CMB Final

L9 Ch8 4c: When the population increased for the second time, how did the amount of lactose decrease?

Once the bacteria turn on expression of the Lac operon, then the proteins needed for consumption of lactose are transcribed and translated, so present.

L12 Ch17 1: What are the three types of protein filaments that make up the cytoskeleton?

One filament type is the intermediate filament. There are numerous types of intermediate filaments. Each type is built from a different subunit: Lamins, Keratins, Neurofilaments or Vimentins. A second type of filament is the microtubule that is built from subunits of α/β-tubulin heterodimers. The third filament type is the actin filament that is built from subunits of actin monomers.

R5 1b: Is Leu-248 part of the helix αC, the DFG-motif, the P-loop or the N-lobe?

P-loop

L11 Ch 15-5b: What would you observe if the plasma membrane fragments were from a prokaryotic cell?

Prokaryotes lack endocytosis, so presumably lack proteins like cargo receptors, so the adaptins added to the preparation would not bind to any proteins at the membranes, so no vesicles would form, but it sure would be interesting if a vesicle-like structure did form.

L10 Ch15 2a: How is import of a protein into the nucleus directional?

Protein import into the nucleus is directional because the Nuclear Import Receptor (NIR) only binds to proteins of the nuclear pore after the NIR binds to a protein's NLS and changes conformation (shape). The NIR binds to the nuclear pore in way that moves the NIR and its cargo into the nucleus, but once in the nucleus the NIR cannot move out through a pore until the NIR binds to Ran-GTP. Binding to Ran-GTP causes the NIR to release the cargo and change conformation so that NIR can bind to nuclear pore proteins in a way that moves the NIR and Ran-GTP out of the nucleus, but the cargo stays in the nucleus.

L10 Ch15-15: Consider a protein that contains an ER signal sequence at its N-terminus and a nuclear localization sequence in its middle. What do you think the fate of this protein would be? Explain your answer

Protein synthesis begins with the N-terminus. As the first part of the protein is synthesized, a Signal-Recognition Particle will bind to the ER signal sequence, and then to the SRP receptor at the ER. The protein will then be synthesized into the ER lumen. Once inside the ER lumen, the protein will not have access to the nucleus, so its nuclear localization sequence serves no function.

L9 Ch8 7: What are three common DNA-binding motifs?

Homeodomain, zinc-finger and leucine zipper.

L9 Ch8 1a: During development, cells differentiate to specific cell types. The process of differentiation involves cutting out and getting rid of genetic information, so that different cell types have different sets of genes in their genomes.Do you agree with this statement? Explain your answer

I disagree with this statement. The process of differentiation involves changing the chromatin state of selected regions of chromosomes to being accessible or inaccessible for expression, so that a given cell type can express a subset of the total genes in its genome. Different cell types express different subsets of genes, although there is overlap because all cells express a set of housekeeping genes.

L7 Ch5 3: Imagine a protein that binds tightly to DNA. Would you expect the amino acids that are oriented toward the DNA backbone to be acidic (e.g. glutamic acid) or basic (e.g. lysine)? Explain your answer

I would expect the amino acids oriented toward the DNA backbone to be basic, because a basic side-chain has a full positive charge that can form ionic bonds with the negatively charged, repeating phosphates that are part of the sugar-phosphate DNA backbone.

R5 5: Suggest a mutation of Ile-293 that would likely reduce or prevent binding of Gleevec to Abl. Explain how your chosen mutation reduces or prevents binding.

I would mutate Ile-293 to a glutamic acid, which has a negative charge and is bigger than isoleucine. The bulkier side chain would make it difficult for Gleevec to fit properly, if at all, in the binding pocket, and the full negative charge would interfere with the hydrophobic force, also disfavoring Gleevec being in the binding pocket.

L9 Ch8 1b: Describe an experiment that demonstrates that differentiated cells do not get rid of genetic information.

If differentiated cells got rid of genetic information, then transferring the genomes from a mature cell type back into an egg cell would not be able to direct development, because genetic information would be missing. Researchers conducted experiments where they used high doses of ultraviolet radiation to destroy the genome in a frog egg cell. The UV causes damage to DNA (particularly cytosines and thymines; pyrimidines), which initiates repair processes that cut the DNA. At high doses of UV, the genome gets cut up into too many pieces and is essentially destroyed. The researchers separated adult cells from the frog skin and used a micropipet to transfer a nucleus from an adult, differentiated cell into the frog egg cell lacking a nucleus (destroyed along with the DNA). The frog egg cell with the adult nucleus was able to run the genetic programs needed for the development of a normal frog.

L12 Ch17-3c: What would happen if only GDP, but no GTP, were present in the solution?

If only GDP was in the solution, then there would be no α/β-tubulin-GTP and microtubules would not be able to grow.

L12 Ch17 3: In a cell, the concentration of α/β-tubulin subunits are typically lower than the amount needed for the spontaneous assembly of a microtubule. How does a microtubule start to assemble in a cell?

In a cell (in vivo), there are stable rings of -tubulin, which are typically located on the surface of the centrosome. A stable γ-tubulin ring serves as nucleation site to which α/β-tubulin subunits can bind to and begin the growth of the 13 protofilaments that make up a microtubule.

L9 Ch8-1b: What would happen in scenarios (1), (2) and (3) if the cells, in addition, produced normal tryptophan repressor protein from a second, normal gene?

In scenarios (1) and (2), the normal tryptophan repressor molecules would completely restore the regulation of the tryptophan biosynthesis enzymes. In genetic terms, the mutant would be a recessive allele, if we were talking about a diploid organism/cell. In contrast, expression of the normal protein would have no effect in scenario (3), because the tryptophan operator would remain permanently occupied by the mutant protein. In genetic terms, the mutant would be a dominant allele, if we were talking about a diploid organism/cell.

L11 Ch 15 2: How do vesicles take the correct cargo?

Proteins in the lumen of the ER or golgi sacs, or just outside the cell, are recognized by transmembrane proteins referred to as cargo receptors. Cytosolic proteins called adaptins bind to the cytosolic domains of cargo receptors and help gather the cargo to a patch of membrane. Coat proteins then bind to adaptins and assemble into a cage-like structure that bends the membrane into a vesicle that contains the gathered cargo.

R5: Figure 4

Several of Abl's amino acids make noncovalent bonds with Gleevec. Threonine (OH), glutamic acid (COOH) and asparagine (NH) side-chains make hydrogen bonds. Methionine makes a hydrogen bond using a backbone amine (NH). From Wikimedia Commons.

R5: Figure 2

Some amino acids. Note that all the amino acids have the same 'backbone', which includes an amine (NH) and a carbonyl (C=O).

L11 Ch 15 6: How does the structure of a Golgi body relate to the order of sugar modifications found on glycoproteins?

