Bacterial Genetics (trans 5)

Question 1

Definition of Terms

Answer 1

 Genetics – study of genes, heredity, and variation in living organisms
 Chromosome – threadlike structure of nucleic acids and protein carrying genetic information in the form of genes
 Genome – total genetic content contained in a haploid set of chromosomes in eukaryotes, in a single chromosome in bacteria, or in the DNA or RNA of viruses.
 Genes – distinct sequence of nucleotides forming part of a chromosome that codes for a known cellular function or process

Question 2

DNA vs RNA
DNA
Superstructure: Double-stranded; Double-helical, anti-parallel, double-stranded & supercoiled Pentose-phosphate backbone: Deoxyribose (no hydroxyl group attached to the pentose ring in the 2’ position)
Base pairing: Complementary base to adenine is thymine
Base pairs present: Cytosine (C), Guanine (G), Adenine (A), Thymine (T)

Answer 2

RNA
Superstructure: Single-stranded, helical (However, RNA can, by complementary base pairing, form intra-strand double helices, as in tRNA)
Pentose-phosphate backbone: Ribose (with hydroxyl groups that make RNA less stable and more prone to hydrolysis)
Base pairing: Complementary base to adenine is uracil, which is an unmethylated form of thymine
Base pairs present: Cytosine (C), Guanine (G), Adenine (A), Uracil (U)

Question 3

DNA AND RNA FUNCTIONS
Replication
 Bacterial DNA synthesis
 Semi-conservative & bi-directional starting at ori(origin) locus (at replication fork) and ending at ter(termination) locus
 Origin – contains an AT-rich region adjacent to the sequences that are recognized by the DNA-binding protein, DnaA.
 DnaA binds to the DNA boxes to initiate a process that leads to an opening of the AT-rich region
 The replication bubble in the DNA molecule is opened further by helicase, and the replication forks at each end of the bubble are expanded
 Single-strand binding proteins (SSBPs) then attach to each single stranded DNA molecule to stabilize the open bubble and subsequently topoisomerases bind to relieve the supercoiling of the double strand DNA created by the unwinding of the strands

Answer 3

 DNA primase – synthesize short RNA primers
 Single primers are synthesized in the origin to make the two leading strands
 DNA Polymerase III – extends each primer by synthesizing a daughter strand of the DNA in 5’→3’ direction as it moves towards the replication fork
 Each lagging strand is made away from the replication fork
 DNA polymerase I – extends the primers by synthesizing a short segment of the daughter strand called an Okazaki fragment (5’→3’ direction) until it reaches the RNA primer from the previously made Okazaki fragment
 DNA polymerase I removes the primer and fills in the vacant region with DNA.
 DNA ligase – ligates two adjacent Okazaki fragments

Question 4

DNA AND RNA FUNCTIONS
Transcription
 Bacterial mRNA synthesis
 Process by which the information in a strand of DNA is copied into a new molecule of messenger RNA (mRNA)
 Operon – a group of adjacent genes controlled by a set of proteins interacting with specific sequences in DNA molecule; leads to initiation of transcription or termination of transcript
 Transcription – initiated just beyond the promoter region; terminated at the termination site
 Promoter region – a short sequence on the DNA that PROMOTES binding of proteins required for initiation. This is NOT where initiation starts
 The sequence of the promoter at positions 35 and 10 nucleotides upstream from the transcription start site are critical to initiating transcription.
 Core RNA polymerase and sigma factor form a holoenzyme, which binds to the promoter region of the DNA molecule forming a closed complex between RNA polymerase and DNA molecule
 The breaking of the H bonds between the bases in the double helix will convert the closed complex into the open complex

Answer 4

 The formation of the RNA transcript from NTPs then takes place as RNA polymerase is forming the new transcript starting at the 5’-end of the transcript.
 New nucleotides are added onto the free 3’-end. The initiation of transcription is completed when sigma factor is released.
 As elongation of the transcript then continues the RNA polymerase moves down the DNA molecule, and the next sequence in the chain is opened up.
 As RNA polymerase moves down the DNA strand the DNA molecule entering RNA polymerase unwinds worming the open complex, while the DNA exiting RNA polymerase rewinds to form double helix.
 Ribonucleotide triphosphates are added to the chain, and the growing RNA transcript continues to elongate as new DNA sequence enters the open complex region.
 The new RNA transcript initially forms an RNA-DNA hybrid complementary to the template strand, corresponding to the sequence of the coding DNA strand.
 In this way the DNA double helix is opened, transcribed, and reclosed with minimal stress on the DNA molecule
 The mRNA has a nucleotide sequence complementary to a template strand in the DNA double helix if read in the 3′–5′ direction. Thus, an mRNA is oriented in a 5′–3′ direction

