Unit 5 Objectives

Question 1

DNA Replication:

Bacteria vs. Eukaryotes

Answer 1

• Bacteria:
  Location: cytoplasm (lack a nucleus),
  Begins: origin of replication. The DNA polymerase synthesizes new DNA in the 5’ to 3’ direction. Primase lays down an RNA primer, and replication proceeds in a bidirectional manner from the origin. DNA ligase connects Okazaki fragments on the lagging strand.
• Eukaryotes:
  Location: nucleus
  Begins: Eukaryotes have multiple origins of replication on each chromosome. The process is more complex due to larger genomes, requiring more replication factors. DNA polymerase and primase work similarly, but helicases and other proteins are involved to maintain chromatin structure during replication. Telomerase is involved in the replication of chromosome ends (telomeres).

Question 2

Transcription

Bacteria vs. Eukaryotes

Answer 2

• Bacteria:
  Location: cytoplasm.
  Begins: RNA polymerase synthesizes mRNA directly from the DNA template. Bacterial mRNA is often polycistronic, meaning it can encode multiple proteins from a single mRNA strand. Transcription is coupled with translation (as soon as the mRNA is synthesized, ribosomes begin translating it).
• Eukaryotes:
  Location: nucleus
  Begin: RNA polymerase II synthesizes mRNA. Eukaryotic mRNA is monocistronic, meaning each mRNA typically codes for a single protein. Eukaryotic mRNA also undergoes splicing, capping, and polyadenylation before being transported to the cytoplasm for translation.

Question 3

Translation

Bacteria vs. Eukaryotes

Answer 3

• Bacteria:
  Location: cytoplasm on ribosomes (30S and 50S subunits).
  Begins: The mRNA is translated by ribosomes, which assemble amino acids into proteins using tRNA and the genetic code. The start codon is formylmethionine (fMet).
• Eukaryotes:
  Location: cytoplasm, but eukaryotes have 80S ribosomes (60S and 40S subunits).
  Begin: The start codon is methionine (not fMet), and the process is more regulated and complex. Translation initiation involves several initiation factors and the recognition of a 5’ cap on the mRNA.

Question 4

How is DNA replication, transcription, and translation similar in bacteria and eukaryotes?

Answer 4

• Both bacteria and eukaryotes use the central dogma of molecular biology: DNA → RNA → Protein
• Both processes involve DNA replication, transcription, and translation using ribosomes, RNA polymerases, and tRNAs to synthesize proteins.

Question 5

How is DNA replication, transcription, and translation different in bacteria and eukaryotes?

Answer 5

• Location: Bacteria carry out all three processes in the cytoplasm,

while eukaryotes perform transcription in the nucleus and translation in the cytoplasm.

• Complexity: Bacterial processes are simpler and faster due to their smaller genome and lack of a nucleus and organelles.
• Eukaryotes have more complex regulation, involving chromatin remodeling, splicing, and post-transcriptional modifications.
• RNA Processing: Eukaryotic mRNA undergoes significant processing (splicing, capping, and polyadenylation), which does not happen in bacteria.

Bacteria often have polycistronic mRNA,

while eukaryotes typically have monocistronic mRNA.

• Ribosome Differences: Bacterial ribosomes (70S) are smaller than

eukaryotic ribosomes (80S)

Question 6

Can We Use These Differences to Develop Antibacterial Drugs?

Answer 6

Yes, the differences between bacterial and eukaryotic processes can be exploited to develop selective antibacterial drugs.

For example:
- Ribosome inhibitors target bacterial 70S ribosomes, inhibiting protein synthesis without affecting eukaryotic 80S ribosomes.

• DNA replication inhibitors target bacterial DNA gyrase and topoisomerase, enzymes that are distinct from those in eukaryotes.
• RNA polymerase inhibitors target bacterial RNA polymerase, which is different from eukaryotic RNA polymerase.