The Golgi body is organized into a stack of sacs (cisterna) that do not share a continuous lumen. Thus, each sac, from the cis sac to the middle sacs to the trans sac, can hold different modifying enzymes in each of their lumens. As a glycoprotein first enters the cis sac, it can be modified, and then sequentially modified as it passes, by vesicle transport, into and out of each subsequent sac.

L10 Ch15-23: What would happen to proteins bound for the nucleus if there were insufficient energy to transport them?

The Nuclear Import Receptor would move some proteins into the nucleus. However, the Nuclear Import Receptors require binding to Ran-GTP to release their cargo and move back to the cytosol. With insufficient energy, the transport of proteins into the nucleus would stop once all the Nuclear Import Receptors moved into the nucleus.

L10 Ch15 6b: How does the protein get into the nucleus?

The Nuclear Transport Receptor selectively binds to proteins with a nuclear localization signal, and then binds to the nuclear pore and facilitates transport of its cargo, in this case the transcription regulator, into the nucleus.

L9 Ch8 4a: When the population increased for the first time, how did the amount of lactose remain the same?

The bacteria first consume the glucose. When glucose is present, the CAP protein does not bind DNA, so the Lac operon is not expressed and the bacteria are not able to consume the lactose.

L9 Ch8 4b: How did the population stop increasing for 10 minutes before increasing again?

The bacteria perceive the reduced amounts of glucose. In response, cAMP increases and binds to the CAP protein, which then is in a conformation that binds DNA and activates expression of the Lac operon. Time is required for transcription and translation to make the enzymes required to metabolize lactose, so for a short period of time the bacteria lack the carbon and energy needed to grow and divide. Once the bacteria have the ability to metabolize lactose, the population begins increasing again.

L8 Ch7 2: When a RNA polymerase binds to a promoter, what determines which strand of DNA will be the template?

Two things determine which strand of DNA will be the template. 1) The sigma factor (or TATA-binding protein in eukaryotes) binds to the asymmetric promoter in a specific orientation, which gives the RNA polymerase a specific orientation, and 2) the polymerase only catalyzes synthesis in the 5' to 3' direction. Once oriented and the DNA helix is opened, the strand that has 3'-5' polarity serves as the template.

L9 Ch8 10: Transcription regulators typically bind DNA as dimers. Do you expect a dimer to bind with higher affinity than a monomer? Explain your answer

Yes, a dimer is likely to form more noncovalent bonds because it has a more surface area in contact with the DNA than a monomer.

L8 Ch7 3: Compare how mRNA is processed in eukaryotes to processing (or lack of) in prokaryotes.

mRNA synthesized in a prokaryotic cell can immediately be bound by a ribosome and translation started even before completing mRNA synthesis. In contrast, in eukaryotes the mRNA must leave the nucleus before being translated by a ribosome, which only happens after the mRNA is processed. The mRNA is given a 5' Cap that is a 7-methylguanosine, introns are spliced out of the RNA and the 3' end is cut-off at a specific sequence and then a series of Adenines are added to yield a poly-A tail. Splicing out of introns is catalyzed by the spliceosome, which is a complex of both RNA and protein. The RNA component selectively binds the splice junctions (by complementary base pairing) and catalyzes the breaking and reforming of the phosphate backbone of the RNA.

L9 Ch8 3c: If a mutant cell has a Lac repressor that is unable to bind lactose, then

the operon will NOT be expressed under any of the four conditions.

L9 Ch8 3d: If a mutant cell has a CAP protein that binds DNA in the absence of cAMP and has a Lac repressor that cannot bind DNA, then

the operon will be expressed under all four conditions.

L9 Ch8 3a: If a mutant cell has a Lac operator that is unable to bind the Lac repressor, then .

the operon will be expressed when glucose is absent and lactose is either absent or present.

L9 Ch8 3b: If a mutant cell has a CAP protein that binds the DNA in the absence of cAMP, then

the operon will be expressed when lactose is present and glucose is either absent or present.

L12 Ch17-3a: What must happen at the end of the microtubule in order for it to stop shrinking and to start growing again?

α/β-tubulin-GTP heterodimers must assemble onto the end if a shrinking microtubule is to be 'rescued' before complete disassembly.

L10 Ch15 1: Describe in general how each compartment gets a distinct set of proteins. Include the following in your description: signal sequences, pores or translocators, vesicle transport.

1) The nucleus gets a set of proteins that each has a nuclear localization signal (NLS). The nuclear proteins are synthesized to completion on ribosomes in the cytosol. The Nuclear Import Receptor binds to a protein's NLS, then binds to proteins of the nuclear pore and moves into the nucleus. Once inside, a GTP-bound Ran protein binds the receptor causing it to change conformation and release the cargo, so delivering the protein into the nucleus. 2) The cytosol gets its proteins by their complete synthesis on ribosomes in the cytosol, and then the proteins stay in the cytosol because they lack signal sequences. 3) The matrix of the mitochondria gets a set of proteins that each has a signal sequence that is recognized by a receptor protein embedded in the mitochondrial outer membrane. After complete synthesis in the cytosol, the signal sequence binds to the receptor and is then transferred to a protein translocator, which is in two components: one spans the outer and the other the inner membrane. The protein is then unfolded and pulled through the protein translocator. The chloroplast and peroxisimes get their proteins by a similar mechanism. 4) The endoplasmic reticulum (ER) gets a set of proteins that have an ER signal sequence that is recognized by a protein called the Signal-Recognition Particle (SRP). The SRP binds to a protein receptor embedded in the ER membrane. Binding brings the partially synthesized protein and the ribosome to the ER. Synthesis of the protein is then completed at the ER: synthesis pushes the protein through a translocator and into the lumen of the ER or if there is an internal signal sequence, then transfer stops and results in a transmembrane alpha-helix. Multiple internal start-stop transfer signals results in multiple transmembrane alpha-helices. 5) Proteins synthesized into the ER lumen, or into the membrane, remain at the ER if they have ER retention signal sequences, or are packaged into vesicles and delivered to the Golgi apparatus, which is how the golgi gets its proteins. Some types of proteins stay in each sac of the golgi, while other proteins move beyond the golgi. Many proteins are packaged into vesicles at the trans-golgi and delivered to the endosome or to the plasma membrane. The former is how the endosome, and once mature, lysosome, get their proteins, and the latter is how the extracellular matrix gets proteins and how the plasma membrane gets proteins.

L9 Ch8 14a: What are three mechanisms for epigenetic inheritance?

1. Copying of histone tail modifications during DNA replication so that chromatin state in daughter cells will be the same as the parent cell. 2. Positive feedback loops that involve a transcription regulator that turns on the expression of numerous cell-type specific genes and also turns on expression of its own gene. This type of positive feedback loop results in daughter cells with the same transcription regulators as the parent cell. 3. Copying of cytosine methylation during DNA replication. Some cytosines in the genome are methylated and are recognized by some chromatin-remodeling complexes that then help chromatin condensation, hence inhibiting gene expression. Copying of cytosine methylation helps daughter cells keep the same chromatin state as the parent cell.

