Exam 3 Chapter 6, 7
What percentage of base pairs of human DNA are used for transcription?
1.5
Translation, 4-step cycle Polycistronic RNA in prokaryotes Eukaryotic proteins are translated on polyribosomes or polysomes Antibodies and targets and consequences
27, 28
Genetic Diversity In a DNA world Coding and template strands
9, 10
Somatic cell mutations •Definitions: mutations that occur during the life of an organism •Can cause unchecked cell division: cancer •Current U.S. cancer statistics from the American Cancer Society: •Both men and women have ~1 in 3 risk of developing cancer and a ~1 in 5 chance of dying from it
CANCER
RNA: Engages in intra-strand conventional base-pairing (A-->U) Unconventional base hydrogen bonding interaction (A--->G and C->U, unconventional but normal for RNA) Adopts the sequence-specific tertiary structure (Some bases hang off to the side or interact in the _____) Note double-helical structure
Core
Proteasome Protein machines composed of 1.A central cylinder of proteases 2.Stoppers on both ends to bind proteins marked for degradation Ubiquitylated Proteins ATP drives degrading proteins into amino acids in the central cylinder ~10 subunits Amino acids & peptides Proteases
Cylinder contains proteases
Nonhomologous end joining A double-stranded break is pieced back together quickly to avoid DNA fragments drifting apart. Repair enzymes "clean" the broken ends; may digest a few nts in the process. (BETTER TO HAVE A FEW MISSING NUCLEOTIDES THEN A WHOLE LOSS OF A CHROMOSOME) Result: Repair may cause a small deletion in the sequence, GENETIC MATERIAL LOST Double-strand break --> End binding with end-binding proteins ---> End bridging---> Protein crossbridge (HELD TOGETHER BEFORE DRIFTING OFF IN THE NUCLEUS) ---> Recruitment of additional proteins and end processing ---> Proteins for DNA processing ----> Gap repair synthesis and ligation
End joining
To translate an mRNA, the 4-letter nucleic acid language code, A C T and G, is converted into the ___________, a ________- letter amino acid language
Genetic code; 20
The reason behind the ribose sugar 2' hydroxyl group allows RNA splicing (even self-splicing) Why is DNA is deoxy 2'? ____________________________________________ 5' end, sugar-phosphate backbone --> OH on 3' carbon of sugar, ribose and 3' hydroxyl on 3' end, phosphate on the 5' end
In order to prevent the splicing of DNA , it cannot splice itself or cut anything out
Protecting Telomeres DNA ends may look like a double-stranded break, targeted by DNA repair machinery, it is a necessity for the double-stranded DNA double helix to stay _______ to provide genetic transfer (KEEP GENETIC CODE INTACT) Possible solution: form a loop
Intact
Eukaryotic mRNA Modification •A cap and tail of extra nucleotides are added to ends of mRNA 1.Facilitate export of mRNA from nucleus 2.Protect mRNA from degradation by enzymes 3.Help ribosomes bind to mRNA DNA, mRNA -> Transcription, nuclear envelope, nuclear pore
Into protein
Bacterial versus eukaryotic genes Coding region (5' -->3') (stretch of coding regions all in ONE place) bacterial gene Coding regions (exons) 5' -->3' (interspersed with non-coding regions -___) Noncoding regions (introns) Eukaryotic gene
Introns
Gamma Radiation and X-rays •Oxygen: a potent mutagen (reason for life, O is a two edged sword, give life or damage life) • Takes a regular guanine and make 8-hydroxyguanine (8-oxoG) •one of the most common DNA lesions •Often caused by ________________ ________________ •Results in _________________: causes G----T and A----C substitutions
Ionizing Radiation Transversions ionizing (Gamma Radiation and X-rays, some ultraviolet) (Radio, microwave and infared and visible are non-ionizing (not damaging))
Eukaryotic transcription initiation •General transcription factors (set of accessory proteins) •Transcription factors assemble at the promoter with polymerase 1.General transcription factor has the subunit, TATA-binding protein (TBP). What does it do? ______________________________________________________ 2.TF11B, followed by other transcription factors and RNA polymerase II, assemble Comprises the Transcription Initiation Complex, THERE IS A TRANSCRIPTION INITIATION COMPLEX TF11H 1.Acts as a ___________; unzips DNA to allow transcription start, using ATP 2.____________________ RNA polymerase tail; releases it from rest of transcription factors (then transcription proceeds) TATA binding protein TATA box ---> Start of transcription (DNA) ---> (TBP, TFIID) ---> (TFIIB) ---> (TFIIE, TFIIH) (TFIF, other factors, RNA polymerase II) ---> Ribonucleoside triphosphates (UTP, ATP, CTP, GTP) ---> RNA ---> Transcription
Recognizes the TATA box, the binding then begins, a bend occurs so TBP can bind and initiate the process Helicase Phosphorylated
Depurination Consequence: Frame-shift Mutation (TWO BASES REMOVED) Water came in and removed entire guanine from the DNA strand, leaving a depurinated DNA strand with an OH, with O added from the water to the sugar for another base to be added. Sugar phosphate after depurination Can be repaired by the base excision method
Missing teeth --> Depurinated A ---> Mutated (old strand, and new strand an A-T nucleotide pair has been deleted Or Missing teeth ----> DNA replication--> New strand, old strand unchanged.
Many molecules of RNA can be rapidly synthesized because
Multiple polymerases use the same gene consecutively
Many molecules of RNA for both prokaryotes and eukaryotes can be rapidly synthesized because
Multiple polymerases use the same gene consecutively
Fidelity of DNA Replication •The majority of mutations do neither harm nor good. •Natural selection removes most of the harmful ones: •Early death before reproduction (evolutionarily appropriate) •Decreased fertility •Favorable changes persist and spread throughout a population •Profoundly faithful preservation of genetic message over tens of millions of years.
Tragic for family, but evolutionarily appropriate
NONSENSE MUTATION Application: Effect of translational mistake Duchenne Muscular Dystrophy Original DNA code for an Amino acid sequence ---> DNA bases ---> Amino acid --> Replacement of a single nucleotide --> Protein, incorrect sequence causes shortening of a protein Definition: A mutation to the genetic code that results in a premature _______ codon
stop. MOST DAMAGING
Adapter proteins: Transfer RNAs (tRNAs) D: Dihydrouridine Ψ: Pseudouridine (additions of different hydrogens) Anticodon tells the tRNA which amino acid to hold GAA - Phe Attached amino acids (Phe) (high energy bond) at 3' end that holds an amino acids ===> Anticodon loop, anticodon ---> A clover leaf
tRNA
Codon-anticodon connections •Anticodons pair with their associated codons in an antiparallel fashion. AAG ----> UUC Amino end on 3' end Phenylalanine (tRNA Phe) ---> Phenylalanine anticodon ---> Phenylalanine codon --> 3' mRNA Proline (tRNA Pro) ---> Proline anticodon ---> Proline codon ---> 3' mRNA
tRNA anticodon twists when it loops
Which of the indicated carbon positions on the ribose sugar (below) contains the high-energy phosphates in nucleotide triphosphate building blocks?
the 5' Carbon
Deamination of 5-methylcytosine If deamination results in a ________________, DNA repair cannot determine if the guanine in the opposite strand was incorrect, or the thymine is incorrect. (because both are natural bases) G-Tor C-T ? Causes ______________ for mutations (because the determination will not be determined by DNA machinery 5-methycytosine ----> Thymine
thymine hotspots
Spontaneous DNA damage Damage caused by _______ •Irony: The proper structure of the double helix depends on an aqueous environment 1.Deamination (cytosine to uracil, change to the code itself) 2.Depurination (KEEPS PHOSPHODIESTER BOND INTACT, Frameshift)
water
RNA in cells differs from DNA in that ___________________.
it is single-stranded and can fold into a variety of configurations
In addition to the repair of DNA double-strand breaks, homologous recombination is a mechanism for generating genetic diversity by swapping segments of parental chromosomes. During which process does swapping occur?
meiosis
What do you predict would happen if you created a tRNA with an anticodon of 5′-CAA-3′ that is charged with methionine, and added this modified tRNA to a cell-free translation system that has all the normal components required for translating RNAs?
methionine would be incorporated at some positions where leucine should be
Transcription is similar to DNA replication in that ___________________.
nucleotide polymerization occurs only in the 5′-to-3′ direction.
Pyrimidine dimers: Application 2 •Xeroderma pigmentosa: condition due to defective nucleotide excision repair enzymes (ONCE THERE IS THYMINE DIMERS in the STRAND, THERE IS NO METHOD TO REPAIR THIS) •Key points •_________________________________________ •______________________________ •______________________________ Xeroderma pigmentosum, is a genetic pathological condition of the autosomalrecessive form in which the body loses its ability to repair damage caused to the body by the ultraviolet rays of the sun
- The inability to repair UV damage - Strong susceptibility to skin cancer (MORE MUTATIONS, MORE PROBLEMS, MORE TUMORS) - Failure to repair thymine dimers (genetically transferred) Xeoderma pigmentosum is genetically transferred DIRECT REPAIR BASE EXCISION REPAIR DEPURINATION DEAMINATION ALL DUE TO INTERNAL METABOLISM
The figure below shows a ribose sugar. RNA bases are added to the part of the ribose sugar pointed to by arrow _____.
1' Carbon
Base excision repair Transitions and transversion Spontaneous DNA damage
1, 2
Translation 3-letter codons Reading frames
23, 24
Aberrant Splicing Mutations that disrupt normal splicing are estimated to account for up to ______________of all disease-causing mutations Proteins encoded by these abnormal transcripts are often truncated or missing whole domains •May alter protein function •May confer new functions •May have genome-wide effects, altering gene expression in cancer-associated pathways •Genes abnormally spliced in cancer cells: >________________ Resisting cell death, deregulating cellular energetics, sustaining proliferative signaling, evading growth suppressors, Avoiding immune destruction, Enabling replicative immortality, Tumor-promoting inflammation, Activating invasion and metastasis, inducing angiogenesis, genome instability and mutation.
1/3 Over 30%
Translation errors and disease •Nonsense mutations account for more than 10% of annotated disease-causing mutations. •Small deletions and insertions account for more than 20% of documented mutations. •May be a causative factor in the pathology of multiple sclerosis and amyotrophic lateral sclerosis (ALS) (BREAKS DOWN NEUROLOGY)
10% 20%
RNA transcription is not as accurate as DNA replication. The error rate is one mistake per every _________ nts.
