Exam 1

Question 1

Who was the father of taxonomy and binomial nomenclature?

Answer 1

Linnaeus

Question 2

What limits are there to binomial nomenclature?

Answer 2

It is based on observable traits and not how the organisms are related. Not suffieicnet for abcteria who only have a few observable traits (morphology, gram stain, sporulation, motility)

Question 3

Who used 16s rRNA to make phyologenies?

Answer 3

Carl Woese

Question 4

What makes 16s rRNA a good molecule for constructing phylogenies?

Answer 4

-Found in all organisms
-Fucntionally equivalent
-Homologous (derived from same ancestor)
-Not subject to lateral transfer
-Evolves slowly

Question 5

What can you learn from a universal tree of life?

Answer 5

-Most life is microbial
-Chloroplasts and mitochondria arose from symbiosis with Cyanobacteria and Proteobacteria,
-The Last Common Ancestor (LCA) was probably a thermophilic, anerobic microorganismrespectively

Question 6

How do you validate a universal tree of life?

Answer 6

1. By comparing it to observable traits
2. By using other molecules like ATPase, 23s rRNA, and ribosomal proteins

Question 7

List three physiological/structural features that distinguish the Bacteria, Archaea and Eukarya from one-another.

Answer 7

-Eukarya have a nucleus while the other two don’t.
-Eukarya has multiple linear chromosomes while the others only have one circular chromosome.
-Eukarya have introns.
-Archaea are ether-linked lipids while the others have ester-links.

Question 8

List three physiological/structural features that can be used to argue that Archaea and Eukarya belong on the same branch of the Tree to the exclusion of Bacteria.

Answer 8

Neither has peptidoglycan. Both have histones and both use methionine for tRNA initiation while bacteria uses f-met

Question 9

What are some limitations of 16s rRNA analysis?

Answer 9

-Cannot differentiate between closely related organisms
-Cannot account for horizontal gene transfer

Question 10

How do you calculate dilution rate?

Answer 10

F/V flow rate/volume of culture vessel

Question 11

How do you calculate the specific growth rate?

Answer 11

M=D or ln2/td (doubling time) needs to be in hours

Question 12

What can you do in a chemostat but not batch culture?

Answer 12

Study physiology of slow growing
cells
* In the lab, E. coli doubles in ~30 min, but in the intestine ~12 hours!
* Mimic nature
* Feed toxic compounds
* Create microcosms (e.g., predator-prey dynamics with bacteria and
phage)
* Select certain mutants, like high affinity for a substrate.

Question 13

What does growth rate effect?

Answer 13

cell size, gene expression, and cell structure

Question 14

How do you calculate doubling time?

Answer 14

take 2 points on the semilog graph. minutes/# doublings

Question 15

Why is doubling time important?

Answer 15

By comparing td under different growth conditions, you can figure
out what your organism prefers (37°C or 20°C? Glucose or
cellulose? Ammonia or alanine?)
* This gives clues to how the organism makes a living.
* By comparing WT to a mutant you can learn about the function of
specific genes.
* By keeping track of td you can tell whether your growth conditions
are same this month as last month, or the same in your lab as in
another lab.

Question 16

Advantages of colony counts

Answer 16

Distinguishes live from dead cells
➢ Can detect low numbers of bacteria

Question 17

Disadvantages of colony counts

Answer 17

But note that CFU is not necessarily the same as number of cells!
➢ Large errors if CFU used for bacteria that grow in clusters or chains

Question 18

Advantages of direct count

Answer 18

Works for clumps/chains

Question 19

Disadvantages of direct counts

Answer 19

➢ Poor detection limit: You need at least 106 cells/ml to see any in the
chamber, so not suitable for dilute cultures.
➢ Slow and laborious.
➢ Does not distinguish live cells from dead ones

Question 20

Disadvantages of OD

Answer 20

• OD measures mass, not number, so if turbidity doubles
  you could have twice as many cells or the same number of
  cells but they got twice a big.
• OD detection limit about 107 cells/ml, so not suitable for
  dilute cultures (low population density)
• OD increase is no longer linear at OD600 above approx. 0.8
• Does not distinguish live versus dead cells

Question 21

Batch culture pros

Answer 21

• Easy
• Reproducible (concept of “balanced growth”)
• Can get large samples of cell material for experiments/measurements

Question 22

Batch cultures cons

Answer 22

• Limited time window to catch your samples
• Hard to study growth under conditions of nutrient limitation

Question 23

You suspect that a bacterial culture contains approximately 108 cells/ml. How would you dilute and
plate this culture to obtain a countable plate in CFU assay?

