Microbial Genetics Flashcards

Question

Describe the components of a Genetic map of the chromosome of E.coli. What are ORF?s What are short tandem repeats and what do they contain? What do the numbers represent in the genetic map? REVIEW

Answer 1

Genetic Map of the Chromosome of E.coli 1. ORFs: Open reading frame in DNA sequence that lacks termination codons and so can potentially translate as a polypeptide chain The entire genome does NOT consist of back to back genes (there are DNA between genes that do not code for anything) 2. Short Tandem Repeats (STRs) -Short DNA sequences (usually 2-5 base pairs) that are repeated numerous times in a head-tail manner. They are important for use in DNA fingerprinting ( bacterial and humans) The repeats will be same, but number of repeats will be different. -The polymorphisms in STRs are due to the different number of copies of the repeat element that can occur in a population The numbers on the genetic map (0-100) represent **Conjugation minutes** (in eukaryotes, introns throw you off in open reading frame; so to help with issue, make CDNA)

Answer 2

DNA structure -Polymer of nucleotides: Adenine, thymine, cytosine, and guanine -Double helix -"Backbone is deoxyribose-phosphate -Strands are held together by hydrogen bonds between AT and CG -Strands are antiparallel

Answer 3

Get rid of thiamine, replace with Uracil; ADD hydroxyl (OH) group here (to make ribose sugar; 2 OHs)

Answer 4

Transcription goes from 5' to 3' 5' refers to phosphate and 3' refers to OH on sugar. always extend 5' to 3'

Answer 5

Semiconservative Replication -1) The double helix of the parental DNA separates as weak hydrogen bonds between the nucleotides on opposite strands break in response to action of replication enzymes 2) Hydrogen bonds form between new complementary nucleotides and each strand of the parental template to form new base pairs 3) Enzymes catalyze the formation of sugar-phosphate bonds between sequential nucleotides on each resulting daughter strand

Answer 6

DNA replicates semi conservativiely. Strand opens up, new nucleotides that complement prexisting nucleotides of template come in and form double bonds enzymes link sugar phosphates together. before known that DNA replication was semiconservative, it was thought replication was Conserative (1 parent made intact, entirely new double stranded molecule) or Dispersive model

Answer 7

Semi conservative model : produces 2 helices in which each helix contains one New strand and One Old (parental ) strand Conservative Model : produces 2 helices in which one helix contains entirely OLD DNA and other helix contains entirely NEW DNA. Dispersive model: produces 2 helices in which each strand contains alternating segments of Old and New DNA -Messelson and Stahl discovered DNA replication was semi-conservative

Answer 8

DNA Synthesis -DNA Polymerases only add new DNTPs to the 3' end -Energy is supplied from phosphates of DNTP's (deoxynucleoside triphosphate) -When a nucleotide triphosphate bonds to the sugar, it loses two phosphates. -Hydrolysis of the phosphate bonds provides the energy for the reaction

Answer 9

nucleoside: Base and sugar Nucleotide: base, sugar and phosphate (either 1, 2 or 3 phosphates)

Answer 10

DNA Synthesis: DNA is copied by the DNA polymerase -in the 5' --> 3' direction -Leading strand is synthesized continuously -Lagging strand is synthesized discontinuously (Okazaki fragments) and initiated by an RNA primer -RNA primers are removed by a DNA polymerase and Okazaki fragments joined by DNA ligase -Overall error rate of E.coli DNA polymerase is low (10^-5) then further reduced to a 10^-9 bases due by PROOFREADING

Answer 11

Steps of DNA Synthesis : 1) Enzymes unwind the parental double helix 2) Proteins bind stabilize the unwound parental DNA (single stranded binding proteins ; ssbps) 3) The leading strand is synthesized continuously by DNA polymerase 4) The lagging strand is synthesized discontinuously. Primase, an RNA polymerase, synthesizes short RNA primer, which is then extended by DNA polymerase 5) DNA polymerase digests RNA primer and replaces it with DNA polymerase 6) DNA ligase joins the discontinuous fragments of the lagging strand

Answer 12

Important enzymes in DNA Replication, Expression and Repair: 1. DNA Gyrase: relaxes supercoiling ahead of replication fork; (Generates (-) supercoils) 2. **DNA Ligase: makes covalent bonds to JOIN DNA strains; Okazaki fragments and new segments in excision repair 3. **DNA polymerases: Synthesize DNA; proofread and facilitate repair of DNA 4. Endonucleases: Cut DNA backbone in a strand of DNA; facilitate repair and insertions 5. **Helicase: Unwinds double stranded DNA 6. Methlyase: adds methyl group to selected bases in newly made DNA 7. Photlyase: uses visible light energy to separate UV-induced pyrimidine dimers 8. **Primase: An RNA polymerase that makes RNA primers from a DNA template 9. Ribozyme: RNAs that possess catalytic activity (can catalyze chemical reaction) 10. **RNA Polymerase: Copies RNA from a DNA template 11. SnRNP: RNA protein complex that removes introns and splices exons together 12. Topoisomerase or Gyrase: Relaxes superocoild ahead of replication fork; Separates DNA circles at the end of DNA replication 13. Transposase: Cuts DNA backbone, leaving a single-stranded "sticky ends"

Answer 13

BOTH DNA Gyrase and Topoisomerase RELAX Supercoiling TYPE I Topoisomerase; that RELAXES supercoiling and cuts ONE DNA strand . -DNA Gyrase is like a form of type II Topoisomerase that RELAXES supercoiling (NEGATIVE Supercoiling) and REDUCES supercoiling. It cuts TWO strands of DNA

