Exam 3

What do human mitochondria genome encode? What are they encoded by?(L.15)

human mitochondria genome encodes only a few genes, including ribosomal RNAs and tRNA and a few mitochondrial specific proteins Most mitochondria proteins are encoded by nuclear genes and imported into the mitochondria.

Image: The human genome composition. (L.15)

"miscellaneous unique sequences"- include the regions 5' to 3' if each transcription unit that contain cis regulatory elements. When added to introns and protein coding sequence, about 39% of human genome is a unique sequence. the rest is some sort of repetitive DNA

Lagging Strand synthesis (L.16)

- As in leading strand synthesis, primase makes RNA primers and DNA pol III adds dNTPS - Synthesis proceeds with new dNTPs added to the 3' end of the new strand. This strand is elongated away form the replication fork. - one asymmetric DNA Pol III complex's synthesizes both strands! How? The DNA of the lagging strand loops around.

Review of DNA polymerases synthesizing DNA. (L.16)

-DNA polymerases always synthesize the new strand of DNA from the 5' direction to the 3' direction (shown in orange) That means that the polymerases travels down the template strand (shown in blue) in the 3' to 5' direction. - We also learned that DNA polymerases can only extend DNA that has an existing 3' hydroxyl group on a primer sequence. - we know that replication is semi conservative and that both daughter strand (shown in orange) are synthesized at the same time.

Since Pol III sub complex can polymerize DNA but has much poorer processivity than needed to replicate the genome, how does it compensate? (L.16)

-The complete pol III holoenzyme includes sets of donut shaped dimers which form B(beta) sliding clamps that greatly increase processivity. -Shown in purple and grey in the figure. On the right, are two views of one of these clamps, showing that t wraps completely around the DNA, which is shown in the center. The clamps hold Pol III holoenzyme on the DNA, giving it much higher processivity than either pol I or pol III.

How many micro RNA genes do humans encode?(L.14)

1,800 microRNA

What are the two parts of the histones in the nucleosome? (L.15)

1. (Core) One part contributes to the compact core structure around which the DNA wraps. -in the figure that is the protein surrounded by DNA. There are many alpha helices within this structure 2. Histone tails -amino terminal part, not part of the core. - unstructured protein domains. They extend out from the core and can adopt a variety of structures depending on the interactions with proteins nearby. -the histone tails in the image are shown in an extended form but in reality they will undergo a variety of motions and/or interact with various other proteins.

What is required for the replication of E.Coli genome? (L.16)

1. Replisome 2. primase

Important information from the figure.(L.15)

17% of the human genome os compromised of many copies [repeat sequences] of this 7 kb LINE transposon, and 13% of this 300 bp SINE transposon.("All repeat")

How many codons code for amino acids? What are the rest? (L.18)

61 of the 64 codons code for an amino acid. The other three codons are called stop codons. These stop are used to indicate the ends of a sequence that codes for a protein. As mentioned earlier, the codon AUG indicates where the reading frame should start. AUG codes for the amino acid methionine. So all proteins start with the same amino acid, methionine, though in a number of cases the met is subsequently removed. You can see in the bottom left, the first two nucleotides of this mini mRNA are not read as they do not contain an AUG. The frame that is read starts with AUG. The reading frame codes for two additional amino acids, Phe and Ser, and then there is a stop codon. So any nucleotides 3' of the stop codon do not code for protein. An open reading frame is the stretch of sequence that starts at a start codon and ends at a stop codon. So the sequence that codes for Met, Phe and Ser is an open reading frame. In a random DNA sequence a stop codon will occur about once ever 22 codons in each of the three reading frames (64/3 = 21.3). A reading frame that is significantly longer than 21 codons is likely to be an open reading frame that is translated into protein. Most polypeptides are more than 100 amino acids long. Thus one can usually find protein coding regions just by looking for long open reading frames.

How does a DNA polymer react to a change the number of bp/turn? How does the reaction to the change help the structure?(L.15)

A change in the number of bp turn means that the topological state of that DNA is changing from relaxed to super coiled.. And so we must remember that the 3D structure of the double helix is flexible. And so the day that it reacts to a change in the number of bp/turn is to adopt a higher order structure called supercoils. The supercoils relive the torsional strain caused bu under or over winding of the helix.

What are the three proteins included in recombinational DNA repaid? What do they do? What are chi sites? (L.16)

A complex of three proteins called the RecBCDenzyme binds to double stranded DNA breaks. Using helicase and 5'to3' and 3'to5' exonuclease activities, both strands of the DNA is unwound and degraded. Once the degrading complex reaches an 8 nucleotide specific sequence element called a chi site, however, the 3'to5' exonuclease activity is reduced and the 5'to3' activity is enhanced. As a result, a single stranded segment of DNA is produced that has a chi site at its 3' end. I should explain more about the chi sites. There are about 1,000 of them in the E. coli genome spread around somewhat at random. They have a specific sequence, which is 5'GCTGGTGG3'. This sequence is bound by the RecCprotein once the RecBCDcomplex reaches the chi site. It is this protein DNA interaction that induces the change in the activities of the exonucleases. The single stranded DNA becomes bound by multiple copies of another protein: Rec A‚—shown in dark green at the bottom. You should be asking, why does RecAbinds to this single stranded DNA in preference to the usual SSB protein. The answer is because RecAis recruited by its interaction with the RecBCDcomplex. So if there is single stranded DNA and RecBCDthe cell knows it need to perform recombinational repair. Single stranded DNA with no RecBCDprotein is bound by SSBs.

What is a nucleosome? (L.15)

A region of DNA wound around histone proteins.

How do super coils help strand separation, energetically ?(L.15)

Although strand separation is not as favorable as supercoiling, super coils that unwind the helix make strand separation more favorable. Cells in DNA are slightly under wound to make it easier for enzymes that need to separate strands such as RNA and DNA polymerases to work.

What is another way other than splicing, that cells use to increase the diversity of transcripts? (L.17)

Another related approach that cells use to increase the diversity of transcripts is to use more than one poly A site. In the example shown, if the site closest to the 5' end is used, the cleavage reaction will remove the sequences further downstream, giving the shorter mature mRNA on the left. If the first poly A site is ignored for some reason and the second site is used instead, the longer mRNA on the right would be produced.

What happen to transcription when lactose is absent? (L.19)

As a result, whether [glucose] is high or low, if lactose is absent the lac repressor stays bound. Therefore there is at most only very weak transcription irrespective of the presence or absence of CAP-cAMP. FIGURE 28-18a,b Positive regulation of the lac operon by CRP. The binding site for CRP-cAMP is near the promoter. The combined effects of glucose and lactose availability on lac operon expression are shown. When lactose is absent, the repressor binds to the operator and prevents transcription of the lac genes. It does not matter whether glucose is (a) present or (b) absent.

How are the topological states of DNA changed from relaxed to super coiled or from supercoiled to relaxed? (L.15)

At least one of the DNA strands has to be broken and the DNA is either unwound to reduce the number of bp per helical turn, or it is over wound to increase the number of bp per helical turn. The DNA strands are then rejoined.

What are the combined effects of glucoses and lactose on the lac operon? (L.19)

By requiring two regulators to get strong transcription of this and other operons, E. coli is able to make sophisticated choices about which genes to switch on or off.

Aminoacylation of tRNA(L.18)

Creation of amino acyl intermediate: Amino acids are covalently coupled to tRNAs by a set of enzymes called Amino acyl tRNA synthetases. There is a specific enzyme for each of the 20 amino acids. These enzymes fall into two classes based on similarities within a class in structure, on amino acid sequence, and on mechanisms. There are Class I and Class II enzymes. Both act via a two step mechanism. The first step is shown and the top. The specific amino acid is reacted with ATP be to form an aminoacyl adenylate, which remains bound to the active site. In this reaction, the COO-of the amino acid attacks the phosphate of ATP to create an aminoacyladenylateintermediate. The pyrophosphate (PPi) that is released is also cleaved to Pi. So the reaction is driven forward by two phosphoanhydride bond cleavages. The fate of the aminoacyladenylateintermediate varies between class I and class II synthetases. Transfer of amino acyl group to tRNA: In both classes the aminoacyl group is transferred to the tRNA. For class I enzymes, the aminoacyl group is transferred initially to the 2'-hydroxyl group of the 3'-terminal A residue, then to the 3'-hydroxyl group by a transesterification reaction. For class II enzymes, the aminoacyl group is transferred directly to the 3'-hydroxyl group of the terminal adenylate.

Suppose you have a relaxed DNA that has Lk=200. What would happen if the DNA is unwound by 2? What would happen if the DNA is overwound by 2? Give the values for Tw and Wr for both cases.(L.15)

DNA is unwound by 2: So Tw=-2, The DNA compensates by introducing a negative supercoil and so the Tw=+2 of the negative supercoil. Therefore the total Tw=0. BUT the Wr=-2. DNA is overwound by 2: So Tw=+2, DNA compensates by introducing positive supercoil so Tw=-2. But Wr=+2.

What are Topoisomers? What is required to to convert between topoisomers? (L.15)

DNA molecules that differ only in their topology. The have the same number of bp, the same sequence, but different degrees of supercoiling. They have a. different linking number. Conversion between topoisomers requires a DNA strand break.

What happens during transcription in RNA polymerase in E.coli? (L.17)

During transcription RNA polymerase unwinds about 17 bp of DNA. This unwound bubble of separated DNA strands moves down the DNA as synthesis proceeds. The figure shows the two strands of DNA. The template strand is in red and the non template or coding strand in blue. The synthesized strand of RNA is shown in green. The transcription runs left to right. Notice that—by definition—the polymerase tracks down the template strand in the 3' to 5' direction while it synthesizes the RNA strand in the 5' to 3' direction. At any one time, the 3' most 8 nucleotides of the most recently synthesized part of the RNA is hybridized with the template. As synthesis continues, the 5' most nucleotides of RNA peel off from the template DNA strand—shown at the bottom of the figure.

Aminoacyl- tRna sythetases(L.18)

Each enzyme binds to a specific amino acid and the matching RNA Most cells contain 20 difference aminoacyl-tRNA syntheses, one for each amino acid To repeat, each aminoacyl-tRNA synthetases binds to a specific amino acid and to the matching tRNA(s) Most cells contain 20 different aminoacyl-tRNA synthetases, one for each amino acid

What does the phrase "the DNA sequence is collinear with the amino acid sequence" used to describe regarding how bacteria protein is encodes? (L.15)

Each protein in bacteria is encoded by a continuous stretch of DNA. The information encoding each adjacent amino acid in a protein is present in an adjacent codon of 3 nucleotides in the genome

How can eukaryotic genome have a negative superhelical density if they do no have a topoisomerase that can introduce negative super coils in relaxed DNA (like prokaryotes which have gyrase)?(L.15)

Eukaryotes use a. combination of nucleosomes and topoisomerase that can relax supercoils to generate a negative super coil. [the underwinding occurs without a strand break, and therefore there must be a compensatory positive (+) supercoil formed in the DNA. on either side of the histone. The positive supercoil is relaxed by a type II topoisomerase, leaving the DNA with a net negative supercoil.] This shows us that nucleosomes do more than just compact DNA. By helping unwind the genome, they make separating DNA strands for transcription and replication energetically more favorable.

Control of gene expression (L.19)

Gene expression is controlled or regulated to vary the abundances and activities of RNAs and proteins. In this lecture we will look at how gene expression is controlled.

rho(p)-independent termination (L.17)

Genes whose transcription is terminated by the rho independent mechanism have two sequence features. One is a specific sequence just 3' of the protein coding region of the gene. The sequence is shown top right: it contains at least 3 Ts on the coding strand, which will lead to three Us in the RNA. The other feature is a self complementary set of sequences that will form a hairpin. Remember we looked at hairpin structures before. It does not matter what the sequences are, as long as they form the hairpin that is about 15 to 20 nucleotides from where transcription should terminate. Once the hairpin forms it tends to pull the transcript out of polymerase. U=A base pairs are weaker, so the three UUU residues makes it easier for the hairpin to pull the transcript out. Once the transcript is off the polymerase, that is it. The RNA cannot be fed back into the enzyme, so transcription is terminated

How does an adaptor tRNA interact with the codon? (L.18)

Here is a closer, more accurate look at how adaptor tRNAs interact with the codon. You can see in this example that the tRNA encodes a three nucleotide sequence that is complementary to the codon. This complementary triplet is referred to as the anti codon. The tRNA binds to the mRNA via hydrogen bonds. Notice that the alignment of the codon and anticodon is antiparallel, just as the two strand of DNA are antiparallel.

What is the leucine zipper motif? pt 2(L.19)

Here you can see how the conserved amino acids fit in the protein / DNA complex. The leucine residues hold the alpha helices of the dimer together. The arginine and lysine residues play a role in the interaction with DNA. Other non conserved amino acids also play a role in distinguishing specific base pairs as well, but are not shown here. Lehninger shows examples of several other DNA binding domain families. That is more detail than we need to learn, but the fact that these are all shown is an indication of the importance of these structures to understanding the control of gene expression.

How any chromosomes do humans have? Give information about them. (L.15)

Human somatic cells (non-sex) have 46 chromosomes There is a maternal and a paternal copy of each chromosome. There are 22 pairs of autosomes plus a pair of sex chromosomes, either XX or XY. (total of 23) Each pair of autosomes contains a unique set of genes

In the topological structure of DNA, there is a small double trended plasmid DNA. Where is it located? If the there are 10.5 bp per turn of the helix, what state is the plasmid in? What shape does the plasmid take when in the relaxed state? (L.15)

In the center is a small double stranded DNA. In the plasmid, the DNA strands are. continuous all the way around the molecule with no breaks. If there are 10.5 bp per turn of the helix , then the plasmid is defined as being in the relaxed state. the plasmid take on the shape of a circular DNA molecule. Since it is circular there are no single stranded breaks or nicks in the strands a therefore it is called a closed circular DNA.

Describe the process of SPLICING. What happens before? (L.15)

In the figure: coding sequences are in wide green bars. the Introns are orange bars. 5' of most exon is gold. (*5' UTR). Blue is the 3' most exon (*3' UTR). 1. RNA polymerase transcribes the gene to synthesize a long single stranded RNA primary transcript. 2. The introns are removed by splicing. the exon are joined end to end to form the mature mRNA transcript *5' UTR: untranslated region, remains as part of the final mRNA and is not translated *3' UTR: 3' end of the untranslated region. *They are both part of the first and last exon.

Where does splicing and transcription occur? Where does the mature mRNA go after? (L.15)

In the nucleus. The mature mRNA is then exports to the cytoplasm to be translated.

What does the superhelical density (σ) or the specific linking difference determine? How do you calculate σ? What is the superhelical density of most DNA cells? (L.15)

It determines the density of supercoiling. σ= (ΔLk)/(Lk0) Most cellular DNAs have σ= -0.05 to -0.07. [think of it as being 5% to7% unwound compared to relaxed DNA]

Is the degree of computation of DNA in eukaryotic cells larger or smaller than in bacteria or phage? (L.15)

MUCH LARGER

What is the MW range of histones? What two amino acids are they rich in and why does it make sense that it would be those two? What percentage of those amino acids are they made of ?(L.15)

MW is between 11,000 and 21,00. They are rich in arginine and lysine which makes sense because these proteins bind to the negatively charged phosphate backbone of DNA. 24% to 40% of histone amino acid residues are compromised of these two amino acids.

What are the different types of DNA repair systems in E.coli? Which ones use one stranded DNA to repair? (L.16)

Mistmatch Repair: fixes mismatches that arise from the occasional incorporation of wrong nucleotides during replication Base Excision Repair: chemically abnormal bases, such as the methylated guanine Nucleotide Excision Repair: large changes that alter the shape of the double helix are detected and removed Direct repairs: chemically removes the covalent Couling between adjacent bases. SOS repair and recombinational Repair: repair cases where the information on both strains of DNA have been lost.

What are the reason for protein degradation? (L.18)

Moving on beyond translation, proteins do not last forever. They are degraded for a number of reasons. One reason is that over time they become chemically damaged or denatured. These disabled molecules need to be removed. A second reason for protein degradation is that for the cell to control (change) the abundances of proteins in response to environmental changes, cells need to be able to reduce the levels of specific proteins at certain times. If proteins have an infinite life time, it would not be possible to reduce their concentrations in cells.

Are eukaryotic chromosomes always tightly organized? What phase are they tightly organized?(L.15)

No. most of the time they are less tightly organized than at metaphase but are still quite condensed.

Where fo most covalent modifications of histones occur (L.15)

On the histone tails.

What is enzyme activity determined by? (L.19)

Please realize that changing the concentration of proteins or changing the activity of protein molecules that are already present in the cell are both ways to control the expression of genes. However, the phrase "the control or regulation of gene expression" tends to be used to refer most often to the four steps that control protein abundances rather than to the action of signal transduction pathways.In this lecture, we will limit ourselves to learning about steps 1 -5 shown on this slide.

What is processivity? What is the polymerization rate? (L.16)

Processivity: The number of nucleotides added before dissociation - the processivity of different DNA polymerases varies widely from a 10s-500,00 nucleotides Polymerization rate: the number of nucleotide added per second -polymerization rates cary from 1 to 1,000 nucleotides/ second - Each specific polymerase has its own processivity and polymerization rate

What do eukaryotic genomes contain a lot of ?(L.15)

REPETITIVE DNA

What is important about Rec A monomers? (L.16)

Rec A monomers bind cooperatively to single stranded DNA forming a helical structure. At the top you can see an EM micrograph of a portion of unbound DNA and—in darker staining—DNA bound by Rec A. At the bottom you can see the structure of at higher resolution. Each different shade of green is a different RecAmonomer. Rec A is the central protein in homologous recombination. It is these protein coated single stranded DNAs that invade the undamaged chromosome to promote strand exchange.

Recombination DNA repair is an example of _________recombination.(L.16)

Recombination DNA repair is an example of homologous recombination.

What is the DNA metabolism. Give a little description of each step.(L.16)

Replication: a new copy of DNA is synthesized with high faithfulness before each process. Repair: errors that arise during or after DNA synthesis are constantly checked for and corrected. Recombination: segments of DNA are also rearranged either within a chromosome or between two DNA molecules by recombination, giving offspring a novel sequence.

How is the genetic code structured? (L.18)

Shown at the top, living organisms use a nonoverlapping mRNA code with no punctuation. Each amino acid is encoded by a three letter word called a codon, or sometimes called a triplet. The code or the words are directly adjacent to one another, with no gaps between them. One feature of this sort of code is that it can be read in three ways. Depending on which nucleotide is used to start the first codon, there are three possible reading frames. Depending on which reading frame is chosen, the sequence of amino acids that would be produced would be very different. The code has a special three letter word .... AUG ,,,, to indicate which the first amino acid is. This is called the start or initiation codon and it defines which of the three reading frame will be chosen.

What is an example of. topological structure? (L.15)

Supercoiling. In the case of the telephone cord example, the cord has many small coils. In addition, there are higher order supercoils, which are the larger scale twists where the coils twist around each other.

