Information Flow and Genetics

TWiM #188: Turducken Antibiotics

Podcast and Annotation Information
  • Annotation by Leonardo Baumgartner, Martin Leyhe, Triston Walsh, Kanwal Alvarez, Gwendowlyn Knapp, and Rebecca Seipelt-Thiemann.
  • Podcast audio by TWiM: Listen to TWiM #188 Podcast
  • Podcast transcript by Sarah Morgan: Access TWiM #188 Transcript
  • Papers Discussed:
    • Liu R, Miller PA, Vakulenko SB, Stewart NK, Boggess WC, Miller MJ. 2018. A Synthetic Dual Drug Sideromycin Induces Gram-Negative Bacteria to Commit Suicide with a Gram-Positive Antibiotic. Journal of Medicinal Chemistry. 61(9):3845-3854. https://doi.org/10.1021/acs.jmedchem.8b00218
    • Mehta AP, Wang Y, Reed SA, Supekova L, Javahishvili T, Chaput JC, Schulz PG. 2018. Bacterial Genome Containing Chimeric DNA-RNA Sequences. Journal of the American Chemical Society. 140(36):11464-11473. doi: 10.1021/jacs.8b07046 

1. Paper Abstracts

1.1. Snippet paper; discussion starts at 13:42 minutes

The Most Interesting Things (according to students)

Combining antibiotics with siderophores allows antibiotics to be effective against bacteria that would otherwise be resistant. 

This article is not licensed for Creative Commons use. Thus, the abstract cannot be copied here. You can access the abstract on the journal’s website: https://pubs.acs.org/doi/10.1021/acs.jmedchem.8b00218

1.2. Main paper; discussion starts at 36:30 minutes

The Most Interesting Things (according to students)

Chimeric genomes, containing both DNA and RNA, may have once existed in organisms as an evolutionary stepping stone from genomes composed of RNA to those solely of DNA.  

This article is not licensed for Creative Commons. Thus, the abstract  cannot be copied here.  You can access the abstract on the journal’s website: https://pubs.acs.org/doi/10.1021/jacs.8b07046

2. Vision and Change Core Concepts and 2024 ASM Fundamental Statements

Snippet Main
Vision and Change Topics
  • Structure and Function (V&C_SF)
  • Metabolic Pathways (V&C_MP)
  • Evolution (V&C_E)
  • Information Flow and Genetics (V&C_IFG)
ASM Fundamental Statements
  • Fundamental Statement 6 (ASM_6): The distinct structures and processes in microbes can be targets for interspecies competition, antimicrobial treatments, and host immunity.
  • Fundamental Statement 7 (ASM_7): Microbes have evolved structures adapted for specific functions that are often associated with a fitness advantage in a particular environment..
  • Fundamental Statement 13 (ASM_13): Intrinsic factors, such as genotype, metabolism, and cell structures, impact the survival and growth of microbes.
  • Fundamental Statement 14 (ASM_14): Extrinsic factors, such as abiotic and biotic interactions in the environment, can impact survival and growth of microbes.
  • Fundamental Statement 1: (ASM_1): All cells, eukaryotic organelles (e.g., mitochondria and chloroplasts), and major metabolic pathways evolved from early progenitor cells.
  • Fundamental Statement 16 (ASM_16): Genetic variation can influence microbial structures and their functions.
  • Fundamental Statement 26 (ASM_26): Humans leverage microbes and their products to address problems and improve quality of life.

