October 22 nd, 2012 Name: CLID: Score: Chem 317 Exam II There are 17 multiple choices that are worth 3 points each. There are 4 problems and 1 bonus problem. Try to answer the questions, which you know first, and then solve the problems, which you are not sure about. Here is the summary of the total 100 points and 8 bonus points. Carefully read the questions. Good luck! Multiple choice (17 total) 3 points each, total 51 points Problems #1 12 points #2 12 points #3 9 points #4 16 points #5 Bonus 8 points Total 108 points You could use your calculator if you need. No notes or books of any sort may be used during the exam. No cell phones. I have neither given nor received aid on the exam. Signature and date 1
I. Multiple Choice (3 points each) 1. Restriction endonucleases type II (or restriction enzymes) are produced by microorganisms as part of the defense system to protect themselves against invasion of foreign DNA by cleaving. The first four nucleotides of the 8-base recognition cleavage site of restriction enzyme Not I are GCGG. The complete sequence of this 8-nucleotide recognition cleavage site is? a) double-stranded DNA; 5 -GCGGCCGC-3 b) double-stranded DNA; 5 -GCGGGGCG-3 c) single-stranded DNA or double-stranded DNA; 5 -GCGGCCGC-3 d) single-stranded DNA or double-stranded DNA; 5 -GCGGGGCG-3 2. A DNA polymerase is an enzyme that catalyzes the polymerization of DNA into a DNA strand using a DNA template. RNA polymerase, also known as DNA-dependent RNA polymerase, is an enzyme that produces RNA using a DNA template. DNA polymerase requires, to initiate DNA synthesis from the direction of and RNA polymerase does not require a primer for start of RNA synthesis from the direction of. a) a DNA primer, not a RNA primer; 5 3 ; 5 3 b) a DNA primer or a RNA primer; 5 3 ; 5 3 c) a DNA primer, not a RNA primer; 3 5 ; 3 5 d) a DNA primer or a RNA primer; 3 5 ; 3 5 3. A cdna synthesis starts with the mrna extracted from a specific tissue or cell type from an organism. In order to generate a cdna from a mrna, four enzymes are generally required. The correct sequence of these four enzymes in generating cdna using mrna as a template is. a) reverse transcriptase DNA polymerase RNase H S1 nuclease b) reverse transcriptase RNase H DNA polymerase S1 nuclease c) reverse transcriptase RNase H S1 nuclease DNA polymerase d) RNase H reverse transcriptase S1 nuclease DNA polymerase 4. A virus X that infects roaches is purified and the nucleic acid fraction is hydrolyzed and analyzed. The nucleotide composition of virus X is found to be 22% A : 28% C : 28%G : 22% T. A virus Y that infects zebrafish is purified and nucleic acid fraction is hydrolyzed and analyzed too. The nucleotide composition of virus Y is found to be 32% A : 18% C : 32% G : 18% U. Based on the information, what are the forms of the genomes of the viruses X and Y respectively? a) single-stranded DNA; single-stranded RNA b) double-stranded DNA; single-stranded RNA c) single-stranded DNA; double-stranded RNA d) double-stranded DNA; double-stranded RNA 2
5. DNA is typically a double stranded helix. The linkage between two adjacent nucleotides is through a. What is the nucleotide sequence of the DNA strand that is complementary to 5 -ATTCCGGACT-3? a) 3 to 5 phosphodiester bond; 5 -AGTCCGGAAT-3 b) 3 to 5 phosphodiester bond; 5 -TAAGGCCTGA-3 c) 5 to 3 phosphodiester bond; 5 -AGTCCGGAAT-3 d) 5 to 3 phosphodiester bond; 5 -TAAGGCCTGA-3 6. You are using a dideoxy sequencing method to sequence a DNA fragment. A sequence gel gives the following data. The sequence of the DNA fragment is and the nucleotide mixture in the reaction dda contains. gel electrophoresis Reaction dda Reaction ddg Reaction ddc Reaction ddt a) 5 -TAGAACGCT-3 ; datp, dgtp, dctp, dttp, ddatp b) 5 -TAGAACGCT-3 ; ddatp, dgtp, dctp, dttp c) 5 -TCGCAAGAT-3 ; datp, dgtp, dctp, dttp, ddatp d) 5 -TCGCAAGAT-3 ; ddatp, dgtp, dctp, dttp 7. Two proteins are said to be homologous if they have been derived from a common ancestor. are homologs that are present within one species with functions; are homologs that are present within different species with functions. a) Orthologs; similar or identical; Paralogs; similar or different b) Orthologs; similar or different; Paralogs; similar or identical c) Paralogs; similar or identical; Orthologs; similar or different d) Paralogs; similar or different; Orthologs; similar or identical 3