L7 Ch5 13: Treatments

4 different treatments are used in Step 2 (Treatment A = Lane A; Treatment B = Lane B and son on): Treatment A: No nuclease treatment. Treatment B: Remove associated proteins with a strong solution first, then brief (30sec) nuclease digestion of naked DNA. Treatment C: Extended (20min) nuclease treatment. Treatment D: Brief (30sec) nuclease treatment.

R5: Figure 3

A close up showing Gleevec in Abl's binding pocket. Some of Abl's amino acids are labeled. The spheres represent van der Waal surfaces. From Seeliger et. al. (2007) Structure v15, p299-311.

L11 Ch 15 1: How do vesicles form?

A vesicle forms by coat proteins binding to a patch of adaptins, which are bound to cargo receptors. In addition to binding to adaptins, coat proteins bind to each other, so assemble into a cage-like structure that bends the membrane into a sphere. The vesicle remains continuous with the membrane by a constricted region. The GTP-binding protein, dynamin, then 'pinches' the constricted region to release the vesicle. 'Pinching' happens by dynamin changing conformation as it binds GTP and catalyzes hydrolysis of GTP.

L8 Ch7 1: Describe the basics of how a eukaryotic cell initiates and transcribes a gene. Use the following terms: promoter, general transcription factors, TATA-binding protein, TFIID, TFIIH, preinitiation complex, phosphorylation of the RNA polymerase tail, DNA template, complementary base-pairing, 5' to 3', terminator.

A eukaryotic cell prepares to transcribe a gene by assembly of a preinitiation complex at the promoter. A promoter is a nucleotide sequence that typically includes 5'-TATA-3' and is recognized (selective binding) by the TATA-binding protein that is a subunit of the TFIID transcription factor. After binding of TFIID to the promoter, several other general transcription factors and RNA polymerase bind to TFIID, and each other, to assemble the initiation complex. Start of transcription depends on input from numerous transcription regulators that bind DNA regulatory sequences. If more positive regulators are present, then TFIIH, a protein kinase, switches to an active conformation and transfers phosphate groups onto the RNA polymerase tail. Phosphorylation of the tail causes the polymerase to dissociate from the initiation complex and begin transcription. The polymerase uses the 3'-5' oriented strand as template and catalyzes 5'-3' synthesis of the messenger RNA (mRNA). The RNA sequence is determined by complementary base pairing to the DNA template: adenine with thymine, cytosine with guanine, guanine with cytosine, and uracil with adenine. RNA chain elongation (synthesis) continues until the polymerase reaches a specific DNA sequence, called a terminator, which causes the RNA polymerase to stop synthesis and disassociate (fall off) from the template.

L7 Ch5 5: What is a gene? Describe, or draw a picture of, the different parts of a gene.

A gene is a stretch of DNA that encodes the instructions for the synthesis of a protein, or the synthesis of a catalytic or regulatory RNA. The instructions include exons, which are the DNA sequences that code for the amino acid sequence [exons are separated by noncoding introns], and a promotor where the RNA polymerase assembles in preparation for transcription. Instructions also include regulatory DNA sequence to which transcription regulators bind and either promote or inhibit the start of transcription. The binding of transcription regulators determines when the gene is expressed, in which cell types (where in a body) the gene is expressed and how many RNA copies are made.

L10 Ch15 4: Describe the mechanism of how a protein gets into the ER lumen compared to how a protein gets embedded in the ER lipid bilayer.

A protein that gets synthesized into the ER lumen has a single, N-terminal ER signal sequence. The signal sequence binds to the protein translocator, after binding to the SRP, and continued synthesis pushes the polypeptide chain through the translocator into the lumen. An enzyme, called a signal peptidase, catalyzes hydrolysis that cuts off the signal sequence, releasing the new protein into the lumen. In contrast, a protein that gets embedded into the lipid bilayer has either NO N-terminal signal sequence, but one or more internal signal sequences, or has a N-terminal signal sequence in addition to one or more internal signal sequences. Each internal signal sequence in turn binds to the translocator, displacing the previous signal sequence, which, if internal, floats into the lipid bilayer as a transmembrane alpha-helix.

L10 Ch15 3: How does a ribosome get located to the endoplasmic reticulum?

A ribosome gets located to the ER if the protein it is translating has an ER signal sequence. As soon as the signal sequence is synthesized, the Signal-Recognition Particle binds to it, and also binds to a receptor embedded in the ER membrane, thus attaching the ribosome to the ER.

R5: Figure 1

A simplified pathway showing a few parts that connect an input growth factor to the output of cell division. Abl kinase transduces the signal by transferring a phosphate from ATP onto a transcription factor, which then turns on many genes encoding proteins that guide cell division.

L9 Ch8 13: What is a transcription factory? How does a transcription factory form?

A transcription factory is a volume of space, in the nucleus, where many different proteins needed for transcription and RNA processing are concentrated. A transcription factory forms from numerous different proteins having affinity for several scaffolds within the volume of space where the factory forms. Histone tails, DNA regulatory sequences and the RNA polymerase 'tail' act like scaffolds; proteins bind to the scaffolds, so spend more time in that volume of space where those scaffolds are located. Other proteins bind to the proteins that bind the scaffolds, so those other proteins also spend more time in that volume of space. Other proteins bouncing around the nucleus that do not bind to the specific scaffolds, and concentrated proteins, of a factory do not spend extra time in the factory (their diffusion is not slowed down by binding interactions). However, those other proteins might have interactions in a different factory elsewhere in the nucleus, so spend more time there (slowed diffusion rate).

L9 Ch8 6: How can a transcription regulator bind to one stretch of DNA, but not to other stretches with different sequences?

A transcription regulator binds a specific DNA sequence by placing amino acid side chains into the major groove, and if the shape and charge distribution of the amino acids matches the edges of the nitrogenous bases that make up the specific sequence, then noncovalent bonds form and binding happens. If the shape and charge distribution of the amino acids does not have complementarity to the shape and charge distribution of the edges of the nitrogenous bases making up a stretch of DNA sequence, then noncovalent bonds do not form and binding doesn't happen.

L9 Ch8 8: Some transcription regulators can cause specific regions of chromatin to decondense and make the DNA accessible to general transcription factors and RNA polymerase. How can a transcription regulator cause specific regions of chromatin to decondense?

A transcription regulator binds to a stretch of DNA that has a specific nucleotide sequence. Some transcription regulators, after binding DNA, have a second site that selectively binds to a histone acetylase (also called acetyltransferase), which catalyzes addition of acetyl onto lysine of the histone tails. Acetylated histone tails favor decondensation. Chromatin-remodeling complexes also bind to the acetylated histone tails and then slide DNA to increase linker DNA and further decondense the chromatin. Some transcription regulators can bind directly to chromatin-remodeling complexes.

L11 Ch 15 4: How do vesicles fuse with the target membrane?