Answer 14

Yes, DNA Gyrase is Prokaryotic specific: only bacteria snRNPS; (small nuclear ribonucleotides proteins) are only eukaryotic (the rest of enzymes are found in both)

Answer 15

E.coli DNA replication is BIDIRECTIONAL it has **single origin of replication* Process: The replication initiates at origin of replication and Polymerase moves in 2 different directions (bidirectional) -2 replication forks that will continue until you have gone all around chromosome and separate two loops with Topoisomerase Then, two genomes get attached to cell membrane at opposite poles, so that when cell divides you segregate chromosomes into different cells **After duplication, each copy of the origin binds to the membrane at OPPOSITE POLES** bacterium E.coli requires 40 minutes to finish replication time for chromosome -but the doubling time of organism is as fast as 20 minutes (2 replication forks, one continuous side, other side discontinuous)

Answer 16

because when organism initiates replication every 20 minutes; after first initiation starts at origin of replication, it makes a complete genome every 20 minutes. It initates a new set every 20 mins, so hence completes another chromosome every 20 minutes (1 is halfway done)

Answer 17

Bacterial Transcription: -DNA is transcribed to make RNA(mRNA, tRNA, rRNA, etc) -Transcription begins when RNA polymerase binds to the PROMOTER sequence -Transcription proceeds in the 5' to 3' direction -Trancription stops when it reaches a TERMINATOR sequence (RNA polymerase is bound to DNA)

Answer 18

Process of Transcription in Prokaryotes: 1) RNA polymerase binds to the promoter, and DNA unwinds at the beginning of a gene 2) RNA is synthesized by complementary base pairing of free nucleotides with the nucleotides bases on the template strand of DNA 3) The site of synthesis moves along DNA; the DNA that has been transcribed rewinds. 4) Transcription reaches the terminator 5) RNA and RNA polymerase are released the DNA helix re-forms.

Answer 19

Prokaryotic RNA polymerases: single polymerase with 1 transcriptional factor (sigma factor that regulates the polymerase) Eukaryotic RNA polymerases: 3 types of RNA polymerases, and lots of transcription factors

Answer 20

Bacterial Rho dependent Termination : Nucleotide sequences are recognized by Rho protein, which bind sequence and cause polymerase to come off Rho independent Termination: sequence dependent; NO proteins involved in sequence. When RNA is made, it will form stem loop structure. Stem is G-C rich (strong stem ) and stem loop is followed by a string of U's (uracils) causing polymerase to fall off

Answer 21

mRNA Transcription and Processing in Eukaryotes: 1. In the nucleus, a gene composed of exons and introns is transcribed to RNA by RNA polymerase 2. Processing involves snRNPS (small nuclear ribonuclear proteins) in the nucleus to Remove the Intron-derived RNA and Splice together the Exon-derived RNA into mRNA 3) After further modification, the mature mRNA travels to the cytoplasm, where it directs protein synthesis **Further Eukaryotic mRNA modifications** Not shown: -5' Methyl guanosine Cap at 5'end (beginning of sequence) -Poly A tail added at end of sequence Introns are removed in the nucleus before mRNA gets transported out of the cytosol (where translation occurs)

Answer 22

Translation (process of converting mRNA into protein) -mRNA is translated into CODONS (three nucleotides) -Translation of mRNA begins at the START codon: AUG -Translation Ends at NONSENSE Codons; **UAA, UAG, UGA**

Answer 23

The Genetic Code -61 sense codons on mRNA encode the 20 amino acids (total of 64 codons; if include 3 nonsense/stop codons) -The genetic code is DEGENERATE (several codons can code for same amino acid) EXCEPT AUG (Methionine) has 1 codon -tRNA carries the complimentary ANTICODON (anticodon is the trinucleotide sequence that is complementary to codon in mRNA) tRNA charged with Amino acid to match mRNA

Answer 24

Degeneracy enables SILENT Mutations (if change the last codon ,you still have same amino acid) ex; Serine has 4 different codons that code for same aa (can change 2 of nucleotides in codon) -in bacteria, the start codon encodes for N-formylmethionine (specific tRNA carries N-formylmethionine )

Answer 25

Codon Bias: The probability that a given codon will be used to code for an amino acid over a different codon ,which codes for the same amino acid.

Answer 26

NO, because most of them (start codons: Nformlymethioine) become processed and removed. Although they start with the codon that encodes for N-fomylmethionine, they have enzymes that will remove it

Answer 27

Necessary components for initating translation: - 2 ribosomal subunits (large and small_ -tRNA, -Anticodon (charged with amino acids) -Methionine (start codon) mRNA

Answer 28

Recognize correct AUG (start) codon: in Prokaryotes, In Eukaryotes, Uses SCANNING Model: Cap (at 5' end) is recognized by the ribosome and when it binds to the ribosome, it scans it for the first AUG. ( CAP protects 5' end and is needed for ribosomes to translate ) In prokaryotes, use Shine Dalgarno Sequence. Upstream of correct AUG, a few nucleotides up, there is a short sequence called shine Dalgarno. This sequence is complementary to sequence in ribosomal RNA, which allows Ribosome to recognize where it should start. (shine dalgarno sequence matches sequence in ribosome, to align it and be translated.) prokaryotes have Shine dalgarno sequence allow ribosome to align with AUGs, also have operon: 3 different start codons and 3 different stop codons

Answer 29

Operon: group of genes that are under control of single operator. Multiple genes on a single RNA transcript (operons are bacteria specific) (eukaryotes doe NOT have operons; since they cannot internally load)