10^4
Transcription of rRNA genes captured in a transmission electron micrograph Prokaryotic Transcription Bacterial transcription Note Asymmetry of the promoter, this determines which strand is used for template
11, 12
Transcription can occur on both strands of DNA for both prokaryotes and eukaryotes Eukaryotic transcription features Eurkaryotic mRNA modification
13-14
Addition of the mRNA Cap Addition of the Poly-A tail Untranslated regions (UTRs) Roles of UTRs
15,1 6
Eukaryotic and bacterial genes Spliceosome Alternative splicing
17, 18
Nuclear export of properly processed mRNA is mediated by nuclear Pore complexes RNA Transcription (prokaryotic vs. eukaryotic Circular dilemma
19, 20
How many genes? }Humans: ___________ genes } } }Nematode worm: 20,470 } } }Water flea: 31,000 genes } } Tomato: 31,760 genes
19,000
On the sugar-phosphate backbone of an RNA-molecule, splicing is possible due to the ____________
2' Hydroxyl group
Depurination is most efficient at a pH of ________
2.5
RNA and the origins of life Autocatalysis RNA as its own template Molecular fossils
21, 22
Adapter proteins: Transfer RNAs (tRNAs) Codon-anticodon connections Wobble rules Initiation of translation
25, 26
How puromycin terminates translation Recycling proteins Translation errors
29, 30
Reading Frames 3 of these codons do NOT code for amino acids (UGA, UAA, and UAG) AUG is the stop (methionine) 61 coding for amino acids 3 coding for stop •There are ______________ potential reading frames for each mRNA molecule •Start from one of three nts •ALWAYS read from the 5'-3' end (start from one of three nucleotides) ONLY __________ of the reading frames specifies the correct protein
3 1, if a nucleotide is added or taken away the protein changes, or shuts down completely
In transfer RNA, the ________ end of the RNA molecule is covalently attached to the ________ end of the amino acid, which it carries.
3', carboxyl
Depurination Deamination
3, 4
mRNA prep •Human genome: ~____________________base pairs of DNA •98.5% are not transcribed. •Average gene size: 30,000 bp •Many genes are larger than 100,000 bps, with the largest known gene, dystrophin, being approximately 2.4 million base pairs •_______% of human genes undergo alternative splicing (AS) •Two mechanisms: 1.Use of the spliceosome, a complex of small nuclear ribonucleoproteins (snRNPs) 2.Self-splicing introns, or _________________, capable of catalyzing their own excision from their parent RNA molecule
3.2 Billion 95% Ribozymes
How many base pairs of DNA are in the human genome?
3.2 billion
Nucleotide substitution - silent mutation Nucleotide substitution - missense mutation Frameshift mutations Application: Effect of translational mistake - Nonsense mutation
31, 32
Translation errors and disease Non sense mutation
33
Mismatch DNA repair mechanism Double-strand breaks Non-Homologous end joining Homologous recombination
7, 8
Translation: 4-Step Cycle Ribosome speed: •Bacteria: 20 amino acids/second •Eukaryotes: 2 amino acids/second Large ribosomal subunit (E site, P site, A site) Small ribosomal subunit (mRNA binding site) Step 1 ---> Growing polypeptide chain ---> Newly bound charged tRNA ---> Step 2 , releases ---> Step 3, large subunit translocates ---> Step 4 small subunit translocates ---> Newly bound charged tRNA Mediated by Elongation Factors (eEFs) that move the ribosome along the mRNA: GTPase proteins are required (USING A DIFFERENT FORM OF ENERGY) P site starts it, A site always has a new amino acid coming in and they are attached to the one in the P site and the growing polypeptide the tRNA is ejected and a new one comes in
4-step The job of this mRNA is to carry the gene's message from the DNA out of the nucleus to a ribosome for the production of the particular protein that this gene codes for. There can be several million ribosomes in a typical eukaryotic cell. These complex catalytic machines, use the mRNA copy of the genetic information to assemble amino acid building blocks into the 3-dimensional protein that are essential for life. Let's see how this works, the ribosome is composed of one large and one small subunit that assembles around a messenger RNA which then passes through the ribosome like computer tape. The amino acid building blocks are carried into the ribosome attached to specific transfer RNAs, which is referred to as tRNA. The small subunit of the ribosome positions the mRNA so that it can be read in groups of three letters known as a codon, each codon on the mRNA matches a corresponding anticodon on the base of a tRNA molecule. The larger subunit of the ribosome removes each amino acid and joins it onto the growing protein chain. As the mRNA is ratcheted through the ribosome, the mRNA sequence is translated into an amino acid sequence. There are three locations inside the ribosome, designated the A site, the P site, and the E site. The addition of each amino acid is a three-step cycle, first, the tRNA enters the ribosome at the A site, and is tested for a codon-anticodon match with the mRNA. Next, provided there is a correct match, the tRNA is shifted to the P site and the amino acid it carries is added to the end of the amino acid chain. The mRNA is also ratcheted on three nucleotides or one codon. Thirdly, the spent tRNA is moved to the E site and ejected from the ribosome to be recycled. As the protein synthesis proceeds, the finished chain emerges from the ribosome and folds up into a precise shape determined by the exact order of amino acids. Thus, the Central Dogma explains how the four-letter DNA code is quite literally turned into flesh and blood.
Given the codon sequence for glutamine is 5'-CAG-3', what is the corresponding tRNA anticodon sequence?
5'-CUG-3'
5-methycytosine Deamination thymine Generic DNA Repair mechanism Mismatch DNA repair
5, 6
Severe consequences of DNA repair failure: •Single nucleotide change: sickle cell anemia Single DNA strand of normal beta-globin gene, GLU, single DNA strand of nutant beta-globin gene, VAL, single nucleotide changed (mutation) Glutamate amino acid -- > valine Primary structure: normal hemoglobin, Secondary structure and tertiary structure: beta subunit Quaternary structure: Normal hemoglobin (top view) with 2 alpha and 2 beta subunits Function: Molecules do not associate with one another: each carries oxygen Primary structure: Sickle-cell hemoglobin, Secondary structure and tertiary structure: Exposed hydrophobic region, beta subunit Quaternary structure: Sickle-cell hemoglobin (top view) with 2 alpha and 2 beta subunits Function: Molecules interact with one another and crystallize into a fiber: capacity to carry oxygen is greatly reduced.
A SINGLE NUCLEOTIDE CHANGE ON THE BETA GLOBIN GENE
What is the consequence of an unrepaired deaminated cytosine in the next round of DNA synthesis?
A for a G substitution
Autocatalysis (to make more of itself) To eventually reach the origin of life, molecules need to be able to make more of themselves. Earliest cell composition •_________________________________________ •_________________________________________ •_________________________________________ Catalysis ----> Making more copies Ribozyme ----> Substrate RNA ---> Base-pairing between ribozyme and substrate --> Substrate cleavage ---> Product release ---> Cleaved RNA + Ribozyme
A simple membrane (seperate inner components from outside world) Self-replicating molecules (make more of themselves) Components to provide material and energy
Initiation of Translation eIF2 brings pre-initiation complex to the ribosome (initiation factor) •All proteins start with a Methionine initially, using initiator tRNA •This is different than regular Met-tRNA •Initiator tRNA binds tightly to the P-site on the small ribosomal subunit Initiator Met is frequently removed later by specific proteases FIRST THE P SITE AND THEN THE A SITE Transcription initiation factors, Met (PAIRS WITH AN AUG), initiator tRNA, small ribosomal subunit with translation initiation factors bound ---> mRNA binding ---> Small ribosomal subunit, with bound initiator tRNA, moves along mRNA searched for first AUG ---> Once Met and AUG meet, translation initiation factors dissociate and THE Large ribosomal subunit binds --> Charged tRNA binds to A site ---> First peptide bond forms CHARGED tRNA binds to the A SITE, the first peptide bond forms, Met is released form the initiator tRNA and binds to the new amino acid that just came in on the A site
A site
During DNA replication, a new strand incorrectly pairs an adenine with a guanine. Without repair, what base pairs will result when the two strands undergo replication again?
A-T and G-C
Adding amino acids using two tRNAS Method: 1.Amino terminus is synthesized before carboxyl terminus 2.The two tRNA are brought close together on the ribosome 3.The growing polypeptide chain is transferred from the peptidyl-tRNA to the aminoacyl-tRNA, forming a new peptide bond in the reaction: peptidyl transferase reaction 4.No new energy required; uses energy from breaking the original bond that "charged" the aminoacyl-tRNA Polypeptide chain with peptidyl-tRNA attached and aminoacyl-tRNA ---> New peptide bond PEPTIDYL- tRNA holding on to the carboxyl terminus ready to receive a new amino acid AMINOACYL-tRNA brings in a new Amino acid
Adding amino acids
Non-enzymatic depurination of nucleic acids Efficient depurination of nucleic acids may occur in acidic gastric juice and some acidic organelles (such as lysosomes), which then shows effects on digestion and assimilation of nucleic acids Adenine nucleotide becomes protonated, with one or two protons, the entire base is removed and we have an apurinic site Non-Enzymatic Depurination of Nucleic Acids: Factors and Mechanisms. Ran An et al
Adenine nucleotide in DNA ---> Monoprotonate nucleotide --> Doubly protonated nucleotide ----> Oxocarbenium ---Oxocarbenium --> Apurinic site in DNA
Refer to the diagram below. What amino acid will the tRNA carry, based on the anticodon? 5' to 3' Anticodon = AGC
Alanine (GCU)
Direct repair Example 1.Guanine acquires an _________* group which disrupts normal pairing with cytosine. During replication, the G would pair with a T, causing a mutation 2.An ____________ transfers the alkyl group to itself, restoring the base to normal. (allowing cytosine to attach to guanine) *Alkyl group: functional group having the formula CnH2n+1
Alkyl Enzyme Alkyl group takes up space on oxygen and thymine comes in with two hydrogen bonds Direct repair with a damaged nucleoside ---> Alkylated guanine, thymine---> Enzyme carries off alkyl group --> Guanine and cytosine
In Direct DNA repair, (a) an ________ group attached to a base is removed by (a) an _______
Alkyl, enzyme
Untranslated regions (UTRs) occur at the beginning and end of a transcript. The role of the UTRs includes
All of the above Transport out of the nucleus Control of mRNA translation Translation efficiency
Animation: Transcription The first step in protein synthesis, is transcription from RNA to DNA. The portion of the DNA that is transcribed into an RNA molecule is called a transcription unit, an enzyme called RNA polymerase carries out transcription with the help of transcription factors, it attaches to the beginning of a region of DNA called the promoter pries the DNA apart and untwists a short portion of the DNA helix, RNA polymerase moves along the DNA pairing up RNA nucleotides , with their DNA complements, adding nucleotides to the end of the growing RNA molecule. Here is a close up view of the elongation, only one DNA strand, the template strand, serves as a template for RNA synthesis. RNA Polymerase moves along the DNA strand in the 3'---> 5' direction, adding nucleotides to the 3' end of the RNA chain, note that U in RNA pairs with A in DNA. Once transcription is complete, RNA polymerase releases the completed RNA and detaches from the DNA.
Animation (DNA is not impacted, the DNA simply opens up, allowing a small portion to be transcribed into RNA by RNA polymerase, and it is released allowing the DNA to be reanea led. )
An aminoacyl-tRNA synthetase must recognize specific nts on both the _______ and the _________ accepting arm of the correct tRNA.
Anticodon; amino acid
The piece of RNA below includes the region that codes for the binding site for the initiator tRNA needed in translation. 5′-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3′ Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome?
Arginine
Addition of the mRNA Cap Modifies the 5' end of the transcript with methylated guanine attached in a 5'-5' triphosphate bridge (which is unusual that it is not 5' to 3' ) When is the mRNA cap added? _________________________________________________________ DNA --> RNA polymerase II ---> Capping factors, splicing factors, polyadenylation factors ----> RNA processing begins ---> mRNA 5' Cap (7-methylguanosine, 5' to 5' triphosphate bridge) 5' end of the initial RNA transcript
As soon as the transcript exits the polymerase, therefore, it is capped right away, to prevent any degradation what so ever.