A vesicle fuses with a target membrane when its Rab proteins bind to tethering proteins on the target membrane. Vesicles also have a protein called v-SNARE that binds to a protein called t-SNARE that is on target membranes. The binding of SNAREs brings the vesicle in contact with the target membrane so that the lipid bilayers flow together completing fusion. Note that the above is for the constitutive secretion pathway (and for vesicles fusing to the internal membranes like the cis-golgi). Regulated vesicle fusion involves an additional protein called Synaptotagmin. In the case of regulated fusion, binding of v-SNARE to t-SNARE docks the vesicle, but fusion happens after Synaptotagmin binds to Ca2+, changes conformation and binds to a t-SNARE called Syntaxin. The binding of Synaptotagmin to Syntaxin leads to vesicle fusion.

L11 Ch 15-5a: What would you observe if you omitted (A) adaptins, (B) clathrin, or (C) dynamin?

Adaptins bind cargo receptors and coat proteins. In the absence of adaptins, the initial forming of a vesicle cannot happen. In the absence of clathrin (a coat protein), the initial forming of a vesicle does not happen. Coat proteins are needed to deform the membrane into a sphere (vesicle). Dynamin is needed to 'pinch' off a vesicle from the membrane. In the absence of dynamin, vesicles would form but remain attached to the membrane by a narrow, constricted region.

L9 Ch8 2: State a similarity and a difference between the expression of house-keeping genes, cell type specific genes and induced genes.

All three categories of genes are transcribed by the same machinery; general transcription factors plus the RNA polymerase assembled into a preinitiation complex at a promoter. Initiation of transcription is by activation of the protein kinase, TFIIH, which phosphorylates the RNA polymerase 'tail'. The RNA polymerase then synthesizes the messenger RNA. The categories of genes vary in when the genes are expressed and in which cell types. House-keeping genes are accessible for expression and are expressed daily (constitutive expression) in all cell types of a body. House-keeping genes code for the basic machinery all cells need to survive and perform basic functions like core metabolism, secretions, gene expression and others. Cell type specific genes are accessible for expression in only a subset of all cell types. Some of the cell type specific genes are expressed constitutively and contribute to the cell type's specific organization. Induced genes are accessible for expression but are not expressed constitutively, but only expressed in response to a specific signal, such as a hormone like cortisol.

L12 Ch17-3b: How would a change in the tubulin concentration affect this switch?

An increase in concentration of α/β-tubulin-GTP in the cytosol would increase the rate at which α/β-tubulin-GTP is assembled onto the plus-end of a microtubule.

L12 Ch17-2: Why do you suppose it is much easier to add tubulin to existing microtubules than to start a new microtubule from scratch? Explain how γ-tubulin in the centrosome helps to overcome this hurdle.

As long as the α/β-tubulin at the end is GTP-bound, the end of a microtubule provides a stable, surface for heterodimers to bind to even if adjacent heterodimers have not yet bound. In the absence of the stable surface, the probability is low that α/β-tubulin-GTP heterodimers will bind to each other laterally in a sufficient number to form a ring before any of the heterodimers catalyzes hydrolysis of GTP, which weakens lateral interactions. In contrast, γ-tubulin can form a stable ring because the subunits form more lateral, noncovalent bonds with each other and do not change shape, so do not loosen. The γ-tubulin ring then provides a stable surface for α/β-heterodimers to bind to even if adjacent heterodimers have not yet bound.

R5 3: Look at Figure 4. The dashed lines show noncovlanet bonds between some of Abl's amino acids and Gleevec. Suggest an explanation for how cancerous myeloid cells with a Thr-315 to Ile (isoleucine) mutation in Abl are resistant to Gleevec.

Cancerous myeloid cells with Abl-Ile-315 are resistant to Gleevec, because the Abl-Ile-315 protein does not form enough noncovalent bonds with Gleevec. Gleevec does NOT bind tight enough to the Abl protein. Thr-315 has a polar side-chain, which includes a -OH (hydroxyl). The oxygen is electronegative and the hydrogen is electropositive. When Gleevec is in the binding pocket, an electropositive hydrogen, of a Gleevec, polar amine group (-NH), is oriented towards the electronegative oxygen so that a hydrogen bond forms. In contrast, isoleucine has a nonpolar side-chain, so the hydrogen bond is lost. Isoleucine's side-chain is also slightly bigger than Thr-315, so Gleevec may not fit properly and additional noncovalent bonds may be lost.

L7 Ch5 10: What is the relationship between DNA packaging and cell differentiation?

Cell differentiation is a process that involves changing the organization of a cell so that it can perform a specialized function. The expression of a specific set of genes leads to changing a cell's organization. A cell undergoing differentiation covalently modifies histone tails. The modifications vary across the genome: some stretches of chromatin have histone tails modified in a way that results in condensing the DNA so it is not accessible, while other stretches are modified so that the DNA will be accessible for expression. The accessible DNA codes for a set of proteins that when expressed determine the cell's specialized organization and function. Cell differentiation starts with a cell receiving position-specific signals (such as hormones or growth factors or other signals) that bind to receptors that initiate signal transduction that activate transcription regulators. Active transcription regulators then bind to specific DNA sequences. There are many different transcription regulators (coded for by different genes) that each bind different DNA sequences. Some transcription regulators, after binding to DNA, selectively bind enzymes that covalently modify histone tails, so these transcription regulators target specific regions of a chromosome for having histone tails modified, which then determines whether that region will condense or decondense by mechanisms discussed for Q9 above. Examples of enzymes that modify histone tails are histone methyltransferases and demethylases, and histone acetyltransferases and deacetylases.

L9 Ch8 14b: How is epigenetic inheritance important for the process of development (building a body out of cells)?

Cells receive position-specific signals during development that tell the cell what parts of the genome to condense and decondense - what parts will be inaccessible and what parts will be accessible for gene expression. After a cell makes the changes to its chromatin, the cell has a particular identity (cell-type; particular organization and behavior; particular pattern of gene expression). To build a developing tissue, the cell-type needs to divide many times, but the original signal is not continually present. To continually give rise to daughters that keep the same behavior, the same chromatin states have to be passed from the parent cell to its daughters. How are chromatin states passed from parent to daughter cells? By the epigenetic inheritance mechanisms (1) and (3) described above.

L7 Ch5 9: Describe the mechanisms that regulate the chromatin state of a region of DNA.

Chromatin state refers to the degree of packaging (condensation) of a region of DNA. There are several mechanisms that regulate the degree that chromatin is condensed. One is a direct effect on packaging by covalent modification of histone tails. For example, adding a methyl group to lysine 9 of a histone tail allows nucleosomes to be packed close together, which is needed for the 30-nm fiber. Adding an acetyl group to lysine 9 has the opposite effect; close packing of core particles is unfavorable so the 'beads-on-a-string' state is favored. A second mechanism is an indirect effect of covalent modification of histone tails. Different chromatin remodeling complexes (multiple proteins forming a quaternary structure) bind selectively to certain histone tail modifications and then couple ATP binding and hydrolysis to sliding nucleosomes. Some complexes slide the DNA in a way that separates the core particles, thus increasing the length of linker DNA so that it is accessible to proteins such as RNA polymerase. Other complexes slide the DNA so the amount of exposed linker DNA is reduced and the nucleosome core particles are close together. A third mechanism is the binding and bending of DNA by histone H1, which allow nucleosomes to pack together into a 30-nm fiber. There are additional mechanisms we did not discuss that pack together loops of 30-nm fibers.