Answer 30

Translation 1. On the ribosome, when ribosome binds DNA, there are three sites, E site (Exit) P site (peptidyl) and A site (Aminoacyl) 2. When you first initiate translation , the first tRNA goes to the P site (only time get amino acid in P site is at initiaton of translation) (subsequently tRNA goes to A site). So have start codon (formylmethioine in bacteria) and next tRNA 3) Another tRNA goes to A site (carrying Leucine) , peptide bond formed between Methionine and leucine 4) methionine will then be released from tRNA and ribosome will move along mRNA in 5' to 3' causing tRNA (that had start codon) to move to exit site and tRNA carrying leucine (attached to Met) to P site. Also another tRNA carrying with another anticodon (that codes for GGU) another Amino acid (glycine) will enter A site. --Transfer leucine to glycine to form peptide bond 5) 3 tRNAS in the E (Aug start codon), P site (leucine with Methionine attached) and A site (Glycine). This process keeps going (growing polypeptide chain) on until reach end of gene. Then, Nothing enters A site because of STOP codon and polypeptide is released from tRNA. 6) Ribosome comes off and new protein is made.

Answer 31

In Prokaryotes, transcription and translation are COUPLED (RNA polymerase (catalyzing transcription) and Ribosome (catalyze translation) can bind to same RNA transcript As soon as polymerase starts making mRNA, polymerase transcribes as soon As AUG code in shine dalgarno sequence are exposed, they are bound by ribosome and start translating before whole mRNA is made As soon as Ribosome moves past AUG , and moves farther down transcript and Aug is exposed again another ribosome will bind hence Polycistronic ( mRNA that encodes for multiple different protein products) Transcription and translation occur at the same time (they are coupled)

Answer 32

Transcription and translation are NOT coupled in Eukaryotes. This is because in eukaryotes, Transcription occurs in Nucleus; mRNA is processed in multiple different ways only when mRNA is fully processed, its exported to cytosol for translation . Hence translation and transcription cannot occur at the same time in eukaryotes.

Answer 33

Transcriptional control in bacteria -Constitutive genes (60-80%) are expressed at a FIXED rate (not regulated in any way; expressed at same level) -Other genes are expressed only as needed -Repressible genes: transcription rate decreased by a regulator protein called a repressor. Default position is ON -Catabolite repression: the INHIBITION of synthesis of alternative catabolic enzymes by a preferred carbon source (ex; glucose) (if you already have glucose, you shut off transcription of other enzymes, since glucose is preferred) Inducible genes: transcription rate Increases by a regulatory proton called an Inducer. Default position is OFF. (protein binds promotor and increases transcription rate)

Answer 34

Lac Operon has both negative and positive regulation structure: *I region is the repressor. It is NOT regulated and has its own promoter *The operon genes for lac operon: ZYA is expressed from a single promoter and has operator operator: a sequence that a repressor binds (control region structural genes: PO ZYA) -P is promoter -O is Operator

Answer 35

Negative regulation of the Lac operon: The repressor mRNA is produced and converted to protein. That repressor binds the operator sequence and prevent RNA polymerase from moving forward and hence BLOCKING transcription **Repressor Active, Operon OFF**

Answer 36

Induction of Lac operon: A small molecule called Allolactose (inducer) binds to repressor, it will cause lac operon to come off -Allolactose: disaccharide, similar compound to lactose that is generated when you bring lactose into the cell. (lactose itself does not affect repressor) beta galactosidase : enzyme that cleaves disaccharide to glucose and galactose. It cleaves a beta bond between the two sugars **Repressor Inactive, Operon still OFF** permease : transport enzyme that brings in lactose Transacetylase; enzyme that puts acetyl group on lactose lactose: disaccharide of glucose and galactose If you have lactose around, you can use enzymes to degrade lactose. operon is still off, since you need something for RNA polymerase to bind. Just removing repressor does not turn on operon (need activator)

Answer 37

To turn on Operon: 1. you have to remove repressor 2. activate repressor

Answer 38

Catabolite Repression: the growth of cells on two different carbon sources; Lactose and glucose -The growth is FASTER if give cell Glucose, hence it is the preferred carbon source. if you mix lactose and glucose together, you find RAPID growth initially, until all of the glucose is used up. Then you have a lag time, and the lactose gets used. ***You do not turn on lactose operon and use lactose until you have used up all of glucose**

Answer 39

To turn on Lactose operon: 1. You must have lactose 2. You do NOT have glucose

Answer 40

Positive regulation of the lac operon when have Lactose and glucose present; Remove the Repressor, BUT RNA polymerase CANNOT bind, because you need Catabolite Activator Protein (CAP) to be active. Under these conditions, CAP is NOT active, so polymerase does not bind and have no products Lactose present, NO glucose: CAP binds the promoter and facilitates loading of the RNA Polymerase. Now you do NOT have repressor and polymerase can transcribe, make products. (Active CAP), inactive lac repressor)

Answer 41

To activate transcription: 1. Activator must be present 2. NO repressor You activate CAP by producing cyclic AMP (cAMP) Glucose induces expression or activation adenylate cyclase. The more adenylate cyclase you have, the more cAMP you have cAMP is a coactivator. The more adenylate cyclase you have (made by glucose), the more cAMP The more glucose, LESS cAMP. (hence cAMP is indirectly proportional to glucose)

Answer 42

1. Presence of Glucose and presence of lactose: **Operon is OFF, because CAP is Not Bound** (also have no repressor) (CAP not bound since glucose is present ) 2. Presence of glucose, absence of lactose: **Operon is OFF both because Lac repressor is BOUND and because CAP is NOT bound** (lac repressor bound, since no lactose present) 3. Absence of Both glucose and Lactose: **Operon is OFF because lac repressor is bound** (since glucose is not present, CAP is bound) 4. Absence of glucose; presence of lactose: ***OPERON is ON** ( becuase CAP is bound, allowing RNA polymerase to bind to promoter and start transcription) (also, Lac repressor is not bound since, lactose is present) -

Answer 43

The Trp Operon: synthesizes Tryptophan -It is normally an INACTIVE Repressor **Repressor inactive, Operon on** if you do not have tryptophan, the repressor is inactive and operon is on (allow biosynthesis of tryptophan) .

