Molecular Biology

Front 1

What are the major differences between prokaryotic and eukaryotic transcription?

Back 1

RNAP in prokaryotes require the use of the sigma subunit and there is only one kind of polymerase. There are 3 in eukaryotic transcription and many TF are involved in the initiation complex. Eukaryotes can have more than one origin of transcription.

Front 2

Which hypothesis regarding eukaryotic RNAPs was proven with -amanitin and actinomycin D (be specific)?

Back 2

There are more than one eukaryotic RNAP, discovered by seeing the behaviour of polymerase under different conditions and sensitivity to amanitin.

actinomycin D (antibiotic) represses RNA elongation

alpha-Amanitin is an inhibitor of RNA polymerase II:

RNA polymerase I is insensitive, RNA pol II is highly sensitive, and RNA pol III is slightly sensitive.

Front 3

Which genes are transcribed by RNAP I? RNAP II? RNAP III?

Back 3

RNAPI makes 28S, 18S, 5.8S rRNAS
RNAPII (The main one) – mRNA & snRNAs
RNAPIII - tRNA

Front 4

What does CTD stand for? Explain the role of CTD tail in eukaryotic gene expression?

Back 4

Carboxyl terminal domain, a series of 7 amino acid repeated multiple times that is responsible in initiating transcription, found in RNAPII. When its not phosphorylated, the CTD tail initiates transcription, so in areas of high transcript only phosphorylated forms are found.

Front 5

How would you define enhancers? What are their characteristics? What is the difference(s) between enhancer and upstream control element?

Back 5

Enhancers allow regulatory proteinst o bind which sitmulate transcription (or in the case of silencers, down-regulate transcription). They can be located very far away from the genes (up to 50kb away), and is orientation independent. Upstream control elements reside close to the promoter (such as the GC Box) and influences transcription rates.

Front 6

Explain the use of reporter genes for estimation of promoter strength.

Back 6

Reporter genes can be inserted following a promoter of interest to determine promoter strength. One can delete parts of the promoter and analysis how this deletion affects the gene expression of the reporter gene (which usually codes for an easily detected product).

Front 7

Explain briefly 5’ deletion series. What kind of information do they reveal?

Back 7

5’ deletion series is where you attach a reporter gene to a promoter of interest and delete parts of the promoter and see how this impacts the expression of the reporter gene. This gives you information about which part of the promoter is critical to gene expression and which isn’t.

Front 8

Explain the modular nature of RNAP II promoters.

Back 8

Modular nature of RNAP II promoters mean that different additions/substraction of regulatory elements such as enhancers, boxes, etc alter the promoter (can be mixed and matched).

Front 9

Explain the tissue (cell type) specificity of eukaryotic cis elements.

Back 9

All cells contain the same DNA and the same cis elements, but the absence/presence of certain TF which binds to certain cis elements change the cell type and tissue specificity. (controls which genes is expressed nad which isn’t in the cell)

Front 10

You have discovered base changes in the promoter region of the operon in a bacterial chromosome. Would you expect these changes to act in trans on another copy of the operon? Explain your reasoning.

Back 10

Changes in the promoter region are cis-acting and not trans-acting because the promoter is a binding site, and does not have a gene product that can move to affect another copy of the operon.

Front 11

What are cis- elements? What are trans- factors? Give an example from the Trp operon (or form the Ara-operon).

Back 11

Cis elements (cis-acting sites; specific nucleotide sequences – precise distribution of A and D) regulate expression of genes on same strand – binding sites
•Sites or sequences on DNA E.g. DnaA box, OriC in E. coli

Trans factors (trans-acting functions; diffuse through the cell/nucleus) - regulate genes distant from the gene from which they were transcribed:proteins,recognize cis-elements and bind to them
• DnaA, DnaB, Dna C

Front 12

4. You have isolated a protein that binds to DNA in the region upstream of the promoter sequence of the gene of interest. If this is a positive regulator (activator) which would be true:
A) Loss of function mutation in the gene encoding this DNA binding protein would cause constitutive expression
B) Loss of function mutation in the gene encoding this DNA binding protein would result in lower or no expression.

Back 12

Activators increase the affinity of RNAP to the promoter, so if the activator is not available the gene of interest would be expressed less.

Front 13

Discuss why lac Oc mutants arecis-acting.

Back 13

Mutations in the operator are cis-acting because it only affects the structural genes downstream from it.

Front 14

Discuss why lac I+ mutants are trans-acting.

Back 14

Since I codes for the repressor protein, I mutants are trans-acting: the protein can diffuse throughout the cell and affect the gene expression of both copies in the merodiploid cell. For I+ the repressor protein cannot bind to lactose so even if lac is present it will always bind to both copies of the operator and prevent the expression of structural genes.

Front 15

Discuss positive and negative regulation of L-ara operon.

Back 15

positive regulation: if arabinose (inducer) is present it binds to AraC, this changes the conformation of the protein, so AraC binds to ara I1 and ara I2 and not to ara O2; leaving PBAD promoter open for RNA polymerase to bind & transcribe the operon. CAP+cyclic AMP need to bind to ara I for activation as well, so activation depends on the presence of arabinose and cAMP.
negative regulation: if no arabinose is present AraC binds to ara O2 and ara I1, bends DNA into a loop that hides PBAD promoter from RNA polymerase; enzymes are not synthesized

Front 16

Regarding the regulation of Trp operon, what do we call the amino acid tryptophan? Why?

Back 16

Tryptophan is a corepressor because when it binds to TrpR, transcription of Trp operon is inhibited, since Trp does not need to be synthesized.

Front 17

What is meant by polycistronic mRNA? Give an example.

Back 17

Cistron – a sequence of DNA that codes for one polypeptide chain and its control regions
So, polycistronic mRNA codes for more than one protein e.g. lac operon – structural genes transcribed from one mRNA

Front 18

What is catabolite repression? What is the role of Catabolite Activator Protein? Explain its action.

Back 18

catabolite repression; once known as “glucose effect” ; when E. coli is grown in a medium containing a preferred energy source transcription of a variety of catabolic operons is repressed (they aren’t needed)

Catabolite Activator Protein: a general/pleiotropic activator: starving cells produce cAMP, which binds to CAP. That complex binds to the CAP binding site upstream of RNAP binding site in 100+ genes/operons & induces their transcription (big catabolic response)

Front 19

Define: repressor, co-repressor, aporepressor and inducer.

Back 19

Repressor: a protein molecule that binds to DNA and deactivates transcription of a gene

Aporepressor: a protein molecule that needs to be activated by a co-repressor binds to the operator and prevents transcription at the promoter

Inducers – inactivate repressor -> induce transcription (lac operon)

Co-repressors – activate
aporepressors -> stop transcription (Trp operon)

Front 20

Define effector and inducer.

Back 20

effectors; small molecules; their binding changes protein (repressor) conformation
Inducers – inactivate repressor -> induce transcription (lac operon)

Front 21

What are activators? What are enhancers?

Back 21

Enhancers are elements upstream of the promoter region: when activators (proteins that perform positive regulation) bind to enhancers, transcription rates are increased.

Front 22

What is the role of auxiliary operators?

Back 22

Auxillary operators are multiple operators. 3 are in the Lac operon: the more operators that bind the repressor, the less likely RNAP is to bind the promoter due to DNA looping. (The repressor is a tetramer and could bind all three). Having multiple operators near the functional operator increases the local concentration of the repressor

Front 23

Discuss the type of regulation of gene expression by two-component regulatory systems in bacteria?

Back 23

2 components: Sensor-transmitter and response regulator

1) Sensor- usually trans-membrane, transmitter usually a kinase – change in env’t leads to conformation changes in the sensor that activates the transmitter domain, which usually autophosphorylates (moves gamma phosphate from ATP to itself)
2) same phosphate is then transferred to the receiver domain of regulator, conformation change activates effector domain
3) Response regulator binds to DNA regulatory sequences in gene(s) – it is a trans-factor

Front 24

Glutamine and arginine in DNA-binding proteins tend to make what kind of bonds with DNA?