Question 5

DNA AND RNA FUNCTIONS
Translation
Bacterial amino acid synthesis
o Process in which cellular ribosomes create proteins
 In prokaryotic cells, translation is initiated by formation of an initiation complex consisting of the 30S ribosomal subunit, formyl-methionyl(f-Met) tRNA,and messenger RNA. The 50S ribosomal subunit then joins the complex.
 The70S ribosome has two sites to which tRNA-carrying amino acids can bind. One is called the peptidyl or P site and the other is called the acceptor or A site. The exit or E site is where tRNAs are released.
 The initiating tRNA, carrying f-Met, binds to the P site. A tRNA that recognizes the next codon and carries the second amino acid then moves into the A site
 The f-Met carried by the tRNA in the P site is then joined to the amino acid carried by the tRNA that just entered the A site by a peptide bond

Answer 5

 The ribosome now advances a distance of one codon and the tRNA that carried the f-Met is released at the E site.
 A tRNA carrying the next amino acid now moves into the A site where the anticodon on the tRNA matches the codon on the mRNA.
 The ribosome shifts down by a distance of one codon. As the shift occurs, the two amino acids on the tRNA in the P site are transferred to the new amino acid and the second tRNA is released from the E site.
 The ribosome continues to move along the mRNA and new amino acids are added to the growing polypeptide chain.
 Elongation of the polypeptide is terminated when a stop codon moves into the A site. A stop codon does not specify an amino acid and does not have a corresponding tRNA.
 The ribosome dissociates into the 30S and 50S subunits and the mRNA and protein are released.

Question 6

PROKARYOTIC CHROMOSOMES

Answer 6

A. Haploid
B. Replicons
C. Pathogenicity Island
D. Plasmids
E. Housekeeping Genes (Jawetz)
F. Transposons
G. Bacteriophages

Question 7

PROKARYOTIC CHROMOSOMES

Haploid

Answer 7

 Since bacterial genes are haploid (with few exceptions), they have single copy of gene thus mutations are easily expressed
 Majority of prokaryotic genomes (>90%) consist of single circular DNA molecule containing 580 kbp to more than 5220 kbp of DNA
 Few bacteria have genomes consisting of two circular DNA molecules

Question 8

PROKARYOTIC CHROMOSOMES

Replicons

Answer 8

 Covalently closed DNA circles (such as the entire bacterial chromosome and plasmids), which contain genetic information necessary for their own replication

Question 9

PROKARYOTIC CHROMOSOMES

Pathogenicity Island

Answer 9

 Specific genes for pathogenic determinants that are often clustered together in the DNA
 Genomic island that converts a “harmless” bacterium to a pathogen when they are integrated into the bacteria’s genome
 Distinct region of DNA which is present in pathogenic bacteria but absent in nonpathogenic strains of the same species.
 Reason why some bacterial species and subspecies are efficient at causing disease in higher organisms compared to other from the same genus.

Question 10

PROKARYOTIC CHROMOSOMES

Plasmids

Answer 10

 Small DNA molecule that is physically separate from chromosomal DNA.
 Can replicate independently from chromosomal DNA
 They are most commonly double stranded
 Plasmids carry genes with independent evolutionary origins. They have the following specialized functions:
- Mediate own transfer from one organism to another
- Genetic acquisition
- DNA rearrangement
- Confer antibiotic resistance and virulence factors

Question 11

**Examples of metabolic activities determined by plasmids

Answer 11

Organism:activity
Pseudomonas: degradation of camphor, toluene, octane, salycylic acid
Bacillus stearothermophilus: a-amylase
Alcaligenes eutrophus: Utilization of H2 as oxidizable energy source
E.coli: sucrose uptake and metabolism, citrate uptake
Klebsiella species: nitrogen fixation
Streptococcus (group N): lactose utilization, galactose phosphotransferase system, citrate metabolism
Rhodospirillum rubrum: synthesis of photosynthetic pigment
Flavobacterium species: nylon degradation

Question 12

PROKARYOTIC CHROMOSOMES

Housekeeping Genes

Answer 12

 These are genes associated with special functions essential for growth.
 These are contained in both bacterial chromosomes and plasmids.

