Exam 4: Microbial Genetics II DNA Repair & Gene Transfer

Question 1

When DNA polymerase III makes a mistake, does this always become a mutation? Explain.

Answer 1

DNA polymerase III incorporates incorrect nucleotides at a low but measurable rate. Most of these mistakes are corrected by proofreading. Most mistakes are caused during DNA synthesis.

Question 2

Briefly describe the process of proofreading.

Answer 2

DNA polymerase recognizes the error, removes it, and repolymerizes this error.

Question 3

Briefly describe the process of base-excision repair. How does it differ between eukaryotes and bacteria such as E. coli? Which polymerases and ot ligases is invoved?

Answer 3

Enzymes recognize abnormal bases such as uracil or hypoxanthine. Erroneous base is excised from the DNA backbone and another enzyme “nicks” the backbone.. In E.coli DNA polymerase I removes and replaces a few nucleotides in the region of the abnormal base. Eukaryotes only remove and replace the abnormal base. DNA ligase covalently links the “old” and “new” DNA strand.

Question 4

Briefly describe the process of mismatch repair. How does this differ from base-excision repair? Which polymerase is involved?

Answer 4

Mismatch repair enzymes find mismatched bases directly following DNA replication. Segments of strand surrounding the mistake are removed. Enzyme replaces the non-methylated strand. DNA polymerase III synthesizes the new segments.

Question 5

When a mistake in DNA replication is corrected via mismatch repair, how is it determined which strand has the correct base and which has the incorrect base?

Answer 5

The enzyme replaces the non methylated strand. The new strand is not methylated yet, but the old strand is. So it bases its correction on the original strand and removes a chunk of the new strand.

Question 6

Name and describe the two different processes by which pyrimidine dimers are efficiently corrected. How are these processes similar, and how are they different? Which polymerase and or ligase are invoved.

Answer 6

Light repair of dimers: ultraviolet light forms pyrimidine dimers, DNA photolyase can break the inappropriate covalent bond forming the dimer. The enzyme used is activated by visible light.

Dark repair dimers “nucleotide excision repair”: uses a different repair enzyme. Process functions in both the light and the dark (no light activation needed). Section of DNA strand including the dimer is excised, gap is repaired by DNA polymerase I and DNA ligase.

Question 7

What is SOS response? How efficient and accurate is this process? Under what conditions is it activated?

Answer 7

aka “error prone repair”. Regular repair systems cannot cope with extensive DNA damage e.g very numerous pyrimidine dimers or damage to both DNA strands. It’s a hail mary to try and fix a bad problem. In E.Coli activates SOS response as last-ditch effort to repair enough DNA to survive, novel DNA polymerase are reproduced. DNA repaired with little regard for base sequence so many mutations occur but a few bacteria survive.

Question 8

Describe the process of direct selection. What types of mutants are typically identified by this method? Give example.

Answer 8

Involves identifying mutants by eliminating wild-type individuals, like finding a needle in a haystack by burning the haystack. Can be used to isolate antibiotic resistant mutants from a wild-type population.

Example; wild type is penicillin sensitive and mutant is penicillin resistant. Grow culture in the presence of antibiotics on plate with antibiotics, only antibiotic resistant mutants will survive.

Question 9

Describe the process of indirect (negative) selection. What types of mutants are typically identified by this method?

Answer 9

Mutants defective in the production of a nutrient cannot be detected via positive selection, negative selection is needed. This is the process used to isolate neurospora carssas. Grow Neurospora carassa on a plate that contains arginine. Both wild and mutant will grow. Use the sterile velvet to take print of growth on that plate. Transfer onto a plate that does not contain arginine. On this plate only wild tye will grow. Compared to plates, the negative space on the new plate represents mutants on the initial plate. Can now pluck those specific colonies for growing a supply of mutants.

Question 10

Describe the process of penicillin enrichment. Is this used to identify penicillin-resistant mutants? Explain.

Answer 10

Can be used to make negative selection of auxotrophic mutants, only works for bacteria. Not used to identify antibiotic-resistant mutants.

Looking for a his- E.Coli (cannot make histamine amino acid like the wild type). Grow bacteria for a short time in minimal media (no histamine present) with penicillin. Penicillin kills dividing cells, auxotrophic mutants survive because they cannot divide and are therefore not damaged. Since no histamine present the hist- cells don’t have what they need to divide and grow, only the wild type are growing and dividing. The auxotrophic mutants can survive since they aren’t in the penicillin very long. Add penicillinase to destroy the penicillin, placed onto enriched media (contains histamine), now you can do indirect selection with the velvet.

Question 11

How does horizontal gene transfer differ from vertical gene transfer?List the 3 types of horizontal gene transfer.

Answer 11

Vertical gene transfer is cell replication which creates genetically identical daughter cells. However horizontal gene transfer occurs when genes are required for microbes of the same generation, it’s the sharing of genetics between adult bacteria.

Horizontal gene transfer occurs via 3 mechanisms: transformation, transduction, conjugation.

