*On the left is the actual electron micrograph. On the right is a schematic of the micrograph where DNA is labeled red and RNA blue.

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Download "*On the left is the actual electron micrograph. On the right is a schematic of the micrograph where DNA is labeled red and RNA blue."

Lewis Stokes
5 years ago
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LS1a Probem Set #6 Due Friday 11/10 at noon in your TF s drop box on the 2 nd foor of the Science Center a questions incuding the (*extra*) ones shoud be turned in 1.

1 LS1a Probem Set #6 Due Friday 11/10 at noon in your TF s drop box on the 2 nd foor of the Science Center a questions incuding the (*extra*) ones shoud be turned in 1. (18 points) Transcription in both eukaryotes and prokaryotes generates a compementary strand of RNA to a gene on the tempate strand of DNA. However, eukaryotic and prokaryotic genes are organized differenty. One key difference can be inferred from this eectron micrograph, which shows newy formed eukaryotic mrna hybridized to the strand of genomic DNA that it was transcribed from: *On the eft is the actua eectron micrograph. On the right is a schematic of the micrograph where DNA is abeed red and RNA bue. a) Expain the presence of this oop of DNA. (4 points) This oop resuts from an intron being present in the DNA. It has been removed from the mrna so the DNA bases from this intron region do not have a corresponding sequence in the mrna to hybridize with. b) If this experiment was repeated using RNA transcribed from a prokaryotic gene and the prokaryotic genomic DNA, woud you expect the resut to be the same or different? Expain. (4 points) The resut shoud be different. Prokaryotic RNA transcripts do not have introns and do not require processing. Thus the sequence of the RNA is virtuay identica to the DNA coding sequence (with the exception of U s repacing the T s), and the RNA woud annea perfecty with no oops to the tempate (non-coding) DNA strand. c) List three differences between prokaryotic and eukaryotic mrnas. (6 points) Eukaryotic mrnas have a 5 cap, 3 poya tai and have spicedout introns. Prokaryotic RNAs do not have any of these features. 1

2 d) Based on your understanding of how genes are organized in the HIV genome, what resut woud you expect if this experiment were performed using HIV mrna hybridized to a strand of the integrated DNA genome? Expain. (4 points) The resut woud depend upon which HIV RNA transcript is used. If an eary HIV transcript that has been spiced into the 4 kb form was used, the resut woud ook ike the efthand image above as the HIV provira DNA woud have to oop out where the introns are. If a ate, unspiced HIV RNA is used, it woud annea to the provira DNA without any oops as a of the introns are sti present in the RNA. For more accurate pictures, see the answer to part e. 2

3 e) (*extra*) Using the map of the HIV genome from Rob s sides (Shown beow). Draw the hybridization of fuy processed po, rev, tat, env mrnas to a strand of the integrated HIV genome. Note: The purpose of this question was to make students appreciate how compex the HIV genome is. They do not need to know which RNAs encode which genes. 3

4 2. (18 points) After hearing Rob discuss how HIV enhances the transcription of its RNA through the use of TAT and TAR, you attempt to utiize this feature to increase the transcription of other genes as we. Beow is the RNA sequence of TAR: 5 - GGGUCUCUCUGGUUAGACCAGAUCUGAGCCUGGGAGCUCUCUGGCUAACUAGGGAACCCA - 3 a) You have a company synthesize a DNA oigonuceotide containing the TAR sequence (shown beow). The sequences in bod have been added. The addition of these sequences does not prevent TAR from functioning propery. Briefy expain why the bod sequences do not prevent TAR from functioning. 5 - GAATTCGGGTCTCTCTGGTTAGACCAGATCTGAGCCTGGGAGCTCTCTGGCTAACTAGGGAACCCAGAATTC - 3 TAR (4 points) These bod sequences do not disrupt the basepairing of the TAR RNA to form a hairpin oop structure. Additionay, they can aso basepair and coud contribute to the formation of the hairpin oop. b) You connect the 3 end of the synthesized DNA oigonuceotide (shown above) to beads. You then add purified TAT protein and determine if the protein wi bind to the DNA oigonuceotide coated beads. You find that no TAT protein binds to the TAR DNA sequence. Expain. (4 points) TAT ony binds to the TAR sequence in RNA. It does not bind to the TAR sequence in DNA. c) You insert the synthesized DNA oigonuceotide containing the TAR sequence at the 5 -end of DNA encoding GFP. When this fusion gene is expressed in TAT expressing ces versus ces that do not express TAT, an increased eve of GFP fuorescence is seen in the TAT expressing ces. Why is TAT abe to bind and enhance transcription in these ces? Was this expected in ight of the resut in part b). (5 points) When this fusion gene is expressed in ces that contain TAT, RNA Po II transcribes the TAR sequence into RNA. As soon as the TAR sequence is transcribed and forms the hairpin oop structure, TAT can bind to the TAR eement in the RNA and increase transcription of the fusion gene. Yes, this was expected in ight of part b) since now the TAR RNA is present. 4