Answer 23

-A countable plate has about 100 colonies.
-Dilute 1 ml culture into 9 ml fresh media. Now the concentration is 107 cells/ml.
-Repeat several times for a total of 5 dilution steps. You are down to 103 cells/ml.
-Spread 0.1 ml on a plate, incubate overnight, count colonies the next day. Alternatively, you can do a
further dilution and plate 1 ml.

Question 24

Theme 1 of metabolism

Answer 24

A reduced compound gets oxidized, releasing energy
that is captured as ATP

Question 25

Theme 2 of metabolism

Answer 25

Cellular redox reactions require electron carriers
* NAD+ is a diffusible electron shuttle in the cytoplasm
* Quinones are a diffusible electron shuttle in membranes
* Iron-sulfur clusters, heme and flavins are incorporated into
proteins (so not diffusible unless the protein diffuses).

Question 26

Theme 3 in metabolism

Answer 26

Energy is captured in two ways
* Ion gradient across a membrane (e.g., proton motive force)
* “High energy bonds” as in ATP
These can be used directly but are also interconvertible

Question 27

What is the role of resonance stabilization in energy?

Answer 27

Products are more stable than reactants (ATP) giving them a much lower energy and resulting in a large release of energy
Ex: NAD+ is a key electron carrier in metabolism
NAD+ has more resonance stabilization that NADH
Therefore, it is “hard” to put electrons into NAD+, but once they are
there it is “easy” for NADH to donate those electrons to something else,
like an electron transport chain.
One can say NADH is a carrier of “high energy electrons”

Question 28

Them 4 in metabolism

Answer 28

Two mechanisms of ATP synthesis
➢ Substrate level phosphorylation
➢ Harnessing a proton motive force (PMF)

Question 29

What is glycolysis

Answer 29

converts glucose to pyruvate while producing a little ATP and a
little NADH.

Question 30

What is the most common pathway for glycolysis in bacteria?

Answer 30

mbden-Meyerhof
Pathway (EMP) the most common pathway for converting
glucose to pyruvate.

Question 31

Steps in glyoclysis

Answer 31

-Phosphorylate glucose twice and
isomerize it to fructose
-Split it down the middle to produce
triose phosphates
-Oxidize the triose phosphates with
concomitant formation of high-energy
phosphate bonds
-Harvest ATP from the doubly
phosphorylated triose phosphates

Question 32

Fermentation themes

Answer 32

1. Recovery of NAD+ that was reduced elsewhere in pathway
2. Stepwise reduction of carbonyl groups to alcohols
3. Sometimes generate an ATP if possible, usually via a CoA intermediate
   (SLP)

Question 33

Conserved things in fermentation

Answer 33

Conserved features of these pathways are
* Redox balance
* Carbon redox patterns (e.g., acid to aldehyde to alcohol)
* A site for energy conservation by substrate level
phosphorylation
* Variable features of these pathways are the specific
substrates, intermediates and end products

Question 34

What is the Stickland Reaction?

Answer 34

(fermentation of amino acids by some Clostridia)
Ferment amino acid pairs
One acts as the e-
donor, the other as the
e- acceptor
Alanine/glycine pair is
textbook exampleAlanine oxidized to acetate + C02
Glycine reduced to acetate
Themes: recovery of NAD+, R-CoA
intermediate allows synthesis of ATP by
SLP.

Question 35

Why can’t glycolysis end with 2 ATP and 2 NAD?

Answer 35

NADH needs to be regenerated

Question 36

TCA or Krebs cycle

Answer 36

1. Pyruvate from glycolysis
   converted to acetyl-CoA
   and fed into TCA cycle
2. Two carbons will be lost
   as CO2 as the cycle turns
3. Two turns per glucose
4. Direct synthesis of ATP
   (via Succinyl-CoA
   intermediate and GTP)
5. Generate 1 FADH2
6. Generate 3 NADH + H+
   Each FADH2 will make 2 ATP in the
   ETC, while each NADH will make 3

Question 37

Electron acceptors in respiration

Answer 37

O2, SO4, CO2, NADP

Question 38

During glycolysis by the Embden-Meyerhof pathway a total of 4 ATP are produced, but the pathway
is said to yield only 2 ATP per glucose. Explain this discrepancy.

Answer 38

Two ATP are invested at the start of the pathway, so the net gain is 2 ATP.

Question 39

Which of the following produces CO2?
Glycolysis
TCA Cycle
ETC with O2 as final electron acceptor
Lactic acid fermentation
Stickland fermentation of an alanine-glycine pair

Answer 39

Which of the following produces CO2?
Glycolysis -NO
TCA Cycle-YES
ETC with O2 as final electron acceptor - NO
Lactic acid fermentation – NO
Stickland fermentation of an alanine-glycine pair - YES

Question 40

Consider a hypothetical E. coli mutant that that cannot transfer electrons from NADH to O2 because it
lacks a functional ETC. Would such a mutant grow in minimal media containing glucose as the energy
source? Explain your reasoning.

