COPEG

Question 1

state coding DNA

Answer 1

DNA sequences which code for a specific sequence of amino acids in a polypeptide or a functional RNA molecule (parts of a gene are known as exons)

Question 2

state what are non-coding DNA

Answer 2

DNA sequences which do not code for amino acids in proteins or functional RNA (between genes: repetitive DNA, within genes: introns)

Question 3

describe regulatory regions

Answer 3

• promotor: non coding DNA sequence which regulates rate of transcription, contains TATA box: binding site of general transcription factor (TFIID). after TFIID binds, other general transcriptor factors bind, recruiting RNA polymerase to bind at the promoter, assembling to form the transcription initiation complex, located at the start of the gene, upstream of the transcriptional start site
• enhancers are non-coding DNA sequences that activators (specific regulatory proteins) bind to, to increase the rate of transcription; may be located upstream/downstream of the transcriptional start site, distant from the promoter
• silencers are non-coding DNA sequences that repressors (specific regulatory proteins) bind to, to decrease the rate/inhibit transcription; may be located upstream/downstream of the transcriptional start site, distant from the promoter

Question 4

describe introns

Answer 4

• non-coding sequences that are part of a gene
• transcribed into RNA but are excised before translation begins

Question 5

give examples of repetitive dna

Answer 5

telomeres and centromeres

Question 6

state what a transcriptional unit composes of

Answer 6

exons and introns, and terminator sequence

Question 7

describe exons and introns

Answer 7

• coding regions (exons) interrupted by non-coding regions (introns)
• after transcription, introns are removed while exons are joined together during RNA splicing
• each exon codes for a particular amino acid sequence of the polypeptide encoded by the gene
• after exons are transcribed into mRNA, parts of exons will not be translated further into polypeptide sequence (UTR)

Question 8

describe terminator sequence

Answer 8

causes RNA polymerase to dissociate from the DNA template strand

Question 9

describe the location and binding site of the 5’ UTR

Answer 9

• found between the 5’ modified guanosine cap of mRNA and start codon
• serves as a binding site for ribosome and other regulatory proteins which control the rate of translation

Question 10

describe the location and function of 3’ UTR

Answer 10

• found after stop codon
• contains sequences which regulate polyadenylation of mRNA, which in turn controls rate of translation

Question 11

describe the structure of centromeres

Answer 11

• repetitive DNA sequences made up of tandem repeats
• found as heterochromatin

Question 12

state the functions of centromeres

Answer 12

• adhesion point for sister chromatids in a chromosome
• site of assembly of the kinetochore, a protein complex which attaches to the nmicrotubules of the mitotic/meiotic spindle
• essential for the equal segregation of sister chromatids during mitosis and chromatids in meiosis II, and segregation of homologous chromosomes during meiosis I to opposite poles, and hence to daughter nuclei
• helps organise chromatin within interphase nucleus

Question 13

describe the structure of telomeres

Answer 13

• dsDNA sequences which form the ends of linear DNA of eukaryotic chromosomes
• made up of hundreds/thousands of copies of a short repeated sequence arranged in tandem repeats

Question 14

describe the structure and function of telomerase

Answer 14

• RNA component: forms complementary base pairs with the 3’ end of the DNA, providing a template for the complementary base pairing of new deoxyribonucleotides to form repeats of the telomere sequence
• protein component: acts as reverse transcriptase; binds to part of the DNA at the 3’ overhang at its active site and uses the RNA template to synthesise multiple DNA repeats at 3’ end of telomere via complementary base pairing. it catalyses the formation of phosphodiester bonds between adjacent deoxyribonucleotides

Question 15

outline the role of telomeres

Answer 15

• protect genes at the end of linear chromosomes from being eroded. telomeres shortened due to DNA polymerase not being able to replicate DNA at the ends of linear chromosomes
• protect the ends of chromosomes from fusion with other chromosomes
• protect ends of chromosome from enzyme degradation
• telomere shortening can trigger apoptosis/telomeres act as signals for cells to enter replicative senescence

Question 16

explain how gene expression can be regulated at chromatin level

Answer 16

condensation of chromatin tends to prevent gene expression by preventing RNA polymerse and transcription factors from gaining access to the promoter of a specific gene, thus inactivating transcription of that gene

Question 17

explain how causing histones to be less tightly bound to DNA increases rate of gene expression

Answer 17

• acetylation of histones neutralises their positively-charged R groups of lysine amino acid residues on histone tails
• decreases affinity of histone octamer for DNA
• chromatin more diffused/less compact
• RNA polymerase and transcription factors can bind to promoter
• these genes are more easily transcribed (catalysed by histone acetyltransferase)
• acetylated histones: transcriptionally active genes