Back 24

H - bonds

Front 25

List different ways of control of prokaryotic transcription initiation and give one example for each of them (this question could be separated in few smaller questions; make sure you understand review slides and that you have at least one example for each of the “control ways”).

Back 25

negative regulation: Lac operon w/repressor, Tryptophan operon w/aporepressor

positive regulation: L-arabinose operon both activator & repressor, catabolite repression is +ve regulation of Lac operon by cAMP-CAP complex

Two-component regulatory system: E.Coli osmoregulation system

Front 26

Define constitutive and regulated proteins. What is the difference between expression of constitutive and expression of regulated proteins?

Back 26

Constitutive proteins are always expressed (housekeeping genes)
Regulated proteins are not expressed all the time and are under the control of mechanisms so they are only expressed when needed or due to environmental change.

Front 27

What is the most important characteristic of binding sites for prokaryotic regulatory proteins (for example lac operon operator)? How are those binding sites different from the RNAP binding site (promoter)?

Back 27

The binding sites for prokaryotic regulatory proteins are inverted repeats since most proteins bind as dimers. RNAP binding sites (promoters) are asymmetrical so only one strand will be transcribed.

Front 28

Describe the most common structural motif found in a DNA binding domain of prokaryotic regulatory proteins.

Back 28

Helix-turn-helix motif, recognizes and binds to DNA, a 20 aa long structure that contain 2 helices and a turn.

Front 29

Apart from the DNA binding domain, we have mentioned other two domains found in prokaryotic binding proteins. What are they?

Back 29

Effector binding domain & dimerization domain

Front 30

DnaA

Back 30

-initiation - promotes the unwinding or denaturation of DNA at oriC - only if DNA is negatively supercoiled

Front 31

DnaB

Back 31

(opens the replication fork during DNA replication + activates primase) and DnaC- “clamp loader” - negative regulator, forms PRIMOSOME by associating with primase so knows where to put primer

Front 32

Single-strand binding proteins (SSB)

Back 32

prevents premature reannealing

Front 33

Primase (DnaG)

Back 33

with RNA polymerase synthesizes a short RNA primer so polymerase works

Front 34

RNase H

Back 34

cuts out the RNA primer, allows completion of new DNA

Front 35

DNA polymerase I

Back 35

DNA repair: 5'->3'(Polymerase) activity and 3'->5' (Proofreading) exonuclease activity

Front 36

DNA polymerase III

Back 36

main polymerase also 3'->5' exonuclease proofreading ability

Front 37

DNA ligase

Back 37

ligates Okazaki fragments

Front 38

Topoisomerase II

Back 38

DNA gets tangled as helicase moves forward Þ TopII converts +ve supercoils in front of growing fork into –ve

Front 39

Topoisomerase IV

Back 39

(belongs to topo II family) responsible for decatenation in vivo

Front 40

Gamma complex

Back 40

load/unload beta clamp onto DNA polymerase III

Front 41

Tau subunits

Back 41

kept 2 core DNA polymerases together so replication of 2 strands is synchronized

Front 42

Beta clamp

Back 42

forms ring around * slides down DNA to hold polymerase to it – higher processivity

Front 43

What is meant by replication being bidirectional? Semiconservative? Continuous and discontinuous?

Back 43

Bidirectional: proceeds in both directions away from the origin – 2 replication forks

Semiconservative: parent strands separated and complements synthesized for each so new double helix has one strand of parental DNA and one strand of newly synthesized DNA

Continuous: uninterrupted replication of DNA in the 5' to 3' direction using a 3' to 5' template

Discontinuous: replication of DNA in short 5' to 3' segments using the 5' to 3' strand as a template while going backward away from the replication fork (lagging strand)

Front 44

Contrast the role of DNA polymerase I and III in E. coli DNA replication.

Back 44

DNA polymerase III - synthesis of DNA
DNA polymerase I - 5' to 3' exonuclease activity – can start at a single-stranded break and progressively remove nucleotides and replace them, removes RNA primer & replaces with deoxyribonucleotides, DNA repair

Front 45

Which subunit of DNA polymerase III provides processivity? Which protein complex loads this subunit onto the DNA?

Back 45

Gamma complex - load/unload beta clamp onto DNA polymerase III
Beta clamp - forms ring around * slides down DNA to hold polymerase to it – higher processivity

Front 46

How can discontinuos synthesis of the lagging strand keep up with continuous synthesis of the leading strand?

Back 46

Tau subunits (DNA polymerase) – kept 2 core DNA polymerases together so replication of 2 strands is synchronized
g complex includes two copies of subunit t

Front 47

Why is decatenation required after replication of circular DNAs?

Back 47

Catenanes - early completed daughter helices of circular DNA are looped together – need to be separated

Front 48

Why do eukaryotes need telomeres but prokaryotes do not?

Back 48

Prokaryotes have circular DNA's [plasmids] so they don't have telomeres. Telomeres are the ends of linear DNA segments - aid in prevention of loss of genetic material .

Front 49

What would be the components necessary to make DNA in vitro by using DNA polymerase I?

Back 49

– DnaA-initiation
– DnaB- helicase and DnaC- “clamp loader”
– Single-strand binding proteins (SSB)
– Primase (DnaG)
– RNase H
– DNA polymerase (I and III)
– DNA ligase
– Topoisomerase

Front 50

What is telomerase and why is it important?

Back 50

Telomerase- modified reverse transcriptase
- Adds nucleotides to 3’ OH end of lagging strand template to prevent shortening – it polymerizes deoxyribonucleotides directed by RNA template (part of the enzyme, complementary to telomeric repeats) - protection from degradation
- Size of telomere regulate cell’s lifetime
- Chromosomes lose about 100 base pairs from their 5’ ends in each mitosis b/c 3’ end is unprotected – can’t fill in gap

Front 51

What properties would you expect an E. coli cell to have if it had a temperature – sensitive mutation in the gene for DNA polymerase I?

Back 51

If the cell was temperature sensitive, then within a certain temperature, DNA polymerase I would fail to remove the RNA primer & fill in Okazaki fragments so the cell wouldn’t be able to replicate.

Front 52

Compare and contrast major eukaryotic and prokaryotic DNA polymerases.

Back 52

In eukaryotes, more than one polymerase is involved in replication, but only DNAPIII is involved in the replication in prokaryotes

Front 53

What is telomerase and why is it important?

Back 53

Telomerase is a protein that contains an RNA template, used to extend the 3’ overhang by one repeating unit to prevent losing genetic information after each replication round.

Front 54

What is the major difference between bacterial and eukaryotic replication that allows an eukaryotic cell to replicate its DNA in a reasonable amount of time?

Back 54

Eukaryotes can have multiple origins of replication going on at the same time, while prokaryotes only have one.

Front 55

Is the following statement true or false: Regardless of whether a gene is expressed in a given cell type, it will replicate at the same, characteristic time during S phase. Explain your reasoning.

Back 55

NOT TRUE – because some genes are euchromatic structure (they’re expressed), replication starts here (easier access), but other genes are in heterochromatic (facultative) in other tissues.

Front 56

Describe the events that occur at an origin of replication during initiation of replication in E. colii?

Back 56

1. 10-20 initiator protein DnaA binds to four DnaA boxes (9 mers) in oriC form initial complex
2. results in a local unwinding of an AT-rich sequence within oriC that serves as a loading zone for the replicative helicase, a double-hexamer of DnaB and DnaC.
3. DnaB/C delivered to this unwound region by DnaA
4. DnaB activated by release of DnaC, moves in 5'-3' direction & interacts with primase DnaG
5. Primase synthesizes a short RNA primer for DNA polymerase III holoenzyme.