3.  Potential Learning Objectives for the Podcast

The student will be able to: Paper1 Order2
  • Identify the function of a siderophore.
  • Compare and contrast gram-negative and gram-positive bacterial cell membranes.
 S L
  • Evaluate hypothetical Minimum Inhibitory Concentration (MIC) data to make conclusions about antibiotic effectiveness.
 S H
  • Recall advantages of new antibacterial products.
  • Describe the advantages and disadvantages of each nucleic acid as a genome.
 M L
  • Evaluate data for a hypothetical experiment based on the podcast discussion to make conclusions about the ability of a DNA binding protein to bind an RNA genome.
 M H

1 Papers: Snippet (S) or Main (M)

2 Learning Objectives: Lower Order or Higher Order (H)

4. Techniques Described (with Time Stamps)

Here is a link to a bio-dictionary that has many, but not all definitions if you need a definition: Explore Biology Bio-Dictionary

4.1. Snippet Paper

  • Minimum Inhibitory Concentration (MIC) (26:06–26:25: 29:00–29:30): The MIC for a drug is the lowest concentration that prevents the visible growth.  Here, the MIC was used to determine the efficacy of “turducken” antibiotic.

4.2. Main Paper

  • Metabolic Engineering (ME) (38:25–39:05): Metabolic engineering is when you modify a cell to increase metabolic pathways to produce a particular product.  Here, the bacterial cells were mutagenized to and those with reduced dCTP available were identified.
  • Mass Spectrometry (MS) (41:47; 45:33; 47:17): Mass spectrometry is an analytical technique that is used to identify and quantify molecules based on mass and charge.  Here, it was used to determine nucleic composition, and if the RNA nucleotides were in one strand or both strands.
  • Mutagenesis (42:00–43:05): Mutagenesis is when the DNA is changed through treatment or naturally.  Here, mutagenesis was induced using a chemical treatment (nitrosoguanidine) and was used to obtain morphologically distinct colonies.
  • Genome Sequencing (43:05; 45:55): Sequencing of the nucleic acid material can be done for an entire genome.  Here it was used to confirm the morphologically distinct colonies were not contaminants.
  • Gel Electrophoresis (54:11; 55:15–55:50): Electrophoresis is a chromatography technique to separate molecules based on size and change.  Here, it was used to determine large pieces of DNA that had RNA incorporated.

5. Connections to General Microbiology Processes/Concepts (with Time Stamps)

5.1. Snippet Paper

  • Novel Antibiotics (13:45–13:55, 17:00–20:05, 21:55–22:33, 24:20–24:45): “Turducken” antibiotics have been shown to be effective against gram-negative and acid-fast bacteria.  The name comes from a Cajun dish where a chicken is stuffed inside a duck, which is then stuffed inside a turkey.
  • Iron Acquisition (14:18–15:12; 17:45–18:30): Iron acquisition is a necessary process for bacterial metabolism.
  • Cell Structure (15:13–17:00): Gram-negative and positive cell wall structures were compared.
  • Antibiotic Targets (16:15–17:00, 18:45–19:20): Beta-lactams, a class of antibiotic, inhibit cross-linking of peptidoglycan.

5.2. Main Paper

  • Phage Therapy (33:13–35:00): Bacteriophage (phage) therapy was used against an antibiotic-resistant bacteria infection.
  • DNA and RNA (37:20–38:20; 39:15–39:30; 46:28–48:00): Evolutionary history, chemistry, and the creation of a chimeric DNA-RNA strands were discussed.
  • Polymerases (44:30–45:05; 49:28–50:45): DNA and RNA polymerase specificities were compared.