Question 18

explain how histones being more tightly bound to DNA decreases rate of gene expression

Answer 18

deacetylation of histones
* restores affinity of histone octamer for DNA
* chromatin more compact
* RNA polymerase and transcription factors cannot bind to promoter and control elements
* these genes are less easily transcribed (catalysed by histone deacetylases)

methylation of histones leads to condensation of chromatin
* methylated histones recruit other proteins that keep chromatin tightly packed
* RNA polymerase and transcription factors have decreased accessibility to genes in a methylated region
* these genes are less easily transcribed (catalysed by histone methyltransferase)

Question 19

explain how DNA modification (methylation) can decrease rate of gene expression

Answer 19

addition of methyl groups to cytosine in DNA within CpG islands, reaction catalysed by enzyme DNA methyltransferase, making chromatin structure more compact

genes not required for expression are not expressed when methylated

Question 20

explain how enhancers increase the rate of transcription

Answer 20

causes DNA to bend to stabilise the transcription initiation complex and increase affinity of promoter for RNA polymerase, increasin rate of transcription of gene

Question 21

explain how silencers decrease the rate of transcription

Answer 21

• cannot facilitate correct positioning of transcription initiation complex/block assembly of transcription initiation complex/recruits proteins that bind and promote histone deacetylation or histone methylation, promoting chromatin condensation
• transcription initiation complex is not formed at promoter
• transcription is inhibited/very low rate of gene being transcribed

Question 22

5’ capping

Answer 22

• modified guanosine nucleotide added to 5’ end of mRNA, protecting mRNA from exonucleases, prolonging lifespan of mRNA

Question 23

RNA splicing

Answer 23

• takes place while RNA still being synthesised during transcription
• spliceosomes recognise splice sites at introns and then excise introns and join together exons
• alternative splicing using different splice sites, resulting in different combination of exons becoming linked together in mRNA, generating different mature mRNA molecules from the same pre-mRNA molecule
• allowing more than one type of polypeptide to be coded for by a singel gene

Question 24

3’ polyadenylation

Answer 24

• proteins recognise the poly(A) signal sequence near the 3’ end of the primary RNA transcript and recruit the enzyme poly (A) polymerase to add 50-250 adenine nucleotides at the 3’ end
• poly(A) binding proteins binds to the 3’ poly A tail to slow down the degradation of the 3’ end of mRNA by exonucleases, promote export of the mRNA from the nucleus and facilitate the initiation of translation

Question 25

how does mRNA exit the nucleus?

Answer 25

* mature mRNA which is ready to be exported out of the nucleus have proteins bound to it at the 5' cap and the 3' poly(A) tail * these proteins help to direct the mature mRNA to the nuclear pores, where the mRNA will exit the nucleus into the cytoplasm

Question 26

regulation at translation level using half-life of mRNA

Answer 26

* mRNAs with long half-lives are translated into many more protein molecules than do those with short half-lives (greater gene expression) * stability of mRNA affected by length of poly(A) tail (the longer the poly(A) tail the more stable the mRNA) * length of poly(A) tail regulated by particular nucleotide sequences within mRNA (3' UTR): contain binding sites for specific proteins that increase/decrease rate of poly(A) tail shortening

Question 27

degradation of mRNA

Answer 27

* enzyme deadenylase shortens poly(A) tail of mRNA, triggering enzymes that remove the 5' modified guanosine cap * once cap is removed, exonucleases rapidly degrade mRNA

Question 28

regulating translation of all mRNA

Answer 28

* translation initiation factors (responsible for binding of small ribosomal subunit and recruitment of aminoacyl-tRNA to convey it to the ribosome) when inactivated/activated, will affect translation of all mRNA in a cell * allowing cell to shut down translation if environment conditions are poor/until appropriate conditions exist

Question 29

regulating translation of specific mRNA

Answer 29

* initiation of translation is blocked by translation repressor proteins that bind to the mRNA at the 5' UTR. binding of repressor prevents small subunit of ribosome from binding to ribosome binding site at 5' UTR of mRNA * initiation of translation prevented bc mRNAs lack poly(A) tails of sufficient length. at an appropriate time, a poly(A) polymerases will add more A nucleotides, allowing translation of these mRNAs to begin

