Bacteria

Question 1

What distinguishes bacteria cells from eukaryotes

Answer 1

They are prokaryotes and thus lack a true nucleus and membrane-bound organelles.

Question 2

Describe the cell wall of a backteria

Answer 2

Petidoly glycan cell wall

Question 3

what is a gram-positive bacteria cell wall

Answer 3

Gram-positive bacteria have:

• thick, multi-layered, peptidoglycan cell walls that are exterior to the membrane.

The peptidoglycan in most Gram-positive species is covalently linked to teichic acid, which is essentially a polymer of substituted glyrcerol units linked by phosphodiester bonds.

• The techoic acids are major cell surface antigens

Question 4

What are gram-negative bacterias

Answer 4

These bacteria have 2 membranes

1. An outer membrane
2. Inner (cytosplasmic) membrane

The peptidoglycan layer is located between the two membranes in what is called the periplasmic space. It contains enzymes and other substances

In contrast to Gram-positive cells the petidoglycan layer of Gram-negative cells is thin, and the cells are consequently more susceptible to physical damage.

Question 5

Some further stuff on the outermembrane of Gram-negative cells

Answer 5

The outermembrane is distinguished by presence of barious embedded, lipopolysaccharides.

• The polysaccharide portion (O-polysaccharide) is antigenic and can, therefore, be used to identify different strains and species.
• The lipid portion (called lipid A) is toxic to humans and animals. Since lipid A is an integral part of the membrane, it is called an endotoxin, as opposed to exotoxins, which are a secreted substance

Question 6

How can gram-positive and negative be differentiated through staining?

Answer 6

• A dye crystal violet is used.
• The cells are then treated with iodine as a mordant, which helps the Gram-positive cells retain the crystal violet.
• Thus when alchohol is used the Gram-negative cells lose their colour, but the Gram-positive cells retain the violet dye.
• As the now colourles Gram-negative cells would be invisible under a light microscope. they are counter-stained with the red dye safranin.

Question 7

How does the bacterial chromosome differ from eukaryotes

Answer 7

1. Genes are grouped into operons where multiple genes come under the control of the same promoter.
2. Genes in one operon also come under control of the same regulatory elements
3. Prokaryotic genes also lack introns and thus do no require splicing after transcription

Question 8

What are plasmids?

Answer 8

Non-essential pieces of DNA called

• exist as small, circular, double-stranded extrachromosomal DNA molecule which may be passed on to the cells of the same generation or to its offspring
• They are capable of replication independent of the bacterial chromosome since they possess their own origin of replication and cells may therefore contain more than one plasmid
• They contain beneficial genes which confer protective traits such as antibiotic resistance, toxin synthesis and enzyme production. This allows them to confer a selective advantage to bacteria.

Question 9

What is binary fission

Answer 9

It is a form of asexual reproduction in which two equal-sized genetically identical daughter cells are produced from a single parent cell.

• The transmission of genetic material from a bacterial cell to its offspring occurs by binary fission
• It includes the replicaion of the bacterial chromosome and thus is unable to give rise to genetic variation in a bacterial population

Question 10

Describe step 1 of binary fission

Answer 10

The bacteria chromosome is attached to the plasma membrane before DNA replication

Question 11

Describe step 2 of binary fission

Answer 11

• DNA replication begins at the single origin of replication, where the replication bubble is first formed when the 2 DNA strands separate. Each parental strand is used as a template for the synthesis of the daughter strand in semi-conservative DNA replication
• The replication bubble grows bidirectionally away from the origin of replication until the entire bacterial chromosome is replicated

Question 12

Describe step 3 of binary fission

Answer 12

After DNA replication is completed, cell growth occurs. Each circular DNA molecule is attached to the cell membrane

Question 13

Describe step 4 of binary fission

Answer 13

The cell elongates, and membrane growth caudses the two chromosomes to be moved apart

Question 14

Describe step 5 of binary fission

Answer 14

• Cell division in bacteria is controlled by the septal ring, a group of proteins which directs the assembly of the septum. The septum eventually separates the 2 daughter cells
• The septum extends as the cell membrane invaginates as new cell membrane and cell wall materials (like peptidoglycan) are added to it

Question 15

Describe step 6 of binary fission

Answer 15

The invaginating cell membrane, together with the newly formed septum, splits the cell into 2 genetically identical daughter cells by cytokinesis.