What is the issue to the clamp in lagging strand synthesis? (L.16)

The DNA has to pass through one side of the clamp, then the clamp needs to close. The clamps loader senses where the RNA primer is and allows the DNA to pass through as shown in the center. Then there is a ATP driven conformational change in protein structure, as a result of which the clamp closes around the DNA and disassociates from the clamp loader.

What are histone octamers? What do they contain?what is their length?(L.15)

The Octobers are the c colored disks in the figure. Each octamer contains two molecules each of histone H2A, H2B, H3 and H4. The length of DNA wrapped around the octamer is 146 bp and it wraps 1.67 times around.

Ribsome structure in bacteria. (L.18)

The cartoon on the left re-emphasizes that the two subunits of the ribosome combine to make the 70 S ribosome. The lengths of the rRNAs are given. The longest, the 23S rRNA, has 3,200 nucleotides. On the right is shown the 3D structure solved by X-ray crystallography. The protein is shown by the thin spaghetti, the thicker regions show the RNA. This helps you see how RNA dominates the structure. Notice the cleft between the 50S and 30S subunits. This is where protein synthesis occurs.

What happens in the. first round of splicing in group I splicing reaction? (L.17)

The electron pushing arrows show the nucleophilic attack on the left.The resulting cleavage of the phosphodiester link between the intron and the exon is shown on the right. A new phosphodiester bond is formed with the guanosine residue, making the whole reaction approximately energy neutral. .

The timing matters during the four steps for regulating protein concentrations in a cell matter. (L.19)

The four steps on the last slide take anywhere from minutes to hours to change the abundances of proteins. That is fairly slow and is much longer than some of the other processes we saw that change the activities of enzymes such as covalent modification by signal transduction cascades. Processes such as covalent modification can take less than a second.

What is the lac operon inducted by? (L.19)

The key to understanding control by the lac repressor is that the activity of the protein is regulated by binding of a small metabolite called allolactose. When lactose is present in cells, most lactose molecules are cleaved by b-galactosidase to form glucose and galactose, but a small portion of the lactose is isomerized into allolactose. So when lactose concentrations are high, allolactose concentrations are high. Allolactose binds to the lac repressor and acts as an effector. Allolactose binding causes a conformational change in lac repressor that greatly reduces the repressors affinity for DNA. The lac repressor thus dissociates from the operator, which leads to a dramatic increase in transcription. Allolactose is called an inducer because its presence leads to an increase in transcription.

What is the parent strand, what is the daughter strand? (L.16)

The original helix to be replicated is the parent molecule and each of its two polymer strands are called the parent strand. (blue in figure) The newly synthesized strands are c called the daughter strands. (red in figure)term-132

what doe the direct repair of pyrimidine dimers do? How does it do it? (L.16)

This fourth pathway is devoted to repairing only pyrimidine dimers. It employs a single enzyme called photolyase which essentially reverses the crosslinks induced by the UV light. This enzyme uses the energy of visible light to drive the reaction. The light absorbed by a pair of chromophores, one of which we have already come across: FADH. The light is absorbed, creating a high energy electron that after a short elector transfer process is transferred to the pyrimidine dimer. The somewhat complex election pushing rearrangements shown in 4 to 5 reverse the covalent links between the bases.

In the experiment, the mutant strain could be rescued by either all 20 amino acids in the minimal media or just arginine present in the media. Based on these results what does it tell you about the mutant?(L.14)

This tells me that the mutants lacked the enzyme needed to synthesize the specific metabolite that rescued their ability to grow on minimal media +the metabolite.

How will the DNA most commonly compensate when there is a lot of strain from under wound DNA? (L.15)

To relieve the strain, the DNA will adopt a supercoil shown in c. The DNA will try to get as close to a normal double helix (achieved by the relaxed state) as possible since it is the most energetically favored.

What are topoisomerases? Why are they important?(L.15)

Topoisomerases are enzymes that change Lk. They are required for DNA unwinding and rewinding during transcription & replication.

What are the two components of Lk?(L.15)

Twist (TW) and Writhe(Wr). Lk=Tw+Wr

What are the four steps for regulating protein concentrations in a cell. (L.19)

We saw earlier that there are four major steps at which the amounts of protein expressed by a cell can be controlled. The rate at which the gene is transcribed. The rate at which the mRNA is degraded. The rate at which the mRNA is translated. AND. The rate at which the protein is degraded. In principle, you can vary the abundance of a protein by changing any of these rates. However, if you do not want to express a particular protein in a cell, energetically it makes most sense not to transcribe it. If instead you transcribed the gene, then translated the mRNA and removed the protein by degradation, as you now know, that would be energetically very costly. For that reason much of the control of gene expression is at the level of transcription. There are times though when these other three steps are used to modify quantitatively the exact levels of protein. Please also realize that while the slide focuses on protein expression, the rate of transcription and degradation of functional RNAs is also controlled.

What is the process of the lagging strand synthesis ?(L.16)

We know that the replication fork moves right to left. In the picture the bottom strand in each panel is the lagging strand. (a) shows a DNA helices called DnaB that has 6 sub units. The complex encircles a single strand of DNA and process in the direction of the replication fork, using ATP to unwind the energy. Notice: at any location where there is a significant length of single stranded DNA, the DNA is bound bu single stranded DNA binding protein (SSB)-shown with green blobs. The binding energy of the SSBs helps keep the DNA strands from reannealing. also at any location where there is a type II DNA topoisomerase gyrase just ahead of the replication fork. This protein relaxes the positive supercoils that are generated by the helicase. (b) the primase makes occasional protein/protein interactions with the DnaB helices, which then allows the primase to initaiate synthesis of a primer. The primase move in the opposite direction to the helicase. This panel shows a single pol III roe synthesizing a lagging strand Okazaki fragment from an RNA primer (c) shows an additional pol III core had been recruited and it is synthesizing a second Okazaki fragment. Notice how the SSBs are displaced and the lagging strand becomes double stranded. A wholesaled series of Okazaki fragments are generated.. (d) shows a structurally more accurate picture. Th three pol III cores are in fact linked together. The leading and lagging strand cores are able to move in opposite directions along their respective template strands while physically moving together with the replication fork. This is accomplished by the lagging strand looping out. -it also shows the locations of the clamps (e) bottom slide shows how the Okazaki fragments are joined to make a continuous strand of DNA. Nick translation by DNA pol I is used to remove the RNA primers or in some cases a specialized RNase called RNase H is used. Most RNases degraded single stranded or double stranded RNA. RNase H, by contrast can only degrade RNA when it is in a duplex with a strand of DNA nick is then sealed by DNA ligase

What happens to transcription when lactose is present? (L.19)

When lactose is present, transcription depends on glucose concentration.The lac repressor dissociates from the promoter, but transcription is only stimulated significantly if cAMP rises and CAP-cAMP binds to the promoter.

What is RNA dependent synthesis or more commonly know as reverse transcription? 2(L.17)

While most DNA and RNA is synthesized from a DNA template, in some cases RNA can be used as the template instead. When RNA is used to make DNA the process is called RNA dependent DNA synthesis, or more commonly reverse transcription.

Image: Electron micrograph of E coli cell. Describe the image. (L.15)

White blob is outer water of E coli cell. The cell's main chromosome is most of the spaghetti (it has been spilled out of the lysed cell. the white arrows pointing to small white circles are pointing to plasmids.

What is the Lk a function of?(L.15)

a function of the length of DNA

What is a silent mutation?(L.16)

a mutation that has no effect on the phenotype or the gene function

What was the experiment that showed that replication initiations at discrete locations: origins? (L.16)

bacteria and phage have an organ. Inman and collaborators made use of the fact hat AT rich stretched of DNA denature more readily than others. (the picture shows that regions that are AT rich are denatured while regions rich in GC are not). Inman and collaborators studies a 49 kb bacteriophage called lambda (λ) Replication forks were visualized on the election microphraphs and their location was identified bas on the AT bubbles. The length of the replication fork can be used to determine how far away it is from the place that the region replicated. From their data, they were able to show that for every lambda (λ) genome they looked at, replication had initiated at the same point as opposed to starting at some different random point.

What shape are the chromosomes in bacteria?what shape are they in eukaryotes? (L.15)

bacteria: circular eukaryote: linear

Why is DNA polymerase I better at repairing DNA than the other polymerases? (L.16)

because it has a second exonuclease activity: a 5' to 3' exonuclease

Why did Waston and Crick propose the semiconservative model?(L.16)

because it made direct use of the complementary structure and hydrogen bonding between A=T and between G=C.

What are the parts of the eukaryotic genome that code for a single protein? What are they interrupted by? Define them.(L.15)

exons(blue): portions of the genome that code fro the protein as well as the parts that code for the messenger RNA that lie 5' or 3' of the protein coding sequences Introns(gray): interrupting regions. Regions of genes that are transcribed but not translated. they do not encode polypeptide sequence

How are we using gene names in this class?(L.16)

gene names for e.coli are lower case and have three letters that are associated with a biochemical function. They may be followed by a fourth upper case letter to distinguish polypeptides that. act in the same process. For example, ropA, rpoD Proteins usually have the same name as the gene, except italics are not used and the first letter is capitalized.

What are the features of protein synthesis ?(L.18)

here are several features of translation that it is important to keep in mind. One. There is a high energy cost. Two high energy phosphoanhyridebonds are hydrolyzed for each tRNA that is amino acylated. A GTP is cleaved to GDP and Pi during the first elongation step and a second during translocation. Thus the equivalent of 4 NTPs are consumed. In kJ/mol that is 4 x 30.5 kJ/mol per peptide bond, i.e. ∆G'0= 122 kJ/mol. The peptide bond has a ∆G'0of hydrolysis of -21kJ/mol. Thus the net standard free energy for formation of each peptide bond is 101 kJ/mol. Two. Translation can rapidly produce many proteins from a single mRNA molecule because clusters of ribosomes called polysomescan form on each mRNA. As shown in the figure in the top right. Three. In bacteria, translation is said to be tightly coupled to transcription. In other words, translation of an mRNA can begin before transcription is finished

What does I mean for DNA when they have either a positive or negative ΔLk? How would you compare negative and positive supercoils?(L.15)

if (-ΔLk) then the DNA is under wound and is negatively super coiled. if (+ΔLk) then the DNA is over wound and has positive supercoils. Negative and positive super coils are mirror image of each other. The high order structures form in the opposite directions.

What does Supercoiling affect?(L.15)

it affects the shape of the DNA as shown by the image.

What does a nucleosome consist of? (L.15)

it consists of DNA wrapped around a complex of eight histone molecules, which are histone octomer.

What is an example of what gene does?(L.14)

it encodes part of the information for which cell tissues specific gene products like muscle myosin should be expressed in.

What types of strains of Neurospora can not grow on minimal media? Why?(L.14)

mutant strains. Because they require more complex media to grow.

What do genes encode?(L.14)

polypeptides. or a functional RNA as well as some of the information needed to correctly express it in the organism

What is meant by semiconservative replication ?(L.16)

refers to replication where the two strands of a double helix are separated and each of these strands is used as a template to make a new strand.

What are centromeres? What are the essential for?(L.15)

regions where the two daughter chromosomes are held together during mitosis. Centromeres have thousands of copies of 5-10 bp repeat. Think of them as physical anchor points on the chromosome to which the cell division machinery attached and separates the two sister chromatids. They are essential for equal distribution of chromosomes sets to daughter cells.

What is the result of the accumulation of mutations in eukaryotic cells? (L.16)

result is strong correlation with cancer; most carcinogens are also mutagens

rho(p)-dependent termination (L.17)

separate set of genes are terminated by the rho dependent mechanism. These genes have a specific DNA sequence that is rich in C and A nucleotides towards the 3' end of the gene. This element is called a rho utilization site or a rut site because it is recognized in the nascent RNA transcript by the rho protein. Rho is an ATP dependent helicase that specifically unwinds DNA/RNA hybrids. Once Rho binds the rut site, it travels along the RNA until it reaches an elongating polymerase, where it releases the RNA, terminating transcription.

What did Arthur Kornberg show with his experiment on DNA polymerase? (L.16)

showed that DNA synthesis required only one protein, a 103 kilo Dalton enzyme called DNA polymerase. He showed that in cells there are actually many other enzyme required to accurately replicate the genome.

What is the lowest energy structure that DNA can adopt to deal with under or over winding? (L.15)

the SUPERCOIL

Describe the genome in the mitochondria and, in the case of plants, also in the chloroplasts. What is the length of mitochondrial genome in humans vs plants? (L.15)

the genomes are double stranded circular DNA's humans: 16.5 kp Plants: 2.5 Mb

What happens to the exon and introns after transcription?(L.15)

the introns are removed after transcription and the exon mRNA sequences are joined or spliced together. the gray parts are removed and the blue parts joined end to end to form the mature mRNA transcript.

What is the diameter of a nucleosome?(L.15)

the nucleosome is about 10 nm. Which is why it is sometimes referred to as the 10 nm filament

What is topology ?(L.15)

the study of the properties of an object that do not change due to continuous deformations, twisting, or stretching of the object. Any tearing will change the topology.

What deformations of DNA do not alter topology?(L.15)

thermal or protein induced bending f DNA or separating of the two strands where no strand breakage occurs.

What is unique about histones' aminino acid side chains (their R groups)? (L.15)

they are covalently modified to an unusually high extent. For example, some of their amino acids are often methylated, phosphorylated, or acetylated.

What do plasmids encode (L.15)?

they encode genes that degrade antibiotics i.e. the plasmid conveys antibiotic resistance to the bacteria they encode proteins that convey a selective advantage to a bacteria under some conditions. they do not encode any essential genes

What is the transcription unit? What will it be synthesized by? What is the primary transcript?(L.15)

transcription unit: the stretch of the genome that includes all of a genes introns and eons. The enzyme RNA polymerase will synthesize a single stranded RNA copy of the entire transcription unit. primary transcript: the initial transcript

What are simple sequence repeats (SSR), also known as satellite DNA? Where are they found?(L.15)

very short sequences of less than [<] 10 bp that occur in tandem (line up) one after the other many many times. They are found at the centromeres and at tellers, though the sequences of the repeats differ between the two regions. *3% of the human genome consists of them

What species have the smallest chromosome and and is their length? (L.15)

viruses (1 kilobase (kb) to 300 kb in length).

What are samples of 3 examples of antibiotics that target translation? (L.18)

Examples of other antibiotics are shown here. Tetracyclines, block the A site on the ribosome Chloramphenicol and cycloheximide block peptidyl transfer. Streptomycin causes the genetic code to be misread & inhibits initiation at high concentrations Studies of the ribosome complexed with these and other antibiotics are fueling attempts to develop new or better antibiotics to help with the arms race against antibiotic resistant bacteria. The study of the mechanisms of translation has had and will continue to be important for human health and medicine.

F1: homozygous white (w/w) and homozygous violent (v/v) will result in progeny having violet flowers. Since (w/w) is recessive. If F1 progeny were crossed with each other what would be the result of the F2 phenotype and genotype? (L.14)

Genotype: 1/4 of the progeny would be (w/w) Phenotype: I/4 of progeny would be white.

What is the difference between human/eukaryote haploid cells and bacteria haploid cells? (L.15)

Humans and many other eukaryotes have two copies of a gene. Bacteria only contain one copy. Some Eukaryotes, however, do for part of their life cycle have haploid genomes and other s have more than two copies of each chromosome.

What is the most compact stage for chromosomes, after the DNA has replicated and just before the cell is to divide? What results from the replication of a single chromosome?(L.15)

Metaphase result is two sister chromatids

Does the linking number of a closed circular DNA change?(L.15)

No. But it can, only if one or both strands of the DNA are cut and then resealed

Compare the mechanism of transposition of DNA to that of a retrotransposon. (L.17)

On the left is the mechanism we saw in the last lecture. A bacterial transposase enzyme cuts the genome, excises the transposon, cuts another location in the genome, then ligates the transposon at the new location. On the right is how a eukaryotic retrotransposon works. The host cell RNA pol II first transcribes the retrotransposon. The retrotransposon proteins are then translated, including a reverse transcriptase and an integrase. The reverse transcriptase generates a double stranded DNA version of the retrotransposon from the RNA transcript using the same approach used by the retrovirus enzyme. The integrase then inserts the retrotransposon in a new location in the genome. Notice one difference between the two transposition events. The eukaryotic retrotransposon is replicative. It makes a new copy of itself, leaving the old copy in the genome. The bacterial transposon moves but does not replicate.

What is the function of the enzyme RNA polymerase? (L.15)

RNA polymerase is the enzyme that transcribes DNA, making single stranded copy of RNA. RNA polymerase must part the two strands of DNA in order to use one strand as a template to synthesize the primary RNA transcript.

DNA polymerase I structure(L.16)

The core of this and most DNA polymerases is shaped like a human hand that wraps around the active site.

How is the linking number determined based on just the picture? How can the Lk be changed from 1 o 6 as shown in the figure?(L.15)

The linking number is determined by the number of times that the plane is pierced. the Lk can be changed from 1 to 6 but cutting one of the ribbons in the top figure and threading it around the other ribbon 5more times and the ribbon was then rejoined.

What causes the binding affinity of nucleosomes for DNA to be very high?(L.15)

The many ionic interactions between the positively charged amino acid residues in histones and the DNA phosphates

What are the two requirements for strongest induction of the lac operon? (L.19)

There are thus two requirements for strong induction of the lacoperon. One. Lactose must be present to form allolactose to bind to the repressor and cause it to dissociate from the operator. This reduces repression. Two, [Glucose] must be low so that cAMP can increase; cAMP then binds to CAP; and CAP-cAMP can bind near the promoter. This causes activation.

What are the two types of termination in E.coli? (L.17)

There are two ways that transcription terminates in E. coli. One is dependent on a protein called rho and one that is independent of rho. This slide show the rho independent mechanism

What is a gene according to Mendel?(L.14)

a discrete unit that does not change from generation to generation. (true in Mendel's experiments but not tru in general life)

What is a Klenow fragment?(L.16)

a distinct domain that can be separated by protease treatment

Where do genes reside? (L.14)

chromosomes

one gene-one enzyme hypothesis (L.14)

each gene contains the information needed to make an enzyme

What does teach gene encode according to Beadle and Tatum? (L.14)

each gene encodes the information needed to make a single enzyme or protein

What varies more: genomes size or gene number?(L.15)

genome size

How does the DNA fit inside the virion?(L.15)

it is tightly coiled in association with special circus proteins

What are telomeres? (L.15)

short repeated sequences. These sequences can occur around 1,500 tines in tandem repeats at the end of each chromosome. So in human the ends of the chromosome have the sequence 5' TTGGG.....

What is the range in the length of DNA in human chromosomes? (L.15)

the length of chromosomes varies about 25x. The total length of the human diploid genome is 6,500 Mb. Which is equivalent to ~2m of DNA.

How much does the wrapping of DNA in a nucleosome compact DNA? What is the overall compaction of DNA in the nucleus?(L.15)

the wrapping of DNA in a nucleosome compact DNA 7 fold. the overall compaction of DNA in the nucleus is 10,000 fold.

one gene-one polypeptide hypothesis (L.14)

there is one gene that codes for one polypeptide.

