Unit 2 - Active Recall Flashcards

Question

What are the three eukaryotic DNA polymerases you need to recall?

Answer 1

DNA polym α: initation of nuclear DNA synthesis and DNA repair (has primase activity) DNA polym δ: lagging strand synthesis of nuclear DNA, DNA repair, and translesion DNA synthesis DNA polym ε: leading-strand synthesis

Answer 2

Controlled initiation: 1) An origin must be selected or “licensed” by replication licensing factors. 2) Origin is then activated and replication begins. 3) Once activated/replicated an origin is deactivated. Note: - Multiple origins are used during DNA synthesis - Origins of replication are activated in clusters (20-80 at a time). Each cluster is known as a replication unit.

Answer 3

- Eukaryotic DNA is packaged into chromatin. - Need to disassemble, produce more histones and reassemble nucleosomes.

Answer 4

PROBLEM Telomeres: - are the ends of linear chromosomes. - are made up of G-rich short repeated sequence. - stabilize chromosomes. Note: - Each round of replication leaves up to 200 bp DNA unreplicated at the 3' end. SOLUTION Telomerase - Specialized reverse transcriptase. - Extends the end of the parental DNA by RNA-templated DNA synthesis. - Responsible for the replication of the chromosome ends - Conclusion: Telomerase extends the DNA, filling in the gap due to the removal of the RNA primer

Answer 5

RNA is usually single-stranded but it can fold into complex secondary structures called hairpin-loops and stem-loops. Loops = bases don't match

Answer 6

1. Initiation does not require a primer: chain synthesis begins de novo 2. Ribonucleotides are added to the 3’ - OH group of the growing RNA chain. 3. DNA unwinds at the front of the transcription bubble and 4. rewinds Notes: - SELECTIVE synthesis of RNA. - Not all DNA is transcribed in a given cell. - Synthesis is complementary and antiparallel to the DNA template strand.

Answer 7

1) DNA template 2) RNA nucleotides (rNTP’s) 3) RNA polymerase and other proteins.

Answer 8

- RNA is synthesized from one of the two DNA strands. - Either DNA strand can be used as the template strand. ➢Template is always read in the 3’ -> 5’ direction. ➢RNA is synthesized in the 5’ -> 3’ direction.

Answer 9

- Region of DNA that codes for an RNA molecule and the sequences necessary for transcription. - Three critical regions: i. PROMOTER (upstream of start site, adjacent to gene) ii. RNA CODING REGION (downstream of start site) iii. TERMINATION SITE (downstream of start site) - Only the RNA coding region is transcribed.

Answer 10

The PROMOTER is a DNA sequence that is recognized and bound by the transcription apparatus (RNA polymerase plus other proteins). The promoter indicates the direction of transcription. Binding of the RNA polymerase to the promoter orients the enzyme towards the start site.

Answer 11

1.Initiation – assembly of transcription apparatus on the promoter and begins synthesis of RNA 2.Elongation – DNA is threaded through RNA polymerase, unwinds the DNA, adds new nucleotides to the 3’ end of the growing RNA strand 3.Termination – the recognition of the end of transcription

Answer 12

- Many bacteria have multiple types of sigma factors, which help in the recognition of multiple classes of promoters. - Without sigma, core enzyme initiates transcription randomly - Holoenzyme is the complete enzyme complex composed of the core RNA polymerase and the sigma factor.

Answer 13

- Promoters contain short stretch of DNA that are conserved among promoters of different genes. These are called consensus sequences. - Most common encountered sequences (or elements) are at -10 (Pribnow box) and -35 nucleotides upstream of the start site . - Binding of transcription apparatus to these sequence orients the RNA polymerase towards the start site. - 35 and -10 elements are not identical in all promoters. - Each is a variation on a theme, i.e. consensus sequence. - Variation affects the strength of the promoter. (strength = frequency of transcription) - recA is a strong promoter.

