Topic 9 - Genetics - Clickers Flashcards
d) histidine
Explanation: In the genetic map for Hfr strain KL98, histidine synthesis genes (his) are located near the region transferred early during conjugation. By omitting histidine from the growth medium, only recombinants that have successfully received and expressed the donor his genes will grow. This makes histidine omission an effective selection marker for identifying successful recombinants.
a) glycine;
b) tyrosine;
c) cysteine
Explanation: In the genetic map for Hfr KL16, the genes encoding the synthesis of glycine (glyA), tyrosine (tyrA), and cysteine (cysC) are located early in the region transferred during conjugation. By omitting these amino acids from the growth medium, only recombinants that successfully receive and express these genes will grow. This makes glycine, tyrosine, and cysteine omission effective selection markers for identifying recombinants in the recipient F- strain.
d) F’
Explanation: The strain shown is an F’ strain, which contains an F plasmid that has incorporated specific chromosomal genes (in this case, thr and leu) through improper excision from the bacterial chromosome. This leads to a plasmid that carries both F factor genes and chromosomal genes, enabling the strain to transfer these additional genes to recipient cells during conjugation. This distinguishes F’ strains from Hfr (which transfer chromosomal DNA directly) and F+ (which carry only the F plasmid without additional chromosomal DNA).
c) threonine
Explanation: The F’ plasmid in this strain carries genes for threonine synthesis (thr). By omitting threonine from the medium, only recipient cells that have successfully acquired and expressed the F’ plasmid (containing the thr gene) will grow. This makes threonine omission an effective selection marker to identify recombinants from the auxotrophic recipient F- strain.
b) transposon
Explanation: This structure represents a transposon, as indicated by the following features:
Transposase genes (tnpA and tnpB): These encode enzymes responsible for the movement (transposition) of the element.
Inverted repeats (38 bp): These flanking sequences are hallmark features of transposons, necessary for recognition and excision.
Additional genes (bla): This transposon contains a β-lactamase gene, which provides antibiotic resistance, a common characteristic of transposons that carry genes beyond those required for transposition.
This distinguishes it from an insertion sequence, which lacks additional genes, and from plasmids or operons, which do not contain the inverted repeats and transposition machinery.
b) replicative transposition
Explanation: This transposon includes a resolvase gene, which is essential for replicative transposition. During this process:
- The transposon is copied into a new DNA site while remaining in its original location, effectively increasing the number of transposon copies.
- The resolvase enzyme resolves the cointegrate structure that forms during transposition, allowing the transposon to integrate into the target DNA site without leaving the donor site.
Non-replicative transposition, in contrast, involves a “cut-and-paste” mechanism where the transposon excises from the original site and moves entirely to a new location, which does not require resolvase.
Which of the following mutant phenotypes would be readily amenable to selection?
a) Virus resistant
b) Pigmentless
c) Cold sensitive
d) Nonmotile
e) Auxotroph
a) Virus resistant
Explanation: Virus resistance can be selected for by exposing the population to a virus; only resistant mutants will survive.
A wild-type is met- pro+. How would you BEST recover a met+ pro+ mutant?
a) defined medium with methionine and proline
b) defined medium with methionine but without proline
c) defined medium without methionine but with proline
d) defined medium without methionine nor proline
d) defined medium without methionine nor proline
Explanation: To recover a met+ pro+ mutant, the defined medium lacking both methionine and proline ensures that only mutants capable of synthesizing both will grow.
A wild type is met- pro+. You have recovered a met+ pro+ mutant. Was this screening or selection?
a) Screening
b) Selection
b) Selection
Explanation: Selection allows growth of only those mutants with a specific phenotype, such as met+ pro+ in this scenario.
Which of the following mutant phenotypes would be readily amenable to selection?
a) Antibiotic resistant
b) Virus-sensitive
c) Auxotroph
d) Pigmented
a) Antibiotic resistant
Explanation: Antibiotic resistance can be selected for by adding an antibiotic to the medium; only resistant mutants will survive.