Polyribosomes or polysomes on eukaryotic RNA Chaperones help the protein fold correctly as it emerges from the ribosome Many more proteins can be made with multiple ribosomes working simultaneously ~80-90 nucleotides between ribosomes mRNA circularizes, aiding in polysome formation Stop codon, growing polypeptide chain, start codon, mRNA
As the mRNA messenger molecule is transported from the nucleus to the cytoplasm. Ribosomes translate the sequence into amino acids. Typically many ribosomes translate the mRNA simultaneously, each ribosome begins at the 5' end of the mRNA, and progresses steadily toward the 3' end. New ribosomes attach at the 5' end at the same rate the previous ones move out of the way. These multiple initiations allow the cell to make much more protein from a single message then if one ribosome had to complete the task before another one to begin. When a ribosome reaches a stop codon, the ribosome and the new protein dissociate from each other and from the mRNA. This electron micrograph depicts a membrane bound polyribosome from a eukaryotic cell
Effect of dimers during transcription Method of correction: ________________________________________________ Requires five enzymes: 1._________________________ 2._________________________ 3._________________________ 4._________________________ 5._________________________ Pyrimidine dimer, thymine dimer (ultraviolet light) ---> Thymine dimer DNA/RNA polymerase movement is stalled due to the thymine dimer (Bend in DNA)
Base excision repair 1. Glycosylase (breaks apart DNA backbone) 2. Endonuclease (free hydroxyl group) 3. Exonuclease (free new phosphate group) 4. DNA polymerase (come in and replace the damaged area with a correct nucleotide) 5. DNA ligase (seals it up) FASTER THAN WE CAN SNAP OUR FINGERS
Cancer and aging •The incidence of cancer increases with age •Greater passage of time → greater accumulation of mutations •Incidence higher in people with defective DNA repair machinery Incidence of colon cancer per 100,000 women, age (years)
Bodies not designed to last forever
Transcription can occur on ______________ strands of DNA for both prokaryotes and eukaryotes. RNA, like DNA synthesis always proceeds in the _____________ direction. Template strand is read 3' to 5' *asymmetry determines orientation Promoter --> Gene A --> RNA transcript from Gene A Gene B --> RNA transcript from Gene B ---> Promotor
Both 5' to 3' Gene B is transcribed 5' to 3' Gene A is transcribed 5' to 3'
Nucleotide substitution—missense mutation Missense mutation: change in one of the codons Gly GGC--> Ser AGC
Can or cannot be damagign
Which of the folllowing is the cause of most DNA mutation?
Cellular metabolism
Coding and template strands ________________________: segment of DNA that codes for a protein ________________________: COMPLEMENTARY base pairs with the coding strand (is what directed the process into mRNA) The resulting mRNA will be an exact copy of the coding strand except for _________ (T) will be replaced with _______ (U). Coding strand, Template strand
Coding Strand Template Strand Thymine, Uracil mRNA strand exactly copies the coding strand except the T is replaced with U
To transcribe a gene, helicase must open the double-stranded helix to expose the correct sequence. DNA instructions for the desired protein will be located on the
Coding strand
3-letter codons Arginine = 6 Aspartate = 2 Leucine = 6 •General information •__________________(three consecutive nts) specify one amino acid •Most amino acids have more than one codon: _____________________ •Most redundant codons have the same nts at the first and second positions and the third one at a different position (________________ position) (CHANGE AT THE THIRD POSITION)
Codon Redundant Wobble
Competition •To "survive," early molecules would divert raw materials to make more of themselves Ø___________________________________________________ Ø___________________________________________________ Ø___________________________________________________ Upset balance→Decay to chemical equilibrium "________________"
Compete for space Compete for resources (various elements) A certain temperature and a minimum of new components Death, end of the early molecule and they could not self replicate
Restoring splice site recognition (cystic fibrosis) •A histone deacetylase inhibitor promotes exon inclusion in CFTR, restoring functional CFTR channels that are defective in cystic fibrosis •The channel transports negatively charged chloride ions into and out of cells. Outside cell, mucus = Mutant CFTR channel = Does not move chloride ions, causing sticky mucus to build up on the outside of the cell Inside the cell = Chloride ions --- Normal CFTR Channel moves chloride ions to the outside of the cell (release chloride ions to prevent a mucus build-up Examples of splicing-based therapeutic approaches Disease Human Target gene. Therapeutic Stage Bardet-Biedl Syndrome BBSI U1/U6 snRNA Pateint cells Bata-thalassemia HBB. PTM. Minigene Cancer. BRCA1 and PTCH1 ASO Minigene Cystic Fibrosis. CFTR U1 snRNA Minigene
Cystic fibrosis
Most deamination changes a (an) ___________ to a _________
Cytosine to a uracil
The repair of mismatched base pairs or damaged nucleotides in a DNA strand requires a multistep process. Which choice below describes the known sequence of events in this process?
DNA damage is recognized, the newly synthesized strand is identified by an existing nick in the backbone, a segment of the new strand is removed by repair proteins, the gap is filled by DNA polymerase, and the strand is sealed by DNA ligase.
After a primer is aded to the lagging strand at the end of a DNA molecule, _____________- adds complementary nucleotides and _______ seals the backbone.
DNA polymerase, ligase
DEAMINATION Deamination, cytosine --> H2O, in, NH3 out--> Uracil THE O FROM H2O is a double bond O = uracil ) Deaminated C (U)----> DNA replication --> Mutated old strand and new strand, a G has been changed to an A to bind to the U. Deaminated C (U)---> C is fixed, DNA replication ---> New strand, old strand, unchanged (GC) Consequence: Point-Mutation Most deamination changes a cytosine to uracil. Can be very damaging affecting the proteins Is this process pathological?
Deamination
Why should an individual with xeroderma pigmentosa avoid direct sunlight?
Defective nucleotide excision repair enzymes
•Depurination One trillion (1012) purine bases are lost from the DNA in human cells ~every 5 seconds by spontaneous ________. •A purine base is deleted from a nucleotide •The phosphodiester backbone is left INTACT Result: lesions like missing teeth Depurinated A
Depurination
Alternative Splicing DNA ---> Exon 1 --> Exon 2 --> Exon 3 ---> Exon 4 ---> Exon 5 RNA ---> Exon 1 ---> Exon 2 ---> Exon 3 ---> Exon 4 ---> Exon 5 Alternative splicing mRNA --- 1,2,3,4,5 Translation ---> Protein A mRNA --- 1,2,4,5 Translation ---> Protein B mRNA --- 1,2,3,5 Translation ---> Protein C
Different exons = different proteins
Radiation and X-rays II •Ionizing radiation also causes double-strand breaks in DNA •_________________: attacks the deoxyribose in the DNA backbone •_________________: generates reactive oxygen species (ROS) which subsequently react with deoxyribose
Direct effect Indirect effect
In an RNA world, the RNA molecule performed all of the following functions except
Directly synthesis RNA using genes
DNA is double-stranded RNA is single-stranded but can also combine in a helical structure Sugar differences Ribose (sugar with hydroxyl on 2' C and 3' C) used in RNA Deoxyribose (sugar without hydroxyl on 2' C, has a hydroxyl group on the 3' C) used in DNA Base Differences Uracil, Used in RNA (demethylated thymine) Thymine, used in DNA Sugar-phosphate backbone ---> Bases, ribose --> 5' end and 3' end, C A U G 2' hydroxyl group allows RNA splicing (even self-splicing) *explains why DNA is deoxy 2' (DO NOT WANT DNA TO BE SPLICED INTO OTHER CONFIGURATIONS)
Do NOT want DNA to splice
What protein in the ubiquitin-proteasome system is responsible for identifying/binding specific proteins targeted for degradation?
E3 Ubq Ligase
Ubiquitin enzymes E1: activates the ubiquitin (Ub) E2: obtains Ub from E1 E2 binds an E3 E3 Ubq ligase binds the protein substrate for transport to the proteosome
E3 Ubq Ligase takes the protein to the proteosome
Translation errors •Translation into proteins is highly error-prone. Misincorporations of amino acids occur ~once in every 1,000 to 10,000 codons translated, i.e., 15% of average-length protein molecules will contain at least one misincorporated amino acid
Errors
aka "primary transcript" Eukaryotes Nucleus --> DNA introns and exons ----> Transcription, pre-mRNA ----> 5' capping RNA splicing 3' polyadenylation ---> RNA cap on mRNA with AAA tail --> Exported out of the nucleus ---> mRNA with AAA tail --> Translation --> Protein Prokaryotes DNA ---> Transcription --> mRNA ---> Translation (Protein)
Eukaryote transcription is more complex then prokaryote
Eukaryotic mRNA splicing Recognized by complementary bases in the U1 and U2 snRNPs Sequences required for intron removal 5' exon 1, intron 3' exon 2, portion of pre-mRNA ----> Intron removed --> Exon 1 on 5' and exon 2 on 3' portion of the spliced mRNA Note: R stands for A or G; Y stands for C or U; N stands for any nucleotide Intron sequence, exon 1, on 5' and Intron sequence in the middle, exon 2 on the 3' end in the portion of pre-mRNA ---> 5' end with OH on the other side is added ----> Lariat + Portion of spliced pre-mRNA
Eukaryotic genes typically contain introns, which have to be removed after transcription. Before the RNA transcript leaves the nucleus, the cell splices out the intron sequences, a few short nucleotide sequences provide the cell cues of what to remove. The elaborate machine that carries out this task is called the spliceosome. A branch point binding protein, BBP, and a helper protein U2AF recognize the branchpoint site within the intron, and an RNA and protein complex called an ssNRP, recognizes the 5' splice site by forming base-pairs with it. Next, another ssNRP base-pairs with the branch site, displacing the bound proteins. Additional ssNRPs, now come into play, and several RNA rearrangements occur to break apart the U4 U6 base-pairs and allow the U6 ssNRP to displace U1 at the 5' splice junction. Now in position, a conserved adenine nucleotide in the intron attacks the 5' splice site, cutting the sugar-phosphate backbone of the RNA. The end of the intron covalently bonds to the adenine nucleotide, forming a lariat structure. The spliceosome rearranges to bring together the exons, allowing the 3' hydroxyl group of the first exon to react with the 5' end of the other. After the two exons are joined into a continuous, the lariat is released and degraded
Translation AGCT (4-letter Nucleic Acid Language) 3.8 Billion years ago, the first cells with DNA _________________: 20-letter Amino Acid Language Almost universal... Slight differences in •mRNA of mitochondria •Some fungi •Some protozoa
Genetic Code
Generic Mismatch Repair Mechanism NO GLYCOSYLASE Exonucleases: enzymes that cleave nucleotides one at a time from the end (Exo) of a polynucleotide chain. Endonuclease: cleaves phosphodiester bonds in the middle (endo) of a polynucleotide chain. Damage to top strand ---> Step 1. Excision of a segment of the damaged strand (Nucleases: Endo-, Exo-)---> Step 2. Repair DNA polymerase fills in missing nucleotides in top strand using the bottom strand as a template (Repair DNA Polymerase) ----> DNA ligase seals nick(DNA Ligase) ---> Net result, repaired DNA
Exo
8-oxog repair 1.DNA glycosylase excises the modified base by the base excision pathway BEFORE DNA replication. 2.________________________can remove the incorrect "A" if repair is completed AFTER DNA replication. 3.DNA glycosylase uses the second opportunity to remove the 8-oxoG. Oxidation ---> Base excision repair, replication ---->Fail safe glycosylase removes 8-oxoG
Fail-safe glycosylase
Conversion of RNA to Protein 3*10^13 cells in the human body •Thousands of proteins made every second in every cell of the human body. •Expression of genes are dependent on need •Each gene can be transcribed at a different rate Gene A ---> Transcription ---> RNA ---> Translation ---> Protein Gene B (DNA) ----> Transcription ---> RNA ---> Translation (Protein) (Gene B is a little gene, does not require a lot of protein when the cells are being used up faster, if a protein of epidermis, needs to make a lot of new cells, while for an organ, it does not need to have a lot of cells made)E
Faster
Oxidation damage: Rat vs. human Rat: •100,000 oxidative lesions per day in each cell. Human: •10,000 oxidative lesions per day in each cell. The smaller the animal, the faster the metabolism The shorter lifespan for a mouse compared to a human Mouse has a _______ heartrate
Faster
Example of homework question •The piece of RNA below includes the region that codes for the binding site for the initiator tRNA needed in translation. • •5′-GUUUCCCAUGACUGCCGGGGGC-3′ • •Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome? ACU Thr
Find start codon AUG, in P site, ACU = Trp
In a DNA World •Segments of DNA called ________ are used to direct transcription of _______ which is then translated into ____________. • •Central Dogma (Francis Crick, 1956) DNA ----> (Transcription) ----> RNA ----> (Translation) ---> Protein ---> (Replication) DNA Gene, DNA ----> RNA synthesis, transcription -----> Protein synthesis (translation )----> Amino acids (protein)
Genes, RNA, Protein
Reverse Transcription in Eukaryotes •When do humans use reverse transcription? •_____________________________ •To preserve RNA as cDNA using the polymerase chain reaction (PCR) Translocation: Telomerase moves 6 nucleotides to the right and begins to make another repeat, The complementary strand is made by primase, DNA polymerase, and ligase (RNA primer)
For telomerase when we deal with telomeres
To support the RNA hypothesis as the earliest form of nucleic acid that could support life, the molecule ___________ can be used to produce the ribose sugar
Formaldehyde
Double-strand BREAKS (EXTREMELY SERIOUS) •Causes: Radiation; chemicals (face creams, cover-up creams, lipstick) •Particularly dangerous •Can fragment DNA: broken pieces can become separated •Can lead to a loss of genes •Two basic strategies evolved: 1.Nonhomologous end joining 2.Homologous recombination Single-strand break and double-strand break
GENES
GTPase for elongation •Elongation factor EF-Tu hydrolyzes GTP and "escorts the tRNAs to the ribosome IF there is the correct codon-anticodon match. •Must exchange GDP for GTP before new round of elongation Polypeptide chain, factor-binding center ---> EF-Tu- GTP, aminoacyl-tRNA ---> EF-Tu-GDP Hydrolysis of GTP promotes 90 degree twist of domains 2 and 3. Incorrect amino acid will not tolerate the strain, leading to dissociation. Site of tRNA binding, domain 2, domain 1, GTP-binding site, switch helix, domain 3 ---> GTP hydrolysis, release of tRNA, bound GDP
GTPase
In the translocation step, eukaryotic elongation factors mediate ribosomal movement along the mRNA. What kind of proteins are eEFs?
