Case Study 7- Mechanisms Of Action corrections

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Utah Valley University *

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Jan 9, 2024

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Having brushed up on the basics of Prokaryotic Genome Organization and replication, you thought that you were ready to start conducting lab work. However, looking over the planned project, you realize that while you’ve spent all this time focusing on the bacterial genome, gene expression (transcription and translation) is what really matters when it comes to what a cell is actually doing. While Dr. Saxe thinks you’re ready, you decide to cement your knowledge of transcription and translation before you get into the lab. Basics of Transcription 1. Where in the eukaryotic cell does transcription take place?  Would be within the nucleus. 2. Where in prokaryotic the cell does transcription take place?  Will take place in the cytoplasm. 3. The image above shows a cycle that occurs during transcription initiation and elongation. a. What is represented by the blue circle?  The Sigma Factor b. What is represented by the red oval?  The RNA polymerase c. What is represented by the red line?  The messenger RNA. d. Describe what is happening in this image.  What is seen in this picture is the beginning of transcription in a prokaryotic cell. The sigma factor reacts with the RNA polymerase to create a holoenzyme RNA. This is bonded to the promoter region of DNA which starts transcription. Modified from ASM: Collection of Class Activities

-10 Region -35 Region Components of a Gene to be Transcribed. Shown below is a nucleotide sequence of a fragment of double-stranded DNA from E. coli. 5'..TCTACGCATCTAGCGGGCTCTTGACA(17bases)TATAATGCTCAATTGTCGCTAAGGAGGTGCTTGATGATCTGGCGAC 3'..AGATGCGTAGATCGCCCGAGAACTGT(17bases)ATATTACGAGTTAACAGCGATTCCTCCACGAACTACTAGACCGCTG Assume that the E. coli consensus promoter is: 5'...TTGACA...(any 17 bases)...TATAAT...3' 3'...AACTGT...(any 17 bases)...ATATTA...5' 1. Overall, what is the purpose of transcription?  To create RNA from DNA. 2. Where is the consensus promoter sequence found – DNA or RNA? DNA 3. In which strand(s) of DNA is the promoter sequence found – template/non-coding, coding/non-template, or both?  BOTH strands of DNA contain the promoter region. 4. How many different variations of promoter sequence are found in a cell’s genome – One or multiple?  Multiple 5. What is meant by the “-35” and “-10” regions of the consensus promoter sequence?  The upstream portion of the promoter, which indicates the start of transcription. 6. What would happen if there was a mutation in the consensus promoter sequence that rendered it completely unusable?  Sigma factor could NOT work with RNA polymerase AND transcription would not happen. Note: Many mutations from the “Wild Type” sequence don’t render it unusable – sometimes they can even increase the functionality! However, for the purposes of our class, we often assume zero functionality. 7. What protein recognizes and binds to the promoter region?  Sigma Factor Modified from ASM: Collection of Class Activities

Modified from ASM: Collection of Class Activities

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-10 Region -35 Region 8. What enzyme carries out transcription of DNA (i.e., makes the mRNA molecule)?  RNA polymerase 9. In which direction does transcription take place (left or right on the diagram above)? What determines the direction of transcription?  Left to Right  The lower DNA strand is the Template/Non-Coding strand. Only the RNA that matches the coding strand is produced, meaning the templates strand must be read from.  Because the newly transcribed strand is antiparallel to its template. The newly made RNA will have its’ 5’-end on the left and be synthesized by adding new nucleotides to the 3’- end on the right. o Like DNA polymerase, RNA polymerase can only add to the 3’-OH end of the transcript. 10. Which DNA strand will be actively transcribed from (i.e., is the template)? What determines which DNA strand gets copied?  The coding strand. o The bottom strand. Recognition of the promoter region by the RNA polymerase + sigma factor will essentially “ point the RNAp in the right direction” by aligning based on the -35 and -10 regions. Practicing Transcription Shown below is a nucleotide sequence of a fragment of double-stranded DNA from E. coli. 5'..TCTACGCATCTAGCGGGCTCTTGACA(17bases)TATAATGCTCAATTGTCGCTAAGGAGGTGCTTGATGATCTGGCGAC 3'..AGATGCGTAGATCGCCCGAGAACTGT(17bases)ATATTACGAGTTAACAGCGATTCCTCCACGAACTACTAGACCGCTG Assume that the E. coli consensus promoter is: 5'...TTGACA...(any 17 bases)...TATAAT...3' 3'...AACTGT...(any 17 bases)...ATATTA...5' 11. Is the start of transcription a precise location?  There is no definitive start location. Modified from ASM: Collection of Class Activities