Question 7

Operon:

Answer 7

An operon is a cluster of functionally related genes in bacteria that are transcribed together under the control of a single promoter and regulated as a unit

Question 8

Structural Gene:

Answer 8

A structural gene encodes a protein or RNA molecule that is involved in the cell’s structure or function

Question 9

Operator:

Answer 9

Where a repressor or activator protein binds to control the transcription of the operon

Question 10

Promoter:

Answer 10

The promoter is a region of DNA where RNA polymerase binds to initiate transcription of the operon

Question 11

Repressor:

Answer 11

Binds to the operator to block RNA polymerase from transcribing the operon, preventing gene expression

Question 12

Activator:

Answer 12

A protein that binds to the promoter or operator to enhance the transcription of the operon, promoting gene expression

Question 13

Gene Regulation in Bacteria

Answer 13

Bacteria regulate gene expression primarily through operons

Inducible operons, like the lac operon, are turned on in response to specific environmental signals (e.g., the presence of an inducer like lactose).

Repressible operons, like the trp operon, are typically turned on, but are repressed when their product (e.g., tryptophan) is abundant.

Question 14

Lac Operon (Inducible)

Answer 14

The lac operon regulates the metabolism of lactose in bacteria and is an example of an inducible operon. The operon consists of three structural genes (lacZ, lacY, lacA), a promoter (P), an operator (O), and a repressor protein (LacI).

When lactose is present, it acts as an inducer, binding to the LacI repressor, which changes its shape and prevents it from binding to the operator.

This allows RNA polymerase to bind to the promoter and transcribe the structural genes (lacZ, lacY, lacA), enabling the bacterium to metabolize lactose.

Question 15

Diagram of the Lac Operon:

Answer 15

[Promoter]–[Operator]–[lacZ]–[lacY]–[lacA]
↑ ↓
(LacI Repressor) (Lactose as inducer)

Question 16

Trp Operon (Repressible)

Answer 16

The trp operon regulates the synthesis of the amino acid tryptophan in bacteria and is an example of a repressible operon. It consists of several structural genes (for tryptophan synthesis enzymes), a promoter, an operator, and a repressor (TrpR).

When tryptophan is present in the cell, it acts as a corepressor, binding to the TrpR repressor and activating it.

The activated repressor then binds to the operator, preventing RNA polymerase from transcribing the operon, thus stopping the synthesis of tryptophan.

Question 17

Diagram of the Trp Operon

Answer 17

[Promoter]-[Operator]-[trpE]-[trpD]-[trpC]-[trpB]-[trpA]

(promoter) ↑ (operator) ↓
(TrpR Repressor) (Tryptophan as corepressor)

Question 18

Briefly describe horizontal gene transfer mechanisms: transformation, transduction & conjugation

Answer 18

Transformation: Uptake of naked DNA from the environment

Transduction: Gene transfer via a bacteriophage

Conjugation: Direct transfer of DNA between bacterial plasmid through a pilus.

Question 19

Describe specialized and generalized transduction

Answer 19

Generalized Transduction: Transfer of random bacterial DNA by a bacteriophage.

Specialized Transduction: Transfer of specific bacterial DNA located near the phage integration site

Question 20

Define mutation

Answer 20

A mutation is a permanent change in the DNA sequence of an organism’s genome, which can occur spontaneously or as a result of external factors, and can lead to changes in the organism’s traits

Question 21

What is a plasmid?

Answer 21

A small, circular piece of DNA found in bacteria (and sometimes in other organisms) that is separate from the chromosomal DNA and can replicate independently, often carrying genes that confer advantages like antibiotic resistance

Question 22

Making Recombinant DNA:

Purpose

Answer 22

The goal of making recombinant DNA is to combine DNA from different sources to create a new genetic sequence, often to;
- study genes
- produce proteins
- or modify organisms

Question 23

Making Recombinant DNA:

Restriction enzymes

Answer 23

aka restriction endonucleases are used to cut the DNA at specific sequences

This allows the isolation of a desired gene from a source organism and also creates compatible ends for inserting the gene into a vector

Question 24

Making Recombinant DNA:

Vector

Answer 24

A DNA molecule (like a plasmid or virus) used to carry the inserted gene into a host cell

The vector is cut with the same restriction enzyme(s) as the target gene, creating compatible sticky or blunt ends, which allows the gene to be ligated into the vector

Question 25

Making Recombinant DNA:

Cloning

Answer 25

The recombinant DNA is introduced into a host cell (usually a bacterium like E. coli) through processes like transformation.