L10 Ch15 6a: In what cell compartment is the protein synthesized?

Cytosol on free ribosomes.

L7 Ch5 2: Describe the structural features of deoxyribonucleic acid (DNA) using the following terms: nucleotide nitrogenous base phosphodiester bond sugar-phosphate backbone chemical polarity of DNA strands 5' phosphate 3' hydroxyl base pairs hydrogen bonds anti-parallel double helix major groove minor groove

Deoxyribonucleic acid (DNA) consists of two long strands that twist to form a double helix that has a major groove and a minor groove. The subunits of each strand are nucleotides. Each nucleotide consists of a nitrogenous base, a deoxyribose sugar and a monophosphate. Each strand has chemical polarity: at one end is a 5' phosphate and at the other end a 3' hydroxyl. The two strands of a DNA helix are oriented with opposite polarity to each other and are thus said to be anti-parallel. The two strands are held together by hydrogen bonds that form between nitrogenous bases: adenine and thymine form complementary base pairs, and guanine and cytosine form complementary base pairs. When DNA is being synthesized, the 5' phosphate of an incoming nucleotide (in the triphosphate form) is covalently bonded to the 3' hydroxylat the end of a growing, new strand. The bond linking two adjacent nucleotides in a strand is called a phosphodiester bond. The repeating sugar-phosphate of each nucleotide make up the sugar-phosphate backbone of a strand of DNA.

L11 Ch 15 5: How is it that oligosaccharides covalently bonded to proteins (glycoproteins) are exposed on the outer surface of the plasma membrane and not the cytosolic surface?

Enzymes that catalyze addition of sugars to proteins are located in the lumen of the endoplasmic reticulum and the lumen of the golgi sacs, but not in the cytosol. As transmembrane proteins are transported from ER lumen, to the golgi sacs and on to the plasma membrane, the lumen facing domains of the proteins are modified by the ER and golgi enzymes. When the vesicle transporting the proteins fuses with the plasma membrane, the lumen face always becomes the outer surface of the plasma membrane.

L7 Ch5 11: Describe the mechanism of somatic epigenetic inheritance.

Epigenetic inheritance is the passing of the pattern of histone tail modifications, across a genome, from a dividing cell to its daughter cells. Because histone tail modifications regulate packaging of DNA, the daughter cells will have access, or no access, to the same gene set as the parent cell. Somatic refers to the cells that build the body in contrast to germline cells that give rise to egg or sperm. During S-phase, DNA replication doubles the amount of DNA so the histones are distributed between the two copies of DNA and new histones have to be added to nucleosomes. The new histones lack tail modifications. The epigenetic mechanism involves proteins that selectively bind to specific histone tail modifications and then 'copy' those modifications onto the tails of adjacent histones.

L7 Ch5 13: Evidence for the packing of DNA in the eukaryotic nucleus originally came from experiments, such as the ones described in the next few problems. Researchers first postulated that nucleosomes existed in the eukaryotic chromatin, but then they had to prove it experimentally. One of the most powerful experiments that helped to prove the presence of nucleosomes was the nuclease digestion experiment, in which they used micrococcal nuclease to digest DNA. This enzyme is a large bacterial enzyme that can only cut exposed DNA (i.e. 'naked' DNA). It cannotcut DNA that is tightly bound to proteins.

Experimental Protocol: 0. Starting point: 30nm chromatin fiber 1. Wash chromatin with weak salt solution to remove H1 histone. 2. Digest 11nm fiber with micrococcal nuclease (according to treatments listed below) 3. Remove associated proteins with a strong salt solution. 4. Separate DNA fragments using gel electrophoresis

R6: Evidence for the packing of DNA in the eukaryotic nucleus originally came from experiments, such as the ones described in the next few problems. Researchers first postulated that nucleosomes existed in the eukaryotic chromatin, but then they had to prove it experimentally. One of the most powerful experiments that helped to prove the presence of nucleosomes was the nuclease digestion experiment, in which they used micrococcal nuclease to digest DNA. This enzyme is a large bacterial enzyme that can only cut exposed DNA (i.e. 'naked' DNA). It cannot cut DNA that is tightly bound to proteins.

Experimental Protocol: 0. Starting point: 30 nm chromatin fiber 1. Wash chromatin with weak salt solution to remove H1 histone. 2. Digest 11 nm fiber with micrococcal nuclease (according to treatments listed below) 3. Remove associated proteins with a strong salt solution. 4. Separate DNA fragments using gel electrophoresis 4 different treatments are used in Step 2 (Treatment A = Lane A; Treatment B = Lane B and son on): Treatment A: No nuclease treatment. Treatment B: Remove associated proteins with a strong solution first, then brief (30sec) nuclease digestion of naked DNA. Treatment C: Extended (20min) nuclease treatment. Treatment D: Brief (30sec) nuclease treatment.

L9 Ch8 12: The expression of many genes in eukaryotes is each regulated by many transcription regulators. Does this mean that there are no simple on-off switches?

For some genes, the chromatin is decondensed and numerous transcription regulators favor assembly of a preinitiation complex at the promoter, but transcription is not initiated. In response to an input signal, one last transcription regulator binds the Mediator, which changes conformation, adjusts the preinitiation complex and initiates transcription. The last regulator, in response to the input signal, acts as an on-off switch; when the input signal stops, then transcription of the gene is also turned off. The textbook uses the example of cortisol, which binds to its receptor, which is a transcription regulator. The cortisol-bound, transcription regulator then binds upstream of numerous genes, which already have other regulators assembled. Addition of the cortisol-regulator completes the needed combination to start transcription, so the numerous genes are expressed. When amounts of the cortisol-regulator complex decrease, expression of the genes stops.

L7 Ch5 6: Is a gene located in heterochromatin more or less likely to be expressed than a gene located in euchromatin? Explain your answer

Genes located in heterochromatin cannot be expressed. Heterochromatin is chromatin in its most condensed state, so transcription regulators and RNA polymerases cannot access the DNA. Euchromatin refers to less condensed chromatin that can be opened up to expose DNA for expression. The most accessible DNA is in chromatin in the 'beads on a string' form (11-nm fiber) and when nucleosomes are separated from each other so that linker DNA is accessible. General transcription factors and RNA Polymerase can assemble at promoter sequences that are within exposed linker DNA.