6. Podcast Questions

  1. What is a siderophore?
    1. A siderophore is an enzyme that produces iron.
    2. A siderophore harvests iron and transports it,
    3. A siderophore is a complex of membrane proteins.
    4. A siderophore harvests ATP using mitochondria.
  2. What are the main differences between gram-negative and gram-positive membranes? Pick all that apply.
    1. Gram-negative microbes has a thinner cell wall than gram positive microbes.
    2. Gram-positive microbes have more internal pressure (80 pounds per square inch).
    3. Gram-positive have efflux pumps to exclude antibiotics while gram negative do not.
    4. Gram-negative microbes have two membranes while gram positive microbes have one.
  3. The podcasters discuss the use of MIC to determine and compare antimicrobial effectiveness. If you tested a set of compounds against three clinical strains and reported the following data as the MIC (g/L) for each, which is likely most effective for clinical treatment?
    1. Compound A
    2. Compound B
    3. Compound C
    4. Compound D
Compound MIC Strain 1 MIC Strain 2 MIC Strain 3
Compound A >500 >500 422
Compound B >500 80 >500
Compound C >500 >500 35
Compound D 36 18 22
known antibiotic 100 60 80
  1. What do narrators and authors suggest as a reason for organisms to transition from mostly RNA genome to DNA genomes?
    1. RNA is too stable.  To replicate their genome, a less stable nucleic acid like DNA is required.
    2. RNA is less stable than DNA which retains the genomic information with greater efficiency.
    3. RNA genomes are more stable but unable to be large so they are ideal for prokaryotes.
    4. DNA genomes are more stable but unable to be large so they are not ideal for prokaryotes.
  2. The podcast host mentioned an antibacterial product called Kinnos wipes. What is the main advantage of disinfecting with Kinnos wipes versus traditional antibacterial wipes.
    1. The Kinnos wipes transfer a blue color to the surface that is being disinfected and disappears after 5 minutes.
    2. The Kinnos wipes are copper-based disinfectant relying on the natural antibacterial properties of copper. 
    3. The Kinnos wipes are an ineffective antibacterial product, but are an effective antifungal product.
    4. The Kinnos wipes have a blue color that is activated with a UV light which indicates the surface is clean.   

hypothetical electromobility shift assay results6. The podcasters discuss how proteins that normally bind DNA and how they might not bind the equivalent RNA version of the genome.  So, a researcher decides to test this using an method called an electrophoretic mobility shift assay (EMSA).  You have hypothetical DNA binding protein X, the normal DNA sequence that protein X binds, and a double-stranded RNA equivalent of the DNA sequence.  In this experiment the nucleic acid (DNA or RNA) is labeled, the protein and nucleic acid are mixed and incubated, fractionated using gel electrophoresis. Plain nucleic acid is a control for each.  The electrophoresis results are shown for DNA alone, DNA + Protein X, RNA alone, RNA + Protein X.  What can you conclude about the ability of DNA binding protein X to bind double-stranded RNA?

    1. Protein X binds DNA and RNA equally well.
    2. Protein X binds DNA, but does not bind RNA.
    3. Protein X binds DNA better than it does RNA.
    4. Protein X binds best to a DNA-RNA hybrid.

7. Figure Reading Exercises

The following are two figure reading exercises, one from the snippet paper (Table 2) and one from the main paper (Figure 1).

7.1. First Figure Reading Exercise

7.1.1. Learning Objectives

Students will be able to:

  • Match the function of each drug conjugate or cellular component with its role in this experiment.
  • Define Minimum Inhibitory Concentration (MIC) and its relationship to drug effectiveness.
  • Analyze Minimum Inhibitory Concentration (MIC) data to make conclusions about effective drug conjugates and conditions for effectiveness for specific strains.
  • Predict the effect of altering drug-conjugate structures or bacterial features on drug  effectiveness.
Experimental Background (Liu et al. 2018, Abstract Figure and Table 2)

Resistance to antimicrobial compounds is a serious threat recognized by the World Health Organization, so there is an urgent need to develop additional therapeutics and therapeutic delivery systems.  With this in mind, Liu et al (2018) developed and tested a drug conjugate for gram-negative bacteria, which are particularly difficult to treat.  These researchers tested for the Minimum Inhibitory Concentration (MIC) using their engineered and synthesized drug conjugates.  The final conjugate included three components: 1) a siderophore (to enable active uptake in an iron-depleted environment into the periplasm), 2) cephalosporin (the linker that enables “release” of the antibiotic by beta-lactamase-induced cleavage when in the periplasm), and 3) oxazolidinone (the target antibiotic that cannot normally penetrate the outer membrane of gram-negative bacteria).  Note that cephalosporin is also an antibiotic.  Please see the abstract image for an overview.