The host cell replicates and expresses the recombinant DNA, allowing for the production of the desired protein or further genetic manipulation

Question 26

What are some applications for recombinant technology

Answer 26

1. Gene Therapy: Recombinant technology is used to introduce therapeutic genes into patients’ cells to treat genetic disorders, such as cystic fibrosis or sickle cell anemia.
2. Production of Therapeutic Proteins: Recombinant DNA is used to produce proteins like insulin, human growth hormone, vaccines (e.g., Hepatitis B vaccine), and monoclonal antibodies, which are used in treating diseases.
3. Agriculture and Food Industry: Recombinant technology is used to create genetically modified organisms (GMOs), such as crops resistant to pests, diseases, or environmental stresses, or animals with improved characteristics (e.g., faster-growing fish). It also allows for the development of enzyme-based food production (e.g., rennet for cheese-making).
4. Bioremediation: Recombinant microorganisms are engineered to clean up environmental contaminants, such as oil spills or toxic waste, by breaking down harmful substances more efficiently.
5. Vaccine Development: Recombinant DNA technology enables the creation of recombinant vaccines that use pieces of pathogens (instead of whole pathogens) to stimulate an immune response, such as the HPV vaccine or recombinant Hepatitis B vaccine.
6. Research and Biotechnology: It is widely used in gene expression studies, protein engineering, and drug discovery. It allows scientists to create model organisms with specific genes knocked out or introduced to study diseases and biological processes.
7. Transgenic Animals: Recombinant technology is used to create transgenic animals for research, such as mice models for human diseases, or to improve livestock (e.g., cows that produce human proteins in their milk)

Question 27

Describe Griffith’s general experiment procedure, the results & its importance to biology.

Answer 27

Griffith’s Experiment (1928)

Procedure:
- Griffith studied two strains of Streptococcus pneumoniae (a bacterium that causes pneumonia) in mice:

1. Smooth (S) strain: This strain had a capsule and was virulent, meaning it could cause disease in mice.
2. Rough (R) strain: This strain lacked the capsule and was non-virulent, meaning it did not cause disease.

• Griffith injected mice with the following:
  
  1. Mice injected with the live S strain died (due to its virulence).
  2. Mice injected with the live R strain survived (due to its non-virulence).
  3. Mice injected with heat-killed S strain survived (since the bacteria were no longer alive to cause disease).
  4. Mice injected with a mixture of heat-killed S strain and live R strain died.

Results:
- Mice injected with the mixture of heat-killed S strain and live R strain died, and when their blood was examined, it contained live S strain bacteria (smooth, virulent).

• This showed that the R strain had been “transformed” by the genetic material from the dead S strain, acquiring the ability to make a capsule and become virulent.

Question 28

Describe Griffith’s general experiment procedure, the results & its importance to biology

Answer 28

• Griffith’s experiment demonstrated the phenomenon of transformation, where genetic material (from dead bacteria) could be taken up by living bacteria, leading to a change in their characteristics.
• It was an early indication that DNA was the molecule responsible for genetic inheritance, though this would not be confirmed until later by scientists like Avery, MacLeod, and McCarty and the work of Hershey and Chase.
• Griffith’s work laid the foundation for understanding genetic exchange in bacteria, which is critical for understanding processes like horizontal gene transfer and antibiotic resistance

Question 29

Describe how bacteria could repair mutations such as that from radiation.

Answer 29

Bacteria repair mutations caused by radiation (e.g., UV or ionizing radiation) through various mechanisms:

1. Direct Repair:
   
   • Photoreactivation: Fixes thymine dimers (caused by UV) using the enzyme photolyase with visible light.
2. Excision Repair:
   
   • Nucleotide Excision Repair (NER): Removes and replaces damaged DNA segments using endonucleases, DNA polymerase, and ligase.
   • Base Excision Repair (BER): Fixes small base lesions using DNA glycosylase, DNA polymerase, and ligase.
3. Error-Prone Repair (SOS Response):
   
   • Activates error-prone repair enzymes to bypass extensive DNA damage, allowing survival but potentially introducing mutations.
4. Recombination Repair:
   
   • Homologous recombination repairs double-strand breaks using an undamaged chromosome as a template.
5. Mismatch Repair:
   
   • Corrects mismatched bases from replication or radiation-induced damage using MutS and MutL enzymes.