GTPases
Central Dogma of Molecular Biology Gene --> DNA --> RNA synthesis, transcription, nucleotides --> RNA --> protein synthesis, translation, protein --> Amino acids, COOH Gene expression is controlled at multiple levels Gene on DNA ---> Initiation elongation (transcriptional control --> primary transcript (splicing capping tailing (A)), RNA processing control --- > Nucleus, cytosol, Nuclear export (lots of small G-Proteins, stability/ turnover) RNA transport control ---> Initiation elongation (translation control) Protein, stability/turnover (ubiquitin-proteasome )
Gene expression is controlled on multiple levels
Base excision repair (URACIL CAN NOT BE ADDED TO A DNA Molecule, must be excised •Steps 1.______________________ removes the damaged base by hydrolyzing the glycosidic bond (2 sugars or a sugar and another) 2._______________________ cuts DNA backbone at 5' position to leave a 3' OH 3.________________________ cuts at the 3' position to leave a 5' phosphate 4.___________________________fills in the gap (with the correct nucleotide) 5.___________________binds the open ends
Glycosylase Endonuclease Exonuclease Repair DNA polymerase DNA ligase Normal base, uracil, free uracil, glycosylase cleaves here ---> Glycosylase ---> AP endo/exonuclease---> AP endo cleaves here (makes 3'-OH for Pol 1) Exonuclease removes to here
Reverse Transcription AIDS is a retrovirus •1970—Reverse transcriptase discovered by David Baltimore and Howard Temin •1975—They share the Nobel Prize in medicine •AIDS epidemic erupts in the mid-to-late 1970s Retrovirus infection and reverse transcription Host cell, retrovirus (with capsid) , cell membrane ---> Endocytosis ---> Uncoating ---> Viral RNA --> Reverse transcriptase ---> Integrase (brings it into the nucleus) ---> Integration of proviral DNA---> DNA transcription, mRNA ---> viral RNA --> Assembly ---> Budding ---> New virion
HIV virus is a typical retrovirus and has an outer envelope and in the center, it has two copies of RNA and an enzyme, reverse transcriptase, which will turn RNA into DNA. The virus with its outer envelope protein, actually directly infects helper T Cells. The way that it does this, is as it comes up to the cell surface it uses receptors specific to T-helper cells, which are CD4 receptors = which defines T-helper cells, a surface receptor that binds to the envelope protein. THat causes a conformational change and allows a second receptor to grab ahold of the envelope, called the chemokine coreceptor (CCR5), the stalk of the envelope protein, pierces through into the host cell, and starts to draw the cell membrane and the viral membrane together. What ultimately happens, is a fusion between the two membranes, and the viral genetic material is injected into the cell and the envelope protein is left at the cell surface. The virus has a matrix and a cap protein that is digested when it enters the cell. The viral enzymes and the viral RNA is released. Reverse transcriptase takes the viral RNA and uses host nucleotides to convert the viral RNA into a single strand of DN. While it does this, it makes random errors, which is typical of Reverse transcriptase, it has very poor proofreading activity. That single-strand DNA is again reverse transcribed, into a double-stranded DNA. At that point, another enzyme that has come in with the virus at the beginning, called integrase, grabs hold of the double-stranded DNA and carries it through a nuclear pore into the nucleus of the cell. Within the nucleus of the cell, it finds the host chromosome, and integrase makes a nick in the host DNA and allows HIV to insert itself into the host chromosome, and that is what causes life-long infection. RNA polymerase comes along and makes messenger RNA. Those mRNA's encode for different viral proteins, they end up associating with ribosomes at the surface of the rough endoplasmic reticulum, and here is a piece of mRNA that is making envelope protein, which is directly produced into the endoplasmic reticulum, and it is shuttled through the endoplasmic reticulum and taken to the cell surface, where at the cell surface, it becomes embedded in the cellular membrane and at this point coalesces with other envelope proteins that have been produced, you have this cluster of envelope proteins on the surface of this infected cell. At the same time, there are other mRNA's being produced, that allows for the translation of other viral proteins. So, here are additional viral proteins being made, which are going to be used to make up the key components that the virus is going to need. These are transported to the cell surface to the area where the envelope proteins are and a strand of RNA, as well as some of the enzymes, are part of that complex. This then buds off at the cell surface at this point, but is still not a mature virion, because the polyprotein chain needs to still be digested into its component parts, that is done by the enzyme called protease, protease breaks up the polyprotein chains and ultimately allows for them to coalesce and form the mature structure that makes up the final virion, and now you have a mature and infectious virion that can go on now and infect other cells.
Which of the following amino acids can be synthesized by bacteria that has reverted to normal due to a mutagen?
Histidine
CpG islands are sequences bearing many cytosine-guanine pairs along the DNA backbone. These islands are often _________________ and a related to __________
Hypermethylated, repair gene suppression
Application 1 What effects would you expect if ddCTP were added at 10% of the concentration of the available dCTP? Deoxycytosine triphosphate (dCTP) ---> Dideoxycytosine triphosphate (ddCTP) Answer There's a one in ten chance than the incorrect nucleotide will be incorporated into the DNA, but... _______________________________________________________________. Used by the drug, AZT, to block HIV infection in AIDS patients. Metal ion A, Metal Ion B, Asp, Asp
It can be added on because the phosphates are there, but You would not be able to continue the DNA strand without the 3' hydroxyl group in the ddCTP, which means the strand will not be able to be continued and lengthened. DNA extension would be terminated. (NUCLEOTIDES WOULD NOT BE ABLE TO BE ADDED)
•Many chemicals are not mutagenic (or carcinogenic) in themselves, but become converted into mutagens and carcinogens as they are metabolized by the body. •The Ames Test is a classic test for chemical mutagenicity.
Metabolism: key to damage
Hotspot mutations •___________________: recently discovered phenomenon in which multiple mutations cluster in a few hotspots in a genome. •Ex. Analysis of 55 spontaneous mutations that occurred within the lacI gene of E. coli revealed that 44 of them involved deaminated methylated cytosine bases. •Kataegis is also observed in ∼55% of ________________________ tumors that results in hypermutation in localized genomic regions. Kataegis: multiple cytosine deaminations resulting in several hotspots in cancer kataegis = Mutation "hotspot" on a chromosome, multiple methylcytosine deaminated to thymine
Kataegis Breast
Final strategy for 3'-5' strand FOR LAGGING STRAND Cannot replicate 5' end Gap remains after primer removal (no way for a primer to fit on that piece, stays unfinished) Solution: long repetitive sequences called telomeres Replication forks reach the end of chromosome ---> Lagging strand, leading strand ---> RNA primers replaced by DNA: Gaps sealed by ligase ----> Lagging strand, leading strand, gap remaining at end of the _________ strand DNA polymerase cannot link these two nucleotides together without a primer, no place for a primer on the 5' end, so it stays unfinished.