Answer 44

Negative Regulation of the Trp operon -if Repressor is active, operon is OFF this occurs if you have tryptophan, you activate repressor and shut off biosynthesis of tryptophan (prevent production of tryptophan) -Tryptophan is a co-repressor, since the presence of tryptophan will allow it to bind to repressor protein and activate it. This activation will then turn OFF operon and shut off biosynthesis of tryptophan

Answer 45

Mechanisms of transcriptional regulation in bacteria You have activators and repressors that are regulated by a small LIGAND lac repressor: Allolactose is the ligand that regulates its binding Trp operon: ligand is tryptophan -CAP is regulated by cAMP

Answer 46

Negative Regulation (bound repressor protein Prevents protein transcription) - Usually a bound repressor protein to promoter will turn gene off. However, if you ADD a ligand to repressor protein, it will switch gene ON, by removing the repressor protein - Also, a repressor protein that is bound to ligand and attached to promoter can turn gene off. However, if you REMOVE ligand from repressor protein, it can switch gene ON, by removing repressor protein. Positive Regulation (Bound Activator protein Promotes transcription) -a bound activator protein to promoter region usually allows RNA polymerase to bind and turn gene on. However, when ADD a ligand you switch gene OFF, by removing activator protein - Also, if you have a activator protein with ligand bond to gene , it allows RNA polymerase to bind to promoter region and turn gene on. However, if you remove ligand, it switches gene OFF, by removing activator protein

Answer 47

it depends on the PLACEMENT of the PROMOTER. - When the protein binds near promoter region, it can recruit polymerase to bind. Hence, the protein has to have ability to bind sequence and bind polymerase. if protein binds upstream from promoter, it will not be able to bind polymerase, it will just block. repressors stop transcription by blocking. activators enhance transcription by recruiting promoter or help promoter bind or help initiation of transcription.

Answer 48

Epigentic control (ex; DNA methylation): heritable changes in gene expression that occur WITHOUT changes in the DNA sequence -A change in phenotype Without a change in genotype DNA methylation is an example -In eukaryotes, only (C*G) nucleotides are methylated (the C in CG is methylated), causing genes to be SILENCED. These silenced genes (off) are passed to offspring, but can be reversed later -In Prokaryotes, GA*TC is the predominant methylation (A is methylated in GATC), and is important for Restriction Modification and Replication Fidelity. May also influence gene expression (Restriction Modification: protects sites from being cleaved by enzymes)

Answer 49

Additional steps of bacterial gene expression that can be regulated in bacteria: 1) mRNA stability 2) Translation 3) Protein stability

Answer 50

Regulatory RNA molecules -Eukaryotic: have microRNAs (miRNAs) that are approximately 22 nucleotide single-stranded RNAs that are incorporated into a RISC (RNA induced silencing complex). RNAs in RISC serve as a guide to target Specific complementary mRNAS and will either alter their translation or stability (degrade mRNA or inhibit translation) (the miRNAs are modified in nucleus and packaged into complex called RISC). -Bacteria: Bacterial small RNAs (sRNAs) are highly structures 50-500 nucleotide RNAs (bigger than miRNAS) that generally bind to PROTEIN targets and modify function. Alternatively may bind to mRNA targets and regulate translation or stability.

Answer 51

Mutation: A change in the base sequence of DNA -mutations may be neutral, beneficial or harmful Mutagen: Agent that causes mutations Spontaneous Mutations: Occur in the ABSENCE of a mutagen (they occur during DNA replication, when polymerase makes mistakes) -mutagen will increase rate

Answer 52

Different types of sequence changes (mutations) in DNA: A. Base substitutions: substitution of one single nucleotide base for anther 1. Transitions- change in one purine to another purine 2. Transversions- change from purine to pyrimidine (**purines: AG; Pyrimidines: CUT**) B. Additions and Deletions C. Inversion: a deletion and reassertion in the opposite orientation -point mutation, specifically base substitution is most easily reverted spontaneously (RNA polymerase can be corrected mistake)

Answer 53

Point mutation: Change within a gene in which a base pair in the DNA sequence is changed, inserted, or deleted (base substitutions like transitions and transvehrsion are examples of point mutations) -(additions and deletions can be point mutations if single nucleotide involved) inversion-NOT point mutation

Answer 54

DNA sequence changes within ORF (open reading frame) can impact the Amino acid composition of the protein translated -Missense mutants: a base substitution that results in the replacement of one amino acid for another (ex: alanine to glycine) -Nonsense mutants: base substitution that creates an in frame STOP codon -Silent mutations: a base substitution does NOT alter the amino acid composition of the protein (due to degeneracy code) (change last codon and still get same aa) -Frame shift mutants: insertion or deletion of one or few bases which alters the reading frame.