Question 13

PROKARYOTIC CHROMOSOMES
Transposons
 Segments of DNA that include genes that can migrate from one locus to another
 Usually enter the cell by being carried on a plasmid
 Can create insertion mutations
 Do not carry genetic information required to couple their own replication to cell division
 Propagation depends on their physical integration with a bacterial replicon
 Insert in random pattern but favor regions encoding tRNAs

Answer 13

Movement of transposons can either be:
 From plasmid to host (bacterial genome)
 From one site on the genome to another (replicates and leaves a copy at original site)
 From host to plasmid
There are two methods of transposition:
 Copy and Paste Mechanism – a copy of the transposon is left on the original site
 Cut and Paste Mechanism – no copy is left on the original site

Question 14

PROKARYOTIC CHROMOSOMES
Bacteriophages
 Viruses associated with prokaryotes. They are viruses that infect and replicate within bacteria
 Constitute the largest of all viral groups
 Nucleic acid molecules:
- Surrounded by a protein coat
- dsDNA (most often, ssDNA and ssRNA)
- Sometimes contain unusual bases such as hydroxymethicytosine
 Exhibit a wide variety of morphologies
 Usually contain specialized syringe-like structures (tails) that bind to receptors on the cell surface and inject the phage nucleic acid into host cell
 When the phage is loaded with nucleic acid, it takes on a different form than when the nucleic acid is absent

Answer 14

Types of propagation:
 Lytic cycle
 Lysogenic cycle

Question 15

PROKARYOTIC CHROMOSOMES
Bacteriophages - Types of propagation:
Lytic cycle (also called vegetative replication)
- Phages produce many copies of themselves inside host cell
- After repeated replication, the cell lyses and releases all the viruses that replicated inside the cell
- Newly produced viruses infect other cells and repeat the process
- Phage kills infected host cell -> Phage breaks open -> death of host cell
Lysogenic cycle
- DNA material of bacteriophages incorporate with the bacterial (host) chromosome, forming a prophage
- Replication of the host chromosome also replicated the DNA of the virus
- There is no damage to host cells thus higher risk of harming the immune system unknowingly

Answer 15

Depending on the physiologic state of bacteria, it can switch between lysogeny and lytic phage:
 Repression (Non-lytic/temperate phage)
- An established prophage frequently confers a cellular immunity against lytic infection by similar phage so it is in lysogenic phase
 Derepression (released from repression)
- Triggered by different stimuli; prophage undergoes lytic replication, leading to formation of a burst of infectious particles

Question 16

Regulation of Prokaryotic Gene Expression

**Operons: cluster of prokaryotic structural genes that encode a related series of metabolic reactions

Answer 16

a) Activation: initiation of transcription
b) Attenuation: a regulatory mechanism that controls the efficiency of transcription after transcription has been initiated but before mRNA synthesis of the operon’s genes takes place, especially when the end product of the pathway is in short supply.
c) Repression: can be viewed as a course-control mechanism, an all-or-non approach to gene regulation; prevention of transcription

Question 17

BACTERIAL GENE TRANSFER
Mechanisms of Recombination:
- Donor DNA that does not carry information necessary for its own replication must recombine with recipient DNA to become established in a recipient strain.

Answer 17

2 types:

1. Homologous
   - Rec gene(reciprocal transfer)
   - A consequence of close similarity in the sequences of donor and recipient DNA
   - Almost always involve exchange between genes that share common ancestry
2. Non-homologous
   - Gene conversion (non-reciprocal transfer)
   - Result of enzyme-catalyzed recombination between 2 dissimilar DNA sequences
   - Insertion of DNA into a recipient to form a copy of a donor transposon

Question 18

BACTERIAL GENE TRANSFER
Mechanisms of Gene Transfer:
- Exchange of small pieces of genome (a few genes at a time)

Answer 18

3 broad mechanisms mediate efficient movement of DNA between cells:
A. Conjugation
B. Transduction
C. Transformation

Question 19

BACTERIAL GENE TRANSFER
Mechanisms of Gene Transfer - Conjugation
 Requires donor cell-to-recipient cell contact to transfer only one strand of DNA
 Recipient cell completes dsDNA by synthesizing strand that complements the strand acquired from donor cells
 Plasmids are most frequently transferred by conjugation

Answer 19

1. Bacterial conjugation

2. Conjugation: Transfer of Chromosomal DNA

Question 20

BACTERIAL GENE TRANSFER: Mechanisms of Gene Transfer
Conjugation: Bacterial conjugation
 Requires direct contact between the cells. Many, but not all, species of bacteria can conjugate
 Can occur between cells of the same species, or ever between cells of different species.
o F factor (Fertility factor )– small DNA circle or plasmid which is required for conjugation.
o F plus – strains of bacteria containing the F factor
o F minus – strains of bacteria without F factor
 An F plus cell, or donor, produces a structure called a Pilus to connect with another recipient cell.
 To begin conjugation, the F factor is cut at a specific region called the origin of transfer by a protein assembly called the relaxosome.
 Relaxosome associates with the strand to be transferred, or the T-DNA strand.

