a. 9 black: 7 albino b. 13 black: 3 albino c. 15 albino: 1 black 10 d 9 black: 3 albino: 1 cream e. 9 black: 3 cream: 4 albino 12. During Meiosis, a number of important processes occur; place the following events in chronological order: i) Sister chromatids separate; ii) Chiasma visible; iii) Ploidy number reduced; iv) Homologous chromosomes separate; v) Independent assortment a. i, ii, iii, iv, v b. ii, v, iv, iii, i c. ii, i, v, iv, iii d. i, ii, v, iv, iii e. v, ii, iv, iii, i 11 13. _______ -form DNA spirals to the right and is the major form of naturally occurring DNA molecules. a. A b B c. D d. Y e. Z 12 14. You are making a physical map for the unsequenced genome of the Poison Dart Frog. You identify three markers (J, K and L) on chromosome 2. Given the following data for a three point test cross, what is the correct order of these three markers along the chromosome? JKL72 JkL 2 jkl 67 Jkl 103 JKl 557 jKl 1 13 jKL99 jkL 582 a. JKL b. KJL c. KLJ d. LJK e. Two of the above are correct 14 15. Mutations in the GAL80 gene in yeast result in the continuous expression of the GAL1, GAL7, and GAL10 genes, even in the absence of galactose. This genetic evidence suggests that the wild-type GAL80 gene product acts as a(n) a. transcriptional repressor. b. promoter element. c. enhancer element. d. transcriptional activator. e. RNA polymerase. 16. You perform a testcross with a female fly who is heterozygous for three marker genes with known distances; white eyes (w), yellow body colour (Y) and miniature wings (m). Calculate the degree of interference 15 in this region if you observe a double recombination event between markers w and m 1.5% of the time. a. +1.3 b. -0.5 c. -1 d. +0.5 e. +1 16 17. The critical step in the regulation of most bacterial genes occurs _____________. a. during the initiation of transcription. b. at release of mRNA. c. during translation. d. post-translationally. e. None of the choices are correct. 18. You are a researcher comparing the nonhistone and histone chromosomal proteins of chromatin isolated from human fibroblasts and frog fibroblast cells. Which of the following statements would be true of the
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