Answer 55

b/c There's a RANDOM chance of hitting a STOP codon

Answer 56

YES, nucleotide changes outside ORF can impact gene expression because **you can have mutation in promoter (affect polymerase binding),mutation in operator sequence, have mutations that affect transcriptional stops and mutations that affect regulatory aspects of gene expression.

Answer 57

Temperature sensitive mutants: mutation in a protein in which protein is more unstable at a particular temperature Ex: TS: high temperature; raise temp, and protein unfold and protein is no longer active--> conditional lethal mutations if have temperature sensitive mutant that might be lethal at high temp (42 degrees Cfor E coli) but low temp at 40 degrees C, it will be permissive and survive.

Answer 58

-Missense mutation: example change cysteine to thymine; mRNA will change codon and cause amino to change from glycine to Serine -Nonsense mutation change nucleotide from thymine to adenine; mRNA will code for a change to U, so have UAG codon which is STOP codon -Frame shift mutation (Deletion): deletion of A will change frame, and change phenylalanine to Leucine, and glycine to alanine, and randomly generate stop codon

Answer 59

Mutation rate: the probability that a gene will mutate when a cell divides (according to textbook) Mutation rates can be defined in several ways: 1) mutations per base pair per generations/cell division ** 2) mutations per gene per generation/cell division** 3) mutations per genome per generation/cell division** (less common) Typically expressed as powers of 10. (

Answer 60

E. coli Spontaneous mutation rate= 1 in 10^9 replicated base pairs (10^-9) or 1 in 10^6 replicated genes (10^-6) or 1 in 10^3 (10^-3) -Mutagens can INCREASE mutation rate > 1000 fold

Answer 61

Mutations are generally Neutral or detrimental (due to evolution) ***cell division and generation are equivalent for single-celled organisms**

Answer 62

Mutagenic agents: -Base modifiers: ex: Nitrous acid and hydroxylamine (modify bases that are already in DNA sequence) -Base analogs: ex: Bromouracil -Radiation: -Frame shift mutagens : Acridine (orange) -Mobile genetic elements also X-rays, UV (Ultraviolet) Light are mutagenic agents

Answer 63

Base modifiers: **Nitrous Acid** -Base modifiers like Nitrous Acid modifies a preexisting base ***Nitrous acid Deaminates adenines and cytosine, so they base pair with C and A ** So, Adenine that is deaminated will base pair with cytosine (deaminated cytosine base pair Adenine) Result: end up getting a G to C change in SECOND generation (first altered parent, one daughter normal, one altered with cytosine)

Answer 64

Nucleoside Analog: molecules that look like the normal nucleotides but Base pair differently (incorporated into DNA, but once in DNA they will change how they base pair) -can form Tautomers ex; **2-Aminopurine nucleoside that looks like Adenine Ex2: **5-Bromouracil nucleoside looks like Thymine**

Answer 65

tautomeric forms: switch form so they can change base pairing 2 AP (2-Aminopurine nucleoside) can base pair with Thymine; but also reform its structure (be protonated) and base pair with Cytosine (has 2 tautomers) - 5-Bromouracil can be incorporated in DNA (Keto form) and base pair with Adenine; but it can also change to Enol form where it base pairs with Guanine

Answer 66

Ionizing radiation (X rays and gamma rays) Ionizing radiation can damage DNA directly or indirectly -Direct action: X-rays and gamma rays can physically BREAK the sugar phosphate backbone (resulting in single or double-strand break), the bases themselves, or the connecting hydrogen bonds (they can break covalent bonds in backbone, purines, or anywhere) Indirect action: cause the formation of ions (produce hydroxyl radical) that can oxidize bases resulting in errors in replication and repair that causes mutations

Answer 67

UV radiation can damage DNA by causing THYMINE DIMERS (UV radiation is NOT Ionizing radiation) UV Thymine Dimers: covalent bond formed between two adjacent thymines on same strand Repair Mechanisms; 1. Photolyases: Enzymes that separate thymine dimers 2. Nucleotide excision repair (not limited tot thymine Dimers) -The thymine dimers produce bulge that can be recognized by enzymes called Photolyases that can break bond (requires light for enzyme function) Another way to repair damage: process of Nucleotide excision repair: -Exonuclease will cut the DNA (cleave on both strands of thymine dimer) and remove damaged DNA -DNA polymerase will fill the gap by synthesizing New DNA, using intact strand as template (replace nucleotides that have been removed) -DNA ligase will SEAL the gap

Answer 68

NO, Nucleotide excision cannot repair deletions because it needs a TEMPLATE strand (to fill in missing part)

Answer 69

RNA polymerase has a mechanism where it looks for mistakes You must distinguish new strand from old strand ***METHYLATION will help recognize new synthesis from old strand new strand will NOT be methylated yet (Assume new synthesis has mutation)

Answer 70

Frame shift mutagens (cause Deletions/insertions) Examples: -Benzopyrene: present in smoke and soot -Aflatoxin: secreted by a variety of molds that grow on peanuts and grain -Acridine orange: Fluorescent cationic dye These mutagens are planar molecules (Flat) that intercalate (insert) in between the bases, that cause polymerase to insert additional nucleotides or delete nucleotides.

Answer 71

The Base analog Aziodthymidine (AZT) is uses as an Antiretroviral drug -Thymidine analog is used to treat AIDS (one of first drugs used for AIDS) Mechanism of action: it TERMINATES the grown chain (when incorporated into DNA, nothing else can be added to it or elongated) (AZT one of first targets of endoviruses)

Answer 72

Retrovirus is impaired more than the host because: 1. Viral Reverse Transcriptase has a 100 fold greater affinity for AZT than cellular DNA polymerase (viruses recognize RT as being more normal nucleotide than DNA polymerase; so RT inserted more frequently) 2. Cells can quickly repair DNA chain if broken by AZT, whereas the virus cannot.