L10 Ch15 6c: Compare the following two mutants: (1) a mutation in the nuclear localization signal of the transcription regulator that prevents it from binding to the nuclear transport receptor, and (2) a mutation in the nuclear transport receptor that prevents it from binding any protein's nuclear localization signal. State the consequence for each mutant and which mutant is more likely to disrupt cell function.

In the case of (1), the transcription regulator will remain in the cytosol. The subsequent consequence to the cell depends on what genes the transcription regulator normally turns 'on', which in the mutant remain 'off' because the regulator cannot enter the nucleus. In the case of (2), the transcription regulator and all other proteins intended to be in the nucleus remain in the cytosol. No proteins enter the nucleus, at least not efficiently (some may slowly by chance bump their way through a pore). Mutant (2) is more likely to disrupt function, because the consequence includes that of mutant (1) plus the fact that no other proteins can get in the nucleus. Such a mutation would likely be lethal.

R5 4: Find Ile-293 in Figure 3 and 4. Ile-293 does not form hydrogen bonds with Gleevec. What type of noncovalent bond(s) does Ile-293 form with Gleevec?

Isoleucine has a nonpolar side-chain. When Gleevec is in the binding pocket, Ile-293 is adjacent to a nonpolar surface of Gleevec. When two nonpolar surfaces are pushing up against each other, there are moments when fluctuation of electrical charges (due to electrons constantly moving) happens and weak attractions form between adjacent atoms. The weak attractions are called van der waals attractions, a type of noncovalent bond. The hydrophobic force may contribute to the nonpolar surfaces of Gleevec orienting toward the nonpolar surfaces of the binding pocket.

L10 Ch15 5: If a protein with a single transmembrane domain (alpha-helix) is embedded in the ER lipid bilayer and transported to the plasma membrane, will the N-terminus (+H3N-) or the C-terminus (-COO-) be located in the cytosol? Explain your answer

It depends on the protein's signal sequences when it was synthesized at the ER. A protein with a N-terminal signal sequence and an internal stop-transfer sequence will have its C-terminus in the cytosol and the N-terminus will be facing the extracellular matrix. A protein with NO N-terminal signal sequence but an internal start-transfer sequence will have its N-terminus in the cytosol and the C-terminus will be facing the extracellular matrix.

L8 Ch7 4: What are 4 different types of RNA involved in converting the DNA code into a sequence of amino acids?

Messenger RNA (mRNA) is a copy of a DNA sequence. The mRNA sequence (a series of triplets/codons) corresponds to the encoded protein's amino acid sequence based off the genetic code. Transfer RNAs (tRNA) are the genetic code. A tRNA is covalently linked to an amino acid and also includes the anti-codon that complementary base pairs with an mRNA codon during translation. Ribosomal RNA (rRNA) forms the structural core of ribosomes, the binding surfaces for both the mRNA and tRNAs, and the active site that catalyzes peptide bond formation. Splicing RNAs (spRNA) that are part of snRNPs (small nuclear RiboNucleoprotein Particles). spRNA binds to (by base pairing) exon-intron junctions and catalyzes intron splicing. Note: rRNA and spRNA are catalytic; they are RNA-enzymes, so called ribozymes.

L12 Ch17 3: Describe how microtubules grow and describe how microtubules rapidly shrink, which is called dynamic instability.

Microtubules begin growing by addition of α/β-tubulin heterodimers to a γ-tubulin ring (α-tubulin binds to the γ-tubulin), and then more α/β-heterodimers assemble simultaneously into 13 protofilaments, which collectively make up a single microtubule. The β-tubulin is a G-protein. α/β-heterodimers in the cytosol are bound to GTP, so the α/β-heterodimers that assemble onto a microtubule are in the GTP-conformation. The GTP-conformation forms noncovalent bonds with heterodimers in adjacent protofilaments. Some time after being added to a filament, the β-tubulin catalyzes hydrolysis of GTP to GDP. The GDP-bound conformation has less noncovalent bonds with heterodimers of adjacent protofilaments. As long as the plus-end of the microtubule has GTP-bound heterodimers (a GTP cap), the microtubule won't shrink, but as soon as the plus-end loses the GTP cap, the microtubule rapidly falls apart because of the weak binding between protofilaments. Notes: (1) The minus-end is where α-tubulin is bound to the γ-tubulin ring (γ = gamma). The plus-end is the opposite end of a microtubule and its surface is the β-tubulin. Going from a minus-end to a plus-end the order is γα/βα/βα/βα/βα/βα/β . . . . . . . (2) A microtubule grows when the rate of adding α/β > the rate of hydrolysis, and a microtubule will suddenly shrink when rate of adding α/β < rate of hydrolysis.

L12 Ch17 2: What are the cellular functions of each filament type?

Microtubules provide tracks for vesicle traffic and moving and positioning organelles. Microtubules also help support cell polarity and cell remodeling, provide a mechanism for chromosome segregation, and support cilia and flagellum. Moving and positioning organelles: Motor proteins attach to cargo and attach and walk along microtubules. The motor proteins move vesicles and organelles along the microtubule tracks. Moving organelles positions the organelles. Motor proteins constantly pull on the endoplasmic reticulum to hold the network extended outward from the nucleus. Motor proteins constantly pull on the sacs of the golgi body to position the golgi body near the centrosome and nucleus. Motor proteins constantly move mitochondria to different positions in the cell and move vesicles between the compartments of the endomembrane system. Cell polarity and cell remodeling: The microtubules can be dynamic, growing (assemble) and shrinking (disassemble), or can be stabilized by a Capping protein binding the plus end of the microtubule (the minus end is typically bound to a ring of γ-tubulin, located on the centrosome, which prevents shrinking from the minus end). Microtubules that are captured by a Capping protein can give a cell structural polarity because the stable microtubules result in vesicle traffic moving towards one side of the cell, and organelles being positioned to one side of a cell. However, in response to signals, the Capping proteins can release the microtubules, which then shrink and new microtubules grow out from the centrosome in new directions: a set of microtubules oriented to one side of the cell can be disassembled and the subunits used to build another set of microtubules oriented towards the opposite side of the cell. The remodeling leads to repositioning organelles and redirecting vesicle traffic. Chromosome segregation: During mitosis, microtubules attach to chromosomes and provide the mechanism for pulling and pushing chromosomes to opposite poles. Just prior to mitosis, interphase microtubules are disassembled, and the dynamic growing and shrinking of new microtubules increases the likelihood that a microtubule will grow into a chromosome's kinetochore, which is a protein complex that includes Capping proteins. Microtubules also provide structural support to cilia and flagella, and the mechanism, along with motor proteins, for generating the motions of cilia and flagella.

L9 Ch8 9: How can transcription regulators that bind DNA 10,000 nucleotides distant from a promoter influence transcription from that promoter?