The researchers tested different combinations of their final drug conjugate against clinical strains of Acetinobacter baumanii as well as two other gram-negative bacteria (Escherichia coli and Pseudomonas aeruginosa).  Minimum Inhibitory Concentrations (MIC) are reported in micromolar in Table 2 (Liu et al. 2018).

  • The figure and table from this paper cannot be copied due to licensing restrictions. Since this paper is not freely accessible, you can request this paper from your library via interlibrary loan (or similar) or contact the corresponding author.  You can access the abstract on the article’s web page: https://pubs.acs.org/doi/10.1021/acs.jmedchem.8b00218

7.1.2. Questions

  1. Match the component with its function in this experiment. (1 = siderophore; 2 = cephalosporin; 3 = oxazolidinone;  4 = siderophore transporter; 5 = iron-depleted medium)
    1. ________ enables active uptake in an iron-depleted environment
    2. ________ releases the antibiotic in the periplasm
    3. ________ is the medium necessary to cause siderophore uptake
    4. ________ is the linker between the siderophore and the target antibiotic
    5. ________ is the target antibiotic
  2. This experiment reports data as a Minimum Inhibitory Concentration (MIC).  What is a MIC, and what does this tell us about drug effectiveness?
    1. MIC is the drug concentration that inhibits 100% of normal growth; it tells us the effective dosage range for an antibiotic.
    2. MIC is the lowest drug concentration that inhibits growth of a microorganism; it tells us whether an antibiotic is effective.
    3. MIC is the drug concentration that inhibits growth at 50% of its normal growth; it tells us whether an antibiotic is appropriate.
    4. MIC is the drug concentration that shows the highest growth inhibition; it tells us how to safely deliver the antibiotic.
  3. What is the main difference between gram-negative and gram-positive membranes that is relevant in this experiment? Pick all that apply.
    1. Gram-positive microbes have a periplasm and gram-negative microbes do not.
    2. Gram-negative microbes have two membranes while gram-positive microbes have one.
    3. Gram-positive microbes have efflux pumps to exclude antibiotics while gram-negative do not.
    4. Gram-positive have more internal pressure (80 pounds per square inch) than gram-negative.
  4. Minimum Inhibitory Concentration (MIC) data for different bacteria are shown in Table 2.  Which species and/or strain shows the greatest sensitivity across any of the drug conjugates tested? (1 = Strain 17961; 2 = Strain 1793; 3= Strain 1797, 4 = Strain 1800; 5 = E. coli; 6 = P. aeruginosa)
    1. All bacteria were equally sensitive to oxazolidinone alone, but none of the other combinations of conjugate.
    2. Bacteria 1 thru 4 were equally sensitive to siderophore + cephalosporin and siderophore + cephalosporin + oxazolidinone.
    3. Bacterium 5 was equally sensitive to siderophore + cephalosporin and siderophore + cephalosporin + oxazolidinone.
    4. Bacteria 5 and 6 were equally sensitive to cephalosporin + oxazolidinone and siderophore + cephalosporin + oxazolidinone.
  5. Based on the data in Table 2, what conclusions can you make regarding A. baumanii strain 17961 (row 1)?
    1. A. baumanii strain 17961 is sensitive to cephalosporin only when delivered via siderophore.
    2. A. baumanii strain 17961 is sensitive to oxazolidinone only when delivered via siderophore.
    3. A. baumanii strain 17961 is sensitive to cephalosporin even when not delivered via siderophore.
    4. A. baumanii strain 17961 is sensitive to oxazolidinone even when not delivered via siderophore.
  6. Based on the data in Table 2, what conclusions can you make regarding A. baumanii strain 1797 (row 3)?
    1. A. baumanii strain 1797 is sensitive to cephalosporin only when delivered via siderophore with a cleavable linker.
    2. A. baumanii strain 1797 is sensitive to oxazolidinone only when delivered via siderophore with a cleavable linker. 
    3. A. baumanii strain 1797 is sensitive to cephalosporin regardless of siderophore delivery and linker presence.
    4. A. baumanii strain 1797 is sensitive to oxazolidinone regardless of siderophore delivery and linker presence.
  7. If the siderophore active transport mechanism was disabled in these bacteria, what effect would you predict for the results in Table 2. Pick all that apply.
    1. The MIC values in the column marked as 3 (oxazolidinone) would be similar to those in column 1 (siderophore + cephalosporin + oxazolidinone).
    2. The MIC values in the column marked as 26 (siderophore + cephalosporin) would be similar to those in column 23 (cephalosporin).
    3. The MIC values in the column marked as 23 (cephalosporin) would be similar to those in column 26 (siderophore + cephalosporin).
    4. The MIC values in the column marked as 1 (siderophore + cephalosporin + oxazolidinone) would be similar to those in column 22 (cephalosporin + oxazolidinone).
  8. If each conjugate was pre-treated with beta-lactamase prior to MIC determination, what effect would you predict for the results in Table 2. Pick all that apply.
    1. The MIC values in the column marked as 22, 23, 26, and 1 would all have small MIC because they all contain cephalosporin, which is activated by beta-lactamase.
    2. The MIC values in the column marked as 26 (siderophore + cephalosporin) would increase to the level of those in column 23 (cephalosporin).
    3. The MIC values in the column marked as 1 (siderophore + cephalosporin + oxazolidinone) would increase to the level of those in column 22 (oxazolidinone + cephalosporin).
    4. The MIC values in the columns marked as 3, 22, 27, and 1 would all have equally large MIC because they all contain oxazolidinone, which is inactivated by beta-lactamase.