Answer 73

YES the mitochondrial RNA polymerase is MORE Sensitive than cellular DNA polymerase

Answer 74

mutations are RARE, even in the presence of a mutagen Strategies to detect mutants: 1) Postiive (Direct) selection detects mutant cells that GROW under conditions their unmutated parents wouldn't 2) Negative (Indirect) selection detects mutants cells that do NOT grow. -replica plating 3) A screen identifies mutant colonies that appear different

Answer 75

It is useful to generate mutants to make more Antibiotics, make a protein more temperature resistant or more active. In research, it helps to ascribe function to genes. (used to understand gene function ) (ex: mutate gene and ask what consequences are of mutant) -Bacterial are useful for generating mutants because: -Large numbers -Reproduce quickly -Most are HAPLOID (beneficial since phenotypes are masked)

Answer 76

Direct Selection: -Grow on media supporting mutant, but NOT wild-type -only rare mutants will survive -ex: antibiotic resistance (trying to find resistance to streptomycin) original cells are sensitive and if put on streptomycin plates, media won't grow. if mutants cells, you have ones that grow streptomycin.

Answer 77

Indirect selection -Used when there is NO medium on which the mutant can grow and the parent cell cannot -Mutants that require a growth factor Not normally present (ex: auxotrophic mutants) -conditional lethal mutants (ex: Ts) -Procedur requires: Replica plating

Answer 78

Replica plating process: 1. Sterile velvet is pressed on the grown colonies on the master plate. -The master plate with medium contains Histidine (amino acid; enriched complex medium) 2. Cells from each colony are transferred from the velvet to new plates -have petri plate with medium containing histidine -have petri plate with medium lacking histidine 3. Plates are incubated 4. Growth on plates is compared. A colony that grows on the medium with histidine, but could not grow on the medium without histidine is Auxotrophic (histidine requiring mutant) (this method is not really how it works, since no stamping really happens, instead you use replica block) in the end you get two copies

Answer 79

Replica block: -put velveteen over top of block, put plate on block, transfer the cells, take the two plates, transfer 1 and then transfer second plate. You can keep transferring a whole bunch of cells and transfer cells from velveteen and makes many copies -the number of cells you transfer will decrease

Answer 80

Direct selection: you can put a billion cells on one plate. Hence Direct selection is EASIER to screen (faster process), and uses less effort with indirect selection, you can only put about 150 colonies on plate.

Answer 81

Phenotypic Screening -look for recognizable differences in colony morphology- Not selection, just screening -pH indicators (change color) -Differential breakdown of material in media Ex; Sugars ex; blood agar ( see if there is zone of clearing if degrade blood)

Answer 82

Ames Test: test used to determine whether a particular compound is mutagenic -uses bacteria Salmonella to test whether chemical compounds can cause mutations in DNA of an organism. (used to identify chemical carcinogens) Process: You have a culture of bacteria (Salmonella strain ) -the Salmonella strain has mutation in it, so it CANNOT make histidine Can have different mutations (point mutation, deletion, transition/transversion) in one of the histidine genes -Add suspected mutagen and rat liver extract, allow cells to grow and then take cells and put media that has no histidine. They will get colonies. Assumption: The colonies grow on media since they have mutation Hence you start with a point mutation, and then look for another mutation that reverts back to wild type (original mutation) ex: change adenine to thymine; you want to change thymine back to adenine 1) How do you know this growth is not due spontaneous mutation? (but due to mutagen) - because there is a control experiment you have to do. No mutagen added, so you have two colonies. (you expect mutagen to increase number) -control: same experiment, except NO mutagen is added -Different mutagens cause different types of mutations. you must test different salmonella stains that have different types of mutations in histidine gene, to determine which mutant it is.

Answer 83

Rat liver extract purpose: trying to mimic what happens in your body. if you ingest compounds that go in liver. The liver will metabolize into something else. You try to Add enzymes from rat liver. increase in colony; demonstrates its a mutagen

Answer 84

Yes, 90% of mutagens are carcinogens

Answer 85

Procedure for Ames Test: 1st you have a strain of Histidine dependent Salmonella that is control (has No suspected mutation added) that will add to media (that lacks histidine). This media will be incubated and will result in only a few colonies of revertant bacteria Then you have experimental sample of smamonella strain (that histidine dependent) and contains suspected Mutagen. You add to media (lacking histidine) and incubate for a couple hours, and result in many colonies of bacteria (due to salmonella mutating to restore histidine gene) The higher number of revertant colonies of bacteria in experimental sample that contains mutagen, suggests that the mutagen will cause mutation in gene, allowing histidine to be produced and increase in colonies to form. -Initially point mutations are made in histidine dependent bacteria (salmonella), causing bacteria to not be able to produce Histidine. Hence bacteria will not grow unless hisitndine is supplied. Once Salmonella (that lack histidine) are cultured in media with mutagen, mutations in gene cause a reversion to amino acid production (reverse mutation)

Answer 86

Vertical Gene transfer: occurs during reproduction between generations of cells Horizontal gene transfer: the transfer of genes between cells of the SAME generation Bacteria can undergo BOTH vertical gene transfer and Horizontal gene transfer (replication correlate to vertical gene transfer; Recombination correlates to Horizontal transfer) (in bacteria, replication occurs first, then recombination)

Answer 87

3 mechanisms of horizontal gene transfer in bacteria I. Transformation II. Conjugation III. Transduction Once DNA gets into the cell, it has to be replicated (or it will be lost when cell divides) -Since the transferred DNA usually does NOT contain an origin of replication, these genes will be passed on to succeeding generations, only if the transferred DNA becomes incorporated in the recipient chromosome.