Front 57

What are cis- elements? What are trans- factors?

Back 57

Cis elements (cis-acting sites; specific nucleotide sequences – precise distribution of A and D) regulate expression of genes on same strand – binding sites
• Sites or sequences on DNA
• E.g. DnaA box, OriC in E. coli

Trans factors (trans-acting functions; diffuse through the cell/nucleus) - regulate genes distant from the gene from which they were transcribed
• Proteins that recognize cis-elements and bind to them
• DnaA, DnaB, Dna C

Front 58

You preformed Cot analysis using genomic DNA samples obtained from a 2 year-old child and a 76 year-old individual. Results of this analysis show that one of the most rapidly reassociating classes of DNA is substantially reduced in the older individual with respect to the 2 year-old. How can you explain this finding? (hint: think about termination of linear DNA replication.)

Back 58

The most rapidly reassociating classes of DNA are highly repetitive. Telomeres are repetitive – the 2 yr. old has longer telomeric sequences because the cells have not divided as many times

Front 59

You have discovered base changes in the promoter region of the operon in a bacterial chromosome. Would you expect these changes to act in trans on another copy of the operon? Explain your reasoning.

Back 59

Changes in the promoter region are cis-acting and not trans-acting because the promoter is a binding site, and does not have a gene product that can move to affect another copy of the operon.

Front 60

What are cis- elements? What are trans- factors? Give an example from the Trp operon (or form the Ara-operon).

Back 60

Cis-element sequences : usually short inverted repeats

Front 61

You have isolated a protein that binds to DNA in the region upstream of the promoter sequence of the gene of interest. If this is a positive regulator (activator) which would be true:
A) Loss of function mutation in the gene encoding this DNA binding protein would cause constitutive expression
B) Loss of function mutation in the gene encoding this DNA binding protein would result in lower or no expression.

Back 61

Activators increase the affinity of RNAP to the promoter, so if the activator is not available the gene of interest would be expressed less.

Front 62

Why are lac Oc mutants cis-acting.

Back 62

Mutations in the operator are cis-acting because it only affects the structural genes downstream from it.

Front 63

Why are lac I+ mutants trans-acting.

Back 63

Since I codes for the repressor protein, I mutants are trans-acting: the protein can diffuse throughout the cell and affect the gene expression of both copies in the merodiploid cell. For I+ the repressor proteincannot bind to lactose so even if lac is present it will always bind to both copies of the operator and prevent the expression of structural genes.

Front 64

Discuss positive and negative regulation of L-ara operon.

Back 64

positive regulation: if arabinose (inducer) is present it binds to AraC, this changes the conformation of the protein, so AraC binds to ara I1 and ara I2 and not to ara O2; leaving PBAD promoter open for RNA polymerase to bind & transcribe the operon. CAP+cyclic AMP need to bind to aria for activation as well, so activation depends on the presence of arabinose and cAMP.
negative regulation: if no arabinose is present AraC binds to ara O2 and ara I1, bends DNA into a loop that hides PBAD promoter from RNA polymerase; enzymes are not synthesized

Front 65

Regarding the regulation of Trp operon, what do we call the amino acid tryptophan? Why?

Back 65

Tryptophan is a corepressor because when it binds to TrpR, transcription of Trp operon is inhibited, since Trp does not need to be synthesized

Front 66

What is meant by polycistronic mRNA? Give an example.

Back 66

Cistron – a sequence of DNA that codes for one polypeptide chain and its control regions
So, polycistronic mRNA codes for more than one protein e.g. lac operon – structural genes transcribed from one mRNA

Front 67

What is catabolite repression? What is the role of Catabolite Activator Protein? Explain its action.

Back 67

catabolite repression; once known as “glucose effect” ; when E. coli is grown in a medium containing a preferred energy source transcription of a variety of catabolic operons is repressed (they aren’t needed)

Catabolite Activator Protein: a general/pleiotropic activator: starving cells produce cAMP, which binds to CAP. That complex binds to the CAP binding site upstream of RNAP binding site in 100+ genes/operons & induces their transcription (big catabolic response)

Front 68

Define: repressor, co-repressor, aporepressor and inducer.

Back 68

Repressor: a protein molecule that binds to DNA and deactivates transcription of a gene
Aporepressor: a protein molecule that needs to be activated by a co-repressor binds to the operator and prevents transcription at the promoter
Inducers – inactivate repressor Þ induce transcription (lac operon)
Co-repressors – activate aporepressors Þ stop transcription (Trp operon)

Front 69

Define effector and inducer

Back 69

effectors; small molecules; their binding changes protein (repressor) conformation
Inducers – inactivate repressor Þ induce transcription (lac operon)

Front 70

What are activators? What are enhancers?

Back 70

Enhancers are elements upstream of the promoter region: when activators (proteins that perform positive regulation) bind to enhancers, transcription rates are increased.

Front 71

What is the role of auxiliary operators?

Back 71

Auxillary operators are multiple operators. 3 are in the Lac operon: the more operators that bind the repressor, the less likely RNAP is to bind the promoter due to DNA looping. (The repressor is a tetramer and could bind all three). Having multiple operators near the functional operator increases the local concentration of the repressor

Front 72

Discuss the type of regulation of gene expression by two-component regulatory systems in bacteria?

Back 72

2 components: Sensor-transmitter and response regulator

1) Sensor- usually trans-membrane, transmitter usually a kinase – change in env’t leads to conformation changes in the sensor that activates the transmitter domain, which usually autophosphorylates (moves gamma phosphate from ATP to itself)
2) same phosphate is then transferred to the receiver domain of regulator, conformation change activates effector domain
3) Response regulator binds to DNA regulatory sequences in gene(s) – it is a trans-factor

Front 73

Glutamine and arginine in DNA-binding proteins tend to make what kind of bonds with DNA?

Back 73

H - bonds

Front 74

List different ways of control of prokaryotic transcription initiation and give one example for each of them (this question could be separated in few smaller questions; make sure you understand review slides and that you have at least one example for each of the “control ways”).

Back 74

negative regulation : regulatory proteins (repressors/aporepressors) in active state turn “off” the expression, often blocking RNAP - Lactose operon w/repressor, Tryptophane operon w/aporepressor

positive regulation : regulatory proteins (activators) in active state turn “on” the expression by binding to DNA near promoter sequence, increasing affinity of RNAP for the promoter - L-arabinose operon both activator & repressor

catabolite repression- positive regulation of lac operon when cell has no glucose
Interaction of Cyclic AMP (cAMP) – CAP complex with RNAP

Two component regulatory systems: sensor-transmitter protein+ response regulator protein - adapts to environmental change – E.Coli osmoregularity system

Front 75

Define constitutive and regulated proteins. What is the difference between expression of constitutive and expression of regulated proteins?

Back 75

Constitutive proteins are always expressed (housekeeping genes)
Regulated proteins are not expressed all the time and are under the control of mechanisms so they are only expressed when needed or due to environmental change.

Front 76

What is the most important characteristic of binding sites for prokaryotic regulatory proteins (for example lac operon operator)? How are those binding sites different from the RNAP binding site (promoter)?

Back 76

The binding sites for prokaryotic regulatory proteins are inverted repeats since most proteins bind as dimers. RNAP binding sites (promoters) are asymmetrical so only one strand will be transcribed.

Front 77

Describe the most common structural motif found in a DNA binding domain of prokaryotic regulatory proteins.

Back 77

Helix-turn-helix motif, recognizes and binds to DNA, a 20 aa long structure that contain 2 helices and a turn.

Front 78

Apart from the DNA binding domain, we have mentioned other two domains found in prokaryotic binding proteins. What are they?

Back 78

Effector binding domain & dimerization domain

Front 79

Compare and contrast Southern and northern blotting technique.

