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51 Cards in this Set
- Front
- Back
- 3rd side (hint)
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working definition of expression
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series of steps that start with DNA resulting in a protein
what percentage of DNA codes for protein? |
1-2% !
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structure of a nucleotide
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base + ribose + phosphate
what types of bases are in nucleic acids? specify which bases. |
pyrimidines: uracil, thymine, cytosine
purines: adenine, guanine |
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how are ribose carbons named? what is attached where in nucleic acids?
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eferred to by prime numbers: 1’, 2’, 3’, 4’, 5’. The purine or pyrimidine base is linked at the 1’ position; a hydroxyl group is at the 3’ position for both deoxyribonucleotides (in DNA) and ribonucleotides (in RNA) and also at the 2’ position for ribonucleotides; the phosphate group(s) is linked at the 5’ position.
where on the BASE does the ribose attach? |
attached at N-9 of purines, N-1 of pyrimidines
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in what direction are nucleotides linked?
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Nucleotides in a nucleic acid chain are linked from a 5’ phosphate to a 3’ hydroxyl, in a phosphodiester bond.
In what order are they read? |
Sequence of a nucleotide chain is, by convention, read in the 3'-5’ direction.
A naturally synthesized chain will always start with a 5’ phosphate and end with a 3’ hydroxyl. |
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Describe structure of DNA
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1. phosphate-sugar invariant; bases variable
2. negatively charged phosphate backbone on outside: hydrophobic bases on inside 3. Strands run “antiparallel” to each other in a double helix. 4. The two strands of double helix DNA are stabilized by hydrogen bonding between G/C base pairs and A/T base pairs, by base stacking, as well as by hydrophobic interactions. Single-stranded RNA can also form intramolecular base pairs. What part is recognized by DNA binding proteins? |
Available H groups of a double-stranded DNA are used for binding by “DNA-binding proteins.”
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Describe nucleic acid melting temperature and factors that affect it
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Double stranded DNA or intramolecular double-stranded RNA structures have an intrinsic melting temperature, at which the two strands separate. The higher the G-C content, the higher the melting temperature. The melting temperature is also a function of ionic strength and pH.
What determines annealing capability? |
Strings of “complementary” (i.e., G/C and A/T or A/U) base pairs can find themselves in solution. Thus “melted” DNA can “reanneal” to the same double-stranded state. The ability of single-stranded DNA to pair with its complementary sequence in solution is the basis of nucleic acid probe technology.
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How is DNA packaged?
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In nucleosomes (DNA + histone proteins) and with the help of polycationic amines (e.g. spermine)
How much DNA wound around each histone core? What is protein-bound DNA called? |
~ 150 base-pairs (bp) of DNA are wound around each histone core. An additional histone monomer, H1, binds to constrain the DNA to the core. Spacer DNA of 20-80 bps links the DNA wound around consecutive histone cores. The DNA-protein unit is called a “nucleosome,” and DNA in its protein-bound form is called “chromatin.”
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What is DNA synthesized from? By what and in what direction?
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Only one strand serves as template. The nucleotide substrates are dATP , dCTP , dGTP , and dUTP . Again, template is read in 3’-to-5’ direction; synthesis is in 5’-to-3’ direction.
Which strand is which? |
Since both the RNA transcription product and the DNA non-template strand are complementary to the DNA template strand, the DNA non-template strand and the RNA have the same
sequence as (except that DNA contains T’s where RNA contains U’s). Hence, the DNA non- template strand is also called the “coding strand.” |
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How to melt DNA?
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High pH or high Temp.
How to melt nucleosomes? |
High [salt]
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Describe hierarchy of chromatin structure
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DNA < beads on a string , 30nm chromatin fiber of packed nucleosome < 50kb loops < 700nm chromosome arm < 1400nm width chromosome
What is condensed chromatin called? What about more permissive chromatin? |
Dense: heterochromatin
Permissive: euchromatin |
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How is chromatin structure modified?
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1. Modify histones: Histones are basic (+ charged), binds negatively charged DNA strongly.