Lagging
Transcription of rRNA genes captured in a Transmission Electron Micrograph 1.Which direction is transcription proceeding? Left to right, or right to left? ________________________ 2.Why are the RNA transcripts shorter than the DNA segments that encode them? ____________________________________ Nuclear envelope, nucleolus, cytosol, nucleus
Left to right ( the transcripts get longer and longer as you go to the right) They have started to fold into their secondary structures, condensing their size
Telomeric repeat sequences GGG GGG, Repeated at the end of the gene, called a telomere ADD NUCLEOTIDES TO START WITH Telomere lengthening is catalyzed by telomerase REVERSE TRANSCRIPTASE (TERT), using an RNA template, USES RNA TEMPLATE Synthesizing a 6-nucleotide repeat over and over, finishes a repeat before it starts another one, once long enough, AN RNA primer will hop on and start adding nucleotides Polymerization: Telomerase synthesizes a 6-nucleotide repeat over and over again Translocation: Telomerase moves 6 nucleotides to the right and begins to make another repeat (finish repeat to begin another) After long enough set of repeats and RNA primer hops on and adds nucleotides. The complementary strand is made by primase, DNA polymerase, and ______
Ligase
Depurination though, Guanines in acidic conditions depurinate very slowly •General depurination info •Increased at ______ pH •Thymine rich sequences depurinate faster than others •Rate for single-stranded DNA (ssDNA) is greater than for dsDNA Doubly protonated nucleotide ---> Oxocarbenium
Low
Mutations during translation
MUTATIONS
Ubiquitin •E3 Ubq ligase enzymes "mark" proteins for degradation with a covalent attachment of chain of small proteins called ubiquitin. PROTEASES DIGEST PROTEIN Central cylinder ( where the proteases), stopper --> Active site (target protein with polyubiquitin chain ---> Ubiquitin recycled ---> Protein degraded
Marks it
Genetic diversity HAPPENS ALL THE TIME IN GAMETES WHEN OFFSPRING ARE PRODUCED •Homologous recombination during Prophase I of _______________. DONE PURPOSEFULLY --> HOW WE GET DIVERSITY AND NEW GENETIC MATERIAL AND SPECIES 2 DNA strands (Up to 4) gets nicked ---> Leads to recombinant chromosomes (invasion). ----> Exchange of genetic information
Meiosis
DNA Damage •_________________: agents that cause mutations (sequence changes) to DNA (removal or addition of nucleotides) •Most DNA damage is due to chemical reactions that occur __________ the cell (most threatened by what is inside our cell •Most is temporary •Correction is immediate due to DNA repair DNA Damage response Cellular metabolism, Viral infection, radiation, chemical exposure, replication errors ----> DNA Damages leading to ---> Cell cycle checkpoint activation, the transcriptional program activation, DNA Repair, Apoptosis
Mutagens Inside
Frameshift mutations Nucleotide deletion Nucleotide inserted
Mutations
Nucleotide substitution—Silent mutation Silent mutation: If a base code results in one of the redundant codons for glycine, there is no change in the polypeptide. Gly GGC
NOT an issue usually if replaced with redundant
Contrasts •Replication errors result in incorrect ________________ nucleotides •For example, adenine to guanine •_________________ reactions (reactions caused by water, to break apart a molecule) result in unnatural alterations to DNA
Natural Hydrolytic
CpG islands in colon cancer •Colon mucosa: •Tumors: 1,734 CpG islands heavily methylated •Adjacent mucosa: no methylation Cancer hijacks the repair systems Normal colon mucosa Colon tumor mucosa
No methylation
Application 2 What effects would you expect if ddCMP were added at 10% of available dCTP? Primer, catalysis, base pairing, template ---> Pyrophosphatase (two phosphates released) Deoxycytosine triphosphate (dCTP) ---> Dideoxycytosine monophosphate (ddCMP)
Not only is the hydroxyl group taken away from the 3' carbon. Two of the phosphates are also removed, meaning a pyrophosphate will no longer be able to be released, which will not lock the nucleotide into place preventing the reversal of the process, and not have the energy to attach the nucleotide to the growing strand ---> Blocks the spread of DNA ---> Which is used to help prevent the spread of cancer (which is uncontrolled cellular growth )
Untranslated regions (UTRs) 5' cap ----> 5' UTR ----> Coding sequene ---> RNA ---> Protein ---> Noncoding sequence ---> 3' UTR-----> Poly-A tail 150-250
Not translated into a protein but has a specific purpose
An mRNA can make proteins as long as it persists in the cell. What determines the lifespan of an mRNA molecule?
Nucleotide sequences in the 3' UTR
Comparison: Prokaryotic and Eukaryotic Transcription Prokaryotic *(Much simpler) •RNA polymerase (one type) •RNA polymerase initiates transcription with ONLY a sigma subunit •Genes lie very close to each other on the DNA strand Eukaryotic *(Much more complex) •RNA polymerase (three types) •RNA polymerase I •RNA polymerase II (initiator) •RNA polymerase III •RNA polymerase II requires general transcription factors, a large set of proteins, to initiate transcription •Genes may have up to 100,000 nt pairs between each one (Much bigger genome many more _________)
Nucleotides
Germ cell mutation ONLY IN SPERM OR EGG •Sickle cell trait: an inherited condition •Homozygous inheritance (two mutated β-globin alleles, one from each parent) causes sickle cell anemia. NEED HOMOZYGOUS RECESSIVE TO HAVE SICKLE CELL •Heterozygous inheritance (one normal and one mutated β-globin allele) results in protection from malaria. •The malarial parasite grows poorly in the sickle-cell form of hemoglobin
ONLY IN SPERM OR EGG
Replication with Mismatch Repair Parent DNA molecule ----> Replication (One way it could go, the top strand is replicated correctly with the original parent strand and the new strand) or replication could go another way (Mistake occurs during replication of the bottom strand, new strand with error, original parent strand ----> Mismatch repair --> Original strand restored *Mismatch Repair Protein Machinery repairs only nucleotides on a newly synthesized strand
ONLY NEWLY SYNTHESIZED STRAND
Polycistronic mRNA in Prokaryotes Definition: one mRNA that encodes two or more proteins Polycistronic mRNA contains multiple open reading frames, each corresponding to a single gene transcript Shine-Delgarno sequence: recruits the ribosome to the mRNA to initiate protein synthesis by aligning the ribosome with the start codon AT THE BEGINNING ONLY. ...BUT each new protein has a new initiation sequence Initiation site, ribosome, FIrst polypeptide, Structural gene 1 --> Termination signal --> second polypeptide---> Structural gene 2---> Termination signal ---> Third polypeptide --> Structural gene 3
ONLY ONE SHINE-DALGARNO ONLY AT THE BEGINNING POLY CISTRONIC in PROKARYOTES = 1 mRNA making SEVERAL proteins Polyribosomal in Eukaryotes = 1 mRNA making 1 PROTEIN
There are three potential reading frames for each mRNA molecule. How many will code for the correct protein?
One
RNA as its own template Original RNA ----> Original sequence serves as a template to produce the complementary sequence ---> complementary RNA ---> Complementary sequence serves as a template to produce the original sequence
Own template
•DNA damage may also result from exposure to polycyclic aromatic hydrocarbons (PAHs). •PAHs: potent, ubiquitous atmospheric pollutants commonly associated with oil, coal, cigarette smoke, and automobile exhaust fumes. (Another source is the forest fires)
PAH
Dissociation and release •At start of transcription, most general transcription factors dissociate from the DNA molecule. •At end of transcription, RNA polymerase (also acts as a helicase for eukaryotic transcriptions) is released •_____________________ are removed from the tail by phosphatases •Only the dephosphorylated form can initiate RNA synthesis RNA transcription Result of eukaryotic transcription: mRNA (Compare to result of prokaryotic transcription: ___________) Ribonucleoside triphosphates (UTP, ATP, CTP, GTP) ---> mRNA ---> Transcription
Phosphates mRNA
Repair proteins There are numerous repair proteins for double-strand breaks. (ATM, a repair protein, recruits other proteins and they are phosphorylated) Mutations in BRCA1 and BRCA2 (ON DOWNSTREAM PROTEINS NEEDED TO REPAIR BREAKS) genes account for 10-15% of hereditary breast cancers, unrepaired breaks lead to loss of genetic material and mutations
Phosphorylated
Nuclear export of properly processed mRNA is mediated by nuclear pore complexes (Transcribed from a coding region of a DNA molecule, added a cap and tail to protect it before released to cytoplasm) Nuclear pore proteins must recognize 1) ____________________; 2) _____________________; and 3) properly spliced proteins called _____________________________ Exon junction complex, poly-A-binding protein, cap-binding protein, 5' cap ---> Nuclear pore complex, nuclear envelope, Nucleus -----> Cytosol, protein exchange ---> Initiation factors for protein synthesis ---> translation
Poly-A binding proteins Cap-binding complex Exon junction complexes
PYRIMIDINES AND DIMERS Direct assaults made on DNA strand due to UV light TWO THYMINES LINKED TOGETHER, THEY SHOULD BE LINKED TO THEIR ADENINES NOT TO EACHOTHER, CAUSES A KINK THE TWO CARBONS COVALENTLY BOND TOGETHER, MAKING IT IMPOSSIBLE FOR THEM TO BOND WITH THEIR ADENINE THE KINKS WILL CAUSE DNA POLYMERASE TO FALL OFF WILL NOT BE STABLE CAUSING SERIOUS DAMAGE CAN HAPPEN WITH CYTOSINE AS WELL Direct problems associated with skin cancer •Pyrimidine dimers can disrupt _______________ and prevent proper replication of DNA. •Pyrimidine dimers have cross-links in both cytosine and thymine residues. UV induces the formation of covalent bonds between consecutive bases near carbon-carbon double bonds 2 thymines, with UV light ---> UV radiation,--> Thymine dimer (thymine links together)(kink and a thymine dimer) Consequences of pyrimidine dimers: Stalled DNA synthesis (DOES NOT PREVENT IT, DNA POLYMERASE CAN HOP BACK ON) Mutations Cancer
Polymerase Polymerase trying to go along strand and comes to a kink, it will fall off, causing serious damage during replication
Roles of UTRs •Modulation of the transport of mRNAs out of the nucleus and of translation efficiency, subcellular localization, and stability • ___________________________ regulation of gene expression •5' UTR: major role in the control of mRNA __________________ •Alternative pre-mRNA ________________________ (Serve a purpose in regulation) 5' cap ----> 5' UTR ----> Coding sequene ---> RNA ---> Protein ---> Noncoding sequence ---> 3' UTR -----> Poly-A tail 150-250 Transcription The central dogma of molecular biology, DNA makes RNA makes protein. Here the process begins, transcription factors assemble, at a specific promoter region along with the DNA, the length of DNA following the promoter is a gene and contains the recipe for a protein. A mediator protein complex arrives, carrying the enzyme RNA polymerase, it maneuvers the RNA polymerase into place. Inserting it with the help of other factors, between the strands of the DNA double helix. The assembled collection of all these factors is referred to as the transcription initiation complex and now it is ready to be activated. The initiation complex requires contact with activator proteins, which bind to specific sequences of DNA known as enhancer regions, these regions may be thousands of base pairs distant from the start of the gene. Contact between the activator proteins and the initiation complex releases the copying mechanism. The RNA polymerase unzips a small portion of the DNA helix, exposing the bases on each strand. Only one of the strands is copied, it acts as a template for the synthesis of the RNA molecule, which is assembled one subunit at a time, by matching the DNA letter code on the template strand. The subunits can be seen here entering the enzyme through its intake hole and they are joined together to form the long mRNA strand snaking out of the top. Note the debris, all the different nucleotide swimming around
Post-transcriptional Translation Splicing
Breast cancer prognosis Individuals with kataegis have a more promising prognosis •Conjecture: kataegis mutations dampen the abnormal expression of neighboring genes (Therefore, the tumor is surronded by tissue responding abnormally, dampens the response, tumor cannot grow)
Prognosis
Note asymmetry of promoter; this determines which strand is used for template A. Promoter --> Start site ----> template strand (DNA)---> Transcription ---> RNA B. Terminator ---> Template strand ---> DNA, stop site ---> transcription ---> RNA _____________________ sequence is not transcribed into the mRNA molecule _______________sequence is transcribed into an mRNA molecule
Promoter is a section of sequence that proceeds upstream, start site 0s +1, -10 upstream site, to -35 upstream site. Has to have this tp transcribe a gene Terminator, Transcription occurs to look like the coding strand, a stop site is reached Promoter Terminator