7.2. Second Figure Reading Exercise

7.2.1. Learning Objectives

Students will be able to:

  • Identify and quantify mutant phenotypes based on microscopy images.
  • Evaluate control high pressure liquid chromatography (HPLC) data to make conclusions about the ability to distinguish RNA and DNA nucleosides.
  • Analyze nucleic acid HPLC data and make conclusions about which microbial genomes possess RNA and/or DNA nucleotides.
  • Evaluate nucleic acid HPLC data and make conclusions about whether contaminating RNA can explain the presence of RNA nucleotides in a genomic lysate.
Experimental Background (Mehta et al. 2018, Figure 1)

Nucleic acids are the genetic material for organisms and viruses. Due to the similarity of DNA and RNA, scientists have speculated that RNA can function as a template for gene expression and encode genotype data.  The authors of this study (Mehta et al 2018) had previously engineered two E. coli strains that exchanged deoxycytosine for a modified RNA-like base.  In this study, they further mutagenized these strains in an attempt to generate E. coli with ribonucleotide components in their genome.   They found bacteria that were distinct in their appearance and confirmed them to be E. coli then set about characterizing this new strain using light microscopy (panel A).  They compared the original strain (DH108), an intermediate mutant strain (hmC-3) and a fully mutant strain (HR-1). Next, they examined the nucleosides present in the original strain (DH108) and the fully mutant strain (HR-1) using high pressure liquid chromatography (HPLC) (panel B).  Finally, they assayed for the possibility of RNA contamination in their genomic DNA samples by first spiking a genomic DNA lysate with either E. coli RNA or different amounts of RNA nucleotides and then continuing through the DNA isolation procedure and examining the purified genomic DNA for ribonucleotides by HPLC (panel C).