Answer 14

down mutations - base substitutions that make the sequence less similar to the consensus sequences reduce the rate of transcription up mutations - sequence becomes more similar to the consensus sequences

Answer 15

- Transcription ends after a terminator sequence is transcribed. - Two major types of terminators in bacteria. i. Rho-dependent (requires Rho protein) ii. Rho-independent (also called intrinsic terminator)

Answer 16

1) Rho binds to RNA upstream of the terminator. 2) RNA polymerase pauses when it reaches the terminator sequence and Rho catches up. 3) Rho unwinds DNA-RNA hybrid using helicase activity.

Answer 17

CORE PROMOTER - Extend upstream/downstream of transcription start site. - Minimal sequence required for accurate transcription initiation. - Includes a number of consensus sequences (common elements: TFIIB, TATA, Initiator and DCE) for transcription factor binding. REGULATORY PROMOTER - Located upstream of the core promoter, exact location can be variable. - Transcriptional activator proteins bind to consensus sequences and affect the rate of transcription.

Answer 18

1. TFIID binds to TATA box in core promoter 2. preassembled holoenzyme consisting of RNA polymerase, transcription factors, and mediators all bind to TFIID 3. transcription activator proteins bind to sequences in enhancers 4. DNA loops out, allowing proteins bound to enhancer to interact with basal transcription apparatus 5. activator proteins bind to sequences in regulatory promoter and interact with basal transcription apparatus through mediator

Answer 19

- Transcription does not end at a specific sequence. - Termination requires cleavage of the mRNA at a specific site. - A 5’->3’ exonuclease degrades the remaining mRNA terminating transcription.

Answer 20

1. Condenses chromatin to supress transcription, mRNA is not made. 2. The mRNA is destroyed before it is translated into protein. 3. Inhibit translation, protein is not made.

Answer 21

- The precursors of miRNAs, called primary miRNA (pri-miRNA), are encoded by the genome. The relevant genomic regions are transcribed by RNA Pol II. - In the nucleus, Pri-miRNA is cleaved by Drosha (RNAse III enzyme) into pre-miRNA, a stem-loop structure - In the cytosol, Dicer cleaves the pre-miRNA into 19–25 nucleotide miRNA:miRNA duplex with no stem-loop

Answer 22

siRNA - Double-stranded RNA that contains up to 100 nucleotides - Structure 21-23 nucleotide RNA duplex with 2 nucleotides 3‘ overhang - Fully complementary to mRNA - only one mRNA target - Endonucleolytic cleavage of mRNA miRNA - Precursor miRNA (pre-miRNA) that contains 70-100 nucleotides with interspersed mismatches and hairpin structure - 19-25 nucleotide RNA duplex usually with 2 nucleotides 3'overhang - Partially complementary to mRNA, typically targeting the 3' untranslated region of mRNA - Multiple mRNA targets (could be over 100 at the same time) - Translational repression.. Degradation of mRNA.. Endonucleolytic cleavage of mRNA (when there is a high level of complementary between miRNA and mRNA)

Answer 23

- Another role of RNAi is to direct localized repressive chromatin formation. - Small RNA (siRNA) duplexes are loaded into a nuclear form of RISC, called the RNA-induced transcriptional silencing (RITS) complex. - Similar to RISC, RITS are effector complexes that are targeted to homologous sequences by base-pairing interactions involving the guide strand of the small RNA. - RITS mediates gene silencing via heterochromatin formation.

Answer 24

- Heterochromatin is the condensed inactive form of chromatin. - Euchromatin is the open and active form of chromatin. - When the chromatin structure is open, transcription factors access the DNA and initiate transcription. - Histone modification occurs at multiple sites by acetylation, methylation and phosphorylation. Note: - siRNAs in RITS can base-pair with RNAs that are transcribed from their target DNA. RITS acts as a recruiter of modifying enzymes to specific DNA regions. - Enzymes, now catalyse the methylation of histone tails or cytosine bases in DNA

Answer 25

- Approximately, 30% of human genes thought to be regulated by RNAi; thousands of genes code for miRNAs. - Very important for regulating gene expression during embryo development. - Important research tool for ‘knocking out’ (silencing) particular genes. - Lead to major advances in the treatment of human diseases.

