bacteria

Question 1

what are the 3 groups of bacterial cell shapes?

Answer 1

1. coccus (spherical)
2. bacillus (rod-shaped)
3. spirillum/spirochete (spiral)

Question 2

what structures are present in bacterial cells?

Answer 2

bacterial cells
- are small and unicellular
- have a peptidoglycan cell wall
- contains circular DNA
- has 70S ribosomes
- lack membrane-bound organelles

bacterial cells, being PROKARYOTIC, lack a true nucleus and membrane-bound organelles

Question 3

what are cell surface structures of bacterial cells?

Answer 3

1. capsule (FYI) ⮕ thick, viscous layer made up of various polysaccharides
2. cell wall ⮕ made up of peptidoglycan (more in other flashcards)
3. cell wall ⮕ similar in composition and function to those of eukaryotic cells, aka they are phospholipid bilayers in which specific proteins are embedded
4. flagellum ⮕ long, filamentous appendage that aids in cell movement
5. pilus ⮕ long, thin appendages that can be used to attach one bacterial cell to another during conjugation

Question 4

describe the peptidoglycan cell wall

Answer 4

all bacterial cells posesses a cell wall composed of peptidoglycan.
- the glycan component is a linear polymer of alternating monosaccharide subunits, N-acetylglucosamine (NAG) and acetylmuramic acid (NAM). the β (1,4) glycosidic bond between the monomers making up the polysaccharide are cleaved by the anti-bacterial enzyme lysozyme
- the peptido portion of the polymer is a short string of amino acids that serves to cross-link adjacent polysaccharide strands at the NAM subunit of the backbone, forming a network with high tensile strength

Question 5

describe the peptidoglycan cell wall for gram-positive and gram-negative bacterial cells

Answer 5

gram-positive bacteria
- have thick, multi-layered peptidoglycan cell walls that are exterior to the membrane
- peptidoglycan in most gram-positive species is covalently linked to teichoic acid, a polymer of substituted glycerol units linked by phosphodiester bonds. teichoic acids are major cell surface antigens

gram-negative bacteria
- have two membranes: an outer membrane and an inner (cytoplasmic) membrane
- the peptidoglycan layer is located between the 2 membranes in the periplasmic space
- the peptidoglycan layer is thin, so gram-negative bacterial cells are more susceptible to physical damage
- the outer membrane is distinguised by the presence of various embedded lipopolysaccharides. the polysaccharide portion (O-polysaccharide) is antigenic and can be used to identify different strains and species. the lipid portion (lipid A) is toxic to humans and animals, and is called an andotoxin as it is an integral part of the membrane

Question 6

what are cytoplasmic structures of bacterial cells?

Answer 6

1. 70S ribosomes ⮕ different in structure compared to eukaryotic 80S ribosomes and are therefore smaller. they are found in large numbers in the cytoplasm

metabolites & enzymes in a bacterial cell are not enclosed within membrane-bound organelles but are found in the cytoplasm

Question 7

not in LO

what is the endosymbiont theory?

Answer 7

the endosymbiont theroy is an evolutionary theory that explains the origin of eukaryotic cells from prokaroytes. it states that mitochondrai and plastids of eukaryotes originated as a result of symbiosis between single-celled organisms.
serisl endosymbiosis: according to this theory,
- an early ancestor of eukaryotic cells engulfed an oxygen-using non-photosynthetic prokaryotic cell
- eventually, the engulfed cell formed a relationship with the host cell in which it was enclosed, becoming an endosymbiont (a cell living within another cell)
- over the course of evolution, te host cell and its endosymbiont merged into a single orgranism, a eukaryotic cell with a mitochondrion
- although all eukaryotes have mitochondria or remnants of these organelles, they do not all have plastids, so the hypothesis of serial endosymbiosis hypothesizes that mitochondria have evolved before plastids through a sequence of endosymbiotic events, forming the ancestor of eukaryotic cells that contain chloroplasts

Question 8

not in LO

what evidence is there for the endosymbiont theory?