Promoters and CpG Islands (ON THE SAME STRAND OF DNA NOT ACROSS FROM EACH OTHER) •Definition: Sequences heavy in "C" and "G" along the DNA backbone (versus the C-G bond in the DNA interior). The cytosine is 5' to the guanine. •Typically 300-3,000 base pairs in length •Found in or near 70% of the ___________ of human genes. •CpG island hypermethylation described in almost every tumor type. Where? ____________________________________________________________________ •CpG islands affect •DNA repair •Cell cycle •______________________________ •______________________________ Detoxification Cell won't be able to destroy itself when the CpG island is methylated Unmethylated ---> Gene, 5-methylcytosine Methylated CpG Island, CpG island ----> Gene Expression repressed, Gene
Promoters CpG island is methylated. Found on the promoters, in the TATA box, where transcription is needed, but in this case, it is halted due to the CpG methylation Apoptosis Cell adhearance
Spliceosome -Pre-catalytic spliceosome (U5 RNA, Intron, U1 RNA, 5' exon, 3' exon, U2 RNA, U6 RNA, U4 RNA) Catalytically activated spliceosome, (U5 RNA, Intron, 5' exon, 3' exon, U2 RNA, U6 RNA) •Eukaryotic mRNA splicing is carried out by small nuclear ribonucleoproteins-- a substance composed of both __________________ and __________________ acid. (snRNPs) forming RNA-protein particles •Spliceosome splicing snRNPs complexes include: U1, U2, U4, U5 and U6 snRNPs (Splice out the introns you dont need)
Protein Ribonucleic
Comparison: DNA replication and RNA transcription DNA replication (making more of itself) •New strand remains H-bonded to the DNA template strand (one leading strand and lagging strand come back together to get the daughter) •Length of DNA molecules in a human chromosome: up to 250 million nucleotides long RNA transcription (using the code of DNA to transcribe RNA from DNA) •The RNA strand is displaced from the DNA molecule after the formation (does not stay attached, has work to do, emitted from the nucleus and expelled into the cytoplasm) •Length of RNA molecules: up to a few thousand nucleotides long (not as long as the DNA molecule) (Transcription --> RNA) RNA reads the code on the DNA molecule and then ejected to do work by forming _______ (DNA ---> RNA polymerase
Proteins
Spontaneous damage 1 •Depurination: Removes a _____________ base from a nucleotide •At apurinic sites the covalent structure of DNA becomes more susceptible to damage •Potential mutagenesis, carcinogenesis and cellular aging •Approximately 2,000-10,000 DNA purine bases are released in each human cell every day
Purine
How puromycin terminates translation 1.Puromycin binds to the A site, substituting for an tRNA. 2.Puromycin is small compared to a tRNA; cannot retain the polypeptide chain on the ribosome 3.The polypeptide dissociates from the ribosome, incomplete Peptidyl-tRNA in the P site ---> Peptidyl transferase center ---> Puromycin in the A site ---> Peptidyl puromycin
Puromycin CANNOT BE USED IN HUMANS BECAUSE IT DESTROYS ALL TYPES OF CELLS
Recycling proteins •Protein life-span •Metabolic proteins: (enzymes, growth regulators, etc.)—days, hours or seconds •Structural proteins: (bone, muscle)—months or years •Protein breakdown: proteolysis •Accomplished by proteases, enzymes to degrade proteins •Short peptides to individual amino acids •Proteolytic pathways 1.Rapidly degrade proteins, as needed 2.Remove proteins that are damaged or misfolded
RECYLCING ENZYMES AMINO ACIDS ARE CONSERVED AND REUSED
Telomere lengthening is catalyzed using a(an) ________ template
RNA
From DNA to protein, the entire point of having a long set of DNA molecules is to direct the activities of the entire organism, keeping it pure and unobstructed, in areas to be transcribed and areas to be silenced An "RNA World" (Possible that all life forms started with RNA) One evolutionary theory for the origin of life states _____ was the only nucleic acid. RNA: a linear polymer made of four different nt subunits linked together by phosphodiester bonds Many functions of RNA: 1._____________________________ 2._____________________________ 3._____________________________ 4._____________________________ In a "DNA World": Segments of DNA called genes are used to direct the synthesis of RNA. RNA world ---> RNA ---> Protein A DNA world Central Dogma ---> DNA ---> RNA ---> Protein
RNA 1. RNA synthesizes proteins (a major part of all cells function and structure) 2. Stores information and remembers what it needs to have to continue processing proteins 3. Catalyzes chemical reactions 4. Regulate gene functions Approximately 30% of our genome is read by a cell, only certain genes are needed at any time
Types of RNA produced in cells Type of RNAs Function Messenger RNAs (mRNAs) ---> Code for proteins Ribosomal RNAs (rRNAs) ----> Form the core of the ribosome's structure and catalyze protein synthesis (made in the nucleolus inside the nucleus of the cell) MicroRNAs (miRNAs) ----> Regulate gene expression Transfer RNAs (tRNAs) ---> Serve as adaptors between mRNA and amino acids during protein synthesis Other noncoding RNAs ---> Used in RNA splicing, gene regulation, telomere maintenance, and many other processes
RNA is essential to life
Circular dilemma Nucleic acids needed to direct synthesis of protein ...but proteins are required to synthesize nucleic acids (primases, polymerases, ligases...) Solution: an RNA world DNA ---> Transcription RNA ---> Translation Protein ---> Replication, DNA
RNA preceded DNA
How many bases = 1 codon? Crick and Brenner's reading frame experiments (made a set of mutations in one gene that lacked 1, 2 3, or 4 bases, etc.) Original WHY DID THE BAT EAT THE FAT NAT Delete 1 WHY IDT HEB ATE ATT HEF ATN AT Delete 2 WHY DTH EBA TEA TTH EFA TNA T Delete 3 WHY THE BAT EAT THE FAT NAT Delete 1 or 2 bases, makes nonsense. Deletion (or addition) of exactly 3 bases makes sense, because it restores the proper __________ ___________. (STILL HAVE DAMAGE) →Will it provide the correct protein? THE BIG QUESTION Conclusion: One codon contains exactly 3 bases
Reading frame
Hyper/hypo-methylated promoters in cancers •DNA repair genes are frequently _________________ in cancers due to hypermethylation of CpG islands within their promoters •Head and neck squamous cell carcinoma: At least 15 DNA repair genes have frequently hypermethylated promoters Promoter hypermethylation of the DNA repair gene MGMT (used to restore mutated guanine) occurs in: 93% of bladder cancers 88% of stomach cancers 74% of thyroid cancers 40%-90% of colorectal cancers 50% of the brain cancers
Repressed
The Ribosome Large ribosomal subunit (Contains E, P, and A) Small ribosomal subunit (mRNA binding site) E site = Exit (release of what has been growing on a polypeptide attached to the growing strand of peptides) P site = Peptidyl tRNA (where all beginnings of translation occur) A site = Aminoacyl tRNA (tRNA bringing new amino acids to the A site)
Ribosome
The ribosome is a Ribonucleoprotein Can be connected to the endoplasmic reticulum (create proteins in the nucleus) or free ribosomes in the cytosol (create proteins for anything else) •Includes a large subunit and a small subunit •Most of the subunits are composed of protein Large subunit = 60s, 49 ribosomal proteins and 3 rRNA molecules, MW= 2,800,000 (Catalyzes peptide bond formation) ---> Small subunit = 33 ribosomal proteins + 1 rRNA molecule, MW= 1,400,00, 40s, (Matches tRNA anticodons to mRNA codons) ---> Complete eukaryotic ribosome = 80s, large unit and small subunit, 82 different proteins + 4 different rRNA molecules , complete eukaryotic ribosome = MW= 4,200,000
Ribosome
Molecular Fossils •Ribozymes: RNA catalysts (work by themselves with NO proteins) •Still work in ____________________ and ___________________________ RNA-based systems ---> RNA ---> Evolution of RNAs that can direct protein synthesis ----> RNA and protein-based systems ----> RNA -> Proteins ---> Evolution of new enzymes that synthesize DNA and make RNA copies from it --> Present-day cells DNA---RNA ---Protein---> DNA
Ribosomes and RNA splicing machinery
mRNA splicing allows greatly expanded gene expression of the 19-20,000 genes of the human genome. Splicing catalysis is completed by __________
Ribozymes
Polycistronic RNA in Prokaryotes The Shine-Dalgarno Sequence: a ribosomal binding site about 8 base pairs upstream of the AUG start site in bacterial and archaeal mRNA. It helps recruit the ribosome to the mRNA to initiate protein synthesis by aligning the ribosome with the start codon. Compare Shine-Dalgarno Sequence with the start site for eukaryotes. The small ribosomal subunit must scan the mRNA to find the AUG start site. mRNA, 165 rRNA --> Shine-Dalgarno sequence ---> Start codon
SHINE-DALGARNO IS DIRECTLY ON THE mRNA
Mismatch repair •Definition: a system for recognizing and repairing erroneous insertion, deletion, and incorrect bases, as well as for repairing some forms of DNA damage. •Removes______________________ _______________________________ •Requirement: Double-helical structure of DNA to provide two copies of the genetic information (always have a template) Compare: base-excision repair is only for small, non-helix-distorting lesions, while in mismatch repair is for helical distorting lesions Single-strand break, mismatch, the damaged base
Sections of nucleotides (not just one nucleotide)
Repair summary DNA Repair Mechanisms
See video
Prokaryotic transcription requires a dissociable subunit of RNA polymerase called the
Sigma factor
Homologous recombination Mechanism of homologous recombination conserved in virtually all organisms on earth. Double strand break --> 2 double-stranded Daughter DNA molecules (one from mother and one from father)---> Nuclease digests 5' ends of broken strands ---> Strand invasion by complementary base-pairing--- Repair polymerase synthesizes DNA (Green) using undamaged complementary DNA as template ---> Invading strand released: broken double helix re-formed ---> DNA synthesis continues using complementary strands from damaged DNA as template ---> DNA ligation ----> Double-strand break is accurately repaired. PREFERRED METHOD OF DOUBLE STRAND BREAK REPAIR, NO GENETIC MATERIAL LOST Repair must happen due to a lot of the mutations happening from _______impact
Solar
Eukaryotic mRNA splicing The splicing catalysis (making and breaking bonds) is carried out by RNA, not by protein. (remove the introns) It base pairs with sequences that signal splicing First example of a ribozyme (That scientists ever discovered) Exon 1 on the 5' end with U1, RNA portion of snRNP base-pairs with sequences that signal splicing, along with U2 on Exon 2 on the 3' end (portion of pre-mRNA) ------(Binding of Additional snRNPs; Assembly of spliceosome ----> ( U4/U6, U1, U5, U2 ) ---> Splicing Splicing ----> U1 Splicing ---> Excised intron in form of a lariat (A, U2) (lariat is the excised intron) Splicing ----> 5' exon 1, 3' exon 2, portion of spliced mRNA
Splicing
Ionizing and non-ionizing radiation Non-ionizing radiation (Cell phones, indirect) ---> Ionizes Water ---> Free radical damages DNA Ionizing radiation (______) directly damages DNA
X-rays
Benefits of DNA vs. RNA •The deoxyribose phosphate backbone is much more __________________, allowing long strands of DNA without breakage (NOT TRUE FOR RNA) •Can use parent strand as a ____________ for repair •Deamination of ____________ to _____________ alerts the repair system •Proteins provide a much richer source of structural components and enzymes (MAKE MORE ADVANCED SYSTEMS DUE TO MORE ADVANCED CHEMICAL GROUPS)
Stable Template Cytosine to Uracil
The Ames test for mutagens Use Typhimurium: a type of salmonella causing typhoid fever and food poisoning to test out whether the body can take a potential mutagen and alter the pathology Dr. Bruce Ames, an American biochemist. Developed the Ames test in the 1970s. The Ames test includes a mixture of liver enzymes to mimic the effect of metabolism on the chemical. In order for a bacteria to grow, must have all 20 of the amino acids, if one is removed or not functioning it can not replicate. Absolutely require histidine for growth. But we have to add something to show that the body itself can cause a change from a regular strain of Typhimurium, to a potent and lethal one. LIVER ADDS MUTAGENICITY PROBLEM IS METABOLISM CHANGING EFFECT OF MOLECULES Take out a rat liver extract, in the liver of mammals, that the change can take place to add mutagenicity, put rat-liver extract along with the Typhimurium that lacks histidine, the mixture of liver enzymes mimics the effect of metabolism on the chemical, incubated. Colonies of mutated Typhimurium that revert, are now histidine positive, are shown in this test once the liver extract is added. The body itself can cause a change to make histidine available caused by metabolism. The problem ended up being metabolism, taking in molecules, and changing its effect.
