iclicker Question #28B - after lecture Shown below is a diagram of a typical eukaryotic gene which encodes a protein: start codon stop codon 2 3

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1 Bio 111 Handout for Molecular Biology 4 This handout contains: Today s iclicker Questions Information on Exam 3 Solutions Fall 2008 Exam 3 iclicker Question #28A - before lecture Which of the following statements are true? (A) Introns are found in prokaryotes and eukaryotes. (B) Introns are removed from DNA during replication. (C) Introns are removed from mrna following transcription. (D) Exons are removed from mrna following transcription. (E) None of the above are true. iclicker Question #28B - after lecture Shown below is a diagram of a typical eukaryotic gene which encodes a protein: 1 start of transcription start codon stop codon 2 3 exon 1 intron 1 exon 2 transcription Changing one base pair at which of the numbered locations above is likely to result in a mutant gene that is unable to produce a functional protein? (A) Location 1 (B) Location 2 (C) Location 3 (D) Location 4 (E) More than one location. 4 Beaming in your answers 1. Figure out your answer and select the appropriate letter (A-E). 2. Turn on your iclicker by pressing the ON/OFF button; the blue POWER light should come on. If the red LOW BATTERY light comes on, you should replace your batteries soon. 3. Transmit your answer as follows: Press the button corresponding to the answer you ve selected (A thru E). The STATUS light will flash green to indicate that your answer has been received. If the STATUS light flashed red, your answer was not received; you should re-send it until you get a green STATUS light. Molecular Biology 4-1

2 Bio 111: Information for Exam III Basic Facts The exam will be held in Lipke and another room on {see syllabus for date} from 12:00 to 12:50. The exam will cover from Cell Biology I through Molecular Biology 7 The exam will consist of approximately 4 questions. These will not be multiple choice; they will be problem-solving. A typical problem starts with a simple question, and then gets harder. For a rough idea of what the exam might be like, you should work through all the problems assigned in APAIB except: Ch 3: 1.all, and (although these are good practice). You need to know: the parts & differences between prokaryotic, plant, & animal cells from lecture DNA/RNA structure (not the specific structures, but be able to recognize the following parts: 5, 3, bases, sugars, phosphates and know their roles in structure and polymerization) Chargaff s ratios (%A %G %C %T) - the principle, not any specific set of ratios DNA/RNA rules DNA replication & leading & lagging strand Transcription & which strand is made mrna splicing & processing (introns & exons) Translation & start codon & stop codon & reading frame Mutations - types from lecture by name as well as other ways to break a gene Parts of a gene (differences between prokaryotic genes & eukaryotic genes) the viral life cycle in general & the life-cycle of HIV at the level of detail presented in lecture. how to explain the effect of an anti-aids drug given how it works the functions of Transpeptidase, peptidoglycan, penicillin, and β-lactamase You do not need to know: the structures of the base-pairs Enzymes & details of DNA replication other than DNA polymerase trna & other details of translation the structure of any particular gene the specific effects of any anti-aids drug You will be given: structures of any relevant molecules a table of the genetic code You may bring in a single sheet of (8 1/2 x 11 inch) paper with any notes you want. You may write on both sides. Molecular Biology 4-2

3 Bio 111 Fall 2008 Exam III Question 1: DNA, Transcription, & Translation (23 points) a) Chargaff s ratios. The table below lists the percentage of each type of base in a sample of double-stranded DNA. Complete the table below by filling in numbers as appropriate for double-stranded DNA: (8 pts) %A %C %T %G %U 20% Shown below is the sequence of a small hypothetical gene from a prokaryote: First nucleotide of mrna pairs here direction of transcription 5 AACTACGATGAGTGATTGACTTCAT3 3 TTGATGCTACTCACTAACTGAAGTA5 b) Give the first 5 nucleotides of the mrna produced by this gene. Be sure to indicate the 5 and 3 ends. (5 pts) c) What is the amino acid sequence of the protein produced by the gene shown above? Be sure to indicate the amino and carboxyl ends. If no protein is made, write No protein made. A table of the genetic code can be found on page 10 of this exam. (10 pts) Molecular Biology 4-3