Answer 8

• the inner membrane of both mitochondria and plastids have enzymes and transport systems that are similar to those found in the plasma membrane of living prokaryotes
• mitochondria and plastids replicate by a splitting process similar to certain prokaryotes
• mitochondria and plastids contain circular DNA which are not associated with histones, resemblin chromosomes of bacteria
• mitochondria and plastids possess cellular machinery, including ribosomes, needed to transcribe & translate their DNA into proteins
• ribosomes of mitochondira and plastids are more similar to prokaryotic ribosomes in terms of size, RNA sequences, and sensitivity to antibiotics compared to cytoplasmic ribosomes of eukaryotic cells

Question 9

describe the circular bacterial chromosome

Answer 9

• the bacterial chromosome is typically a single, circular, double-stranded DNA molecule which contains essential genes required for survival
• the bacterial chromosome must be compacted approximately 1000-fold, by associating the DNA with positively charged histone-like proteins that compact the DNA into looped domains. further compaction is achieved by supercoiling the bacterial DNA, aided by other specific bacterial proteins
• genes are grouped into operons, where multiple genes come under the control of the same promoter and the same regulatory elements
• prokaryotic genes lack introns so they do not require splicing after transcription (unlike eukaryotes)

Question 10

what are plasmids and what do they do?

Answer 10

• plasmids contain other non-essential pieces of DNA (bacterial chromosomes contain the minimal genetic requirement for bacterial survival)
• plasmids exist as small, circular, double-stranded extrachromosomal DNA molecule which may be passed on to cells of the same generation or to its offspring
• plasmids are capable of replication independent fo the bacterial chromosome as they possess their own origin of replication, so cells may thus contain more than one plasmid
• plasmids contain beneficial genes which confer protective traits like antibiotic resistance, toxin synthesis and enzyme production -> confers a selective advantage to bacteria

Question 11

what are the roles of genetic material in bacteria?

Answer 11

1. replication
2. gene transfer
3. gene expression

Question 12

what is binary fission?

Answer 12

transmission of genetic material from a bacterial cell to its offspring occurs by binary fission. binary fission is a form of asexual reproduction in which 2 equal-sized, genetically-identical daughter cells are produced from a single parent cell. it includes the replication of the bacterial chromosome.

binary fission is unble to give rise to genetic variation in a bacterial population.

Question 13

describe, in detail, the process of binary fission.

Answer 13

1. the bacterial chromosome is attached to the plasma membrane before DNA replication
2. DNA replication begins at the single origin of replication, where the replication bubble is first formed when the two 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. (results in 2 identical chromosomes)
3. after DNA replication is completed, cell growth occurs. each circular DNA molecule is attached to the cell membrane.
4. the cell elongates, and membrane growth causes the 2 chromosomes to be moved apart
5. 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 (grows inwards) as new cell membrane and cell wall materials (like peptidoglycan) are added to it.
6. the invaginating cell membrane, together with the newly formed septum, splits the cell into 2 genetically identical daughter cells by cytokinesis

Question 14

what are the 3 ways genetic material can be transferred from one cell to another?

genetic material exchange & genetic variation occurs!!

Answer 14

1. transformation
2. transduction
3. conjugation

Question 15

what is homologous recombination?

Answer 15

homologous recombination is when DNA is integrated into the chromosome of the recipient cell. the transferred portion of the donor chromosome will be exchanged with a portion fo the chromosome of the recipient cell that is very similar in sequence (highly homologous). the segment of the recipient chromosome that is exchanged for the donor chromosome is excised and degraded

occurs for transformation and transduction, but NOT for conjugation.

Question 16

what is transformation?

Answer 16

transformation is the process by which a recipient cell takes up small fragments of naked DNA from the surrounding environment. (this DNA can originate from a donor bacterial cell which lyses and releases its DNA into the surrounding environment or artificially constructed plasmids)
only competent bacterial cells are naturally able to undergo transformation. competence depends on teh presence of competence factors produced by the bacterial cell, which are cell surface proteins that bind to DNA fragments and aid in their uptake

Question 17

describe, in detail, the process of transformation.

Answer 17

1. the donor bacterial cell lyses and releases naked DNA fragments (donor DNA fragment)
2. a competent recipient cell takes up one or more of the donor DNA fragments into its cytoplasm via its competence factor
3. homologous recombination of the donor DNA fragment takes place with a homologous section of the recipient cell’s chromosome
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 is excised and degraded. the recipient cell is now known as a recombinant cell

transformation can be artificially induced in the laboratory by electroporation or treatment of bacteria with calcium chloride followed by heat shock

Question 18

what is transduction?

Answer 18

in transduction, bacteriophages carry bacterial genes from their first host cell to their second host cell due to errors in the phage reproductive cycle. the DNA fragment of the donor cell may be incorporated into the genome of the recipient cell via homologous recombination

2 types of transduction: generalized transduction & specialized transduction

Question 19

what is generalized transduction?

