M2M LO Unit1 Flashcards
Distinguish purines and pyrimidine bases, ribose and deoxyribose, ribo- and deoxyribo nucleosides, nucleotides, nucleoside di and triphosphates.
Purines and pyrimidines make up the two groups of nitrogenous bases,including the two groups of nucleotide bases.Two of the four deoxyribonucleotides and two of the four ribonucleotides,the respective building blocks of DNAand RNA,are purines (G/A).
Purines: Fifty percent (50%) of the bases in nucleic acids,adenineand guanine,are purines. In DNA,these bases form hydrogen bondswith their complementarypyrimidinesthymineand cytosine,respectively. This is called complementary base pairing.
Pyrimidine: Three nucleobases found in nucleic acids,cytosine(C), thymine(T), and uracil(U), are pyrimidine derivatives.
Ribose is a five-carbon sugar that’s the primary building block of ribonucleic acids. Deoxyribose has been de-hydroxylated at the 2’ position.
A ribonucleoside is a type of nucleoside including ribose as a component.
A deoxyribonucleoside is a type of nucleoside including deoxyribose as a component.
Nucleosides are the central ribose sugar and a base attached to it at the 1’ position.
A nucleoside with one phosphate group attached to the 5’ position of the ribose is called either a nucleotide or a nucleoside monophosphate.
If there’s a chain of two phosphates tagged on at the 5’ position, it’s a nucleoside diphosphate
If the chain is three phosphates long, it’s a nucleoside triphosphate.
Discuss the relative solubility of the components of nucleotides and the diseases related to their insolubility.
Phosphates are highly hydrophilic
Purine is less soluble than pyrimidine.
Base is less soluble than nucleoside.
Nucleoside is less soluble than nucleotide.
Phosphates>Nucleotide>Nucleoside>Base
Diseases related to purine insolubility, leading to build up of derivatives in tissues and kidney issues:
Gout- build up of uric acid in joints as a precipitant of purines
Lesch-Nyhan Disease- Causes severe neurologic symptoms
Identify the chemistry in the phosphodiester linkage of DNA and RNA polynucleotide strands and discuss the reason for 5’ -3’ polarity.
Each nucleotide has an open hydroxyl group at the 3’ ribose position and a phosphate at the 5’ position (as well as a base stuck onto the 1’ carbon). It’s the binding of one nucleotide’s phosphate group to another’s OH group that makes the “chain” of DNA/RNA. Thus, at the end of each DNA or RNA chain, there’s either a ‘spare,’ unlinked phosphate group (what’s called the 5’ end) or a ‘spare,’ unlinked hydroxyl group (called the 3’ end). The phosphodiester linkage is the link between phosphate and hydroxyl.
Discuss the important experiments that helped to establish DNA as the genetic material.
Avery, McCloud and McCarty: established DNA as the genetic material with their Pneumococcus experiments (Smooth strain killed mice, Rough strain did not. DNA from heat-killed S cultured with R then killed mice).
Franklin and Wilkins: x-ray diffraction suggesting a helical structure.
Watson and Crick discover definitive double-helical structure.
Describe the implications of Chargaff’s rules.
The molar ratios of total purines and total pyrimidines are roughly equal (G+A=C+T)
The molar ratios of adenine to thymine, and guanine to cytosine, are roughly equal. (G=C, A=T)
G+C / A+T ratio is different for different organisms
Describe the Watson-Crick three-dimensional model for DNA structure and recognize the major and minor grooves, the phosphodiester backbone, and the base pairs.
DNA is a right handed double helix. The sugar phosphate backbone is hydrophilic and is located on the outside of the molecule. The bases are paired and stacked on the inside due to their relative hydrophobicity. The major groove is the larger groove seen on the double helix and constitutes the ‘gap’ between the curves on the same single strand of DNA. The minor groove is the ‘gap’ between the backbones of complementary DNA strands. The relative distance crossed between AT bonds and GC bonds is pretty much the same. There are about 10 base pairs per turn of the helix.
Describe the chemical basis for the stability of the double helix DNA in solution.
Electrostatic repulsions between phosphate groups (a) are mostly neutralized by positively charged species in the cell(magnesium); (b) the base pair linkages give the helix a lot of stability; and (c) adjacent base pairs “stack” on top of each other, providing additional delocalization options for the electrons and giving, often, more stability than comes from the base pairing itself. The stacking interactions are stronger between G/C than A/T.
Increased salt concentrations will increase the stability of the DNA molecule (thus increase its melting temperature Tm)
Extremes of pH alter ionization of bases which form H-bonds and thus decreases stability
Increase in DNA length will increase stability
The higher the GC content the more stable the DNA (more H-bonding/delocalization potential).