4 Question 2: Mutation (25 points) Consider a small gene that produces the following mrna; the mrna encodes the protein aligned below the mrna. Spaces have been included to make it easier to read. Consider each mutation separately. A table of the genetic code can be found on page 10 of this exam CCAGA AUG UUC AGA UAA UUGA-3 N-met phe arg-c a) Suppose there were a mutation in the DNA such that the U indicated by (+) were changed to a A in the mrna. i) What would be the amino acid sequence of the protein produced by this mutated gene? Be sure to indicate the N and C termini. If no protein would be produced, write No protein. (4 pts) ii) What type of mutation would this be? Circle one (2 pts). missense nonsense frameshift b) Suppose there were a different mutation in the DNA such that the U indicated by (+) were deleted. What would be the amino acid sequence of the protein produced by this mutated gene? Be sure to indicate the N and C termini. If no protein would be produced, write No protein. (6 pts) Molecular Biology 4-4

5 Question 2, continued: The sequence of the mrna is shown again below. This is the same as the mrna on page CCAGA AUG UUC AGA UAA UUGA-3 N-met phe arg-c c) Give a single mutation that would cause the resulting mrna to produce no protein at all. Indicate your mutation by circling one appropriate base in the mrna above and filling in the blanks below as appropriate. There are many right answers here; give only one. (6 pts) Original base New base produced. Explain briefly how the mutation you described causes no protein to be 5 -CCAGA AUG UUC AGA UAA UUGA-3 N-met phe arg-c d) A single base-pair change in the gene encoding this mrna results in an mrna that encodes the following protein: N-met-phe-C Which base in the mrna was altered as a result of this mutation? Indicate your answer by circling the altered base in the mrna and filling in the blanks below as appropriate. (7 pts) Original base New base Explain briefly how the mutation you described causes the new protein to be produced. Molecular Biology 4-5

6 Question 3: Gene Structure & Mutation (26 points) Consider a small eukaryotic gene. Treat each of the parts of this question SEPARATELY. a) On the gene below, use an arrow to indicate a location where a mutation could result in a gene that produces a protein that has the same amino acid sequence as that produced by the non-mutated gene. Explain how a mutation at this location could result in the production of an identical protein. If more than one answer is possible, give only one. If it is not possible, write not possible and explain why it is impossible. (5 pts) 50 base-pairs Explanation: start codon stop codon Exon 1 intron 1 Exon 2 b) On the gene below, use an arrow to indicate a location where a mutation could result in a gene that produces no mrna but still produces the same protein as the non-mutated gene. Explain how a mutation at this location could result in the production of normal protein but no mrna. If more than one answer is possible, give only one. If it is not possible, write not possible and explain why it is impossible. (5 pts) 50 base-pairs Explanation: start codon stop codon Exon 1 intron 1 Exon 2 Molecular Biology 4-6

7 Question 3, continued: c) On the gene below, use an arrow to indicate a location where a mutation could result in a gene that produces a mature mrna that is longer than that produced by the non-mutated gene. Explain how a mutation at this location could result in the production of a longer mature mrna. If more than one answer is possible, give only one. If it is not possible, write not possible and explain why it is impossible. (5 pts) 50 base-pairs Explanation: start codon stop codon Exon 1 intron 1 Exon 2 d) On the gene below, use an arrow to indicate a location where a mutation could result in a gene that produces neither protein nor mrna. Explain how a mutation at this location could result in no production of mrna or protein. If more than one answer is possible, give only one. If it is not possible, write not possible and explain why it is impossible. (5 pts) 50 base-pairs Explanation: start codon stop codon Exon 1 intron 1 Exon 2 Molecular Biology 4-7

8 Question 3, continued: e) On the gene below, use an arrow to indicate a location where a mutation could result in a gene that produces a protein that is shorter than that produced by the non-mutated gene. Explain how a mutation at this location could result in the production of a shorter protein. If more than one answer is possible, give only one. If it is not possible, write not possible and explain why it is impossible. (6 pts) 50 base-pairs Explanation: start codon stop codon Exon 1 intron 1 Exon 2 Molecular Biology 4-8