Answer 19

during the reproduction of VIRULENT phages, a small fraction of the virions produced during the lytic cycle contain a random fragment of the bacterial genome instead of phage DNA.
- this occurs due to the accidental incorporation of a random fragment of DNA from its first host cell into the phage capsid
- when such a defective phage (that carries additional host genes) infects a second host cell, the DNA of the donor bacterium is inserted into the recipient bacterium
- this is followed by the integration of the donor genes into the recipient cell’s genome by homologous recombination

note: each portion of the bacterial genome has approximately the same probability of being transferred from donor to recipient bacteria (any bacterial gene can potentially be transferred)

Question 20

describe, in detail, the process of generalized transduction

Answer 20

1. the virulent phage injects its DNA into its first host bacterial cell (donor bacterium) and the first host cell’s chromosome is degraded
2. the phage uses of the host’s DNA replication machinery to synthesise more phage DNA, and uses the host cell’s gene expression machinery to synthesise more phage proteins (eg capsid proteins)
3. occasionally, a small piece of the first host cell’s degraded DNA is accidentally packaged within a phage capsid in place of the phage genome duing the assembly stage of the lytic cycle. this phage is known as a defective phage. the first host bacterium is lysed, and the new phages are released into the environment.
4. the defective phage progeny, which contains the first host cell’s DNA fragment, may infect a second host cell and inject the DNA fragment acquired from the previous host cell into it
5. the donor DNA is incorporated into the second host cell’s genome by homologous recombiantion, in which the donor DNA replaces the homologous region of the recipient cell’s chromosome. the recipient cell is now known as a recombinant cell

Question 21

what is specialized transduction?

Answer 21

during the reproduction of TEMPERATE phages, the phage genome is integrated into the first host cell’s chromosome as a prophage during the lysogenic cycle. upon induction, the phage genome is excised and the cell is switched to the lytic cycle.

• the excision of the prophage is sometimes imprecise, resulting in segments of phage DNA that lack part of the normal phage genome and contain part of the bacterial chromosome located adjacent to the prophage attachment site.
• EITHER when such a phage infects a second host cell, the DNA of the donor bacterium is inserted into the recipient bacterium along with a portion of the phage genome
• OR the integration of the donor gnes into the recipient cell’s genome by homologous recombination can occur

only specific portions of the bacterial genome, genes near the prophage insertion site on the host chromosome, have a high probability of beingt ransferred from donor to recipient bacteria.

Question 22

describe, in detail, the process of specialized transduction.

Answer 22

1. the genome of a temperate phage integrates as a prophage into the chromosome of the first host bacterium at the prophage insertion site
2. 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 bacterial DNA that was adjacent to the prophage insertion site
3. the phage DNA reproduces itself using the first host cell’s DNA replication and gene expression machinery. each newly formed phage now contains part of the first host cell’s DNA
4. the first host bacterium is lysed, releasing phages into the environment. the phages infect a second host cell and inject the DNA fragment acquired from the previous host cell into it
5. the donor DNA is incorporated into the second host cell’s genome by prophage integration (if phage DNA transferred contains genes required to enter lysogenic cycle) or homologous recombination (if otherwise). the recipient cell is now known as a recombinant cell

Question 23

compare the differences between generalized transduction & specialized transduction

Answer 23

type of phage
- generalized transduction: virulent phage
- specialized transduction: temperate phage

phage reproductive cycle
- generalized transduction: lytic cycle
- specialized transduction: lysogenic cycle, convertin to lytic cycle upon induction

transfer of donor DNA
- generalized transduction: any donor bacterial gene can be accidentally packaged in phage capsid and can be transferred to the recipient cell
- specialized transduction: only donor genes near the prophage insertion site can be transferred to the recipient cell due to imprecise excision of prophage from the donor cell’s chromosome

Question 24

what is conjugation?