Back 79

Southern blotting is a technique used for DNA detection while northern is used for RNA detection. In both, endonucleases cut up molecules, fragments electrophoresed and transferred to membrane where complementary probe hybridizes to sequence of choice. In Southern, the DNA must first be denatured while in northern, denaturation is not necessary because RNA is already single stranded. In both cases, a probe is used (DNA probe can be used in northern)

Front 80

What kind of information can we obtain from a northern blot?

Back 80

– northern blotting: RNA detection
– steady-state level of a specific transcript in a certain RNA mixture : abundance of specific mRNA at certain time, under certain conditions

Front 81

What of information can we obtain from a Southern blot?

Back 81

Southern blotting: DNA (capital S) detection
– estimating the # and position of gene copies in the genome, restriction mapping of genomic fragments, detection of cloned sequences, detection of transgenes, detection of homologous sequences in different genomes , detection of repetitive sequences

can identify specific restriction fragments in a complex mixture of fragments)

Front 82

Compare the information obtained by northern analysis with the information obtained by microarray experiments.

Back 82

Northern analysis shows the steady state level of RNAs – one gene one experiment. Microarrays show how multiple genes interact together under different circumstances. Northern is useful when trying to isolate a certain mRNA, while microarray is more useful when trying to see what genes are expressed under what conditions.

Front 83

How do we treat a DNA gel prior to Southern blotting? Explain why.

Back 83

-gel treated with an alkaline solution (sodium hydroxide) to denature the double-stranded DNA: improves binding of negatively charged DNA to a positively charged membrane, separates it into single DNA strands for hybridization to the probe & destroys residual RNA

Front 84

Thinking question: you have made a short probe (50 nucleotides) from a certain genomic DNA (note: genomic DNA includes both exons and introns). Are you sure that you would be able to use this probe for northern hybridization? Explain your reasoning.

Back 84

Although DNA probes can be used in northern blotting because it can bind to RNA, not enough information is known about the source of the genomic DNA from which the probe was created from. If the probe was made from an intron part in the genomic DNA, then it would not be useful in northern blotting since RNA only contains exons.

Front 85

Distinguish between a template, primer and a probe.

Back 85

A template is a single DNA strand that serves as pattern for building a new second strand - can be the target DNA or RNA that a probe binds to. Probes (DNA or RNA) are labelled for easy identification and are used to hybridize to a target sequence and later isolated. Primers are short RNA/DNA sequences used to initiate the synthesis of DNA.

Front 86

Why is it important to know the exact start site of transcription?

Back 86

The start site of transcription gives important information about the location of the promoter, and thus giving an idea to how the promoter affects transcription.

Front 87

What does SDS-PAGE stand for? Explain the roles of SDS in SDS-PAGE (keep in mind - two major roles).

Back 87

Sodium dodecyl sulfate polyacrylamide gel electrophoresis

• SDS – negatively charged detergent; binds to hydrophobic protein regions & UNFOLDS protein
• Proteins bind SDS in constant ratio to their mass –gives them a negative charge proportional to their mass

Front 88

Compare the information you could obtain by SDS-PAGE with the information you could obtain by using Western blotting.

Back 88

SDS-PAGE - separates proteins according to their size

Western blotting - detects a specific protein out of those already sorted by SDS-PAGE

Front 89

General knowledge: distinguish between antibody and antibiotic.

Back 89

ntibody – protein used by the immune system to identify and neutralize foreign objects

Antibiotic – chemical that inhibits or abolishes the growth of bacteria

Front 90

How does salt concentration influence hybridization process between target DNA or mRNA and probe during nucleic acid hybridization step performed in northern (or Southern) blotting method?

Back 90

Salt concentration affects the stringency of the hybridization process: low [salt]/buffer concentration is high stringency

The stricter the conditions for binding, the better matched the probe and target must be in order to be seen.

Front 91

How does temperature influence hybridization process between target DNA or mRNA and probe during nucleic acid hybridization step performed in northern (or Southern) blotting method?

Back 91

High temperature = high stringency

Front 92

How does size of the probe affect the hybridization process between target DNA or mRNA and probe during nucleic acid hybridization step performed in northern (or Southern) blotting method?

Back 92

The length of the probe determines the melting temperature of DNA and so determines the ideal hybridization temperatures. Also, the longer the probe is the more likely that mismatches aren’t grouped together so hybridization will be stronger

Front 93

probe’s % of identity with the target sequence*

Back 93

% of sequence identity between target sequence and a probe determines what hybridization conditions we need, if probe and target are not very similar, cannot use high stringency conditions

Front 94

Distinguish between the terms “mutation”, “DNA repair” and “recombination”.

Back 94

Mutation - permanent change in DNA
DNA repair – occurs before & after replication: if fails, get mutations
Recombination – when there is no undamaged complementary strand (template) available recombination repairs DNA by filling a gap in one strand of duplex DNA by retrieving a homologous single strand from another duplex (switching alleles)

Front 95

List and briefly explain three major causes for mutation in DNA.

Back 95

Replication errors: mistakes made during replication not detected & corrected by DNAP I/III proofreading mechanisms

Spontaneous changes in DNA: depurinations (hydrolysis of a purine base (Adenine or Guanine) from the deoxyribose-phosphate backbone leaving an –OH), deaminations of C into U + A → hypoxanthine (which pairs with C, not T)

External factors:

radiation, mutagens, base analogues, spontaneous changes, change in temperature

Front 96

2Explain how errors in DNA replication can lead to mutations.

Back 96

When DNAPIII inserts an incorrect nucleotide that doesn’t base pair with the one on the template, a mutation has been introduced. If not corrected, further replication cycles will incorporate the error and cause changes in protein.

Front 97

Explain how radiation causes mutations.

Back 97

(ionizing=free radical formation

non-ionizing: photochemical fusion of two pyrimidines that occupy adjacent positions in same DNA strand
=pyrimidine dimers (UV)

(gamma+x)> DSB

Front 98

Explain how mutagens cause mutations.

Back 98

alkylation - e.g.nitrosamines adding methyls etc)
intercalating agents (ethidium bromide - makes polymerase skip)

reactive oxygen species -oxoG adduct” – guanine oxidized to 7,8-dihydro-8-oxoguanine
highly mutagenic – base pairs with A and C

Front 99

4.Distinguish between the effects of mutations on the somatic and germ cells of multicellular organism.

Back 99

In a germ cell or germ cell precursor - disease predisposition may be(come) hereditary

In a somatic cell - non-hereditary disease (e.g. cancer)

Front 100

List the various types of DNA repair mechanism (we have mentioned seven).

Back 100

1.DNAPIII repair
2.Direct reversal of damage
3.Base excision repair
4.Nucleotide excision repair
5.Mismatch repair
6.Recombination repair
7.nonhomologous end joining & SOS translesion repair)

Front 101

DNAPIII repair

Back 101

3’ to 5’ exonuclease activity to catch incorrect nucleotodies that DNAPIII put in itself.

Front 102

Mismatch repair

Back 102

protein MutS scans and recognizes mismatched nucleotides, MutL binds to form MutSL complex. It scans until a GATC sequence is found and MutH cuts out the strand that is not methylated, polymerase enters to fill in the gap and ligase to join the pieces back together.

Front 103

Direct reversal repair

Back 103

using photolyase to break the covalent bond that forms between thymine dimers

Front 104

Base excision repair

Back 104

using glycolyase to scan through damaged nucleotide and remove it from the backbone by flipping the base out and cutting it.

Front 105

5Nucleotide excision repair

Back 105

fixes distortions in the backbone instead of single bases

UvrA scans for distortion
UvrB binds to UvrA when distortion is found
UvrC makes a nick in one strand and removed damaged region.

Front 106

How does mismatch repair system know which mismatched nucleotide should be replaced?

Back 106

Dam methylase activity!