How would you retrieve a pro- mutant from a prototroph population?
a) Replica plating on defined media with and without proline
b) Defined medium without proline
c) Defined medium with proline
a) Replica plating on defined media with and without proline
Explanation: Replica plating helps identify pro- mutants by comparing growth on media with and without proline.
How would you retrieve a trp+ mutant from an auxotrophic wild type population?
a) Replica plating on defined media with and without tryptophan
b) Defined medium with tryptophan
c) Defined medium without tryptophan
a) Replica plating on defined media with and without tryptophan
Explanation: Replica plating helps identify trp+ mutants by comparing growth on media with and without tryptophan.
Which of the following statements about plasmids is FALSE?
a) Some have selectable markers
b) Replicate by host DNA polymerase
c) They are usually circular
d) Many carry genes essential to host metabolism
e) They are usually smaller than a chromosome
d) Many carry genes essential to host metabolism
Explanation: Plasmids typically carry non-essential genes, such as those conferring antibiotic resistance or other advantages, rather than genes essential for host metabolism.
Common desirable plasmid traits for easier gene cloning include all EXCEPT:
a) possess OriV
b) large size
c) multiple cloning site
d) selectable marker gene
e) high copy number
b) large size
Explanation: Large size is undesirable for gene cloning as it makes plasmids harder to manipulate and less efficient for transformation.
A wild-type is met+ pro-. Which defined medium should you use to recover a met+ pro+ mutant?
a) Defined medium with methionine and proline
b) Defined medium without methionine nor proline
c) Defined medium without methionine but with proline
d) Defined medium with methionine but without proline
b) Defined medium without methionine nor proline
Explanation: To recover met+ pro+ mutants, this medium ensures growth only for mutants synthesizing both methionine and proline.
Esther Lederberg developed the __________ plating technique.
a) streak
b) pour
c) patch
d) replica
e) spread
d) replica
Explanation: Esther Lederberg developed replica plating, a technique used to transfer bacteria from one plate to another to identify mutants.
With regards to plasmid replication, which of the following statements is CORRECT?
a) Copy number is a function of a cell’s carbon source
b) Larger plasmids are best for gene cloning
c) Multiple plasmids from an INC group may replicate within the same cell
d) Plasmid-encoded genes determine copy number
d) Plasmid-encoded genes determine copy number
Explanation: Genes encoded by plasmids regulate their own replication, affecting the copy number per cell.
On what medium would you select for pro+ recombinants from a cloning experiment with a vector that has a streptomycin-resistance selectable marker?
a) Defined medium without proline nor streptomycin
b) Defined medium with proline and streptomycin
c) Defined medium without proline but with streptomycin
d) Defined medium with proline but without streptomycin
c) Defined medium without proline but with streptomycin
Explanation: This medium selects for pro+ recombinants and ensures only cells with the streptomycin-resistant plasmid grow.
A property common to ALL restriction recognition sites is:
a) they are palindromes that are cut to generate overhanging and complementary sticky ends.
b) they dictate where both strands of the DNA double helix should be cut.
c) they are DNA sequences 6 base pairs in length.
b) they dictate where both strands of the DNA double helix should be cut.
Explanation: All restriction recognition sites specify where the restriction enzymes cut both DNA strands.
Which of these would be essential for E. coli to host a plasmid?
a) antibiotic resistance
b) multiple cloning site
c) oriV
d) cos sites
c) oriV
Explanation: The oriV site is essential for plasmid replication in E. coli and other hosts.
For highest transformation efficiency, which would you choose?
a) chemically competent cells
b) electrocompetent cells
b) electrocompetent cells
Explanation: Electrocompetent cells have the highest transformation efficiency due to increased membrane permeability during electroporation.
For naturally competent cells, how are new alleles acquired?
a) homologous recombination
a) homologous recombination
Explanation: Naturally competent cells integrate new DNA into their genome via homologous recombination.