Answer 24

in conjugation, direct contact between the donor and recipient bacteria leads to establishment of a cytoplasmic bridge between them, followed by transfer of the donor’s DNA to the recipient cell.

the donor bacterial cell possesses an F factor, which exists as a plasmid, the F plasmid, or as a segment of DNA integrated into the bacterial chromosome. genes on the F factor are responsible for the synthesis of the sex pillus and to transfer the F factor from a donor cell to a recipient cell

the F factor has its own origin of replication and replicates as it is transferred, so after conjugation, both the donor and recipient cells contain an F factor.

cells with an extrachromosomal F factor are known as F+ cells, and cells lacking the F factor are known as F- cells

Question 25

describe, in detail, the process of conjugation

Answer 25

1. the F+ donor cell uses a sex pilus to attach to an F- recipient cell
2. a temporary cytoplasmic mating bridge is formed between the 2 cells, providing a route for DNA transfer
3. the sugar-phosphate backbone of one strand of the F plasmid in the F+ donor cell is nicked by endonuclease (phosphodiester bond is broken). the nicked single strand DNA separates from its complementary strand and moves to the F- recipient cell through the cytoplasmic mating bridge
4. each parental strand becomes a template for the DNA synthesis of a complementary 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 phosphodiestere bond to close the gap in each F plasmid
5. the cells move apart and the sex pilus breaks, forming 2 bacterial cells that are both F+. (a F+ bacterial cell converts a F- cells to a F+ cell when the two cells conjugate)

Question 26

what are the 3 main differences between prokaryotic and eukaryotic gene expression?

Answer 26

1. degree of compaction of chromosomes ⮕ prokaryotic chromosome tends to be less compact compared to eukaryotic chromosome.
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, so their chromosomes are not surrounded by nuclear membrane unlik eukaryotes

because prokaryotes don’t have a nuclear membrane, simultaneous transcription and translation is possible

Question 27

what are the components of a typical prokaryotic structural gene?

Answer 27

1. promoter
2. coding region
3. untranslated regions (5’ UTR and 3’ UTR)
4. terminator sequence

Question 28

what is the promoter?

prokaryotic gene

Answer 28

• the promoter determines the recognition and binding of prokaryotic RNA polymerase
• the promoter consists of the RNA polymerase recognition site (5’-TTGACA-3’) and the RNA polymerase binding site (5’-TATAAT-3’) / Pribnow box
• the RNA polymerase recognition site is located 35bp upstream of the transcription start site while the RNA polymerase binding site is located 10bp upstream of the transcription start site
• the sequences upstream of the RNA polymerase recognition & binding sites are the non-template strand sequences

Question 29

what is the coding region?

prokaryotic genes

Answer 29

• the coding region begins with the start codon 5’-AUG-3’, the coding sequence for the polypeptide, and ends with the stop codon
• the coding region of prokaryotic genes is not interrupted by introns, unlike eukaryotic genes
• the coding region gives rise to mRNA, which is translated to form polypeptides

Question 30

what are the untranslated regions?

prokaryotic genes

Answer 30

5’ UTR
- the 5’ UTR consists of sequences between the transcription start site and the start codon (before coding region)
- the 5’ UTR contains a sequence which, when transcribed, will give rise to the shine-dalgarno sequence on the mRNA which is required for ribosome binding during translation

3’ UTR
- the 3’ UTR consists of sequences immediately following the stop codon

Question 31

what is the terminator sequence?

prokaryotic genes

Answer 31

the terminator sequence is used to terminate transcription by dislodging the RNA polymerase from the template DNA

Question 32

what are the 3 stages of prokaryotic transcription?

Answer 32

1. initiation
2. elongation
3. termination

same as eukaryotes!

Question 33

describe the initiation stage of prokaryotic transcription

Answer 33

• initiation begins when RNA polymerase binds to the promoter of the gene.
• RNA polymerase is only able to bind to the promoter specifically in the presence of the sigma (σ) factor, which is a protein subunit required by RNA polymerase for the initiation of transcription.
• the σ factor on RNA polymerase recognizes & binds at the -35bp recognition sequence and at the -10bp Pribnow box (binding sequence) in the promoter. the -10bp sequence is A-T rich, making it easier to break the hydrogen bonds, thus unwinding the DNA (since A & T have 2 H bonds)
• RNA polymerase transiently unwinds DNA, forming a transcription bubble, allowing one strand to be used as a template for transcription

Question 34

describe the elongation stage of prokaryotic transcription

Answer 34

• RNA polymerase reads the DNA template strand in the 3’ → 5’ direction. RNA polymerase moves down the template strand as elongation continues
• ribonucleoside triphosphates are added in a complementary manner, in the 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 two strands of DNA upstream of the transcription bubble are re-wound into the double helical structure

Question 35

describe the termination stage of prokaryotic transcription

Answer 35

termination occurs when the core RNA polymerase dissociates from the template DNA, and there are 2 mechanisms for this to occur:

a) intrinsic termination / rho-independent transcription termination
- intrinsic termination uses a terminator sequence within the RNA, usually a palindromic GC-rich sequence followed by 4 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

b) rho-dependent termination
- rho-dependent termination uses a termination factor called the ρ (rho) factor, which is a protein that stops mRNA synthesis as specific sites
- the ρ factor binds at a ρ recognition site on the mRNA strand, and moves along the mRNA towards the RNA polymerase
- when the ρ factor reaches the RNA polymerase, it destablizes the mRNA-DNA hybrid, releasing the newly synthesised mRNA from the elongation complex.