In bacteria, DNA strands are methylated by Dam methylase so immediately after replication, the DNA is hemimethylated. MutS, binds to mismatches in DNA and recruits MutL, which activates the endonuclease MutH. MutH binds hemimethylated GATC sites so only the unmethylated daughter strand is cleaved so it can be resynthesized.

Front 107

How are DSB breaks repaired through recombination repair?

Back 107

– taking the correct copy of the damaged area from the other homologous chromosome during alignment.

Front 108

What is involved in non-homologous end joining?

Back 108

Ku70/Ku80 dimer - recognizes broken ends; forms scaffold to hold ends together

DNA-dependent protein kinase (DNA-PK) – brings and phosphorylates protein Artemis

activated Artemis has exo- and endo- nuclease activities when activated; trims overhangs, cleaves hairpins

unknown DNA polymerase fills in

DNA ligase IV, XRCC4 join double-stranded ends

Front 109

Transleson (SOS) synthesis

Back 109

highly error prone because it introduces mutations but still allows replication to complete by inserting nucleotides to the damaged area independent of the template

Front 110

What are biological roles of DNA recombination?

Back 110

-DNA repair
-Creation of new gene/allele combinations (crossing over during meiosis)
-Formation of new genes (e.g., Immuno-globulin rearrangement)
-Integration of a specific DNA element (e.g. phage genome)

Front 111

List and briefly describe the ways genomic DNA can be rearranged (there are three of them).

Back 111

Through inversions, insertions and deletions (site specific recombination).

Inversions occur where the gene is flanked by repeated sequences and uses recombinase to take out the gene and invert it. This is useful because it can create different products. Ends of all known IS elements show inverted terminal repeats (ITRs).


Insertions and deletions are used for recycling genes.

Front 112

Give the detailed description of the base excision repair process in bacteria.

Back 112

• glycosylase recognizes, removes damaged base by hydrolyzing glycosidic bond
• additional step by AP (for aPu or aPy) endonuclease/phosphodiesterase: removes abasic sugar (sugar phosphate)
• replacement by DNA Pol, ligase
• DNA glycosylases are lesion-specific: e.g., for U, for oxoG

Front 113

Give the detailed description of the nucleotide excision repair process in bacteria.

Back 113

- removes bulky base damage including thymine dimers
- vrA2-uvrB complex moves along the DNA, looking for damage
-UvrA: detects distortion (rather than specific base change)
-UvrB: melts DNA - ss bubble around lesion
-recruits UvrC; UvrA dissociates (req ATP)
-UvrC: joins UvrB → UvrBC makes 2 incisions: 7 nts to 5’ side of damage, 3-4 nts to 3’ side (req ATP)
-UvrD unwinding helps release ss segment
-DNA Pol I excises damaged strand…

Front 114

Describe briefly the mechanism of direct reversal of damage in bacteria

Back 114

Direct reversal - repair of thymine dimers that prevent DNA polymerase from replicating the DNA strand beyond the site of dimer formation

• photolyase: enzyme coded by phr gene in E. coli (not present in humans, present in some Eu, including some animals)
• light-dependent activity breaks covalent bonds between thymines; chemical basis still unknown - photoreactivation

Front 115

Distinguish between the two DSB repair mechanisms we talked about in class.

Back 115

Nonhomologous end joining - two ends of broken DNA are directly ligated together - error prone repair

Recombination repair - Gap is repaired with “stolen” sequence

Front 116

Describe the function of proteins involved in homologous recombination of E. coli.

Back 116

1)RecA – responsible for strand invasion & pairing of homologous DNA

2)Rec BCD –nuclease/helicase - binds to duplex DNA generates single strands for invasion; requires ATP

3)RuvBD – causes branch migration (formation of crossover) & recognizes Holliday junctions

4)Ruv C – endonuclease - resolves holliday junctions.

Front 117

How can recombination (NOT necessarily homologous) occur between the bacterial chromosome and “shorter incoming DNA”?

Back 117

Transformation: cell can absorb and integrate fragments of DNA from their environment

Transduction: viruses transfer genes between prokaryotes.

Conjugation: one cell directly transfers genes (e.g., plasmid) to another cell.

Front 118

What is SOS repair mechanism, when is it used and why is it important?

Back 118

-If SOS repair fails in eukaryotes, then the cell commits suicide by apoptosis

- SOS genes activated after DNA damage by the accumulation of single stranded (ssDNA) regions generated at replication forks, where DNA polymerase is blocked.

Front 119

Describe the role of chi sequences in homologous recombination of E. coli.

Back 119

Chi ( c ) sequences: Crossover - hotspot – instigator
A concensus sequence 5’- GCTGGTGG - 3’
Role: enhances recombination frequency
When RecBCD reaches a chi sequence, degradation of 3’ end stops & degradation of 5’ end increases - created 3’ overhang for RecA to bind to & start recombination


-foreign DNA lacks chi sequences & is destroyed by RecBCD - protects cell

Front 120

List & describe 3 ways DNA moves around.

Back 120

1. General or homologous recombination: exchange between a pair of homologous DNA sequences
2. Site-specific recombination: between sequences with a limited stretch of similarity; involves specific sites (recombination sites)
3. Transposition: mobile DNA element moves from one site to another (donor and target site), usually little sequence similarity involved

Front 121

What are the key steps in single stranded model of homologous DNA recombination?

Back 121

1. Alignment of 2 homologous DNA molecules (RecBCD)
2. break in one strand of each homologous DNA (RecBCD)
3. Strand invasion: ss region from one parental strand pairs with complementary strand from the homologous DNA molecule, extension of 3' ends, connect in cross structure - Holliday junction (RecA), (Rec BCD)
4. Movement of Holliday junction - branch migration (RecBD)
5. Cleavage of Holliday junction - resolution (RecC)

Front 122

Explain the relationship between hybrid duplex and heteroduplex. (You can use diagram.)

Back 122

Heteroduplex is where the two strands of DNA came from different origins. Two homologous heteroduplex forms a hybrid duplex.

Front 123

1What are the roles of homologous recombination in eukaryotes?

Back 123

• required for proper chromosome pairing during meiosis (without it chromosomes fail to align)
• crossing over - gene reshuffling - variability

Front 124

When does the programmed creation of DSBs occur in eukaryotes? In which type of cells? Briefly describe the process.

Back 124

Programmed generation of DSBs occurs during meiosis at recombination hotspots

•Spo 11 (Topo II-like enzyme) generates ds breaks in one of the paired homologous chromosomes
•Mre11 (nuclease with 5’-3’ digestion) leaves 3’ overhangs
•Dmc1 and Rad1 are RecA type strand-exchange (strand invasion) proteins – nicked strands cross over to non-nicked strands
•DNA synthesis occurs, followed by DNA ligation to form double Holliday junction
•DNA strands cut at intersections forming chromosomes with or without crossovers

Front 125

Define gene conversion (use your own words). What is the significance of gene conversion?

Back 125

If the sequence used as a template for repair by homologous recombination is different from the damaged sequence one allele may be converted into another in nonreciprocal transfer of genetic information.
Significance: non-Mendelian inheritance patterns

Front 126

What is the role of mismatch repair mechanism in gene conversion?

Back 126

Mismatch repair system recognizes mispaired bases in heteroduplex (B/b’), excises and replaces one of the strands to restore complementarity
®Converts one allele into another

Front 127

Explain the role of site-specific recombination in infection of E. coli genome by lambda phage.

Back 127

Lambda DNA integrase inserts & deletes lambda DNA from the bacterial chromosome.

Lambda bacteriophage can multiply in E. coli by the lytic pathway OR insert its DNA into bacterial genome and enter a latent prophage state, and shift back to lytic growth later. The prophage’s DNA is replicated along with bacterial DNA.

Front 128

What are potential effects of transposons on the genome?

