Practice Problems 10

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1 Life Sciences 1a Practice Problems Progesterone is a steroid hormone that prepares the body for pregnancy. Progesterone exerts its function by binding to the progesterone receptor in the cell. This binding event in turn leads to the transcriptional activation of various genes. Progesterone production is stimulated by Luteinizing ormone (L), whose production is stimulated by gonadotropin releasing hormone (GnR). This pathway can be illustrated as shown below: GnR L Progesterone a) igh levels of progesterone suppress the release of GnR, and in turn lead to a reduction in progesterone levels. What term is used to describe this type of feedback control? b) If this were an example of the opposite type of feedback control, what effect would high levels of progesterone have upon it s own production? c) The structures of progesterone and cholesterol are shown below. Do you think progesterone requires a special transporter to pass through the plasma membrane? Explain. Progesterone Cholesterol Prior to progesterone binding, Progesterone Receptor (PR) is predominantly found in the cytoplasm. owever after progesterone binds, the hormone/receptor complex moves into the nucleus, binds to DA, recruits transcription factors, and activates transcription. d) If Progesterone Receptor is unable to move into the nucleus, could transcriptional activation occur? Explain?

2 e) The lacz gene can be used as a reporter to study transcriptional activation, because its protein product (β-galactosidase) can be easily measured by a simple enzyme assay. The wildtype and a mutant PR were studied using the DA construct shown below. Progesterone Response Element (PRE) refers to the DA sequence that is bound by the activated PR. PRE lacz The following results were obtained, where the asterisk indicates when progesterone was (or was not) added to the cells. Please explain what has happened to the mutant PR and how the data supports your interpretation.

3 2. In class you learned how Gleevec can inactivate Bcr-Abl. owever it has also been discovered that Gleevec can also inhibit the activity of other tyrosine kinases. ne example is c-kit. c-kit is a receptor tyrosine kinase that is normally only active when it is in the ligand bound state. owever in many cancers a deletion in c-kit has occurred that causes the receptor to be constitutively activated. a) Gleevec inhibits CML by binding to and stabilizing one of three forms of Bcr-Abl, the closedinactive form. ow does this stabilization reduce Bcr-Abl activity with respect to LeChatelier s principle? b) Gleevec inactivation of mutated c-kit differs from the inactivation of Bcr-Abl. Gleevec binds in the adenine-binding pocket between the large and small lobes of the kinase domain. ow would that binding affect its function? Below is a diagram of Gleevec and the key amino acids (numbered) of c-kit with which it forms intermolecular interactions S c) Please identify the four amino acids (640, 670, 673, and 810), and write what type of intermolecular interactions form between them and Gleevec.

4 d) ver time, cancerous cells evolve to become resistant to Gleevec. A common Gleevec resistant mutant occurs where residue 670 mutates into an Isoleucine residue. Please explain two ways that this mutation could reduce Gleevec binding affinity. 3. Bacterial cells can take up the amino acid tryptophan (Trp) from their surroundings, or if there is an insufficient external supply, they can synthesize tryptophan from other small molecules. The Trp repressor is a bacterial gene regulatory protein that shuts off the transcription of genes that code for the enzymes required for the synthesis of tryptophan. Trp repressor binds to a site in the promoter of these genes only when molecules of tryptophan are bound to it (see figure below). a) Why is this a useful property of the Trp repressor? b) What would happen to the regulation of the tryptophan biosynthesis enzymes in cells that express a mutant form of Trp repressor that (1) cannot bind to DA or (2) binds to DA even if no tryptophan is bound to it? c) What would happen in scenarios (1) and (2) if the cells, in addition, produced normal Trp repressor protein from a second, normal gene?