What are the two things that DNA polymerases can do? (L.16)

they can add nucleotides or dissociate

What is the length of plasmids? (L.15)

2,000-10,00 bp but can be 400,000

What is the C value paradox? (L.15)

the disparity between the size of genomes. The fact that the genomes of some organisms, particularly animals and plants, have far more DNA than is needed to code for their proteins. (The C value is the number of pico grams (pg) of DNA per cell, where 1 pg DNA is about 1,000 Mb of DNA.)

What happens to the Lk, if you cut the DNA and untwist two turns at the cut ends , then rejoin the DNA strands? (L.15)

the linking number would drop by 2 since the strand is piercing the plan two times less often. Since the there are two less linings the change in linking number is a negative number. [Lk=200-2=198] [ΔLk=-2] therefore this DNA will adopt a super coil

What is the a difference between how the eukaryotic and bacteria genome is encoded?(L.15)

the parts of eukaryotic genome that code for single protein are often interrupted while for bacteria encoding is continuous.

What is the problem with the chromosome packaging for the bacteria E.coli? (L.15)

the picture shows E.coil compared to the length of its DNA genome. the problem is that the E coli cell is about 850 times shorter than the genome! However the DNA is coiled up in association with proteins in cell the E coli genome I 4,639,675 bp (4.6 megabases Mb) = 1.7mm in length E coli cell is about 2 μm long

What does the presence of histone H1 nucleosomes do? (L.15)

the presence of histone H1 nucleosomes will stack together to forma diner 30 nm wide. Image: to the left is won how the nucleosomes stack to form a helix and in the center is the structure shown with the DNA included in blue. *However this is not what happens in cells. In vivo more complex less well understood higher order structures are formed. The current thinking is that nucleosomal DNA assembles in local clusters who's structures are dynamic. One model call these structures fractal globules (far right). In the figure, parts of DNA. are colored differently. The sea is to avoid knotting, the same parts of the genome cluster together.

What is DNA recombination? (L.16)

the rearrangement of genetic information within or between DNA molecules.

What is the function of a gene?(L.14)

-it carries to the information to make a polypeptide or a functional RNA - it includes addition information for how much of these functional products should be produced and how these amounts should change with different circumstances.

How do you measure DNA topology? (L.15)

the main metric is the Linking Number. (a property of double stranded DNA (dsDNA))

What is the process of Mismatch repair?(L.16)

*The mismatch repair systems assume that replication errors must reside in the unmethylated strand. Early Steps: The unmethylate strand is detected and nick is introduced into it as follows in the new figure newly synthesized DNA is the salmon color strand while the parental strand is blue. 1. A complex of two proteins MutL and MuSt recognize the distortions in helix cause by the mismatched nucleotide pairs and bind at the mismatch. 2. Two molecules of a separate protein, MUtH, binds on either side of this complex and the resulting larger complex threads theDNA on the both sides MutH recognizes GATC sequences. Once it reaches one, it cleaves the back bone on the nonmethylated DNA strand on the 5'-side of the G. THIS IS THE KEY STEP. the ability of MutH to sense the difference between methylated and unmethylated strands of the GATC sequence. -mismatch could be 1,000 bp away from GATC however GATC will over on ave about once every 4^(4)=256 nucleotides, so often the distance is less. -MutH cleaves the nonmethylated DNA strand on the 5'-side of the G Late Steps: In the second phase of the reaction, the DNA on the newly synthesized strand is degraded and the strand is then resynthesizes. The process is slightly different depending on if the backbone nick is on the 3' or 5' side on the mismatch. 3. The DNA in unwound by a helicase and the newly synthesize strand is degraded by an exonuclease enzyme. Exonuclease enzymes move progressively, removing one nucleotide at a time. (much more active and can remove more nucleotides than exonuclease of polymerases) The resulting single stranded DNA is stabilized by binding multiple copies of SSB. 4. Then DNA polymerase III resynthesizes the strand, hopefully without any mistakes. The final nick in the DNA is sealed by ligase

What is the replisome? What are some of the different types of proteins in the replisome? Describe their function. (L.16)

-Replisome: Over 20 other proteins along with the pol III holoenzymes. Types of proteins in the replisome: Helicases, enzymes that use ATP energy to unwind DNA strand, parting them Topoisomerases, relive the stress of supercoiling that is caused by unwinding DNA Binding Proteins, to stabilize separated strands , some of which bind to single stranded DNA to stabilize it preventing the reassoiation of the complementary strands primases, make RNA primers for replication to initiate DNA ligases, seal nicks between successive nucleotides on the same strand, such as those left by nick translation or between (Okazaki fragments).

What are examples of E.Coli promoters? (L.17)

-Two consensus. sequences at -10 (TATAAT) and -35(TTGACA) for sigma subunit binding - AN A-T rich elements between -40 a d - 60 that binds the core and promotes strands separation The figure shows the promoters of six genes in E. coli. In yellow, is the part of the genome that encodes the 5' ends of the genes' transcribed regions. The blue and purple sequences to the left of this are the sequences 5' of that, i.e. the promoter sequences. By convention, the first nucleotide in the transcript is numbered +1, those nucleotides 5' of it have negative numbers counting from -1, going right to left. There is no nucleotide "0" Each sigma unit recognizes a different nucleotide sequence. different groups of genes have promoter sequences recognized by a different sigma fact

What factors contribute tot the C value paradox? (why does only a tiny fraction of animal genomes encode proteins?) (L.15)

1. Eukaryotic gens have introns (that get spliced out) 2. sequences that control which cells a gene is expressed in are much longer and more complex in animals and plants than in bacteria. - there are special sequences 5' to 3' of transcription units and also within some introns that control the rates at which genes are transcribes.

What are the three types of recombination? (L.16)

1. Homologous recombination: the change pf sequences between two DNAs that share an extended region of similar sequence. For example, between maternal and paternal chromosomes. 2. Site-specific recombination: s the exchange of DNA that only occurs at a particularsite within a chromosome. The site is usually determined by the presence of a particular sequence, e.g. ATGGCCATTA 3. DNA transposition: occurs when a special class of short DNAs move from one chromosome to another. These special DNAs are called transposons.

Translation elongation and termination.(L.18)

1. In part D of the lecture, we will continue on with the mechanism of translation. There are three steps to each elongation cycle. Each elongation cycle adds one amino acid. At the top is a reminder of the 70S initiating ribosome. ThefMet-tRNAfMetis bound at the P site and it interacts with the AUG of the mRNA.Since the initiation factors have been removed, the A site is now available and thus can be bound by an incoming amino acyl tRNA. The incoming amino acyl tRNA binds first to one of the elongation factors, GTP associated EF-Tu. Then this combination binds to the A site. The GTP is hydrolyzed, which releases EF-Tu from the ribosome. The GDP bound EF-Tu is regenerated by removal of the GDP and binding of a new molecule of GTP in a reaction mediated by EF-Ts. 2. Step 2 of each elongation cycle generates the peptide bond. There are two amino acids bound to tRNAs positioned for joining. One is on the A site, the other on the P site.For the first elongation cycle, N-formylmethionylgroup is transferred from its tRNA in the P site to the amino acid in the A site. The reaction is catalyzed by the 23S rRNA (ribozyme). Just as we saw in Lecture 3, the amino group of the amino acid in the A site acts as a nucleophile, creating a peptide bond with the fMetat the P site.Note the involvement of the 2'-hydroxyl group of the 3'-terminal adenosine as a general acid-base catalyst in this reaction on the ribosome, though. So the ribosome is a ribozyme (almost). The formation of the peptide bond leaves an "uncharged" (deacetylated) tRNAfMetin the P site. This uncharged tRNA shifts so that its 3' and 5' ends are in the E site, the purple parts. Similarly, the 3'and 5'ends of tRNA bearing both amino acids shift to the P site. The anticodons remain in the P and A sites, however. 3. In step 3 of elongation, the ribosome moves one codon toward the 3'-end of the mRNA. This process is called translocationand uses energy from GTP hydrolysis. The GTP is bound to EF-G, which is also called a translocase. First, shown on the left, GTP bound EF-G binds at the A site, displacing the mRNA and the two tRNA that interact with the mRNA by one codon. This places a new codon at the A site and leaves the A site open for new aminoacyl-tRNA.At the same time, shown on the right, the GTP is hydrolyzed to GDP, which provide the energy for a conformational change in the structure of the ribosome and the translocation of the mRNA. Finally, not shown, the EF-G disassociates from the ribosome taking with it the GDP. And the tRNA in the E site also leaves. The ribosome is would then be ready to accept a third amino acylated tRNA at the A site. Multiple rounds of the elongation cycle then occur until a stop codon arrives at the A site. 4. Either an UAA, UAG, or UGA codon in the A site will trigger the action of termination factors (release factors) RF-1, RF-2, RF-3.First, a release factor (RF-1 or RF-2, depending on which termination codon is present), binds to the A site. This leads to hydrolysis of the ester linkage between the nascent polypeptide and the tRNA in the P site and release of the completed polypeptide. Finally, the mRNA, deacylated tRNA, and release factor leave the ribosome, which dissociates into its 30S and 50S subunits. This process is aided by ribosome recycling factor (RRF), IF-3, and energy provided by EF-G-mediated GTP hydrolysis.The 30S subunit complex with IF-3 is ready to begin another cycle of translation.

What are the three fundamental rules of replication?(L.16)

1. Replication is semiconservative. 2. It begins at a specific location in chromosome called an origin and it proceeds away from each origin in both directions (bidirectionally). . 3. Synthesis of new DNA occurs in the 5'-->3' direction and semi-discontinuous

What are the two way hat DNA will compensate when there is a lot of strain from under wound DNA? (L.15)

1. Super coiling 2. separation of strand for 1 turn of helix (helps DNA replication)

How is DNA terminated? (L.16)

1. There are multiple sequence elements called Ter sequences at which replication can halt. these short sequences are bound by a protein called Tush's which will halt progress of a replication fork. 2. Once on of the two replication fork has stopped, the other replication fork will often continue around the chromosome until it meets the term terminated fork, at which point the replisomes disassemble 3. After the replisomes disengaged, the chromosomes will remain linked due to a few unresolved superhelical turns. These are resolved by a type II topoisomerase which is called top IV. Like all type II topoisomerase, the enzyme cleaves both strand of one of the DNAs and passes the other DNA though the gap religating 4. When two DNAs are linked they are said to be catenated.

Translation: initiation (L.18)

1. To initiate translation at the first AUG codon in an mRNA, a special tRNA is used. This initiation tRNA is only used for this first codon. In bacteria, as well as chloroplasts and mitochondria, the initiation tRNA inserts a modified form of methionine, N-formylmethionine. As a result the initiator tRNA is therefore called tRNAfMet. At any other AUGs in the interior of the reading frame use a normal tRNAMetis to insert met. Thus all other methioninesin proteins are not modified. 2. In step 1, the 30S ribosomal subunit binds to IF-1 and IF-3. There are three sites in the 30S ribosome called the aminoacyl site or A site; the peptidyl site or P site; and the exit site or E site. IF-1 binds the A site and IF-3 binds to the E site. IF-3 acts to keep the 30S and 50S subunits apart. The mRNA binds after IF-1 and IF-3. The initiating AUG codon in the mRNA is guided to the P-site on the ribosome by a short sequence in the mRNA called the Shine−Dalgarno sequence. This sequence is just 5' of the AUG and is complementary to a sequence in ribosomal RNA. In other words, base pairs form between part of the rRNA in the ribosome and the mRNA to align the AUG in the correct position. It is the presence of the Shine−Dalgarno sequence that distinguishes the initiating AUG from the other AUGs in the mRNA. In Step 2: IF-2 and FormylmethioninetRNA join the complex. FormylmethioninetRNA binds to the P site on the ribosome and also interacts via the anticodon with the initiating AUG on the mRNA. fMet-tRNAfMetis the only tRNA that can bind first at the P-site. All other incoming aminoacyl tRNAs bind first at the A site. The presence of IF-1 at the A site prevents the non fMet-tRNAfMetfrom binding to the 30S subunit. IF-2 has a molecule of GTP bound to it. 3. In step 3 of initiation, the large 50S subunit combines with the 30S subunit. IF-2 hydrolyzes GTP and the three disassociate from the complex, forming the 70S initiation complex.

How do we know that synthesis is semi discontinuous?(L.16)

1. synthesis always occurs by adding new nucleotides to the 3' end -OH 2. the leading strand is made continuously as the replication fork advances. -the upper strand of DNA is replicated once a short primer is produced (short red line) a polymerase molecule can simply synthesize a continuous new strand. 3. the lagging strand is made discontinuously in short pieces (Okazaki fragments) that are later joined together - 1,000-2,000 nucleotides long - bottom strand is a bit more complex, the replication fork is moving left to right in the figure, but DNA polymerase can only move right to left on the bottom strand (lagging strand). - the cell's solution to this problem is for the polymerase to synthesize short 1 to 2 kb lengths of DNA called Okazaki fragments. The fragments are generated in turn, with one on the right being the most recently synthesized. After the Okasaki fragments are generation, additional processes join them up to form a continuous strand.

In the human genomes, only ____% of the total genome codes for the linear amino acid sequence of the 20,000 or so human proteins. (L.15)

1.5% The biological significance of the rest of the noncoding sequences is not fully understood.

What are the two important special function repetitive DNA sequences in Eukaryotic chromosome? (L.14)

Eukaryotic chromosomes have two important special-function repetitive DNA sequences: centromeres, which are attachment points for the mitotic spindle, and telomeres, located at the ends of chromosomes.

DNA + protein= ? (L.15)

Chromatin

What is the leucine zipper motif? (L.19)

A second family of transcription factors have so called leucine zipper DNA binding motifs. This family of transcription factor are mostly found in eukaryotes, not bacteria. At the bottom are the amino acid sequences of the DNA binding domains of four of these proteins. Three from mammals, one from yeast. The most important amino acids that are conserved within this domain are highlighted in color. The red amino acids are leucine residues that are spaced every 7 amino acids. This part of the domain forms an alpha helix, so residues separated by 7 amino acids will occur on the same face of the helix. (Remember from Lecture 4 that there are 3.6 amino acid residues per turn of the alpha helix). Leucine zippers form dimers in which the two molecules are held together by hydrophobic interactions via these leucines. The part of the protein that contacts DNA has many arginine and lysine residues. These residues of course are positively charged. In fact almost every family of DNA binding motif includes multiple Arg and Lys residues because the R groups on these amino acids form stabilizing charge/charge interactions with the negatively charged phosphate backbone of DNA.

Why is protein degradation inevitable? (L.18)

As I mentioned in the first lecture, the half-lives of proteins range from seconds to days and in fact in a few cases, such as collagen in your tendons, half lives can even be months. Hemoglobin, for example, is long lived. By contrast, many metabolic regulatory proteins that respond to rapidly changing needs are short lived. Once proteins become defective proteins they are rapidly targeted for destruction. All proteins are eventually degraded.

The second genetic code. (L.18)

Aminoacyl-tRNA synthetases must be specific for both amino acid and tRNA. The matching of each amino acid with the correct tRNA can be viewed as the "second genetic code". This second "code" is in the molecular recognition of a specific tRNA molecule by a specific synthetase. i.e. how the synthase is able to recognize differences in the structure and chemistry of each tRNA. Only a few nucleotides in tRNA confer the binding specificity. Many of these nucleotides are in the anticodon loop and include the anticodon itself. In some cases as many as 10 nucleotides are involved in conferring specificity, in others very few nucleotides are involved. In the most extreme case, the primary determinant in Ala-tRNA is a single G=U in the amino acid arm.

Antibiotics and toxins targeting translation. (L.18)

Because protein synthesis is so central to life, it is the target for many natural toxins, which of course include antibiotics. Natural selection in eukaryotes has favored production of toxins that are specific to bacterial translation without affecting translation in the eukaryote. The antibiotic shown here is called puromycin. It is produced by a mold. The antibiotic resembles the aminoacyl end of a charged tRNA, and it can bind to the ribosomal A site and participate in peptide bond formation. The product of this reaction, peptidyl puromycin, is not translocated to the P site. Instead, it dissociates from the ribosome, causing premature chain termination. So this prevents the bacteria making full length proteins.

Describe Recombination DNA repair: step 2, 3, and 4. (L.16)

At the top is a copy of the structure produced by step 2: where the DNA strands have begun to swop between chromosomes. Once this structure is formed another protein complex called RuvABpromotes the movement of the point where the DNA strands cross, a process called branch migration—shown in step 3. The complex structures shown on the right are called Holliday intermediates after the scientist who first proposed them. All of the single stranded regions shown in the figure are coated with RecA. In step 4, Another enzyme called RuvCthen cuts the DNA strands that cross over. The DNA strands are then ligated, yielding the structure shown on the bottom. As you can imagine, the replisome is reloaded onto this structure, and replication continues in steps not shown in the slide.

mana sequences for translation initiation (L.18)

At the top of this figure is shown the Shine-Dalgarno sequences of several mRNAs. The important nucleotides are highlighted in orange. They occur just upstream of the initiating AUG, shaded in green. At the bottom is the consensus sequence (or the ideal sequence) for the Shine-Dalgarno sequence and just above it is the sequence of the 3' end of the 16S rRNA. Notice the base pairing that is possible between an mRNA and the rRNA. These base pairs align the mRNA in the P-site of the 50S subunit.

Why are telomeres important? (L.15)

B/c they are solutions to overcoming the problem of replicating linear chromosomes. As we have seen DNA polymerases need t have short double stranded region so that the polymerase can extend the primer to generate a copy of the template strand. That means that the most 5' parts of the chromosome cannot be replicated and in fact after each cell division the chromosomes would actually get shorter.

Why are we seeing increasing numbers of bacteria that are resistant to antibiotics (L.15)?

BECAUSE OF Plasmids! Plasmids can replicate and be easily swapped between bacteria (even different species). Continuous selection for antibiotic resistance maintains plasmids in the cells. (if selection is removed , plasmids will over time be lost from the cell).

We know that the transcription of genes for both bacterial and eukaryotic genes use transcription factors. However due to their size they behave differently. Describe how transcriptions factors transcribe for both Eukaryotic gene and bacterial genes and how do the transcription factors compensate for the larger size of the eukaryotic cells? (L.15)

Bacterial Genes: transcription factors in bacteria bing to sequences within 100 bp of transcription units. Eukaryotic Gens: since animals and plants are much more complex their genes are regulated by many more transcription factors that bind over much larger regions of DNA. They compensate for the larger size by having large cis regulatory sequences. which is another reason why eukaryotic genomes are much larger than bacterial genomes.

What is a base excision repair? What are the two ways AP sites arise?(L.16)

Base excision repair is where one of the bases is missing from the DNA, but the sugar phosphate backbone remains intact. Such a site is called and apurinic or aplyrimidinic site or an AP site for short. Two ways apurinic or aplyrimidinic sites an arise: 1. Bases are not 100% stable and so at a low rate, they will deaminate. One of the most common reactions is the deamination of cytosine to uracil. Only one cytosine base will feaminate to uracil every 24 hrs for every 10^(7) bases. However, because of the large number of residues, that is 100 C to U changes per day. Cells have special enzymes that continually scan the genome looking for chemically wrong bases. If there is a uracil bade in DNA or any other non standard base, these enzyme remove the base, creating an AP site. These enzymes are called glycosylases and there are specific glycosylases for removing each kind of non standard base. 2. second way that AP sites arise is due to the instability of the linkage between guanosine and the ribose. The end spontaneously hydrolysis. Like deamination, the rate of depurination is slow, but given the large size of the genome. 10,000 purines fall of the the genome in each one of your cells everyday.