Answer 26

- Combination approach of targeted delivery of siRNA and chemo drugs to the cytoplasm of a cancer cell. - The siRNA-containing nanoparticle is taken into the cell by endocytosis. The nanoparticle unpacks and the siRNA is released.

Answer 27

- Continuous sequence of nucleotides encodes a continuous sequence of amino acids. - Number of nucleotides in the gene is proportional to the number on amino acids in the protein.

Answer 28

A gene includes DNA sequence that codes for all exons, introns, and those sequences at the beginning and end of the RNA that are not translated into a protein.

Answer 29

Prokaryotic protein-coding genes: - usually found in a contiguous array in the DNA called an Operon. Operon operates as a unit utilizing a single transcription start site for multiple genes. Eukaryotic protein-coding genes: - each gene is transcribed from its own start site to yield a pre-mRNA that is processed into a functional mRNA encoding a single protein. Note: In prokaryotic DNA genes contain little to no noncoding gaps (introns) and the DNA is transcribed directly into colinear mRNA, which then is translated into protein while the mRNA is still being produced.

Answer 30

Messenger RNAs (mRNAs), which encode the amino acid sequences of proteins, are modified after transcription. Modifications: 1) Capping of the 5’ end 2) Polyadenylation of the 3’ end 3) Splicing (removal) of introns

Answer 31

- A methylated guanine nucleotide is attached to the 5′-end of the pre-mRNA. - Methyl groups (CH3) are added More detail: - A methylated guanine (G) nucleotide is joined to the pre-mRNA by an unusual 5’-5’ linkage involving 3 phosphate group - Modification is necessary for efficient initiation of translation, transport of mRNA from nucleus, protects mRNA from degradation, and enhances RNA splicing.

Answer 32

- around 50 to 250 adenine (A) nucleotides are added to the 3′-end of the pre-mRNA. - Modification is necessary for efficient translation and protects mRNA from degradation.

Answer 33

- Splicing – removal of introns from pre-mRNA (ie: involves breaking and reforming of bonds) - Requires three consensus sequences in the pre-mRNA: 5’ splice site, 3’ splice site and branch point. - Consensus sequences are used by the SPLICEOSOME to recognize and remove introns. More detail: - Introns are removed in the form of a lariat and exons are spliced together by two successive reactions.

Answer 34

- Splicing – takes place on a SPLICEOSOME - A Ribonucleoprotein complex [300 proteins and 5 small nuclear RNAs (snRNAs)]. - Contains five snRNPs (small nuclear ribonuclear proteins)...: snRNA + protein = snRNP U1, U2, U4, U5 and U6 - snRNPs are central to the activity of the spliceosome. - The snRNAs of U1 and U2 base pair with the consensus sequence at the 5’ splice site and the branch point site of the pre-mRNA.

Answer 35

- Functional coupling of mRNA transcription and mRNA processing by RNA polymerase II (Pol II). - Coupling of events are mediated by the ‘tail’ or C- terminal repeat domain (CTD) of the largest subunit of Pol II. - mRNA processing enzymes are recruited to the CTD of Pol II during transcription.

Answer 36

- CAP is added as soon as the 5’ end of the pre-mRNA emerges from the polymerase. - Capping enzymes are recruited to the C-terminal domain (CTD) of RNA polymerase (Pol) II during the early stages of transcription. - RNAP II mediates the functional coupling of transcription to splicing by directing the nascent pre-mRNA into spliceosome assembly

Answer 37

- Assembly of the spliceosome occur co-transcriptionally, while the RNA polymerase (Pol II) is still actively transcribing the template. - Components of the spliceosome are recruited to the RNA while transcription is occurring. - The CTD of Pol II interacts directly with splicing proteins to recruit them to the RNA.