Answer 40

Yes, such a mutant can grow by glucose fermentation. As noted in one of the slides, E. coli carries out
the mixed acid fermentation, but the question did not specify that you include the name of the pathway.

Question 41

In the reaction H2 + ½ O2 → H2O, what is the electron donor and what is the electron acceptor? What
is the redox state of the H and the O at the start? At the end?

Answer 41

H2 is the donor and its redox state is 0 at the start, +1 at the end.
O2 is the acceptor and its redox state is 0 at the start, -2 at the end.

Question 42

Would you expect Clostridium difficile to grow on minimal media with alanine as the sole source of
energy? What about E. coli? Explain your reasoning. You can assume C. difficile is incubated
anaerobically but the E. coli culture is aerobic.

Answer 42

C. difficile→ No, it requires an electron acceptor such as glycine to ferment alanine.
E. coli → Yes, alanine gets deaminated to pyruvate (look at the C. dif pathway I presented in class), and
pyruvate can enter the TCA cycle.

Question 43

E. coli can respire glucose with either oxygen O2 or nitrate (NO3-) as terminal electron acceptor. If
given a choice, which do you think would be preferred? Use the Redox Tower from the lecture slides to
justify your answer.

Answer 43

O2 is preferred because the energy drop is larger. The E0’ for the Nitrate/nitrite couple is +0.42 but for
the O2/H20 couple it is +0.82.

Question 44

Recall that the PMF powers not only ATP synthesis but also many transporters and flagella. How
does a bacterium like E. coli generate a PMF when it is growing by fermentation?

Answer 44

Some of the ATP from fermentation is hydrolyzed by the F1F0 ATP “synthase” to generate a PMF. In
other words, the F1F0 ATP synthase runs backwards.

Question 45

How many CO2 are released per pyruvate in the TCA cycle? How many NAD+ and how many FAD are
reduced?

Answer 45

2 CO2, 3 NADH and 1 FAD

Question 46

What is the nernst equation

Answer 46

ΔG°‘ = -nFΔE0‘
= -2 x 96.48 kJ/V x 1.24 V
= -239 kJ

Question 47

What change in energy is minimally required for an organism

Answer 47

The cutoff for growth is about -30 kJ/mol

Question 48

How do black smokers work

Answer 48

Superheated water
(350°C) comes up
from the mantle
carrying dissolved
minerals. The
minerals ppt when the
hot water mixes with
cold sea water (2°C),
creating black
“smoke”. Sea water
has lots of oxygen,
allowing for bacteria
to respire the
reduced compounds
emerging from the
smoker. Sulfide
oxidation is most
important. The
bacteria also fix CO2

Question 49

How do the tube worms live?

Answer 49

Symbiosis with sulfur-oxidizing bacteria. The tubeworms have an internal organ called the
“trophosome” that houses vast numbers of
sulfur-oxidizing bacteria

Question 50

How does the tube worm symbiosis work?

Answer 50

• CO2, O2 and H2S absorbed into blood in
  brachial plume; blood delivers these inorganic
  compounds to bacteria in trophosome (shown in
  blue)
• Bacteria oxide H2S to produce ATP (an
  example of aerobic respiration).
• The ATP (and reductant) are used to fix CO2
• Major N source is nitrate (NO3-), which the
  bacteria reduce to ammonia (NH3) and
  incorporate into amino acids.
• Some of the organic molecules made by the
  bacteria pass into the blood; the tubeworm
  uses these for energy and biosynthesis

Question 51

Describe the rumen’s bacterial content

Answer 51

100-150 Liters
39°C
pH 5.5-7
Anaerobic
Well-mixed
Typical cow rumen
contains ~400
bacterial species at
~1010 bacteria per ml.

Question 52

Rumen biochemistry

Answer 52

cellulose is converted to glucose using cellulase from bacteria. Feremnting bacteria then turn glucose into volatile fatty acids which go into the cows blood stream as energy. Methanogenic archaea take the leftover CO2 and H to make methane

Question 53

Why are methanogens important?

Answer 53

Without methanogens the
pathway grinds to a halt
and the ruminant starves
because the build up of H2
prevents fermentation.
The H2 is an electron sink
that the fermenters need
to recover their NAD+.

Question 54

How does salmonella produce inflamamtion

Answer 54

Neutrophils migrate into the lumen of
the intestine, where they release
reactive oxygen species (ROS)
* ROS oxidizes thiosulfate to
tetrathionate
Bottom line: Salmonella provokes host inflammation to create its own
private electron acceptor, thereby gaining a huge growth advantage
over the “healthy” gut bacteria!

Question 55

What are VFAs, how are they made in the rumen, and how does the ruminant benefit from them?