How does the enzyme RNA polymerase alter supercoiling? (L.15)

Because RNA polymerase must part the two strands of DNA in order to use one strand as a template to synthesize the primary RNA transcript it has two implications for the DNA topology: 1. the more unwound or negatively super coiled the DNA the easier it is for the polymerase to unwind the DNA. 2. The action of this enzyme pulls the strands apart and thus will locally affect supercoiling. They will tend to unwind the DNA behind the enzyme and over wind the DNA ahead of the enzyme. Since the DNA in chromosomes is physically tethered to proteins and thus cannot spin to undo this winding effect, there is a transient change in the local super coiling. ~Note: the topology (the linking number) of the whole chromosome does not change as no DNA strands have been broken. However, there are local effects on the density of supercoils~

Why is it important that the correct bases are incorporated in to the primer strand?(L.16)

Because most mutations are deleterious to an organism (reduce its fitness) and life could not exist if the frequencies of errors were too high. Error rate in E.coli is very low: only one mistake is mace for every 10^(9) to 10^(10) bp replicated. aka one mistake forever 1,000 t0 10,000 cell divisions

How can a mutation in some genes be rescued? What does this do for the mutant? (L.14)

By added a a specific metabolite to the media. It allows the mutant to grow.

How are we able to tell that replication of circular DNA is bidirectional? (L.16)

By an experiment by John Cairns. DNA was radioactively labelled by adding thymidine base that had been label with tritium ((3)^H) to the media that cells were growing in. The cells were labelled for one and a half times the time it takes the cells t to divide which is sufficient time for at least one of the two strands of DNA to be radioactively labelled and for the most recently synthesized DNA to be labelled on both strands. The cells were then lysed (ruptured) on a photographic media so that the length and shapes of the newly synthesized DNAs could be visualize. The bottom shows two region that are strongly labelled and there is a region in between them that is less strongly labelled. The interpretation is that the chromosome is being replicated at two locations, which have come to be know as Replication Forks. Since there are two forks, replication must be proceeding around the chromosomes bidirectionally.

What is the alternative semiconservative replication? (L.16)

Conservative Replication. the alternative would be process by which the parent molecule remained intact and somehow a new double strand molecule of DNA was synthesize and had the same sequence of bases as the parent.

How are traits encoded? Give an example (L.14)

Could be encoded by differentiations versions (or alleles) of the same gene or encoded by different genes. ex. Two forms of a single gene encode the information for flower color: One form code white, the other violent. * Pod color is coded by a separate gene whose two alleles encode either green or yellow pods.`

What is bigger the linear dimension of DNA or the dimensions of virions? (L.15) What are the differences in length?

DNA T4 bacteriophage has 60 kb genome which corresponds to a length of about 22 μm, whereas the viral coat is only 0.2 μm. About the 300th the length of DNA . *DNA IS A LOT LARGER THAN THE SIZE OF THE VIRION yet dna is still in there

Which DNA polymerase is more efficient in replicating the whole genome? Why is it better and why is the other one worse? (L.16)

DNA polymerase III is the more efficient at replicating the genome. It has a polymerization rate of up to 100 times faster than DNA polymerase I. It also has a processivity of greater than 500, 000 nucleotides. DNA polymerase I: It's the most abundant but it it not ideal for replicating the whole genome. Polymerization rate is only 600 nucelotides per minute . Has low processivity, synthesizing no more than 200 nucleotides before disengaging from DNA.

Let's talk about the structure of poly III. What is special about the functional parts of the polymerase? What parts of the complex are most like DNA pol I? (L.16) What are the green sections?

DNA polymerase III is: - much larger, more complex enzyme than polymerase - form shown on the left of the enzyme is called DNA pol III holoenzyme. Each of the different colored sections is a different functional part of the enzyme and some of these parts contain more than one polypeptide. Most functional parts are present three times. -Parts of the complex which are most like DNA pol I are the yellow segments. These are called the pol III cores. In green, is a five subunit complex to which all three cores are attach via T (tau) subunits, section is called the clamp loader. The yellow and green sections together form what is called DNA pol III.

What are the two different topological states of DNA?(L.15)

DNA relaxed and DNA super coiled

Image shows the importance and complexity of DNA metabolism. (L.16)

E. coli DNA metabolic processes are highly conserved in all species.

Describe how replication begins at the origin.(L.16)

E.Coli origin of replication is a specific sequence of DNA of 245 bp in length. Two critical parts of this sequence: 1. RT sites 2. DNA unwinding element (DUE)[AT rich regions] 1. R sites contain a specific DNA sequence that is recognized and bound by an ATP binding protein called DnaA. DnaA binds to the variants of the following 9 nucleotide sequence: T, T, A or T, T, any nucleotide, C, A, C, C. The binding of DnaA at these high affinity sequences promotes cooperative DNA binding of additional molecules of DnaA to lower affinity sites throughout the origin. These lower affinity sites will not be a precise match to the consensus sequence. 2. the DnaA protein molecules on the DNA form the helical structure shown in the center, which bend the DNA around, introducing positive supercoils in this region. 3. the resulting torsional strain induces the AT rich DUE to melt into two separate strand of DNA-shown in the center in orange. This bubble of single stranded DNA is recognized by a complex that compromises helicase DnaB and another protein, DnaC, whose sole role is to load DnaB onto the DNA. 4. Two separate Dnab complexes are loaded, one for each replication fork * the first primer generated at the origin one DnaB will be for the leading strand in one direction while for the second primer made by the same DnaB molecule on that same template strand will be for the first Okazaki fragment on the lagging strand.

Describe RNA synthesis: (L.17)

Enzymes that catalyze reaction are DNA dependent RNA polymerases (but often just called RNA polymerases) 1. The NTP adds to the 3' end of growing RNA strand 2. growing chain is complementary to the template strand in DNA 3. The synthesis is catalyzed by RNA polymerase and inlaces 2 Mg^(2+) ions our catalysis During synthesis, triphosphate nucleotides are incorporated one at a time into the growing RNA strand. The reaction involves nucleophilic attack by the 3'-OH whose nucleophilicity has been increased my nearby Mg2+ ions. The 3' OH of the RNA strand attacks the alpha phosphate of the incoming nucleotide, releasing pyrophosphate. ase pairing and base stacking provides the specificity to ensure the correct base is (usually) added. All of this is just like DNA replication. RNA polymerases can initiate synthesis without a primer, whereas DNA polymerases cannot.

What other major roles does recombination play in meiosis in germ-sin cells? (L.16)

Homologous recombination plays other important roles. We will not get into details, but probably one of its most important other roles is recombination between homologous pairs of chromosome at meiosis, i.e. between maternal and paternal copies of chromosomes. Just to remind you that at prophase, homologous chromosomes pair. And then at some frequency parts of the maternal and paternal chromosomes swap. On the figure you look carefully at the end of the red arrows. There are darker shaded segments of chromosomes that come from the homologous chromosomes. This process occurs via homologous recombination, which initiates at double strand breaks that are induced in one of the chromosomes. The entire process is very similar to that in Recombinational repair

What is critical for gene expression? (L.19)

From the previous section it should be clear that the ability of proteins to recognize some sequences of DNA with high affinity and others with much lower affinity is critical for the control of gene expression. Much work over the last 40 years has gone into determining how proteins discriminate different DNA sequences. One key finding is that there are several families of proteins that each have evolved different ways of contacting DNA. We'll briefly look at a couple of examples in this section.

What is the adapter hypothesis? (L.18)

How is the linear sequence of nucleotides in mRNA converted into a linear sequence of amino acids. Let's first take stock of the problem. There are 20 amino acids that make up protein polymers. There are four bases that make up the mRNA polymer. Obviously there cannot be a simple code where one base specifies one amino acid. So we need to have a code that uses more than one base to define an amino acid. In fact, we will need a code that uses at least 3 bases. A two base code such as AA for one amino acid, AG for another amino acid, AC for a third etc. will not work as there are only 16 possible combinations. A three base code will work, however, as there are 64 possible combinations such as AAA, AAG, AAC etc. It turns out that life uses a three letter code. The way that the code is read was first proposed by Francis Crick in his so called "Adaptor Hypothesis". He realized that there needed to be a molecule that could base pair with the mRNA and that also had attached to it an amino acid. We now know that these adaptors exist and are called tRNAs.

Genes are not all expressed at the same levels. (L.19)

Genes are not all expressed at the same levels. The slide shows the number of protein molecules per mammalian cell for a number of genes. The most abundant protein is many cells is actin, which in present at 150 million molecules per cell and constitutes 5% of all protein in a cell by mass. There are over a million molecules per cell of some of the protein complexes we have learned about. For example, there are about 8 million ribosomes per typical mammalian cell and 5 million molecules of the ATP synthase complex. However, the median abundance of a protein is about 150,000 molecules per cell. Many regulatory proteins such as transcription factors can be present at as few as 1,000 molecules per cell, though the median transcription factor is present at 60,000 molecules per cell. These differences in abundances are crucial. If a transcription factor were expressed at the same abundances as ribosomes the transcription factor would bind to every part of the genome and cause aberrant transcription of every gene. If there were only 1,000 ribosomes in a cell, the cell would only be able to make less than 0.1% of the protein it needs. Abundances matter.

What are the four classes of introns? Describe them? (L.17)

Group I and group II introns are self-splicing. Theyrequire no additional proteins or ATP. They are an example of a catalytic RNA and are sometimes called ribozymes. However, note a self spicing intron is not catalytic in the same way that an enzyme is; an enzyme can be reused to catalyze multiple substrate molecules but group I and II introns catalyze the reaction of just one molecule. Other RNA catalysts are more like enzymes and can catalyze repeated reactions, as we will see later. Group I and II introns are found in some nuclear, mitochondrial, and chloroplast rRNAs, tRNAs and mRNAs. Group I and II introns differ mainly in the exact details of their splicing mechanism. The next class of introns are Spliceosomal introns. Theseare spliced by enormous complexes called spliceosomes that contain both RNA and protein components. Spliceosomal introns are much the most common introns and are found in protein-coding genes in eukaryotes. tRNA introns are spliced by protein-based enzymes. The tRNA primary transcript is cleaved by an endonuclease and the exons are joined by an ATP-dependent ligase.

What are the major proteins, RNA's and metabolites involved in each of the four steps in E.coli? (L.18)

Here are a list of the major proteins, RNAs, and metabolites involved in each of the four steps in E. coli. For stage 1, I'll just note that the minimum number of tRNAs needed to recognize the 64 codons is 32 given the wobble rules and the need to code for 20 amino acids, the 3 stop codons, and for reasons I'll explain shortly two different tRNAs for the AUG codon. For stage 2, in addition to the two ribosomal subunits, three protein initiation factors (IF 1 -3) are required. In terms of RNAs, the mRNA is needed and an amino acylated tRNA for AUG that is used only for the first AUG in the protein. This special AUG tRNA is not used for other methionine codons in the rest of the reading frame. Translation uses GTP not ATP as the energy carrier. For stage 3, the ribosomal subunits have formed the 70S ribosome. Three elongation proteins are required (EF-Tu, EF-Ts, EF-G). Amino acylated tRNAs for every possible amino acid codon and the three stop tRNAs are also required during elongation. For stage 4, another group of proteins called release factors are needed to stop translation and recycle the 70S ribosome to create the 30S and 50S subunits.

Measurement error mist be accounted for. (L.19)

I promised at the start of the first lecture to tell you a bit about my research. One of the questions my collaborators and I have addressed was .... is the error in the data large enough to change the conclusions of studies such as Schwanhausser's. Without getting into details, the scatter of data you see on either side of these regression lines is an estimate of the error in Schwanhausser'sdata. There is a standard statistical technique that allows you to take this error into account. When we applied this statistical technique to Schwanhausser'sdata, it suggested that most of the poor correlation between the mRNA and protein abundance data was due to this error, not to differences in the rates of translation.

What is the mechanism of degradation? (L.18)

In E coli, there are several systems that degrade proteins. Each system targets subsets of proteins depending on their structure or subcellular location. For example, Lon (for "long form") is an ATP-dependent protease that hydrolyzes defective or short lived peptides. This protease places a protein molecule in a chamber, unfolds it, and then degrades it within the chamber. Other proteases may further degrade large peptides that released from the chamber.

What is another way that bacterial transcription is controlled? (L.19)

In addition to control by transcription factors like the lac and trprepressors, we have previously learned about another way that bacterial transcription is controlled. We learned in lecture 17 that there are several types of sigma factor that each recognize different sets of sequences just 5' of the RNA start site. The sigma factor recruits RNA polymerase to bind to those promoters that have that sigma factors specific recognition sites.

genes also control the expression of same protein between cells (L.19)

In addition to controlling differences in the expression levels of proteins within a cell, expression of the same protein differs between cells. I n the example shown, a regulatory transcription factor called ftzis expressed in only a subset of cells in the early embryo of a fruit fly. This embryo is about half a millimeter long and has about 5,000 cells around the outer surface. The small grey ovals, one of which is arrowed, are nuclei. The very dark back areas show those cells which express mRNA transcribed from the ftzgene. The other cells do not transcribe this mRNA. Differences in which cell this and other transcription factors are expressed determine the the way that these early undifferentiated cells will subsequently divide. The difference between the parts of animals, such as head vs legs, originate due to differences in the expression of regulatory transcription factors in early embryos. This is true in all animals. Indeed, surprisingly some of the transcription factors that control differences between parts of animals are conserved across all animals.

RNA processing (L.17)

In bacteria, mRNAs are transcribed and degraded very rapidly, within 1 to 2 minutes. The cell has no nucleus. The mRNAs are translated into protein as they are being transcribed. There is not much time or need for complex processing of mRNA transcripts in bacteria. In contrast, in eukaryotes mRNAs are transcribed in the nucleus and then transported to the cytoplasm where they are translated to make proteins. Most eukaryotic mRNAs are much more stable than prokaryotic messages. Eukaryotic mRNAs have an average half live of 10 hours in rapidly dividing cells and 100 hours in non dividing cells. For this and other reasons, eukaryotic mRNAs are modified much more extensively than prokaryotic mRNAs. So in this section, we'll mostly consider eukaryotes.

What is the mechanism of a DNA ligase reaction? (L.16)

In each of the three steps shown, one phosphodiester bond is formed at the expense of another. Steps 1 and 2 lead to activation of the 5' phosphate in the nick. An AMP group is transferred first to a Lys residue on the enzyme and then to the 5' phosphate in the nick. In step 3, the 3' hydroxyl group attacks this phosphate and displaces AMP, producing a phosphodiester bond to seal the nick.

What is the strategy for degradation for eukaryotes? (L.18)

In eukaryotes a different strategy is used. Proteins that are identified for degradation are first linked to the protein ubiquitin. Ubiquitin is a 76 amino acid protein that has nearly an identical sequence in eukaryotes as diverse as humans and yeast. The name comes from the fact that the protein is ubiquitous (at least in eukaryotes). Ubiquitin is linked to target proteins via an activating enzyme E1, a conjugating enzyme E2, and ligating enzyme E3. We do not need to learn these details, though. After one Ubiquitin molecule has been added, additional cycles produce polyubiquitin, a covalent polymer of ubiquitin subunits that targets the attached protein for destruction in eukaryotes. Ubiquinatedproteins are cleaved by large protease called the 26 proteasome complex.

Does just one way of alternate splicing or are there many possibilities? (L.17)

In fact, a wide series of possible alternate splicing patterns are possible, some of which are shown here. Take the time to look carefully at the various possible splice alternatives. Look at each row and make sure you understand how the mature mRNAs on the right are produced from the spicing patterns shown on the left.

How does nucleosomal DNA behave in the scaffold attachment site? (L.15)

In metaphase chromosomes there is a protein scaffold won which Nucleosomal DNA is attached. There ware specific parts of the genome that attach to this scaffold (scaffold attachment sites). The nucleosomal DNA in between the seafood attachment sites loops out, and perhaps for something like the proposed factual globules in the last slide.

In order for DNA to wrap around the histone core, what does it need to do? Why? Where does it get the engird to do this?(L.15)

In order for DNA to wrap around the histone core the DNA must remove one helical turn. It requires this removal because it is engertically more favorable for the double helix to slightly unwind in order for the DNA to roll around the octomer core. Since the histones are tightly bound to the DNA backbone.

What does the Nick translation do for the enzyme in the process of DNA? How does this help in DNA repair and synthesis?(L.16)

It effectively moves the nick along the DNA as the enzyme progresses. This process has arose in DNA repair and in the removal of RNA primers during replication. The strand of nucleic acid to be removed is shown in purple, the replacement Brandon red . DNA synthesis begins at a Nick. A nick remains where eDNA polymerase I eventually dissociates, and the nick is later sealed by another enzyme called DNA ligase

What does mismatched repair rely on? What happens that causes mismatched base pairs needs mismatch repair? How does the enzyme know which is the correct base and which one the wrong one in Coli? (L.16)

It relies on methylation to determine which of the two bases is correct. DNA replication occasionally inserts incorrect bases on the newly synthesized strand leading to mismatched base pairs. These pairs distort the helix structure as the pair is the wrong size. The cells have enzymes that can detect these small distortions. In E.coli , DNA becomes methylated several minutes after it is synthesized. An enzyme called Dam methylase inserts CH3 groups in this adenines that occur in the GATC sequence. No other adenines are methylated because the methylase binds only to the sequence GATC. All parental strands of DNA are thus methylated, but for the first few minutes after DNA synthesis the daughter strands are not. The mismatch repair systems assume that replication errors must reside in the unmethylated strand.

Is mitochondrial genome more like bacteria genome or eukaryote genome? Why?(L.15)

It's more like bacteria genome because its origin is bacterial. Mitochondria and chloroplasts originated as endosymbionts that invaded photo-eukaryotic cells.

What is the most important role in regulating a protein? (L.19)

Jacob and Monod's discovered how expression of the lac operon was controlled in 1960. For the 30 years after that discovery most attention focused on transcription. However, increasingly there have been a number of discoveries that suggest that differences in the rate of these other three steps also play a role. Some groups have attempted to quantitate the degree to which each of the four steps control the differences in protein abundances. This slide summarizes work from one highly cited study by Schwanhausseret al. T his group concluded that post transcriptional control was more important that transcriptional control. Their evidence for this is largely that in mouse tissue culture cells measured levels of mRNA only correlate poorly with mass spectrometry measurements of protein abundance for thousands of genes. The number of RNA molecules per cell are plotted on the x axis and the number of protein molecules per cell are plotted on the y axis. Each data point is the abundances of one gene. Highly abundance genes data is to the top right, low abundance genes to the bottom left. Notice that axis are scaled in log 10. Realize that if protein and mRNA abundance correlated perfectly, that would suggest that there was no difference in the rates of translation or protein degradation. Both translation or protein degradation rates would have to be the same for every gene. If that were the case, this then the amount of each protein produced would be proportional to the number of mRNA molecules. So instead of this messy scatter of points you can see, all of the data points would lie in a line along that curved red regression line. But instead Schwanhausseret al.'s measurements of protein and RNA levels do NOTfall in a line. They see this large scatter of data points. The degree of this scatter Schwanhausseret al.'s took to be a measure of the combined differences in the rates of protein degradation and translation. Schwanhausseret al also separately measured the half lives of proteins and mRNAs using a method I'll not describe. Using these data together with the protein and RNA abundance data, they calculated the fraction of protein expression that is explained the variance in the rates of each of the four steps. As you can see, they felt that there was more variation in the rates of translation of mRNA than there was any other step. In their view, differences in transcription rates played only a minor role in determining the abundances of protein. Many other groups reached a similar conclusion based on related experiments. However, there is another reason why you might have a poor correlation between protein and mRNA abundance. If the mass spec data and or the mRNA abundance data are not accurate, the data will not correlate well for that reason.

