Exams 1-3 Q+A

Question 1

What bacterial system did we cover benefits from being a very small cell?

Answer 1

Chemotaxis

Question 2

What is the process of carbon fixation? What kind of microbes can accomplish this?

Answer 2

Carbon fixation is the process of adding a CO2 to an organic molecule. Microbes that can do this are called autotrophs

Question 3

What molecules make up peptidoglycan?

Answer 3

N-acetyl glucosamine, d-alanine, n-acetylmuracim acid, A diamino acid

Question 4

Why is the presence of lysine or diaminopimelic acid important to the integrity of peptidoglycan?

Answer 4

They allow for cross-linking of adjacent peptidoglycan chains

Question 5

What phenotype would you expect to be associated with a population of bacteria without functional autolysin enzymes?

Answer 5

They will still be metabolically active, motile, and chemotactic, however they would never divide into separate daughter cells. They would also be resistant to beta-lactam antibiotics (ex: penicillin) and glycopeptide antibiotics (ex: vancomycin)

Autolysin is the enzyme that nicks the cell wall, allowing a new peptidoglycan molecule to fill in the gap.

Question 6

What is one of the most common mechanisms of acquired resistance to beta-lactam antibiotics?

Answer 6

Enzymatic digestion of the antibiotic

(acquisition of genes for beta-lactamases)

Question 7

What would your gram stain look like if you forgot to do the safranin counterstain on a mixture of both gram positive and gram negative bacteria?

Answer 7

You’d see purple bacteria only

Question 8

Why won’t mycobacteria stain in a gram staining procedure?

Answer 8

Because their arabinogalactan repulses the stains

Remember that TB is a mycobacteria, and for a long time we couldn’t prove that TB was caused by a bacterium bc I just didn’t show up in gram stains. Instead, we had to develop Acid-Fast stains to get the dye past the mycolic acid coat and visualize the fuckers

Question 9

Why were Fred Griffith’s rough Strep. pneumoniae mutants unable to kill the mice?

Answer 9

Because the rough mutants lacked capsules, they couldn’t avoid phagocytosis by the mouse WBCs.

Question 10

What happens to CheA when an MCP is engaged with it’s cognate chemoattractant?

Answer 10

CheA will decrease phosphorylation of its histidine

When the chemoattractant is present, MCP inactivates CheA, preventing it from phosphorylating its histidine, preventing the start of the phosphorylation cascade, allowing the flagellar basal body to keep rotating

Question 11

What happens to CheY-PO4 levels when CheA decreases phosphorylation of its histidine?

Answer 11

CheY-PO4 levels decrease

Question 12

What kind of flagellar rotation is stimulated when MCPs are engaged with their cognate chemoattractant?

Answer 12

The tumbling response is suppressed, CCW rotation

Question 13

What is the mechanisms of TcdA and TcdB toxins in C. diff?

Answer 13

Inactivation of a host protein using a glucose molecule

Question 14

How do bacteria perform chemotaxis? (don’t mention Che proteins or MCP in this answer!)

Answer 14

Bacteria suppress their tumbling response in the presence of a chemoattractant.

Question 15

What is the inherent chemical difference in peptidoglycan structure between Staph. aureus (gram positive coccoid) and E. coli (gram negative)?

Answer 15

Staph aureus is crosslinked with a 5-glycine pentapeptide interbridge between the L-lysine of each peptidoglycan molecule.

E. coli is crosslinked with a singular peptide bond between the diaminopimelic acids of each peptidoglycan molecule

Question 16

Why must some proteins fold in the periplasm rather than the cytoplasm?

Answer 16

The disulfide bonds necessary for proper folded structures cannot be formed in the cytoplasm’s reducing environment

Question 17

Why is the constant movement of motile bacteria important in the process of chemotaxis?

Answer 17

Bacteria sense chemical gradients temporally, and need to move constantly through their environments to sense and respond to chemical signals

Question 18

What is the phenotype of an E. coli with the following protein deletion: CheZ

Answer 18

Nearly constant tumbling

CheZ the phosphatase that removes CheY’s phosphate, disengaging it from the basal body and halting the induced tumble. Without CheZ, CheY wouldn’t have the short half-life necessary to regulate the tumbling response.

Question 19

What is the phenotype of an E. coli with the following protein deletion: CheA

Answer 19

No tumbling

CheA is the histidine kinase that starts the phosphorylation cascade that induces tumbling when chemoattractants are absent. Without CheA’s phosphorylation activity, no tumbles would ever be induced.

