4. Growth and Environment

Question 1

What are the four phases of growth?

Answer 1

Lag phase, exponential phase, stationary phase, and death phase.

Question 2

In the equation describing log growth, what does g represent and what does k represent?

Answer 2

G equals the time it takes for a population to double, and K equals one over G, which is the growth rate (number of generations per unit time).

Question 3

Describe the direct cell count technique of measuring population growth.

Answer 3

It involves using a microscope to count cells in unknown volume of sales suspension period is also known as a total cell count because it can’t tell the difference between dead and living cells.

Question 4

Describe the indirect cell count/the plate count method of measuring population growth.

Answer 4

Unicellular organisms are put on the surface of an agar solidified nutrient medium. Following an incubation period, the colonies are counted. This provides an indirect measure of the number of cells in the cell suspension before it was plated. A cell suspension has to be diluted before plating in order to ensure that at least some plates yield an appropriate amount of colonies for counting. This method only counts living cells.

Question 5

Describe the growth of filamentous prokaryotes.

Answer 5

They do get bigger as a result of cell divisions and the growth can be followed using a microscopic count or by measuring biomass. It cannot use plate count because it will always yield a single colony.

Question 6

Describe the growth of hyphal prokaryotes.

Answer 6

They get bigger largely by cell elongation rather than division. Only measuring biomass is a good method of measuring its growth.

Question 7

What are psychrophiles, mesophiles, thermophiles , and extreme thermophiles?

Answer 7

Psychrophiles: 0-15 degrees Celsius. Ocean, snow.
Mesophiles: 15-50 degrees Celsius. Soil, water, warm blooded animals.
Thermophiles: 50-80 degrees Celcius. Hot springs, compost piles.
Extreme thermophiles: 80-100 degrees Celcius. Mud pots, hot springs, hydrothermal vents.

Question 8

What are nonhalophiles, moderate halophiles, extreme halophiles, and halotolerate nonhalophiles?

Answer 8

Nonhalophiles: do not require NaCl and grow optimally at conc. less than that of seawater.
Moderate halophiles: require NaCl and grow optimally near the concentration characteristic of seawater.
Extreme halophiles: require a minimum concentration of 10% NaCl in the environment.
Halotolerant nonhalophiles: do not require NaCl. But it tolerates it.

Question 9

What are some adaptations halophiles and halotolerant organisms exhibit to prevent plasmolysis?

Answer 9

The salt in strategy: involves transporting salt into the cytoplasm from the environment to balance.
The compatible solute strategy: involves making organic solutes that accumulate to high concentrations in the cytoplasm to balance.

Question 10

How does environmental pH affect the cell?

Answer 10

It can affect the folding of proteins and the ability of membrane phospholipids and glycolipids to form a bilayer structure.

Question 11

What are acidophiles, neutralophiles, akaliphiles?

Answer 11

Acidophiles grow at a environmental pH of 2 to 5, neutralophiles grow at 5 to 8, and akaliphiles grow at 8 to 11.

Question 12

What is the effect of environmental oxygen concentration?

Answer 12

It can become reduced to the hydroxyl radical. First reaction provides the superoxide anion, the second reaction provides hydrogen peroxide, and the third one provides the hydroxyl radical.

Question 13

What are the enzymes that catalyze the destruction of reactive oxygen species?

Answer 13

Super oxide dimutase (SOD): catalyzes the destruction of super oxide anion.
Superoxide reductase: catalyzes the destruction of super oxide anion.
Catalase: catalyzes the destruction of hydrogen peroxide.
Peroxidase: catalyzes the destruction of hydrogen peroxide.

Question 14

What is the effect of environmental UV radiation?

Answer 14

Bases of nucleic acids and aromatic side chains of certain amino acids absorb UV light and become chemically reactive. It can form directly valid points between two adjacent pyrimidine bases.

Question 15

What are some ways of treating damage done by UV light?

Answer 15

Photoreactivation: an enzyme called photolyase binds to the area of distorted DNA and uses the absorption of UV light for energy to break the covalent bonds between the bases.
Nucleotide excision repair: enzymes remove a portion of the strand of DNA containing the dimer and use the other undamaged strand as a template to replace the DNA.
Recombinational repair: The replication machinery skips over the damaged area. This leaves a gap. The gap is filled in by the RecA protein which binds to the damaged strand and catalyzes homologous recombination between it and a properly replicated strand.
Error prone repair: the mutations are the result of the cell doing the best they can to repair its DNA.

Question 16

What are the effects of gamma radiation?

Answer 16

Hydroxyl radicals are created when gamma rays strike water molecules. This is solved using a protein-based recombinational repair system.

Question 17

Describe antibiotic producing organisms.