Back 128

Transposons can cause spontaneous mutations – in humans can cause Haemophilia A, primary breast cancer

•Insert into genes/coding sequences
•Insert into regulatory sequences (induce change in gene expression)
•can also form chromosomal rearrangements and relocate genes

Front 129

List and briefly describe three mechanisms by which genetic elements are able to move from one site in the genome to another.

Back 129

1)DNA movement by cut-and-paste or replicative pathways using transposase (DNA-ONLY TRANSPOSONS)

2)Moves via RNA intermediate produced by promoter in long terminal repeats using reverse transcriptase and integrase (RETROVIRAL-LIKE RETROTRANSPOSONS)

3)RNA intermediate produced from a neighboring promoter using reverse transcriptase and endonuclease (NONRETROVIAL RETROTRANSOPSONS)

Front 130

Who was Barbara McClintock and what was her major scientific contribution?

Back 130

Nobel prize winner - discovered transposition: mobile DNA element s in corn, at that time nobody believed in eukaryotic transposable elements.

Front 131

Retrotransposons

Back 131

(retroviral-like) carry inverted terminal repeat sequences for recombinase binding and two genes important for recombination

Cannot move from cell to cell; move only to new DNA within the cell

code for enzymes:
reverse transcriptase (RT), use RNA template to synthesize DNA (to be integrated).
and integrase (the transposase)

Front 132

Nonretroviral retrotransposons
or polyA retrotransposons

Back 132

“Target site primed reverse transcription”

LINEs (long-interspersed sequences)encode two ORFs, which are transcribed as a bicistronic mRNA composed of ORF1 (RNA binding protein, etc.) and ORF2 (endonuclease and reverse transcriptase activities)

Contain poly-A tails for priming reverse-transcription

Front 133

DNA-only transposons

Back 133

Example: insertion sequence (IS) elements


Encode only genes for mobilization and insertion (host replication machinery used for replication).

Ends of all known IS elements show inverted terminal repeats (ITRs).

Front 134

What are purple spots in colorless (yellow) corn kernels the result of?

Back 134

c/c = yellow kernels and C/- = purple kernels

If reversion of c to C occurs in a cell, cell will produce purple pigment (and a purple spot).

c” allele results from a non-autonomous transposon called “Ds” inserted into the “C” gene (Ds = dissassociation).
Autonomous transposon “Ac” controls “Ds” transposon (Ac = activator).

Front 135

Nonhomologous recombination

Back 135

transposable (mobile, transposon) elements insert into DNA that has no sequence homology with the transposon.

Front 136

What are the major differences between prokaryotic and eukaryotic transcription?

Back 136

RNAP in prokaryotes require the use of the sigma subunit and there is only one kind of polymerase. There are 3 in eukaryotic transcription and many TF are involved in the initiation complex.

Front 137

Which hypothesis regarding eukaryotic RNAPs was proven with -amanitin and actinomycin D (be specific)?

Back 137

There are more than one eukaryotic RNAP, discovered by seeing the behaviour of polymerase under different conditions and sensitivity to amanitin.

Front 138

Which genes are transcribed by RNAP I? RNAP II? RNAP III?

Back 138

RNAPI makes 28S, 18S, 5.8S rRNAS
RNAPII (The main one) – mRNA & snRNAs
RNAPIII - tRNA

Front 139

What does CTD stand for? Explain the role of CTD tail in eukaryotic gene expression?

Back 139

Carboxyl terminal domain, a series of 7 amino acid repeated multiple times that is responsible in initiating transcription, found in RNAPII. When its not phosphorylated, the CTD tail initiates transcription, so in areas of high transcript only phosphorylated forms are found.

Front 140

How would you define enhancers? What are their characteristics? What is the difference(s) between enhancer and upstream control element?

Back 140

Enhancers allow regulatory proteinst o bind which sitmulate transcription (or in the case of silencers, down-regulate transcription). They can be located very far away from the genes (up to 50kb away), and is orientation independent. Upstream control elements reside close to the promoter (such as the GC Box) and influences transcription rates.

Front 141

Explain the use of reporter genes for estimation of promoter strength.

Back 141

Reporter genes can be inserted following a promoter of interest to determine promoter strength. One can delete parts of the promoter and analysis how this deletion affects the gene expression of the reporter gene (which usually codes for an easily detected product).

Front 142

Explain briefly 5’ deletion series. What kind of information do they reveal?

Back 142

5’ deletion series is where you attach a reporter gene to a promoter of interest and delete parts of the promoter and see how this impacts the expression of the reporter gene. This gives you information about which part of the promoter is critical to gene expression and which isn’t.

Front 143

Explain the modular nature of RNAP II promoters.

Back 143

Modular nature of RNAP II promoters mean that different additions/substraction of regulatory elements such as enhancers, boxes, etc alter the promoter (can be mixed and matched).

Front 144

Explain the tissue (cell type) specificity of eukaryotic cis elements.

Back 144

All cells contain the same DNA and the same cis elements, but the absence/presence of certain TF which binds to certain cis elements change the cell type and tissue specificity. (controls which genes is expressed and which isn’t in the cell)

Front 145

If you know the binding site for certain transcription factor (TF), outline experiments you would use to purify this TF and to assay its activity.

Back 145

You can use DNA-affinity chromatography, in which the column contains the DNA binding sites for a certain TF. Put the TF through the DNA and you can isolate it and purify it through DNase footprinting. To assay its activity, you can use yeast 2-hybrid system. In which 2 plasmids are inserted inot a host cell that lacks the that TF. One plasmid contains the gene that codes for the TF, and the other contains a TF binding site and a reporter gene. In the host cell, the TF will be expressed, which will bind to the binding site and cuase expression of the reporter gene. Mutation or deletion of the gene or bidning site of the TF gives valuable information about its usage.

Front 146

Distinguish between the function of promoters and enhancers in transcriptional regulation.

Back 146

Promoters are areas where the initiatoinal complex and GTF can bind. Enhancers are cis elements that can be located very far from the promoter, and stimulate transcription (by making it stronger or weaker).

Front 147

Distinguish between the function of general transcription factors and transcription activators in transcriptional regulation.

Back 147

General Transcription factors (GTF) are involved in the initation complex of transcription, while transcription activators are those that bind to regulatory elements to stimualte transcription

Front 148

List and briefly explain four major domains in eukaryotic transcription factors.

Back 148

The first one is DNA binding domain, where zinc fingers are involved in interacting with the major/minor groove of the DNA to allow TF to bind. The second is transcription activiation domains which are independent of DBD and helps with the activiation of transcription. Next is ligand binding domain, involved in the C4 zinc finger which binds to a ligand and changes the conformation of the TF. Last is the dimerization domain, which allows for dimerization of different TF to create combinatorial control (different combinations of TF create different functions).

Front 149

List most frequent structural motifs in eukaryotic DNA binding domains.

Back 149

the zinc finger, homeodomain proteins, leucine zipper and HTH

Front 150

List three classes of transcription activation domains in eukaryotic transcription factors.

Back 150

Acidic activation domain, glutamine rich domain and proline rich domain.

Front 151

What is achieved by the ability of some transcription factors to form heterodimers (basically two things/players in regulation)?

Back 151

Greater complexity and different combinations which allow for different functions. Also can create flexibility in what DNA binding sites those TF can bind to.

Front 152

What is meant by the independence of the DNA-binding and transcription-activating domain of a transcription factor?

Back 152

The functions of DBD and AD is independent of each other which means they don’t’ affect each other.

Front 153

What is the role of the TATA box? What happens when TATA box is removed from the RNAP II promoter?

Back 153

TATA box is critical in the initiation of transcription because it allows for TBP to bind, which causes the rest of the initiation complex (and RNAPII) to bind. When the TATA box is removed, transcription could be conducted, unless through initiator and enhancer elements.

Front 154

What is combinatorial control of transcription?

Back 154

Different combinations of TF via the dimerization domain allows for different functions and sites to which the TF can bind to.

Front 155

What is the role of TFIIE and TFIIH in transcription initiation by RNAP II?