8. Bioinformatics is to study DNA, RNA and protein sequences through a combination of biochemistry, computer science and statistics approaches and to convert sequence information into knowledge. The bioinformatics concepts we have discussed in class include convergent evolution (A), divergent evolution (B), orthologs (C) and paralogs (D). Match all their possible example(s) (1, 2, 3) with each of the bioinformatics concepts (A, B, C, D). For example: A:1 means that 1 matches with A. Bioinformatics Concepts A: Convergent evolution B: Divergent evolution C: Orthologs D: Paralogs Examples 1: Human Ribonuclease and Human Angiogenin 2: Chymotrypsin and Subtilisin 3: Human Ribonuclease and Bovine Ribonuclease a) A: 1; B: 2 and 3; C: 1; D: 3 b) A: 2; B: 1 and 3; C: 1; D: 3 c) A: 1; B: 2 and 3; C: 3; D: 1 d) A: 2; B: 1 and 3; C: 3; D: 1 9. In Chapter 6, we have discussed protein sequence and structure relations. Match all possible example(s) (1, 2, 3, 4, 5) with each of the following statements on protein sequence and structure (A, B, C). For example: A:1 means that 1 matches with A. Statements on Protein Sequence and Structure A. Two proteins have very similar amino acid sequences, and both have similar 3D structures. B. Two proteins have very different amino acid sequences, however, both have similar 3D structures. C. Two proteins have very different 3D structures, however, both have similar active sites Examples 1. Actin and Hsp70 2. Chymotrypsin and Subtilisin 3. Human Ribonuclease and Human Angiogenin 4: Human Hemoglobin and Human Myoglobin 5: Human Ribonuclease and Bovine Ribonuclease a) A: 3, 4 and 5; B: 1; C: 2 b) A: 3, 4 and 5; B: 2; C: 1 c) A: 4 and 5; B: 1; C: 2 and 3 d) A: 4 and 5; B: 2; C: 1 and 3 4
10. Bioinformatics study is to protein relation. Below are segments of the sequences of cytochrome C from species A, B, C, D and E. These sequences correspond to the C-terminal 25 residues of the proteins. The different amino acids are underlined. MVFVGLKKPEERADLIAYLKKATSS (species X1) MVFVGLLKPEERADLIAYLKKATSS (species X2) MVFVGLEKPEERADLIAYLKKATSS (species X3) MVFVGLRKPEERADLIAYLKKATSS (species X4) MVFVGLFKPEERADLIAYLKKATSS (species X5) The order of the closeness from high to low of species of X2, X3, X4 and X5 to species X1 is. These sequences are examples of C, (choose all that apply). A. orthologs B. paralogs C. homologs D. convergent evolution E. divergent evolution a) X4 > X2 > X5 > X3; A, and E b) X4 > X2 > X5 > X3; B, and D c) X4 > X3 > X2 > X5; A, and E d) X4 > X3 > X2 > X5; B, and D 11. A virus causes Acquired Immunodeficiency Syndrome (AIDS). It was first exposed in 1981 in central Africa. More than 25 million HIV patients were dead by 2006 and 45 million people are estimated to be infected by HIV. What is the correct order of viral proteins in HIV life cycle from the first to last step? a) protease reverse transcriptase DNA polymerase integrase b) reverse transcriptase DNA polymerase integrase protease c) reverse transcriptase integrase DNA polymerase protease d) reverse transcriptase DNA polymerase protease integrase 12. Transcription is the process where a portion of the DNA molecule is copied. This coded information is then translated to create proteins the cell can use. Transcription process of eukaryotic cells differs from prokaryotic cells in several ways. Conversion of heterogeneous RNAs (hnrnas) of eukaryotic cells to mature mrna requires several post-transcriptional modifications (Modifications occur after hnrnas have been generated) while prokaryotic RNAs do not have those modifications. The correct order of post-transcriptional modifications of eukaryotic RNAs are. a) 5 -capping by m 7 G A poly A tail at the 3 -end RNA splicing b) A poly A tail at the 3 -end 5 -capping by m 7 G RNA splicing c) RNA splicing 5 -capping by m 7 G A poly A tail at the 3 -end d) RNA splicing A poly A tail at the 3 -end5 -capping by m 7 G 5