Question 36

what are the 3 steps of prokaryotic translation?

Answer 36

1. initiation
2. elongation
3. termination

note:
- translation & transcription occur simultatneously due to the absence of a nuclear membrane
- since translation is initiated while the transcript is still being formed, no post-transcriptional modification is possible to prokaryotic mRNAs. splicing is unnecessary due to the lack of introns, and there is no addition of 5’-methylguanosine cap or 3’ poly(A) tail, making prokaryotic mRNA very unstable
- bacterial cells have fewer types oftRNA than eukaryotic cells, although the same number of codons exist and code for the same amino acids

Question 37

describe the initiation stage of prokaryotic translation

Answer 37

• iniiation of translation begins when the 30S small ribosomal subunit, aided by translation initiaiton factor, recognizes & 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, tRNAᶠᴹᵉᵗ, then binds at the P site of the large ribosomal subunit with the aid of another translation initiation factor
• a 3rd translation initiation factor aids in the binding of the large subunit to the small subunit
• the translation initiation complex is formed, and all initiation factors are released.

Question 38

describe the elongation stage of prokaryotic translation

Answer 38

• before translation is initiated, tRNA activation by aminoacyl-tRNA synthtase occurs
• elongation is carried out by the 70S ribosomes of prokaryotes in the same manner as eukaryotes, but prokaryotes use fewer elongation factors compared to that of eukaryotes

Question 39

describe the termination stage of prokaryotic translation

Answer 39

• termination of protein synthesis is carried out by the binding of release factors, which recognizes the 3 different stop codons
• the entry of the release factors into the A site results in the release of the polypeptide chain and the dissociation of the subunit of the ribosome

Question 40

what is an operon?

Answer 40

• bacterial structural genes with related functions are generally located adjacent to each other and are under the control of the same promoter and regulatory regions. this generates polycistronic mRNAs that encode multiple proteins
• the grouping of genes of related function forms a transcription unit, known as an operon
• genes in an operon are transcribed together and thus a single regulatory mechanisms can control a whole cluster of functionally related genes. this ensuresthat the required proteins are produced in sync as necessary

Question 41

what are the 2 major modes of regulation in a prokaryotic cell

Answer 41

1. transcriptional control ⮕ controls the amount of a protein. regulation of the amount of protein synthesized occurs at the transcriptional level (how much mRNA is produced)
2. post-translational control ⮕ controls the activity of pre-existing proteins. regulation of the activity of a protein can occur by adding functional groups or cleavage

Question 42

what does an operon consist of?

Answer 42

1. promoter for structural genes: the promoter lies upstream of the structural genes and provides a site for RNA polymerase to bind to and initiate transcription
2. operator: an operator is a segment of DNA which regulates the rate of transcription of the sturctural genes by interacting with a specific repressor protein
3. structural genes: a structural gene is a region of DNA that codes for any RNA or protein product other than a regulatory protein. structural genes code for protein or RNA molecules that form part of a cellular structure or have an enzymatic function.

regulatory genes lie outside the operon, and code for a specific protein that regulates the expression of the structural genes.

Question 43

distinguish between structural genes and regulatory genes

Answer 43

proteins coded
- structural genes code for a functional RNA, structural protein, enzyme or any functional protein within the cell
- regulatory genes code for regulatory proteins whose sole function is to regulate the transcription of structural genes

position found
- structural genes are usually found as part of an operon
- regulatory genes lie outside the operon

Question 44

define inducible and repressible operons

Answer 44

inducible operons
- inducible operons are normally not transcribed
- inducible operons are turned on/induced by the substrate of the enzyme for which the structural genes code, which acts as an inducer
- this ensures that enzymes are produced only when the substrate is present
- inducible operons usually code for enzymes involved in catabolic pathways (breaking down of substances)
- an example of an inducible operon in the lac operon

repressible operons
- repressible operons are normally transcribed
- repressible operons are turned off/repressed upon accumulation of the prdouct of the metabolic pathway, which functions as a corepressor
- by suspending production of an end-product when it is already present in sufficient quantity, the cell can allocate its organic precursors and energy for other uses
- repressible operons usually code for enzymes involved in anabolic pathways (synthesis of complex substances from simpler ones)
- an example of a repressible operon is the trp operon