ACETYLATION - addition of a negatively charged acetyl group on basic amino acid residues to introduce electrostatic repulsion - e.g. Lysine, K+ ---HAT---> K-acetyl (neutral) --less positively charged histones binds DNA less strongly 2. Chromatin remodeling complexes 3. Aliphatic cations - highly positively charged ---e.g., spermine, a polycationic amine ---allows phosphate backbone to be packaged What part of the histone is modified? |
Histone tail
"tails in the wind" |
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What are common histone modifications?
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K9 Methylation
K4, K9 Acetylation S10 Phosphorylation, K14 Acetylation What does each mean? |
K9Meth: heterochromatin formation, gene silencing
K4,K9Ac: gene expression S10Phos, K14Ac: gene expression |
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What defines a transcriptional unit?
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Point where initiation of txn takes place is defined as the “+1” site. RNA polymerase synthesizes a transcription product, termed a “pre-mRNA,” which terminates well downstream at a “terminator” site.
How is the pre-mRNA processed? What is in an mRNA? |
The pre-mRNA is subsequently processed in several ways to arrive at the final mRNA product. Splicing of pre-mRNA results in loss of introns (intervening sequences) and retention of exons (expressed sequences). A sequence within the mRNA is termed the “coding sequence,” which contains the nucleotide triplet codons that the ribosome will read to make a polypeptide chain. The mRNA also contains untranslated 5’ and 3’ regions (5’ UTR and 3’ UTR), a 5' Cap and a 3' poly-A Tail
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How many RNA polymerases are there in human cells?
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Three multisubunit protein complexes in human cells:
RNA pol I transcribes mostly rRNA genes. RNA pol II transcribes protein-coding genes. RNA pol III transcribes mostly tRNA genes. What is the initial codon for translation? The stop codon? |
Start = AUG (Met)
Stop = UAA |
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What defines a +1 site for transcription?
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1. Core promoter elements
2. General transcription factors Define each. |
The core promoter is a DNA sequence located in the immediate vicinity of the +1 site, at which RNA pol II will bind, directed by numerous other protein factors. Some promoters contain “consensus” sequences, i.e., similar sequences are found in the promoter regions of many genes. One example of the TATA box will be cited.
Protein complexes called TFIIs (transcription factor for RNA polymerase II) bind DNA and each other to direct proper localization of RNA pol II for initiation of txn at +1. For example, upstream of the +1 site of some genes a subunit of TFIID, named TBP (TATA-box binding protein) binds to and bends the DNA, which is the first step in txn initiation. Another large protein complex called “Mediator” is also involved in txn initiation. The complete ensemble of protein complexes at the txn initiation site is called the “pre-initiation complex” (PIC). |
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What categories of elements are involved in transcription?
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Cis and Trans elements
In addition to the core promoter, a number of DNA “boxes” function as binding sites for proteins that will work to increase the rate of txn initiation. Consensus sequences on the DNA itself are called “cis elements” whereas proteins that bind such sequences or each other are called “trans elements.” cis elements that are bound by “constitutive” (= unregulated) transcription factors are generally located close to the core promoter and are called “promoter proximal elements.” cis elements that are bound by regulated transcription factors are called “enhancers,” and these elements can be located quite far from the promoter. What does a txn factor consist of? |
A txn factor consists of several protein domains, including a DNA-binding domain that recognizes a specific DNA sequence (often direct or inverted repeats), an activating domain that interacts with other proteins to enhance the txn initiation rate, and a dimerization domain that allows formation of homo- or hetero-dimers (or tetramers) of txn factors.
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What determines the activity of a transcription factor?
