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PRINCIPLES OF GENETICS (BSCI222)
O’Brochta Spring Semester, 2012 EXAMINATION III (from Transposable Elements – Genetic Engineering) (200 points) Name_________________________________ UID _____________________________ Discussion Section______________
Instructions:
1. Put your name on each page of the exam.
2. Read each question carefully.
3. Place all answers in the space provided (in pen, for re-‐grade
considerations).
4. You have 75 minutes for this exam.
GOOD LUCK!
Honor Pledge: I pledge on my honor that I have not given or received any unauthorized assistance on this examination.
Signature and date Total Pts. ________ v
Questions 1 - 25 are worth 4 pts each. Write your choice of answer in the space provided. ____ 1. What is the function of dideoxynucleotides in Sanger DNA sequencing? A) They act as primers for DNA polymerase. B) They act as primers for reverse transcriptase. C) They cut the sequenced DNA at specific sites. D) They allow only the specific sequencing of the RNAs of a genome. E) They stop synthesis at a specific site, so the base at that site can be determined. ____ 2. Which of the following is a process whereby cancer cells travel to other sites in the body and establish secondary tumors? A) oncogenesis B) angiogenesis C) malignancy D) secondary tumorigenesis E) metastasis ____ 3. The mutation shown in the diagram below can best be described as a _____________mutation. A) missense B) nonsense C) silent D) frameshift E) reverse ____ 4. 454 sequencing technology relies on A) the use of dideoxynucleotides B) autoradiography C) detecting pyrophosphate production during phosphodiester bond formation D) mechanisms independent of DNA synthesis
____ 10. The antibiotic resistance genes present in virtually all plasmid cloning vectors are there to A) permit the unlimited production of antibiotics for pharmaceutical purposes B) counteract the modification system present in bacteria to defend themselves against phage infection C) help indicate when a DNA fragment has been successfully inserted into the plasmid D) enable the selection and growth of only bacteria that contain the plasmid cloning vector ____ 11. A non-autonomous Class II transposon would A) have two Long Terminal Inverted Repeats (LTRs) B) lack a functional transposase-coding region C) would express functional reverse transcriptase but not integrase D) has a functional transposase-coding region ____ 12. Mouse transformation (creation of transgenic mice) by pronuclear injection of DNA A) results in the integration of the transgene in predetermined locations B) results in random integration of the transgene C) involves growing and transferring embryonic stem cells D) would not work without the Ti plasmid. ____ 13. Illumina sequencing technology depends on A) the use of DNA ligase B) capillary electrophoresis C) light emission from luciferase/luciferin D) fluorescence of terminal nucleotides that block further elongation after each step in the sequencing reaction ____ 14. Deletions can be a problem if they involve a gene showing haploinsufficiency. Haploinsufficiency A) is when a gene is only expressed when there is one copy of the gene B) is when having only one copy of a gene results in a mutant phenotype C) is seen frequently in polyploids D) is when a cytogeneticst has insufficient information to interpret a karyotype ____ 15. Ti plasmids are A) found in Agrobacterium and parts of the plasmid can become integrated into plant genomes. B) useful cloning vectors for constructing gDNA libraries C) use reverse transcriptase during the integration reaction D) usually hold about 500 bp of foreign DNA
____ 16. Insertion or removal of one or two nucleotide base pairs in a protein coding region of DNA usually results in a ____________ mutation. A) transition B) frameshift C) reversion D) transversion E) suppressor ____ 17. Huntington disease can strike at an earlier age and bring about a more rapid degeneration and death in successive generations within a family. This phenomenon can be explained by which mechanism? A) Presence of a transposable element in the gene B) Chronic exposure to mutagens in the environment C) Expansion of a trinucleotide repeat in the coding sequence of the gene D) Presence of an extra chromosome in the germ line E) Absence of a gene product that is involved in DNA repair ____ 18. Golden Rice A) is a transgenic variety expressing glyphosate resistance genes B) was created by direct injection of DNA into the pronucleus of a fertilized egg C) contains transgenes that permit beta carotene production in the endosperm D) could have been made using conventional breeding practices. ____ 19. A transposable element is found to use RNA as an intermediate in transposition. On the basis of this information, which of the following would you also expect? A) The transposable element also probably makes transposase. B) The transposable element will require reverse transcriptase activity in order to integrate. C) The transposable element will be unable to undergo lateral gene transfer. D) The transposable element probably contains short inverted repeats at each end of the element E) The transposable element is moving by a cut-and-paste mechanism ____ 20. Which of the following best describes the Activator ( Ac ) and Dissociation ( Ds ) transposable elements in maize (corn)? A) Ac elements cannot transpose unless a Ds element is present. B) Ac contains a functional transposase gene; Ds lacks a functional transposase gene. C) Ds contains a functional transposase gene; Ac lacks a functional transposase gene. D) Both Ac and Ds have functional transposase genes. E) Neither Ac nor Ds contain functional transposase genes.
