Lecture 11: Chromosomal Mapping

Question 1

1. What are three reasons bacteria are more amenable to genetic analysis than elephants?

Answer 1

1. Relative genomic simplicity
   - fewer genes
   -more gene-dense
2. Haploidy allows for mutation effects to be seen more directly without complications of dominance
3. short generation times allow inheritance to easily be studies
4. large number of progeny
5. ease of propagation

Question 2

Are all cells in a bacterial colony genetically identical? Why or why not?

Answer 2

Yes they are all derived from the same singular cell as they reproduce asexually through binary fission.

Question 3

Define prototroph and auxotroph.

Answer 3

Phototroph: Uses light for energy- can grow on minimal media
Autotroph: can produce its own food for energy-cannot grow on minimal media

Question 4

You isolate an E. coli colony that grows on complete media but not minimal media. You find that the colony can grow if it is supplemented with alanine and leucine but not each one separately. You also
find that this colony cannot grow in lactose media even when supplemented with alanine and leucine.
What is the genotype of this colony? Explain your answer.

Answer 4

The colony is alanine-, leucine- , and lac-

If it is autrophic for a nutrient genotype is nutrient-, if it can grow in lactose it is lac+, if it cannot grow in lactose it is lac-

Question 5

What is a plasmid? Why is it typically not considered part of the bacterial genome?

Answer 5

circular DNA carries nonessential genes. that are important in specific environments (e.g antibiotic resistance)

not considered part of the bacterial genomes as they are extrachromosomal and can be gained and lost easily

can be transferred horizontally/laterally (two unrelated individuals ) allowing the transfer of antibiotic resistance

Question 6

Is bacterial recombination always a one-way process? Explain your answer.

Answer 6

Bacterial recombination is a one way process between a donor and a recipient cell.

It can occur via three ways:
Conjugation: transfer via conjugation pilus
transformation: uptake of exogenous DNA
transduction: gene transfer mediated by virus

Question 7

What did the U-tube experiment show? Why could this inference be made? Why does this make
sense given what you know about conjugation?

Answer 7

The u- tube experimented tested whether physical contact was necessary between cells for the transfer of DNA

Pure Culture of Y-10 and 58-161 added separately to media and no colonies observed. MIxed cultures then adding to media resulted in photorophic colonies

U-tube results: u-tubed used alternting pressure and suction to assure medias are well mixed. ( media can pass through filters but not cells)

Conclusion: when u-tube mixed culture applied to minimal media no colonies were observed–> physical contact is needed for conjugation

Question 8

Explain F plasmid, who discovered and conclusions resulting from it

Answer 8

WIlliam Hayes discovered that in conjugation two strains do not contributes equally like in a cross of eukaryotes

F plasmids are F+ and can be donors of genetic material
cells lacking F plasmid can be recipients

only donor cells initiate conjugation

F plasmid encodes 36 genes and contains insertion sequences that can undergo homologous recombination

Question 9

explain whole process of conjugation use terms such as, relaxosome, relaxase, coupling factor, t stranding, rolling circle replication, exconjugant.

Answer 9

Exporter complex: exporter and pilus proteins make bridge for DNA to be transferred

Relaxasome: cuts one strand of donor DNA at OriT site
relaxase is left,
coupling factor bind relaxase and relaxase facilitates the single strand from the 5’ end to move to the exporter complex

AT THE SAME TIME replication to replace the Transfer strand occurs su h that the donofr remains F+

rolling circle replication: In the donor cell replication circle replication occurs and the t-strand is spooled into the recipient

F plasmid is reconstituted fully in donor cell
the recipient now has a different genetic reconstitution ( F+) and is an exconjugant.(both cells end F+)

Question 10

Why can two donor cells not conjugate with one another?

Answer 10

expression of proteins on the cell surface encoded on the F plasmid leading to surface exclusion

Question 11

What is the significance of oriT, and why is it important that the 5’ end of the T strand always be
transferred?

Answer 11

It is important for the 5’ end to be transferred because that leaves a free 3’OH such that replicatino of the T strand can occur and the F+ plasmid can be reconstituted in the donor cell

Question 12

In conjugation between an F+ cell and an F recipient, how does the exconjugant differ from the
original recipient cell?