What is the mechanism for splicing in group I introns? (L.17)

Just to repeat and be really clear, group I and II introns splice themselves. The RNA itself does it all. This slide shows the mechanism for group I introns. The exons are in green and the intron is in yellow. The mechanism involves two rounds of nucleophilic attack by 3' -OH groups to phosphates in the sugar backbone of the RNA. In the first round, the 3'-OH of a free guanosine such as GMP or GDP is used as the nucleophile. This attacks the phosphodiester bond between a U and A residue at the 5' end of the intron, releasing the 5' end of the intron and exposing the 3'-OH of the upstream (5') exon. In the second round, the 3'-OH of this U-ending upstream exon then attacks the downstream intron/exon boundary to join the two exons. The spliced intron is eventually degraded. The secondary structure of the mRNA, such as hairpin loops etc., plays an important part in determining the exactly where the splicing happens. In other words, the RNA fold up in a way that brings the reacting nucleotides together. I'll show you an example of secondary structures role soon..... for group II introns

What can the lac repressor do to DNA? (L.19)

Lac repressor is a tetramer. More exactly it is a dimer of dimers. The protein is shown in the pale red cartoon at the bottom, and from an X ray crystallography study to the right. Each half dimer binds to one half of the inverted repeats shown in the last slide. Dimers bound at O1and either O3or O2then come together, which loops out the DNA between the two operator sites. This whole looped DNA structure strongly reduces the ability of RNA polymerase to bind the promoter. It reduces transcription of the lac operon 1,000 fold, but transcription occurs at a low, basal rate, even with the repressor bound. This is important as a low basal activity of lac permease is required at all times. One of the key pieces of evidence that led Jacob and Monod to propose their model for the control of gene expression is that mutations in the operator sequences caused transcription of the lac operon to be high. In cells containing these mutant genes, the lac operon was not repressed even when lactose was absent. This suggested that some protein was binding to the operator in normal cells and repressing transcription. But in bacteria with mutant operator sites, it was proposed the repressor could no longer bind to the operators and thus no longer repress transcription. Mutations in the lac I gene that prevented expression of the lac repressor also led to derepressionof the lac operon. i.e. also led to transcription of the lac operon.

Transcription by RNA pol II involves many proteins. (L.17)

Lehninger describes how this collection of proteins assembles in order on the promoter. At the top you can see the location of the TATA box at -30. However, while this ordered reaction takes place in test tube experiments, in cells at many promoters it seems that often all of these proteins are recruited at once along with a much larger collection of proteins. For that reason, we'll not learn this ordered reaction. In the last lecture, we'll learn about the more complex reality in cells. Like bacteria, eukaryotic RNA pol II first binds to the promoter as a closed complex. Then it unwinds the DNA to create an open complex. Elongation factors then bind to the polymerase and—as I mentioned—the CTD is phosphorylated by pTEFb. The CTD is an extended domain, that looks like a dog or cats tail sticking out of the polymerase. It is comprised of between 27 -52 copies of a seven amino acid repeat ......YSPTSPS. As you might guess from this sequence, the CTD is a disordered protein domain, and it serves as a binding site for a variety of other proteins

What is the difference between DNA lesions vs mutations? (L.16)

Lesions is some damage t DNA and becomes a mutation if not reparied. Mutation: can be substitutions (point mutations) deletions or insertions. they are permanent there are thousands of lesions occurring everyday but few become mutation thanks to DNA repair

Give 3 examples of Lesions. (L.16)

Lesions: alterations in DNA Mutagens: alter the chemical make up of one or more bases. 1. some mutagens methylate bases, altering their properties. in example on the top left, a Guanine base has been methylated, blocking its ability to correctly hydrogen bond with cytosine. Instead now it pair with thymine. 2. example in bottom, UV light is absorbed by adjacent pairs of thymine or cytosine bases to form radicals that react, covalently coupling adjacent bases in a way that distorts the shape of the helix. 3. DNA is less chemically stable than you think and so some bases may simply fall off.

What three roles does RNA play in all living cells? (L.17) What additional functions does RNA have?

Messenger RNAs (mRNAs): encode thiamin cid sequences of all the polypeptides found in the cell. Only function is toprovide the information for the amino acid sequence of its corresponding protein. mRNAs are the RNAs that the central dogma refers to and indeed most genes do code for mRNAs and proteins. FUNCATIONAL RNAs: made form genes that only make RNA Transfer RNAs (tRNA): match their anticodon to the mRNA whole carrying a specific amino acid used for protein synthesis Ribosomal RNAs (rRNAs): are constitutions of the large and small ribosomal subunits additional: MircrosRNAs (mirRNAs): regulated the expression of eukaryotic genes, possibly via binding to specific nucleotide sequences. -human genome encodes over 1,500 micro RNA genes -miRNAs are short 22 nucleotide RNAs that bind to mRNAs from subsets of genes. Ribozymes: catalytic RNA molecules that act as enzymes Small nuclear RNAs (snRNAs): help splice out introns -removal of introns from primary transcripts of protein coding genes involve 5 RNAs of 100 to 200 nucleotides

What two other structures in eukaryotes other than the nucleus also contain DNA?(L.15)

Mitochondria and chloroplasts (in plants)

What is happening in mitosis? What is happening in the: S phase? G1 phase? G2 phase? G0 phase? quick overview(L.15)

Mitosis: when the cells actually split in two S phase +G1 phase (interphase): cells are undergoing repeated division cycles -during interphase the chromosomes are more amorphous (less tightly organized) G2: DNA is replicated G0 phase: when cells in your body are not dividing , terminally differentiated. -during G0 phase chromosomes are in the same amorphous state as G1

What are the are other person and transcription factors? (L.19)

Most genes in E. coli are organized into operons. Commonly each operon contains a group of genes that have a similar function. In the example shown here, the genes in the trpoperon are needed to synthesize tryptophan. In other cases, genes in an operon may not have a common function but they are needed at similar times or conditions. i.e. the cells needs certain sets of genes to be switched on and off together. Many operons are controlled by a mix of activators and repressors. Importantly, each regulator binds to a different operator sequence. Operator sequences tend to be fairly long .... in E coli, say 15 to 17 bp, and are only present in a few locations in the genome. Operators are mostly found at the 5'ends of the operons they control. Thus the sequence specificity of these regulatory proteins is a major mechanisms that determines how they act. In the example shown the trprepressor binds to an operator sequence when tryptophan levels in the cell are high. Tryptophan binds to the trprepressor, acting as an effector that causes a conformation change in the protein, increasing its affinity for DNA. When the concentrations of typamino acid are low, trpdisassociates from the repressor, the repressor disassociates from the DNA the operon is transcribed. The enzymes produced as a result synthesize tryptophan from precursor metabolites. Which will of course increase the amount of trpin the cell.

Where are introns found (L.17)

Most genes in vertebrates have introns and a minority of genes in yeast have introns as well. In vertebrates, Exons are usually <1,000 bp in length whereas introns can be anywhere from 50 bp to 700,000 bp in length. The average length of an intron is ~1,800 bp. Some animal genes have dozens of introns. The human genome has more than 200,000 introns spread across ~20,000 genes

Are Histone tails the same throughout all of the genome? If not why are they different?(L.15)

No. they are not! Histone tails in different parts of the genome are differently modified and these modifications affect the way that the histones compact DNA and recruit other proteins to the DNA. The differences in the histone tail are associated with differences in rates of transcription of the genes they are associated with, suggesting they may be involved in controlling gene expression.

Why do nucleosomes have small DNA sequence preference? Why?(L.15)

Nucleosomes will bind to any DNA sequence but some combination of base pairs bend more easily than others. Therefore nucleosomes occur with a higher frequency in short sequences of 2-3 AT base pairs because they allow the minor grove of DNA to be compressed more. Nucleosomes also prefer to bind to sequences that have a few AT bps every 10 or so nucleotides and in between these tend to have GC bps. This sequence preference though small tents to affect exactly where a nucleosome sits, not whether it binds to DNA at all.

What problem do the clamps create in the Lagging strand synthesis? (L.16)

Problem: The processivity clamp keeps polymerase on DNA for >500,000 bp but Okazaki fragments are only 1,000 - 2,000 bp. The polymerase cores must be disengaged after synthesis of each fragment. Solution: The B(beta) sliding clamp is left behind at the end of each Okazaki fragment to disengage itself. The clamps are then removed in a separate reaction. The core Pol III enzyme "jumps forward" and a new clamp is placed on DNA location of each new RNA primer by the clamp loader. The leading strand B(beta) sliding clamp never dissociates. (The pol II complex swings from clamp to clamp like an orangutan)

What is Proofreading? How does it work?(L.16)

Proofreading is a correction mechanism in the synthesizing of DNA to avoid mutations since mutations can be deadly. It uses and activity intrinsic to many DNA polymerases. These enzymes have an active site for polymerization and a separate active site located in the palm that can remove nucleotides in 3' to 5' direction. [Exonuclease Activity] The way that it works is, 1. if the enzyme incorporated a wrong base into the primer, the steric constraints we saw prevent translocation of the enzyme. The kinetic pause allows the movement of the template from the polymerase active site to the 3' to 5' exonuclease site. The exonuclease then excises one or more bases. (only a few nucleotides are excised before the template relocates to the polymerization active site and synthesis recommences. Reaction cause the Net error rate of enzyme to be (1/10^(6)) to (1/10^(8))

How do you determine the linking number for: Relaxed closed circular DNA(w/ 2100 bp). Under wound DNA (w/ 2 turns) Nick strand? (L.15)

Relaxed closed circular DNA(w/ 2100 bp): Lk0=(#bp int the plasmid)/ (bp per turn.) [so for a plasmid of 2,100 bp it would be Lk0= 2,100 bp/10.5= 200] *the more bp there are in a molecule, the bigger the linking number will be!* Under wound DNA (w/ 2 turns): ΔLk= Lk-Lk0 Nick strand: because DNA is nicked and therefore must be relaxed there is no linking number

Retrotransposons are defective retroviral genomes. (L.17)

Retrotransposons are derived from retroviruses. They are mobile genetic elements found in eukaryotes, but not in bacteria. These transposons encode an enzyme with homology to the reverse transcriptase of retroviruses and many are flanked by LTRs, just as retroviral genomes are. Retrotransposons insert at new positions in the genome via RNA intermediates using their reverse transcriptase to make DNA from RNA. This distinguishes them from bacterial transposons, which only move directly from DNA to DNA. Retrotransposons are a form a parasitic DNA. They are in effect defective viral genomes have lost the ability to make virion particles and so they are stuck in the cell. 8% of the human genome is made up of LT R retro tra ns po s o ns . The figure shows the genome structures and the genes encoded by retrotransposons found in yeast and the fruit fly, Drosophila. You should notice that they do not have the envprotein needed to make infectious virus.

What is the mechanism of retroviruses? (L.17)

Retroviruses have genomes of ssRNA. These RNA genomes encoded several genes including the enzyme reverse transcriptase. The virus particle that infects cells includes molecules of the reverse transcriptase protein. When the virus enters the host cell, the reverse transcriptase synthesizes first one strand of DNA from the RNA, then the same enzyme uses an RNase activity to degrade the RNA strand, then it synthesizes a second strand of DNA. Once converted to a double-stranded DNA, the DNA enters the nucleus and is integrated into the host genome. The integration is catalyzed by a virally encoded integrase. Integration of viral DNA into host DNA is mechanistically similar to the insertion of a bacterial DNA transposon into a bacterial chromosome.The virus then uses the host cells enzymes to transcribe and translated its genes to make more viral proteins and single stranded viral RNA genomes, which eventually are packaged to make viral particles that leave the cell to infect new cells.

What disease can retrovirus cause? How? (L.17)

Retroviruses include viruses that that cause cancers in animals, though no retroviruses directly cause human cancers. The HIV virus that causes AIDS is also a retrovirus. The figure shows the genome structure and the genes encoded by HIV, at the bottom, and a virus that causes cancer in chickens: the Rous sarcoma virus (rsv), at the top. Retroviruses have repeated sequences at each end of the genome called long terminal repeats (LTR). These are used by the integrase to insert them into the genome. Interestingly, the HIV reverse transcriptaseis unusually error prone. This a major reason why it has proven so difficult to develop a vaccine for AIDS. The sequences of its proteins keep changing, meaning that antibodies to one HIV strain are not effective against another strain of the virus. The AIDS therapeutic cocktail has saved the lives of many millions of people. You have probably heard that this cocktail includes a protease inhibitor. That inhibitor acts by inhibiting the protease that cleaves the polyprotein.

How is the genetic code structured? (L.18)

Shown at the top, living organisms use a nonoverlapping mRNA code with no punctuation. Each amino acid is encoded by a three letteterm-240r word called a codon, or sometimes called a triplet. The code or the words are directly adjacent to one another, with no gaps between them. One feature of this sort of code is that it can be read in three ways. Depending on which nucleotide is used to start the first codon, there are three possible reading frames. Depending on which reading frame is chosen, the sequence of amino acids that would be produced would be very different.The code has a special three letter word .... AUG ,,,, to indicate which the first amino acid is. This is called the start or initiation codon and it defines which of the three reading frame will be chosen.

What are some similarities between DNA replication in eukaryotes and in E. coli? What are the main differences? (L.16)

Similarities: many of the enzymes are related. They have sufficiently similar sequences that we can conclude they have a common ancestor. Other enzymes may not be homologous by sequence but have similar functions. Differences: because of the larger size of the eukaryotic chromosome, it would take too long to replicate a sinful chromosome starting from a single origin chromosome. So to overcome this Eukaryotic chromosomes have multiple origins saved about 30 to 300 kb apart. The origins in animals and plants are less well defined than in E.coli and tend to be larger

How many helical turns would be in the relaxed state? How would the DNA look?(L.15)

Since the DNA has 84 bp and there are 10.5 bp per turn you would have a total of 8 helical turns (84/10.5). Since the DNA is in a relaxed state the shape would be that of a circular DNA and would not have any supercoils. Would jus be a circle (0).

What are plasmids? (L.15)

Small, circular DNA molecules found in bacteria *bacteria have both plasmids and Chromosomes

How do U1 and snRNP and U2 snRNP bind? What does is affect? (L.17)

Spliceosomal introns contain a GU at their 5'-end and AG at their 3'-end. You can see this at the bottom of the slide, where the intron is in yellow. Several other nucleotides tend to occur at the other locations nearby—as shown in the figure. For example, there is often an AG at the 3' end of the upstream exon. Two of the snRNAs pair with the exon/intron junctions, bringing their associated proteins with them and—in the case of U2—the other three snRNPs as well. U1 pairs with the 5' junction, U2 with the 3' junction. The pairing of U2 distorts the backbone of the intron, displacing an A residue (the one highlighted in pink). This distortion increases the nucleophilicity of the 2' -OH on the A residue. This nucleophile attacks the phosphodiester bond at the 5' boundary of the intron to form a lariat structure, just like we saw for group II introns.

What is alternative splicing and how is it done? (L.17)

Splicing probably has several benefits to the organism, but one major one is that it allows for several variants of each protein to be expressed through alternate splicing. A single gene can yield different peptides depending on RNA processing. Particular regions may be retained or removed, yielding different mature transcripts. In the figure you can see that there are two alternate 3' splice sites and one 5' splice site. Depending on which 3' spice site is used a different mature mRNA will be produced. Depending on the sequence coded by the exons, the longer mRNA could encode a longer version of the protein, for example. At least 95% of human genes are alternatively spliced, so this mechanism can substantially increase the diversity of proteins expressed.

What is the mechanism for Topoisomerase I? (L.15)

Step 1: Type I topoisomerases covalently attach to one of the strands of DNA and as part of the same reaction step cleave the DNA strand Botton of this slide shows that the DNA cleavage occurs when a tyrosine residue in the active site of the enzyme undertakes a nucleophilic attacks of a phosphodiester bond in the backbone of the target DNA. Step 2: Enzyme changes to an open conformation. Step 3: The unbroken DNA strand passes through the break in the first strand and the broken strands Step 4: After the unbroken strand has been passed through the gap in the DNA strand the 3' OH from the broken DNA end attacks the phosphotyrosine bond, reforming the DNA helix and releasing the enzyme.

Describe the Topoisomerases Type I mechanism steps 2 and step 3.(L.15)

Step 1: Type I topoisomerases covalently attach to one of the strands of DNA and as part of the same reaction step cleave the DNA strand. Step 3: The unbroken DNA strand passes through the break in the first strand and the broken strands get re-ligated at the same time as the covalent coupling of the DNA to the enzyme is reversed.

Describe the process of a base-excision repair in process. (L.16)

Step 1: a DNA glycosylase recognizes a damaged base (uracil) and cleaves between the base and deoxyribose in the backbone. Step 2: an endonuclease called an AP endonuclease cleaves the phosphodiester backbone near the AP site. Step 3: DNA pol I initiatives nick translation repair synthesis from the free 3' hydroxyl at the nick using its 5'-3' exonuclease activity. This replaces a portion of the damaged strand. Step 4: the nick remaining after DNA polymerase I has dissociated is sealed by DNA ligase. other glycosylases create AP sites where there are 8-hydroxyG, hypoxanthine, 3-methyladenine, and other wrong bases. Ap sites that result from depurination are repairs by the same pathway, except that there is none for glycosylase to remove the base as it is already gone.

What is the pathway for nucleotide excision repair in E.coli? (L.16)

Step 1: an enzyme called an excinuclease binds to DNA at the site of a bulky lesions nand cleaves the damaged DNA strand on either side of the lesion. In E.coli , this enzyme is a complex called the ABC exinuclease. Step 2: The DNA segment of 13 nucleotides removed with the aid of a helicase. Step 3: The gap is filled in by DNA polymerase I. Step 4: the remaining nick is sealed with DNA ligase

How many polypeptides are there in a complex?(L.16)

THERE ARE MANY as shown by the table but all you need to know is that there are MANY POLYPEPTIDES IN A COMPLEX.