Question 20

What is the phenotype of an E. coli with the following protein deletion: CheY

Answer 20

No tumbling

CheY is the response regulator that, when phosphorylated, jams the basal body to make it stop rotating, inducing a tumble

Question 21

What is the phenotype of an E. coli with the following protein deletion: CheW

Answer 21

Chemotaxis is much less efficient

CheW is the adaptor protein that connects all the MCPs together and integrates their signals.

Question 22

What is the phenotype of an E. coli with the following protein deletion: CheB

Answer 22

Chemotaxis is much less efficient

CheB is the response regulator methylesterase that resets CheR’s methylation of the MCP by chewing methyl groups of MCP glutamic acids when CheA transfers it a phosphate. Without CheB the cell wouldn’t be able to reset the methylation and respond to low levels of chemoattractant again

Question 23

What is the phenotype of an E. coli with the following protein deletion: CheR

Answer 23

Chemotaxis is unaffected

CheR puts the methyl groups onto the MCPs glutamic acids, allowing the cell to accommodate and respond to higher and higher levels of chemoattractant. Without CheR chemotaxis would be unaffected, but the process of accommodation would be.

Question 24

What is the process of accommodation in the context of bacterial chemotaxis?

Answer 24

Accommodation allows a bacteria in an intermediate chemoattractant concentration to seek out and respond to higher chemoattractant concentrations

Overtime, as CheA remains inactivated from attachments bound to their respective MCPs, CheR methylates the MCPs glutamic acids, changing the receptors’ conformation and making it more difficult for the attractant to bind. The MCPs now require higher concentrations of the attractant to inactivate CheA at the same rate, pushing CheA to activate and induce a CW tumble

Question 25

Give three examples of unusual aspects of the peptide portion of peptidoglycan

Answer 25

1. Alternating L and D isoforms of amino acids. (eukaryotes and archaea can use only L isoforms of amino acids 2. Contains one diamino acid, leading to branching. (this kind of branching isn't seen in other peptides) 3. The proteins aren't synthesized ribosomally

Question 26

Would a vancomycin resistant staph aureus infection be sensitive to penicillin/ampicillin? Remember that vancomycin resistant bugs have replaced the D-alanyl-D-alanine with two lactic acids at the terminus of their 5 amino acid peptide in peptidoglycan.

Answer 26

If the D-alanyl-D-alanine has changed, the transpeptidases would have to change also to allow the transpeptide bonds to be formed. This altered transpeptidase likely will not bind to beta-lactam antibiotics, rendering drugs like penicillin and ampicillin ineffective

Question 27

Why or why not would you treat MRSA with augmentin?

Answer 27

MRSA is resistant to beta-lactam antibiotics because it has altered its transpeptidase such that it binds to D-alanyl-D-alanine without binding to any beta-lactams. Augmentin is designed to target a different antiotic mechanism: beta-lactamases. Its job is to bind to beta-lactamases defensively excreted by the bacteria as a distraction so other beta-lactam antibiotics can bind to transpeptidase and inhibit crosslink formation. Augmentin is useless against any other antibiotic resistance mechanisms.

Question 28

How is energy directly generated in the EMP glycolytic pathway?

Answer 28

Substrate level phosphorylation

Question 29

Where is reducing power generated in the EMP glycolytic pathway?

Answer 29

Generated when NAD+ is reduced to NADH. This happens when glyceraldehyde-3-P is changed into 1,3-bisphosphoglycerate

Question 30

What are the most likely fates of the electrons associated with NADH+H+?

Answer 30

Placed on pyruvate in a fermentation reaction or Placed onto a dehydrogenase and into electron transport

Question 31

What are the most likely fates of the electrons associated with NADPH?

Answer 31

Use in anabolic processes NADPH isn't involved with fermentation or the ETC, it's only involved with anabolic processes

Question 32

Under which circumstances is it advantageous to utilize the ED path rather than the EMP?

Answer 32

When the bug is in need of anabolic processes. ED produces 1 NADH, 1 NADPH, and 1 ATP. That NADPH can go into anabolic processes. EMP produces only 2 NADH and 2 ATP, neither of which can go into anabolic processes

Question 33

What does oxidative phosphorylation use as its direct energy source?

Answer 33

PMF. The correct answer is NOT NADH because NADH doesn't directly generate energy. It reduces the first protein in the ETC cascade, which allows those proteins to generate PMF further down the line. The last protein, ATP synthase, uses that PMF to create ATP, the actual energy source.

Question 34

What is the goal of all fermentation?

Answer 34

To oxidize NADH to regenerate NAD+ pools

Question 35

True or false: the concentration of acyl homoserine lactone inside the bacterium is equal to the concentration outside the cell

Answer 35

True. Thank you acyl side chain.