Answer 55

VFA = volatile fatty acids. Short-chain compounds like acetate, propionate, butyrate. They are
produced by bacteria that ferment sugars released from cellulose and other complex
carbohydrates derived from plant material. The VFAs diffuse across the rumen wall and enter
the cow’s blood stream, ultimately to become a major source of energy for the cow which would
otherwise be unable to digest cellulose.

Question 56

How does the microbial consortium in the rumen allow herbivores to get energy from complex plant
polysaccharides?

Answer 56

Cellulose degrading microbes convert cellulose to free sugars, which are then taken up and
fermented to VFAs by fermenting bacteria. The VFAs provide energy to the cow, as described
above (ultimately they enter the TCA cycle).

Question 57

Explain what is meant by producer, primary consumer and secondary consumer. Provide an example
of each from the hydrothermal vent ecosystem.

Answer 57

Producer = autotroph. Lithoautotrophic bacteria like the one that lives in Riftia trophosomes and
gets energy by oxidizing sulfide to sulfate is an example.
Primary consumer = an organism that eats the producers. Sea vent crabs eat bacteria.
Secondary consumer = an organism that eats the primary consumers. An octopus that eats the
crabs is an example.

Question 58

Diagram Riftia pachyptila and use your cartoon to explain how an animal with neither a mouth nor an
anus can obtain “food.”

Answer 58

H2S, O2, CO2 and NO3- are taken up by the worm in the gill plume, which is highly vascularized
and exposed to sea water. These chemicals are abundant in the vicinity of deep sea
hydrothermal vents because they are dissolved in superheated water that comes up from
magma below. (Except for O2, which dissolves into sea water at the ocean surface.) The
chemicals are of no use to the tube worm (except for O2—the tube worm is an aerobe with a
metabolism like ours). The worm’s circulatory system transports the chemicals to the
trophosome, where they diffuse into cells called bacteriocytes. Bacteriocytes are packed with
symbiotic Proteobacteria that get their energy by oxidizing H2S with O2 as terminal electron
acceptor. While most of the electrons liberated from H2S are used to make a PMF and thus
ATP, some are diverted to biosynthesis, such as reduction of CO2 to make sugars. This is
possible because the bacteria are autotrophs and have a Calvin Cycle like plants do.
Meanwhile, the bacteria also reduce nitrate to ammonia, which they incorporate into glutamate
and glutamine. From there the N can be transferred to various other N-containing molecules.
Some of the carbon compounds and amino acids are exported to the Riftia. In summary, Riftia
“feeds” the bacteria small molecules (H2S, O2, CO2 and NO3-). The bacteria are amazing
biochemists who can turn these simple compounds into all the complex molecules that make up
their cells. In return the bacteria “feed” their host sugars and amino acids.

Question 59

The bacteria found as intracellular symbionts in bacteriocytes of Riftia pachyptila can be isolated and
grown in pure culture in the lab. If you were designing a growth medium to isolate these organisms,
what would you use as the energy source? As the carbon source? Would you incubate your plates
aerobically or anaerobically? Explain your reasoning.

Answer 59

• Energy source = H2S, the preferred electron donor
• C source = CO2 because the organism is an autotroph and fixes CO2 using the Calvin
  Cycle like plants do. So, would not add glucose (or the like) to the plates!
• Incubate aerobically because O2 is the terminal electron acceptor for sulfide oxidation.
  Contrary to popular belief, there is plenty of O2 in seawater. Our lungs are not designed
  to extract dissolved O2, so we can’t breathe under water, but the O2 is there.

Question 60

Salmonella in the gut can obtain energy from fermentation of carbohydrates or from respiring
carbohydrates with tetrathionate as terminal electron acceptor. Where does the tetrathionate come
from?

Answer 60

Anaerobic bacteria in the lumen of the small intestine make sulfide, H2S, as a byproduct of their
metabolism. Because sulfide is toxic, our cells oxidize it to the harmless form thiosulfate,S2O32-.
In response to a Salmonella infection our body mobilizes neutrophils, which transmigrate across
the intestinal epithelium and release “reactive oxygen species” (ROS) in the lumen of the small
intestine. The ROS is intended to kill invading pathogens, but some of it reacts with thiosulfate,
converting it to tetrathionate, S4O62-.

Question 61

magine that you infect a mouse orally with WT and ttrA mutant Salmonella (ttrA encodes one of the
subunits of tetrathionate reductase). You observe that the animals dosed with WT Salmonella develop
diarrhea, whereas those that received the ttrA mutant do not. How do you explain these findings?

Answer 61

he ttrA mutant Salmonella probably can’t colonize the gut because this mutant would not be
able to use tetrathionate as terminal electron acceptor. Salmonella can grow by fermentation,
but it is not very competitive against the dedicated fermenters (mostly Bacteroides and
Clostridia) that make up our normal intestinal microbiota.