12. Write out the mRNA that will be transcribed from this gene (same as on the first page, just copied here for your convenience), include 5' and 3' ends.  5’- AUUGUCGCUAAGGAGGUGCUUGAUGAUCUGGCGAC-‘3 Hint: The middle of the -10 region of the promoter (TATAAT) occurs approximately 10 bases upstream from the actual start of transcription. Count about 10 bases to the right (downstream) to find the first base that is transcribed. 13. The end of this mRNA molecule is not depicted but could be accomplished by one of two mechanisms. Briefly describe each of these two mechanisms: Rho dependent termination  Rho binds to RNA and kicks out RNA polymerase from DNA. Rho independent termination  RNA creates a hairpin loop which creates a series of UUUUUU. Forcing RNA polymerase to fall off. 14. Does the mRNA in prokaryotic cells get processed prior to translation?  No it does not. 15. What does this enable in prokaryotic cells that is not found in Eukaryotic cells?  Translation  Coupled transcription-translation. 16. What happens to mRNA after termination?  It falls off and gets recycled.  It gets degraded by RNAases Modified from ASM: Collection of Class Activities

This was hard… while you still need to learn translation, you decide to stop here even though these two processes are often coupled in prokaryotic cells. Modified from ASM: Collection of Class Activities

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Gene Expression: Translation You are now practicing your knowledge of gene expression by considering how translation take place. Below is the mRNA that you (should have) produced from the worksheet on Transcription. 5’ – CGCUAAGGAGGU GCUUGAUGAUCUGGCGAC – 3’ 17. On the above diagram of the mRNA, indicate where the 30S subunit of the ribosome will initially bind  AGGAGGU 18. How many different variations of the Shine Dalgarno sequences are found in a cell’s genome – One or multiple?  Multiple  One- unlike the promoter sequence, this is highly conserved. 19. What is this sequence called (write both names)?  Shine Dalgarno Sequence 20. On your diagram of the mRNA, indicate the position of the first translated codon. What amino acid is encoded by this codon?  AUG- Methionine 21. What is the codon for the second amino acid in this polypeptide?  AUC-Isoleucine 22. What is the sequence of the anti-codon region of this tRNA? Include 5' and 3' ends.  5’-UAG-‘3  3’-UAG-‘5 23. What is the role of tRNA in translation?  Make sure the accurate amino acid is placed in the ribosome. Modified from ASM: Collection of Class Activities

More closely examining the relationship between mRNA and translation Shown below is a partial sequence of an RNA molecule. 5' GUACUAAGGAGGU G 1 UGUGAUGAA C 2 CAAGUCUAGGG... 24. What types of RNA are not translated?  rRNA and mRNA  Anything that isn’t mRNA – e.g., rRNA, tRNA. 25. What determines whether a piece of RNA will be translated?  Shine Dalgarno Sequence.  + Start Codon 26. Is this likely mRNA or another form of RNA? Explain your answer.  mRNA  Definitely an RNA and you know it is mRNA as it has a ribosome-binding site and an AUG start codon. 27. How is the “frame” set during translation (i.e., what determines the sets of nucleotides associated with each codon)  A set of three bases creates the start codon and allows the translation to occur. 28. Would the frame change if you deleted the base marked G 1 on the sequence above? Why or why not?  It would not change. 29. Would the frame change if you deleted the base marked C 2 on the sequence above? Why or why not?  Yes, it mutates the future codons. Modified from ASM: Collection of Class Activities

30. What would happen if there was a mutation in the Shine Dalgarno sequence?  No translation occurs.  The small subunit of the ribosome would not be able to bind, and no translation would take place. Modified from ASM: Collection of Class Activities

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Putting it All Together to Help Dmitri Finally allowed into the lab, Dr. Saxe has three different bacterial isolates that were obtained from TB patients (labeled A-C below). Phil has exposed each isolate to an isoniazid E-test to determine the Minimal Inhibitory Concentration (MIC), required for each strain. It is key to note that MIC concentration just inhibits the organism growth, a MLC or minimum lethal concentration must be determined to know the concentration that will kill an organism. One of the genes that develops resistance-causing spontaneous mutations is inh A. This gene encodes the enzyme NADH-dependent enoyl-ACP reductase, which is required for formation of mycolic acids that make up the mycobacterial cell wall. It is believed that the antibiotic isoniazid can bind to the InhA enzyme (produced by the inhA gene) and inhibit this function. Your research project is to compare the DNA sequence of inhA in the resistant isolates to the isoniazid sensitive organism. Once you identify what spontaneous mutation(s) have occurred, you are to determine how that caused resistance to develop. Dr. Saxe tells you that he wants an explanation by the end of the day, so you get right to work! 31. Based on the E-test MIC results, rank the three isolates in order from most resistant to least resistant (most sensitive).  Most Resistant-B o There is no zone of inhibition.  Resistant- C o A small zone  Less Resistant- A o A large zone Modified from ASM: Collection of Class Activities A B C When interpreting these, the dark band around the strip indicates a lack of bacterial growth.