Answer 20

 Accessory proteins of the relaxosome are released, but a portion of the relaxosome called the relaxase remains attached to the T-DNA.
 This T-DNA/relaxase complex is recognized by a coupling factor and transferred to the exporter, a complex in the F plus cell that is contiguous with the pilus.
 The exporter pumps the T-DNA relaxase complex into the recipient cell. Once the entire T-DNA molecule is transferred to the recipient cell, relaxase joins the ends to make a circular DNA molecule.
 As the T-DNA is transferred to the recipient cell, it is replicated to become double-stranded. In the donor cell, the F factor DNA was also replicated to become double-stranded. This actually occurred as the T-DNA was being transferred to the recipient cell. In the end, both cells wind up with a complete, double-stranded copy of the F factor.
 Their connection through the pilus is released, and each is now an F plus cell that can go on to conjugate with other cells.

Question 21

BACTERIAL GENE TRANSFER: Mechanisms of Gene Transfer
Conjugation: Transfer of Chromosomal DNA
 Sometimes, F plasmid becomes integrated into the host cell genome. The host cell is now referred to as HFr (high frequency of recombination) which is able to transfer some of the host genes to the recipient.
 The sex pilus contacts the recipient F minus cell and pulls the cells together. The donor chromosome is transferred as single-stranded DNA starting at the origin of transfer. Gene 1, which is the closest to the origin, is transferred first. Segments of the integrated plasmid are at the beginning and the end of the DNA being transferred

Answer 21

 It is theoretically possible for the complete genome and the F plasmid to be transferred to the recipient cell. However, this does not happen because the donor and recipient cells will separate prior to the complete transfer of the donor chromosome. The transferred DNA becomes double-stranded
 The donor DNA is integrated into the recipient cell’s DNA by homologous recombination. The recipient now carries transferred genes, but remains F minus, whereas the donor cell remains HFr.

Question 22

BACTERIAL GENE TRANSFER: Transduction
 Donor DNA is carried in a phage coat and is transferred into the recipient by the mechanism used in phage infection
 In generalized transduction, a segment of DNA is carried from one bacterial cell to another by a bacterial virus called bacteriophage or phage. The phage attaches to the bacterial cell and injects its nucleic acid into the host cell.
 A phage enzyme is produced that breaks down the host DNA into smaller fragments. Phage DNA is replicated and phage coat proteins are produced.

Answer 22

 During formation of the mature phage particles, a few phage heads may surround fragments of bacterial DNA instead of phage DNA. The phage particle carrying the bacterial DNA infects another cell, transferring the bacterial DNA to the new cell.
 When the bacterial DNA is introduced into the new host cell, it can become integrated to the bacterial chromosome, thereby transferring genes to the recipient. This cell then multiplies and carries new genetic material.

Question 23

BACTERIAL GENE TRANSFER: Transformation
 Direct uptake of “naked” donor DNA by the recipient cell
 DNA Transformation involves the transfer of naked DNA into a recipient cell.
 In the first step, double-stranded donor DNA binds to specific receptors on the surface of a competent cell. One strand of the donor DNA is degraded by nucleases while the other strand enters the cell.

Answer 23

 The single-stranded donor DNA pairs with a homologous region on the recipient DNA and is integrated into the recipient genome by a breakage and reunion mechanism called homologous recombination.
 If there are any differences between the nucleotide sequences of the donor and recipient DNA’s, the mismatch repair system comes into play. The repair system removes either the donor or the recipient strand, and replaces it with the complementary sequence. Since either strand may be repaired, some cells contain the new donor DNA and others have the original DNA sequences. In the laboratory, cells are plated on selective media so that only the transformants will grow

Question 24

MUTATION AND GENE REARRANGEMENT

Answer 24

A. RESTRICTIONS ON GENE TRANSFER

B. SPONTANEOUS MUTATION

Question 25

MUTATION AND GENE REARRANGEMENT: RESTRICTIONS ON GENE TRANSFER
 Remember that not all genes can be transferred from one bacterium to another, sometimes there are restrictions and limitations wherein the genetic material of a bacterium is not transferred to the other.
 These restrictions are the following: restriction endonucleases and plasmid incompatibility.

Answer 25

1. Restriction Endonucleases (watch VIDEO15) – cleavage of donor DNA before becoming part of the recombinant replicon. Sometimes, due to these enzymes (i.e. some restriction endonucleases) of the recipient cell, the incoming genetic information gets cleaved (i.e. the genetic information gets destroyed) before it can be incorporated into the cell.
2. Plasmid Incompatibility – incompatible plasmids result in one plasmid lost at a higher than normal rate when cell replicates. Remember: PLASMIDS are the ones that go in and out of the cell to transfer genetic information but sometimes they are not compatible to the recipient bacteria.