All three share these 3 characteristics: transformation is unidirectional (donor to recipient), onlay a subset of the donor DNA is transferred (recipient is different form its donor and original self), and homologous genes are replaced in the recipient (new genes are integrated and any duplicates are removed, no net change in gene number).

Question 12

What is the benefit of horizontal gene transfer?

Answer 12

Creates genetic diversity that evolution can act on and increases the dispersion of antibiotic resistance and virulence genes.

Question 13

Explain why horizontal gene transfer should be considered a form of sexual reproduction.

Answer 13

The transformed cells are genetically unique just like offspring are genetically unique, the overall result is a genetically new individual. New combination of alleles created.

Question 14

Explain why horizontal gene transfer should not be considered a form of sexual reproduction.

Answer 14

No net growth in generation size.

Question 15

Describe the process of bacterial transformation.

Answer 15

DNA is picked up from the environment by the recipient and integrated into their DNA. Donor cell lyses (dead), chromosome is released, chromosome is partially degraded. Inorder for this to occur the recipient cells must be competent.

Single stranded donor DNA enters the host cell. DNA integrates into the recipient chromosome ( breakage and reunion, replaces homologous sequences). Mismatched repair “fixes” DNA, and can make new alleles permanent. Transformed cells are genetically unique.

Question 16

What changes are present in a bacterium that is “competent.”

Answer 16

Near the end of the log phase, cell wall changes, and receptor proteins are made. Occurs natural in several species e.g. streptococcus and bacillus. Also can be induced in the lab on a large number of species e.eg E.Coli.

Question 17

Describe the process of generalized transduction.

Answer 17

Viral-mediated horizontal gene transfer.

Bacteriophage (virus) infects host bacteria, converting the host into a factory for producing viruses. Host DNA is degraded, viral components are produced. Host DNA fragments are sometimes mistakenly packaged into phage capsid instead of viral DNA. Forms a Transducing phage.
 
Transducing the phage binds to a new host cell it will insert its DNA. However its dNA is not viral DNA its bacterial DNA. This DNA replaces homologous genes and the recipient bacterium is now recombinant. End result is a healthy bacteria that has DNA different then it use to have.

Question 18

Describe the process of specialized transduction.

Answer 18

Viruses can integrate into the host chromosome, forms prophage. Viral gene expression is largely repressed, provirus is passed to daughter cells via cell division.

The virus circularizes and integrates into the bacterial chromosome, and does not negatively affect the bacteria. When the bacteria divides into two daughter cells the prophage is replicated and transferred as well. Majority of the time the prophage will stay in place generation to generation daughter cell to daughter cell. But it has the ability to pop out and initiate the lytic cycle (generally under detrimental conditions). 

The provirus can excise from a host chromosome. Sometimes perfectly but sometimes imperfect excision occurs. Imperfect excision removes host DNA with viral DNA. New viruses are assembled, DNA is packaged into each new virus. Imperfect excision forms a transducing phage because it has hot DNA present. Transducing phage “infects'' new host and bacterial DNA is introduced and retained. Novel genes added to host.

Question 19

How are generalized transduction and specialized transduction similar? How are they different?

Answer 19

Both are the same in that a virus can be used as a vehicle to transport genes from one bacteria to another. In generalized there’s just the accidental packaging that’s purely bacterial in a tiny fraction of the virtual releases. In specialized many copies of this virus are made again but they are all identical and they all have a small piece of bacterial DNA.

Question 20

Describe the structure and function of a fertility factor.

Answer 20

Also called F plasmid and F factor. Fertility factor is a plasmid gene that directs formation of sex pilus and the ability to transfer DNA. Hand full of genes not present in all plasmids.

Question 21

How does an F+ cell differ from an F- cell?

Answer 21

F+ have an F factor while F- do not have an F factor. F+ have the ability to be donors but not recipients while F- are recipients and not donors.

Question 22

How does an F+ cell differ from an Hfr cell?

Answer 22

F+ plasmid exists as its own circular chromosome. In the Hfr cell the F factor is integrated into the host cell chromosome. Hfr stands for high frequency recombination. Hfr cell posses and F factor so it can produce a sex pilus and can be a donor for F- cells. However a Hfr cannot turn a F- into an F+, the F- is now recombinant.

Question 23

How does an F+ cell differ from an F’ cell?

Answer 23

F’ is produced when excision of the F factor from the chromosome converse Hfr back to F+. When this is done imperfectly a modified F+ cell is produced called F’. F” Possesses chromosomal genes inserted into the F factor.

F+ has separated plasmid containing F factor from its chromosomal DNA.

Question 24

Describe the process of conjugation between an F+ cell and an F- cell. Do you think this transfer commonly benefits the recipient? Why does this process exist? (This is a tough question.)

Answer 24

Sex pills draw cells together, direct physical connection. Single strand of F plasmid is “nicked”, and the covalent bond is broken. Nicked strand is transferred linearly into the recipient through sex pilus. Will circularize in the recipient cell and cell synthesizes complementary strands so plasma is now double stranded. Recipient cell is now F+.

