2.5

Question 1

Obligate aerobes-

Answer 1

(respiration only) require O2 to grow – Mycobacteriumtuberculosis(infectslungs)

Question 2

Obligate anaerobes-

Answer 2

(fermentation only) can’t grow when O2 is present

Question 3

Facultative anaerobes

Answer 3

anaerobes-(contains genes for fermentation and/or respiration) can grow with or without O2

Question 4

Aerotolerant-

Answer 4

(anaerobic metabolism-fermentation, but can tolerate
oxygen)
– Streptococci-mostfermentsugar,butcantoleratesomeoxygen (aren’t killed by O2 and can inactivate toxic free radicals)
– S.mutans-cantolerateoxygen(incompleteTCAcycle)fermentssugar

Question 5

Doubling time (Td)-

Answer 5

time it takes a single bacteria to divide in two.
– Rate is determined by environmental conditions (i.e. available nutrients)
– Limited by expression of genes needed to metabolize nutrients and replicate DNA.
– Some bacteria are more efficient than others • E. coli Td = 20 min -rich media at 37°C
• P. gingivalis Td = 9 hr

Question 6

Lag phase

Answer 6

time when bacteria adapt to environment -genes are turned on to begin metabolism of nutrients

Question 7

Exponential(log)phase

Answer 7

bacteriagrowlogarithmically (slope of line= growth rate -Td)

Question 8

Stationary phase

Answer 8

Where growth stops because nutrients are exhausted or toxic by-product buildup.

Question 9

DNAgyrase-

Answer 9

introduces supercoils

Question 10

TopoisomeraseI-

Answer 10

relaxessupercoils

Question 11

Steps to chromosomal replication:

Answer 11

1. Initiation- depends on growth rate 2. Elongation

3. Termination

Question 12

Base substitution

Answer 12

When a nucleotide is changed

Question 13

Insertion

Answer 13

When a nucleotide is inserted

Question 14

Deletion

Answer 14

When nucleotide is deleted

Question 15

Nonsense

Answer 15

When a nucleotide change results in a stop codon

Question 16

Missense

Answer 16

When a nucleotide change results in an amino acid change

Question 17

Duplication

Answer 17

When DNA sequence is abnormally copied

Question 18

Natural transformation

Answer 18

(ability to take up DNA from the

environment)

Question 19

Conjugation

Answer 19

(transfer of plasmid DNA through sex pili),

Question 20

Transduction

Answer 20

(bacteria phage [virus] accidentally inject bacterial DNA into host bacterial cell).

Question 21

Transposable elements

Answer 21

(DNA that can insert or excise its self from the chromosome or plasmid)

Question 22

Once DNA is taken into the cell:

Answer 22

Can be degraded into pieces (may serve as raw materials)
– If homologous sequence is present, DNA can be incorporated into the chromosome (homologous recombination)
–Acquisition of new genes –Mutations in existing genes

Question 23

Natural transformation

Answer 23

Ability to take up DNA from the environment

Mediated by quorum sensing- peptide pheromones produced by other bacteria are sensed by bacteria and turn on genes for competence.

Question 24

Transduction

Answer 24

transfer of bacterial DNA by a phage to recipient bacteria

Question 25

Promoter elements

Answer 25

RNA polymerase- binds to promoter located upstream of gene start codon

Question 26

Operon-

Answer 26

Multiple genes arranged together on the same mRNA transcript expressed from the same promoter. These genes are said to be co-transcribed.

Question 27

Virulence factors

Answer 27

genes directly related to the ability of | bacteria to cause disease

Question 28

Random mutagenesis

Answer 28

random mutations are introduced creating a mutant library that can be screened

Question 29

Positive selection-

Answer 29

only those cells in a population with the trait are selected.

Question 30

Negative selection-

Answer 30

the absence of a particular mutant after a screen.

Question 31

Allelic replacement-

Answer 31

a gene is replaced by an antibiotic marker.

Question 32

Complementation-

Answer 32

a gene from a pathogenic organism is expressed in a non-pathogenic strain to see if the strain becomes virulent (i.e. host cell invasion).

Question 33

Positive selection-

Answer 33

only bacteria with the desired phenotype (ability to invade) are selected in this screen 1. Non-invasive bacteria containing plasmid library with potential genes required for invasion are grown in broth. 2. Mammalian cells are exposed to library of different strains for 2 hrs 3. Mammalian cells are washed and treated with Gentamicin to kill external bacteria (bacteria that didn’t invade) 4. Cells are lysed and internal bacteria are recovered on agar plates containing antibiotics. 5. The genes present on plasmid responsible for invasion of host cells are determined (DNA seq.) 6. Virulence factors are verified

Question 34

Negative selection-

Answer 34

selection of mutants unable to grow under certain conditions Signature-tagged mutagenesis to identify virulence genes 1. DNA seq tagged transposon mutant library is created 2. Each well on microtiter plate represents a separate mutant, which are replica plated onto a membrane- lysed and DNA denatured 3. Tn5 mutants are pooled and injected into mouse 4. Bacteria able to invade are then recovered from spleen. 5. Input pool and recovered pool are compared (PCR followed by hybridization to blot containing Tn library DNA) 6. The mutants which are not in the recovered pool are avirulent. 7. Location of Tn5 is determined by DNA seq.