Stain of S typhimurium which is His Neg + Rat liver extract ---> They required histidine for their growth ---> Culture plate with growth medium (lacks histidine) ----> Colonies of mutated S typhimurium- revert and is not His positive
The following DNA CODING sequence includes the beginning of a sequence for a protein. What would be the result of a mutation that changed the underlined G to a T? 5′-AGGCTAATGATCgAGACTGCGAGCCC...3'
Stop
RNA and the origins of life •Hypothesis: RNA served two purposes that allowed it to stand alone 1.___________________________________ 2.______________________________________________________________________ Support: Ribose is readily formed from _________________, a principle product on primitive earth: CH2O (possible to have an RNA world) Formaldehyde and Ribose
Stored genetic information Catalyzed chemical reactions in primitive cells Formaldehyde
One of the benefits of having a DNA-based world versus an RNA-based world is greater stability of its
Sugar-phosphate backbone
On the coding strand of a eukaryotic DNA molecule, the transcript includes a core promoter with a transcriptionaal start site and often a
TATA box
Eukaryotic Transcription Eukaryotic transcription features •Regulatory DNA sequences •Has a core promoter that includes •______________: feature of many eukaryotic promoters •Usually located 25 nts upstream from the transcription start site •_________________________(+1) •Must "unpack" DNA from nucleosomes and other chromatin structures *(become euchromatin and availible) Downstream is where transcription is going to occur to give us the mRNA Coding-strand sequences ---> Core promotor --> Transcriptional start site, TATA box (Upstream <---): Transcription downstream -->
TATA box Transcription start site (+1)
DNA instructions for the desired protein will be located on the coding strand. The mRNA molecule is, therefore, made by base-pairing with the _______________
Template strand
Which of the carbon positions indicated by different letters on the ribose sugar (below) is the main biochemical difference between DNA and RNA building blocks?
The 2' carbon
The figure below shows a ribose sugar. The part of the ribose sugar where a new ribonucleotide will attach in an RNA molecule is pointed to by arrow ____.
The 3' Carbon
Holliday junction •Holliday junction •Holliday junction results
The Holliday junction, an important intermediate structure in homologous DNA recombination, is formed when two homologous double-stranded DNA molecules reciprocally exchange DNA strands, this junction can be visualized directly with an electron microscope. In the cell, this junction is formed and stabilized by a specific group of helicase proteins, seen here in the background, which use ATP hydrolysis to move the junction up and down the DNA as shown in this animation. Holliday junction is when two homologous duplexes are linked by the reciprocal exchange of a pair of DNA strands. The formation of a Holliday junction is initiated by the formation of a D-loop, between homologous duplexes in which a strand of one duplex, in grey, invades the other duplex, in pink. Regions of heteroduplex DNA are formed in which complementary strands from the two different duplexes are paired. Cleavage of the D-loop and subsequent ligation results in a Holliday junction that links the two duplexes. Holliday junctions adopt a planar unfolded structure when they interact with recombination proteins. Notice that both of the duplexes contain both homoduplex (grey with grey, pink with pink), and heteroduplex (grey with pink). Also, notice that the Holliday junction involves four DNA strands, there are two axes of symmetry in the Holliday junction, one horizontal and one vertical. Two strands of the four-strand Holliday Junction, are cleaved by a symmetrical nix made by a specialized recombination enzyme called the resolvase, this cleavage can occur across either the horizontal or vertical axes. Note the very different consequences of the two cleavages, upon cleavage along the horizontal axis, the parental segments, Big A and Big B, and little a and Little b, are still covalently linked with each other. Cleavage along the vertical axes, however, leaves Big A in grey and little b in pink and little a in pink and big B in grey, covalently linked to each other. Ligation of the cleaved strands generates pairs of progeny duplexes, Holliday junction cleavages along the horizontal axes result in product duplexes in which the parental segments remain covalently linked they have not been exchanged. These duplexes are called non-recombinant products. By contrast, Holliday junction cleavages along the vertical axes result in product duplexes in which parental duplexes have been exchanged, in other words, fragment segments from each parent are covalently linked in the product duplexes, for example, Big A in grey is now joined to little b in pink, the progeny duplexes are called recombinant products. Both pairs of progeny duplexes also contain regions of heteroduplex DNA in which complementary strands of each parental duplex are paired. Thus, formation and resolution can result in the permanent exchange of DNA signals between homologous and parental duplexes.
In the case of cancer, what is the outcome of mutated p53 due to pyrimidine dimers?
The cell cannot self-destruct
Pyrimidine dimers: Application 1 p53 prevents mutations and cancer P53 PROTEIN IS THE MOST PROTECTIVE ANTI TUMOR MOLECULE IN THE ENTIRE BODY , PROTECTS BODY FROM CANCER P53 TRIGGERS CELL SUICIDE, THE PYRIMIDINE DIMER PREVENTS THE DIMER FROM KILLING ITSELF We make 3*10^11 new cells per day. •Pyrimidine dimers are the most frequent DNA mutations found in the p53 protein in skin cancers. • What is the outcome of mutated p53? _____________________________ p53 tumor suppressor protein triggers cell suicide Normal cell ---> p53 protein, excessive DNA damage ---->Cell suicide (apoptosis) Normal process
The cell will not preform apoptosis, tumor growth will occur and continuous replication of the damaged cell can continue Cancer disables the protective systems in our body so it can proliferate
Application: Telomerase and Cancer •Telomerase activity is DECREASED after birth, leading to shortening of telomeres as aging progresses •In cancer, there's an increase in telomerase activity. (CANCER CELLS WANT CELL DIVISION CONTINUOUSLY, telomerase helps that ) •In lab studies, cancer cells treated with drugs that inhibit telomerase often stop dividing. (Provision to halt cancer cells in their tracks •Should cancer treatment include telomerase inhibitors? If we inhibit telomerase in cancer cells, we would have to direct this to cancer cells only, other healthy cells life spans would decrease as well) •Telomere lengthening
The ends of linear chromosomes pose unique problems during DNA replication because DNA polymerases can only elongate from a free 3' hydroxyl group, the replication machinery builds the lagging strand by a back-stitching mechanism. RNA primers provide 3' hydroxyl groups at regular intervals along with the lagging strand template. Whereas the leading strand elongates continuously in the 5' to 3' direction all the way to the end of the template, the lagging strand stops short of the end. Even if a final RNA primer was built at the very end of the chromosome, the lagging strand would still not be complete. The final primer would provide a 3' hydroxyl group to synthesize DNA, but the primers would later need to be removed the 3' hydroxyl groups on adjacent DNA fragments provides starting places for replacing the RNA with DNA, however at the end of the chromosome, THERE IS NO 3' hydroxyl group available to prime DNA synthesis. Because of this inability to replicate the ends, chromosomes would progressively shorten during each replication cycle. This end replication problem is solved by the enzyme telomerase. The ends of chromosomes contain a G-rich series of repeats called a telomere. Telomerase recognizes the tip of an existing repeat sequence. Using an RNA template within the enzyme, telomerase elongates the parental strand in the 5' to 3' direction an adds additional repeats as it moves down the parental strand. The lagging strand is then completed by DNA PRIMASE ALPHA, which carries a DNA primase as one of its subunits (AFTER THE RNA PRIMER IS ADDED). In this way the original information at the ends of linear chromosomes is completely copied in the new DNA
Lifetimes of mRNA One mRNA can make proteins as long as it persists in the cell. mRNA lifetimes vary •Bacteria: ~3 minutes •Eukaryotes: varies from <30 minutes to >10 hours (e.g., mRNA encoding β-globin •How is the lifetime controlled? •_________________________ •General rule: Small protein amounts needed = shorter lifespan and vice versa Gene A --> Transcription RNA ---> Translation Protein
The nucleotides at the 3' UTR
In the snRNPs of the "spliceosome", what actually mediates identification of splice sites and excision/ligation?
The small RNA molecules
A different dilemma: 5-methylcytosine •Definition: a naturally-occurring base invertebrates What would be the purpose of this base? METHYL GROUP SILENCING → Transcriptional silencing Cytosine --> 5-methylcytosine
Transcriptional Silencing
All mutations involve changes to the DNA sequence. Not all mutations, however, are damaging to a genome. These are called "silent" mutations. What type of mutation can result in a silent mutation?
Transition
Transitions and Transversions _____________________: point mutation in DNA that substitutes a pyrimidine (single-ring) for another pyrimidine or a purine (double-ring) for another purine. •Can result in a "silent substitution" (may not cause damage to the genetic code in anyway, because still purine for a purine, and may be able to be still used) •______________________: point mutation that substitutes a pyrimidine for a purine •Can result in damage to the genetic code for proteins
Transition Transversions Prymidine ---Pyrimidine Purine to purine Transition T--A A ---G Transversions Pyrimidine to purine T to A or G C to G or A Purine to pyrimidine A to T or C G to C or T
Translation Ends at a Stop Codon •Protein release factor (PRF) mimics the overall shape and charge distribution of a tRNA molecule (PRF is a protein, mimicking a tRNA molecule). •The presence of a release factor in the A-site of the ribosome causes peptidyl transferase to catalyze the addition of a WATER molecule to the peptidyl-tRNA in the P-site. •This reaction frees the carboxyl end of the peptide so that it detaches from the tRNA in the P-site and is released into the cytoplasm (all parts detach and you get the free polypeptide) WATER ENDS THE POLYPEPTIDE GROWTH Terminal portion of mRNA --> Binding of release factor to the A site ---> H2O comes in Released polypetide chain ---> Termination ---> Ribosome dissociates The completed polypeptide is terminated and released by being hydrolyzed from the tRNA
Translation
Steps to complete mismatch repair 1.Remove the damaged DNA Cleave covalent bonds that join the nucleotide(s) to the rest of the strands 2. Repair DNA polymerase binds to 3'-hydroxyl end of cut strand. Fills in gap with complementary nts based on the _____________________ strand •3. After the gap is filled in, the DNA backbone is sealed by DNA ligase
Undamaged
Emergence of DNA •First step: formation of U-DNA (_____________________________) U-DNA may have first appeared in a virus, making this first U-DNA organism resistant to the RNAses of its host (Some modern viruses have a U-DNA genome) •Second step: __________________________(modification of dUMP into dTMP) Big Bang ---> Time (billions of years ago ---> Solar system formed ---> ( RNA world) First cells with DNA --> First mammals --> Present
Uracil DNA Selection of the letter T
The DNA profiles of an ape and human are nearly 99 percent the same
Very close
Sex-determination genes Whale Human
Very close T for a C A to a G A to a G G to an A Only 4 changes
Wild-type hemoglobin DNA CTT, GAA --> mRNA GAA--> normal hemoglobin Glu Mutant hemoglobin DNA CAT, GTA --> mRNA GUA--> sickle-cell hemoglobin Val
Wild-type = normal Mutant = bad A mistake in just one base leads to GUA instead of GAA, causes a valine instead of Glu === Misshapened blood cells.