Question 30

Describe Direct Selection

Answer 30

• Description: Direct selection is used to identify mutants by growing them in conditions where only the mutant can survive or grow.
  • Example: If a bacterium is treated with an antibiotic, only antibiotic-resistant mutants will grow, while the non-resistant ones will be killed.

This allows for the selection of the mutant directly from the population by placing it on a medium containing the selective agent

Question 31

Describe Replica plating

Answer 31

• Description: Replica plating is a method used to identify auxotrophic mutants (organisms that cannot synthesize certain compounds) by transferring colonies from one plate to another with different media.
  • Mutants unable to grow on the selective medium (e.g., if they cannot synthesize a specific nutrient) will be identifiable by the lack of colony growth in specific locations on the new plate.

Question 32

Describe F factor and its importance

Answer 32

Description: A type of plasmid that carries genes responsible for conjugation

Enables a bacterium to transfer genetic material to another bacterium through a pilus.

Importance: Allows for the transfer of antibiotic resistance genes, virulence factors, or other beneficial traits

Question 33

Describe R factor and its importance

Answer 33

Description: A type of plasmid that carries antibiotic resistance genes.

Can provide bacteria with resistance to one or more antibiotics

Importance: contributes to antibiotic resistance

Question 34

Describe Virulence plasmid and its importance

Answer 34

Description: contain genes that enhance the pathogenicity of a bacterium, such as genes for toxins, adhesion factors, or enzymes that help the bacteria invade host tissues.

Importance: crucial for the ability of certain bacteria to cause disease

Question 35

Describe the conjugation process between F+ & F- strains

Answer 35

1. F+ Cell (Donor):
   
   • The F+ cell contains an F plasmid (F factor), which carries the genes required for conjugation. It has a sex pilus (a hair-like appendage) that allows it to connect with an F- cell.
2. Pilus Formation:
   
   • The F+ cell extends its sex pilus to make contact with the F- cell. The pilus attaches to the surface of the F- cell, and the cells begin to draw closer together.
3. Transfer of the F Plasmid:
   
   • The F plasmid in the F+ cell begins to replicate. The replication starts at a specific origin of transfer (oriT), and a single strand of the plasmid DNA is transferred through the pilus into the F- cell.
   • Simultaneously, the donor F+ cell retains a copy of the plasmid.
4. Completion of Transfer:
   
   • The single-strand of F plasmid DNA that enters the F- cell is then replicated inside the F- cell, converting it into an F+ cell.
   • The F- cell now contains a complete copy of the F plasmid, becoming an F+ cell capable of further conjugation.
   • Both cells now have the ability to transfer the F plasmid to other F- cells via conjugation.

Question 36

Describe the conjugation process between Hfr & F- strains

Answer 36

1. Hfr Cell (Donor):
   
   • An Hfr cell is a bacterial cell in which the F plasmid is integrated into the bacterial chromosome. This integration occurs at a specific site on the chromosome.
   • The Hfr cell has a sex pilus that allows it to connect to the F- cell.
2. Pilus Formation:
   
   • The Hfr cell extends its sex pilus to make contact with the F- cell, and the cells are drawn closer together.
3. Transfer of DNA:
   
   • Hfr conjugation begins with the replication of the F plasmid that is now part of the chromosome.
   • The replication starts at the origin of transfer (oriT), which is located on the F plasmid.
   • A single-stranded DNA copy of the F plasmid and a portion of the chromosomal DNA adjacent to the integration site are transferred through the sex pilus into the F- cell.
4. Incomplete Transfer:
   
   • The DNA transfer typically begins from the origin of transfer and proceeds in a linear fashion along the chromosome.
   • Hfr conjugation often results in incomplete transfer of the bacterial chromosome because the entire chromosome is too large to be transferred in one round of conjugation. Only a portion of the chromosome (along with the F plasmid) is transferred before the conjugation process is interrupted.
5. Recombination in the F- Cell:
   
   • After transfer, the single-stranded DNA that entered the F- cell is converted into double-stranded DNA and integrates into the recipient’s chromosome through a process called recombination.
   • The F- cell does not become F+ because it only received part of the F plasmid and not the full plasmid.