26. 12 pts. Concerning the movement of Class I transposable elements a) What chemical reaction does the reverse transcriptase of an LTR-type retrotransposon perform? Your answer should describe the substrate(s) of the reaction and the product(s) of the reaction. b) Explain the role of this specific reaction and reaction products in the movement of these transposable elements c) What other LTR-type retrotransposon-encoded enzyme is required to integrate the reaction products of the reverse transcriptase reaction into chromosomes. ANSWER a) RT is an RNA dependent DNA polymerase. Using RNA as a template RT polymerizes a complementary copy of DNA. b) This reaction is essential for LTR retrotransposon movement because once the cDNA is converted to double stranded DNA (by DNA polymerase) the transposable element can be integrated into a chromosome. c) element-encoded integrase will integrate the cDNA into a new target site. 27. 14pts. Retinoblastoma (RB) protein is important in regulating the cell cycle. A frameshift mutation in Exon 1 results in premature termnation of translation and no RB protein. a) Explain briefly how RB regulates the cell cycle. Make sure you indicate what part of the cell cycle is being regulated. b) Rank the predisposition to cancer of individuals that are homozygous for wild-type RB allele, heterozygous for mutant allele, homozygous for mutant allele. Explain the basis of your ranking. c) Would you consider RB to be an oncogene or a tumor suppressor gene. why? a) should describe the phosphorylation cycle of RB and RB's relationship to a transcription factor. G1 - S transition b) wild-type least predisposed - normal G1-S transition; heterozygote somewhat perdisposed - has on good copy but a single mutation in remaining wildtype copy will promote oncogenesis. Homozygotes highly predisposed - no RB therefore misregulation of G1-S transition. c) tumor suppressor gene - recessive - RB normally keeps cell cycle in check. Removing it leads to misregulation of cell cycle
28. 12 pts. The mutagen EMS converts guanine (G) to O - 6 - ethylguanine (G). O - 6 - ethylguanine (G) forms base pairs with thymine (T) instead of cytosine (C). Suppose that exposure to EMS damages a DNA molecule in a cell in G1 of the cell cycle. Here is what the DNA sequence looked like immediately before and immediately after the mutagenic event during the G1 phase of the cell cycle. Assume the cell harboring this mutation is a normally dividing cell. a) Diagram the steps required for the incorporated G* to induce a stably inherited mutation. Your diagram should include all necessary rounds of replication. b) Characterize the mutation induced by EMS as a transition , transversion, or frameshift. ANSWER a) b) The mutation is a CG to TA transition. 29. 10 pts. Explain briefly how defects in DNA-repair mechanisms can lead to the development of cancer and explain what cancer geneticists are referring to when discussing "driver" and "passenger" mutations? ANSWER Defects in DNA-repair mechanisms increase mutation rates, increasing the probability of a mutation that could lead to cancer. Driver mutations are those that contribute to oncogenic transformations while passenger mutations are those mutations found in tumors cells but have only gone along for the ride and are not involved in oncogenic transformation.
32. 10 pts. A DNA sequence encodes a protein with the amino acid sequence Met-Leu-Ser-Ile-Met-Ala. A single mutation occurs in the DNA sequence so that now it encodes a truncated protein with the amino acid sequence Met-Leu-Val. Use the codon table provided as a reference and a) Propose an explanation for the type of mutation that produces the new amino acid sequence and support your answer with a diagram showing the sequence before and after the mutation. Indicate where the mutation occurred. b) Give an example of a second single mutation that occurs in the gene for the truncated protein (Met-Leu- Val, above) that would produce the following amino acid sequence: Met-Leu-Val-Ile-Met-Ala and support your answer with a diagram showing the sequence before and after the mutation. a) A deletion of the first nucleotide (A) in the original serine codon, which changes the reading frame: AUG-CUC-AGU-AUA-AUG-GCC → AUG-CUC-GUA- UAA (stop codon) b) Insertion of an A residue just in front of the stop codon to change the reading frame: AUG-CUC-GUA-UAA (stop codon)-UGG-CC → AUG-CUC-GUA-AUA-AUG-GCC 33. 10 pts Total (2 pts each). Species I is diploid (2 n = 6) with chromosomes AABBCC; a related Species II is diploid (2 n = 6) with chromosomes MMNNOO. For each part, rewrite the entire genome (all chromosomes) of the appropriates species to reflect the presence of the chromosomal mutations indicated. a) tetrasomic for chromosome N b) an autotetraploid of Species II c) monosomic for chromosome B d) a double trisomic for chromosomes A and C e) a nullisomic for chromosome O a) MMNNNNOO b MMMMNNNNOOOO c) AABCC d) AAABBCCC e) MMNN
34. 12 pts. You are interested in a particular segment of rhinoceros DNA and would like to clone it into a particular cloning plasmid. You have the genome sequence of the rhinoceros but you only have a few rhinoceros cells from which you will only be able to obtain picograms of genomic DNA. In order to clone into a plasmid you will need at least hundreds of nanograms of the DNA of interest and plasmid DNA. You have milligram quantities of the plasmid DNA. You have the following restriction map of the region that includes the DNA of interest and the particular plasmid you are using (E = EcoR I, H = Hind III, X = Xba I, S = Sph I, N = Not I). Beginning with the rhinoceros cells in a tube describe the main steps you would take to accomplish this cloning task. Your task will be completed when you have bacteria containing your newly constructed plasmid. When appropriate specifiy the restriction enzyme(s) would you choose to clone most easily the DNA of interest into the cloning vector such that you insert all of the rhinosceros "DNA of interest" and indicates the substrates for those reactions. ANSWER Isolate gDNA Amplify region of genome including DNA of interest and restriction sites X and N by PCR Digest PCR products with the enzymes X and N. Digest the plasmid with X and N Ligate X-N PCR fragment with X-N-cut plasmid. Introduce into E. coli