Question 16

What is transformation

Answer 16

The process by which a recipient cell takes up small fragments of naked DNA from the surrounding envrionment. This DNA can orginate either from:

1. a donor bacteria cell which lyses and releases its DNA into the surrounding envrionment
2. artificially constructed plasmids

Only competent bacteria cells are naturally able to undergo transformation. Competence depends on the presence of competence factors produced by the bacteria cell.

Question 17

Whar are competence factors

Answer 17

Cell surface proteins that bind to DNA fragments and aid in their uptake.

Question 18

Describe step 1 and 2 of transformation

Answer 18

Step 1:

• The donor bacteria cell lyses and releases naked DNA fragments (Donor DNA fragment)

Step 2:

• A competent recipient takes up one or more of the donor DNA fragments into its cytoplasm via its competence factor

Question 19

Describe step 3 and 4 of transformation

Answer 19

Step 3:

• Homologous recombination of the donor DNA fragment takes place with a homologous section of the recipient cell’s chromosome

Step 4:

• This results in the homologous segment of the donor cell’s DNA being incorporated into the recipient cell’s chromosome
• and the homologous segment of the recipient cell’s chromosome being excised and degraded.
• The recipient cell is now known as recombinant cell

Question 20

What are the 2 ways to artificially increase competency in bacteria

Answer 20

1. Treatment of bacteria with calcium chloride followed by heat shock
2. Electroporation

These artificial methods of transformation utilize plasmid DNA instead of linear DNA fragments and can be used in bacteria species which do not naturally transform. Thus, transformation can be widely used in recombinant DNA technology.

Question 21

What is transduction

Answer 21

When bacteriophages carry bacterial genes from their first host cell (donor) to their second host cell (recipient) due to erros in the phage reproductive cycle.

The DNA fragment of the donor cell may be incorporated into the genome/chromosome of the recipient cell via homologous recombination.

Question 22

What are the 2 forms of transduction

Answer 22

1. Generalized transduction
2. Specialized transduction

Question 23

What is generalized transduction?

Answer 23

Transduction mediated by virulent phages, due to each portion of the bacterial genome having approximately the same probability of being transferred from donor to recipient bacteria.

Ie. any bacteria gene can potentially be transferred by this method

Question 24

Describe step1 and 2 of generalized transduction

Answer 24

Step 1:

• The virulent phage injects its DNA into its first host bacteria (donor bacterium) and the first host cell’s chromosome is degraded.

Step 2:

• The phage makes use of the host’s DNA replication machinery to synthesise more phage DNA. It also uses the host cell’s gene expression machinery to synthesise more phage proteins, eg. capsid proteins.

Question 25

Describe step3 of generalize transduction

Answer 25

Step 3:

• Occasionally, a small piece of the host cell’s degraded DNA is accidentally packaged within a phage capsid in place of the phage genome during the assembly stage of the lytic cycle.
• This phage is known as a defective phage.
• The first host bacteria is lysed, and new phages are released into the environment

Question 26

Describe steps 4 and 5 of generalized transduction

Answer 26

Step 4:

• The defective phage’s progeny which contains the first host cell’s DNA fragments may infect a second host cell and inject the DNA fragment acquired from the previous host cell into it.

Step 5:

• The donor DNA is incorporated into the second host cell’s genome by homologous recombination, in which the donor DNA replaces the homologous region ofthe recipient cell’s chromosome. The recipient cell is now knwon as a recombinant cell.

Question 27

What is specialized transduction?