Much of the intervening, spacer DNA, between the distant regulatory DNA (enhancers) and promoter, can be condensed and the regulatory DNA, with bound transcription regulator, can be 'looped' so that the protein is in close proximity to where the transcription pre-initiation complex assembles on the promoter. The proximal transcription regulator then binds to a Mediator complex, which is in contact with the pre-initiation complex at the promoter. The binding of the regulator influences the conformation of the Mediator, which then influences the assembly-disassembly rate of the preinitiation complex or turns on TFIIH.

R5 6: Resistance to Gleevec is most frequently found to be due to mutation of Thr-315 to isoleucine, however, there are numerous other amino acids (such as Ile-293) that contribute to the binding surface. Propose an explanation for why Gleevec-resistant cancer cells are not frequently found with mutations other than T315I?

Mutations changing amino acids at the Abl-Gleevec binding surface can also change the ability of Abl to bind to ATP and catalyze its hydrolysis has part of Abl's normal function. Mutations other than T315I may prevent ATP binding and hydrolysis so result in a loss of Abl function, which would result in cells that do not divide frequently so would be out competed by Abl-gain of function, mutant cancer cells that divide frequently.

R5 1a: Is Thr-315 part of the helix αC, the DFG-motif, the P-loop or the N-lobe?

N-lobe

L9 Ch8 5: If you repeated the experiment above but added the antibiotic tetracycline [look up what tetracycline inhibits] just before the first population increase subsided, would the second phase of population growth happen? Explain your answer.

No, tetracycline binds to the ribosomal A-site preventing tRNA binding and hence inhibits translation. The Lac operon will be expressed, but only the mRNA will be made, NOT the encoded proteins. Without the enzymes needed to metabolize lactose, the cells lack the input of energy and carbon needed for growth and division.

R5 2: Do both Thr-315 and Leu-248 form part of the interface between Abl and Gleevec? Explain your answer.

No. Thr-315 does, but Leu-248 does not. Thr-315 makes up part of the surface of Abl's binding pocket, so when Gleevec is in the binding pocket, Thr-315 contributes to the interface between the two partners. Leu-248 does not contribute to the surface of Abl's binding pocket. In the absence of Gleevec, and after Abl is phosphorylated, ATP fits into the binding pocket.

L9 Ch8 11: Many transcription regulators can form both homodimers and heterodimers. Do you expect a heterodimer to bind the same DNA sequence as a homodimer? Explain your answer

No. A heterodimer has two different transcription regulators, each with its own distinct DNA-binding domain-amino acid sequence. A homodimer has two of the same transcription regulators. A homodimer might bind a DNA sequence such as GTAC-N10-GTAC, while a heterodimer (with one protein the same as the homodimer) might bind GTAC-N10-TTGG, where 1/2 the binding sequence is different because the heterodimer has two distinct partners.

L7 Ch5 7: What are two implications for regions of different chromosomes being in close proximity to each other?

The close proximity of DNA from different chromosomes can contribute to rearrangement mutations such as reciprocal translocations. When a double strand break happens, the repair of the break can involve using a near-by unbroken piece of DNA as a template for repair. The repair process can result in recombination between the two DNA fibers (the broken DNA and the unbroken DNA). If the two DNAs are different chromosomes, then the result can be a reciprocal translocation. For example, if recombination is between Chromosome 4 and Chromosome 6, then the result will be part of Chromosome 4 attached to 6, and part of 6 attached to 4. A second implication is that accessible, looped out genes from the different chromosomes occupy a shared volume of space so that when the genes are being transcribed, the many proteins needed for transcription and RNA processing are concentrated within the shared volume of space. The accumulation/concentrating of those proteins increases the efficiency of transcription; proteins that need to bind and interact with each other, bump into each other more frequently because they are concentrated in that shared space. We will talk a bit more about transcription factories in the lecture on regulating gene expression. Here is a figure to help visualize the description above about looped out DNA sharing a volume of space in which transcription proteins accumulate.

L7 Ch5 13b: What proteins remain associated with the chromatin after step 1?

The histones of the nucleosome core particle remain associated with the chromatin.

R6 2: What proteins remain associated with the chromatin after step 1?

The histones of the nucleosome core particle remain associated with the chromatin. Histone H1 does not remain associated. The low salt treatment causes H1 to dissociate from the DNA.

L9 Ch8-1c: Imagine the two situations shown in Figure Q8-12. In cell I, a transient signal induces the synthesis of protein A, which is a transcriptional activator that turns on many genes including its own. In cell II, a transient signal induces the synthesis of protein R, which is a transcriptional repressor that turns off many genes including its own. In which, if either, of these situations will the descendants of the original cell "remember" that the progenitor cell had experienced the transient signal? Explain your reasoning.

The induction of a transcriptional activator protein that stimulates its own synthesis can create a positive feedback loop that can produce cell memory. The continued self-stimulated synthesis of activator A can in principle last for many cell generations, serving as a constant reminder of an event that took place in the past. By contrast, the induction of a transcriptional repressor that inhibits its own synthesis creates a negative feedback loop which ensures that the response to the transient stimulus will be similarly transient. Because repressor R shuts off its own synthesis, the cell will quickly return to the state that existed before the signal.

L7 Ch5 8: The length of the human genome, if all 23 DNA fibers were laid end to end, is ~ 1 meter. How do these long DNA fibers fit into a nucleus with a diameter ~ 5-8 μm?

The long DNA fibers fit into a nucleus by being packaged into condensed chromatin. The DNA is wrapped around a core of 8 histone proteins (two each of H4, H3, H2A and H2B) to form a nucleosome. About 147 base pairs are wrapped tight around the histones and about 50-80 base pairs, called linker DNA, are exposed between two nucleosomes. This first level of packaging is often referred to as 'beads on a string' because of how it looks in an electron microscope. The nucleosomes can slide, with the help of proteins, close together reducing the linker DNA or slide further apart to increase the linker DNA. Nucleosomes can be packed closest together when a protein called histone H1 binds to and curves the DNA exiting the core histone particle. The binding of histone H1 allows the nucleosomes to pack tight together into a 30-nm fiber, a second level of packaging. Thirty nanometer fibers can then be arranged in loops, a third level of packaging. Some loops are then packed close together, a fourth level of packaging, to form the most condensed state of chromatin, called heterochromatin. Euchromatin refers to chromatin that ranges from the 30-nm fibers that are looped out (not packed together) to chromatin that is 'beads on a string' with nucleosomes further apart so that much linker DNA is exposed.

L7 Ch5 13c: How does the length of the nuclease digestion impact the results?

The more time the nuclease has to cut naked DNA, the shorter and more uniform the cut fragments will be. [uniform meaning more similar in size; less smearing]

R6 3: How does the length of the nuclease digestion impact the results?

The more time the nuclease has to cut naked DNA, the shorter and more uniform the cut fragments will be. [uniform meaning more similar in size; less smearing]

L12 Ch17 5: Describe the mechanism of how the motor protein kinesin 'walks' on microtubules.