Answer 38

- Polyadenylation factors are recruited to the CTD of Pol II. - RNA is cleaved at the Poly(A) 3’ cleavage site. - Degradation of the remaining RNA by Rat1 terminates transcription.

Answer 39

A single pre-mRNA can be processed in different ways to produce different mRNA molecules thereby translating to different proteins...Ways to do this…. 1) Alternative splicing: pre-mRNA can be spliced in different ways. 2) Alternative PolyA site: PolyA tail can be added at different 3’ cleavage sites.

Answer 40

- Each mRNA produced has a different combination of exons. - Each mRNA when translated produces a different protein (isoforms of protein). quick overview? 1. Exon skipped 2. Intron retention 3. Alternative 5’ or 3’ splice site 4. Mutually exclusive exons

Answer 41

- Pre-mRNA contains multiple 3’ cleavage sites. - The 3’ cleavage site used determines the length of the mRNA transcript. - Each mRNA when translated produces similar proteins of different size.

Answer 42

- Many genetic disease arise from mutations that affect pre-mRNA splicing. i. Affect use of splice sites ii. Affect the splicing machinery iii. Affect regulators of alternative splicing - About 15% of all single nucleotide mutations (point mutations) that cause diseases alter pre-mRNA splicing. - Mutations that affect the splice sites may…. i. disrupt the consensus sequence so that the spliceosome can no longer recognize the 5’ or 3’ splice site. ii. create a new 5’ or 3’ splice site. iii. initiate usage of an existing cryptic 5’ or 3’ splice site. - Mutations can lead to exon skipping (complete or partial) or intron retention (complete or partial). This result in gene loss of function either by generating a nonfunctional protein or by altering the stability of the mRNA (mRNA is degraded and no protein is made).

Answer 43

Occasionally a gene is found with a sequence of nucleotides that does not exactly match that of its RNA product. Editing alters the coding information of the of mRNA transcripts. Ways to do this……. 1) Substitution Editing (base conversion) 2) Insertion Editing

Answer 44

Chemical alteration of individual nucleotides by specific enzymes. Example C is converted to U.

Answer 45

- Addition of U nucleotides occurs by cleavage of the mRNA, insertion of the U nucleotide, and ligation of the ends. - The reactions are catalyzed by a complex of enzymes under the direction of guide RNA (gRNA). - gRNAs base-pair as best they can with the mRNA to be edited and serve as a template for the addition of nucleotides.

Answer 46

- Eukaryotic RNA is processed before export to the cytoplasm and translation. - A modified nucleotide is added to 5’ end (capping) and a polyA tail is added to the 3’ end (polyadenylation) of the pre-mRNA. - Introns (noncoding sequences) are removed from pre-mRNA and exons (coding sequences) are spliced together by two transesterification reactions to generate the mature mRNA. Splicing takes place in the large ribonucleoprotein complex, the spliceosome. - Alternative processing of a pre-mRNA produces different types of mRNA, resulting in the production of different types of proteins from a single gene. Increases protein diversity. - RNA editing can also alter the protein product of a gene by changing the nucleotide sequence of the mRNA. Nucleotides are substituted, deleted or inserted.

Answer 47

Each amino acid consists of a central carbon atom (C α) attached to.. (1) an amino group (NH3+) (2) a carboxyl group (COO−) (3) a hydrogen atom (H) (4) a radical group, designated R.

Answer 48

- Three consecutive ribonucleotides (CODON) specifies one amino acid. - In DNA the 3 nucleotides are called as Triplet Code - In mRNA the 3 nucleotides are called as CODON Why it had to be a triplet code? - 20 amino acids - 4 nucleic acid bases - Number of possible codons = 4n (n = # nucleotides per codon) --> 1 base not enough, 2 bases not enough, 3 bases → 64 possible codons

Answer 49

61 sense codons; these code for amino acids. - Includes a start codon (AUG) - Initiates translation - Codes for the amino acid methionine (MET). 3 stop or nonsense codons - UAA, UAG and UGA - Terminates translation - Does not code for amino acids.