Answer 17

Antibiotics are made by soil microbes, specifically members of the hyphal actinobacteria. Chemist often modify the structure of natural antibiotics and these are called semisynthetic.

Question 18

What are exospores?

Answer 18

A type of resting cell. The purpose of it is to spread the organism to areas with more favourable nutritional status.

Question 19

Describe the water solubility of antibiotics.

Answer 19

All antibiotics contain both polar and nonpolar groups. Many antibiotics also have polar groups that ionize in water.

Question 20

How do you some antibiotics work?

Answer 20

They disrupt cellular process or structure unique to the target organism, or they disrupt a cellular process or structure in the target organism without disrupting the same process or structure in the producing organism. Most antibiotics interfere with either peptidoglycan synthesis, translation, transcription, or DNA replication. A few disrupt membranes. Antibiotics can be either bacteriocidal or bacterialstatic. Bacterialstatic antibiotics inhibit cell division without killing the cell. Bacteriocidal antibiotic is lethal.

Question 21

Describe beta-lactam antibiotics.

Answer 21

They inhibit peptidoglycan synthesis and they have a beta lactam ring.

Question 22

How do antibiotics inhibit peptidoglycan synthesis?

Answer 22

Enzymes called autolysins Break pre-existing peptidoglycan bonds to create sites for the insertion of new material. They break the covalent oh glycostatic bond between the NAM and NAG sugars. Enzymes like transpeptidase and transglycosylase help the elongation of peptidoglycan. Antibiotics prevent the transpeptidation reaction by binding to and inhibiting the enzyme that catalyzes this. The cell would rupture from osmotic lysis.

Question 23

Describe vancomycin.

Answer 23

It also inhibits peptidoglycan synthesis but it binds to AA4 and AA5 of the disaccharide pentapeptide unit in the periplasm.

Question 24

Describe polymyxin.

Answer 24

The ring portion of the molecule has many positively charged groups. It has an affinity for the LPS core that is at least three orders of magnitude higher than the calcium and magnesium ions that helps stabilize outer membrane structure. The cytoplasmic membrane is disrupted through the insertion of polymyxin’s hydrocarbon tail into the membrane.

Question 25

What are some barriers to antibiotic penetration?

Answer 25

Capsules, s layers, peptidoglycan, cytoplasmic membrane, and gram-negative outer membrane.

Question 26

Why are some bacteria more resistant to antibiotics than others?

Answer 26

Antibiotic exclusion, gram negative versus gram-positive bacteria, gram negative bacteria, antibiotic modification, target site variation, and antibiotic efflux.

Question 27

How do bacteria acquire resistance to an antibiotic?

Answer 27

Alteration of target site, acquisition of enzyme that chemically modified and inactivates the antibiotic, change in the cell envelope so that its antibiotic exclusionary properties are enhanced, and acquisition of a cellular membrane protein that transports antibiotics out of the cell.

Question 28

What is the issue with antibiotic resistance?

Answer 28

Bacteria acquire antibiotic resistance through natural processes, and the widespread killing of sensitive bacteria by the overuse of antibiotics means that there is a more likely chance an infection will be caused by a pathogenic bacteria that is also antibiotic resistant.

Question 29

Describe vegetative cells and resting cells.

Answer 29

Vegetative cells are active cellular forms capable of reproduction, and resting cells are dormant cellular forms that aren’t dead but are not capable of reproduction.

Question 30

What are the two types of resting forms?

Answer 30

Exospores and endospores.

Question 31

How is the endospore formed?

Answer 31

It begins with copying of the genome that yields to membrane-bound compartments within one cell wall unit: a larger mother cell and a forespore which ultimately becomes the endospore. The mother cell membrane engulfs the forespore and then they both cooperate to make a cell wall made of peptidoglycan. The mother cell, now called the sporangium, is degraded by enzymes and that releases the mature spore.

Question 32

What are exotoxins?

Answer 32

Extra cellular proteins made in the bacterial cytoplasm and transported into the environment. They contribute to food poisoning. They are associated with a selfish genetic element like a prophage or plasmid.

Question 33

What is the difference between food borne infection and foodborne intoxication?

Answer 33

Food borne infection is eating food containing bacteria and then the bacteria grow in the intestine and produce and exotoxin. Foodborne intoxication is eating food that already have bacteria growing and producing an exotoxin.

Question 34

Describe C. Botulinum.

Answer 34

A type of foodborne intoxication. It is killed by exposing it to moist heat. Endospores are more resistant to moist heat killing because the low water content of the core means that essential core proteins must be very poorly hydrated. Dried proteins are more resistant to heat denaturization then proteins in aqueous solutions.