BIO211 Practice Questions Winter 2022 PDF

BIO211 – Practice Questions Winter 2022 Short answer (10 points): Every short answer question on the exam is going to be testing your knowledge on a specific molecular process or pathway that was presented in one of the preceding lectures and how it relates to cell behavior and development. The qu...

BIO211 – Practice Questions Winter 2022 Short answer (10 points): Every short answer question on the exam is going to be testing your knowledge on a specific molecular process or pathway that was presented in one of the preceding lectures and how it relates to cell behavior and development. The question will usually ask you to explain how a specific perturbation (i.e. chemical inhibition or protein mutation) will affect the mechanism in question and ultimately cell dynamics and embryonic development. Read the question carefully to see if you can identify precisely the molecular mechanism that the question is asking about. Explaining the mechanism by presenting the ordered steps in the process, along with the molecular details (e.g. protein names, phosphorylation, ubiquitylation, activation, and/or degradation) will typically be worth six points. The mechanism can be presented using words and/or a diagram. Two points will be awarded if you correctly predict how cell behavior and embryonic development are affected. The final two points will be given if you can explain why cell behavior or embryonic development will be affected. 1. The drug blebbistatin inhibits myosin II activity by preventing the hydrolysis of ATP by the myosin II heavy chain. Explain how the myosin II conformational cycle will be affected by the addition of blebbistatin (6 points). Predict what will happen if blebbistatin is added to an 8-cell xenopus embryo as it enters anaphase (2 points) and explain why (2 points). 2. Imagine that a DNA replication error at the first cell division in a newly fertilized oocyte (in other words an embryo) introduces a point mutation into the talin protein that prevents it from unfolding in response to force. Describe how this mutation will affect the formation of integrin-actin complexes (focal adhesions) (6 points). Predict how embryonic cell migration will be affected (2 points) and explain why (2 points). 3. Ciliobrevin D inhibits the dynein minus end microtubule motor. Explain how dynein functions to maintain the mitotic spindle (6 points). What will happen to the mitotic spindle positioning and chromosome segregation if this drug is added during prophase (2 points)? Provide the rationale leading to your prediction (2 points). 4. What will happen to the cell cycle if M-cyclin is no longer able to be ubiquitylated (6 points)? What would the consequences be for an 8-cell xenopus embryo just pass the G2/M checkpoint (2 points) and why (2 points)? Answers on the next page… Answers: 1. The drug blebbistatin inhibits myosin II activity by preventing the hydrolysis of ATP by the myosin II heavy chain. Explain how the myosin II conformational cycle will be affected by the addition of blebbistatin (6 points). Predict what will happen if blebbistatin is added to an 8-cell xenopus embryo as it enters anaphase (2 points) and explain why (2 points). ATP hydrolysis is a critical step in the myosin II cycle leading two a conformational change in the lever arm of the myosin II which swings the head forward to grab the next actin monomer within the actin filament. Release of the inorganic phosphate is what triggers the power stroke, loss of the ADP and tight-binding to the actin filament. The tight-binding is relieved when the myosin II head binds to a new molecule of ATP. Thus, upon addition of blebbistatin, ATP will no longer be hydrolyzed and the myosin II head will remain unbound to the actin filament (6 points). The 8-cell xenopus embryo will be able segregate its chromosomes via the mitotic spindle, but will unable to go through cytokinesis because myosin II will not be able to generate the force necessary to pinch the two daughter cells apart using the contractile ring (2 points). The end result will be each of the 8 cells in the embryo will contain two nuclei (2 points). 2. Imagine that a DNA replication error at the first cell division in a newly fertilized oocyte (in other words an embryo) introduces a point mutation into the talin protein that prevents it from unfolding in response to force. Describe how this mutation will affect the formation of integrin-actin complexes (focal adhesions) (6 points). Predict how embryonic cell migration will be affected (2 points) and explain why (2 points). The talin protein is essential for connecting the activated integrin receptors that are engaged by extracellular matrix to actin filaments within the cytoplasm. Once integrins become activated talin and kindlin are recruited to the cytoplasmic tails of the integrin receptors. Talin than binds to the vinculin adapter protein which then connects the complex to pre-existing actin filaments. When force is applied, the talin protein is pulled apart to reveal additional vinculin binding sites. In this way, the application of force to the complex leads to a positive feedback loop and the engagement of additional vinculin adapters and recruitment of additional F-actin filaments. If talin is no longer able to change its conformation in response to force, then this positive feedback loop is unable to take place (6 points). The focal adhesion complexes will remain small and not able to generate the force necessary to retract the trailing edge of migrating cells or allow the leading edge to protrude using the molecular clutch mechanism (2 points). Embryonic cell migration will be completely stopped as a result (2 points). 3. Ciliobrevin D inhibits the dynein minus end microtubule motor. Explain how dynein functions to maintain the mitotic spindle (6 points). What will happen to the mitotic spindle positioning and chromosome segregation if this drug is added during prophase (2 points)? Provide the rationale leading to your prediction (2 points). During metaphase, dynein connects astral microtubules to the actin cytoskeleton underneath the plasma membrane. The motor domains of dynein are bound the plus ends of the astral microtubules, while the cargo binding domain is connected to the actin cortex underneath the plasma membrane. When active, the dynein motor will attempt to walk towards the minus end of the microtubule, towards the centrosome. Because the cargo binding domain is held in place by the actin cortex, the end result is the dynein motor will pull the centrosome and the rest of the mitotic spindle towards the edge of the cell. This force is a balanced by the same mechanism occurring on the other side of the spindle and the net result is the mitotic spindle is positioned towards the middle of the cell (6 points). The addition of ciliobrevin D will cause the astral microtubules to no longer be pulled apart to opposite ends of the cell and the mitotic spindle will no longer be positioned in the mid-point of the cell (2 points). This means when the chromosomes are pulled apart in anaphase, they may not be equally distributed between the daughter cells (2 points). 4. What will happen to the cell cycle if M-cyclin is no longer able to be ubiquitylated (6 points)? What would the consequences be for an 8-cell xenopus embryo just pass the G2/M checkpoint (2 points) and why (2 points)? Cdc20 will still be able to get activated by the proper positioning of mad2 and attachment of the kinetochores to the kinetochore microtubules. Cdc20 will bind and activate APC, but APC will no longer be able to catalyze the addition of poly-ubquitin chains to cyclin M in cooperation with the E1 and E2 ligases. Thus, cyclin M will not be degraded by the proteasome and its expression will persist in the cytoplasm, and its associated cyclin- dependent kinase will remain active (6 points). Overall the embryo will remain stuck at the metaphase- anaphase transition and unable to segregate the daughter chromosomes and complete the cell cycle (2 points) since it’s the loss of M cyclin that is one of the critical molecular switches that allows the cell cycle to move through this checkpoint (2 points).