Study Guide Ch. 8 DONE

Question 1

Review the structure of a DNA and an RNA molecule and how they differ from each other as well as the rules of base pairing in a DNA molecule.

Answer 1

o DNA: Double stranding (Helix)
o RNA: Single Stranded
o Base Pairing Rules:
 DNA: A-T, C-G
 RNA A-U, T-A, C-G

Question 2

Describe the different steps of DNA replication in the bacterial chromosome.

Answer 2

o Occurs right before cell division (binary fission)
o Replicates in a semi-conservative (part old and part new strand) manor catalyzed by the enzyme DNA polymerase
o Other proteins and enzymes are involved in this process
- Gyrase
- DNA-stabilizing proteins
- Helicase
- Primase
- Ligase
o Most bacterial DNA replication is bidirectional
o Each offspring cell receives one copy of the DNA molecule
o Starts at the Original Chromosome

Question 3

Explain how information flows from gene to protein.

Answer 3

Transcription and Translation

Question 4

Distinguish between transcription and translation and where do they occur in the cell.

Answer 4

o Transcription:
- Process of making a copy of RNA from the DNA genetics
- Occurs in the nucleus of a cell
o Translation:
- mRNA is translated into the “language” of proteins
- Occurs in the cytoplasm of a cell

Question 5

Define codon, and explain what relationship exists between the linear sequence of codons on the mRNA and the linear sequence of amino acids in a polypeptide

Answer 5

o Codon: the sequence of nucleotides in groups of three
o Each codon on the mRNA codes for a specific amino acid

Question 6

Explain the processes of transcription and translation

Answer 6

o Transcription:
 Synthesis of a complementary mRNA strand from a DNA template
 Template DNA strand is the one transcribed (copied by complementary base-pairing)
* Coding strand is not transcribed but is complementary to the template strand. Therefore, it has the same sequence as the mRNA transcribed off the template strand, except its thymine bases are uracil bases in the mRNA
 Begins when the RNA polymerase binds to the promoter sequence on DNA
 Stops when it reaches the terminator sequence on DNA

o Translation:
 When mRNA is translated into the language of proteins
 Groups of three mRNA nucleotides that code for a particular amino acid
 Begins at the start codon AUG
 Ends at nonsense codons
 tRNA molecules have an anticodon that base pairs with the codon

Question 6

Distinguish among mRNA, tRNA and rRNA.

Answer 6

o mRNA: messenger RNA
o tRNA: transcription RNA
o rRNA: Ribosomal RNA

Question 7

Describe the structure of a ribosome and of a tRNA and explain how their structure is related to their function.

Answer 7

o Ribosome: Large and Small Subunits, made of proteins and rRNA
o tRNA: anticodon that is complementary to a codon on the mRNA strand.
- Related: The tRNA binds to the mRNA at the ribosome and ensures that the correct amino acid is added to the growing polypeptide chain.

Question 8

Explain what the genetic code is and list the three stop codons and the one start codon.

Answer 8

o Genetic code is the table that translated codons to amino acids
o Start: AUG
o Stop: UAA, UAG, UGA

Question 9

Given a sequence of bases in DNA, predict the corresponding codons transcribed on mRNA, the corresponding anticodons of tRNA and the corresponding amino acid coded.

Answer 9

o DNA: ATCGGA
o mRNA: UAGCCU
o tRNA: AUCGGA
o aminoa: lle gly

Question 10

Explain the inducible operon model for the control of protein synthesis in bacteria.

Answer 10

o Inducible operon:
 Structural genes are not transcribed unless an inducer is present
 Structural genes include the enzymes necessary to breakdown lactose.
 In absence of lactose, repressor binds to the operator, preventing transcription
 In presence of lactose, metabolite of lactose allolactose (inducer)- binds to the repressor and the repressor cannot bind to the operator and transcription occurs.

Question 11

Explain the Repressible operon model for the control of protein synthesis in bacteria.

Answer 11

o Repressible Operon:
 Structural genes are transcribed until they are turned off
 Excess tryptophan is a corepressor that binds and activates the repressor to bind to the operator, stopping tryptophan synthesis.
 Opposite of inducer: bar is up all the time except when there’s lot of tryptophan because it binds to the repressor and changes shape and bar goes down.

Question 12

Define mutation, induced and spontaneous mutations, mutagenic agent, wild type and mutant strain.

Answer 12

o Mutation: a permanent and heritable change in the base sequence of DNA
o Induced:
o Spontaneous mutations: occur in the absence of a mutagen (DNA polymerase can make mistakes and mismatch base pairs
o Mutagenic agent:
o Wild type: Non-mutated; most common strand
o Mutant strain: shows variance in a characteristic

Question 13

Physical mutagenic agents and explain how they work

Answer 13

Physical:
- Radiation:
- Ionizing radiation (e.g., X-rays, gamma rays) causes the formation of ions that can break the covalent bonds in the DNA backbone. This results in breaks or modifications in the DNA structure, which can lead to mutations during repair.
- Non-ionizing radiation (e.g., UV light) causes thymine dimers to form between adjacent thymine bases in DNA. These dimers distort the DNA structure, leading to replication errors if not repaired by mechanisms like photolyase repair.