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1. txn factor properties: cellular concentration, location, post-tln modifications
2. chromatin, histone modifications: addition/removal of acetyl, phosphoryl, mthy groups; or substitution of core histones with alternative histone proteins that influence the accessibility of target DNA sequences for txn initiation. 3. DNA Methylation: Methylation of C residues in CpG dinucleotides can serve as a recognition site for binding of proteins that, in turn, recruit enzymes that alter the properties of local chromatin. One example is MeCP2, a protein that binds to 5-methyl cytosine and recruits a histone deacetylase. 4. Repressor binding: The activity of a txn factor may be regulated by an oppositely-acting repressor protein that binds to the txn factor directly or binds to cis sequences and interferes with the txn factor binding or PIC formation directly. |
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Steps of transcription
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Txn
1. Formation of pre-initiation complex (PIC) 2. Initiation TFIIH (serine protein kinase phosphorylates serine CTD) Helicase unwinds DNA template strand Topoisomerase relieves tension 3. RNA polymerase doesn't need a ds-primer: cant start strand de-novo. First nucleotide is a tri-phosophate nucleoside. 4. Termination: CTD dephosphorylated |
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Types of DNA elements, their Location, What proteins they bind
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1. Core promoter: +1 to -30 : TFIIs
2. Promoted proximal: -40 to -200: Txn factors "constitutive' 3. Enhancers : "anywhere" down or upstream : Txn factors, regulated (Txn activating proteins) What are coactivators? Difference between them and TFs? |
Coactivators are protein complexes that do not interact with DNA directly, as do txn factors. Coactivators bind to the txn factor activation domain and recruit either the Mediator or chromatin remodeling complexes. It is now believed that control of virtually all eukaryotic transcription is in some part via the protein complex called “Mediator.” Input from transcription factors bound at enhancers signals to the mediator complex, which, in turn, recruits general transcription factors and stimulates transcription initiation.
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What are transcriptional repressors? Example of repression?
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Transcriptional repressors have similar domain structures to transcriptional activators in that they have a DNA binding domain and a repressor domain that interacts with a corepressor,
Methylation of cytosine in a CpG dinucleotide generally acts to repress transcription. The mechanism is by recruitment of HDAC activity, as shown at right. What is a corepressor? Example? |
Recruitment of the corepressor has the opposite effect of a coactivator, making the DNA inaccessible to binding of factors required for transcription initiation.
One type of corepressor is shown at right, a histone deacetylase (HDAC), which does the reverse reaction of a histone acetylase. |
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What is epigenetics?
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Epigenetics: somatically heritable changes in gene expression that do not involve changes in nucleotide sequence.
Examples? |
Some epigenetic processes: - DNA methylation - histone modifications - binding of variant histones - nucleosome remodeling
- non-coding RNAs |
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What is PEPCK, what does it do?
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phosphoenolpyruvate carboxykinase, an enzyme that is important in making glucose (= gluconeogenesis). The activity of PEPCK is strictly proportional to its concentration in the cell, which is believed to be controlled solely by the rate of txn.
How is PEPCK txn activated? |
PKA --> nucleus -(+P)-> CREB
PKA-(+P)-> SIK1 -(+P)->TORC; TORC --> nucleus -binds-> CREB CREB --> CRE --> ^PEPCK txn CREB --> CBP-HAT --> ^PEPCK txn PKA also enters the nucleus and phosphorylates CREB, a txn factor that is bound in its inactive form to a DNA element called the CRE (cyclic AMP response element) at -80 of the PEPCK gene. Phosphorylated CREB recruits a histone acetylase (part of CBP) that makes the promoter region more accessible, and PEPCK txn levels rise, resulting in a higher cellular concentration of PEPCK and more gluconeogenesis. Transcription of the PEPCK gene also requires a second coactivator named TORC (transducer of regulated CREB activity). TORC is found in the cytoplasm and can only enter the nucleus when it is unphosphorylated. Activation of PKA results in phosphorylation and inactivation – in the cytoplasm – of a kinase (SIK1) that normally phosphorylates TORC, preventing it from entering the nucleus. Phosphorylation of SIK1 by PKA inhibits its activity, resulting in unphosphorylated TORC that enters the nucleus, binds to CREB, and acts as a coactivator for PEPCK txn. |
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What are the body's responses to short-term low blood glucose?