13. Sickle-cell anemia arises from a mutation in the gene for the β chain of human hemoglobin. The change from GAG codon to GTG codon in the mutant eliminates a cleavage site for type II restriction enzyme Mst II, which recognizes the target sequence CCTGAGG. These finding form the biochemical basis of a diagnostic test for the sickle-gene. A rapid procedure for distinguishing between the normal and the mutant genes has been developed. The procedure involves three major steps A, B and C: A. Enzyme Mst II cleavage B. Gel electrophoresis C. PCR The correct order of those steps for this procedure is. You expect to see DNA fragment(s) in normal people and DNA fragment(s) in sickle-cell disease patient from gel electrophoresis. a) Enzyme Mst II cleavage Gel Electrophoresis PCR; two; one b) Enzyme Mst II cleavage Gel Electrophoresis PCR; one; two c) PCR Enzyme Mst II cleavage Gel Electrophoresis; two; one d) PCR Enzyme Mst II cleavage Gel Electrophoresis; one; two 14. Insulin is a hormone secreted by the pancreas in response to lower blood glucose levels. Diabetes patients need insulin. You are in the process to express human insulin in bacterial E. coli cells. Based on your knowledge, the correct sequence of the major steps and reagents for this process is. a) mrna extraction PCR cdna generation restriction enzyme DNA ligase expression system construction b) mrna extraction cdna generation restriction enzyme DNA ligase PCR expression system construction c) mrna extraction cdna generation PCR restriction enzyme DNA ligase expression system construction d) cdna generation mrna extraction PCR expression system construction restriction enzyme DNA ligase 15. Genetic information flows from DNA to RNA and then from RNA to protein. The following duplex DNA is transcribed from right to left and is printed here. 5 -TCTGACTATTCAGCTCTC-3 3 -AGACTGATAAGTCGAGAG-5 The template strand of this duplex DNA is and the amino acid sequence of the polypeptide that this DNA sequence encodes is? Assuming that transcription starts from the first nucleotide. The genetic codes are in the last page of the test. You can tear it off for your convenience. a) 5 -TCTGACTATTCAGCTCTC-3 ; Ser-Asp-Tyr-Ser-Ala-Leu b) 3 -AGACTGATAAGTCGAGAG-5 ; Leu-Ala-Ser-Tyr-Asp-Ser c) 5 -TCTGACTATTCAGCTCTC-3 ; Leu-Ala-Ser-Tyr-Asp-Ser d) 3 -AGACTGATAAGTCGAGAG-5 ; Ser-Asp-Tyr-Ser-Ala-Leu 6
16. Nucleases can be divided into several classes. The nucleotide cleavage sites (the arrows) of the enzymes A and B are shown in the following diagram. Enzyme B A G C T G P P P P P OH Enzyme A Both enzyme A and B are DNases. Enzyme A is also and enzyme B is too. a) an endonuclease, a nuclease a; an exonuclease, a nuclease b b) an endonuclease, a nuclease b; an exonuclease, a nuclease a c) an exonuclease, a nuclease a; an endonuclease, a nuclease b d) an exonuclease, a nuclease b; an endonuclease, a nuclease a 17. DNA double helix model was proposed by James Watson and Francis Crick. They both obtained Nobel Prize in 1953 by discovering DNA double helix model. The following is a list of specific non-covalent interactions. A. Base stacking through π-π interactions B. Electrostatic interactions between positive ions and phosphate groups C. Hydrogen bonds between base pairs D. Hydrogen bonds between bases and water molecules E. Hydrogen bonds between sugar-phosphate backbond and water molecules DNA double helix structure in a water solution is maintained by. a) A, B, C, D, and E b) A, B, C and D, but most likely not E c) A, B, C and E, but most likely not D d) Only C 7
II. Problems 1. DNA sequencing: the following DNA fragment was analyzed by the dideoxy method. The primer is used for DNA sequencing is shown below. primer 5 3 3 GCTAACCT-5 (template strand) A sample of the DNA was reacted with DNA polymerase and each of the nucleotide mixtures (in an appropriate buffer) list below. Concentration of each nucleotide in the mixture is the same. 1. datp, dttp, dctp, dgtp, ddttp 2. ATP, TTP, CTP, GTP, dgtp 3. datp, dttp, dctp, ddctp; ddgtp 4. datp, dttp, dctp, dgtp The resulting DNA was separated by electrophoresis on an agarose gel and the bands on the gel were located. Draw each band in lane 1, 2, 3 and 4. You are not required to explain your answers. (12 points) 1 2 3 4 gel electrophoresis 8