Distinguish between linear and circular and relaxed and supercoiled forms of DNA.
Linear vs circular: Linear DNA is large and the segments can not distribute twisting so they can become supercoiled. Circular DNA is shorter and bound to itself (in prokaryotes) so can’t really become supercoiled.
Relaxed vs supercoiled: Relaxed DNA is essentially a straight ribbon of DNA with appropriate twisting while supercoiled DNA has an additional level of twists and coils on the DNA that influences its shape… important for DNA packaging in eukaryotes.
Describe the chemical modifications of bases in DNA including different forms of DNA damage (methylation, deamination, depurination, UV cross linking) and their significance to disease.
Methylation: DNA methylationin vertebrates typically occurs at CpG sites(cytosine-phosphate-guanine sites, that is, where a cytosineis directly followed by a guaninein the DNA sequence). This methylation results in the conversion of the cytosine to 5methylcytosine.The formation of Me-CpG is catalyzed by the enzyme DNA methyltransferase.Human DNA has about 80%90% of CpG sites methylated, but there are certain areas, known as CpG islands, that are GC-rich (made up of about 65% CG residues), wherein none are methylated. These are associated with the promoters of 56% of mammalian genes, including all ubiquitously expressed genes. One to two percent of the human genome are CpG clusters, and there is an inverse relationship between CpG methylation and transcriptional activity.
Deamination: C can lose an amine to become U, which results in mutated DNA. Hopefully a repair enzyme would recognize this, but not always. Additionally, deamination of guanine can cause C-G to be turned to A-T, and vice versa for deamination of Adenine.
Depurination: Depurination is an alteration of DNAin which the purinebase (adenineor guanine)is removed from the deoxyribosesugar by hydrolysisof the beta-N-glycosidic link between them. After depurination, the sugar phosphate backbone remains and the sugar ring has a hydroxyl(-OH) group in the place of the purine. It’s usually caught by DNA repair enzymes.The problem is that it significantly weakens the phosphodiester backbone at the depurination site—so if you have a couple of these nearby, it can break the backbone.
Ultraviolet light: can covalently bond thymines, which distorts the DNA helix and can block replication enzymes. Generally repaired by nucleotide excision repair
Alkylating agents: Nucleophilic attack of bases on nucleotides (mustard gas, cisplatin)
Explain the chemistry of DNA polymerization and how nucleoside analogues are used as drugs.
DNA polymerization is covered in the next section. Essentially, you can use molecules that are very similar to nucleosides to block replication of virally infected cells by having the replicating DNA (which recruits free-floating nucleosides as it replicates) incorporate the analogues into the growing chain. The analogues are different enough from actual nucleosides to ensure that the resulting DNA chains are nonfunctional.
Notice you can also use differences in the viral reverse transcriptase pathways to design nucleoside analogues preferentially incorporated by reverse transcriptase pathways.
More-specific nucleoside analogues usually used against retroviruses. Less specific nucleoside analogues usually used as chemotherapy against cancer.
Attacking DNA metabolism can occur through 4 methods
Blocking synthesis of precursors
Intercalation (getting in the middle)
Covalently binding bps
Attack topoisomerases which relieve supercoiling
Describe how DNA melting temperature/annealing to complementary sequence is negatively affected if there is a mismatch and how this can be taken advantage of in diagnostic techniques to distinguish the presence of a particular unique sequence using a “probe” that is completely complementary to that sequence.
Using base pairing as a diagnostic tool: because perfectly complementary single strand DNA molecules pair with higher stability, a “probe” DNA single strand can be used to test whether a sample has the perfect complementary DNA sequence. The probe is tested for “hybridization” to an unknown DNA sample. Even a single mismatch can be distinguished. Variations on this phenomenon have been used for many diagnostic purposes in identifying genetic abnormalities that can arise from as little as one base pair change compared to unaffected individuals.
Define the major similarities and differences between DNA and RNA.
DNA: has no hydroxyl group at the 2’ position of the ribose, which makes it more stable and less prone to hydrolyzation by nucleophilic attack at the 2’ location. DNA binds cytosine, guanine, adenine, and thymine as its bases.
RNA: hydroxylated at its 2’ ribose position; binds uracil instead of thymine. RNA is usually single-stranded, although it can form double-stranded loops (often called ‘hairpin loops’) with itself.
Define 3 classes of RNA in a human cell.