Answer 27

Transduction mediated by temperate phages. This is due to only specific portions of the bacteria genome ( eg. genes near the prophage insertion site on the host chromosome) having high probability of being transferred from donor to recipient bacteria

Question 28

Describe step 1 of specialized transduction

Answer 28

The genome of a temperate phage integrates as a prophage into the chromosome of the first host bacterium (donor cell) at a specific site known as the prophage insertion site

Question 29

Describe step 2 of specialized transduction

Answer 29

Upon induction, the phage genome is excised from the first host cell’s chromosome.

• Due to imprecise excision, the phage DNA sometimes takes with it a small region of the bacterial DNA that was adjacent to the prophage insertion site.

Question 30

Describe step 3 of specialized transduction

Answer 30

The phage reproduces itself using the first host cells’ DNA replication and gene expression machinery.

Each newly formed phage now contains part of the first host cell’s DNA

Question 31

Describe step 4 of specialized transduction

Answer 31

The first host bacterium is lysed, releasing phages into the environment. The phages infect a second host cell (recipient bacterium) and inject the DNA fragment acquired from the previous host cell into it

Question 32

Describe step 5 of specialized transduction

Answer 32

The donor DNA is incorporated into the second host cell’s genome by

1. Prophage integration (if the segment of phage DNA transferred contains the genes required to enter the lysogenic cycle) OR
2. Homologous recombination (if the segment of phage DNA transferred does not)

the recipient cell is now knwon as a recombinant cell

Question 33

What is conjugation

Answer 33

Direct contact between the donor and recipient bacteria leads to establishment of cytoplasmic bridge between them, followed by transfer of the donors’ DNA to the recipient cell.

Question 34

What is the F factor

Answer 34

The donor bacterial cell possesses it (F stands for fertility btw).

• The F factor exists as a plasmid known as the F plasmid OR
• as a segment of DNA integrated into the bacterial chromosome.

Genes on the F factor are responsible for synthesis of the sex pilus abd trabsfer of the F factor from a fonor cell to a recipient cell.

The F factor has its own origin of replication and replicates as it is transferred, resulting in both the donor and recipient cell containing an F factor after conjugation is completed

Question 35

What is F-, F+?

Answer 35

Cells with an extrachromosomal F factor (ie. F plasmid) are known as F+

Cells lacking the F factor are known as F- cells.

Question 36

Describe step 1 in conjugation of a F+ x F- mating

Answer 36

The F+ donor cell uses a sex pilus to attach to an F- recipient cell

Question 37

Describe step 2 in conjugation of a F+ x F- mating

Answer 37

A temporary cytoplasmic mating bridge is formed between the 2 cells, providing a route for DNA transfer

Question 38

Describe step 3 in conjugation of a F+ x F- mating

Answer 38

The sugar-phosphate backbone of one strand of the F plasmid in the F+ donor cell is nicked by an endonuclease (ie. a phosphodiester bond is broken)

The nicked single strand DNA separates from its complementary strand and moves to F- recipient cell through the cytoplasmic mating bridge.

Question 39

Describe step 4 in conjugation of a F+ x F- mating

Answer 39

Each parental strand becomes a template for the DNA synthesis of a complementrary daughter strand in the donor and recipient cells. This occurs by semi-conservative replication, catalysed by DNA polymerase.

Upon completion of replication, DNA ligase catalyses the synthesis of a phosphodiester bond to close the gap in each F plasmid.

Question 40

Describe step 5 in conjugation of a F+ x F- mating

Answer 40

The cells move apart and sex pilus breaks, forming two bacteria cells that are both F+.