Template vs coding strand (L.17)

Template strand: serves as template for RNA polymerase Coding strand: the non-template strand; has the same sequence as the RNA transcript The DNA template strand for RNA synthesis is defined in the same way that the DNA template strand is defined in DNA synthesis, and that is how I have shown it in the last couple of slides. In transcription, the non-template DNA strand is called the coding strand. In case that name sounds counter intuitive, look at the sequences shown in the bottom half of the slide. You can see that the non-template strand has the same sequence as the RNA transcript, except that at every location where there is an A in the DNA there is a Uracil base in the RNA. For this reason, the non-template strand is usually called the coding strand. The coding strand runs 5' to 3' in the same direction as the RNA sequence. It is a widely used convention that when only one strand of DNA sequence is written, the sequence given is that of the coding strand.

What is the the 5' cap (L.17)

The 5' cap is added enzymatically to the 5' most nucleotide of the synthesized primary transcript of mRNAs. The cap is a guanosine nucleotide that has a methyl group on the Nitrogen atom of the base—shown by the red arrow. This nucleotide is linked to the transcript via an unusual 5' to 5' tri phosphate linkage. The guanosine is added unmethylated and is subsequently methylated once it has been added to the transcript. The first two nucleotides in the synthesized transcript may also be methylated on the 2' -OH of the ribose residue. The methylguanosine cap protects the mRNA from degradation by 5' to 3' RNA exonucleases that are present in the cytoplasm. Any mature mRNAs in the cytoplasm that have their 5' cap's removed are rapidly degraded. The cap is added when only 20 -30 nucleotides of RNA have been synthesized. The cap is added by a protein complex of called the Cap synthesizing complex that is tethered to the CDT. Once the cap is added, it is bound by a second complex called the cap binding complex (CBC) that also associates with the CTD, forming the structure shown at the bottom right. The 5' cap remains joined to RNA polymerase II during much of the elongation phase. The cap binding complex remains attached to the mRNA when it is released from pol II and serves to protect the mRNA from degradation in the cytoplasm. Capping synthesizing factors do not bind to the RNA Pol I or Pol III enzymes. As a result transcripts made by these polymerases—i.e. rRNA and tRNAs—are not capped. The presence or absence of the 5' cap is one way that cells distinguish which RNAs are mRNAs. For example, the cap helps the translation machinery bind specifically to mRNAs.

What does the 5' to 3' exonuclease activity do? How does its' position help it?(L.16)

The 5' to 3' exonuclease activity resides in a separate domain position at the from of the enzyme as it moves down the template. As a result the enzyme degrades any DNA strand ahead of the protein while simultaneously synthesizes a new strand behind it. DNA polymerase I promotes a reaction called nick translation, where a nick is a broken phosphodiester bond that leaves a free 3' hydroxyl and a free 5' phosphate on one strand of the DNA.

What experiment was done in order to differentiate between conservative vs semiconservative replication? Describe the Experiment. What do the results tell you?(L.16)

The Meselson-Stahl Experiment 1. Cells grown on only (15)^N Medium Grew E. coli in a medium where are the nitrogen isotopes were (15)^N isotope [is heavier of the two nitrogen isotopes] called the heavy nitrogen. After many generations of cell growth and division all of nitrogen cells would be (15)^N. - ONE band DNA in centrifuges CsCl. When DNA is added to a concentrated solution of Cesium Chloride (CsCl) and spun at very high speed for many hours the DNA will migrate to a set location in the centrifuge tube. The CsCl will become more concentrate at the bottom of the centrifuge tube and so the solution will be more dense at the bottom. The position in the tube where DNA migrates depends upon its density. DNA migrates to where the CsCl solution has the same density as the DNA. DNA from cells gown in heavy nitrogen migrates as bands towards the bottom of the tub (panel a). 2. Cells switched to (14)^N medium and allowed to divide once. -ONE band at higher position than (15)^N DNA, but lower than all (14)^N DNA. They grow cells in heavy nitrogen for many generation, and the transferred the cells to a medium containing (14)^N [light nitrogen] for a short period that allowed for cells to divide once and therefore long enough for the genome to replicate just once. The DNA from those cell was spun in ultra centrifuge in a separate tube and found to migrate asa single band that was higher up the centrifuge than that of the heavy nitrogen DNA, where the density of CsCl is lower. (panel b) 3. Cells divide once more -Two bands one with all (14)^N DNA, one hybrid They repeated the second experiment but allowed the cells to grow in the light nitrogen medium for twice as long (until cells had divided one more time). The DNA from this experiment produced two bands of DNA. Shown in panel c. One band corresponded to the density of the DNA from the second experiment, the other ran at a lower density (higher up the centrifuge tube) The results are consistent with semiconservative replication and inconsistent with conservative replication. Because semiconservative replication predicts the densities of DNA seen by Meselson and Stahl. You can see this from the predicted densities shown fro the DNA molecules on the right. Conservative replication would never produce the result seen in panel b. For experiments b and c it would give two bands, one at the all (15)^N and one at the all (14)^N.

What is the dominant step at determining protein levels? (L.19)

The best estimate we were able to make suggests that differences in transcription rates between genes dominate in determining the difference in protein abundances between genes. The comparison between our estimates and Schwanhausser'sis shown. I show you this data to make the point that this is an area where the field does not agree. Other researchers have agreed with our conclusions and indeed were wrote a review in Science describing our work and a figure summarizing our conclusions is included in the undergraduate textbook Molecular Biology of the Cell. But other workers do not accept our conclusions. In this course, you have mostly learned about research conducted many years ago that is now accepted as true. But there are many areas under current investigation where different ideas are being tested and debated. This is a part of science you will learn more about in more advanced courses.

What is the mechanism of slicing of group II introns? (L.17)

The chemistry of group II splicing is similar to that of group I splicing, except for the identity of the nucleophile in the first step and formation of a lariat(a lasso) like intermediate. The Lariat is shown in the center and has a 2',5'-phosphodiester bond. Think about that: A phosphodiester bond is formed between the 2' position on one ribose and the 5' position on another ribose. The nucleophile is a 2'-OH of an A residue within the intron—shown at the top of the loop. This attacks the phosphodiester bond at the junction between the upstream (5') exon and the intron—between the U and the G. The backbone is cleaved and a new phosphodiester bond is formed that links the A-2'OH to a G-5'-phosphate, producing the lariat structure shown in the center. The newly exposed 3'-OH of the upstream exon then attacks the downstream intron/exon phosphodiester linkage, releasing the intron as a lariat and joining the two exons. The lariat will eventually be degraded.

How is the code written? (L.18)

The code is written in the 5' to 3' direction. Every one of the 64 possible three letter codons is used to mean something. The code is degenerate in that most amino acids are coded by more than one codon. The third nucleotide in each codon generally carries less information and, as we'll see is less important for binding to the tRNA. The table shows a common way to present the code. The first letter of a codon is shown in the left, with each row containing that nucleotide. The Second letter of the code is show at the top, with each column containing that nucleotide. The third letter of each triplet is written with the codon. The amino acid that each triplet codes for is also indicated next to each codon. As examples, you can see in the top left that the amino acid Phe is coded by two codons, UUU and UUC. Some amino acids are coded by four of more codons. For example, Gly is coded by GGU, GGC, GGA, GGG. Notice that generally the first two letters are the same for a given amino acid, with the degeneracy being due to differences in the third position.

What does the lac operon reveal about gene regulations> (L.19)

The discovery of how the lac operon is controlled was made in 1960 by two French researchers at the Pasture institute: Jacob and Monod. These two showed that three genes for the metabolism of lactose are regulated together as an operon. These genes are: b-galactosidase (or lacZ), which cleaves lactose to yield glucose and galactose; lactose permease (or galactosidepermease; lacY), which transports lactose into the cell, and thiogalactosidetransacetylase (lacA) The rate of transcription of the lac operon relies in part on negative regulation by a repressor called the lac repressor or lacI

What is the mechanism of a eukaryotic type IIα topoisomerase?(L.15)

The enzymes bind to two different segments of DNA and make a double strand cut in one of them. Step 1: First one subunit of the enzyme binds to one segment of DNA. (light blue DNA) Step 2: A second segment of the same DNA molecule is bound by the second subunit (dark blue) at the N gate. Step 3: A protein conformational shift closes the second subunit around the DNA, trapping it. The first DNA segment is cleaved by the first subunit, forming two 5' phosphotyrosine linkages not unlike that formed Step 4: the second DNA passes through the cut DNA an then is released

What is the final step on RNA processing? (L.17)

The final step in RNA processing is degradation. Unlike DNA, which lasts until the cell dies, RNAs are being constantly degraded. The lifetime of biomolecules is measured by their half life (t1/2). This is the time taken for 50% of the molecules present at time zero (t=0) to be degraded. So if you have 100 molecules at t=0 and 5 minutes later you only 50 molecules remaining intact, the t1/2= 5 mins. RNA half-lives vary from seconds to many hours depending on the gene and the organism. In an organism at steady state—for example a bacteria or a yeast growing in a nutrient rich media with no change in temperature or other conditions—the rate of degradation of the RNA will match the rate of synthesis of the RNA. As a result, the concentration of the RNA will not change in the cell. At steady state in a population of cells that are constantly dividing, the average t1/2for vertebrate mRNAs is ≈10 hrbut thereare a wide range of values; for some genes it can be as little as 8 mins. whereas for other gene itcan be > 24 hr. In non-dividing vertebrate cells the mean t1/2= 100 hr. In bacteria, the mean t1/2is ≈1.5mins. Degradation occurs via the action of ribonucleases. In bacteria, hairpin secondary RNA structures in bacterial mRNA can extend the half-life by protecting against degradation. In eukaryotes, the 5'cap and 3' poly (A) tail together with the cap binding and poly A binding proteins associated with them aid the stability of the mRNA. These special structures at the ends of eukaryotic mRNAs are protective because many of the ribonucleases in the cytoplasm are exonucleases. The control of the rate that mRNAs are degraded is based on the controlled removal their special methylguanosine 5' caps or their 3' poly-(A) tails. When the 5' cap is removed by an enzyme called DCP2, a highly active 5' → 3' exonuclease called Xrn1 quickly degrades the mRNA, starting from the 5' end—this is shown by the orange Pacman in the left hand part of the figure. Alternatively, if the poly A tail is removed by a deadenylase enzyme, a 3' → 5' exonuclease called the exosome degrades the RNA. In some special cases, an endonuclease is targeted to make a single cut within the mRNA. This usually happens when the translation machinery stalls on the mRNA and cannot be removed, so the mRNA is targeted for destruction. Once the cleavage has occurred, the free 3'-OH and 5' ends on either side of the cut are attacked by Xrn1 and the exosome.

Where are clusters of bacterial genes transcribes? (L.19)

The first example of the regulation of gene expression was the lac operon in E. coli. Bacterial genes occur in clusters where the protein coding regions of several genes occur next to each other. These clusters are called operons. The genes within a single operon are transcribed together in what is called a polycistronicmRNA. The transcription unit has a single promoter where the RNA polymerase binds. The promoter is just 5' of the region coding for the first protein in the operon. Not shown, but once the mRNA has been transcribed, each protein coding region (here labelled A, B, C ) has a Shine-Dalgarno sequence at its 5' end. The ribosome initiates translation for each gene separately. Within the promoter region are binding sites for RNA polymerase, which we saw in an earlier lecture. Promoter regions also have binding sites for sequence specific transcription factors. Specific transcription factors bind to particular DNA sequences that tend to be from 8 to 18 bp in length. Some transcription factors will increase the rate at which RNA polymerase initiates transcription of the operon. These transcription factors are called activators. Other transcription factors will decrease the rate at which RNA polymerase will transcribed an operon. These transcription factors are called repressors.

Is the genetic code universal? What are the exceptions? (L.18)

The genetic code is universal. It is the same code in all organisms. This is further evidence that modern day life forms all evolved from a single common ancestor. There are some minor changes in the code in few cases, though. For example, vertebrate mitochondria use a slightly different code. A UGA encodes Trp in mitochondria (instead of STOP). AGA/AGG encodes STOP in mitochondria (instead of Arg). Vertebrate mitochondria genomes only have about 16 protein coding genes, so it would be less destructive to change what a codon says in such a case. In contrast, if the code for one of your codons was changed to another amino acid, it would change the sequence of most of your proteins, which would be lethal.

What were the approaches used to decipher the genetic code? (L.18)

The genetic code was deciphered in the 1960s. A number approaches were used to crack the code. The first experiments are summarized here. Scientist learned how to synthesize RNA polymers that had simple nucleotide sequences. In the three examples shown, these synthetic RNAs only containing one nucleotide repeated many many times. E coli bacterial cells were broken open to provide a crude extract containing all of the proteins and other contents of the cell. When high concentrations of synthetic simple sequence RNAs were incubated in these extracts, large amounts of protein were produced that had only one amino acid repeated many times. In the case of a synthetic RNA made of only Us, the resulting protein only contained Phe. From this it was concluded that UUU codes for Phe. Similar arguments can be made for the two other synthetic RNAs shown. More of the code was cracked by using slightly more complex synthetic RNAs, such as these shown. You should be asking. Hey wait a minute. You said you needed at AUG to tell the system when to start synthesizing a protein. That is true in the cell under normal conditions. However, it turns out that these early experimenters did not know that and they got lucky. The conditions in the cell free extracts are unusually permissive and allow translation to occur without starting at an AUG. In cells, an AUG codon is absolutely required to start translation. With increasingly precise synthetic polymers it was possible to break the genetic code in just a few years.

What are spliceosome introns removed by? Describe their structure. (L.17)

The overwhelming majority of introns in protein coding genes are removed by the spliceosome. This large complex includes both protein and RNA components. The spliceosome contains five RNAs called small nuclear RNAs or snRNAs. The snRNAs are about 100 -200 nucleotide length. The snRNAs are U1, U2, U4, U5, and U6 and are each bound by collections of proteins to form snRNPs, pronounced "snurps"for small nuclear ribonuclear proteins. The U1 thru U6 snRNPs together constitute the spliceosome.

How is secondary RNA structure used to align the reactive parts of RNA together? (L.17)

The intron sequences are in yellow and salmon. Parts of the upstream and downstream exons are in green. The 5' exon base pairs with part of the reactive sequences for the intron, bringing them together so that the correct phosphodiester bond is cleaved. We'll ignore the rest of the complexities shown in the figure.

What is the heix-turn-helix motif? (L.19)

The lac repressor and many other of the bacterial transcription factors belong to the helix turn helix motif family. The helix turn helix motif is about 20 aa long. It has two alpha helices separated by a beta turn. One ahelix recognizes the DNA and is called the recognition helix. This is shown in dark red in panel a. The bturn sets the angle of the second ahelix, shown in dark blue in panel a, as shown. The recognition helix sits in the major groove. As I'll show in the next slide, there are specific hydrogen bonds between the recognition helix and specific base pairs in the DNA. At the bottom is shown the structure of the whole lac repressor tetramer. The Four DNA-binding helix-turn-helix motifs are shown in a different color in this panel: gray. The parts of a transcription factor that bind to DNA are called the DNA binding domain. Each of the different transcription factor families have a different DNA binding domain. DNA binding domain are examples of functional domains that we learned about in Lecture 4.

How is the synthesis of tellers done? (L.17)

The last example of reverse transcription I'll describe is the synthesis of telomers. If you recall from earlier, telomers are found at the ends of chromosomes. They consist of ~1,500 copies of a simple short repeated sequence. In the example in the figure, that sequence is TTGGGG. Telomers protect the ends of chromosome from getting progressively shorter after each cell division, which would happen otherwise as DNA polymerase needs a primer to initiate replication. The telomer repeats are added to the ends of the chromosome by an enzyme called telomerase. Telomerase uses a cleaver approach to add sequences: it provides its own template. The protein part of telomerase binds an 150 nucleotide RNA that includes the complement to one and a half copies of the TTGGGG sequence (i.e. AACCCCAAC). The figure explains the process quite well. The ends of the chromosome have a single stranded region. The telomerase lines up as shown at the top, synthesizes the second half of the template sequence using its reverse transcriptase activity, then the enzyme translocates and repeats the process. This extends the single stranded region. The replication RNA primase synthesizes a RNA primer towards the end of the newly added DNA. Then DNA polymerase extends this to create create double stranded DNA. A special pair of proteins bind at the very end of the telomer to maintain a short single stranded region (TRF1 and TRF2 in the figure).

How is the polyA tail added? (L.17)

The poly A tail is around 30 to 100 nucleotides long and serves as a binding site on mRNA for various proteins, including some proteins in the cytoplasm that are involved in translation. Like the 5' cap, the poly A tail also helps protect the mRNA from degradation. If the poly A tail is removed, the mRNA will be quickly degraded. RNA polymerase synthesizes a transcript beyond where the 3' end of the mRNA is encoded in the genome. A cleavage and poly A addition site element is located 10 to 30 nucleotides upstream of the 3' end of the mRNA. The sequence of this element is AAUAAA and is highly conserved. The AAUAAA element is bound by an enzyme complex that includes an endonuclease, a polyadenylate polymerase, and several other multi subunit proteins involved in sequence recognition, cleavage, and regulation of the length of the poly(A) tail,. This whole slew of proteins are tethered to the CTD of pol II. In step 1, the RNA is cleaved by the endonuclease at a point 10 to 30 nucleotides 3'of the AAUAAA element. In step 2, an unusual polymerase called polyadenylate polymerase synthesizes a poly(A) tail 80 to 250 nucleotides long, beginning at the cleavage site. This polymerase does not need a template strand. It just keeps adding As to the 3' end of an RNA. Once the poly A tail has been added it is bound by multiple copies of a protein called poly A binding protein (PAB). PAB only binds to pure A sequences of RNA and its presence protects mRNAs from degradation.

If you cut the circle of DNA and unwind the strands partially and then you join them back together. You unwind the DNA for several helical turns such that one turn is removed for every 8 helical turns in the relaxed state in a molecule that has 84 bp. What would this result in for the structure of DNA? Would this structure be energetically favorable or unfavorable? (L.15)

The result will be that there are 7 turns for every 84 bp. You would end up with Under wound DNA. This new structure would be highly unfavorable energetically because it would be in such a strained state.

The ribosome is key in protein synthesis. (L.18)

The ribosome is a massive molecular machine and is highly abundant. There are about 25,000 ribosomes per E coli cell and they makes up ~25% of the dry weight of bacteria. The ribosome is comprised of ~65% rRNA and 35% protein. We saw earlier that the bacterial ribosome has 3 RNAs. The small subunit includes the 16S rRNA and the large subunit the 5S and 23S rRNAs. These rRNAs form the core of the ribosome and the RNA does the catalysis of peptide bond formation. The 30S subunit has in addition 21 protein components and the large subunit 36.