Answer 50

Methionine (Met) and Tryptophan (Trp)

Answer 51

Some amino acids are specified by more than one codon - degeneracy. - Codons that specify the same amino acids are synonymous codons. - The code is degenerate but NOT AMBIGUOUS!!! A codon NEVER specifies more than one amino acid.

Answer 52

1. Partial – changing the third base in a codon from a purine to a purine (G <-> A), or from a pyrimidine to a pyrimidine (C <-> U). 2. Complete – changing the third base in a codon to any of the four bases, e.g. GGU, GGC, GGA, or GGG, all code for glycine.

Answer 53

1. Isoaccepting tRNA: tRNAs bind (‘accept’) same amino acid, but recognize different codons (by using different anticodons). - 30 different tRNAs, but only 20 amino acids! - All tRNAs are used! So different tRNAs accept the same amino acid -> ISOACCEPTING. - Example: tRNAs with anticodon 3’ AGG 5’ and 3’ AGU 5’ both accept serine. 2. Wobble effect: allows same aa-tRNA to pair with more than one codon. - 61 sense codons, but only 30 anticodons (30 tRNAs). - All 61 codons are used! So the same anticodon base pairs with different codons. - Example: tRNA with anticodon 3’ AGG 5’ base pair with codons 5’ UCC 3’ and 5’ UCU 3’.

Answer 54

- Wobble occurs between the 1st (5′) base of an anticodon of t-RNA (Wobble Position) and the 3rd (3′) base of a codon of mRNA. - The 1st (5’) base of a anticodon is allowed to move slightly “wobble” from its normal position to form a non-Watson & Crick base pair with the codon. - Wobble at the 5' end of the anticodon allows variable nucleotide at the 3' end of the codon, therefore more than one codon for an amino acid.

Answer 55

- Inosine is an intermediate in the metabolism of purine. - Inosine is commonly found in tRNAs and is essential for proper translation of the genetic code in wobble base pairs.

Answer 56

- Reading frame refers to protein-coding region of the mRNA. - Specifies a single protein starting and ending at internal sites within the mRNA. - All RFs begin with AUG, therefore all proteins begin with Methionine (M).

Answer 57

1) Codons consisting of three nucleotides are read 5’ to 3’. 2) Codons are not overlapping, each base is part of only one codon. 3) No gaps, each base in the coding region of an mRNA is part of a codon. 4) Message is translated in a reading frame set by the initiator codon, AUG.

Answer 58

- An open reading frame is a portion of a RNA molecule that, when translated into amino acids, contains no stop codons. - Presence of stop codons in 2 of the above reading frames means they are not open reading frames (ORFs). A long open reading frame is likely part of a gene.

Answer 59

- Six possible reading frames in DNA sequence (3 top, 3 bottom) - The correct frame is the reading frame which goes from start to stop

Answer 60

Frameshift mutations – Insertion and Deletion Point mutations or base substitution - alter a single nucleotide

Answer 61

- Process of converting genetic information stored in nucleic acid sequences into proteins. - Sequences of mRNA are translated into unique sequence of amino acids in a polypeptide chain (linear order is preserved throughout!)

Answer 62

- Translation of an mRNA takes place on a RIBOSOME. - The mRNA is read in the 5’ to 3’ direction. - Amino acids are attached to the carboxyl (C) terminus of the growing polypeptide chain; polypeptide is made in the amino (N) to carboxyl (C) direction. Note: - In bacteria, multiple ribosomes can simultaneously translate the same mRNA molecule –polyribosome (or polysome)

Answer 63

1. mRNA template - Polycystronic, i.e. multiple ORFs) 2. tRNAs (transfer RNAs) - All tRNA's from all organisms have a similar structure & attachment site - Serves as a link between the genetic code in mRNA and the amino-acid sequence of a protein. - Sequence is always 5’-CCA-3’ 3. Amino acids 4. Ribosomes - RNA-protein complex: Ribonucleoprotein - Prokaryote version has 3 RNA molecules and 2 sub-units 5. Many accessory proteins 6. Energy provided by GTP hydrolysis

Answer 64

- Directs the synthesis of proteins. - Synthesis takes place in the cavity between subunits. - Small subunit holds mRNA - Growing polypeptide exits through the tunnel in the large subunit. - Adds 20 amino acids per second. - 3 tRNA binding sites A = Aminoacyl binding site P = Peptidyl binding site E = Exit site - tRNA binding site spans large and small subunits, so tRNA anticodon loop can make contact with mRNA.