Back 155

TFIIE binds to the TFIIF/RNAPII complex to the already bound TFIIB element over the start site and generates the required energy. TFIIH has helicase ability (and 9 subunits) that allows for promoter clearance and phoshporlyation of CTD tail.

Front 156

What are the roles of TFIID in transcription initiation by RNAP II (be as specific as possible; have to talk about TBP and TAFs)?

Back 156

TFIID has a TBP subunit which binds to the TATA Box (first TF involved initiation) and contains TAF – it’s the foundation fo the complex and prevents nucleosome from forming. TAF is able to coactivate with enhancer bound elements and determine whether TFIID stays at the promoter

Front 157

What are the roles of TFIIB in transcription initiation by RNAP II?

Back 157

Bends DNA and orients RNAPII on it. The C terminal binds to the DNA & TBP, N terminal touches the start point.

Front 158

What are TATA-less promoters? How could transcription be initiated at TATA-less promoters?

Back 158

Transcription can still be initiated at TATA-less promoters via initiators or enhancers. In initiator elements, TAFII150 and 250 can bind to the initiator element and bring in the TFIID, once TFIID is in place, the rest can proceed. A similar method is for enchancer elements except sp1 does the job.

Front 159

Describe the role of histone acetylation/deacetylation in regulation of transcription.

Back 159

Transcription factors affect the acetylation/deacetylation of histones. When histones are acetylated, the DNA area is opened up and allows for proteins to interact and have access to the promoter. Deacetylation of histones is involved in gene expression.

Front 160

Describe the role of chromatin remodelling complexes in regulation of transcription.

Back 160

Chromatin structure must be rearranged so you have nucleosome free areas so RNAP can access the promoter

Front 161

Explain the role of enchancesomes and architectural proteins in regulation of transcription initiation?

Back 161

Enhanceosomes are formed when activators bind with enhancers. Architectural proteins change the shape of the DNA to allow more proteins to bind and contact the promoter. Both work together to stimulate transcription.

Front 162

Explain the role of insulators in regulation of transcription initiation.

Back 162

Insulators prevent the interaction between enhancers with the wrong promoters since enhancers can be located quite a distance away from the promoter its suppose to interact with.

Front 163

What is the role of methylation in regulation of transcription?

Back 163

Methylation PREVENTS transcription, genes must not be methylated if they’re to be expressed. Since TF can’t bind to promoters that are methylated.

Front 164

9. Describe the role of TBP during transcription (think about promoters for all three eukaryotic RNAPs and TATA-less RNAPII promoters – how do RNAPs bind to them; also, think about coordination of activities of all three polymerases).

Back 164

TBP is involved in transcription with all 3 polymerases. In RNAPI, TBP is associated with the SL1 complex. In RNAPII it’s involved in TFIID and in RNAPIII Its involved with TFIIIB. TATA binding protein helps position the RNAP in the right location and is essential.

Front 165

What is the role of Sp1 protein? What is the role of SL1 protein? What do they have in common?

Back 165

SL1 protein in involved in RNAPI’s initiation complex and contains the TBP subunit that binds to the TATA box. It ensures proper positioning of RNAPI

Front 166

Explain the mechanism of attenuation of the Trp operon. What is the importance of this mechanism for a bacterium?

Back 166

Attentuation is a way to regulate the expression of the structural genes in the Trp operon by ending transcription early if there is an abundance of Trp. When there is an abundance of Trp, the genes do not need to be transcribed, so when the leader sequence is transcribed, it is translated by the ribosome. There is a Trp-rich area in the peptide, but since there is plenty of Trp, translation continues and a stem loop forms between area 3 and 4 and a poly U sequence tha’ts transcribed signals the end of transcription before the RNAP can reach the structural genes.
When there is a lack of Trp, the ribosome stalls while trying to translate the leader sequence as it waits for tRNA that has Trp. The pausing of this creates a stem loop in regions 2 and 3, so that 3 and 4 can’t bp and termination doesn’t occur.

Front 167

Describe two distinctly different ways in which the trp operon is controlled by the overall availability of tryptophan.

Back 167

Trp operon can be regulated via negative regulation using trp as a co-repressor molecule, which binds to the repressor when there is an abundance, and the repressor binds to the promoter and halts transcription. It’s also involved in the regulation via attenuation.

Front 168

Describe rho-independent transcription termination

Back 168

Rho-independent termination involves the transcription of a poly U sequence and the formation of a step loop in the mRNA. The loop is formed by a GC rich area that base pairs with itself to form the loop, and the series of U creates a very weak bond with the DNA, which can easily be melted by the stalling of RNAP

Front 169

19. Which proteins are involved in mRNA cleavage and polyadenilation? Describe their order of association with pre-mRNA and their role (don’t forget the role of CTD tail).

Back 169

CTD tail in RNAPII recruits all the necessary enzymes for polyA. There is a concensus sequence located on the RNA, and a GU rich region downstream which is later cut off and PolyA occurs there. The enzyme CPSF binds to the sequence while CStF binds to the GU region. CFI & II cleaves the RNA at the GU region and CStF falls off. PAP (a polymerase) comes in an puts in a series of A at the 3’ end while PABI proteins bind to the tail.

Front 170

How is 5’ cap added to the nascent RNA?

Back 170

Via the enzyme mRNA guanyltransferase, which is recruited by the CTD tail.

Front 171

What is the relationship between hnRNA and mRNA?

Back 171

hnRNA is made up of pre-mRNA and snRNAs, together forming hnRNP. The hnRNA is covered with proteins during transcription.

Front 172

What are the general steps in processing of a pre-mRNA into a mRNA?

Back 172

The pre-mRNA has a 5’ cap put on it, a 3’ PolyA tail added to it, splicing to take out all the introns and transport to take it out of the nucleus before it’s translated.

Front 173

What is the role of snRNAs in the spliseosome?

Back 173

It assists splicing by associated with Sm proteins to form snRNPs. They bind to the pre-mRNA with other proteins to create the splicesosome and activate it later on.

Front 174

List different mechanisms of post-transcriptional control of gene expression

Back 174

1) Localizing mRNA into certain areas
2) RNA editing (U-deletion/insertion)
3) PTGS
4) Natural antisense RNA
5) Translational control switch
6) mRNA longetivity

Front 175

Explain the role of mRNA stability in control of gene expression.

Back 175

1) mRNA longevity – there are stabilizing elements in the 3’UTR of the mRNA that determines its longevity. Presence of such elements allow for faster degradation of mRNA

Front 176

Explain the role of the translational control switch in control of gene expression.

Back 176

2) Translational control switch (Fe control) – Two genes can be controlled by a single switch, through the presence of an IRE (introns response element), and IRE binding proteins which can encourage the translation of one gene and discourage the translation fo the other.

Front 177

Explain the role of PTGs in control of gene expression.

Back 177

DICER is a protein that can cleave dsDNA into siRNA or miRNA (those with hairpins). These RNA is then incorporated into RISC and will bind to the target mRNA and stop translation, induce degradation or cleave RNA.

Front 178

Explain the role of mRNA localization in control of gene expression.

Back 178

the mRNA can be delivered to different areas of the cell during development, which will affect which proteins are expressed where.

Front 179

Explain the role of RNA editing in control of gene expression.

Back 179

U-deletion/insertion – using small guide RNAs (gRNA), deleting or inserting a series of uridine into the sequence

Front 180

3. Explain the connection between pre m-RNA splicing and transport of mRNA from the nucleus

Back 180

In some viral genes, certain genes are transcribed into a mRNA, but not all are translated at the same time. Those that are spliced are transported out of the nucleus and translated into the rev protein, which re-enters the nucleus, binds to a site on the unspliced genes and transports them out. This allows unspliced genes to exit the nucleus and be packaged into viral proteins.