Question 30

post-translational level: biochemical modification

Answer 30

* cleavage: cutting polypeptides into smaller, active final products. hormone insulin is synthesised in pancreatic cells as one long inactive polypeptide (pro-insulin). after translation, a large centre portion is cleaved, leaving 2 shorter chains that constitute the active insulin molecule * chemical modification of protein (glycosylation, addition of lipids, methylation, acetylation, phosphorylation and dephosphorylation) can take place during post-translational chemical modification in RER/GA

Question 31

post-translational level: protein degradation

Answer 31

* selective breakdown of proteins * lifespan of intracellular protein varies -> allowing a cell to adjust the kinds and amounts of its proteins in response to chances in its environment * proteins marked for destruction by addition of a protein called ubiquitin * enzymes in the cytoplasm attach ubiquitin molecules to the protein targeted for degradation, ATP expended * ubiquinated protein recognised by proteasomes * protasome degrades ubiquinated protein into short peptides, ATP expended, ubiquitin released and can be recycled

Question 32

structure of lac operon

Answer 32

* length of DNA comprising regulatory elements and coding sequences for three structural genes * promoter serves as binding site for RNA polymerase to initiate transcription * operator serves as binding site for lac repressor to inhibit transcription * lacZ gene codes for beta-galactosidase, which hydrolyses lactose to glucose and galactose, and convert lactose to allolactose * lacY gene codes for lactose permease to actively transport lactose into the cytoplasm of bacterium * lacA gene codes for transacetylase that transfer an acetyl group from acetyl CoA to lactose * transcription termination sequence marks the end of the operon * regulatory gene/lacI gene coding for lac repressor

Question 33

presence of lactose with high glucose (lac operon)

Answer 33

* lac operon is transcribed only when lactose is present * entry of small amount of lactose into bacterial cell where it is converted to allolactose by beta-galactosidase * allolactose acts as an inducer and binds to allosteric site of lac repressor, changing the 3D conformation of the repressor * lac repressor can no longer bind to the operator * RNA polymerase binds to promoter and transcribes the three structural genes lacZ, lacY, lacA, leading to synthesis of enzymes needed for the utilisation of lactose such as beta-galactosidase, permease and transacetylase * high concentration of glucose inhibits adenyl cyclase, resulting in low cAMP concentration * catabolite activator protein (CAP) is not activated and does not bind to CAP binding site on promoter * lower affinity of RNA polymerase for the promoter * basal level of transcription of structural genes even in the presence of lactose

Question 34

what happens to the lac operon in the presence of lactose with low concentration of glucose?

Answer 34

* lac operon is transcribed only when lactose is present * entry of small amount of lactose into bacterial cell where it is converted to allolactose by beta-galactosidase * allolactose acts as an inducer and binds to allosteric site of lac repressor, changing the 3D conformation of the repressor * lac repressor can no longer bind to the operator * RNA polymerase binds to promoter and transcribes the three structural genes lacZ, lacY, lacA, leading to synthesis of enzymes needed for the utilisation of lactose such as beta-galactosidase, permease and transacetylase * low concentration of glucose results in high concentration of cAMP, cAMP binds to and activates CAP * activated CAP then binds to the CAP binding site within the lac promoter * increasing affinity of RNA polymerase binding to the lac promoter and hence rate of transcription of structural genes in the presence of lactose

Question 35

what happens to lac operon in the absence of lactose?

Answer 35

* regulatory gene lacI codes for the allosteric repressor protein, lac repressor * lac repressor synthesised in active form that binds readily to the operator and blocks transcription by RNA polymerase * in the absence of lactose, the lac repressor binds to the operator and the RNA polymerase is unable to bind to the promoter, thus structural genes are not transcribed

Question 36

what happens to the trp operon in the absence of tryptophan?

Answer 36

* regulatory gene trpR codes for the allosteric trp repressor * trp repressor is synthesised in the inactive form, which has little affinity for the trp operator * in the absence of tryptophan, trp repressor does not bind to the operator hence, RNA polymerase is able to bind to the trp promoter and the structural genes are transcribed

Question 37

what happens to the trp operon in the presence of tryptophan?

