Several common antibiotics affect some strains of bacteria's ability to carry out transcription and/or translation. For example: Rifamycin inhibits prokaryotic RNA polymerase Chloramphenicol blocks the transfer of the peptide from the P to A site. a) For each of these drugs, identify at what point it could affect the process of DNA->RNA->protein. Be as specific as possible. b) Why do you think these drugs kill bacteria but spare animal cells? (Hint: remember bacteria are prokaryotes)

Several common antibiotics affect some strains of bacteria's ability to carry out transcription and/or translation. For example: Rifamycin inhibits prokaryotic RNA polymerase Chloramphenicol blocks the transfer of the peptide from the P to A site. a) For each of these drugs, identify at what point it could affect the process of DNA->RNA->protein. Be as specific as possible. b) Why do you think these drugs kill bacteria but spare animal cells? (Hint: remember bacteria are prokaryotes)

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 20QP: If an extra nucleotide is inserted in the first exon of the beta globin gene, what effect will it...

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Many antibiotics are effective as drugs to fight off bacterial infections because they inhibit protein synthesis in bacterial cells. Using the information provided in the following table that highlights several antibiotics and their mode of action, discuss which phase of translation is inhibited: initiation, elongation, or termination. What other components of the translational machinery could be targeted to inhibit bacterial protein synthesis? Antibiotic Action 1. Streptomycin Binds to 30S ribosomal subunit 2. Chloramphenicol Inhibits peptidyl transferase of 70S ribosome 3. Tetracycline Inhibits binding of charged tRNA to the A site of the ribosome 4. Erythromycin Binds to free 50S particle and prevents formation of 70S ribosome 5. Kasugamycin Inhibits binding of tRNAfMet 6. Thiostrepton Prevents translocation by inhibiting EF-G
For the experiment shown Fig. 6.30 and 6.31, what is the advantage to use a labeled incoming nucleotide (α-32P-UTP or α-32P-CTP), not labeled reagent 1 itself? Please draw a diagram to help explain your answer by contrasting the RNA polymerase labeled with α-32P-Reagent (Panel A) vs α-32P-UTP (Panel B).
Why can it be said that transcription of thebacteriophage T7 genome requires two enzymes?
For each of the following initiation factors, how would eukaryotic initiation of translation be affected if it were missing? A. eIF 2 B. eIF4 C. eIF5
Consider the following sequence of DNA: 3'-TTA CGG-5'What dipeptide is formed from this DNA after transcription and translation? b. If a mutation converts CGG to CGT in DNA, what dipeptide is formed? c. If a mutation converts CGG to CCG in DNA, what dipeptide is formed? d. If a mutation converts CGG to AGG in DNA, what dipeptide is formed?
For each of the following initiation factors, how would eukaryoticinitiation of translation be affected if it were missing?A. eIF2B. eIF4C. eIF5
Which of the following statements concerning the sequence of events involved in the initiation of transcription by E. coli RNA polymerase are correct? ( can bemore than one) A.)Initiation starts with a pyrimidine nucleotide triphosphate. B.)The sigma factor (σ) functions during initiation of transcription. C.)Initiation starts with a purine nucleotide triphosphate, which is destined to become the 5'-end of RNA. D.)Initiation starts with a purine nucleotide triphosphate, which is destined to become the 3'-end of RNA. E.)RNA polymerase binds to the DNA at a point approximately 35 bp upstream of the start of transcription, at a sequence TTGACA. F.)The sigma factor does not dissociate until RNA elongation is complete.
Consider the following DNA template: 5’-AAGAGGTTCCAATGCAGGCACTCACCAACTCTTAAATAAA-3’ 3’-TTCTCCAAGGTTACGTCCGTGAGTGGTTGAGAATTTATTT-5’ If the bottom DNA strand is used as template to transcribe mRNA, predict the amino acid sequence that would result from the process of translation. Met-Ala-Leu-Thr-Gln-Glu-Gly Met-Gly-Ser-Leu-Asn-Ser-Gln Met-Thr-Asn-Ser-Leu-Ala-Gln Met-Gln-Ala-Leu-Thr-Asn-Ser Met-Glu-Ala-His-Trp-Ser-Tyr
Transfer RNA in eukaryotic cells is synthesized by which of the following enzymes (sensitive to high concentrations of the fungal poison a-amanitin, but not to low concentrations)? RNA polymerase I DNA polymerase I RNA polymerase II DNA polymerase II RNA polymerase III Which of the following processes of genetic information flow can occur under laboratory conditions, but has never been observed to occur under natural conditions (either in living cells or in viruses)? transcription of RNA from a DNA template (using DNA-dependent-RNA-polymerase) self-replication of RNA from an RNA template (using RNA-dependent-RNA-polymerase) direct-translation of protein from a DNA template (using special ribosomes) self-replication of DNA from a DNA template (using DNA-dependent-DNA-polymerase) translation of protein from an RNA template (using ordinary ribosomes)
Since RNA polymerase has an error rate of 1 / 10^4 nucleotides, and the DNA polymerase has an error rate of 1 / 10^7 nucleotides, can cells tolerate errors made in transcription in comparison to errors made during DNA replication?
The 5′ region of the TPP riboswitch in Bacillus subtilis is very similar to the TPP riboswitch in E. coli. Even so, the riboswitch in B. subtilis regulates transcription, whereas the one in E. coli regulates translation. What is the role of the 5′ region in both riboswitches? How can one riboswitch regulate transcription while the other regulates translation?
Consider the Rho-dependent terminator sequence 5’CCCAGCCCGCCUAAUGAGCGGCCUUUUUUUU-3’. What affect would a point mutation at any one of the bolded and underlined nucleotides disrupt termination of transcription? Group of answer choices Mutation in one of these nucleotides would disrupt base pairing, preventing the formation of the hairpin and disrupting termination. Mutation in one of these nucleotides would have no affect on base pairing, so the termination hairpin is formed and termination proceeds. Mutation in one of these nucleotides would not disrupt base pairing, but would prevent the formation of the hairpin and disrupt termination. Mutation in one of these nucleotides would disrupt base pairing, but not affect the formation of the hairpin and termination proceeds.