Structural
rRNA (ribosomal RNA)- make up ribosomes and translate tRNA (transfer RNA)- move RNA around
snRNA (small nuclear RNA) and snoRNA (small nucleolar RNA) for a variety of in-cell modifications such as splicing.
Regulatory
miRNA (mirco RNA) and siRNA (small interfering RNA) to downregulate gene expression.
Information-containing
mRNA (messenger RNA) to be translated into proteins.
Explain how puromycin mimics amino-acyl tRNA to terminate translation.
Puromycin is an antibiotic that mimics the “acceptor” 3’ end of a tRNA that is charged with an amino acid. Puromycin can bind in the ribosome as it is translating and covalently attach to a growing polypeptide chain, preventing the completion of translation.
Discuss the meaning of these terms in the context of DNA Replication: “semi-conservative”, “bidirectional”, “Okazaki fragments”, “origin”, “fork”.
Semi-conservative: Each DNA strand is preserved as one half of a new double DNA strand; thus each new DNA strand has one-half original material and one-half newly synthesized material.
Bidirectional: Means that when the replication machinery attaches to the DNA double helix, replication proceeds in both directions along that helix at once.
Okazaki fragments: Small stretches of DNA synthesized during replication in the 5’ to 3’ direction (on the lagging strand). Since the synthesized strand is running in the 3’ to 5’ direction, and since new dNTPs can only be added at the 3’ hydroxyl group, the DNA synthesis process takes a kind of “leapfrogging” approach whereby small segments on that strand are copied 5’ to 3’ and then melded together later.
Origin of replication, aka replication origin: Specific sequences for recognition by binding proteins. Usually contain multiple short repeats, as well as an A-T rich streak. There are hundreds per chromosome, 1 in a prokaryote.
Replication fork: where the DNA helicases have unwound the double helix. Effectively the H-bonds of the base pairs have been split apart and the two strands are peeled away from each other, thus forming a “fork” in which the replication machinery sits and synthesizes complementary strands.
Describe the functions of the following proteins during DNA replication: Origin binding proteins, Helicases, Single-strand binding proteins, Primase, DNA polymerase I, DNA polymerase III, DNA ligase, Sliding clamp, Topoisomerase/gyrase, Telomerase/Reverse transcriptase.
Origin binding proteins: Proteins bind to the origin and become part of the complex, also recruits Pol III.
Helicases: enzymes that catalyze the breaking of H-bonds between base pairs and the subsequent ‘unwinding’ of the helix.
Single-strand binding proteins: bind to the melted strands of original DNA to prevent them from re-annealing or getting messed up. More important for Okazaki fragments as they spend more time single stranded (you really don’t want single stranded stuff hanging around).
Primase: Enzyme that catalyzes the addition of the RNA primer to begin replication.
Pol I: DNA polymerase I, “distributive.” Versitile enzyme that replaces the RNA primers using 3 different functions: DNA polymerase, 3-5’ exonuclease activity and 5-3’ exonuclease activity. Does not have the sliding clamp so it is slow and distributive.
Common to both Pol III and Pol I: “proofreading” activity, or 3’ to 5’ exonuclease activity. If the wrong dNTP is added during DNA synthesis, the synthesis stops and ‘backs up’ slightly (in the 3’ to 5’ direction) and chops off the last nucleotide added.
Pol III: DNA polymerase III, “processive.” Synthesizes DNA strand from its complement on both leading and lagging strand. High processivity due to a sliding clamp mechanism that holds the polymerase tightly to the DNA. No 5’ to 3’ exonuclease activity (thus can’t be used to remove RNA primers).
Processivity clamp: the aforementioned ‘sliding clamp’ mechanism. Present in Pol III.
DNA ligase: enzyme responsible for sealing Okazaki fragments together once the RNA primers have been replaced by Pol I.
Telomere: sequence at the ends of chromosomes, consisting of a large number of repeating segments. Gets consistently shorter every time the chromosome is replicated, since the RNA primer on the very last O. fragment can’t be replaced by Pol I (Pol I needs to have a nearby 3’ OH from the next fragment to bind and replace the RNA primer). After a certain point, the telomeres get short enough that the cell becomes unstable and is destroyed.
Telomerase: Enzyme responsible for ensuring that the telomeres of chromosomes in certain immortal structures, such as germ cells, never shorten. Effectively, they act as reverse transcriptases, binding to the ends of DNA sequences and adding on some extra dNTPs. The reason this works is that the telomeres have more or less a uniform repetitious sequence.
Topoisomerase/gyrase: Enzyme responsible for relieving torsional strain in the DNA helix in the region ahead of the replication fork. It does this by clipping the phosphodiester backbone in selected places and then putting it back together without strain. Topogyrase is specific to prokaryotes.