Hence, a F+ bacterial cell converts a F- cell to a F+ when the two cells conjugate

Question 41

What are the 3 main differences between prokaryotes and eukaryotes in gene expressions

Answer 41

1. Degree of compaction of chromosomes
   
   • Prokaryotic chromosome tends to be less compact as compared to eukaryotic chromsome
2. Presence of operons
   
   • prokaryotic genes are grouped into a cluster under the control of one promoter while eukaryotic genes have a promoter for each gene
3. Presence of nuclear membrane
   
   • Prokaryotes do not have membrane bound organelles, hence their chromosomes are not surrounded by nuclear membrane unlike eukaryotes

Question 42

What is an operon

Answer 42

Bacterial structural genes with related functions are generally located adjacent to each other and are placed under the control of the same promoter and regulatory regions.

The grouping of genes of related function forms a transcriptional unit, which is an operon

• Genes in an operon are transcribed together and hence a single regulatory mechanism can control the whole cluster of functionally related genes.

Question 43

Describe initiation of transcription in prokaryotes

Answer 43

Carried out by a prokaryotic RNA polymerase which can only bind to the promoter specifically in the presence of the sigma (weird symbol) factor.

• The sigma factor on RNA polymerase recognises and binds at the -35 recognition sequence and the -10 pribnow box in the promoter. The -10 bp sequence is A-T rich, making it easier to break the hydrogen bonds.
• Similar to eukaryotic transcription, RNA polymerase transiently unwinds DNA to form a transcription bubble, allowing one strand to be used as a template for transcription

Question 44

Describe stage:2 elongation of transcription in eukaryotes

Answer 44

• RNA polymerase reads the DNA template strand in a 3’-5’ direction. The RNA polymerase moves down the template strand as elongation continues.
• Ribonucleoside triphosphates are added in a complementary manner in a 5’→3’ direction by the core enzyme of RNA polymerase.
• As elongation of the mRNA continues, single-stranded mRNA trails out of the transcription bubble and the 2 strands of DNA upstrean of the transcription bubble are re-wound into their double helical structure

Question 45

Describe intrinsic termination (also called rho-independent transcription termination)

Answer 45

• It utilizes a terminator sequence within the RNA, usually a palindromic GC-rich sequence followed by four or more U residues.
• The GC rich region forms a hairpin loop structure via complementary base pairing that causes RNA polymerase to dissociate from the DNA template

Question 46

Describe rho-dependent termination

Answer 46

• Rho-dependent termination uses a termination factor called the p (rho) factor, which is a protein that stops mRNA synthesis at specific sites.
• The p factor binds at a p recognition site on the mRNA strand, and moves along the mRNA towards the RNA polymerase
• When the p factor reaches the RNA polymerase, it destabilizes the mRNA-DNA hybrid, thuus releasing the newly synthesized mRNA from the elongation complex.

Question 47

this few pages i dint really make

Answer 47

pages 37 to 41

Question 48

Describe the initiation of translation in prokaryotes

Answer 48

• Initiation of translation beginds when the 30S small ribosomeal subunit, aided y translation initiation factor, recognizes and binds to the Shine-Dalgarno sequence in the 5’ UTR of mRNA.
  
  • The 16s rRNA in the small ribosomal subunit aids in positioning the ribosome such that the start codon AUG is located in the P site of the ribosome.
• The initiator tRNA (tRNA^fMet) then binds at the P site of the large ribosomal subunit with the aid of another translation initiation factor
• A third translation initation factor aids in the binding of the large subunit to the small subunit. The translation initiation complex is thus formed and all the initiation factors are released.

Question 49

Describe elongation portion of translation in prokaryotes

Answer 49

Elongation of the nascent polypeptides is carried out by the 70S ribosomes of prokaryotes in the same manner as in eukaryotes.

However, prokaryotes utilize fewer elongation factors as compared to the much larger number of elongation factors that aid eukaryotic polypeptide elongation.

Question 50

I cant really understand this page

Answer 50

page 43. use this page to memorize elongation portion

Question 51

Describe termination of translation in prokaryotes

Answer 51

Carried out by the binding of release factors, which recognise the 3 different stop codons. The entry of these release factors into the A site results in the release of the polypeptide chain and the dissociation of the subunits of the ribosome.