How would you make chemically competent cells?
a) treat with calcium cations
b) put the peeps in the chilly pot
c) diggety-dang-a-dang
d) respect the drip Karen
e) sksksksksksk and I oop
f) uh oh stinky
a) treat with calcium cations
Explanation: Treating cells with calcium cations makes their membranes permeable to plasmid DNA.
What can be an episome? Choose all that apply.
a) insertion sequences
b) plasmid
c) virus
d) transposon
All answers are correct.
Explanation:An episome is any genetic element capable of integrating into a host genome or existing independently in a cell.
a) Insertion sequences can integrate into DNA, acting as episomes in certain cases.
b) Plasmids often function as episomes when they integrate into a host chromosome.
c) Viruses, especially temperate phages, integrate into host genomes during their lysogenic cycle.
d) Transposons are mobile genetic elements that can also exist independently or integrate into DNA.
Which of the following statements about Hfr strains is/are CORRECT? Choose all that apply.
a) can transfer genomic DNA from a donor to a recipient
b) can help map chromosomes
c) can produce F’ bacteria, which can conjugate
d) can have multiple insertion sites dependent on homology
All answers are correct.
Explanation:
a) Transfer genomic DNA: Hfr strains transfer part of the donor’s genomic DNA to the recipient during conjugation.
b) Help map chromosomes: The time it takes for different genes to transfer during conjugation can be used to map chromosome locations.
c) Produce F’ bacteria: Hfr strains can excise from the chromosome, sometimes taking host DNA with them, creating an F’ plasmid capable of conjugation.
d) Multiple insertion sites: Hfr formation depends on homology between the plasmid and chromosome, allowing it to insert at various sites with shared sequences.
Which of the following statements about transposons is FALSE? (choose all that apply)
a) transposons can be carried between cells by conjugation, transformation, or transduction
b) transposition occurs infrequently
c) transposons encode only what is needed for moving
d) replicative transposition increases transposable element count by 1
e) transposons insert randomly at target sites
c) transposons encode only what is needed for moving
Explanation: Transposons often carry additional genes, such as those for antibiotic resistance, beyond what is required for movement. This makes them versatile genetic elements.
All of these are advantages of being naturally competent except for one – find it.
a) Nutrition, especially under starvation conditions
b) Obtaining an auxotrophic phenotype from a related mutant strain
c) Repair, especially in response to UV damage
d) Recombination to provide new traits
b) Obtaining an auxotrophic phenotype from a related mutant strain
Explanation: Natural competence is advantageous for acquiring beneficial traits, not for obtaining an auxotrophic phenotype, which is typically detrimental.
Which of the following is FALSE of generalized transduction?
a) some phage DNA in each transducing particle
b) very few transducing particles generated per cell
c) random packaging of host DNA
d) dependent on the lytic pathway
a) some phage DNA in each transducing particle
Explanation: Generalized transduction involves random packaging of host DNA into transducing particles, which do not necessarily contain phage DNA.
In a cross involving a streptomycin-sensitive wild-type Hfr donor and an F- recipient that is a streptomycin-resistant leucine auxotroph (leu-), an appropriate selection for leu+ recombinants would be…
a) defined medium with leucine and streptomycin
b) defined medium without leucine or streptomycin
c) defined medium with leucine but without streptomycin
d) complex medium containing streptomycin
e) defined medium lacking leucine but with streptomycin
e) defined medium lacking leucine but with streptomycin
This selection medium ensures that:
- Streptomycin presence: Only the F- recipient, which is streptomycin-resistant, can grow. The Hfr donor, which is streptomycin-sensitive, will not survive, eliminating false positives.
- Lack of leucine: Only recombinants that have acquired the leu+ gene from the Hfr donor through conjugation will grow. Auxotrophic (leu-) F- recipients will not survive due to the absence of leucine.
This combination allows for the selection of leu+ recombinants while excluding both non-recombinant recipients and the donor strain.