5 d) Next you insert the synthesized DNA oigonuceotide containing the TAR sequence (in red) into another gene (in back beow). When you express this modified gene in TAT expressing and non-expressing ces, you do not detect any difference in eves of the protein expressed in the TAT expressing ces. Furthermore when you purify the TAR containing mrna from these ces and try to bind TAT protein to it, TAT does not bind. Expain. 5 AACCAGAGAGACCCTTGCTACGAATTCGGGTCTCTCTGGTTAGACCAGATCTGAGCCTGGGAGCTCTCTGGCTAACTAGGGAACCCAGAATTC ATG - 3 TAR (5 points) When this fusion gene is transcribed into RNA the back sequence at the 5 end can form a hairpin oop with the 5 -end of the TAR sequence (5 -AACCAGAGAGACCC GGGTCTCTCTGGTT.3 ). The formation of this new hairpin oop prevents the formation of the hairpin oop that TAR normay forms and therefore TAT cannot bind. 5

6 3. (30 points) Beow is the DNA sequence of EF-Tu from the bacterium Bacius stearothermophius. The sigma factor binding sequences and the transcriptiona start site are highighted in bod. The 5 and 3 ends of the genes are shown, but for the sake of simpicity the midde sequence ( , designated by ) has been omitted in this diagram CGAATAATTGATTTCTCTTGCTAGTTCCGCTATAAATACTTATGTAAGTAAGACTTTT-3 3 -GCTTATTAACTAAAGAGAACGATCAAGGCGATATTTATGAATACATTCATTCTGAAAA AAGGAGGATCTTTCTCATGGCTAAAGCGAAATTTGAGCGCACGAAACCGCACGTCAACATT -TTCCTCCTAGAAAGAGTACCGATTTCGCTTTAAACTCGCGTGCTTTGGCGTGCAGTTGTAA ACAGTTGGCGCTGGTTCCGTATCGGAAATCATCGAGTAATCAAGAAAAAGGATGTCCAATC- TGTCAACCGCGACCAAGGCATAGCCTTTAGTAGCTCATTAGTTCTTTTTCCTACAGGTTAG GTTGGACATCCTTTTTCTTTATCCTGGCT-3 -CAACCTGTAGGAAAAAGAAATAGGACCGA-5 a) Pease write the first 18 bases of the mrna that woud be transcribed, 5 to 3. (3 points) 5 -GUA AGU AAG ACU UUU AAG-3 b) What wi be the first five amino acids of the protein derived from the mrna (pease denote N and C ends of the peptide and give your answer in both the three etter code and singe etter code) (5 points) 5 -ATG GCT AAA GCG AAA-3 N- Met- Aa- Lys- Aa- Lys-C N-MAKAK-C c) What wi be the ast five amino acids of the protein derived from the mrna (pease denote N and C ends of the peptide and give your answer in both the three etter code and singe etter code)? (5 points) 5 -TCG GAA ATC ATC GAG TAA-3 N-Ser- Gu- Ie- Ie- Gu-C N-SEIIE-C d) What is the ength of this protein (in amino acids)? (1.5 points) 395 Amino Acids e) Is the ength of the protein the same as the ength of the mrna? Briefy expain why or why not. 6

7 (4 points) No, the start and stop sites for transcription are not the same as the start and stop sites for transation. f) For each of the foowing nuceotide changes, pease determine their effects on the codon, amino acid specified, and on the resuting protein (truncation, singe amino acid change, radicay different protein, or no change). (7.5 points tota-1.5 for each row) Mutation Change in Change in Effect on Protein Codon Amino Acid Exampe: T# deetion UUU UUG Phe Leu Radicay different protein T37 deetion GCU GCA Aa Aa Radicay different protein A1201 G1201 GUA GUG Va Va No change C65 A65 CAC AAC His Asn Singe aa change A59 T59 AAA UAA Lys stop truncation A1223 T1223 AGA UGA none Not in coding region g) EF-Tu is essentia for bacteria surviva. Antibiotics that act on EF-Tu are used to combat infections caused by Bacius stearothermophius. One antibiotic, kirromycin, freezes EF-Tu in the GTP position (GTP cannot be hydroysed to GDP). Briefy expain how this woud effect transation? (4 points) The hydroysis of EF-Tu-GTP to EF-Tu-GDP is necessary for EF- Tu to dissociate from the aminoacy-trna. EF-Tu needs to dissociate from the aminoacy-trna so that the aa-trna can fuy occupy the A-site and add its amino acid to the growing poypeptide chain. 7

