You are studying the M-cyclin. You treat mitotic cells with an inhibitor of the proteasome and find that M-cyclin is no longer degraded and that this prolongs mitosis. You also find that in the presence of the inhibitor, M-cyclin is now running slower/larger in a Western than you have previously observed. In 1-2 sentences, explain why this might be happening.
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- You are studying the M-cyclin. You treat mitotic cells with an inhibitor of the proteasome and find that M-cyclin is no longer degraded and that this prolongs mitosis. You also find that in the presence of the inhibitor, M-cyclin is now running slower/larger in a Western than you have previously doserved. In 1-2 sentences, explain why this might be happening.Write the word TRUE if the statement is correct and if false, replace the underlined words with the correct one. 1. Heterochromatin is typically gene-rich and composed of loosely-packed DNA that is transcriptionally active. 2. Molecular switches regulate the expression of genes by directing where and when genes should be turned on or off depending on the cell signal. 3. DNA Methylation and acetylation of histones are inverselycorrelated 4. Specialized cells produce specialized proteins derived from luxury genes which are constitutively expressed. 5. In mammals, methylation patterns that regulate DNA structure and gene expression become evident from the gastrula stage.You are studying Protein X which plays a role in promoting the G1/S phase transition in eukaryotic cells. You design an experiment using wild-type yeast cells to measure the amount of Gene X MRNA and activity levels of Protein X during the cell cycle. The results from your experiment are shown in the graph below. From the data, which of the following could Protein X be? Protein X activity levels Gene X MRNA levels Relative Units G1 G2 M cyclin inhibitor protein O cyclin dependent kinase O Helicase Cyclin
- You are studying cancer progression in mice. Your results show the following pathway in which two proteins play important roles in how mouse cell division is regulated. nutrients cell division Prot 1 Prot 2 You isolate a mutant strain of mice that have a Prot 1 mutation (Prot 1-). These mutant mice frequently develop cancer as young adults. You isolate a mutant strain of mice that have a Prot 2 mutation (Prot 2-). These mutant mice frequently develop cancer as young adults. Which of the following can you conclude from these observations? Select all that apply O Function of wild-type Prot 2 is to promote cell division when nutrients are present Function of wild-type Prot 2 is to promote cell division when nutrients are absent Function of wild-type Prot 1 is to inhibit cell'division when nutrients are absent Function of wild-type Prot 1 is to inhibit cell division when nutrients are presentThe retinoblastoma protein (RB) suppresses human cell division by arresting cells in the G₁ phase of the cell cycle and preventing progression to the next phase. It accomplishes this task by binding to another protein, E2F, a transcription factor needed for further progression through the cell cycle. Normal progression through the cell cycle is accomplished when cyclin-dependent kinases (CDKs) phosphorylate RB, preventing its binding to E2F. Many viruses can induce abnormal exit from G, using viral proteins that bind to RB at a motif at the N-terminal called LXCXE. An example is the E7 papilloma protein, which causes the excessive proliferation of cells in warts. The site at which LXCXE proteins bind is called the pocket domain and is highly conserved on RB and related proteins in plants and animals. The configuration of the pocket domain is well established. Mutant experimental RB proteins are available with alterations in the conserved amino acids of the pocket domain. A simple…The retinoblastoma protein (RB) suppresses human cell division by arresting cells in the G₁ phase of the cell cycle and preventing progression to the next phase. It accomplishes this task by binding to another protein, E2F, a transcription factor needed for further progression through the cell cycle. Normal progression through the cell cycle is accomplished when cyclin-dependent kinases (CDKs) phosphorylate RB, preventing its binding to E2F. Many viruses can induce abnormal exit from G, using viral proteins that bind to RB at a motif at the N-terminal called LXCXE. An example is the E7 papilloma protein, which causes the excessive proliferation of cells in warts. The site at which LXCXE proteins bind is called the pocket domain and is highly conserved on RB and related proteins in plants and animals. The configuration of the pocket domain is well established. Mutant experimental RB proteins are available with alterations in the conserved amino acids of the pocket domain. A simple…
- Mutants of cyclin B that are resistant to degradation by the cyclin B protease have been generated. How would the presence of these cyclin B mutants affect the events at the metaphase to anaphase transition?CYCLINs are regulatory proteins active for only portions of the cell cycle. Their transcription is activated at specific points during the cell cycle, for example, for CYCLIN1 (a CYCLIN B) of Arabidopsis during the G2 phase of the cell cycle. Since specific CYCLINs are required at different phases of the cell cycle, the stability of CYCLIN proteins is also highly regulated. This is accomplished by targeted degradation of CYCLIN proteins, also at specific times during the cell cycle, and which is mediated through a specific 9 amino acid sequence called the 'destruction box'. The CYCLIN1:GUS construct includes all the transcriptional elements required for CYCLIN1 regulation and a destruction box is translationally fused with the GUS coding sequence. How would the expression pattern you observed be altered if the destruction box was not included?To identify genes controlling the cell cycle in budding yeast, a genetic screen was carried out. In this screen, haploid yeast cells were exposed to a DNA damaging agent to introduce random mutations in the genome. By culturing cells at an elevated temperature (e.g. 37 degrees), where many mutated genes lose their function, scientists identified yeast mutants that showed growth defects and arrest at specific stages of the cell cycle (e.g. in mitosis with large buds). In this screen, mutants of the cyclin-dependent kinase were identified, but not mutants of cyclins. Explain the reason for this outcome.
- Concerning the Tools of Genetics Box Analysis ofCell-Cycle Mutants in Yeast:a. Describe how you would use replica plating ofmutagenized, haploid yeast cells to identifytemperature-sensitive (ts) mutations in essentialgenes needed for yeast growth and survival.b. Among the many ts mutations you found in part(a), how would you distinguish mutations in genesneeded for cell-cycle progression from those ingenes needed for other aspects of the life of yeasts?c. If you had a large collection of yeast cell-cyclemutants, how would you determine which of themutations are in the same gene and which are indifferent genes?You have identified five genes in S. cerevisiae that are induced when the yeast are grown in a high-salt (NaCl) medium. To study the potential roles of these genes in acclimation to the growth in high-salt conditions, you wish to examine the phenotypes of loss- and gain-of-function alleles of each. How will you do this?In an experimental motile fibroblast cell line, Nucleation Promoting Factor (NPF) was mutated in such a way that it no longer could bind to the G-actin monomers. Which of the following would you observe in these cells as a consequence of the mutation at the molecular level? None of the other is true. The cell will not be able to form branching of the actins using Arp2/3 complex at the leading edge and causing movement arrest The filamin protein will not be recruited for crosslinking the actins and will cause the arrest of movement. The Arp2/3 and filamin will not be able to dissociate from actin in the absence of NPF and the cells will be restricted to move in one direction only