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Glucagon --> PKA --> PFK+(P), PEPCK txn ^
1) Enzyme phosphorylation by PKA 2) Txn factor phosphorylation by PKA Explain each. |
1) pancreatic α cells--> glucagon,
glucagon --> PKA PKA -(+P)-> PFK (gluconeogen. enzyme) 2) PKA --> nucleus -(+P)-> CREB PKA-(+P)-> SIK1 -(+P)->TORC; TORC --> nucleus -binds-> CREB CREB --> CRE --> ^PEPCK txn CREB --> CBP-HAT --> ^PEPCK txn |
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What are the body's responses to chronic low blood glucose?
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1) Txn activation of PEPCK by glucocorticoids
Glucocorticoid binds GR (receptor and txn factor); GR --> nucleus; GR -binds-> PGC-1; PGC-1 a) binds HAT protein b) interacts with Mediator to promote PIC formation at PEPCK promoter Connection between glucagon and glucocorticoid pathways? |
expression of PGC-1 itself is dependent on binding of phosphorylated CREB upstream of the PGC-1 gene
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What are the body's responses to high blood glucose?
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Insulin --> CBP(+P) , TORC (+P)
phosphorylation of CBP such that it is no longer able to form a CEBP-CBP complex activation of a kinase that phosphorylates TORC, such that it can no longer enter the nucleus |
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What steps are involved in post-transcriptional processing?
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1. 5' capping - 7-methyl guanosine
2. Splicing 3. Poly A Tail What is the function of each? |
1. Only mRNAs (and not other cellular RNAs) are capped, and the cap serves as a recognition site for binding of proteins that recruit a ribosome. In addition, the cap functions to protect pre- mRNA and mRNA from degradation by exonucleases that chew in the 5’-to-3’ direction. Uncapping is one of the first steps in mRNA decay, which is now being appreciated as an important step in regulation of gene expression (and is studied in the Bechhofer lab).
2. Multiple mRNA products from the same gene; exon shuffling; combinatorial expression 3. The poly(A) tail is bound by PABP, poly(A) binding protein. This protects the 3’ end from rapid degradation by 3’-to-5’ exonucleases and, as we will see, enhances translation via an interaction between the 3’ end and the 5’ end of the message |
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Describe the 5' cap
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The “cap” at the 5’ end of an mRNA consists of a guanosine nucleotide that has been modified to contain a methyl group at the 7 position of the purine ring structure. This is linked to the 5’ end of the transcript in an unusual 5’-to-5’ phosphodiester bond.
Draw it. |
The cap structure consists of a GTP molecule attached in an unusual 5’-to-5’ linkage to the 5’- terminal nucleotide of the pre-mRNA. The purine ring of this GTP is then methylated at the 7 position, giving 7- methylguanosine (abbreviated m7G). The first and second 2’-OH groups are also often methylated. The cap structure is required for translation initiation.
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Describe steps of splicing
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- contents of exons
The first exon starts with the +1 nucleotide and includes the start codon. Upstream of the start codon is the 5’ UTR. The coding sequence spreads out over multiple exons; splicing must be absolutely precise in order not to disrupt the coding sequence reading frame. The stop codon at the end of the coding sequence is followed by the 3’ UTR, which may be spread over several exons. The final exon contains the polyadenylation signal (see below). Exons are generally shorter than introns; the average exon is 150 nts, and the average intron is 3,000 nts. Splicing is catalyzed by the spliceosome, a large complex of nearly 200 proteins and 5 snRNPs (small, nucleolar ribonucleoprotein particles). Splicing itself consists of two transesterification reactions, i.e., a reaction in which one phosphodiester bond is replaced with another. We will present the splicing reaction in lecture, only to highlight certain ideas (but not that you need to know the reaction pathway). How does the splicesome know where to start and stop splicing? |
- conserved signals
Almost universally, introns start with a GU dinucleotide and end with an AG dinucleotide. These dinucleotides are recognized in the context of loose consensus sequences that include exon and intron sequences. Introns contain a “branch point” near the 3’ splice site, which is an A residue in the context of a loose consensus sequence, as well as a run of pyrimidines also near the 3’ splice site. - exon definition Exons contain ESE sequences, which are exonic splicing enhancer sequences. These RNA boxes are bound by a class of proteins known as “SR proteins” because they are rich in serine and arginine. As such, the correct 5’ GU and 3’ AG dinucleotides can be recognized specifically by the splicing machinery – in the background of a large intron that has many of these dinucleotides – on the basis of their proximity to exons in which SR proteins are bound. This is the constitutive function of SR proteins. They also have a regulatory function; |
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What allows docking of post-transcriptional machinery to the ribosome?
