
Concept explainers
Antibiotics and Protein Synthesis
Antibiotics are molecules produced by microorganisms as defense mechanisms. The most effective antibiotics work by interfering with essential biochemical or reproductive processes. Many antibiotics block or disrupt one or more stages in protein synthesis. Some of these are mentioned here.
Tetracyclines are a family of chemically related compounds used to treat several types of bacterial infections. Tetracyclines interfere with the initiation of translation. The tetracycline molecule attaches to the small ribosomal subunit and prevents binding of the tRNA anticodon during initiation. Both eukaryotic and prokaryotic ribosomes are sensitive to the action of tetracycline, but this antibiotic cannot pass through the plasma membrane of eukaryotic cells. Because tetracycline can enter bacterial cells to inhibit protein synthesis, it will stop bacterial growth, helping the immune system fight the infection.
Streptomycin is used in hospitals to treat serious bacterial infections. It binds to the small ribosomal subunit but does not prevent initiation or elongation; however, it does affect the efficiency of protein synthesis. Binding of streptomycin changes the way mRNA codons interact with the tRNA. As a result, incorrect amino acids are incorporated into the growing polypeptide chain, producing nonfunctional proteins. In addition, streptomycin causes the ribosome to randomly fall off the mRNA, preventing the synthesis of complete proteins.
Puromycin is not used clinically but has played an important role in studying the mechanism of protein synthesis in the research laboratory. The puromycin molecule is the same size and shape as a tRNA/amino acid complex. When puromycin enters the ribosome, it can be incorporated into a growing polypeptide chain, stopping further synthesis because no peptide bond can be formed between puromycin and an amino acid, causing the shortened polypeptide to fall off the ribosome.
Chloramphenicol was one of the first broadspectrum antibiotics introduced. Eukaryotic cells are resistant to its actions, and it was widely used to treat bacterial infections. However, its use is limited to external applications and serious infections. Chloramphenicol destroys cells in the bone marrow, the source of all blood cells. In bacteria, this antibiotic binds to the large ribosomal subunit and inhibits the formation of peptide bonds. Another antibiotic, erythromycin, also binds to the large ribosomal subunit and inhibits the movement of ribosomes along the mRNA.
Almost every step of protein synthesis can be inhibited by one antibiotic or another. Work on designing new synthetic antibiotics to fight infections is based on our knowledge of how the
Questions
Why are antibiotics ineffective in treating the common cold and other virus infections?

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Chapter 9 Solutions
Human Heredity: Principles and Issues (MindTap Course List)
- Please indentify the unknown organismarrow_forwardPlease indentify the unknown organismarrow_forward5G JA ATTC 3 3 CTIA A1G5 5 GAAT I I3 3 CTIA AA5 Fig. 5-3: The Eco restriction site (left) would be cleaved at the locations indicated by the arrows. However, a SNP in the position shown in gray (right) would prevent cleavage at this site by EcoRI One of the SNPs in B. rapa is found within the Park14 locus and can be detected by RFLP analysis. The CT polymorphism is found in the intron of the Bra013780 gene found on Chromosome 1. The Park14 allele with the "C" in the SNP has two EcoRI sites and thus is cleaved twice by EcoRI If there is a "T" in that SNP, one of the EcoRI sites is altered, so the Park14 allele with the T in the SNP has only one EcoRI site (Fig. 5-3). Park14 allele with SNP(C) Park14 allele with SNPT) 839 EcoRI 1101 EcoRI 839 EcoRI Fig. 5.4: Schematic restriction maps of the two different Park14 alleles (1316 bp long) of B. rapa. Where on these maps is the CT SNP located? 90 The primers used to amplify the DNA at the Park14 locus (see Fig. 5 and Table 3 of Slankster et…arrow_forward
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- For short answer questions, write your answers on the line provided. To the right is the mRNA codon table to use as needed throughout the exam. First letter U บบบ U CA UUCPhe UUA UCU Phe UCC UUG Leu CUU UAU. G U UAC TV UGCys UAA Stop UGA Stop A UAG Stop UGG Trp Ser UCA UCG CCU] 0 CUC CUA CCC CAC CAU His CGU CGC Leu Pro CCA CAA Gin CGA Arg CUG CCG CAG CGG AUU ACU AAU T AUC lle A 1 ACC Thr AUA ACA AUG Mot ACG AGG Arg GUU GCU GUC GCC G Val Ala GAC Asp GGU GGC GUA GUG GCA GCG GAA GGA Gly Glu GAGJ GGG AACASH AGU Ser AAA1 AAG Lys GAU AGA CAL CALUCAO CAO G Third letter 1. (+7) Use the table below to answer the questions; use the codon table above to assist you. The promoter sequence of DNA is on the LEFT. You do not need to fill in the entire table. Assume we are in the middle of a gene sequence (no need to find a start codon). DNA 1 DNA 2 mRNA tRNA Polypeptide C Val G C. T A C a. On which strand of DNA is the template strand (DNA 1 or 2)?_ b. On which side of the mRNA is the 5' end (left or…arrow_forward3. (6 pts) Fill in the boxes according to the directions on the right. Structure R-C R-COOH OH R-OH i R-CO-R' R R-PO4 R-CH3 C. 0 R' R-O-P-OH 1 OH H R-C-H R-N' I- H H R-NH₂ \H Name Propertiesarrow_forward4. (6 pts) Use the molecule below to answer these questions and identify the side chains and ends. Please use tidy boxes to indicate parts and write the letter labels within that box. a. How many monomer subunits are shown? b. Box a Polar but non-ionizable side chain and label P c. Box a Basic Polar side chain and label BP d. Box the carboxyl group at the end of the polypeptide and label with letter C (C-terminus) H H OHHO H H 0 HHO H-N-CC-N-C-C N-C-C-N-GC-OH I H-C-H CH2 CH2 CH2 H3C-C+H CH2 CH2 OH CH CH₂ C=O OH CH2 NH2arrow_forward
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