Question 14

Explain the difference between insertion sequences, transposons and plasmids.

Answer 14

o Insertion Sequences: about 1,000bp just containing the gene for transposase
o Transposons: Larger and carry more genes, includes insertion sequences
o Plasmids: Extra chromosomal pieces of DNA that don’t contain essential genes

Question 15

Differentiate between resistance, auxotrophic, morphological, behavioral, & conditionally expressed mutants.

Answer 15

o Resistance: Resistance against antibiotics, phages, chemicals etc.
o Auxotrophic: New growth is required (additional nutrients required)
o Morphological: loss of a cell component
o Behavioral: change in behavior like loss of motility
o Conditionally expressed mutants: Only expressed sometimes like heat or cold sensitive

Question 15

Explain why are mutations readily evident in bacteria and know the frequency at which spontaneous mutations may occur in bacteria.

Answer 15

o The frequency at which spontaneous mutations occur is 1 mutation per 10^4 replication cycles
 Typically have 10^9 cells, expected to have at least 1 mutant per colony
o They are readily evident in bacteria because they increase the rate of mutation.

Question 16

Explain the different kinds of mutations including replacement (transition and transversion), microdeletion, microinsertion, deletion and insertion.

Answer 16

o Replacement:
o Microdeletion: Removal of 1 nucleotide
o Microinsertion: Addition of 1 nucleotide
o Deletion: Removal of 2 or more nucleotides
o Insertion: Addition of 2 or more nucleotides

Question 17

Explain the consequences caused the different kinds of mutations (mis-sense, non-sense, silent, & frameshift).

Answer 17

o Silent Mutations: Base substitution changed the normal codon into a codon that codes for the same AA. Protein has a normal sequence
o Missense Mutation: Base substitution changes the codon for one that codes for a different amino acid. Changes amino acid sequence
o Non-Sense Mutations: Base Substitution changed a coding codon into a nonsense (stop) codon. Cuts protein short
o Frameshift: Shifts the translational “reading frame” of the mRNA, is the result of insertion or deletion of 1 nucleotide pair. Effects functionality of the protein

Question 18

Describe the different mechanisms of DNA repair such as mismatch repair and excision repair.

Answer 18

o Mismatch Repair: Repairs point mutations, DNA polymerase has a proofreading function that allows them to detect the mismatched base, excising it ad repairing it by adding the correct nucleotide
o Excision Repair: Repairs thymine diners by cutting and removing them from the DNA molecule (excise), replacing the gap by complementary base pairing and sealing the nick

Question 19

Describe the Ames test and explain their respective purposes.

Answer 19

• Used to identify protentional mutagens
• Exposes nutant bacteria to mutagenic substances to measure the rate if reversal of the mutagen
• EX: auxotrophic His- mutant strain exposed to the chemical agent becomes His+ and can grow in a minimal medium after exposure to agent (causes a mutation that reverses a mutation)

Question 20

Describe the replica plating technique and explain their respective purposes.

Answer 20

• Use to isolate specific mutants
• Isolated colonies are obtained by serial dilution and growth in complete medium
• Imprint of the master plate us made on sterile velvet surfaces
• Medium is manipulated to have the environment conductive to growth of mutant
• Transfer the colony from the velvet surface to the experimental plate

Question 21

Differentiate between vertical and horizontal genetic transfer.

Answer 21

o Vertical Gene Transfer: Transfer of genes from an organism to its offspring (humans to babies)
o Horizontal Gene Transfer:
 Transfer of genes between cells of the same generation.
 Chloe giving BC curly hair gene
 Transformation, Conjugation, Transduction
 Involves interaction of DNA from a donor cell with the DNA of recipient cell
 Last step is DNA recombination

Question 22

Explain Griffith experiments with the smooth and rough strains of Streptococcus pneumonia that marked the discovery of transformation.

Answer 22

o Capsules: Made of carbohydrates looks smooth and shiny
o Wild Type: Smooth, S-strain, Virulence factor
o Mutant Strain: Rough, R-Strain, harmless
o He boiled bacteria (killing it) and the mouse was healthy
o Boiled again and mixed it with the living bacteria. Mouse got sick and died
o Proposes that the boiling of the cells released the DNA into the environment and the living cells gained the ability to make capsules (recombination)
o They were transformed
o Example of Horizontal gene transfer

Question 23

Describe the steps in the mechanism of transformation.