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Phosphorylation of the CTD.
What factors are bound? |
Polyadenylation factors, splicing factors, capping factors
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What types of splicing errors can occur?
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3' splice site mutations, exonic substitutions, 5' splice site mutations, deep intronic mutations
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Describe steps of polyadenylation
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- poly(A) signal
At end of the 3’ UTR sequence, a polyadenylation signal (most often AAUAAA) signals for endonucleolytic cleavage 10-30 nts downstream. The free 3’-hydroxyl is acted upon by poly(A) polymerase, a specialized RNA polymerase that adds 100-200 A residues in a template- independent fashion . - poly(A) binding protein The poly(A) tail is bound by PABP, poly(A) binding protein. This protects the 3’ end from rapid degradation by 3’-to-5’ exonucleases and, as we will see, enhances translation via an interaction between the 3’ end and the 5’ end of the message |
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Northern blotting: description and how to do it
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- detection of a specific mRNA
1) order radioactive oligo probe complementary to 20-30 nt of desired mRNA 2) get rid of nucleus, use polyA tail to purify mRNA on poly-dT column 3) gel electrophoresis, blot, wash extra probe, expose to X-ray film to get band pattern What information can you get from Northern blotting? |
Since the amount of probe that will bind is a function of the number of target molecules, a Northern blot can give quantitative information about the level of expression. Other information includes tissue specificity, regulated expression, alternative splicing, deletion/insertion mutations, defective splicing.
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Micro-array: description and how to do it
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To compare transcriptomes of two cell types (e.g., normal and cancer), poly(A)+ mRNA is transcribed by reverse transcriptase to give a set of “cDNAs” (complementary DNAs) that is synthesized in the presence of nucleotides that are tagged with a different color fluorescent dye for each of the cell types. The cDNA pool is washed over the glass slide, and cDNAs stick to their complementary oligos imprinted at set positions on the slide. The color of each dot is determined by the amount of differentially labeled cDNA that binds there.
How different from Northern blotting? |
Use of a unique probe in a Northern blot experiment yields information on a single mRNA. Use of a microarray allows probing for all mRNAs expressed in a cell at one time. The concept is the same as for Northern blotting, except here the “probe” consists of thousands of different oligos spotted onto a glass slide, each dot representing coding (non-template) DNA sequence of a specific gene.
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Experimental use of cDNA
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Reporter Constructs
A cDNA copy of a specific mRNA can be used to construct an in-frame fusion between a protein coding sequence and a “reporter gene,” which is a protein moiety that fluoresces a certain color (e.g., green fluorescent protein or GFP, and many other colors). Level of expression, location in a cell, differential expression in tissues can all be followed by assaying not the protein of interest (which may be difficult) but the level of GFP, which is simple. Clinical use of cDNA? |
expression of recombinant protein
A cDNA copy of a specific mRNA can be hooked up to expression signals (promoter, enhancers) such that the mRNA is transcribed at a high level, leading to useful yields of the protein product. Human proteins that are of therapeutic value but are in low abundance naturally can be expressed in a heterologous system, from which it can be easily purified. |
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What are the components of translational machinery?
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1) tRNA
2) ribosome 3) protein factors 4) aminoacyl tRNA synthetases describe each. |
1) The link between mRNA and protein is a small nucleic acid that has a 3-nt complement to a codon in the middle of the molecule (the “anticodon”) and an amino-acid residue covalently linked at the 3’ end. Each amino acid has a cognate tRNA, and some amino acids have more than one tRNA to which they can be attached. A special form of methionyl tRNA is used for translation initiation. The enzyme that attaches an amino acid to its cognate tRNA (the process of tRNA “charging”) is called an aminoacyl tRNA synthetase and this is accomplished at the expense of two high-energy phosphate bonds. The third codon position (“wobble” position) is variable for some of the amino acids.