Answer 12

They both have the F+ genetic material in the end

Question 13

How is an Hfr chromosome generated? What sequences does this rely on?

Answer 13

Hfr chromosome generated when the F factor is recombined into the bacterial chromosome. This relies on insertion sequences

Question 14

Why can the whole bacterial chromosome not be transferred in Hfr conjugation?

Answer 14

partial transfer due to there being limits of how much DNA can get through a pilus before pilus breaks

Question 15

Are the exconjugants in Hfr conjugation F+ or Hfr? Why or why not?

Answer 15

The exconjugants are not F+ because the entire DNA sequence would have to be transferred to the recipient for the F+ to circularize.

Question 16

Why is the Hfr transferred fragment always linear?

Answer 16

since there is no full oriT segment the Transferedd DNA cannot circularize and will remain linear

Question 17

Why can interrupted matings be used for time of entry mapping?

Answer 17

by breaking the pilus and disrupting conjugation will allow the relative gene order and distance to be determined as transfer begins at the oriT and transfer of DNA is linear over time

Question 18

You perform time of entry mapping on the genes A,B,C,D, and E. You know that in your Hfr strain
C is transferred very quickly. Your donor is A+
,B+
,C+
,D+
,E+ strs
and your recipient is A-
, B-
,C-
,D-
,E- strr
You perform time of entry mapping experiments and test for transfer of the + alleles among C+ strr
exconjugants. Given the following results, infer the relative gene order and relative distance between
genes

Question 19

be able to provide cicrular map given hr derivatives

Question 20

What is the difference between an F+ cell and an F’ cell?

Answer 20

Hfr bacterial chromosome ends up on the F plasmid forming a F’ cell

difference: carrying a bacterial chromosome or not

Question 21

What is a partial diploid? How are they generated?

Answer 21

partial diploid is an exconjugant bacterium that has two alleles of a specific gene only

Question 22

How is an F’ cell generated?

Answer 22

when in inexact excision causes F factor to contain Hfr chromosomal genes

Question 23

Explain the process of transformation. Use heteroduplex and strand invasion in your answer

Answer 23

1. exogenous DNA binds a receptor site on competent cells, once it enters cells one strand is degraded
2. strand invasion: ssDNA pairs with a homologous region in the bacterial chromosome
3. one strand of DNA is displace, Heteroduplex ( two partly mismatched strands derived from two different parent molecules)

Question 24

If genes A and B are never co-transformed with one another, what does that tell us about these
genes?

Answer 24

these genes are not linked and have to much of a distance between them

Question 25

Why can bacterial DNA be readily packaged into a phage particle?

Answer 25

as long as the bacterial DNA fragment is close in size to the phage’s genome it can be incorporated into the phage particle.

Question 26

What is cotransduction? How can it be used to map genes?

Answer 26

cotransduction is transfer of two genes into phage. those closer to the marker of the donor genotype will have higher cotransduction rates

Question 27

If two strains of phage have separate mutations in the same gene, what would be the expected result in a complementation test? Are there any exceptions to this result? Explain.

Answer 27

if we coinfect bacteria with phages consisting of different mutations there would be progeny of the mutation so you would see no plaques

however in rare cases there would be a crossover and the wildtype would be restored so you would see placque

Question 28

If two strains of phage have mutations in the same position in the same gene, what would be the
expected result in a complementation test? Are there any exceptions to this result? Explain.

Answer 28

you would not for any plaques form as there is no crossover occur as they are in the same position therefore the wildtype is not rescued

Question 29

Explain why the use of antibiotics selects for resistance. Why is this particularly problematic in
bacteria?

Answer 29

bacteria have selective pressure to develop mutations to resist the antibiotics however because lateral gene transfer is so frequent this could lead to it spreading so fast that the harmful bacteria will become resistant to the antibiotic

Question 30

If a bacterium is only sensitive to a concentration of a drug that exceeds what can be tolerated by
a patient, is that bacterium treatable by that drug? Why or why not?

Answer 30

the bacteria is treated but the patient would die

Question 31

Why does not taking a full course of antibiotics yield a greater risk of antibiotic resistance being a
problem?

Answer 31

by taing the antiobitic for part of the course you have developed bacteria that are highly resistance to the pathogen so if the infection comes back you will either need a bigger dose or a different drug