Question 26

MUTATION AND GENE REARRANGEMENT: RESTRICTIONS ON GENE TRANSFER
 Restriction endonucleases are enzymes that cleave DNA at specific nucleotide sequences. The sequence recognized is often four to six nucleotides long. For example, the restriction endonuclease EcoRI recognized the sequence, GAATTC.
 The nucleotides at one end of the recognition sequence are often complementary to those at the other end. The two strands of the DNA duplex have the same nucleotide sequence running in opposite directions for the length of the recognition sequence.
 Because the same recognition sequence occurs in both strands of the DNA duplex, the restriction endonuclease can bind to and cleave both strands of the DNA molecule.

Answer 26

 Because the bond cleaved is typically not positioned in the center of the recognition sequence, and the DNA strands are anti-parallel, the cut sites are offset from each other.
 After cleavage, each DNA fragment has a single stranded end a few nucleotides long and the single stranded ends of the two fragments are complementary to each other. These single stranded ends can pair with each other (sticky ends).
 Once their ends have paired, the two fragments can be joined together with the enzyme DNA ligase, which re-forms the phosphodiester bonds of DNA.
 Significance: Any two fragments of DNA produced by the same restriction endonuclease can be joined together. Restriction endonucleases are fundamental tools in genetic engineering.

Question 27

Different Types and Activities of Restriction Enzymes
Type I: Cleaves DNA at random sites far from its recognition sequence
Type II: Cleaves DNA at defined positions close or within its recognition sequence
Type IIG: Cleaves outside its recognition sequence with both Rease and Mtase enzymatic activities in the same protein
Type IIP: Cleaves symmetric targets and cleavage sites

Answer 27

Type IIS: Recognizes asymmetric sequences
Type III: Cleaves outside its recognition sequence and require 2 sequences in opposite orientations within the same DNA
Type IV: Cleaves modified DNA (ex: methylated)

Question 28

MUTATION AND GENE REARRANGEMENT: SPONTANEOUS MUTATION
- This includes base substitution, insertion & deletion (frameshift), rearrangement & duplication, inversions, transpositions.

Answer 28

a) Base Substitution
b) Insertion and Deletion (Frameshift)
C. Reversion
D. Suppression

Question 29

MUTATION AND GENE REARRANGEMENT: SPONTANEOUS MUTATION - Base Substitution
– consequence of mispairing between complementary bases during replication. It results to MISSENSE (i.e. when coding a different protein) and NONSENSE (i.e. when no protein is produced, nothing is produced).
 If there are mistakes or mutations, there are ways to repair them. The bacteria also have repair mechanism similar to humans. For example, some bacteria have the characteristic to withstand intense heat because they can secrete a cover, if they lose that property, they will die, and thus, this is not beneficial to these bacteria. What they do is to repair.

Answer 29

 These repair mechanisms are mismatch repair and SOS response.
 Mismatch Repair – a mechanism that corrects the mismatch. For example, the mismatch A-T pair is corrected and replaced by C-G pair to form the normal protein.
 SOS Response – mechanism involving a very large damage.

Question 30

MUTATION AND GENE REARRANGEMENT: SPONTANEOUS MUTATION
 A mutation occurs by base substitution when an incorrect base is incorporated into DNA. Some base substitutions occur because purines and pyrimidines exist in two structural forms.
 The most common form results in base-pairing between adenine and thymine, and between guanine and cytosine.
 However, the hydrogen atoms can move to form a base with altered hydrogen bonding properties, creating a tautomeric shift. When a tautomeric shift occurs in adenine, the adenine can bond to cytosine.

Answer 30

 A tautomeric shift in thymine allows it to bond to guanine.
 This error in DNA replication is passed on to the cell’s progeny. The change in a single nucleotide in the DNA results in a change in the corresponding nucleotide in messenger RNA.
 The change in the codon can result in a different amino acid being incorporated into the protein.

Question 31

MUTATION AND GENE REARRANGEMENT: Insertion and Deletion (Frameshift)
- Introduction or removal of a single base pair from DNA.
 The nucleotide sequence in DNA determines the nucleotide sequence in mRNA and consequently, the sequence of amino acids in a protein.
 A mutation in the DNA can result in a change in the amino acid sequence of a protein.
 Insertion: One type of mutation that can occur either spontaneously or as the result of a mutagen is the addition of one or more nucleotides during DNA replication.
 Because translation of a gene begins with a specific codon and proceeds one codon at a time, the addition of an extra nucleotide shifts the codons in the mRNA. This is called frameshift mutation.