Example of E.coli promoters (L.17)

The six genes in the figure are all bound by s70, not by the other sigma factors. s70—the orange blob in the figure—binds to the sequences shown in purple in the -35 region and the -10 region. These two sequences are each bound by a different part of the sigma factor and are separated by a set number of nucleotides that allows the sigma factor to bind each sequence at the same time. This spacer region is 17 nucleotides long. A further spacer of 6 nucleotides separates the -10 region from the +1 start of transcription. The positioning of the sigma factor on the DNA determines where the RNA polymerase holoenzyme will initiate transcription. You'll notice that the sequences in the -10 and -35 regions are not all identical between the six promoters, but they are similar. The top row shows a so called consensus sequence, which is the optimum sequence that the sigma factor binds most strongly too. The protein will bind less strongly to those promoters that have sequences that are most different from the consensus sequence and as a result, those genes will be transcribed less frequently. In other words, the affinity (Kd) of sigma for a promoter is used to control the amount of mRNA and protein produced by the gene. The core of RNA polymerase itself can also recognize specific DNA sequences via its alpha subunits. The alpha subunits bind to a region called the UP element that is present in some genes and not others. Those genes that have the UP element tend to transcribed more highly as they more efficiently recruit polymerase. This element is present in one of the six genes above, to the left. The two alpha subunits are shown in purple. Note that the other polypeptides in the polymerase core are bound in a complex with the alpha and sigma subunits. They have been omitted from the figure for simplicity. Just imagine they are there. When the cell changes the abundances of the different sigma factors due to alterations in environmental conditions, the amounts of RNA polymerase holoenzymes bound to different sets of genes will change. Therefore, promoters affect the rate at which genes are transcribed and how much of each protein the cell expresses. This is a major difference between DNA synthesis and transcription. DNA synthesis must copy the genome uniformly. Transcription by contrast is all about copying different part of the genome to different degrees. We'll learn more about the control of gene expression in the last lecture.

What is the fundamental point of recombination repair? What are the first two steps? (L.16)

The top part of the slide reminds you what we saw a few slides ago. If there is an unrepaired break in one strand of DNA ahead of a replication fork, when the fork reaches the unrepaired break the fork will collapse, leading to a double stranded DNA break in one of the two chromosomes. Notice that in the figure, the arrowheads represent 3'ends. The first two steps of recombinational repair are shown below. The fundamental point of recombinational repair is that it takes part of one of the intact strands of DNA from the non damaged chromosomes and swops it with one of the stands from the damaged chromosome. So after that strand swop both chromosomes have one intact strand of undamaged DNA, which can then be repaired by process similar to those we have already seen. Step 1: generates a single stranded region of DNA for the DNA strand with a 3' -OH group. Step 2 in a remarkable process, the part of the other chromosome that is homologous (i.e. has the same sequence) to the new single stranded region unpairs from the undamaged chromosome and crosses over and pairs with the single stranded region. Notice that this will only work if the sequences are either identical or very similar. If the sequences were very different, the two strands would not be able to come together and form a double helix.

What are the maturation/processing of mRNA in eukaryotes steps? (L.17)

The various steps in processing a eukaryotic mRNA are sometimes called maturation. Remember that eukaryotic genes have intronswhich are removed by splicing. Remember also that the part of the genome that is transcribed is called the transcription unit;and the initial RNA that is produced is called the primary transcript. The top part of the figure shows RNA pol II transcribing a gene. Several of the RNA processing enzymes bind to the CTD, ensuring that they are in close proximity to the nascent transcript. The first step in maturation is the addition on an unusual nucleotide at the 5' end of the mRNA that is called the 5' cap. Then either during or after the primary transcript has been synthesized, the introns are removed. Finally the 3' end of the primary transcript is cleaved and a long stretch of A nucleotides is added to form the poly A tail. All of these maturation/processing steps occur in the nucleus. We'll look at each in turn in more detail next.

How many different sigma factors are in E.coli? (L.19)

There are seven different sigma factors in E. coli. To remind you, sigma factors are not all expressed at the same time. Different combinations of these proteins are present in cells depending on the condition. To remind you of an example I mentioned before, when cells are heated to induce the heat shock response, the amounts of each sigma factors and their associated holoenzymes change. The sigma 32 protein is expressed at high levels only when cells are heated above their normal temperature. This sigma factor directs RNA polymerase to bind to several operons that express chaperonins and other proteins that refold proteins that have been denatured by heat shock. Remember we saw 2D PAGE gels of proteins that whose expression is changed in response to heat shock in Lecture 3. There are many more transcription factors like lac and trprepressor in E coli cells than there are sigma factors. However, we can now see that sigma factors are really a specialized form of transcription factor. The only difference is that sigma factors stably bind with the RNA polymerase off the DNA. But otherwise, sigma factors and transcription factors both regulate subsets of genes and a large part of their specificity comes from their ability to bind specific DNA sequences.

How does DNA footprint work? What does it do? (L.17)

The way that we know where proteins bind on DNA is through a technique called DNA footprinting. It is worth understanding this method as variants of it are widely used in many contexts. This and similar footprinting method are based on the premise that the part of a DNA or RNA that is bound by a protein will be protected from either chemical modification of enzymatic cleavage, whereas the unbound parts of the nucleic acid will not be protected. On the right is a schematic of the method. First a several hundred bp DNA fragment that is thought to contain a binding site is isolated. Today one would most likely PCR amplify a region of DNA, but historically one would generate a restriction fragment of DNA that had been cloned in a plasmid. Next one radioactively labels one end of the DNA, by means we'll skip over. The labelled DNA fragment is divided into two reactions, one containing the protein of interest, a second, negative control one lacking the protein. The protein will bind to DNA in the first reaction. Both the protein binding reaction and the negative control reaction are then treated with some agent that will mark the DNA. In the example shown, an endonuclease called DNase I is added for a short time under conditions where it only cuts each DNA on average once. This will lead to the collection of DNA fragments shown in the center. Notice that the fragments that do not include the radioactive label are not shown, even though they exist in the reaction. The DNA fragments are then separated on an acrylamide electrophoresis gel and the radioactively labelled DNA bands are visualized by autoradiography. Obviously the unlabeled DNA is not visible, which is just as well as it would obscure the result. At the bottom right is a cartoon of the expected result and to the left is the data for E. coli RNA polymerase binding to a promoter. Because all of the DNA fragments have one common end and a second end that differs due to the location of the enzyme cut site, the distance that the DNA migrates in the gel corresponds to the location of the cut site. The lanes labelled -are from the negative control. The lanes labelled + contained the protein. You can see on the left that the polymerase protects a large region from about +1 to -55 from digestion. Thus one can conclude that the polymerase is binding over this region.

How many DNA polymerases in E.coli? What are the main ones? Why? (L.16)

There are 5 DNA polymerases. DNA polymerase I: it's a DNA replication protein who's main function is DNA repair and general genome maintenance. DNA polymerase III: principle polymerase that replicated the genome of DNA the other three are involved in DNA repair

Proteins Required for Initiation of Transcription at the RNA Polymerase II. Promoters of Eukaryotes(.L.17)

There are a complex host of other proteins involved in the initiation, elongation and termination of transcription by each of the three eukaryotic polymerase. This slide shows a small subset of those involved in transcription by RNA polymerase II. The proteins shown are the minimum set needed for initiation and elongation in addition to RNA pol II. We are not going to learn all of their names. But we can break them down into two groups. A set that are required for initiation shown at the top and those for elongation below. I've highlighted two proteins that are worth learning about in red. The TATA binding protein (tbp) binds to a sequence about 30 nucleotides upstream of the initiation of transcription. The most strongly bound promoters have the sequence TATAAAAat that location, which is called the TATA box.Some promoters lack the TATA box but tbpis still recruited to the promoter because of its interaction with other proteins that bind to other parts of the promoter. The pTEFb protein is a kinase that phosphorylates Pol II on a domain called the C terminal domain or CTD. This phosphorylation converts the polymerase from its initiating form to its elongating form, somewhat analogous to the change from promoter clearance to elongation seen in bacterial transcription.

RNA polymerase core complexes with one of several sigma factors. (L.17)

There are multiple types of holoenzyme, each of which has a different sigma subunit. The table shows the amounts of each holoenzyme type in a cell under a particular set of growth conditions. Note there are 700 molecules per cell of sigma70, yet there are less than 10 molecules of sigma 32. Under other conditions, such as when nutrients are not present or when cells are heated to induce the heat shock response, the amounts of the sigma factors and their associated holoenzymes change. The sigma subunit directs the holoenzyme to a region on the gene at the 5' end of each transcription unit called the promoter.

What are the gene products of an integrated retrovirus? (L.17)

There are said to be three genes encoded in a retroviral genome: gag, pol and env. However, some of these "genes" yield several proteins. The viral genome that has been integrated into the host genome is transcribed by pol II using promoter sequences present in a region called the long terminal repeat (LTR) at one end of the genome. The RNA is translated to produce long polyproteins, some of which include all three genes. The polyprotein is cleaved by a protease encoded by the virus. The products include ..... Gag proteins that form the core of the virus and envelope proteins that sit in the membrane that surrounds the virus (see figure on last slide). The reverse transcriptase and the integrase are encoded by the pol gene.

How do primases work? What is the problem that they solve and how do they solve it? (L.16)

They overcome the problem that DNA polymerases can only extend an existing strand of DNA that has a 3'-OH . *RNA polymerases do not have this limitation. The can initiate polymerization on a single stranded DNA* Primase enzymes are used to produce short RNA primers of 10 to 60 nucleotides in length. These are extended by the DNA pol III holoenzyme-shown in bottom figure. *Recall that a replication fork has leading strand and a lagging strand. Elongation fo the leading strand is straight forward. A single primer generated at the origin of replication is extended all the way around the genome by the pol III holoenzyme. Synthesis of the lagging strand is more complex

What do transcription factor molecules (purple blobs) do in the DNA?(L.15)

They regulate the rate at which the gene is transcribed in to mRNA. They recognize specific sequences of DNA that are around 6 to 12 bop in length. they can be sound 1 to 50 kb away from the transcription unit.

What are nucleotides in tRNAs recognized by in tRNA synthestases? (L.18)

This figure shows the nucleotide positions in tRNAs that are recognized by aminoacyl-tRNA synthetases. In the left panel, purple dots show positions that contain the same nucleotides in all tRNAs and therefore cannot be used to discriminate one tRNA from another. Positions that are known recognition points for one (shown in orange) or more (shown in blue) aminoacyl-tRNA synthetases. Structural features other than nucleotide sequence are important for recognition by some of the synthetases. The panel on the right shows the locations of recognition nucleotides in three dimensions, with the orange and blue residues again representing positions recognized by one or more aminoacyl-tRNA synthetases, respectively. gYou can imagine how each synthetase is able to recognize a specific tRNA.

What is the structure of the la operon/ (L.19)

This figure shows the structure of the lac operon and also of a second transcription unit just 5' of it. This 5' transcription unit encodes the lacIgene for the Lac repressor. LacIhas its own promoter PI. Transcription of the repressor is independent of transcription of the lac operon. The lac operon is the region from O3to lacA. The lac operon promoter is shown by the arrow. The three protein coding regions for lacZ, Y and A are 3' of it. The lac repressor can bind to three so called operator sites (O1-O3). Operator sites are a specific term given to the DNA binding sites for some bacterial transcription factors. The Lac repressor binds most strongly to the operator O1,which is adjacent to the start site of transcription. The affinity of a dimer of lac repressor for the O1operator is high, the is Kd~10−10M. The sequence of the O1operator is shown. You can see it is an inverted repeat. Binding of the repressor helps prevent RNA polymerase from binding to the promoter, reducing the rate of transcription by 100 fold. The repressor also binds to one of two secondary operators (O2or O3), which reduces transcription a further ten fold.

Cleavage of the Holiday junction. What cuts it and what happens as a result? (L.16)

This slide gives a different new of the Holliday intermediate—shown on the left. And—on the right—how RuvCcuts the DNA to allow the two chromosome to separate. This process is called resolvingthe Holliday intermediate.

Protein Synthesis(L.18)

This slide gives an overview of the process, dividing translation into 4 stages. The first stage we have already discussed is the amino acylation of tRNAs. This stage does not involve the ribosome. The remaining steps do.The ribosome has two subunits, the small or 30S subunit and the large or 50S subunit. (though as we will see the so called small subunit is actually humongous, the large even more so). Stage 2 in this figure is the initiation or translation, in which the small subunit of the ribosome binds near the 5' end of the mRNA to a region where the initiating AUG codon is. A special tRNA specific to the first AUG codon is recruited to the ribosome, with the anticodon and AUG codon base pairing together. Then the large subunit of the ribosome assembles with the small subunit. Stage 3 is the elongation of translation. Successive amino acylated tRNAs are recruited and used to synthesize the polypeptide, one amino acid at a time. Stage 4 is the termination of translation. Synthesis stops when the ribosome reaches the first stop codon in the reading frame. The ribosomal subunits are released and recycled for another round of protein synthesis.

alternatively processed primary transcript example (L.17)

This slide shows an example of an alternatively processed primary transcript. The calcitonin gene from rats is a peptide hormone that is produced in the thyroid gland to regulate calcium levels. The processing of the RNA in the thyroid is shown on the left, and as you can see the mature mRNA includes the exon that codes for the calcitonin peptide. That is the exon shown in dark blue, exon 4. Another protein with a different amino acid sequence is expressed in the brain from the same part of the genome—shown on the right. In the brain, the same promoter is used to initiate transcription but a different poly A site is used that is further downstream than the one used in the thyroid. Thus while the primary transcripts are the same in both tissues, after cleavage and poly adenylation the partially processed transcripts—shown in the center—differ at their 3' ends. It looks like the splice acceptor site at the 5' of exon 5 is preferred over that for exon 4 because in the brain exon 5 is joined to exon 3, eliminating the calcitonin coding sequences. Exon 5 codes for another peptide hormone called calcitonin gene related peptide (CGRP), which is a vasodilator that has some association with migraine attacks. So in this case, two completely different proteins are expressed from the same primary transcript; and the regulation of which is expressed is controlled post transcriptionally. This is a rather extreme case, however. Most alternate splice variants from a given primary transcript produce variants of the same protein, perhaps changing one or two of the domains, but including mostly the same amino acid sequences. In the case shown, in the figure it is unclear if you should call this two genes that share the same primary transcript or one gene that makes two proteins. Personally, I favor the former definition.

What role do h-bonds play in protein/dan binding? (L.19)

This slide shows some of the amino acids that are commonly are found in DNA binding domains and how they hydrogen bond with specific base pairs. On the left you can see that either Glutamine of asparagine can form a specific H-bond with adenine's NH-6 and N-7. Arg can form specific H-bonds with the cytosine-guanine base pair. Notice how in both cases, two hydrogen bonds are made. The amino acid and the bases fit together like pieces in a jigsaw puzzle. The major groove is the right size for the ahelix and has exposed H-bonding groups.

How does RNA polymerase generate a positive super coil? (L.17)

This slide shows the E. coli RNA polymerase. You can see that the enzyme encompasses more of than the 17 bp region where the DNA strands are separated. In total about 35 bp of DNA are enclosed by the enzyme. The DNA runs in a channel through the center of the enzyme, with the channel being wider where the strands are separated. There is a second channel through which the RNA exits, and another through which the NTPs enter. Just like DNA polymerases, RNA polymerases creates positive supercoils ahead of the enzyme and negative super coils behind the enzyme. Just like in replication, DNA topoisomerases relive these supercoils.

How the Wobble Base of the Anticodon determines the Number of Codons a tRNA ca. recognize. (L.18)

This slide summarizes the rules by which wobble pairs act. tRNAs with a C or A in the first position of the anticodon can only make Watson and Crick base pairs with bases in the third position of the codon. i.e. there are no wobbles. tRNAs with a U or G in the first position of the anticodon can make either a Watson and Crick base pair, or a GU wobble pair. Some tRNAs have a modified base called Inosine that can pair with 3 codons. The inosine is chemically modified post transcriptionally by deamination of adenine.

What is the number of codons for each of the 20 amino acids? (L.18)

This table shows the number of codons that code for each of the 20 amino acids. Between 1 to 6 codons are used to code for an amino acid.

What are the three phases of Transciprtion? (L.17)

Transcription can be divided into three phases: Initiation, elongation and termination. This slide shows the first two phases. The initiation of transcription requires the interaction of the RNA polymerase core and the sigma factor with the promoter DNA to form what is called the closed complex. In the closed complex, the promoter DNA is stably bound by the holoenzyme but is not unwound. A 12 to 15 bp region of DNA from within the -10 region to position +2 or +3 is then unwound to form a transcription bubble, creating the so called open complex. The polymerase then synthesizes the first few nucleotides of RNA in process called promoter clearance. The elongation phase commences when thesigmasubunit is released from the core polymerase and is replaced by another protein called NusA. NusAenables the polymerase core to synthesize RNA more efficiently than when it is bound by sigma. The rest of the gene is then transcribed by the NusA/ core polymerase complex. When transcription is complete, the RNA is released; the NusAprotein dissociates; and the RNA polymerase releases from the DNA (step 6). Another σsubunit binds to the RNA polymerase and the process begins again at the promoter.

What are the three RNA polymerases in Eukaryotes? (L.17)

Transcription in eukaryotes is much more complex than in bacteria. I'll summarize some of these complexities in this part B of the lecture. There are three different RNA polymerases in Eukaryotes. RNA polymerase I, whichsynthesizes ribosomal RNAs. RNA polymerase II, which synthesizes mRNAs. This polymerase is very fast. It can synthesize 500-1000 nucleotides/sec and it acts on thousands of promoters RNA polymerase III,which synthesizes tRNAs and some other small RNAs as well. These polymerases all have subunits that show sequence homology to bacterial RNA polymerase, illustrating that the fundamental nature of transcription to living systems.

What are wobble pairs and how are they formed? (L.18)

You might think from what I have said so far that cells must have 64 tRNAs, each with an anticodon sequence that is specific to one codon. In fact that is not the case. The third position does not always use a standard Watson and Crick base pairing. For example, as we saw earlier, G can pair with U. Although this pairing is weaker than that of G with C, it is sufficient to allow recognition. Non Watson and Crick base pairs that allow a single tRNA to recognize more that nucleotide are called wobble pairs. Many organisms have fewer tRNAs than 64. E. coli only has 47.

Ribosome Rescue (L.18)

Translation does not always proceed smoothly. For example, damaged mRNA leads to the formation of an incomplete peptide and a so called "nonstop complex." The ribosome is stalled and cannot be used. In such a case, the ribosome is rescued by a process called trans-translation. A transfer-messenger RNA (tmRNA) and small protein called SmpBbind to the stalled complex in the empty A site. The transfer-messenger RNA is what it sounds like. The 5' end of the RNA very closely resembles a tRNA and is amino acylated. Attached to this at the 3' end are additional RNA sequences that include a stop codon and thus serve as an mRNA. Translation continues until a stop codon is found in the tmRNA. The defective mRNA and polypeptide are both degraded and the ribosome is released for use by the cell.

What are transposons? How doe they move themselves? (L.16)

Transposons Remember these constitute about 50% of the DNA in the human genome. They can be thought of as self replicating, parasitic DNA. Transposons can change their relative position in the genome. They carry genes that encode proteins that are responsible for moving them around the genome. Transposons are usually between a few hundred to a few thousand bp in length. They are often flanked by short sequences that occur at both ends—shown in blue—called terminal repeats. There are several kinds of transposons. For simplicity, I'll describe how just one kind moves itself. DNA transposons in bacteria encode a transposase enzyme that can cut a target region of the chromosome, cutting each strand at different locations to create two single stranded ends. Typically these cuts are only 5 -10 nucleotides apart, so the figure greatly exaggerates their size. The transposon then inserts into the genome using a ligase. Replication by the cells enzymes then repairs the DNA, producing the structure shown at the bottom. Notice that the 5 -10 nucleotide region of the target DNA between the cut sites is duplicated in the genome by the transposition. Many of the transposons in the human genome transpose differently. They use an single stranded RNA intermediate during transposition. I'll leave the description of how they transpose to the Lecture on RNA metabolism.