F factor sometimes integrates into the host cell. Accomplished via a single crossover between the circular plasmid and the circular chromosome. Genes on the plasmid are still expressed. Integrated plasmid is passed vertically to daughter cells.

Question 25

Conjugation

Answer 25

requires direct contact between donor and recipient, donor remains alive. Transfer of SNA through a sex pilus, hollow protein tube constructed by the donor.

Question 26

Describe the process of conjugation between an Hfr cell and an F- cell. Do you think this transfer commonly benefits the recipient? Explain.

Answer 26

Hfr produces sex pills. Strand of the integrated F factor is “nicked” somewhere in the middle. Single strand starts peeling away from the circular strand of the chromosome. Portion of the F factor is transferred first then transfer of the chromosome begins, connection is generally broken before transfer is complete. Would take ~100 minutes for a full transfer. Recipient remains F- since only part of the F- factor is transferred! Chromosomal genes transferred replace homologous gene sin recipient, recipient is now genetically unique “recombinant”

Question 27

Describe the process of conjugation between an F’ cell and an F- cell. Do you think this transfer commonly benefits the recipient? Explain.

Answer 27

F’ can conjugate with F- converting them into a F’ cell. Gets all the benefits of a F+ however that plasmid has some bacterial DNA in it as well so whatever those genes are gets transferred If they confer some kind of benefit like the ability to process lactulose or antibiotic resistance then the F- gene gains that.

Question 28

How is a fertility factor like a virus?

Answer 28

Both can have DNA integration into a bacterial chromosome.

Question 29

What is a transposon? How long is it?

Answer 29

Mobile DNA elements, also called “transposable elements” or “jumping genes”, ~700-40,000 base pairs in length. Able to move from one position to another in the genome.

Question 30

What is transposition?

Answer 30

Transposons that move one from position to another in the genome. Sometimes they pick-up and move, other times they make a copy so the original remains. When they insert themselves into the coding region of a gene they cause a frameshift mutation and stop translation.

Question 31

How is a transposon like a fertility factor?

Question 32

Can transposons cause mutations? Explain.

Answer 32

Yes, transposons are mutagens. They can cause mutations in several ways: If a transposon inserts itself into a functional gene, it will probably damage it.

Insertion can result in partial or complete activation of the gene it inserted itself into. 

Excision if compete can reactivate the gene restoring its function. If imperfect excision occurs partial reactivation can occur.

Question 33

Are there any transposons present within the human genome? Explain.

Answer 33

Found in my organisms including humans. About half human DNA consists of inactivated transposons and transposon-like elements.

Question 34

Describe the basic structure of a transposon.

Answer 34

Vary in nucleotide sequence but all possess genetic palindromes at both ends (read the same way both backwards and forwards). Semi autonomous DNA elements, able to move within a gene, able to insert elsewhere within chromosomes or plasmids, plasmids carry transposons to new cells.

Question 35

What is an insertion sequence?

Answer 35

Simplest transposon. Palendome on the ends and transposable gene in the middle. Gene codes for transposase. Enzyme transposase recognizes these terminal repeats, cuts the DNA, and inserts the transposon elsewhere in the genome. Transposon and target sequence are both duplicated in this process.

Question 36

How does a complex transposon differ from an insertion sequence?

Answer 36

Contain additional genes beyond transposons. e.g genes for antibiotic resistance.

Question 37

In what way are transposons evolutionarily important?

Answer 37

Transposons have the ability to insert themselves on a plasmid and then through horizontal gene transfer can be shared creating a separate advantage and opening a door for evolution.

Question 38

Which of the following is NOT typically used in the process of indirect selection?

• Selective conditions
• Permissive conditions
• Replica plating
• ALL of these are used in the process of indirect selection
• Incubation
• Sterile felt or velvet

Answer 38

ALL of these are used in the process of indirect selection

Question 39

Which of the following is NOT an example of a spontaneous mutation?

• Incorporation of an “enol” tautomer during DNA synthesis
• ALL of these are exmples of spontaneous mutations
• Covalent alteration of DNA by chemical products of fatty acid metabolism
• Imperfect crossing over
• Errors in replicationby DNA polymerase in the presence of ethidium bromide
• Nondisjunction during meiosis

Answer 39

Errors in replicationby DNA polymerase in the presence of ethidium bromide. **Becuase it is a mutagen. **

Question 40

Cytosine is converted into uracil through the process of ___________.

• depurination
• hydroxylation
• beta-oxidation
• deamination
• conjugation
• interchelation

Answer 40

Deamination

Question 41

In bacteria, DNA polymerases IV and V are involved in ____________.

• SOS response
• mismatch repair
• light repair of T-T dimers
• proofreading
• dark repair of T-T dimers
• base excision repair

Answer 41

SOS Response

Question 42

What characteristics do all 3 forms of horizontal transfer share?

Answer 42

All three share these 3 characteristics: transformation is unidirectional (donor to recipient), onlay a subset of the donor DNA is transferred (recipient is different form its donor and original self), and homologous genes are replaced in the recipient (new genes are integrated and any duplicates are removed, no net change in gene number).