2) Translation is performed by a ribosome consisting of two subunits (“large” and “small”) that each contain dozens of proteins and a large quantity of rRNA (hence the name “ribosome” for ribonucleic acid). Ribosomal RNA functions as the actual enzyme of peptide bond formation, with the proteins playing a structural role. Three “sites” on the ribosome are designated for their function in translation vis-à-vis the incoming tRNA. The A site (aminoacyl site) is where the next charged tRNA binds. The P site (peptidyl site) is where the tRNA with the growing peptide chain is located. The E site (exit site) is where the tRNA in the P site goes after its peptide cargo has engaged in peptide bond synthesis. 3) Eukaryotic initiation factors (eIFs) are required to bind the ribosome to the mRNA 5’ end and position the ribosome for correct initiation of translation. eEFs (elongation factors) participate in bringing succeeding charged tRNAs to the site on the ribosome where peptide bond formation takes place, as well as translocation of the ribosome to the next codon. A single eRF (release factor) is required for release of the finished polypeptide from the ribosome upon encountering a stop codon, as well as release of ribosome from the mRNA. |
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Describe initiation of translation
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- 5’ cap and Kozak scanning model
Interaction of eIFs with 5’ cap forms a recognition site for ribosome binding. The ribosome “scans” in the 5’-to-3’ direction for an AUG that is in the correct sequence context to function as a start codon. The start of the codon sequence defines the end of the 5’ UTR. |
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Describe the interaction between the tRNA anticodon and the codon
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Incorporation of the “correct” amino acid in the growing peptide chain depends on the dwell time of a charged tRNA with the codon. A longer dwell time, due to correct complementarity of the anticodon and codon, allows the GTP-bound eEF1A, which brings the tRNA to the A site, to hydrolyze GTP. GTP hydrolysis is required for eEF1A to leave (as the GDP-bound form) and peptide bond formation.
How is eEF1A regenerated? |
Regeneration of eEF1A-GTP: When a charged tRNA is brought to the A site by eEF1A-GTP and stable codon-anticodon pairing occurs, the ribosome itelf functions as a GAP to activate GTP hydrolysis by eEF1A. The inactive eEF1A-GDP complex leaves the ribosome and GTP exchange is catalyzed by eEF1B, a GEF
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What are some energy considerations in translation?
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In addition to the hydrolysis of GTP required to release eEF1A and allow peptide bond formation, the movement of the ribosome down one codon requires hydrolysis of a GTP in the translocase reaction. Thus, incorporation of one amino acid in a growing polypeptide chain requires hydrolysis of four high-energy bonds: two in the aa-tRNA synthetase reaction and two in the elongation reaction.
Why 2 bonds per ATP? |
(two GTP--> GDP, one ATP--> AMP) ~ total four; and about a million high energy bonds for one protein :O
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Describe interactions between 3' and 5' end of mRNA
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A ribosome arriving at a stop codon is released from the mRNA, but the 3’ end and 5’ end of an mRNA connect via the interaction of poly(A)+ binding protein and eIFs that bind the 5’ cap. This allows ribosomes to rapidly bind again at the 5’ end to initiate translation again.
What is a polysome? |
“Polysomes” are formed when multiple ribosomes are translating a single mRNA, and participation of an mRNA in active translation is revealed by its partitioning into the polysome fraction.
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What are some regulatory pathways for translation?
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1) Universal shutdown by eIF2 phosphorylation --> cannot funciton as an initiation factor
2) Protein binding to specific structures in the 5’ and 3’ UTR 3) RNAi Describe 2) and 3) |
2) Regulatory proteins may bind to a site in the 5’ UTR to interfere with ribosome scanning. Such sites can be hairpin stem-loop structures that form by intramolecular base pairing. Regulatory proteins may bind to specific features in the 3’ UTR, recruiting an endonuclease that cleaves the message and exposes a 3’ end for rapid degradation by 3’ exonucleases.