Define Twist (TW) and Writhe(Wr). (L.15)

Twist is the number of times the DNA turns around the helical acid of the DNA. The favorable twist for DNA is 10.5 bp per turn. Writhe is a measure of the higher order shape than the DNA adopts during supercoiling. The easiest way to understand writhe is by realizing Lk=Tw+Wr. To reach favorable twist of one twist/10.5bp, the DNA adds or subtracts writhe of the whole double helix.

What are the two major types of topopisomerases that are found in both bacterial and eukaryotes? What are their functions?(L.15)

Type I: make a transient cut in one DNA strand; they relax negative supercoils and therefore change Lk in steps of +1. -make transient single strand breaks in he DNA before resealing the nick Type II: make a transient cut in both DNA strands, they use ATP energy and they change Lk in steps of 2. Bacterial type II toposiomerases can introduce negative supercoils into relaxed DNA. These are called gyrases. Eukaryotic topoisomerases cannot introduce negative super coils into relaxed DNA, but they can relax both positive and negative supercoils.

What can Type II topoisomerases do to linked circles? Why can they do this? What are these linked circles called?(L.15)

Type II topoisomerases can decatenate (unlink circles) that are joined as the one in the top of the figure. They can do this because of their ability to cut both strands of DNA. When two circular DNAs are linked in this way they are said to be catenane or to be concatenated

How are splice some assembled? (L.17)

U2, U4, U5, and U6 bring at least 50 proteins with them to create a spliceosome—shown in the center of the figure. Interestingly, ATP is required for assembly of the spliceosome but not for the actual cleavage and resealing of the RNA. So while the spliceosome is not a ribozyme—it does need protein—, it has some characteristics of a catalytic RNA. After formation of the lariat structure, the lariat is removed and the two exons are joined in just the same way as in a group II intron. Lehninger says that parts of the spliceosome are attached to the CTD of RNA Pol II and suggests that splicing is coordinated with transcription. However, more recent experiments show that splicing often occurs many many kilobases away from the location of the synthesizing RNA pol II. So it is unclear how well splicing and transcription are coordinated.

How are viruses able to live? Describe them. Give an example of a virus. (L.15)

Viruses are not living organisms. They require a host. (molecular parasites) They can have a single or double stranded genome that be either RNA or DNA and these genomes can be either linear or circular Ex. Genome of COVID 19 virus is about 30 kb of single stranded RNA.

What is the variance in measures translations> (L.19)

We also used data that measured how much translation rates actually vary between genes. The black line the range in rates of translation that schwanhausseret al assume must occur based on their idea that there is no error in the data. The other colored lines are actual measurements of translation rates made by a number of groups. The x axis is the rate of translation, the y axis the frequency. So a gene to the right has a high rate of translation and a gene to the left has a low rate of translation. More genes have an intermediate translation rate than have a very high or very low rate. You can see that the measured experimental data (the colored lines) are distributed over a narrower range than schwanhausseret al assumed.

How can the idea of topology structures be applied to DNA?(L.15)

Well the double helix of DNA has two polymer strands wrapped around each other ( these form a coil). *reminder: there are 10.5 base pairs for each turn of the helix.

What does CAP-cAMP do? (L.19)

What I have said so far is the original simple story that Jacob and Monod figured out. However, operons are often regulated by more than one transcription factor. The interplay between multiple transcription factors can allow more complex regulation. In the case of the lac operon, there is a second transcription factor called CAP, or catabolite activator protein. This protein is only active when glucose levels in cells are low. The activity of CAP is controlled by an effector called cyclic AMP (cAMP). CAP is only active when cAMP is bound to it. cAMP is produced in cells when glucose levels are low. CAP-cAMP binds near the promoter and increases the rate of transcription by 50 x. In the absence of CAP-cAMP, the RNA polymerase open complex doesn't form. CAP-cAMP can only activate when the Lac repressor has dissociated.

What happens to lactose when glucoses is abundant or when its not? (L.19)

When glucose is abundant and lactose is not present, cells make only very low levels of the enzymes needed for catabolism of lactose. Under these high-glucose / low-lactose conditions, transcription of the lac operon is repressed. If on the other hand glucose is scarce and the cells are fed lactose, the cells can use lactose as their energy source. Under these conditions, the cells then express the genes for the enzymes for lactose metabolism. Transcription of the lac operon is no longer repressed. Instead the operon is strongly transcribed by RNA polymerase.

None of our genes are reversed transcribed so what does it do instead? (L.17)

When reverse transcription was first discovered, it forced a revision to the central dogma, shown in the right. An additional arrow had to be added for reverse transcription. However, please realize that there are only a limited set of cases of reverse transcription. None of your genes are reverse transcribed. Instead ..... There are a class of viruses called Retroviruses that use reverse transcription. Some of the retrotransposons that are part of the repetitive DNA in your genomes originated via reverse transcription And telomers at the ends of eukaryotic chromosomes are added by a special reverse transcriptase called telomerase

Are plasmids important for recombinant DNA Technology (L.15)?

Yes! They are important for cloning genes. The vector DNA's we looked at last lecture such as pBR322 are derived from natural plasmids

Can both the template and coding strand code for proteins. (L.17)

Yes! coding information may be on either strand ("top" or "bottom") Each strand codes for a number of proteins Genes are arranged in genomes such that the coding strand can occur on either strand of DNA. In the example shown, each gene is shown with a green arrow indicating the direction of transcription. The particular example shown is the genome of a virus that infects humans called Adenovirus, but as we will see in the remaining lectures, genes are encoded on both strands of DNA in bacterial and animal genomes as well.

How can you separate DNA topoisomers? How can you tell which ones are highly super coiled? (L.15)

You can separate DNA topoisomers on DNA arose electrophoresis gels. Those DNAs are more highly supercoiled, and they travel more quickly towards the anode. The more super coils DNAs are in effect smaller and can work their way through the agarose gel more quickly.

Where are telomeres found? (L.15)

at the ends of chromosomes

How are telomeres added to the DNA?(L.15)

by a special enzyme called telomerase that essentially glues the above short sequences one repeat at a time to the ends of chromosomes to stop them from getting shorter.

How does the cell maintain the average net negative superhelical density of the genome?(L.15)

by maintaining a balance of type I and type II topoisomerase. The opposing action of these enzymes migrate the local increase in supercoiling.

How do mutations in DNA happen? How are these mistakes fixed?(L.16)

chemical reactions or replication errors corrupt the information in DNA, usually on one strand. Usually this corruption occurs by altering the information on one strand of the DNA. In order to fix these mistakes DNA repair pathways use the information on the undamaged DNA strand asa template to correct the error. However some base changes escapse repair and after another round of replication lead to. changed sequence in one of the two chromosomes

Whats did Mendel conclude from his White flower, violet flower experiment? (L.14)

concluded that each trait was defined by a gene.

What is important about the histone tail modifications? (L.15)

hese modifications affect the way that the histones compact DNA and recruit other proteins to the DNA.

We know that SOS repair and Recombinational repair pathways, repair act in cases where the information on both strands of DNA has been lost. How does a DNA lose information on both strands? (L.16)

figure shows. Imagineer an unprepared lesion such as thymine dimer shown on the left op, the black box. or shown on right, too-imagine an unrepaired break that is missing a few nucleotides, perhaps a nucleotide excision repair where pol I has yet to fill in the gap. Id pol III holoenzyme attempts to replicate these damaged parts of the chromosome, the lower parts of the figures show what would happen. in both cases , the polymerase would not be able to get passed the lesion. So one of the two strands of DNA is not replicated. In some cases the parental strand is subsequently destroyed by nucleases (on the right) and in other cases it remains as a single strand of DN (on left). But in both cases, part of the sequence information is lost in both strands.

What does the nucleotide excision repair do? (L.16)

it corrects lesions that cause large distortions in the DNA. common form of lesion is a :pyramid dimer UV light is absorbed by bases, particularly by thymine bases. This results in the formation of free radical on one of the bases that very rapidly reacts with the adjacent base to form one a several possible covalent bonds. Two examples at the bottom right are two of the possible products. This covalent coupling presents the bases offsetting as usual within the double helix structure and results in a quite kink in the sugar phosphate backbone. The subset lesions that do not get rearer result in mutations that can lead to skin cancer.

What does the DNA polymerase active site when it comes to use pairs? (L.16)

it excludes base pairs that have an incorrect geometry. The shape of the active site is crucial! BUT they still inset wrong bases [1/(10^(4))] to [1/(10^(5))] times. There are several DNA repair mechanisms that are employed by the cell to correct these errors and a achieve a final much lower error rate of (1/10^(9)) to (1/10^(10)).

Why does RNA polymerase not have the ability to proof read? (L.17)

it lacks a 3' to 5' exonuclease. has a high error rate. There are also nor RNA repair systems tot fix these errors . synthesized and degraded in cells. Any errors will lead to some protein molecules being made that are inaccurate, but often those do not fold correctly and will be degraded by a system I'll describe later. Plus if a few protein molecules have errors, the cell can live with a little sloppiness. Unrepaired errors in DNA are forever. Errors in RNA are "whatevers". The RNA polymerase holoenzyme has five core subunits (a2bb'w)—whose names you do not need to memorize—plus a sixth called sigma(o)—which you should learn. The figure shows only the five core subunits.

Since nucleosomes are spaced on average one every ____bp, what are the regions between each nucleosome called? What histone protein binds in the linker region?(L.15)

nucleosomes are spaced on average one every 200 bp the regions between each nucleosome is called the linker DNA. A fifth histone protein called histone H1 binds in the linker region. 95% DNA is the eukaryotic nucleus is either bound by a nucleosome or is a short linker sequence.

3D structure of yeast tRNA^(Phe) (L.18)

ou can see how the tRNA folds up in 3D to form a the twisted L shape I mentioned earlier. There is a schematic model on the left and a space filling model on the right. Note the location of the anticodon in yellow and the CCA sequence in purple.

What are transposons? (L.15)

parasitic DNA. They do not code for a host function. Instead, they make copies of themselves and insert those new copies in the parts of the genome. Some were viral genomes but loathe ability to infect new cells. Only make copies of themselves very rarely! Most of the millions of copies of transposons in the genome are inactive and are never able to transpose.

What is the mechanism of DNA polymerases? (L.16)

parental strand serve a template and the reaction uses the trinucleotide form of the four deoxynucelotides and its products are pyrophosphate (PPi) as well as the synthesized DNA. 1. Parental DNA strand is template dNTPS are substrates in synthesis. 2. Nucleophilic 3' OH of the growing chain attacks the alpha-phosphate of the incoming dNTP. - 3'OH is REQUIRED explains why enzyme requires a primer - 3'OH is made a more powerful nucleophile by Mg^(2+) one of the Mg^(2+) ions deprotonates 3'-OH to increase its nucleophilicity. (metal ion catalysis) 3. Energy for reaction from hydrolysis of pyrophosphate ((delta)G= 19 kJ/mol) and base stacking and hydrogen bonding in product. - this synthesis shows very little engird change because high energy nucleotide phosphate bond if first broken then remade. However, the pyrophosphate that is produced is hydrolyzed. Also the newly added base forms hydrogen bond and base stacking interactions in the DNA. All of which provide energy to drive the synthesis reaction forward. 4. The incoming dNTP interacts with the MG ions in a part of the enzyme active site called the insertion site. 5. Once the nucleotides covalently couple to the primer strand the enzyme usually translocates along the template strand by one nucleotide so that the newly incorporated nucleotide sites at a second part of the active site called the postinstertion site. 6. The enzyme is then ready to add a new base at the insertion site.

rRNAS are cleaved from a larger precursor and modifies by snoRNAs(L.17)

rRNAs are also processed. I don't want to go through the details of the processing. I am using this slide mostly to introduce you to ribosomal RNAs. rRNAs are much the most abundant RNAs in a cell, constituting at least 80% of RNA by mass. Together with over 50 ribosomal proteins they form the ribosome, which is the molecular machine that translates mRNAs into protein. There are three rRNAs in all organisms, and fourth that is found only in eukaryotes. The rRNAs are named by their size as determined by the rate that they sediment in a centrifugation experiment. The larger their sedimentation coefficient (S value) the longer the RNA. In vertebrates, three of the rRNAs are transcribed from one of a number of copies of the rRNA gene in a compartment in the nucleus called the nucleolus. A 45S (14 kb) primary transcript is first transcribed. It is then cleaved in two steps to form 5.8S, 18S, and 28S rRNAs. These are then packaged together another rRNA (the 5S RNA) and the set of ribosomal proteins into two subunits: the 40S and 60S ribosomal subunits. Only when they have been assembled are the ribosomal subunits transported from the nucleolus to the cytoplasm for use in translation. There are a number of complexities in this processing that are mentioned in Lehninger but which we'll ignore. These include that a number of specific nucleotides are modified during processing and that a whole class of RNAs are involved in RNA processing that are called small nuclear RNAs (snoRNAs). We'll come back to the ribosome in the next lecture.

What are the characteristics of tRNAs? (L.18)

tRNAs are single stranded RNAs of 73-93 nucleotides in both bacteria and eukaryotes. As we saw in earlier lectures, tRNAs have secondary RNA structures due to the formation of short dsRNA sections. When their RNA secondary structures are draw as shown, they have a characteristic cloverleaf pattern. There are three leaves and a stalk (in the imagination of molecular biologists at least). As I will show in several slides time, in 3 dimensions tRNA structures are like a "Twisted L" shapes. In the figure, the large dots on the backbone represent nucleotide residues; the blue lines represent base pairs. Characteristic and/or invariant residues common to all tRNAs are shaded in light red. Most tRNAs have a G at their 5'-end; and all have CCA at their 3'-end. As mentioned earlier, many of the bases are modified during tRNA maturation. Of the four arms, two are critical for the adaptor function: these two are the amino acid arm and the anti codon arm. The amino acid arm has an amino acid esterified via a carboxyl group to the 2'-OH or 3'-OH of the A of the terminal CCA sequence. The anticodon arm contains the anticodon loop. The other two arms, the D arm and the TψC arm help maintain the overall 3D structure of the tRNA. Some tRNAs have an additional arm of varying length as well. Symbols are: Pu, purine nucleotide; Py, pyrimidine nucleotide; ψ, pseudouridylate; G*, either guanylate or 2'-O-methylguanylate.

What are tRNAs processed by? (L.17)

tRNAs play a special role in the translation mRNAs to make proteins. We'll learn about how they do that in the next lecture, but for now just accept that a tRNA looks like the clover leaf structure shown on the right. Notice that there are several hairpin secondary structures. The interesting thing for us for now is that tRNAs are processed from a longer primary transcript, shown on the left, by a true ribozyme called RNase P. RNase P is an endonuclease that cuts the tRNA primary transcript to remove the 5' most sequences shown in yellow. RNase P is an RNA/protein complex, but the RNA alone can catalyze the cleavage. It is a true ribozyme in that a single molecule of RNase P can catalyze cleavage of many tRNA primary transcripts. The full details of how the ribozyme works are still being figured out, but its mechanism is different from intron splicing in that it requires a metal ion, usually magnesium 2+. The metal ion increases the nucleophilicity of a 2'-OH which attacks a phosphodiester bond at the cleavage site. A number of other ribozymes have been discovered that have other properties. The discovery of ribozymes was a big deal. It is not just an interesting quirk that RNAs can act like enzymes. It gives us a way to think about how life may have first evolved. Think about it. How could you simultaneously evolve proteins, DNA, and RNA, each dependent on the other for their existence. The idea is that before there were cells that used the central dogma, there were proto living systems that used only RNA—a so called RNA world. The RNA would be both the genetic material and the enzymes that catalyzed the reactions, including replicating the RNA. You could imagine that these RNA only proto living systems then developed a way to make proteins, which provide a much broader array of catalytic functions. And separately such systems started using DNA as the genetic material. We do not know for sure that this is how life first evolved, but it is the best rationalization biochemists have come up with. tRNA primary transcripts are processed by other enzymes that cleave at other sites, removing a small intron; add a short CCA sequence at the 3' end; and chemically modify a number of bases by methylation or deamination. These changes and the secondary structure you can see are all needed for tRNAs molecules to function.

If a wild type Neurospora could grow on a minimal media that included only a single carbon source (a sugar), what does this tell you about the wild type Neurospora? (L.14)

tells me that the Neurospora must have anabolic pathways that can generation all the metabolites and biomolecules needed to make this mold using only a sugar carbon source. *if it needed anything else then it could not grow on a minimal media

What is the diameter of nucleus in the human cell and how does the DNA fit? (L.15)

the diameter is 5 to 10 micro... So the DNA in your cells must be compacted about a 200,00 fold to fit in the nucleus

Which form of compensation for a strained DNA is less energetically favorable? Why?(L.15)

the strand separated structure is less energetically favored than the super coils because it has broken multiple h-bonds and disrupted the hydrophobic effect and base stacking in the region where the strands separate.

How will the DNA less commonly compensate when there is a lot of strain from under wound DNA?(L.15)

the strands will separate to allow most of the structure to adopt the classic 10.5 bp per turn helix. As a result the DNA will not show a supercoil.

We know that in order for DNA to wrap around the Histone it requires the removal of one helical turn. How does the under-winding occur?(L.15)

the underwinding occurs without a strand break, and therefore there must be a compensatory positive (+) supercoil formed in the DNA. on either side of the histone. The positive supercoil is relaxed by a type II topoisomerase, leaving the DNA with a net negative supercoil.

What two approaches does the cell use to repair structures who have no information in both DNA strands? (L.16)

two approaches that the cell uses to repair structures such as those on the bottom.: 1. One is to use the information on the second chromosome as a template to repair the damaged chromosome. This is achieved using a process called recombination. 2. the second approach the cell uses is to employ either one of two other DNA polymerases: pol IV or pol V. These proteins are part of the so called SOS response. The SOS response is n emergency response that switches on the expression of these two polymerases as well as that of other proteins to protect the cell when it detects a high amount of DNA damage. For example, the SOS response is induced when cells are irradiated with high levels of UV light. Pol IV and V are able to synthesize a daughter strand opposite a lesion by incorporating any nucleotides without regard to correct base pairing. Often they incorporate the wrong base pairs, but in situations such as the above, on the left, that is better than having no DNA on one strand. These error prone syntheses is called translesionDNA syntheses. The enzyme is able to do this by having a less restrictive space in its active site. If you remember how the grey wavy line—the boundaries of the active site—prevented wrong base pairs from fitting into the active site of pol I and III. In pol IV and pol V, the active site space is larger and can accommodate wrong base pairs and even somewhat misshapen DNA. The SOS response still requires one stand of DNA to have an intact sugar phosphate backbone, however. In contrast, recombinational repair does not even require that. So let's learn about recombination and how that can allow repair as well as other phenomena such as mitotic and meiotic chromosomal exchange.