Answer 88

If DNA in cell is circular and has origin of replication, it can be replicated -if DNA is linear (does not have origin of replication) it must be incorporated into genome (or it will get lost)

Answer 89

Homologous Recombination: The exchange of genes between two DNA molecules -Crossing over occurs when two chromosomes break and rejoin -Recombination is a complex process involving different enzymes, with redundant functions **The RecBCD pathway is the main pathway used in bacteria ** - ***Homologous recombination requires the RecA protein** **sequences that are recombined are homologous ** if DNA is linear and integrates into chromosomes you need 2 crossover events (only parts where crossover occurs must be identical sequence) **If DNA is circular and integrates with chromosome, (with homologous sequence) you only need to have 1 recombination event (1 cross over event)**

Answer 90

Homologous Recombination 1. DNA from one cell aligns with DNA in recipient cell. Notice that there is a nick in the donor DNA 2. DNA from the donor aligns with complementary base pairs in the recipient's chromosome. This can involve thousands of base pairs 3. Rec A protein catalyzes the joining of the two strands 4. The result is that the recipient's chromosme contains new DNA. Complementary base pairs between two strands will be resolved by DNA polymerase and ligase. The donor DNA will be destroyed. The recipient may now have one or more genes -This process differs substantially in eukaryotes

Answer 91

Frederick Griffith's eperiment (1928) deomonstrated GENETIC TRANSFORMATION experiment: 1. Living encapsulated bacteria injected into mouse, the mouse died. The cosines of encapsulated bacteria were isolated from dead mouse 2. Living non- capsulated bacteria (non virulent) injected into the mouse, the mouse remained healthy. A few colonies of non-encapsulated bacteria were isolated from mouse; phagocytes destroyed the nonencapsualted bacteria 3. Heat killed-encapsulated bacteria injected into mouse, the mouse remained healthy . No colonies were isolated from the mouse 4. Living nonencapsulated and heat-killed encapsulated bacteria injected into mouse. Mouse died. Colonies of encapsulated bacteria were isolated from deda bacteria. -1944, Avery, Macleod and McCarty showed that DNA was responsible for converting the harmless streptococcus pneumonia into virulent strain

Answer 92

Transformation: The Direct uptake of DNA by recipient cells -natural transformation is an active process demanding specific enzymes produced by the recipient cell -The natural occurrence of this property is unusual among specific enzymes produced by the recipient cell. -The natural occurrence of this property(naturally take up DNA) is unusual among bacteria (1% in laboratory conditions) and Species Specific (about 15 species in bacteria) -Although natural transformation is only possible for a small percentage of bacterial species, almost All species can be transformed under laboratory conditions

Answer 93

Transformation process: 1. As the DNA fragment enters the cell (recipient) , one strand is degraded 2. Single strand from donor finds homologous sequence on recipient chromosomes 3. Donor sequence replaces recipient sequence, which is degraded.

Answer 94

Cell competence: Cell's efficiency at uptaking DNA from its environment

Answer 95

Bacterial conjugation: transfer of a plasmid from one cell to another by means of cell to cell contact Conjugation mechanism differs between gram-negative and gram-positive species -conjugation among gram-negative bacteria (ex: Escherichia coli) The F. Plasmid is the best studied conjugative plasmid (F plasmid (fertility plasmid) contains transposons, origin of transfer, origin of vegetative replication)

Answer 96

Gram-negative Bacterial Conjugation -F plasmid-encoded hair-like projection termed sex pili contact specific receptor sites on the surface of the recipient cell (The sex pilus makes contact with F cell and forms mating bridge) -Although the sex plus is a hollow tube, its not a conduit for DNA, but Brings the cells in proximity. (F+ cell carries plasmid; F- cell does NOT carry plasmid)

Answer 97

Mechanism of Gram-negative conjugation 1. An F+ and an F- cell make contact and form cytoplasmic bridge 2. One strand of the plasmid is nicked 3. Strand elongation at the 3' end displaces the 5' end into the bridge 4. The single strand is cut after one complete circle is transferred. At the same time, the complementary plasmid strand is synthesized in the F+ cell. 5. The ends are ligated restoring the doubt stranding circular configuration

Answer 98

Conjugation among Gram-positive bacteria (Enterococcus faecalis: pCF10) -Does NOT involve sex pili - Has recipient and donor cells -Adhesion protein: Asc10 -Donor-produced pheromone: cCf10 -recipient produce pheromone : cCF10 -Phermone inhibitor : ICF10 The single arrow indicate positive control The inverted arrows indicate negative control and the double arrows indicated polypeptide synthesis from a plasmid (ICF10, Asc10_ or chromosomal (cCF10) gene

Answer 99

Mobilization of the E.coli chromosome: Hfr's -The F plasmid can integrate into the E.coli genome via homologous recombination, thereby allowing the chromosome the ability to transfer itself to another cell -A cell with an integrated F is called an Hfr cell -Hfr is the acronym for high-frequency recombination , which refers to the fact that some chromosomal genes are transferred from an Hfr cell to an F-cell with considerably higher frequency than from an F+ cell. -conjugative transfer occurs concomitantly (at the same time) with DNA replication, so that the donor cell does not lose genetic information

Answer 100

They BOTH have Sequences elements; TRANSPOSONS

Answer 101

NO, because the bridge breaks; amount of time

Answer 102

NO, it will occur w sequence elements REVIEW

Answer 103

Formation of the HFR strain: Integration of an F plasmid into a bacterial chromosome (E.coli) occurs by recombination within a simple translocation - Now that you have integrated plasmid, conjugational transfer begins at the ori site; thus the two parts of split plasmid become located at either end of the chromosome. The chromosome now is an Hfr chromosome, capable of being transferred to another cell.