Question 36

Why do bacterial mRNA transcripts encoding proteins require a 5' UTR (aka leader RNA)?

Answer 36

Because the 5' UTR contains a Shine Dalgarno sequence

Question 37

You have two distinct and distant operons, A and B, each with the same promoter sequence driving their transcription in an E. coli cell. Operon A has 50x the amount of mRNA as operon B. How do you explain the difference in the volume of transcription from the exact same promoter?

Answer 37

Operon B is subject to attenuation

Question 38

What causes the switch in sabA gene expression in a population of H. pylori from phase off to phase on seen during phase variation?

Answer 38

A frameshift mutation

Question 39

The operators of the LexA regulon have varying affinities for the protein LexA. Why does umuDC have an operator for LexA that has a very high affinity for this protein?

Answer 39

It insures that umuDC is not expressed until the SOS response has been functioning for an extended period of time

Question 40

For what kind of pathway would enzyme induction be an appropriate means to ensure metabolic efficiency?

Answer 40

A catabolic pathway Ex: galactose catabolism, sorbitol catabolism

Question 41

In low levels of tryptophan-tRNA molecules, what kind of stem loop will form and what will happen to the transcription of the tryptophan operon?

Answer 41

A 2:3 stem loop will form and transcription will continue

Question 42

Under what circumstances would G6P be shunted into the PPP instead of ED or EMP

Answer 42

When there are high cellular levels of ATP

Question 43

What is the signal being sensed and responded to in tryptophan biosynthetic operon attenuation?

Answer 43

The presence or absence of the leader peptide

Question 44

What is the advantage of detecting when a quorum has been achieved in Staph. aureus?

Answer 44

Quorum dependent proteins can be made and secreted, and if they attract a host immune response at quorum their pop. is now large enough to avoid being completely wiped out.

Question 45

What is the advantage of catabolite repression in favoring glucose over all other sugars?

Answer 45

Glucose fits directly into the glycolytic pathways. All other sugars require modifications to undergo glycolysis and are therefore less efficient.

Question 46

In the EMP pathway, pyruvate kinase catalyzes the final step of glycolysis (phosphoenol-pyruvate is catabolized to pyruvate). Why is it logical for fructose-1,6-bisphosphate (an earlier intermediate in this path) to mediate this step while its absence prevents this step?

Answer 46

Frutose-1,6-bisphosphate allosterically modifies pyruvate kinase and allows it to catabolize PEP to pyruvate. This ensures that PEP is catabolized only when there's glycolytic activity happening. If no glycolytic activity is happening, PEP is needed to transport more sugars into the cell to generate more ATP, and would be wasted if instead it was all catabolized (it would produce less overall ATP than the import of new sugars would).

Question 47

Attenuation models work well for amino acid biosynthesis. Why do they not work as well for nucleic acid or fatty acid synthesis?

Answer 47

Attenuation of protein biosynthesis relies on differeing levels of amino acids, the monomers of proteins. Attenuataion of nucleic acids wouldn't work as well. Attenuation for lipids definitely wouldn't work, as they aren't made of amino acid monomers. Attenuation works because the end product of the pathway is involved in the attenuation signal's synthesis. Lipids have no part in the signal's synthesis, so attenuation can't regulate their biosynthesis. Nucleic acids are involved, however the attenuation mechanism would need to be altered for it to regulate their biosynethetic pathways

Question 48

You have manipulated the EnvZ/OmpR two component system in a bug such that in low osmolarity environments the system now phosphorylates 0% of OmpR but phosphorylates OmpR appropriately in high solute environments. What is the result of your experiment?

Answer 48

The bug wouldn't express any OmpF.

Question 49

In an E. coli cell with a defective micF gene (no expression at all), what is the effect on the bacterium's ability to acclimate to low solute environments?

Answer 49

No effect

Question 50

You substitute a serine for a histidine at the site of phosphorylation of the histidine kinase EnvZ in an E. coli, but disaster strikes as the bacterium can no longer express either OmpF or OmpC! Why?

Answer 50

Serine is phosphorylated via low energy phosphoester bonds while histidine is phosphorylated as high energy phosphoimidizole bonds, so the phosphorylation from serine to OmpR aspartate does not occur.

Question 51

Earth's atmosphere would become depleted of N2 without...

Answer 51

...denitrifiers

Question 52

What is the role of Type III secretion systems in rhizobium-legume mutualism?

Answer 52

Providing a check on plant defense mechanisms that might otherwise kill the bacteroid

Question 53

What materials flow from the plant to the bacteroid in legume-rhizobium mutualism?