However, the chromosomal genes from the donor may integrate into the F- cell’s genome.

6. Result:
   
   • The F- cell becomes a recombinant with the donor’s chromosomal genes, but it remains F- since it didn’t receive the full F plasmid.

Question 37

Describe how gene regulation works when bacteria are grown on TSI medium (a medium containing glucose and lactose in different concentrations).

(Given a photo or actual TSI tube, interpret the results including whether or not the lac operon is “on” or “off”)

Answer 37

In TSI medium, bacteria regulate gene expression based on sugar availability. The lac operon is repressed when glucose is present, but activated when glucose is exhausted and lactose is available. This regulation ensures that bacteria use the most efficient sugar first (glucose) and switch to lactose when necessary

Question 38

Describe the transformation experiment performed in lab & be able to analyze data of a similar type of experiment. Be able to figure out what may have gone wrong with this type of experiment given a specific result. Define “competent” with respect to this type of experiment. How can cells become “competent”?

Answer 38

In a transformation experiment, bacteria take up foreign DNA (such as plasmids or chromosomal fragments) from their surroundings and integrate it into their own genome. This process is commonly used to introduce new genetic material into bacterial cells

Basic Procedure:
1. Preparation of Competent Cells:
- Bacteria are made competent, meaning they are able to take up foreign DNA. This is typically done by treating the cells with calcium chloride or other chemicals, or by using electroporation (an electric shock).

2. DNA Introduction:
   
   • The competent cells are then exposed to the foreign DNA (usually a plasmid). The DNA can enter the cell through natural porins in the cell membrane or via induced pores.
3. Plating on Selective Media:
   
   • After DNA uptake, the bacteria are plated on selective agar that contains a selective agent (e.g., an antibiotic). Only those cells that have successfully taken up the DNA and express the antibiotic resistance gene (if present) will survive and form colonies.
4. Colony Analysis:
   
   • Transformed cells (cells that successfully integrated the foreign DNA) grow into visible colonies, while non-transformed cells (those that did not take up the DNA) are unable to grow due to the presence of the selective agent.

Analyzing Data and Possible Problems:
- Expected Result: If the transformation is successful, colonies will appear on the plate containing the selective agent (e.g., antibiotic), indicating that the cells have taken up the plasmid and are expressing resistance.

Question 39

Describe the transformation experiment performed in lab & be able to analyze data of a similar type of experiment. Be able to figure out what may have gone wrong with this type of experiment given a specific result. Define “competent” with respect to this type of experiment. How can cells become “competent”?

Answer 39

• Possible Problems:
  
  1. No Colonies: If no colonies appear, the competent cells may not have been properly prepared or the DNA might not have been effectively introduced.
     
     • Possible causes: Incorrect heat shock, old or improperly treated cells, or DNA degradation.
  2. Colony Growth on Control Plate Only: If colonies appear only on the non-selective control plate, it suggests the cells may not have received the foreign DNA or the selective agent was not correctly applied

Question 40

Describe the transformation experiment performed in lab & be able to analyze data of a similar type of experiment. Be able to figure out what may have gone wrong with this type of experiment given a specific result. Define “competent” with respect to this type of experiment. How can cells become “competent”?

Answer 40

Competent Cells:
- Competent cells are bacterial cells that are capable of taking up foreign DNA from their environment.

• How to make cells competent:
  
  1. Chemical Transformation: Treating cells with calcium chloride or other chemicals to make their cell walls more permeable.
  2. Electroporation: Exposing the cells to an electric field to temporarily increase membrane permeability.
  3. Heat Shock: Exposing the cells to a rapid temperature change (e.g., ice to heat) to further increase DNA uptake.