Answer 31

 A frameshift affects all amino acids incorporated beyond the original site at which the addition occurred.
 If the new codon generated by the frameshift is a stop codon, the protein synthesized will be shortened and is often nonfunctional.
 Deletion: Another type of frameshift mutation occurs when a nucleotide is deleted during DNA replication.
 Deletion of a nucleotide in DNA results in a change in the codons in mRNA from the point of the deletion and changes in the amino acids inserted into the protein.

Question 32

MUTATION AND GENE REARRANGEMENT: Reversion

Answer 32

 In some conditions, for example, an enzyme cannot be coded anymore due to lost activity, but the lost activity can be regained through reversion, either phenotypic reversion or genotypic reversion. The bacteria have these mechanisms to revert back their original state because of these repairs.
 Phenotypic Reversion – process of regaining an activity lost as a consequence of mutation.
 Genotypic Reversion – process of restoring the original DNA sequence.

Question 33

MUTATION AND GENE REARRANGEMENT: Suppression

Answer 33

 Suppression is similar to a secondary mutation which helps to restore the original state of the gene. There are two types: intragenic suppression and extragenic suppression.
 Intragenic Suppression – second mutation at a different site within the affected gene restores the structure required for activity.
 Extragenic Suppression – second mutation outside the affected gene restores the structure required for activity.

Question 34

GENETIC ENGINEERING: Genetic Modification

Answer 34

 It is the direct manipulation (DNA recombination) of an organism’s genome to introduce desirable traits using biotechnology
 Site-directed mutagenesis

Question 35

GENETIC ENGINEERING: Cloning
 It is a set of experimental methods used to assemble recombinant DNA molecules and to direct their replication within host organisms

Answer 35

Steps in Gene Cloning:
Step 1: Isolate the desired gene
Step 2: Purify and fragment with a restriction enzyme
Step 3: DNA fragments are incorporated into bacterial plasmids
Step 4: DNA ligation
Step 5: Transformation

Question 36

GENETIC ENGINEERING: Cloning
 Step 1: Isolate the desired gene
 Step 2: Purify and fragment with a restriction enzyme
o Restriction enzyme produces staggered cuts in specific sequences in the DNA, generating fragments with sticky ends
o Each fragments has a single-stranded sequence of nucleotides on its ends that is capable of hybridizing with DNA that has been fragmented with the same restriction enzyme
 Step 3: DNA fragments are incorporated into bacterial plasmids
o The plasmid used for cloning has a single restriction site, and when cleaved by the restriction enzyme, generated the same cohesive ends that are in the fragments of the DNA to be clones
o Generally, plasmid contains:
- Cloning/restriction site – where foreign DNA fragment can be inserted
- Drug-resistance gene – destroys antibiotics to allow selective growth of the host cell
- Replication origin – allows plasmid to replicate the host cell

Answer 36

 Step 4: DNA ligation
o DNA ligase binds the cohesive ends of the plasmid.
 Step 5: Transformation
o The host cells are added to the recombinant plasmids. In transformation, a few cells take up a recombinant plasmid while most other cells do not.
o The replication origin allows the plasmid to replicate by using the host cell enzymes.
o Plasmid replication is independent of host cell division, but plasmids are distributed to each daughter cell when the host cell divides.
o As the plasmids replicate and the host cells multiply, the number of copies of the recombinant plasmid is greatly amplified.
o The multiple daughter cells form a colony/clone.
o Because all the host cells in a colony are derived from a single cell, they all contain copies of the same recombinant plasmid with its fragment of foreign DNA.
o A variety of assay methods can now be used on the bacterial colonies to determine which contains the particular DNA sequence we wish to isolate

Question 37

DNA CHARACTERIZATION - Restriction Mapping

Answer 37

 It is the process of obtaining structural information on a piece of digested DNA by the use of restriction enzymes and then separating the resultant DNA fragments by gel electrophoresis
 Used as a reference to engineer plasmids or DNA

Question 38

DNA CHARACTERIZATION - Sequencing

Answer 38

 It is the process of determining precise order of nucleotides within a DNA molecule
 Displays gene structure
 Deduce the amino acid sequence of gene products
 Analysis reveals regulatory regions that control gene expression & genetic “hot spots” susceptible for mutation.

Question 39

DNA CHARACTERIZATION - Hybridization
Hybridization Probes
a) Southern blot
o Method for detection of a specific DNA sequence in DNA samples
o Hybridization of DNA to DNA
b) Northern blot
o technique to study gene expression by detection of RNA in a sample
o provide quantitative information about RNA synthesis
c) Western blot (protein immunoblot)
o antibodies are used to detect cloned genes by binding to their protein products.

