Which situation would be the most likely result of a mutation in a eukaryotic organism that causes it to produce less p21 protein than normal? Cells with radiation-induced DNA damage would continue through the cell cycle without repairing the damaged DNA. Cells with incomplete DNA replication would continue through the checkpoint. Cell replication would be slowed due to an increase in active G1/S cyclin-dependent kinase. Cells would not respond to growth factors. Cells would have a hypersensitive response to growth factors.
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- Imagine that there are mutations in the CDK genes such that their gene products are nonfunctional. What effect would this mutation have on an immature unspecialized blood cell precursor found in the bone marrow? The cell would not be able to reproduce itself. The cell would complete the cell cycle using cyclins in the absence of CDKS. The cell would be able to replicate its DNA but not translate DNA into RNA. The cell would be able to enter mitosis but not complete it. The cell would still phosphorylate the CDK-associated target proteins, and would do so more quickly.The process of cellular reproduction or divisions is a heavily regulated process for a number of different reasons. In the case of the eukaryotic cell there are a number of different checkpoints that a cell must pass before it can continue on. Please explain what those checkpoints are looking for and where those checkpoints are in the cell cycle. If a cell doesn’t pass any of these checkpoints what happens to those cells?During cell division cycle, cells need to monitor the process of DNA replication and segregation of replicated DNA so that these processes are error-free. Some potential errors that could occur include incomplete DNA replication, DNA damage in genome, and uneven separation of replicated genome. These mechanisms that cells used to monitor these processes are referred as the "cell cycle checkpoint, which can control specific Cdk activity to regulate the progression of cell cycle. For the following checkpoint mechanisms, indicate which Cdk activity is attenuated? Also indicate one of the key proteins or protein complexes involved in the following checkpoints. a) DNA damage checkpoint during S-phase b) Mitotic checkpoint during M-phase
- Which of the following would result in cell cycle arrest? O phosphorylation of Rb by G1-Cdk phosphorylation of p53 O phosphorylation of helicase by S-Cdk phosphorylation of M-Cdk by CAK phosphorylation of APC by M-Cdk"Agent V" is the name of an anticancer (chemotherapy) drug. This drug works against cancer cells by inhibiting the formation of microtubules in sensitive cells. Consider a cell that is sensitive to agent V (in other words, agent V is effective at stopping growth of this cell). Based on this information, agent V would cause the cell to be frozen at which of the major cell cycle checkpoints (G1, G2 or M checkpoint)? Explain"Agent V" is the name of an anticancer (chemotherapy) drug. This drug works against cancer cells by inhibiting the formation of microtubules in sensitive cells. Consider a cell that is sensitive to agent V (in other words, agent V is effective at stopping growth of this cell). Based on this information, agent V would cause the cell to be frozen at which of the major cell cycle checkpoints (G1, G2 or M checkpoint)? Explain your answer in 2- 4 sentences.
- The following figure shows how the activities (y-axis) of Cdk-Cyclin B complex and the APC complex change through the cell cycle. Active Cdk1-Cyclin Inactive Cdk1-Cyclin Active APC A B A B timeThe 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…
- Which of the following is false about cyclin-cdk complexes? OCdk's do not have to bind to cyclin proteins for complex to be active. OCdk/cyclin complexes phosphorylates proteins required to trigger next cell cycle phase. Process acts as molecular brakes to ensure cell is ready to continue with cell cycle. O Cyclin concentrations increase gradually, but cdk must be phosphorylated by specific kinase for complex to be activeWhich of the following changes to the microtubule binding protein, Tau, commonly occurs in Alzheimerâ s and other neurodegenerative diseases? A Tau is hypo-phosphorylated and does not bind to microtubules to destabilize them. B Tau levels are so low in the cell that they only bind to some microtubules. C Tau is hyper-phosphorylated and does not bind to microtubules to stabilize them. D Microtubules are hyper-phosphorylated and do not bind to Tau.In the following study, the investigators wanted to determine the role of cyclin B in controlling the cell cycle. Earlier researchers had found that extracts made from frog eggs (Xenopus) contained all the necessary proteins and machinery required for DNA replication. This included proteins that regulated the mitosis promoting factor (MPF). At the time of this study, cyclin B was show to affect MPF activity and the research group wanted to test using Xenopus egg extract in an assay. In Figure 1 (a) MPF activity was tested for its ability to phosphorylate Histone (H1) in sperm chromatin over a certain period of time. Additionally, the cyclin B concentration in the extract was measured. In figure 1b, the extract was tested after treatment with RNase which degraded only the mRNA and not RNA or FRNA in the extract. Knowing that cyclin B is a short-lived protein, why do you suppose the graph shows the results you see in figure 1b?