Answer 23

o Transformation genes transferred from 1 bacterium to another as “naked DNA”
 No cell-to-cell contact is needed
 By integration of new DNA fragments the recipient may gain a characteristic that it previously lacked
 Griffith Experiment
o Steps:
 Recipient cell competency-requires the expression of specific cellular proteins
* Gram +: Cells can take both homologous and heterologous DNA
* Gram-: Cells can only take heterologous DNA
 Adsorption: reversible, followed by irreversible binding to cell membrane
* Donor DNA sticks to the recipient surface
 Penetration:
* Gram+: one strand enters and the other is digested
* Gram-: DS-DNA is not degraded even though only 1 strand recombines
 Recombination:

Question 24

Explain the process of conjugation

Answer 24

1. F+ form to make pilus to make contact with F-
2. Pilus acts as bridge between two cells (allows for transfer of genetic material)
3. Transfer: 1 strand of F plasmid from F+ is transfered to F- cell
4. Replication: F+ strand is replicated when enters F- cell and now forms complete plasmid, F+ replicated to keep being double stranded
5. Recipient (F-) becomes F+ since it now has the F plasmid and it can make other F- cells F+

Question 25

• Describe and differentiate between the lytic and lysogenic cycles of bacteriophage replication.

Answer 25

o Lytic:
 1. Virus adheres to the host cell (bacteriophage) surface
2. DNA of virus penetrates the cell and goes inside
3. DNA circularizes and replicates
* Many copies of viral DNA made inside bacterial cells
* Made by using the machinery of host cell since viruses don’t have polymerase or anything for replication
4. Assembly/Maturation: Different components of baby viruses (progeny)are made
5. Lysis–> cell bursts releasing progeny
o Lysogenic:
1. Attatch (adheres like in lytic)
2. Penetrate
3. Integration: into the bacterial cells chromosome
* Advantage: Doesn’t kill host cell right away. As the bacterial cells DNA replicates, so does the viruses DNA.
5. Some get triggered so they have the DNA come off the chromosomal DNA (excise) then follows the rest of the lytic cycle
o Different because they can create a population with the DNA living inside them, so they don’t have to lytic cycle (lysis) right away

Question 26

• Describe and differentiate between generalized transduction and specialized transduction.

Answer 26

o Generalized: carried out by lytic viruses
 1. Phage infection: bacteriophage infects donor host cell by injecting it’s DNA
 2. Host’s bacterium’s DNA is degraded to small fragments
 3. Packaging; instead of packing only phage DNA it can accidentally package host cell DNA fragment
* Now you can have defective virus (exogenote), doesn’t have viral DNA but has bacterial DNA
 Now you can have recombinant cell (cell that will have partially their own DNA and another cells DNA
 Generalized because any piece of bacterial DNA has equal chance of getting packaged up
o Specialized: carried out by bacteriophages that replicate via lysogenic cycle
 1. Attach
 2. Penetrate
 3. When it de-integrates sometimes it takes some of the adjacent pieces of bacterial chromosome with it
 4. Replication viral DNA and destruction of bacterial DNA when they get a capsid
* Creates progeny (offspring)
o Specialized because only the genes that were flanking the DNA have the possibility of being transferred to donor cell

Question 27

HGT: Transformation

Answer 27

Optic of naked foreign DNA by a bacterial cell

Question 28

HGT: Conjugation

Answer 28

Cell-to-cell contact for hGT

Question 29

HGT: Transduction

Answer 29

A bacterial virus serves as a vehicle for horizontal gene transfer from one bacteria to other

Question 30

Chemical mutagenic agents and explain how they work.

Answer 30

Chemical:
- Nucleoside analogs: These are chemicals structurally similar to normal nucleotides but with altered base-pairing properties. When incorporated into DNA during replication, they cause base-pair substitutions. For example, 5-bromouracil (an analog of thymine) pairs with guanine instead of adenine, causing base-pair mutations
- Nitrous acid: This agent causes deamination of adenine and cytosine, converting them into bases that pair incorrectly (e.g., adenine becomes hypoxanthine, which pairs with cytosine instead of thymine). This results in point mutations during replication.

Genetic mutagenic agents:
- Transposons (“Jumping genes”): These are genetic elements that can move from one location in the genome to another. As they insert themselves into new locations, they can disrupt normal gene function, causing mutations. Transposons can lead to gene inactivation or altered expression, resulting in phenotypic changes in the organisms

Question 31

Genetic mutagenic agents and explain how they work.

Answer 31

Genetic mutagenic agents:
- Transposons (“Jumping genes”): These are genetic elements that can move from one location in the genome to another. As they insert themselves into new locations, they can disrupt normal gene function, causing mutations. Transposons can lead to gene inactivation or altered expression, resulting in phenotypic changes in the organism.

Question 32

Differentiate between an F+, F-, F’, and Hfr strains.

Answer 32

F+: bacterial cell contains F plasmid

F-: bacterial cell that doesn’t contain F plasmid

F’: forms when F plasmid excises from bacteria chromosome

Hfr: F plasmid is integrated into bacterial chromosome