Reverse transcriptase: Enzyme responsible for copying a base sequence INTO DNA (as opposed to out from it), usually from RNA. This can be endogenous (ie. telomerases) or exogenous (ie. retroviruses).
Describe how DNA polymerase creates the phosphodiester bond during addition of deoxyriboncleotides (dNTPs).
It breaks off a diphosphate group from the dNTP and uses the energy liberated from that reaction to bind the remaining phosphate group to the hydroxyl group of the previous nucleotide on the chain.
Recognize that DNA polymerase requires an RNA primer, that DNA synthesis only occurs in the 5’-to-3’ direction, and that errors during replication are corrected by the 3’-to-5’ exonuclease proofreading activity of the DNA polymerase.
none
Describe the order of events that occur during, the differences between, and coordination of, DNA synthesis on the leading strand and the lagging strand.
Leading strand: pretty simple, relatively speaking:
origin binding proteins bind to origin.
DNA melted apart locally by helicases.
Topoisomerases relieve tension ahead of the replication fork.
Pol III elongates DNA complementary to leading strand.
The two strands, one new, one old, are annealed.
Lagging strand: similar but a little more complicated since DNA synthesis can only occur in the 5’ to 3’ direction. Share first 3 steps with leading strand, then:
Primase attaches RNA primer to lagging strand segment
Pol III elongates DNA from RNA primer back a short ways, forming an Okazaki fragment.
RNA primer is removed and replaced with DNA by Pol I.
Fragments are sealed together with DNA ligase.
The two strands, one new and one old, are annealed.
Describe the “end replication problem” and the activity of telomerase.
The “end replication problem” refers to the fact that the leading stand can be synthesized to the very end, but the lagging strand cannot during DNA replication.
This is because you need an RNA primer to begin synthesis of each piece of the lagging strand DNA, but at the end of the DNA there is nothing for this piece to attach to thus the last section of the lagging strand cannot be synthesized. The result is that the telomeres (i.e. the end of chromosomes) get shorter and shorter as a cell replicates its genomes and divides, until they are so short that they signal for cell death – a normal healthy process in ageing cells.
Telomerase is an RNA-dependent DNA polymerase that maintains chromosomal ends by copying the telomeric repeat sequence from an RNA template. Telomerase activity is repressed in normal somatic cells.
In cancer cells, the telomerase enzyme is de-repressed, restoring the ends of the chromosomes to their full length and therefore blocking the normal cell death of old cells, promoting tumor growth. As such, telomerase is a potential target for anti-cancer drugs.
Describe the relationship between mutations, DNA repair and cancer. Give examples of heritable human diseases that are caused by defective DNA repair pathways.
A mutation or mismatch is needed in a gene associated with cell proliferation
The cell must not notice the change or be unable to fix it to lead to proliferation
Self destruction pathways must be repressed or inactivated
These three criteria lead to a cell being cancerous
Examples of heritable human disease from defective DNA repair: xeroderma pigmentosum and cockayne syndrome.
Describe the sources and nature of damage to DNA, the type of machinery used to repair the damage, and the molecular consequences of failure in DNA repair, e.g. thymine dimers, uracil mis-incorporations, bulky chemical adducts, and double-strand breaks.
Thymine dimers: Good candidate for nucleotide excision repair. Usually caused by UV radiation causing linkage between adjacent thymine residues, causing a bulging deformation of the helical structure. If unrepaired, can cause problems with normal processing due to malformed helix. Also will cause Pol III to fall off and Pol II take over, leading lots of errors
Uracils in DNA: Good candidate for base excision repair. Usually caused by the deamination of a cytosine residue to produce uracil in the DNA. If uncorrected, can cause problems with both the process of replicating/transcribing this portion of the DNA and also with recognition sites of transcription enzymes.
Specifically, if this portion of the DNA is transcribed or replicated as is, it will pair with an adenine residue, not a guanine; thus will have effectively swapped a C for a T.
Bulky chemical adducts: like thymine dimers, except usually caused by chemotoxic binding of large molecules to bases in a DNA helix. (nucleotide excision repair)
Double-strand breaks: Good candidate for either homologous recombination repair or non-homologous end joining (good to know: non-homologous end joining is the major form of double-strand break repair). Caused by a double break of the phosphodiester backbone (not sure what underlying causes are). If unrepaired, since a chromosome is one long DNA sequence, can lose up to half of the chromosome (very bad).