8 4. (10 points) Severa HIV genes are encoded in overapping portions of the genome. For instance, TAT and REV sequences are partiay overapping. Beow is the vira coding strand (non-tempate strand) DNA sequence for TAT beginning with the start codon and ending with the stop codon. 5 - atggaaccag tagatcctaa aatagaaccc tggaatcagc caggaagtcg gcctaagact ccgtgtaccc catgctattg taaaaagtgt tgctatcatt gcccaatatg cttcttaaac aagggcttag gcatttccta tggcaggaag aagcggagac aacgacgaac agctcctcct ggcagtaaga accatcaaga tcctgtatca aagcaacccg tatcccaaac ccaacgggag ccgacaggcc cagagaaaca gaagaaggag atggagagca aggcaacacc agatcgattc gattag-3 a) How many amino acids does Tat mrna (306 nuceotides tota ength) encode for? (2 points) 306/3 = for the stop codon = 101 b) The protein sequence for Tat and Rev are provided beow. Based on these sequences, do they share the same transationa start site? Expain. Tat MEPVDPKIEPWNQPGSRPKTPCTPCYCKKCCYHCPICFLNKGLG ISYGRKKRRQRRTAPPGSKNHQDPVSKQPVSQTQREPTGPEKQK KEMESKATPDRFD Rev MAGRSGDNDEQLLLAVRTIKILYQSNPYPKPNGSRQAQRNRRRR WRARQHQIDSISQRILSPYLGRSTEPVPLQLPPIERLRLDCSED CGNSGTQGVGDPQIPEEPGVLLGTGTKE (3 points) No, they do not since the amino acids sequences they encode are entirey different. c) Based on the Rev sequence provided, pease abe the first Methionine codon for Rev on the above sequence. (2 points) atg (aug in RNA) abeed in red above encodes the first met for Rev d) Expain how this same stretch of nuceotides can encode two different poypeptides with different sequences. (3 points) Two different start sites for transation are used. 8

9 5. (24 points) In order for mrna to be transated by the ribosome it must exit the nuceus of eukaryotic ces and enter the cytopasm. The ce uses a compex system of export machinery to accompish this task. a) What prevents RNA from diffusing out of the nuceus? (3 points) Nucear membrane. Transport into and out of the nuceus is reguated. b) What woud be a potentia consequence if RNA coud freey diffuse out of the nuceus? (4 points) Unprocessed RNAs coud potentiay be transated. This woud resut in the production of unusua proteins as introns may sti be present in some of the RNAs, and code for additiona amino acids or aberrant terminations. c) Normay, a ce ony exports mature or fuy processed mrna out of the nuceus. However, we earned in ecture that HIV can co-opt the ceuar machinery to export intron-containing RNA into the cytopasm. The protein Rev is essentia for this process. In order for Rev to function it must be abe to: be transported into the nuceus, bind the Rev-responsive eement (RRE), and interact with exportin. Briefy describe why these three things are necessary for Rev to function. (6 points) 1. If Rev cannot enter the nuceus, it cannot bind to RRE of unspiced HIV RNAs and mediate their nucear export. 2. If Rev cannot bind RRE, it cannot distinguish the RNAs it is supposed to bind to versus any other RNA. 3. Rev itsef is incapabe of exporting moecues out of the nuceus. It needs to attach itsef to exportin and Ran, which are the proteins that can directy mediate nucear export. d) (*extra*) The region of the REV protein that binds to RNA is rich in the amino acid arginine. Do you think this stretch of arginines is sufficient to specificay export RRE containing mrnas? Briefy expain why or why not. Probaby not. The sidechain of arginine is positivey charged at physioogica ph s and is thus most ikey to interact ionicay with negativey charged moecues, such as phosphates of the RNA backbone. The backbone does not have any sequence specific information, and thus is unikey to be used to bind RNAs in a sequence-specific fashion. e) In studying another virus, the mouse mammary tumor virus (MMTV), scientists ooked at a gene caed MMTV-env. This gene is predicted to be around 2000 base pairs (bp) ong. However, when the MMTV-env gene was introduced into 9

10 uninfected ces, they found that the MMTV-env-derived mrna in the cytopasm was 902 bases (ane 1, red arrow). Why is the size of the cytopasmic mrna different from the size of the predicted primary transcript? (3 points) an intron has been spiced out of the 902 bp mrna. f) Interestingy, when the MMTV-env gene was introduced into ces infected with MMTV, two different cytopasmic mrnas were identified. One was 902 bases (ane 2, red arrow) and the second was 2000 bases (ane 2, bue arrow). What might account for the appearance of this new cytopasmic mrna? (4 points) The 2000 bp transcript was exported into the cytopasm prior to spicing. Therefore MMTV has a mechanism to transport unspiced RNAs out of the nuceus. g) When the 2000 base and 902 base RNA fragments were sequenced and transated, they transated into two different proteins. One of them transated into a protein with a stretch of arginines and a nuceoar ocaization signa. Which band most ikey contains these motifs? Expain. (4 points) The 902 bp fragment probaby has these motifs (which are hamarks of Rev). Normay a RNAs are competey processed prior to nucear export. Therefore upon initia MMTV infection, the 902 bp RNA is the first one to be present in the cytopasm. This RNA coud then code for a protein to hep export incompetey spiced RNAs out of the nuceus, the 2000 bp fragment. If we assumed that the 2000 bp RNA encoded a protein with arginines and a nuceoar ocaization signa then the 902 bp RNA that is present in the cytopasm first woud not be capabe of encoding a REV-ike protein to export the 2000 bp RNA out of the nuceus. 10

Life Science 1A Practice Midterm Exam 2. November, 2006

Name: TF: Section Time Life Science 1A Practice Midterm Exam 2 November, 2006 Pease write egiby in the space provided beow each question. You may not use cacuators on this exam. We prefer that you use