Answer 65

- Small subunit holds mRNA so only 1 codon at a time can be read by a tRNA. - Amino acid at end of tRNA makes contact with catalytic region of the large subunit that joins amino acids together to form peptide bonds. - The arm of the tRNA is positioned in such a way as to allow polypeptide chain to exit through a tunnel in the back of the ribosome. - The growing polypeptide is always attached to a tRNA. Note: - Many antibiotics attack translational apparatus (e.g., binding to tRNA sites, blocking exit tunnel)

Answer 66

1. tRNA Charging 2. Initiation 3. Elongation 4. Termination and peptide release

Answer 67

- Attachment of an amino acid to the tRNA is referred to as tRNA charging. - ALL amino acids are attached to the adenine (A) nucleotide in the 3’ end acceptor stem. - R group = unique side chain specific to a particular amino acid. - Energy required for binding of aa to tRNA supplied by ATP tRNA+aa = aminoacyl-tRNA e.g.: aminoacyl-tRNAAla has alanine bound to tRNA....Could also write Ala-tRNAAla - Aminoacyl-tRNA synthetase correctly attaches an amino acid to the appropriate tRNA. - 20 different aminoacyl-tRNA synthetase. - Each aminoacyl-tRNA synthetase recognizes one amino acid and attaches it to the correct set of tRNA’s.

Answer 68

Key Steps: - Assembly of ribosomal subunits at the translation start site. - Base pairing of initiator-tRNA anticodon with start site codon in the mRNA. Requirements: 1) mRNA 2) Small and large ribosome subunit 3) Initiator-tRNA 4) Initiation factors (IFs) 5) Guanosine triphosphate (GTP)

Answer 69

- IF-3 binds to the small subunit, preventing the large subunit from binding. - Small ribosome subunit bind to the mRNA. - 16S rRNA of the small ribosome subunit complementary base pairs with the Shine-Dalgarno sequence. - Initiator-tRNA anticodon (3’ UAC 5’) base pairs with mRNA initiation/start codon (5’ AUG 3’). - Initiator tRNA is charged with fMet amino acid (N-formylmethionine). - IF-3 proteins keep the small and large ribosomes separated. - IF-1 and 2 facilitates initiator-tRNA binding to the correct site. This forms the 30S initiation complex. - Initiation factor proteins dissociate and the large ribosome subunits binds. This forms the 70S initiation complex.

Answer 70

The key steps in elongation are: 1) entry of aa-tRNA into the A site of the ribosome, 2) peptide bond formation, 3) ribosome translocation, and 4) Exit of tRNA from E-site of the ribosome Requirements - aa-tRNAs (or charged tRNAs) - 70S initiation complex - Elongation factors (EFs) - GTP

Answer 71

1) A-site (aminoacyl) - accepts the incoming aa-tRNA carrying the next amino acid to be added to the chain. 2) P-site (peptidyl) - holds the tRNA carrying the growing polypeptide. 3) E-site (exit) - discharged tRNA's leave the ribosome from this site.