9 Question 4: HIV & AIDS (26 points) a) Listed below are some of the steps in the HIV life cycle. Fill in the boxes with the appropriate missing steps. (4 pts each) Note that the numbering here may not be the same as in lecture. 1) Virus binds to outside of cell. 2) Virus fuses with the cell membrane and is taken into the cell. 3) 4) Reverse transcriptase makes a single-stranded DNA copy of the virus's RNA. 5) Reverse transcriptase makes a second strand of viral DNA base-paired with the first. 6) The double-stranded DNA copy of the virus's RNA goes to the host cell nucleus. 7) The DNA integrates into the DNA of the host cell. 8) The viral DNA remains latent for some time. 9) 10) Viral proteins are made. 11) Virus particles are assembled. 12) Virus particles leave the cell. b) Which of the above step(s) is/are unique to retroviruses? Give the number(s) of the step(s) that apply only to retroviruses and not non-retroviruses; if none apply, write none (4 pts). Molecular Biology 4-9

10 Question 4, continued: c) Consider the drug puromycin, which binds to ribosomes and prevents translation. i) Would you expect puromycin to inhibit the growth of human cells? Yes No No explanation is necessary. (1 pt) ii) Would you expect puromycin to inhibit the replication of HIV? Yes No Explain your reasoning briefly. (4 pts) d) Consider the drug ddi which inhibits reverse transcriptase. i) Would you expect ddi to inhibit the growth of human cells? Yes No Explain your reasoning briefly. (4 pts) ii) Would you expect ddi to inhibit the replication of HIV? Yes No No explanation is necessary. (1 pt) e) Which drug would provide the most safe and effective treatment for a human patient infected with HIV? (4 pts) puromycin ddi Explain your reasoning briefly. Molecular Biology 4-10

11 The Genetic Code U C A G U C A G UCU ser UAU tyr UCC ser UAC tyr UCA ser UAA STOP UCG ser UAG STOP UUU phe UUC phe UUA leu UUG leu CUU leu CUC leu CUA leu CUG leu AUU ile AUC ile AUA ile AUG met GUU val GUC val GUA val GUG val CCU pro CCC pro CCA pro CCG pro ACU thr ACC thr ACA thr ACG thr GCU ala GCC ala GCA ala GCG ala CAU his CAC his CAA gln CAG gln AAU asn AAC asn AAA lys AAG lys GAU asp GAC asp GAA glu GAG glu UGU cys UGC cys UGA STOP UGG trp CGU arg CGC arg CGA arg CGG arg AGU ser AGC ser AGA arg AGG arg GGU gly GGC gly GGA gly GGG gly U C A G U C A G U C A G U C A G Molecular Biology 4-11

12 Solutions 1) a) 30%; 20%; 30%; 20%; 0% b) 5 UACGA3 c) N-met-ser-asp-C 2) a) i) N-met-ile-arg-C ii) Missense b) N-met-ser-asp-asn-C c) Many possible, here is one: A in AUG changed to G - this removes the start codon so no protein will be made d) Only one possibility: the A in AGA to a U - this makes a premature stop codon. 3) There were many possible correct answers to most of these; here are some correct answers: a) In the intron - changes here are spliced out and therefore do not change the mature mrna, so do not change the protein. b) Not possible. An mrna is required to make a protein. c) A mutation in the start intron 1 signal - removing this would mean that intron 1 would not be removed so the mature mrna would be longer. d) Promoter mutation - the would no longer be recognized by RNApol so mrna would not be made; without mrna, no protein will be made e) In the coding region - a new stop codon would cause premature termination of translation resulting in a shorter protein. 4) a) Viral genome enters cell; viral DNA transcribed b) 4, 5, 6, 7, 8 c) i) Yes ii) Yes - viral mrna must be translated for viruses to reproduce d) i) No; human cells don t have or need Reverse transcriptase ii) Yes e) ddi This would inhibit HIV without killing the patient. Molecular Biology 4-12