Front 181

4. What is trans-splicing? Give an example.

Back 181

Trans-splicing is splicing two different mRNA and taking the exons to form a new transcript.
In Trypanosoma, there are many leader sequences that are transcribed (mini exons) which are later trans-spliced together with a protein encoding exon contained in a polycistronic primary transcript.

Front 182

What does S in 16S stand for? What is the numerical value of this constant?

Back 182

S refers to sedimentation rate, the rate at which the protein sediments when centrifuged. 1S = 10-13 seconds.

Front 183

What are the roles of three major RNAs in protein synthesis?

Back 183

mRNA carries the codon to be translated, tRNA carries the amino acid to be incorporated, rRNA is involved in the assembly of the ribosomal complex

Front 184

What is the name of the region of tRNA molecule which attaches to an amino acid?

Back 184

3’ trailer end, the acceptor stem and contains CCA which amino acid binds to.

Front 185

How many different tRNAs are there in an eukaryotic cell? How many different aminoacyl tRNA synthetasis are there in an eukaryotic cell?

Back 185

There are more than 1 tRNA for each amino acid, but there are only 20 enzymes.

Front 186

What are the roles of tRNA in translation?

Back 186

tRNA has 2 roles, to carry the amino acid to the ribosome and to recognize the corresponding codon on the mRNA to its anticodon.

Front 187

What is the wobble position for an anticodon? For a codon?

Back 187

The wobble position for an anticodon (on the tRNA) is the first codon from the 5’ end, and for a codon (on the mRNA) is the 3rd codon from the 5’ end.

Front 188

Explain what does it mean when we say that the code is degenerate? Codons are synonymous?

Back 188

Degenerate means that each amino acid have multiple codons that code for it. Synonymous means that multiple codons can code for the same resulting amino acid.

Front 189

How are ribosomal subunits held together? What could be used to separate subunits?

Back 189

Ribosomal subunits are held together by divalent cations

EDTA can be used to break these

Front 190

1. What is the role of Shine-Dalgarno sequence?

Back 190

In prokaryotes, the shine-dalgarno sequence is responsible for initiation of translation. The 16S RNA in the 30S subunit of the ribosome recognizes the SD sequence, which lies slightly upstream of the start codon. It poisitions the ribosome in a correct position for translation.

Front 191

2. What is the role of Kozak sequence?

Back 191

In eukaryotes, it’s Kozak sequence is for the initiation of translation, where the 40S ribosomal subunit carrying the charged tRNA binds to in the 5’ end and scans until it finds an appropriate start codon.

Front 192

3. What are the major differences in initiation of translation between eukaryotes and prokaryotes (remember: first AA and the way mRNA and tRNA bind to ribosomes)?

Back 192

In prokaryotes, the first amino acid is a modified version of met, which is formyl-met. In eukaryotes, met is the start codon.

Front 193

Describe in detail events during initiation of the Eukaryotic translation.

Back 193

In prokaryotic initiation translation, IF3 binds to 30S to prevent the binding of 50S subunit, IF1 binds to prevent the tRNA from binding to the A site, IF2 carries GTP which is later hydrolyzed for energy. This 30S initiation complex looks for the SD sequence and binds to it. Then tRNA binds to the P site, and IF3 is removed so 50S subunit can bind. IF1 and 2 falls off and GTP is hydrolyzed to create the 70S initiation complex.

Front 194

Describe in detail events during initiation of the prokaryotic translation.

Back 194

In eukaryotic translation initiation, 40S subunit binds to eIF3, eIF4c, eIF1A to prevent the binding of the 60S subunit. A ternary complex consisting of the initiator tRNA, eIF2 + GTP binds ot the 40S. This gives you the 43S pre-initiation complex. A bunch of eIF4 factors prevent the formation of secondary structures in mRNA, and this binds to the 43S by the 5’ methyl cap and scans for the correct Met to start at, which is signalled by a Kozak sequence.
When Met is found, 1A, 3, & 4 are released and the 40S initiation complex is formed. eIF5 dispalces the other factors, hydrolyzes GTP and allows the 60S to bind, creating the 80S initiation complex.

Front 195

What are two cases of RNA having enzymatic capability?

Back 195

ribosomes ARE ribozymes

peptidyl transferase 23S rRNA

Front 196

What is a peptidyl transferase? What is catalyzed by peptidyl transferase?

Back 196

An enzyme that catalyzes the formation of peptide bond during chain elongation in translation. It takes the amino acid residing on the P site and connects it with the new one on the A site.

Front 197

7. How is gene expression controlled at the level of translation (mRNA is in contact with ribosomes or their subunits/translation factors; three things mentioned in the class)?

Back 197

First way of regulation is via phosphorylation of translation factors to decrease translation since P-ing the translation factor makes it so it can’t be recycled back. This occurs due to aa starvation, therefore accumulation of uncharged tRNA activating a protein kinase which P’s the eIF2.

Second way is by having multiple AUG codons, either by having a weak kozak sequence so the ribosome scans and skips to the next one, allowing different proteins to be created with a different N-terminus. The other is via AUG/UGA combination, a small ORF located upstream of the main ORF (uORF), it traps the scanning ribosome and cuases it to drop down from the mRNA

Third way of regulation is by the IRES (internal ribosome entry site), which can be located between 2 AUG codons to allow the ribosome to start scanning at that location. This makes this regulation independent of the 5’ methyl cap (since it is usually needed during initiation).

Front 198

Which two factors greatly influence efficiency of protein synthesis?

Back 198

Polysomes increase the efficency of protein synthesis by having multiple ribosomes translating a single mRNA at the same time.
Recycling rate of the PAB1 increases efficiency having it close to the start site all the time.

Front 199

Explain what happens with polypeptides after translation.

Back 199

Translation is terminated by having 3 proteins, RF1 and RF2 recognizes the stop codon, RF3 releases the completed chain and RRF mimics a charged tRNA but doesn’t have a 3’ amino acid binding site, disassembling the complex.

Front 200

Define cytoplasmic male sterility in plants.

Back 200

Cytoplastmic male sterility refers to the inability for certain plants to produce male gametes since genes outside of the nucleus does not separate according to Mendelian genetics.

Front 201

Explain endosymbiotic theory.

Back 201

A theory which suggests the soruce of DNA in mitochrondia & chloroplasts orginiate from prokaryotes, as the structure of the DNA is much more similar to prokaryotes (lack of histones, circular DNA). Perhaps there was a symbiosis between the prokaryotic and eukaryotic cell where through evolution the two were merged.

Front 202

Which genome encodes for proteins found in mitochondria? Explain.

Back 202

Most of the proteins are coded by chromosomal DNA in the nucleus, while the mtDNA codes mostly for all the RNA involved in protein synthesis.

Front 203

What are the characteristics of mitochondrial transcripts?

Back 203

Mitochondrial transcripts don’t’ have a promoter, the promoters are located in the D-loop (PH and PL). They don’t’ have a 3’ Poly A tail nor do they have a 5’ cap.

Front 204

How is normal tRNA : rRNA ratio maintained during the transcription of mitochondrial DNA?

Back 204

Transcription stops at the 3’ end of the 16S rRNA gene and goes back to start at the PH promoter again. This way you get a lot more rRNA in the transcript, which ensures a normal tRNA:rRNA ratio.

Front 205

Define and explain the importance of heteroplasmy and homeoplasmy.

Back 205

During mitosis, there is imperfect transmission of mitochrondial content between the two daughter cells. Heteroplasmy is where mutation in a mtDNA in a cell leads to a mixture of mutant/normal mtDNA molecules. Given enough time, you’ll get a purely normal cell and a purely mutant cell, which is homeoplasmy.

Front 206

Is mutation rate in mitochondrial DNA high or low? Explain why.

Back 206

The mutation rate is high in mtDNA due to the lack of protective histones and the high rate of replication, and no repair systems (thymine dimers can’t be fixed).