Answer 104

Chromosmes cross to F- cell: 1. a+ enters f- at 5 min 2. b+ first enters F- at 8 min 3. c+ first enters F- at 17 min 4. d+ first enters F- at 30 min The entire chromosome takes 90 mins to cross to F- (this rarely happens because the pairs usually separate before transfer is complete) (note: the whole entire chromosome does not get transformed)

Answer 105

Construction of linkage maps by interrupted-mating experiments F- cells that are str^r are crossed with Hfr cells that are str^s -The F- cells also have a number of mutations that prevent them from carrying out certain metabolic steps REVIEW

Answer 106

The linear order of transfer is dependent on the Hfr strain (insert site and orientation of F plasmid) -Homologous regions were pairing take place are between 2 and b -Where plasmid conjugated depends on where insertions happens -Where the it inserts, dictates where the origin of replication is. The 1 region is transferred last--> the b region b and c regions are direction of transfer and the a-2 region of Hfr are transferred first

Answer 107

Mobilization of the E. coli chromosme: F' plasmids The F plasmid can be correctly excised from Hfr DNA to yield an F+ cell containing the original bacterial chromosome and independent F factor -However, should the excision involve a different bacterial IS element than was originally used for integration, the excised F factor will contain a segment of bacterial DNA. This form is called F' ( genetic map of E.coli chromosome can show approximate regions carried by F-prime factors)

Answer 108

Isolate the plasmid from cell> look at the plasmid REVIEW THIS

Answer 109

Transduction: phage mediated genetic recombinant in bacteria in simplest terms, a traducing particle might be regarded as bacterial DNA (or phage-bacterial DNA hybrid) in a phage coat. Types: -Generalized transduction -Specialized transduction

Answer 110

Generalized transduction: a random fragment of bacterial DNA is accidentally encapsulated in a phage protein coat in place of the phage DNA. The P1 bacteriophage is an example

Answer 111

The frequency of transducing a specific gene from a donor cell to recipient cell is low because: 1. The encapsulation of bacterial DNA is LOW (1/1000) 2. The probability that the transuding particle carries the gene of interest is LOW (depends on genome size and capacity of virus particle) 3. Successful infection of the P1 transducing particle is required 4. The transduced gene must RECOMBINE with the homologous portion of the chromosome

Answer 112

Contrasnduction frequency: the frequency you start with REVIEW

Answer 113

Specialized transduction; Only those genes located on the bacterial genome close to the prophage insertion site can be transduced -A specialized traducing particle contains a hybrid phage and bacterial DNA molecule which integrates into the phage genome as the phage DNA would do in the process of becoming normal in the prophage.

Answer 114

Generalized Transduction -Involves injecting phage and bacterial chromosome, phage DNA -Phage replication and fragmentation of bacterial DNA occur (a+ = any bacterial gene) -Then lysis occurs forming normal nontransducing phages and a rare phage (a+) which has accidentally packaged the a+ gene, and can transduce an a- cell to a+ Specialized Transduction -involves Integrated prophage, UV induction -Includes Normal excision of prophage and RARE abnormal excision of prophage that picks up the adjacent s+ gene (s+ = special bacterial gene) -Then lysis occurs for both processes (Normal excision and Rare excision)/ -Lysis in Normal excision will produce normal nontransducing phage -lysis of Rare abnormal excision will produce phages that carry s+ one and can transduce from s- to s+ hence both generalized and specialized transduction have normal transuding phage, undergo lysis and will make rare phage

Answer 115

Transposable elements: self-mobilizable genetic elements or "jumping genes" that can be transferred from place to place on the chromosme and into and out of plasmids two types: replicative and non replicative transposons

Answer 116

Replicative transposons: leaves a copy of itself at the original location Non-replicative transposon- Does NOT leave a copy of itself -most transposons are replicative

Answer 117

Target site preferences are rarely very stringent or very long. Thus, suitable insertion locations for many IS (insertion sequences) occur thousands of times in each genome. However, insertion is NOT random

Answer 118

3 major kinds of transposable elements: a. Insertion sequence (IS) elements b Transposons (composite and Tn3 type) -carry antibiotic resistance gene c. Certain phages (such as Mu)

Answer 119

Transposes and Resolves are genes that control the movement of transposon Transposon at the original site can jump to other sites, leaving behind a copy itself. it can also jump into and out of plasmid

Answer 120

Insertion of the Transposon Tn5 into R100 1. Transposes cuts DNA, leaving sticky ends 2. Sticky ends of transposon and target DNA anneal

Answer 121

All the previous decscirbe mechanisms permit rapid exchange of genes from one strain to another However, most resistance gene transfer is plasma mediated, but plasma mediated traits can interchange with chromosomal elements -Virtually ALL resistance determinant genes on R plasmids are present at transposons

Answer 122

Escherichia coli O157:H7 Normally E.coli benefits the body by suppressing the growth of harmful bacterial species and synthesizing appreciable amounts of vitamin -A toxin-producing strain of E.coli was first seen in 1982 and is now the Leading cause of Diarrhea worldwide Shigatoxin is the toxin produced by Strain.

Answer 123

Specialized transduction

Microbial Genetics Flashcards

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