Explain the basic steps of mismatch repair, describing the type of damage repaired by this pathway, and understand the marking of the old strand of DNA by methylation in E. coli.
Mismatched base recognized soon after synthesis (before methylation) on new strand (not old strand; old strand recognized by methylation in E. coli, mechanism of recognition unknown in humans). A stretch of DNA behind and in front of the mismatch is clipped by endonucleases, excised by helicase and exonucleases, and replaced with the correct sequence by DNA Pol III (and sealed with ligases).
Describe the basic mechanisms of base excision repair, nucleotide excision repair, double-strand break repair by homologous recombination and NHEJ, and the types of lesions corrected by these DNA repair pathways.
Nucleotide-excision repair (NER): tends to repair more overt modifications that alter the helical pattern of the affected DNA. Process: recognition, clipping the backbone by endonucleases, excision of the affected part, replacing by Pol I, resealing by DNA ligase.
Notice that the recognition pathways here need a transcription factor, TFIIH, to work properly. (needs the helicase to melt the DNA)
Notice also that there’s two kinds of NER:
Transcription coupled NER: the distortion is within a gene being actively transcribed
Global Genome NER: the distortion isn’t within a gene being actively transcribed, goes back over the whole thing.
Base-excision repair (BER): tends to repair subtler modifications, like a mismatched base pair not caught by either proofreading or mismatch repair. Process: recognition, clipping off the inappropriate base by glycosylases, clipping the backbone by endonucleases, chewing off by exonucleases of the affected part, replacing by Pol I, resealing by DNA ligase.
Recombinatorial repair : also called homologous recombination. Repairs double-stranded breaks or crosslinks in DNA. Process: partially degrades sides of the break to create primers for DNA synthesis. Copies intact, homologous sequence from other chromosome that aligns with it. Each strand aligns itself with a strand on homologue and fills in its gap from that strand.
NHEJ [Non-Homologous-End-Joining]: A form of double-stranded break repair that doesn’t involve the homologous chromosomes. Essentially you unwind the two ends with helicases, pair up a few matching bases, and reseal the phosphodiester backbone. Note that this can be inaccurate, as you often lose a few bases off the unpaired strands during the resealing.
Describe the mechanism that enables replication to continue in the face of DNA lesions that other repair pathways fail to remove, and know the unfortunate consequence of this process for the cell.
The mechanism is called lesion bypass polymerization and usually occurs when the cell doesn’t have enough resources to fix all the thymine dimers occasioned by UV exposure.
Big damage stops DNA Polymerase III, if NER cannot occur then two bypass polymerases bind to the complex. This creates a conformational change in the complex placing the bypass polymerases across from the damage where they add nucleotides without proof reading. After the damage the complex changes conformation and returns to DNAP III doing its business with proof reading.
The error rate is 2 to 4 orders of magnitude higher than normal replication, thus frequently results in cancers, etc.
Explain the concept of DNA damage checkpoint and its role in maintaining genome stability.
The cell cycle is paused if damage is sensed so that defects aren’t passed along to future generations of cells. During this arrest, repair machinery is up-regulated. 2 pairs of protein kinases are central to this mechanism (ATR for stalled forks and ATM for double strand breaks). They amplify the signal and phosphorylate Chk1 and Chk2 DNA for repair, arrest, or apoptosis. Chk1 and Chk2 phosphorylate p53 which causes cell cycle arrest or activate cell death program. Chk1 and Chk2 kinases are important to ensure that problems have been fixed. Mutations in any of these checkpoint DNAs lead to genome instability because the cell is now unable to regulate itself and may proliferate into a cancer. The checkpoint is the first step in guarding against cancer.
Describe the chemical reaction catalyzed by RNA polymerase and why it is unidirectional.
RNA polymerase catalyzes phosphodiester bond formation between ribonucleasides in 5’ to 3’ direction, based on template of 3’ to 5’
Need triphosphate nucleoside to make hydrolysis spontaneous
Catalysis means this reaction is done 1 way (unidirectional)
Distinguish five steps in the transcription cycle common to bacterial and eukaryotic RNA polymerases.
1: RNA polymerase binds to promoter sequence on the helical DNA in a “closed complex.”
2: Polymerase melts DNA strands apart near transcription start site, forming an “open complex” aka the “transcription bubble.”
3: Polymerase catalyzes phosphodiester linkage of two initial rNTPs.
4: Polymerase advances 3’ to 5’ down template strand, melting DNA and linking rNTPs.
5: At transcription stop site, polymerase releases completed RNA and dissociates from DNA.
Note that steps 1-3 are called initiation, step 4 is elongation, and step 5 is called termination.