2. Each protein has its unique function and each RNA has its own function. Match the following functions or processes or interactions numbered from 1 to 16 with their corresponding proteins or RNAs listed from A to J. You need to match the number(s) with their corresponding proteins or RNAs by filling out the blanks with your answers. You are not required to explain your answers. It is possible that some functions, or processes or interactions may not have their matches. (12 points) (12 points) Functions or Processes or Interactions 1. Bind promoter containing -25 sequence 2. Bind promoter containing -35 sequence 3. Bind mrna at its 5 -untranslated region 4. Cleave peptide bonds 5. Cleave phosphodiester bonds 6. Function in RNA splicing 7. Function in mrna degradation 8. Function in DNA repair 9. Function in transcription termination 10. Function in translation termination 11. Function in muscle movement 12. Function in assisting protein folding 13. Interact with CD4 receptor 14. Interact with CD4 ligand 15. Is a cytoskeleton component 16. Stimulate blood vessel growth Proteins A: Actin B: Angiogenin C: Eukaryotic RNA polymerase D: HSP70 E: gp120 F: Nucleases G: Rho H: Ribosome I: Small non-coding RNA J: Substilisin Your answers (your match) 9
3. Plasmids are naturally occurring extrachromsomal circular double-stranded DNA. They are important in DNA recombination technology. (total 9 points) (a) What are the features of plasmid? You are not required to explain the features. (3 points) (b) What is(are) requirement(s) for a plasmid to replicate itself in both E. coli cells and yeast cells? Use 25 words or less. (2 points) (c) You have a plasmid with ampicillin resistance in E. coli cells. But, ampicillin resistance seems disappeared when you transformed the same plasmid into yeast cells. List two possible reasons for this observation and justify your answer. (4 points) 10
4. Huntington s disease (HD) is an inherited neurodegenerated disorder characterized by the gradual, irreversible impairment of psychological, motor and cognitive functions. The molecular basis of the disease is becoming better understood. The genetic mutation underlying HD has been traced to a gene encoding a protein of unknown function. In individuals who will not develop HD, a region of the gene that encodes the amino terminus of the protein has a sequence of CAG codon for glutamine (Gln, Q) that is repeated 6 to 39 times in succession. In individuals, with adult-onset HD, this codon is typically repeated 40 to 55 times. A small portion of the amino acid-terminal coding sequence of HD gene is given below. The nucleotide sequence (5 3 ) of the DNA is shown in italic, the amino acid sequence corresponding to the gene is shown under the nucleotide sequence and CAG repeat is highlighted in bold and underlined. (total 16 points) ATGGCGACCCTGGAAAAGCTGATGAAGGCCTTCGAGTCC M A T L E K L M K A F E S CTCAAGTCCTTCCAGCAGCAGCAGCAGCAGCAGCAGCAG L K S F Q Q Q Q Q Q Q Q Q CAGCAGCAGCAGCAGCAGCAGCCGCCACTTCCG Q Q Q Q Q Q Q P P L P (a) Design a sense primer and an anti-sense primer to amply the exact CAG repeat in the diagram above. Each of the primers must be 6-nucleotide long. Write nucleotide sequences of the sense primer and the anti-sense primer from 5 to 3. You are not required to explain it. (4 points) The sequence of the sense primer from 5 to 3 : The sequence of the anti-sense primer from 5 to 3 : (b) What will be the size of PCR products in base pair (bp) for a normal person and a HD patient using 6-nucleotide sense and anti-sense primers? Assuming that the normal person has 23 CAG repeats and the HD patient has 48 repeats. Show your calculations or explain it. (4 points) (c) Assuming that you have 10 copies of double-stranded DH gene and are supplied with Taq DNA polymerase, dntps, sense and anti-sense primers, and all other reagents necessary. How many double-stranded copies of HD gene if you use 10 double-stranded HD genes as templates after 20 cycles? You are not required to explain your answer. But you need to show your calculations. (4 points) 11
(d) Assuming that you have 10 copies of single-stranded HD gene and are supplied with Taq DNA polymerase, dntps, sense and anti-sense primers, and all other reagents necessary. How many double-stranded copies of HD gene if you use 10 single-stranded HD genes as templates after 20 cycles? You are not required to explain your answer. But you need to show your calculations. (4 points) 5. Bonus question (8 points) (a) DNA polymerase has a proofreading activity that correct wrong nucleotides added by mistake of the enzyme. Unlike DNA polymerase, RNA polymerase does not have the proofreading activity. Explain why the lack of proofreading activity of RNA polymerase is not detrimental to the cells. (4 points) (b) What will happen to DNA replication and transcription if the following molecule is added to the cells? Justify your answer. (4 points) 12
13