3) RNAi/miRNA --> RISC --> SLICER, DICER |
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What are some events that can cause global inhibition of translation?
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ER stress/unfolded proteins, Ca2+ release from ER, Heat shock, TNF-α, LPS, Virus infection, GF deprivation, AA starvation, Heme deficiency
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What techniques for DNA replication efficiency does the cell employ?
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1) multiple origins of replication
2) bilateral replication from an ori 3) high processivity difference between DNA and RNA synthesis/polymerases? |
Unlike synthesis by RNA polymerases, DNA synthesis proceeds only from a pre-existing “primer” that provides a free 3’ hydroxyl onto which DNA polymerase adds the next nucleotide.
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What are the steps of DNA replication?
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1) Formation of PIC
2) Initiation 3) Elongation 4) Termination Describe each |
1. PIC Formation ---> O.R.C
2. Initiation: DNA primase, a subunit of DNA pol alpha is an RNA polymerase that lays down a short RNA primer, which is then extended by DNA polymerase. This happens once for the leading strand and multiple times for the lagging strand. 3. Elongation (very processive - doesn't fall off) hands DNA off to DNA-polymerase δ Helicase/Topoisomerase 4. Termination RNA primers displaced RNA cleaved by endonuclease at DNA/RNA junction DNA Ligase:3'-OH to 5' phosphate bond seals the nick Voila! |
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What DNA polymerases are involved in replication?
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DNA pol δ, and DNA pol α.
Describe each |
More than one DNA polymerase enzyme is involved in cellular DNA replication. The major replicative enzyme is DNA pol δ, a highly processive enzyme. It is locked onto the DNA template by a protein clamp that gets loaded onto the template soon after replication initiates. Synthesis of the RNA primer and primer extension to start replication are accomplished by DNA pol α.
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Describe inhibitor of HIV DNA replication
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Selective DNA polymerase inhibitors can be used to stop viral DNA replication while allowing cellular DNA replication. For example, the HIV life cycle includes reverse transcription of genomic RNA into double-stranded DNA that then integrates into the chromosome. Nucleoside analogues that do not have a 3’ hydroxyl onto which a subsequent nucleotide can be added are used as antivirals. Examples are azidothymidine (AZT) and dideoxythymidine.
Why don't they affect human DNA polymerases? |
Km of these substrates for HIV RT is low and high for DNA polymerase |
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How to use PCR to detect carrier mutations?
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1) Detecting a 4bp insertion
Run PCR using custom primers surrounding the 50bp exon on parent's DNA Then run gel to separate products If parent is WT: PCR'd fragments will be 50bp. If one parent is a carrier for TS mutation: PCR fragments will be 50 and 54 bp's 2) Another mutation is G-->C mutation in the GTAAG 5' of Exon 12 that affects its splicing: 3'====[TCTG]GTAAG=====5' use restriction endonuclease DdeI which recognizes ====CTNAG==== which will target the mutation Order PCR primers to amplify 135bp region that includes target sequence Digest with DdeI Run gel If WT: single band 135bp long If heterozygote: bands formed from 135, 80, 55 bp sequences How to detect foreign genome, e.g. HIV? |
HIV detection - small amount of viral genome detected through amplification
HIV - RNA genome: we want to detect 10 particles/ml reverse transcribe the genome using DNA primers to make cDNA duplicate cDNA to make dsDNA use primers to HIV sequences and run PCR and gel band indicates presence of HIV genome; can be quantitative to determine viral load. |
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What are 4 types of DNA errors requiring repairs?
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1) Replication error
2) Base modifications 3) Pyrimidine dimers 4) Double strand breaks Describe enzymes involved in each. |
1) Replication error: DNA pol δ's 3'->5' proofreading exonuclease; mismatch repair by MSH, MLH
2) Base modifications - base excision by glycosylase, cut by AP endonuclease, ligated; or direct repair of base 3) Pyrimidine dimers - cleave affected strand, fill in with non-processive DNA pols η - TLS, L - TLS (translesion synthesis) 4) Double strand breaks - NHEJ by Ku70,etc; homologous recombination by BRCA2, Rec proteins |
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Homologous recombination: what are some outcomes of crossing-over events?