5 4. In the absence of glucose, E. coli can proliferate using the pentose sugar arabinose. The ability of E. coli to utilize the sugar arabinose is regulated via the arabinose operon, depicted in the figure below. The araa, arab, and arad genes encode enzymes for the metabolism of arabinose. The arac gene encodes a gene regulatory protein that binds adjacent to the promoter of the arabinose operon. To understand the regulatory properties of the AraC protein, you engineer a mutant bacterium in which the arac gene has been deleted and look at the effect of the presence or absence of the AraC protein on the AraA enzyme. a) If the AraC protein works as a gene repressor, would you expect araa RA levels to be high or low in the presence of arabinose in the arac mutant cells? What about in the arac mutant cells in the absence of arabinose? Explain your answer. b) Your findings from the experiment are summarized in the following table: Genotype araa RA Levels in the absence of arabinose in the presence of arabinose arac + (normal cells) low high arac (mutant cells) low low Do the results indicate that the AraC protein regulates arabinose metabolism by acting as a gene repressor or a gene activator? Explain your answer. 5. In a series of experiments, genes that code for mutant forms of a receptor tyrosine kinase are introduced into cells. The cells also express their own normal form of the receptor from their normal gene, although the mutants are constructed so that they are expressed at considerably higher concentration than the normal gene. What would be the consequences of introducing a mutant gene that codes for a receptor tyrosine kinase (A) lacking its extracellular domain, or (B) lacking its intracellular domain?

6 6. When activated by the signal, the platelet-derived growth factor (PDGF) receptor phosphorylates itself on multiple tyrosines (as indicated below by the circled Ps; the numbers next to these Ps indicate the amino acid number of the tyrosine). These phosphorylated tyrosines serve as docking sites for proteins that interact with the activated PDGF-receptor. These proteins are indicated in the figure below, and include the proteins A, B, C, and D. The PDGF-receptor is activated when it binds to the signal, PDGF. ne of the cellular responses to PDGF is an increase in DA synthesis, which can be measured by incorporation of radioactive thymidine into the DA. To determine whether protein A, B, C, and/or D are responsible for activation of DA synthesis, you construct mutant versions of the PDGF-receptor that retain one or more tyrosine phosphorylation sites. You express these mutant versions in cells that do not make a PDGF-receptor. In these cells, the various versions of the PDGF-receptor are expressed normally and in response to PDGF binding, become phosphorylated on whichever tyrosines remain. You measure the level of DA synthesis in cells that express the various mutant receptors and obtain the data shown below. a) From these data, which, if any, of these proteins A, B, C, and D are involved in the stimulation of DA synthesis by PDGF? Why? b) Which, if any, of these proteins inhibit DA synthesis? Why? c) Which, if any, of these proteins appear to play no detectable role in DA synthesis? Why?

7 7. You have identified a novel gene (X) in the Drosophila Receptor Tyrosine Kinase (RTK) signal pathway responsible for normal wing development. When the signal pathway is defective, adhesion between cell layers in the wing is disrupted such that they pull apart swelling with air - a phenomenon known as blistering. This pathway is readily identified in adult animals. To identify where in the signaling pathway this newly discovered gene acts, you engineer flies that lack the functional X (X-) and cross them to other fly stocks that contain a number of mutant phenotypes in the RTK pathway. You are searching for mutations that when crossed to your X- line restore wild type wing phenotype. Such rescuing of the X- mutant phenotype identifies components that act downstream of X in the RTK pathway. Analysis of the following set of double mutants yields the following results. Double mutant genotype Wing development phenotype X- ; MAPKK C normal X- ; Ras-P normal X- ; EphR -terminal- blistering X- ; EphR activated normal X- ; Ras lipid - blistering Explanation of genotypes: MAPKK C is a constitutively activated MAP kinase kinase that phosphorylates targets in the absence of any signal; Ras-P is a constitutively phosphorylated form of Ras; EphR -terminal- is an Eph receptor isoform that lacks a ligand binding domain; EphR activated is an Eph receptor isoform that does not require ligand binding for activation; Ras lipid is a mutant Ras that does not get prenylated (lipid group does not get attached) a) Where in the signaling pathway does X act? What is the most likely role of X? b) Does X exert its influence by acting through the MAP kinase pathway? Explain your reasoning.