Answer 37

* in the presence of tryptophan, binding of this corepressor to the allosteric site of the trp repressor alters the tertiary structure of the repressor, activating the trp repressor such that it binds to the operator * operator is now bound by active trp repressor protein and RNA polymerase is unable to bind to the promoter to transcribe the five structural genes, unable to synthesise enzymes needed for biosynthesis of tryptophan

Question 38

state the similarities of the structure and organisation of prokaryotic and eukaryotic genomes

Answer 38

* both genomes are made up of double-stranded DNA * both genomes have the DNA associated with proteins

Question 39

state the differences of the structure and organisation of prokaryotic and eukaryotic genomes (11 pts)

Answer 39

* in prokaryotic genome, DNA is found as a circular DNA molecule while in eukaryotic genome, DNA is found as a linear DNA molecule * in prokaryotic genome, DNA is complex with non-histone proteins while in eukaryotic genome, DNA is complexed with histone proteins * prokaryotic genomes are much smaller while eukaryotic genomes are much larger * in prokaryotic genome, DNA is found in non-membrane bound nucleoid region while in eukaryotic genome, DNA is found in double-membrane bound nucleus * in prokaryotic genome, the genome is mostly located on one main chromosome while in eukaryotic genome, the genome is divided into different chromosomes * in prokaryotic genome, plasmids may be present while in eukaryotic genome, plasmids are usually absent * in prokaryotic genome, telomeres are absent since DNA is circular while in prokaryotic genome, telomeres are present in chromosomes * in prokaryotic genome, most of the DNA in a genome codes for protein, tRNA and rRNA, with a small amount of non-coding DNA while in eukaryotic genome, most DNA is non-coding * introns are generally absent in prokaryotic genome while in eukaryotic genome the gene exists as split. genes -- pieces of coding sequence, the exons, are separated by non-protein-coding segments, introns * in prokaryotic genome the genes are closely packed while in eukaryotic genome, there are large amounts of non-coding DNA between genes * in prokaryotic genome, non-coding DNA is mostly made up of regulatory sequences (promomters, operators, repressor binding sites) while in eukaryotic genome, non-coding DNA is made up of regulatory (promoters, enhancers, silencers) and repetitive sequences

Question 40

outline the difference between prokaryotic control of gene expression with the eukaryotic model (8 pts)

Answer 40

* in prokaryotes, genes involved in similar metabolic functions are grouped as an operon and are under the control of a single promoter and other control elements while in eukaryotes, each gene has its own promoter. some genes of related function may be under the influence of the same control element * in prokaryotes, genes of the operon are transcribed together into a single mRNA while in eukaryotes, each gene is individually transcribed * absence of histones in prokaryotes = absence of chromatin remodelling while in eukaryotes, control of gene expression by chromatin remodelling (acetylation/methylation of histones) is possible * prokaryotes uses mostly negative regulation (repressors that bind to operators) and some positive regulation while eukaryotes uses mostly positive regulation because of DNA stored as chromatin (rendering most promoters inaccessible and so genes are normally silent in the absence of other regulation) * absence of RNA processing in prokaryotes (absence of introns) while in eukaryotes, RNA processing provides opportunities for regulating gene expression (alternative splicing is unique to eukaryotes) * in prokaryotes, control of gene expression by enhancers is rare while in eukaryotes, control of gene expression by enhancers is common * in prokaryotes, transcription and translation are tightly coupled and happen simultaneously while in eukaryotes, transcription occurs first in the nucleus, and translation takes place in the cytoplasm after mRNA is transported out of the nucleus * in prokaryotes, regulation at the translation level is less prominent while in eukaryotes, regulation at the translation level is more prominent

Question 41

define control elements and explain how control elements and othe factors influence transcription

Answer 41

* control elements are non-coding regions of a DNA molecule that has the ability to influence the rate of transcription * promoter: contains a TATA box which is recognised by general transcription factors (TFs), TFs interact with RNA polymerase to form the transcription initiation complex, allowing basal level of transcription of a gene * enhancer: a type of distal control element found few thousands of nucleotides upstream/downstream of start site/even within an intron. activators bind to enhancers, interact directly with RNA polymerase and general TFs to accelerate the assembly of the TIC. activators may assist in recruiting histone acetylase and chromatin remodelling complex to allow greater accessibility to general TFs and RNA polymerase to the promoter region, resulting in increase in rate of transcription * silencer: DNA region bound by repressors. inhibits gene expression by preventing bound activator from binding to TFs, repressing chromatin remodelling complexes and recruting histone deacetylases to make chromatin inaccessible to general TFs and RNA polymerase, transcription rate decreases/stops

Question 42

describe the eukaryotic processing of pre-mRNA

Answer 42

* post-transcriptional modification is required to process pre-mRNA to form mature mRNA and to facilitate the export from nucleus to cytoplasm * 5' end is immediately capped with a modified form of guanosine * at 3' end, where there is polyadenylation sequence present, formation of a poly(A) tail is catalysed by enzyme poly(A) polymerase * signals for RNA splicing are short nucleotide sequences at ends of introns, spliceosome interacts with splice sites at ends of an intron. it cuts at specific points to release the intron, then immediately joins together two exons that flanked the intron