Question 45

describe the action of the trp operon.

repressible

Answer 45

in the absence of tryptophan,
- the trp repressor is in an inactive conformation and cannot bind to the operator as it does not have a complementary conformation to the trp operator sequence
- RNA polymerase can bind to the promoter, allowing the transcription of trp operon genes
- the operon is switched on and the genes are expressed to give enzymes involved in the metabolic pathway
- tryptophan is synthesised

in the presence of excess tryptophan,
- tryptophan functions as a corepressor
- tryptophan binds to the trp repressor, changing it to its active conformation which can now bind to the operator as it is complementary in conformation to the trp operator sequence
- RNA polymerase is blocked from binding to the promoter, and transcription cannot take place
- the operon is switched off and the genes are not expressed to give enzymes involved in the metabolic pathway
- no synthesis of tryptophan takes place

Question 46

what structural genes do the lac operon contain and how is it under dual control?

Answer 46

the lac operon contains:
- lac Z gene coding for the enzyme β-galactosidase, which breaks down lactose to glucose and galactose
- lac Y gene coding for the transport protein galactoside permease, which enables the bacterial cell to take up lactose
- lac A gene coding for transacetylase, whose function in lactose metabolism is still unclear (BRUH)

the lac operon is under dual control:
- negative control by the lac repressor (influenced by the presence/absence of lactose)
- positive control by the catabolite activator protein (CAP) (influenced by the presence/absence of glucose)

Question 47

lac repressor

describe the negative regulation of the lac operon

inducible

Answer 47

in the absence of lactose
- the lac repressor protein is in an active conformation and binds to the operator sequence of the operon as it is complementary in conformation to the lac operator sequence
- RNA polymerase is prevented from binding to the promoter and transcription cannot take place
- the operon is switched off and hydrolysis of lactose cannot occur as β-galactosidase is not produced

in the presence of lactose
- allolactose, an isomer of lactose, is formed in small amounts from lactose that enters the cell
- allolactose acts as an inducer by binding to the lac repressor, switching it to its inactive conformation. the inactive repressor cannot bind to the operator as it does not have a complementary conformation to the lac operator sequence
- RNA polymerase can bind to the promoter and transcription of the lac operon genes can take place
- the operon is switched on and hydrolysis of lactose occurs as β-galactosidase is produced

Question 48

catabolite activator protein (CAP)

describe the positive regulation of the lac operon

repressible

Answer 48

in the presence of glucose
- cAMP concentration falls
- binding between cAMP and CAP does not occur
- CAP assumes an inactive conformation and cannot bind to the CAP binding site in the lac operon as it does not have a complementary conformation to the CAP binding site
- transcription of the lac operon proceeds at only a low rate, even in the presence of lactose

in the absence of glucose
- cAMP concentration rises
- binding between cAMP and CAP occurs
- CAP assumes an active conformation, bindingg to the CAP binding site as it is complementary in conformation to the lac operator sequence
- binding of RNA polymerase to the promoter is enhanced and transcription of the lac operon proceeds at a high rate

bacteria prefers to use glucose over lactose, and won’t use lactose unless all glucose is used up. so bacterial cells only need enzymes for lactose breakdown if glucose is absent and lactose is present!

Question 49

distinguish between inducible and repressible operons

Answer 49

type of metabolic pathway
- inducible operon: catabolic
- repressible oepron: anabolic

usual condition of operon
- inducible operon: switched off
- repressible operon: switched on

usual state of repressor protein
- inducible operon: active conformation -> binds to operator -> prevents RNA polymerase from binding to promoter
- repressible operon: inactive conformation -> does not bind to operator -> RNA polymerase can bind to promoter

molecule bound to repressor protein
- inducible operon: inducer
- repressible operon: co-repressor

conditions under which transcription occurs
- inducible operon: availability of substrate -> ensures that enzymes are only synthesized when needed, conserving available energy and resources
- repressible operon: availability of end product in sufficient quantity -> ensures that resources are not used to synthesize products that are already present in sufficient quantity, conserving available energy and resources