8 1. Progesterone a) negative feedback high levels of progesterone should inhibit additional synthesis, production, or release of additional progesterone b) positive feedback progesterone is stimulating the production of additional progesterone c) progesterone should be able to move through the membrane without a transporter as it is mainly a hydrophobic molecule similar to the membrane component, cholesterol. d) no. If the receptor cannot enter the nucleus, it would be unable to bind to the genomic DA and activate transcription. e) The mutant PR is constitutively active (always on) independent of progesterone binding. ne of the ways transcription is controlled is by keeping transcriptional activators away from the DA. owever if a transcriptional activator is now always by the gene s they activate, such as if the progesterone receptor is always in the nucleus, it may be able to activate transcription all the time. [ote: other answers may be possible and would be given credit if properly explained.] 2. c-kit. a) Bcr-Abl can exist in three states that are in equilibrium with one another, closedinactive, open-inactive, and open-active. Gleevec is able to bind and stabilize the closed-inactive form of Bcr-Abl and increase its concentration in a cell. Since the total amount of Bcr-Abl in the cell is relatively constant, a consequence is that the amount of active Bcr-Abl has to decrease. b) c-kit is a kinase, and kinases require ATP (adenosine triphosphate) to add phosphates onto other molecules. If the adenine-binding pocket is occupied by Gleevec, then ATP is unable to bind to c-kit, and c-kit is unable to phosphorylate other molecules. Additionally kinases must orient ATP and their targets appropriately to catalyze phosphate transfer. This is accomplished through the motion of the large and small lobes of the kinase. If the space between the lobes is occupied by Gleevec, then the motion is restricted, which results in loss of kinase activity c) 640 Glutamic Acid, Glu, E 670 Threonine, Thr, T 673 Cysteine, Cys, C 810 Aspartic Acid, Asp, D all hydrogen bonding

9 d) 1. steric hindrance mutation of Threonine into Isoleucine results in a larger residue being present at that location. The larger size of the sidechain could prevent Gleevec from fitting into its binding site. 2. loss of hydrogen bonding the mutation replaces an amino acid with a polar sidechain for one with a hydrophobic sidechain. This would result in a loss of hydrogen bonding, and a corresponding increase in enthalpy. 3. Alberts Question 8-8. Alberts 8-2, 8-4, 8-9, 8-10 also reinforce important concepts. 4. Ara a) If the AraC protein acted as a gene repressor for the arabinose operon, araa RA levels should be high in the presence or absence of arabinose when there is no AraC protein around. In fact, the araa RA levels should be high all the time, regardless of the presence or absence of arabinose, since the AraA gene should be transcribed under all conditions in the absence of AraC. b) The results are consistent with AraC acting as a gene activator for the arabinose operon. A gene activator must bind to the promoter regions of the arabinose genes in order to stimulate their transcription. Thus, if the gene for the regulatory protein is deleted, the arabinose genes cannot be turned on. 5. Alberts Question Alberts 16-2, 16-3, 16-5, 16-9, 16-10, 16-12, 16-14, 16-22, also reinforce important concepts. 6. PDGF a) Proteins A and D are involved in stimulating DA synthesis. PDGF-receptors that can bind to only A or D (see experiments 2 and 5) can stimulate DA synthesis to about 50% of normal amounts (which is represented by experiment 1). A and D are both needed and are used in an additive fashion; this is evident from experiment 6: when a PDGF-receptor can bind both A and D, DA synthesis levels are close to that obtained with the normal receptor. b) Protein B is an inhibitor of DA synthesis. Consequently, receptors with binding sites for B and D (see experiment 7) stimulate a lower DA synthesis rate than do receptors that only bind D (experiment 5). c) Protein C plays no detectable role in DA synthesis. Receptors that can only bind C (experiment 4) activate DA synthesis about as much as receptors that don t bind any proteins (experiment 9; our negative control). Furthermore, the binding of protein C does not affect the response mediated by protein D when the receptor can bind both C and D (experiment 8). 7. RTK and X a) Gene X acts upstream of MAPKK (because constitutively activated MAPKK rescues the mutant phenotype) and upstream of Ras (because constitutively activated Ras rescues the mutant phenotype). This places the function of gene X as a direct interaction with the receptor. Given that an activated receptor

10 (dimerization of which normally requires ligand binding) rescues the mutant phenotype, but the ligand binding defective mutant does not rescue the mutant phenotype, one could conclude that that the protein product of gene X is likely the ligand for the Eph receptor. b) The data are consistent (although not conclusive) with a role for MAP kinase in this pathway because constitutively activated MAPKK restores a normal phenotype and MAPKK targets MAPK for activation.

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