Examining the Isolates You first look at the sequence of the inh A gene in the most sensitive isolate (Isolate A) in order to determine the wild type (non-mutated) sequence of the gene and protein. The gene sequence below has been transcribed and translated. Dr. Saxe gives you the following details about the genes in Mycobacterium tuberculosis:  The highlighted sequence represents the -10 region of the promoter sequence  The bold nucleotides are the terminator.  The underlined nucleotides encode the Shine Dalgarno sequence.  The first … represents the approximately 10 nucleotides between the promoter and the start of transcription.  The second and third …… regions represent long sequences of nucleotides that are not necessary to examine in this exercise. Isolate A 5’ GA CACAACACAAGGACGCAC…CGTAAGGAGGT CGATGACA……GTGGTGCATTCGATT……CTCTAGGC AT 3’ 3’ CT GTGTTGTGTTCCTGCGTG …GCATTCCTCCA GCTACTGT……CACCACGTAAGCTAA……GAGATCCG TA 5’ 32. Which strand (top or bottom) would be the coding strand, and which would be the template strand? Explain how you would know.  The top strand is coding and the bottom is template strand. 33. What would the mRNA strand look like? You may include the …, but be sure to include all given nucleotides.  5’-CGUAAGGAGGUCGAUGACA…GUGGUGCAUUCGAUU…CUCUAGGCAU-‘3 34. Use the mRNA sequence you produced to produce the amino acid sequence. Be sure to take note of any portions of the mRNA that would not be included in the reading frame. You can assume the first nucleotide after the … is the beginning of a new codon.  Methionine, Thyronine, Valine, Valine, Histidine, Serine, Isoleucine, Leucine. Modified from ASM: Collection of Class Activities

Isolate B Next, you look at the DNA sequence from isolate B, which showed complete resistance. The bold underlined nucleotide indicates the spontaneous mutations that occurred. 5’ GA CACAACACAAGGACGCAC…CGTAAGGAGGT CGATGACA……GTGGTGCAT G CGATT……CTCTAGGC AT 3’ 3’ CT GTGTTGTGTTCCTGCGTG …GCATTCCTCCA GCTACTGT……CACCACGTA C GCTAA……GAGATCCG TA 5’ 35. Did the spontaneous mutation lead to a silent, frameshift, missense, nonsense mutation, or none of the above? Explain by indicating how translation has been affected (you don’t need to transcribe and translate the entire sequence unless you find it helpful). You can assume the first nucleotide after the … is the beginning of a new codon.  It led to a missense mutation causing the amino acid sequence to be changed. You then looked at the DNA sequence from isolate C, which showed resistance but was still sensitive at high levels of the drug. Again, the bold underlined nucleotide indicates the spontaneous mutation which has occurred. Isolate C 5’ GA CA T AACACAAGGACGCAC…CGTAAGGAGGT CGATGACA……GTGGTGCATTCGATT……CTCTAGGC AT 3’ 3’ CT GT A TTGTGTTCCTGCGTG…GCATTCCTCCA GCTACTGT……CACCACGTAAGCTAA……GAGATCCG TA 5’ 36. Did the spontaneous mutation lead to a silent, frameshift, missense, nonsense mutation, or none of the above? Explain by indicating how translation has been affected (you don’t need to transcribe and translate the entire sequence unless you find it helpful). You can assume the first nucleotide after the … is the beginning of a new codon.  None of the mutations fit the criteria for the change seen.  Substitution – no transcription would occur… could be considered nonsense, but I’d say none of the above. You should have recognized that the mutation in isolate C occurred in the promoter region of the gene, and is somehow increasing the resistance to the antibiotic which binds to the protein produced by this gene. To determine how it is causing resistance you must first think about the role of the promoter. 37. What protein binds to the promoter region of DNA? What is the function of this protein?  Sigma Factor o Recruits RNA polymerase to the site of transcription. Modified from ASM: Collection of Class Activities

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A literature search reveals that RNA polymerase holoenzyme binds best when the DNA sequence is 5’CATAACA3’. 38. Based on this information would it bind better to the promoter found in isolate A or isolate C?  Isolate C 39. Based on your answer above, will there be more of the InhA enzyme produced in isolate A or isolate C?  Isolate C 40. Thinking about the function of isoniazid and the InhA protein it interacts with to inhibit, develop a hypothesis explaining why the mutation in isolate C causes resistance at low levels of the drug but not at higher levels.  Isolate C creates too much protein so at low levels, there are still amounts of the Isolate C that are functioning since the drug does not inhibit all of it. For strain B the mutation occurred in the coding region of the inhA gene. There are three main mechanisms that pathogens can use to become resistant. These are listed below. i. Alter the concentration of drug inside the cell (pump it out or prevent it from getting in) ii. Modify the drug’s structure (degrade the drug, for example b-lactamases degrade the beta- lactam ring making bacteria penicillin resistant) iii. Modify the drug’s cellular target so it can no longer interact or inhibit it 41. Which do you think is the likely mechanism that is making strain B resistant? Explain your reasoning.  Option 3- Modify the drug’s cellular target so it can no longer interact or inhibit it. Modified from ASM: Collection of Class Activities

Case Study 7- Mechanisms Of Action corrections

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