Answer 39

Hybridization Techniques
a) Restriction Fragment Length Polymorphism (RFLP)
o Demonstrated whenever the Southern blot pattern (distribution of restriction sites) obtained with one individual is substantially varied from the one obtained with another individual
o Used to trace DNA from a small sample to its human donor.
b) Polymerase Chain Reaction (PCR)
o Biochemical technology used to amplify a single or a few copies of a piece of DNA across several orders of magnitude, generating thousands to millions of copies of a particular DNA sequence

Question 40

The Structure of DNA and rNA
The base pairs are stacked within the center of the DNA double helix, and they determine its genetic information. Each turn of the helix has one major groove and one minor groove. Certain proteins have the capacity to bind DNA and regulate gene expression by interacting predominately with the major groove, where atoms comprising the bases are more exposed.

Answer 40

Each of the four bases is bonded to phospho-2′-deoxyribose to form a nucleotide. Th e negatively charged phosphodiester backbone of DNA faces the solvent. Th e length of a DNA molecule is usually expressed in thousands of base pairs, or kilobase pairs (kbp) . Whereas a small virus may contain a single DNA molecule of less than 0.5 kbp, the single DNA genome that encodes Escherichia coli is greater than 4000 Kbp.

Question 41

A few RNA molecules have been shown to function as enzymes (ribozymes). For example, the 23S RNA in the 50S ribosomal subunit catalyzes the formation of the peptide bond during protein synthesis.

Answer 41

siRNAs are double-stranded RNA molecules, 20–25 nucleotides in length, that play a variety of roles in biology. Some have been shown to function as regulators by either binding near the 5′ end of an mRNA, preventing ribosomes from translating that message, or base pairing directly with a strand of DNA near the promoter, preventing transcription.

Question 42

The prokaryotic genome
Some bacterial species are efficient at causing disease in higher organisms because they possess specific genes for pathogenic determinants. These genes are often clustered together in the DNA and are referred to as pathogenicity islands.

Answer 42

These gene segments can be quite large (up to 200 kbp) and encode a collection of virulence genes. Pathogenicity islands (1) have a different G + C content from the rest of the genome; (2) are closely linked on the chromosome to tRNA genes; (3) are flanked by direct repeats; and (4) contain diverse genes important for pathogenesis, including, antibiotic resistance, adhesins, invasins, and exotoxins. as well as genes that can be involved in genetic mobilization.

Question 43

The prokaryotic genome

Genes essential for bacterial growth (often referred to as “housekeeping genes”) are carried on the chromosome

Answer 43

plasmids carry genes associated with specialized functions

Question 44

The prokaryotic genome
Transposons are genetic elements that contain several genes, including those necessary for their migration from one genetic locus to another. In doing so, they create insertion mutations. The involvement of relatively short transposons (0.75–2.0 kbp long), known as insertion elements, produce the majority of insertion mutations. These insertion elements (also known as insertion sequence [IS] elements) carry only the genes for enzymes needed to promote their own transposition to another genetic locus but cannot replicate on their own. Almost all bacteria carry IS elements, with each species harboring its own characteristic ones.

Answer 44

Transposons do not carry the genetic information
required to couple their own replication to cell division, and therefore their propagation depends on their physical integration with a bacterial replicon. This association is fostered by enzymes that confer the ability of transposons to form copies of themselves; these enzymes may allow the transposons to integrate within the same replicon or an independent replicon. Many plasmids are transferred among bacterial cells, and insertion of a transposon into such a plasmid is a vehicle that leads to the transposon’s dissemination throughout a bacterial population.

Question 45

The Viral Genome
Viruses are capable of survival, but not growth, in the absence of a cell host. Replication of the viral genome depends on the metabolic energy and the macromolecular synthetic machinery of the host.

Answer 45

Therefore, successful propagation of the virus requires (1) a stable form that allows the virus to survive in the absence of its host, (2) a mechanism for invasion of a host cell, (3) genetic information required for replication of the viral components within the cell, and (4) additional information that may be required for packaging the viral components and liberating the resulting virus from the host cell.

Question 46

The Viral Genome
Distinctions are frequently made between viruses associated with eukaryotes and viruses associated with prokaryotes, the latter being termed bacteriophage or phage.

Answer 46

The nucleic acid molecule of bacteriophages is surrounded by a protein coat. Considerable variability is found in the nucleic acid of phages. Many phages contain double-stranded DNA; others contain double-stranded RNA, single-stranded RNA, or single stranded DNA. Unusual bases such as hydroxymethylcytosine
are sometimes found in the phage nucleic acid.

Question 47

Phage
Phages can be distinguished on the basis of their mode of propagation. Lytic phages produce many copies of themselves as they kill their host cell. The most thoroughly studied lytic phages, the T-even (eg, T2, T4) phages of E coli, demonstrate the need for precisely timed expression of viral genes to coordinate events associated with phage formation.