  • The figure from this paper cannot be copied due to licensing restrictions. Since this paper is not freely accessible, you can request this paper from your library via interlibrary loan (or similar) or contact the corresponding author.  You can access the abstract on the article’s web page: https://pubs.acs.org/doi/10.1021/jacs.8b07046

7.2.2. Questions

  1. E. coli typically have a rod-shape (bacillus) when viewed under a light microscope.  Based on the hmC-3 strain’s microscopic image (central panel A), what fraction of this strain has an abnormal shape phenotype and what is that phenotype?
    1. They are round-shaped (cocci) and are represented at 75%
    2. They are round-shaped (cocci) and are represented at 25%
    3. They are rod-shaped (bacillus) and are represented at 25%
    4. They are rod-shaped (bacillus) and are represented at 75%
  2. As a set of controls, the authors used HPLC to identify the retention times of different nucleosides using standards (panel B) containing only deoxyribonucleosides (blue trace) or ribonucleosides (orange trace). What conclusions can you make from these control data? [pick all that apply]
    1. RNA nucleosides can be distinguished from DNA nucleosides.
    2. Only RNA has uracil ribonucleoside and only DNA has thymine deoxyribonucleoside.
    3. Only RNA has uracil ribonucleoside and only DNA has adenine deoxyribonucleoside.
    4. Only guanine and adenine RNA nucleosides can be distinguished from DNA nucleosides.
  3. Having identified the HPLC retention times for the nucleosides, the authors next used HPLC to identify the different nucleosides in two genomic samples (panel B), one from E. coli strain DH10B (black trace) and E. coli strain HR-1 (red trace).  Which of the following statements are true for ribonucleoside presence in the genomes? Pick all that apply.
    1. rCytosine in both HR-1 and DH10B
    2. rGuanine in HR-1, but not DH10B
    3. Uracil in DH10B, but not HR-1
    4. rAdenine in HR-1, but not DH10B
  4. Having identified that RNA nucleosides are present in one of the genome samples, they next wanted to investigate whether the RNA nucleosides they observed might be due to cellular RNA contaminating the genome sample (panel C). To do this, they spiked a genomic lysate with E. coli RNA (green trace) or RNA nucleotides at two different concentrations (blue and orange traces), then performed the remainder of the genomic DNA extraction and analyzed the nucleic acid using HPLC.  What conclusion can you make about RNA contamination of the genomic sample and what data are your evidence?
    1. There is no RNA contamination; The orange, blue and green traces do not show RNA nucleotides (red trace).
    2. There is RNA contamination; The orange, blue and green traces are different from the control (red trace).
    3. There is no RNA contamination; The green trace is different than the orange and blue traces (controls).
    4. There is RNA contamination; The red trace shows RNA and is different than the control traces (orange, blue, green).

8. Paper Information and Licensing

8.1. Snippet paper

  • Liu R, Miller PA, Vakulenko SB, Stewart NK, Boggess WC, Miller MJ. 2018. A Synthetic Dual Drug Sideromycin Induces Gram-Negative Bacteria to Commit Suicide with a Gram-Positive Antibiotic. Journal of Medicinal Chemistry. 61(9):3845-3854. https://doi.org/10.1021/acs.jmedchem.8b00218
  • This article is not licensed for Creative Commons use. Thus, the abstract and figures cannot be copied here.  Since this paper is not freely accessible, you can request this paper from your library via interlibrary loan (or similar) or contact the corresponding author.  You can access the abstract on the article’s web page: https://pubs.acs.org/doi/10.1021/acs.jmedchem.8b00218

8.2. Main paper

  • Mehta AP, Wang Y, Reed SA, Supekova L, Javahishvili T, Chaput JC, Schulz PG. 2018. Bacterial Genome Containing Chimeric DNA-RNA Sequences. Journal of the American Chemical Society. 140(36):11464-11473. doi: 10.1021/jacs.8b07046 
  • This article is not licensed for Creative Commons. Thus, the abstract and figures cannot be copied here.  Since this paper is not freely accessible, you can request this paper from your library via interlibrary loan (or similar) or contact the corresponding author.  You can access the abstract on the article’s web page: https://pubs.acs.org/doi/10.1021/jacs.8b07046

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Podcast Annotation and Resources in Microbiology Copyright © 2025 by Rebecca Seipelt-Thiemann; Nancy Boury; Gwendowlyn S. Knapp; Amaya Garcia Costas; and Patrick Armstrong is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.

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