Name the four cellular RNA polymerases and their main functions.
RNA pol1: makes ribosomal RNA(rRNA)
RNA pol2: makes messenger RNA (mRNA), small nuclear RNA (snRNA), micro RNA (miRNA)
RNA pol3: makes transfer RNA (tRNA), lncRNA (long non-coding RNA for regulatory functions)
Mitochondrial RNApol: makes RNA for the mitochondria
Define a promoter and name sequence elements characteristic of promoters in human genes.
A promoter is a sequence of DNA upstream of the transcription start site that positively affects the expression of the gene. (where RNA Polymerase binds)
Consensus elements: -30 (TATA box), initiator, promote
The TATA box is a frequently conserved TATA sequence about 30 bases upstream from the start site. Mutations in the TATA box often result in reduced expression of the gene (ie beta-thalassemia with B-hemoglobin), because the TATA box binding protein (TBP) helps in assembly of the pre-initiation complex of general transcription factors at the promoter
Initiator is +1 in some but not all eukaryotic gene
Promoter proximal elements are promoter DNA sequences between 30-1000 bp upstream of the start site
Enhancer elements are promoter DNA sequences much farther upstream (10,000-50,000 bp). This acts through DNA looping.
Describe how α-amanitin and rifampicin block transcription.
alpha-amanitin: extremely toxic substance found in death cap mushrooms. Acts by inhibiting the movement of RNA Pol II, binding its bridge substructure so that translocation of the polymerase down the DNA chain can’t happen.
rifampicin: broad-spectrum antibiotic. Acts by binding the beta subunit of bacterial RNA polymerase, plugging up the exit chamber where assembled RNA exits the transcriptional complex. Thus elongation is prevented from going farther than a few base pairs due to having nowhere to go.
Name four components of the RNA polymerase II pre-initiation complex.
RNA pol2, general transcription factors (TF2a,b,d,e,f,h), promoter DNA, mediator
Describe the clinical syndromes caused by mutations in TFIIH subunits.
TF2H is a transcription factor that also serves to repair damages to DNA
Problems with nucleotide excision repair: Cockayne’s syndrome, Trichothiodystrophy
Xeroderma pigmentosum - light sensitivity, abnormal pigmentation, cancer susceptibility, neuorological abnormalities, unscheduled DNA synthesis).
Describe the three major ways in which most pre-mRNA’s are processed
Capping: replacement of the 5’ triphosphate (from the first rNTP to be added) with a backwards, 7-methylguanosine (no phosphate group).
Splicing: excision of introns and desegmentation of exons.
Cleavage/Polyadenylation: cleavage of RNA at 3’ end past the consensus sequence and polyadenylation (> 200 A’s) of cleaved site.
Note that these processing steps take place while the RNA is still being made, not after it’s finished and released. For example, the cleavage/polyadenylation step is partly responsible for Pol II being released from the DNA.
Compare and contrast a pre-mRNA with a mature mRNA.
Pre-mRNA-Has its 3’ phosphate still intact, has large caps of non-coding sequences between coding exons called introns, the 3’ end is un-modified
Mature mRNA-5’ phosphate has been capped by 7-methyl guanisine, introns have been spliced out, poly A tail has been added to the 3’ end
List the functions of the 5’ cap of the mRNA.
It makes the 5’ end resistant to exonucleases (which target the “lone ends” of single D/RNA strands).
It helps with splicing and processing through a cap-binding complex that recognizes the cap. (primes for splicing, 3’ tail, translation)
Translation factor eIF4E (eukaryotic initiation factor 4E) recognizes the cap for transport to the ribosomes.
When the cap is eventually removed, it signals for the mRNA to be degraded.
List the three reactions required to add a 5’ cap to pre-mRNA.
(1) Cut off the last PO4 from the triphosphate group at the 5’ end of the mRNA.
(2) Add guanosine triphosphate (GTP) backwards via guanylyl transferase. It loses two PO4 groups of its own (making GMP) and forms a 5’-to-5’ triphosphate bond (2 PO4 from mRNA, 1 PO4 from GMP) to the end of the mRNA.
(3) Methylate the 7-position of the guanosine cap via S-adenosyl methionine (SAM; nearly universal methyl donor in cell; donates methyl to form homocysteine) to form 7-methyl-guanosine at cap.
Recall the conserved sequences at the 5’ and 3’ ends of most introns and the consensus sequence at the polyA site.
Splice site at 5’ end of intron: GU.
Splice site at 3’ end of intron: AG.
Consensus sequence at poly-A site: AAUAAA.