Answer 72

STEP 1: aa-tRNA binds to A-site - Incoming aa-tRNA bind to the A site. - Elongation factors (EF) guides the incoming aa-tRNA to the correct site. - aa-tRNA anticodon base pairs with the mRNA codon. - Peptidyl transferase catalyses peptide bond formation. - Enzyme activity performed by rRNA in the large ribosome subunit (ribozyme). STEP 2: Peptide bond formation and the release of amino acid from P-site - Peptide bond formation between amino acid attached to P-site tRNA and amino acid attached to A-site tRNA. - Peptide bond formation releases the amino acid from the tRNA at the P-site. - Growing polypeptide now attached to A-site tRNA. - The ribosome translocates in the 5’ to 3’ direction on the mRNA. STEP 3: Translocation and exit of E-site tRNA, A-site now open to accept new aa-tRNA - Translocation facilitated by elongation factors. - P-site tRNA is now located at E-site, this tRNA exits. - A-site tRNA now located at the P site. - A-site is now empty and ready to accept another aa-tRNA.

Answer 73

- Protein synthesis terminates when the ribosome translocates to a stop codon. - NO aa-tRNA enters an A site that has a stop codon. - Release factor (RF) protein binds to the A site and triggers release of the polypeptide from the P-site tRNA.

Answer 74

Comparison: - Prokaryotic mRNA: Polycystronic, i.e. multiple ORFs - Eukaryotic mRNA: Monocystronic mRNA, i.e. single ORF Eukaryotic translation: - Composition of the eukaryotic ribosome complex is different. (60s + 40s = 80s) - No consensus sequence for ribosome binding. - 5’ CAP and 3’ poly(A) tail facilitate ribosome binding. - Initiation complex scans along mRNA until 1st AUG codon is encountered.

Answer 75

- is a group of bacterial structural genes - under the control of a single promoter - they are transcribed together - produce a single mRNA molecule that encodes different proteins. - regulates the expression of genes by controlling transcription PROG? - Promoter, regulator, operon and genes

Answer 76

- A regulator gene helps to control the expression of the structural genes of the operon by increasing or decreasing their transcription. - Although it affects operon function, the regulator gene is not considered part of the operon. - Regulatory Molecule = metabolite (precursor or product of metabolic pathway)

Answer 77

1. Negative control - in which a regulatory protein is a repressor, binding to DNA and inhibiting transcription. 2. Positive control - in which a regulatory protein is an activator, binding to DNA and stimulating transcription.

Answer 78

Inducible Operon - transcription is normally OFF (not taking place) - When Regulatory Molecule binds to Regulatory Protein = Transcription is Turned ON Repressible Operon - transcription is normally ON (taking place) - When Regulatory Molecule binds to Regulatory Protein = Transcription is Turned OFF

Answer 79

they are synthesized only when their substrate (V) is available. also.. - can use the Product (U) to provide negative feedback … turning off the genes involved in synthesis.

Answer 80

- The lac operon of E. coli is an example of a negative inducible operon. - The enzymes β-galactosidase, permease, and transacetylase are encoded by adjacent structural genes in the lac operon and have a common promoter (lacP). - β-Galactosidase - lacZ gene, permease - lacY gene, and transacetylase - lacA gene.

Answer 81

- Jacob and Monod worked out the structure and function of the lac operon by analyzing mutations that affected lactose metabolism. - Partial-diploid strains of E. coli were used. - Mutations on bacterial DNA and the plasmid showed some parts of lac operon are cis acting (able to control the expression of genes on the same piece of DNA), whereas other parts are trans acting (able to control the expression of genes on other DNA molecules) MAIN MUTATIONS: 1. Structural-gene mutations 2. Regulator-gene mutations 3. Operator mutations 4. Promoter mutations

Answer 82

- Mutations on lacZ or lacY structural genes altered the amino acids and affected the structure of the proteins. - Mutations of the lacZ and lacY genes were independent and usually affected only the product of the gene in which the mutation occurred. - β-galactosidase and permease were produced normally in the presence of lactose Note: - The genotype of a partial diploid is written by separating the genes on each DNA molecule with a slash

Answer 83

- The partial diploid lacI+ lacZ−/ lacI− lacZ+ produces β-galactosidase only in the presence of lactose because the lacI gene is trans dominant. - The partial diploid lacIs IacZ+/ lacI+ lacZ+ fails to produce β-galactosidase in the presence and absence of lactose because the lacIs gene encodes a superrepressor.