Answer 53

Organic acids

Question 54

What materials flow from the bacterium to the plant cell in legume-rhizobium mutualism?

Answer 54

NH3

Question 55

What plant molecules are involved in the dialogue between the legume root and rhizobium bacterium?

Answer 55

Flavonoids

Question 56

What bacterial molecules are involved in the dialogue between the rhizobium bacterium and legume root?

Answer 56

Nod factors

Question 57

What are the shared characteristics between the agrobacterium tumefaciens-plant interaction and the rhizobium-legume interaction?

Answer 57

They are both plasmid driven and in both the plance cell provides nutrients for the bacterial cells

Question 58

What is the effect of the flavonoids on Rhizobium bacterial cells in the soil?

Answer 58

They allosterically modify a transcription factor They change bacterial gene expression They facilitate chemotaxis towards the plant

Question 59

A modified version of what process occurs in the process of crown gall tumor development in plants infected by Agrobacterium tumefaciens?

Answer 59

Conjugation

Question 60

What is the material that travels from Agrobacterium tumefaciens into the plant cell?

Answer 60

A nucleoprotein

Question 61

What is the role of bacterial opinse in the nutrition of a crown gall tumor plant cell?

Answer 61

They aren't! Opines are produced by the place cell, not the bacterium, and they feed the bacterium!

Question 62

You accidentally spill a toxic substance into your pure bacteria culture. Despite the toxin, a few bugs survive. Why?

Answer 62

The survivors could have restructured an integron. The survivors could have been living in a biofilm as metabolically inactive cells.

Question 63

What is the role of relaxase in bacterial conjugation?

Answer 63

To nick plasmid DNA at oriT to initiate Rolling Circle Replication

Question 64

How is an Hfr strain of E. coli derived from an F+ E. coli strain?

Answer 64

In an Hfr strain, the F-plasmid is incorporated into the host's bacterial chromosome

Question 65

Why is it advantageous for an integron to be controlled by the SOS response?

Answer 65

Because it gives rise to the integron's expression platform and ensures that genes in the low cost memory cassette reservoir aren't expressed until LexA levels drop very low, when the cell is really stressed.

Question 66

Why must there be a "microaerobic environment" associared with Rhizobia associated with legume root nodules, but not when the rhizobia are living free in the soil?

Answer 66

Nitrogenase can't function in the presence of oxygen, so when the rhizobia aren't undergoing nitrogen fixation and nitrogenase isn't needed, they don't need to create any anaerobic conditions or limit their aerobic conditions to nly the microaerobic pockets (leghemoglobin) necessary to survive.

Question 67

Why does the EnvZ/OmpR two componenet system regulate the expression of OmpC and OmpF? What is the advantage to E. coli of expressiong OmpC vs OmpF in one environment or another?

Answer 67

OmpF and OmpC are porin proteins. OmpF's pore is bigger than OmpC's pore, meaning that it has a higher ion flux rate and allows more stuff to enter the cell. This is useful in poorly osmotically concentrated environments where nutrients are harder to come by, but potentially dangerous in highly osmotically concentrated environments because the risk of uptaking toxins is higher. To mitigate this risk, E. coli switches to use a porin protein with a smaller pore, OmpC. The highly osmotically concentrated environment still allows the E. coli to take in nutrients, but the risk of uptaking toxins is now decreased

Question 68

Explain the paradox of nitrogen fixation being poisoned by O2, yet the bacterial genus rhizobium is an obligately aerobic bacterium.

Answer 68

Rhizobia use O2 as their terminal electron acceptor in their ETC. They will die without it. However, nitrogenase, the enzyme that fixes N2 --> NH3, is incredibly sensitive to O2 and won't function in its presence. This is fine when the rhizobia doesn't need to fix N2 --> NH3, bt a problem when it does. To solve this, the rhizobia uses leghemoglobin, a molecule that't only created when the rhizobia is in symbiosis with a legume. Leghemoglobin encases O2 and delivers it directly to the terminal step of their ETC, without the oxygen interacting with or upsetting the nitrogenase.

Question 69

What is the nature of the material passed from A. tumefaciens into a plant cell during crown gall tumor disease pathogenesis and how does this relate to plasmid replication?

Answer 69

The TDNA portion of the Ti plasmid is passed from the bacterium to the plant cell in Crown Gall Tumor formation. The Ti plasmid undergoes rolling circle replication, but generates a single stranded copy of only the TDNA region, not the entire plasmid. The TDNA strand is then pumped out of the bacterium and into the plant using a Type IV secretion system. This process is essentially a modified version of conjugation, except now it's selective about what genetic information is passed to the plant.