Describe how alternative splicing permits multiple proteins to be produced by splicing defects.
(normal) Retain or remove exons, mutually exclusive exons, exon truncation.extension at 5’ end, exon truncation/extension at 3’ end, intron retained or removed.
Since the 5’ splice site (the one identified by snRNA that recruits the spliceosome) at any given intron is part of the DNA sequence, it’s vulnerable to being corrupted by mutation and being unrecognizable as an intron, in which case the finished mRNA has an extra sequence in it and the final protein winds up significantly different
Give examples of genetic disorders caused by splicing defects.
Marfan’s syndrome: caused by mutations that disrupt splicing of the fibrillin gene transcript (fibrillin is a connective tissue protein that is important for the integrity of the walls of the heart and blood vessels). They are tall and prone to aneurysms.
Abnormal splicing of CD44 (cell-surface glycoprotein) is a predictor of tumor metastasis. Used as diagnostic and prognostic marker.
Describe the function of U1 and U2 snRNA’s in splicing.
U1snRNA binds to the GU 5’ splice site of introns.
U2snRNA binds to the branch point (A) on the pre-mRNA sequence between the 5’ and 3’ splice sites
U2AFsnRNA binds to the 3’ splice site of intron (AG)
Lariat splicing mechanism: the U1snRNP brings the 5’ splice site into proximity to the branch point. The U2snRNP activates the 2’ hydroxyl group at the branch point, which attacks the phosphodiester bond just past the GU 5’ splice site. Now there’s a free 3’ OH on the end of the 5’ exon, which attacks the phosphodiester bond at the AG 3’ splice site– linking the two exons and excising the snRNP/intron complex to be degraded.
Identify on a diagram of a gene the following:
transcription start site: Look for either the +1 position or the little bent arrow coming out of the sequence at this point.
introns: Should be between the GU (5’) and AG (3’) splice sites.
5’ splice sites: Look for GU.
3’ splice sites: Look for AG.
branch points: should be in the introns, between the GU (5’) and AG (3’).
exons: should be between the introns. Mark the introns with GU at 5’ and AG at 3’, then look between them for the exons.
5’ UTR: look for a region between position +1 (start of transcript) and the start codon (see below) on the processed mRNA strand.
3’ UTR: look for a region after the stop codon (see below) until the end of the transcript, including the consensus sequence and the poly-A tail.
initiation codon: also called start codons. Encodes methionine. Look for 5’ AUG.
termination codon: also called stop codons. Look for 3’ UAG, UAA, or UGA.
poly A site: look for consensus sequence AAUAAA
Describe the two reactions that make the mature 3’ end of mRNA’s.
1) recognition of consensus sequence at pre-mRNA’s 3’ end (AAUAAA) and cleavage of the mRNA soon after this sequence.
(2) polyadenylation of free hydroxyl at 3’ end. (not coded for)
Describe the relationship between 3’ end processing of the pre-mRNA and termination of transcription at the end of a gene.
The cleavage and polyadenylation occur while the RNA polymerase continues, it could soon fall off or it could continue to make another mRNA.
The poly-A tail seems to be necessary for the RNA polymerase complex to detach from the DNA (termination).
Describe the two major functions of the mRNA’s poly A tail.
protection from degradation
export of mRNA from nucleus
Give an example of how alternative poly A sites can be used to make more than one protein from a single gene.
two different forms of immunoglobulin M (IgM), membrane-bound and secreted, are formed by alternative poly-A sites in their common gene. This can occur because transcription continues past poly-A site depending on which one is used it will make 2 variants (heavy and light chains).
Discuss the basics of eukaryotic transcription (DNA control elements, transcription factors) and the role they play in human disease and list the different eukaryotic DNA control elements.
DNA control elements: DNA elements that act locally. Binding of transcription factors to these elements controls expression of the gene that the element is associated with.
TATA box/Initiator sequence- This element is generally 25-35 bps upstream of the transcription start site. It determines the site of transcription initiation and directs binding of RNA polymerase II. This is the site at which general transcription factors bind.
Promoter proximal element
Enhancers
Transcription factors: Proteins encoded by one gene that act on other genes to regulate their transcription. Can therefore diffuse around the nucleus and affect transcription
of numerous genes. Can either activate or repress transcription.
Define promoter proximal element and enhancer.
promoter proximal element: usually within 200 bp upstream of start site, about 20 bp long. Bound by transcription factors to regulate transcription.
enhancers: usually much farther upstream, or downstream, than promoters, although still fairly short in themselves (8-20 bp). Can be upstream of the start site, downstream of the last exon, or within introns in the gene itself. Similar function to promoters.