Answer 47

Temperate phages are able to enter a nonlytic prophage (indicating that they have inserted into the bacterial chromosome) state in which replication of their nucleic acid is linked to replication of host cell DNA. Bacteria carrying prophages are termed lysogenic because a physiologic signal can trigger a lytic cycle resulting in death of the host cell and liberation of many copies of the phage. The best characterized temperate phage is the E coli phage λ (lambda).

Question 48

Phage
The dsDNA of many lytic phages is linear, and the first
stage in their replication is the formation of circular DNA. This process depends upon cohesive ends, complementary single-stranded tails of DNA that hybridize. Ligation, formation of a phosphodiester bond between the 5′ and 3′ DNA ends, gives rise to covalently closed circular DNA that may undergo replication in a manner similar to that used for other replicons. Cleavage of the circles produces linea DNA that is packaged inside protein coats to form daughter phages.

Answer 48

The ssDNA of filamentous phages is converted to a circular double-stranded replicative form. One strand of the replicative form is used as a template in a continuous process that produces single-stranded DNA. The template is a rolling circle, and the ssDNA it produces is cleaved and packaged with protein for extracellular extrusion.

Question 49

Phage
The ssDNA of filamentous phages is converted to a circular double-stranded replicative form. One strand of the replicative form is used as a template in a continuous process that produces single-stranded DNA. The template is a rolling circle, and the ssDNA it produces is cleaved and packaged with protein for extracellular extrusion.

Answer 49

ssRNA phages are among the smallest extracellular
particles containing information that allows for their own replication. The RNA of phage MS2, for example, contains (in fewer than 4000 nucleotides) three genes that can act as mRNA following infection. One gene encodes the coat protein, and another encodes an RNA polymerase that forms a dsRNA replicative form. ssRNA produced from the replicative form is the core of new infective particles.

Question 50

Phage
Some temperate bacteriophages, exemplified by E coli
phage P1, can be established in a prophage state as a plasmid. The dsDNA of other temperate bacteriophages is established as a prophage by its insertion into the host chromosome. The site of insertion may be quite specific, as typified by integration of E coli phage λ at a single int locus on the bacterial chromosome. The specificity of integration is determined by identity of the shared DNA sequence by the int chromosomal locus and a corresponding region of the phage genome. Other temperate phages, such as E coli phage Mu, integrate in any of a wide range of chromosomal sites and in this aspect resemble transposons.

Answer 50

Prophages contain genes required for lytic replication
(also called vegetative replication), and expression of these genes is repressed during maintenance of the prophage state. A manifestation of repression is that an established prophage frequently confers cellular immunity against lytic infection by similar phage. A cascade of molecular interactions triggers derepression (release from repression), so that a prophage undergoes vegetative replication, leading to formation of a burst of infectious particles. Stimuli such as ultraviolet (UV) light may cause derepression of the prophage. The switch between lysogeny—propagation of the phage genome with the host—and vegetative phage growth at the expense of the cell may be determined in part by the cell’s physiologic state. A nonreplicating bacterium will not support vegetative growth of phage, but a vigorously growing cell contains sufficient energy and building blocks to support rapid phage replication.

Question 51

Mechanisms of Gene Transfer
The DNA composition of microorganisms is remarkably fluid. DNA can be transferred from one organism to another, and that DNA can be stably incorporated in the recipient, permanently changing its genetic composition. This process is called horizontal gene transfer to differentiate it from the inheritance of parental genes, a process called vertical inheritance. Three broad mechanisms mediate efficient movement of DNA between cells—conjugation, transduction, and transformation.

Answer 51

Conjugation requires donor cell-to-recipient cell contact to transfer only one strand of DNA. The recipient completes the structure of double-stranded DNA by synthesizing the strand that complements the strand acquired from the donor. In transduction, donor DNA is carried in a phage coat and is transferred into the recipient by the mechanism used for phage infection. Transformation, the direct uptake of “naked” donor DNA by the recipient cell, may be natural or forced. Forced transformation is induced in the laboratory, where, after treatment with high salt and temperature shock, many bacteria are rendered competent for the uptake of extracellular plasmids. The capacity to force bacteria to incorporate extracellular plasmids by transformation is fundamental to genetic engineering

Question 52

Mutagens
The frequency of mutation is greatly enhanced by exposure of cells to mutagens. Ultraviolet light is a physical mutagen that damages DNA by linking neighboring thymine bases to form dimers. Sequence errors can be introduced during enzymatic repair of this genetic damage.

Answer 52

Chemical mutagens may act by altering either the chemical or the physical structure of DNA. Reactive chemicals alter the structure of bases in DNA. For example, nitrous acid (HNO2) substitutes hydroxyl groups for amino groups. The resulting DNA has altered template activity during subsequent rounds of replication.