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1) equal crossing over
2) un-equal crossing over 3) inverted repeats on the same chromosome Which can lead to disease, and how? |
- unequal crossing over
Two or more copies of the same gene can be present on a chromosome. If homologous copies of the gene line out of register and undergo recombination, the result will be unequal distribution of gene copies, which may lead to disease. - inverted repeats on the same chromosome 10% of genome involve simple sequence repeats. Thus, the repeat sequences supply regions of homology at which crossing over can occur. We will present the example of hemophilia A, for which almost half the cases are due to a recombination event leading to an inversion of a chromosomal region. |
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Gene therapy concepts and problems
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Allow homologous recombination to occur between faulty chromosomal DNA and an introduced WT DNA.
Problems: 1) Insertional mutagenesis 2) Immunological aspects 3) Wrong placement Explain each |
1) - insertional mutagenesis
A big issue in gene therapy is that, since one cannot know in advance where the introduced DNA will integrate, there is always a risk of insertional mutagenesis, where the inserted DNA sits next to a gene and turns on its expression inappropriately. This can lead to leukemias, as happened in a gene therapy trial in France involving children being treated for adenosine deaminase deficiency. 2) - immunological aspects The most efficient way to get large pieces of DNA into a cell is by using viruses that infect cells and deposit the DNA they contain. However, the downside of this methodology is the possible immune reaction to such viruses. A major setback in gene therapy trials in the US occurred in 1999, when a patient undergoing gene therapy for treatment of a urea cycle enzyme defect died due to immune system reaction to the virus used to carry the inserted DNA. 3) Being in the wrong location almost always results in down-regulation of expression or complete silencing, and the presence of the wild-type gene is not beneficial (and may be harmful, see below). |
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What is transposition?
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Transposition is the movement of a unit of DNA called a “transposon” from one site to a new target site. Unlike homologous recombination, transposition occurs in the absence of homology and the insertion generally occurs at random sites. It is catalyzed by a transposase enzyme, which is often encoded by a gene located on the transposon itself. As uncontrolled transposition would wreak havoc with genome structure, this process is highly controlled via regulation of transposase gene expression.
Difference between bacterial and eukaryotic transposition |
While bacterial transposons move via a DNA-only mechanism, eukaryotic retrotransposons move via an RNA intermediate. The transposon sequence is transcribed by RNA polymerase, followed by reverse transcriptase conversion of the RNA intermediate to double-stranded DNA, which then integrates into the target DNA. The retrotransposon sequence includes the coding
sequence for reverse transcriptase. The mechanism of retrotransposition is similar that of HIV integration. |
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What are some repetitive sequences in the human genome?
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1) LINES - long interspersed elements
2) SINES - short interspersed elements 3) VNTR - variable number of tandem repeats Describe each. |
1. The human genome contains about 500,000 copies of an element termed “L1.” The full-length element is about 6,000 bp long, but the vast majority of L1 elements are truncated. Even of the full-length elements, most are mutated to the point of non-functionality, and there are only about 50 active L1 elements. These code for a protein that has reverse transcriptase activity. Retrotransposition by an L1 element is known to be the cause of disease in several cases. More importantly, an L1 element provides a region of homology at which recombination can take place.
2. The major type of SINE in the human genome is the AluI element, a 300-bp DNA that is present in about a million copies. These contain no coding sequence, and their transposition depends on the assistance of L1 elements. Here, too, movement of an AluI element can cause disruption of a gene and disease. 3. The genome is littered with small DNA sequences (9-80 bps) that can be present in multiple contiguous copies in the same orientation (known as “tandem repeats”). For a tandem repeat at a particular locus, different individuals may have a different number of repeats. The number of repeats can be determined by performing PCR using primers that target unique sequences outside the repeat region. The size of the PCR product will depend on the number of repeats. When this is done for a number of repeat regions simultaneously, a unique personal pattern is generated. This is the basis of identification of individuals in forensics. |