Describe a disease that arises from a mutation in a DNA control element, and how the mutation leads to the disease state.
Beta-thalassemia (from thalassa, Greek for sea): mild inherited anemia (low hemoglobin count). Caused, here, by a mutation in the promoter of the b-globin gene, resulting in lowered rate of production of b-globin protein. (less promotion = less transcription)
Gamma-delta-beta thalassemia: more serious anemia caused by a deletion in the locus control region for the transcription of all globin genes, resulting in the loss of globin transcription.
Hemophilia B Leyden: X-linked disease (usually males) that affects clotting. Again, a problem in the promoter region of a clotting protein gene. Tends to get partially better at puberty.
Fragile-X syndrome: Again, usually a disease of men. Results in mental retardation and atypical development of the face with enlarged testicles (macroorchidism). Caused by an expansion in the CGG count upstream of a particular gene (the FMR1 gene), which results in an abnormally high rate of methylation in that region and transcriptional silencing of the gene.
Describe the role of transcriptional activators and repressors.
They bind to either the DNA control elements (ie, to the DNA itself) or to other factors bound to control elements (ie, to other proteins that are bound to the DNA).
They increase or decrease the rate of transcription of the gene’s protein(s).
List the two classes of activators and repressors.
Those that bind to the control elements in the DNA are called sequence-specific DNA binding proteins or SSDBPs. They usually bind to short (6-8 bp) sequences by inserting their alpha-helices into the major groove of the sequence in question.
Those that bind to SSDBPs are called co-factors. These can both increase or decrease efficacy of transcription factors.
Describe the domains of a sequence specific DNA binding protein.
Generally two domains. Notice that these domains are modular– that is, if you pull one domain off of one protein and attach it to another, the second protein will have the function of the domain you just attached.
First domain: DNA binding domain. Very highly structured and evolutionarily conserved. Binds to the DNA target.
Second domain: activation (or repression) domain. Fairly unstructured and less conserved; they recruit other proteins (either co-factors or general transcription factors) to bind and affect transcription.
Note the difference between co-factors and general transcription factors: co-factors influence the rate of transcription, while GTFs provide the pre-initiation complex needed to begin transcription.
List the four major families of sequence specific DNA binding proteins and describe the means for categorizing the proteins into these families.
These are categorized based on tertiary structure differences:
Homeodomain proteins have a helix-turn-helix structure. They tend to be regulators of development and affect many genes at once.
Zinc-finger proteins have a “finger” made up of two antiparallel beta sheets and an alpha helix, held together by a zinc ion. This finger is what binds with the DNA. The largest family of SSDBPs. Include androgen and estrogen (nuclear hormone) receptors.
Basic leucine zipper proteins (bZIP): (the “basic” here refers to the fact that they have a high-pH region that binds to the DNA) Chop sticks. Hydrophobic residue every 7 amino residues, (like leucine). Dimerizes to bind DNA. (c-fos and c-jun).
Basic helix-loop-helix proteins (bHLH): also has basic region for DNA binding. Muscle group (MyoD, myogenin).
Describe a particular human disorder that arises from a mutation in a sequence specific DNA binding protein, explaining how the mutation leads to the disorder.
Craniosynostosis: premature closure of the skull sutures in infants. Arises from a mutation in the homeodomain protein that causes the protein to bind more strongly, creating a “hypermorphic allele” that activates genes more strongly than it should. (upregulates proteins that close sutures via homeodomain mutation)
Androgen insensitivity syndrome: Feminization or undermasculinization. Indifference of androgen receptors to androgen hormones. Caused by a mutation in the zinc-finger androgen receptor binding domain or ligand binding domain. This downregulates the transcription of genes controlled by male androgens.
Waardenburg syndrome: deafness, pigmentation defects. Associated with mutations in the MITF (
phthalmia-associated transcription factor) gene (which codes for a bHLH binding protein that regulates melanocyte development).
Describe combinatorial control as a mechanism for controlling gene expression.
Combinatorial control refers to the fact that SSDBPs can dimerize, making possible a wide variety of possible DNA binding sequences.
Note that this is always within classes of SSDBPs (e.g. zinc-finger with zinc-finger)
For example:
SSDBP can dimerize into homodimers and heterdimers, the combination of these proteins leads to many different transcription factors. If you have 4 monomers that can mix and match into dimers you have 2^4 transcription factors. Larger complexes, trimers and tetramers, are also possible.
With 1,000s of TF genes you can have 10,000s of TFs (Not 2^1000 - not all TFs can form heterodimers with each other).