The motor protein Kinesin functions as a dimer. Each monomer has a motor head (catalytic core plus neck linker) that binds tubulin and an extended helix that wraps around the partner's helix to form a coiled-coil tail, which binds cargo. The catalytic core binds ATP and catalyzes hydrolysis of ATP. The sequence of a Kinesin walking is as follows. A motor head bound with ADP binds to tubulin, which causes a change in the motor head's shape. The change in shape causes ADP to fall out of the binding pocket. Very quickly an ATP then binds to the pocket and causes the neck linker to move and bind to the catalytic core. Movement of the linker swings the second motor head forward, which binds to tubulin. The ATP-bound motor head then hydrolyzes ATP and releases a phosphate. The ADP-form loosens the linker and allows the motor head to be swung forward to repeat the walking sequence.

L7 Ch5 4: What events are happening in a nucleolus?

The nucleolus is a volume of space in which many copies of ribosomal RNA are synthesized by transcription of many DNA templates, which are the many gene copies coding for ribosomal RNA. Many ribosomal proteins also accumulate and are concentrated in this volume of space. Ribosomal RNA and the proteins self-assemble into the small ribosomal subunit and the large ribosomal subunit. The subunits then diffuse out of the nucleus through the nuclear pores and into the cytosol, where the subunits then get involved in translation.

L12 Ch17 6: If a vesicle is moving towards the centrosome, then is the vesicle being moved by a Kinesin motor protein or a Dynein motor protein? Explain your answer

The vesicle is attached to a Dynein motor. Kinesin motors walk towards the plus-end of a microtubule, while Dynein motors walk towards the minus-end of a microtubule. The minus-end of a microtubule is attached to a ring of γ-tubulin, which is located on the surface of the centrosome. The centrosome is located near the nucleus, so cargo moving by Dynein motors also move towards the nucleus and towards the golgi body, which is kept near the nucleus by Dynein motors. Cargo moving by Kinesin motors is moved away from the nucleus and towards the plasma membrane. Kinesin motors constantly pull on parts of the endoplasmic reticulum to hold the reticulum outward from the nucleus. Kinesin motors also move vesicles from the trans-golgi towards the plasma membrane or endosomes.

L11 Ch 15 3: How do vesicles go to the correct membrane?

There is a gene family coding for Rab proteins and another family coding for tethering proteins. A family means multiple genes coding proteins that share much identity in amino acid sequence, but not complete identity. Some of the differences in amino acids are at binding sites so that each member of the family binds selectivity to different partners. Each membrane, ER, cis-golgi, trans-golgi, endosome and the plasma membrane, has specific Rabs associated with the membrane's cytosolic surface. When a vesicle forms, the Rab proteins are gathered onto the surface of the vesicle. Each type of Rab protein binds to a specific tethering protein. Each membrane has its own distinct tethering proteins, so that a vesicle with a specific Rab will only fuse with a membrane that has the matching tethering protein. After a vesicle fuses with the target membrane, the Rab proteins are recycled back to their original membrane. Rabs are monomeric G-proteins. When on a vesicle surface, a Rab is in the GTP bound state. After binding to a tethering protein at a target membrane, the Rab catalyzes the hydrolysis of GTP, changes conformation and is recycled back to its original membrane.

L9 Ch8-1a: What would happen to the regulation of the tryptophan operon in cells that express a mutant form of the tryptophan repressor that: 1) cannot bind to DNA? 2) cannot bind to tryptophan? 3) binds to DNA even in the absence of tryptophan?

Transcription of the tryptophan operon would no longer be regulated by the absence or presence of tryptophan; the enzymes would be permanently on in scenarios (1) and (2) and permanently off in scenario (3).

L7 Ch5 13a: Which treatment(s) are the controls? What do they tell you?

Treatment A is a negative control, which tells you that in the absence of the nuclease the DNA is stable; not cut into pieces. Treatment B is a positive control, which tells you that the nuclease is active and able to cut naked DNA.

R6 1: Which treatment(s) are the controls? What do they tell you?

Treatment A is a negative control, which tells you that in the absence of the nuclease the DNA is stable; not cut into pieces. Treatment B is a positive control, which tells you that the nuclease is active and able to cut naked DNA.

L10 Ch15 2b: Does directional import require an input of energy?

Yes, for continued directional import of proteins, an input of energy is needed. The NIR can move cargo into the nucleus without an input of energy but cannot move back out of the nucleus without an input of energy. Some cargo is moved directionally into the nucleus, however, transport of proteins (cargo) into the nucleus would stop as soon as all the NIRs had moved into the nucleus, unless there is an input of energy. The input of energy is Ran-GTP. Binding of Ran-GTP to the NIR causes the NIR to change conformation so that the cargo is released and the NIR with Ran-GTP bind to nuclear pore proteins in a way that moves the NIR and Ran-GTP out of the nucleus. The NIR returns to its original conformation after Ran catalyzes hydrolysis of GTP to GDP, which causes Ran to release the NIR. NIR can then bind another protein (cargo). Thus, GTP hydrolysis drives the work of conformation changes in the NIR that results in a continued directional import of proteins into the nucleus. Note that Ran catalyzes hydrolysis after binding to Ran-GAP in the cytosol. Ran-GDP has a conformation that allows binding to nuclear pores and transport into the nucleus. In the nucleus, Ran-GDP binds to Ran-GEF, which causes Ran to release GDP and bind to GTP.

L12 Ch17 4: For some cell behaviors, is it necessary that microtubules grow and shrink? Explain your answer.

Yes, for some cell behaviors, such as cell division, microtubules need to turnover, which means microtubules that assemble need to disassemble so that the subunits can be used to build new microtubules that grow in new directions from the centrosome. For example, to segregate chromosomes each chromosome has to be attached to several microtubules. Attachment requires microtubules to grow from the centrosome to a chromosome's kinetochore. Numerous microtubules growing from the centrosome do not grow directly to the kinetochore, so must be disassembled to provide subunits to increase the chance that a new filament will grow to a kinetochore. The dynamics of microtubules allows microtubules to search for kinetochores: microtubules failing to connect disassemble, while microtubules connecting become stable. To pull the chromosomes to opposite poles during anaphase also requires the kinetochore-microtubules disassemble.

L7 Ch5 13d: Is there experimental evidence confirming the presence of nucleosomes? Explain using the information in the gel to support your answer.

Yes. If there were no nucleosomes (DNA wrapped around proteins), then treatment D would look the same as treatment B; DNA of many different sizes because none of the DNA would be protected by proteins; the nuclease could cut anywhere along the DNA. However, treatment D shows a pattern of fragments that are multiplies of about 200 bp, which suggests that there is a regular spacing of proteins that are blocking and preventing nuclease from cutting within the DNA associated with the proteins. Treatment C shows that when the nuclease is given lots of time to cut what DNA is accessible (the DNA between the regularly spaced proteins), all the cut chromatin is of a uniform size containing DNA of a length that is about 140-150 bp; the DNA that is protected by proteins (the histones)