Replication without Mismatch Repair Parent DNA molecule ----> Replication (One way it could go, the top strand is replicated correctly with the original parent strand and the new strand) or replication could go another way (Mistake occurs during replication of the bottom strand, new strand with error, original parent strand ----> Replication without repair --> Strand with error, newly synthesized strand (mutated DNA molecule AU, both base pairs are wrong DAMAGE) or Newly synthesized strand and original parent strand (normal DNA molecule)
Without
Enzymes for transcription •Bacteria: RNA polymerase •Eukaryotes: •RNA polymerase II—transcribes the majority of genes including all that encode proteins and microRNAs (miRNAs). •RNA polymerase I and III—transcribe genes encoding ribosomal RNA, transfer RNA, and RNAs with structural and catalytic roles •Reverse transcriptase—transcribes DNA from RNA Primase? RNA polynucleotides are started ___________________ a primer Hypothesis: RNA does not need to be as accurate as DNA replication (can be stalled, completely destroyed) Makes one mistake for every _________ nucleotides copied RNA polymerase: Prokaryotic RNA polymerase and Eukaryotic RNA polymerase
Without 10^4
Wobble rules •The _________________ rules enable a single type of tRNA to recognize more than one codon. (WOBBLE POSITION IS THE IN THE THIRD POSITION IN THE CODON) •For example, a tRNA with an anticodon sequence of 3′-AAG-5′, which carries phenylalanine, can recognize a 5′-UUC-3′ and a 5′-UUU-3′ codon. •Humans have 48 different anticodons Mismatch is tolerated in the third position. Proposed reason: it may not have to hydrogen bond precisely. (proteins may be able to be produced without such precision) Location of wobble position revised wobble rules
Wobble
Rapid Synthesis •Synthesis of the next RNA can start before completion of the first RNA from the same gene How?_______________________________________________ •Dozens of polymerases can utilize the same stretch (gene) of DNA consecutively. •Medium-sized gene: ~1500 nt pairs RNA polymerase ---> DNA template strand ---> Newly synthesized RNA transcript ---> Active site of polymerase ---> Ribonucleotide triphosphate tunnel ---> Incoming ribonucleoside triphosphates
You simply add a new RNA polymerase, that attaches to the beginning of the gene as soon it is transcribed into an mRNA
What does the large ribosomal subunit generally do? a. mediate peptide bond formation b. matches anticodon in tRNA to codon in mRNA c. participates in pre-initiation complex d. a and b e. b and c
a.
Deamination: Mistake in metabolism? •Useful for excess protein intake. Why? •Deamination is used to break down _____________________ for energy. The amino group is removed from the amino acid and converted to __________________. •Where does the ammonia go? •Converted to ________ and excreted as urine People in the US consume a lot of meat Serine ---> Deamination (NH3) ---> Pyruvic Acid ---> Citric acid cycle ---> CO2 +H2O + Energy or Pyruvic acid ----> Gluconeogenesis ---> glycogen (ENERGY STORED IN MUSCLE AND LIVER)
amino acids Ammonia Urea
Which one of the following enzymes is responsible for faithful execution of the genetic code"?
amino acyl-tRNA synthetase
Aminoacyl-tRNA synthetases •Most organisms have a different synthetase enzyme for each amino acid •Humans have nearly 500 different tRNA genes but only 48 anticodons represented •Must recognize specific nts in BOTH the ___________ and the amino-acid accepting arm of the correct tRNA
anticodon
When RNA is transcribed off of a eukaryotic DNA template, what determines which strand the RNA polymerase binds to and orients to begin transcription? a. the asymmetric promoter sequence b. the transcription factors that recognize and bind the promoter sequence and the polymerase c. a and b d. transcription factors that recognize and bind the sigma factor e. a, b, and d
c.
Mutant genes "key to long life" Proceedings of the National Academy Sciences, team from Albert Einstein College of Medicine. Ashkenazi Jewish community studied. (FROM EUROPE AND OUTSIDE THEIR ORIGINAL TERRITORY IN THE MIDDLE EAST) Blood samples were from: • 86 very old people, but generally healthy, people with an average age of 97 • 175 of their offspring • 93 other people who were the offspring of parents who had lived a normal lifespan and could therefore make up a control group (78 for men in USA and 83 for men in Japan) Researchers found that the centenarians and their offspring had HIGHER levels of telomerase and significantly longer telomeres than the unrelated people in the control group and that the trait was strongly heritable. Telomerase can REPAIR THE TELOMERES, preventing them from shrinking. ( The older we get our DNA molecules to lose a few nucleotides at the end with every cell division, telomeres shorten as we age) Possible to produce drugs that stimulate the enzyme? Yes, But, as one commentator noted, "giving the cells more chances to divide may increase the chances of damaging mutations developing and causing ______." (Little chance to alter genome and not have negative side effects) BBC News, Nov. 15, 2009, http://news.bbc.co.uk/2/hi/health/8359735.stm
cancer
8-hydroxyguanine (8-oxoG) Guanine and cytosine is a standard GC base pair, however when guanine is turned upside down and oxygen added, guanine can no longer pair with cytosine, instead guanine pairs with adenine, 8-oxoG pairs with Adenine Oxoguanine glycosylase removes 8-OXOG during base excision repair. Without repair: will give rise to a G:C to T:A transversion, one of the most common mutations found in human _______
cancers
What feature of RNA allows it to fold into complex three-dimensional structures? a. complementary base pairing b. flexible single-stranded phosphodiester backbone c. inter-strand hydrogen bonding d. all of the above
d. all of the above
What is the consequence of an unrepaired depurination mutation in the next round of DNA synthesis?
deletion of a base pair
What does the small ribosomal subunit generally do? a. mediate peptide bond formation b. matches anticodon in tRNA to codon in mRNA c. participates in pre-initiation complex d. a and b e. b and c
e.
Aminoacyl tRNA Synthetases Responsible for charging (activate it with energy) the correct AA to a specific tRNA Uses nucleotide sequences in the anticodon and acceptor arm of tRNA Amino acid (tryptophan) ---> tRNA, and tRNA Synthetase (tryptophanyl tRNA synthetase) ---> Linkage of Amino acid to tRNA using the energy from ATP --> High energy bond ----> Anticodon in tRNA binds to its codon in mRNA ---> Anticodon in tRNA base-pairing ---> Codon in mRNA ---> Net result: AMino acid is selected by its codon in an mRNA Uses energy of ATP to join Amino Acid to 3' end of tRNA ANTICODON PAIRS PERFECTLY TO THE CODON IN mRNA Attached amino acid (Phe) 3' end ---> Anticodon loop and anticodon (terminal 3' nucleotide of appropriate tRNA) --> Adenine ---> High energy ester bond ---> Aminoacyl-tRNA
mRNA
CBD •Carboxyl-terminal domain (CBD) on RNA polymerase II •Phosphorylated •Carries the ________________, splicing components, and ______________________ and cleavage factors •Exchanges initiation factors for those linked to elongation and termination Capping enzyme ---> RNA polymerase II ---> RNA ----> Components of splicing machinery --->Polyadenylation and cleavage factors Transcription ---> RNA
poly-A binding proteins, cap-binding complex (necessary for processing of mRNA to be completed)
Addition of the Poly-A Tail •Poly-A "signal sequence" in DNA indicates termination of the RNA •Capping enzyme --> RNA polymerase 2, RNA ---> Components of splicing machinery ---> polyadenlyation and cleavage factors • •Cleavage and polyadenylation factors are transferred to the RNA transcript, indicating point of cleavage •The poly-A tail is added by a different enzyme: _____________________________ Poly-A signal sequence in DNA ---> Poly-A signal sequence in RNA (CstF, CPSF, 5' Cap, RNA) ----> RNA cleavage (CstF) ----> 5' cap (CPSF) -----> Poly-A polymerase (PAP) ---> PAP, 5' CAP ---> Poly-A-binding protein ---> 5' cap ---> additional poly-A-binding protein ----> 5' cap A200
poly-A polymerase adds the tail
The splicing challenge •Initially transcribed mRNA: ________________________. •Final transcript: ___________________ mRNA •The majority of mature mRNAs are 2-4 kb, but primary transcripts can have many introns, each typically 0.5-10 kb (take out more introns, then the amount of exons in the strand) •Question: Why rely on splicing instead of increase the number of genes? •Speculation: it allows an organism to carry fewer genes in its genome (saves energy?), but its the gene amount thats smaller, not the organism
pre-mRNA Mature mRNA
Splicing to rescue protein expression •Two possibilities: 1.Induce aberrant splicing to disrupt gene expression of proteins involved in the pathogenesis 2.Skip exons that have mutations or deletions that disrupt protein-coding Alternative splicing --> 5' 3' splice junctions ---> Splicing repressor ---> A splicing repressor prevents the recognition of a 3'-splice junction. The next 3'-splice junction that precedes exon 3 will be chosen ---> Exon 2 is skipped and not included in the mRNA
protein expression
Alternative RNA splicing expands the repertoire of proteins Proteins are "modular" and made up of related domains Exon 1, 2, 3 on the DNA ----> Transcription (pre-mRNA )---> Alternative PRE-mRNA splicing ----> three alternative mRNAs Family 1 and Family 2 = Single-domain protein families A two-domain protein family
protein families
Prokaryotic Transcription Bacterial transcription: Recognizing the start site *__________________: a dissociable subunit of RNA Polymerase → only used in __________________ Start site ---> Promoter and RNA polymerase ---> Gene, template strand, with terminator and stop site ---> RNA Synthesis begins ----> Sigma factor released ---> Polymerase Clamps down on DNA: RNA synthesis continues ----> Growing RNA transcript ---> Termination and release of Bothe polymerase and completed RNA transcript --> Gene ---> Sigma factor rebinds
sigma factor Bacteria Once the promoter is set, the transcription can being with the sigma factor attached --> Stop site and teminator site is reached ---> Sigma factor is released as the RNA synthesis begins ---> Polymerase clamps down ---> RNA transcript is released along with the polymerase ---> Sigma factor rebinds to the RNA transcript