Answer 84

Most mutations in the operator, the binding site for repressor, lead to lower affinity for the repressor and hence less binding. Thus these mutations allow continued transcription (and thus expression) of the lac operon even in the absence of inducer; this is referred to constitutive expression.

Answer 85

A plasmid is a small DNA molecule within a cell that is physically separated from a chromosomal DNA and can replicate independently.

Answer 86

- Designated lacP−, interfere with the binding of RNA polymerase to the promoter. - E. coli strains with lacP− mutations don’t produce lac proteins either in the presence or in the absence of lactose. - The partial diploid lacI+lacP+lacZ+/ lacI+lacP−lacZ+ exhibits normal synthesis of β-galactosidase - lacI+lacP−lacZ+/ lacI+lacP+lacZ− fails to produce β-galactosidase whether or not lactose is present.

Answer 87

- E. coli metabolize glucose preferentially, even in the presence of lactose and other sugars. They do so because glucose enters glycolysis without further modification and therefore requires less energy to metabolize than do other sugars. - When glucose is available, genes that participate in the metabolism of other sugars are turned off through a process known as catabolite repression. Efficient transcription of the lac operon, for example, takes place only if lactose is present and glucose is absent. Note: - lac Operon exhibits Catabolite Repression by High Glucose Positive (by CAP) Inducible (by cAMP) Control

Answer 88

1. Alteration of DNA or chromatin structure 2. Transcriptional control 3. RNA processing & degredation 4. translational control 5. Post-translational modification

Answer 89

i. Chromatin-remodeling complexes - bind directly to particular sites on DNA and reposition the nucleosomes, allowing other transcription factors and RNA polymerase to bind to promoters and initiate transcription ii. Methylation of histones and DNA / Acetylation of histones - Addition of methyl groups (CH3) to the tails of histone proteins and/or DNA brings about either the activation or the repression of transcription. - and the Addition of acetyl groups (CH3CO) to histones usually stimulates transcription.

Answer 90

i. Transcriptional factors and regulator proteins - Transcriptional activator proteins stimulate and stabilize the basal transcription apparatus. - They interact directly or indirectly through coactivator proteins ii. Enhancers and Insulators - Enhancers are regulatory elements that affect the transcription of distant genes. - They can stimulate any promoter in their vicinity . - Insulators block the effect of enhancers in a position dependent manner.

Answer 91

i. Gene regulation through RNA splicing and multiple 3’ cleavage sites: - Alternative splicing allows pre-mRNA to be spliced in multiple ways, generating different proteins in different tissues or at different times in development. - Multiple 3’ cleavage sites use different cleavage sites to produce mRNA’s of different length. ii. RNA Interference (siRNA, miRNA and methylation) - RNA silencing leads to the degradation of mRNA or to the inhibition of translation or transcription. - Small interfering RNAs (siRNAs) degrade mRNA by cleavage. - MicroRNAs (miRNAs) lead to the inhibition of translation and/or mRNA degradation. - Some siRNAs bring about methylation of histone proteins or DNA, inhibiting transcription.

Answer 92

i. miRNA regulation of translation - miRNAs inhibit the translation of complementary mRNAs. - Researchers suggest that miRNA can inhibit the initiation step of translation as well as steps after initiation, such as ribosome stalling or premature termination. ii. Availability of components for translation - Ribosomes, charged tRNAs, initiation factors, and elongation factors are all required for the translation of mRNA molecules. - The availability of these components affects the rate of translation and therefore influences gene expression. Note for ii: - Virus exposure causes an increase in availability of initiation factors for translation - Rate of translation and protein synthesis is increased

Answer 93

Protein modification and degradation - Many eukaryotic proteins are extensively modified after translation by the selective cleavage and trimming of amino acids from the ends, by acetylation, or by the addition of phosphate groups, carboxyl groups, methyl groups, carbohydrates, or ubiquitin (a small protein). - Control of these modifications, which affect the transport, function, stability, and activity of the proteins.

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