BIO130H1 Lecture Notes - Lecture 18: Myosin, Dynein, Hydrolysis

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23 Mar 2017

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They are made of monomers but they are polar, which means that the ends are different. Polar bonds = one side is more positive and one side is more negative. One end is n terminus while the other end is c terminus. The n"s and the c"s are the plus end. 13 parallel proto filaments to form a hollow cylinder. Myosin ii, plus end is directed to compare the different functions: ensure to compare their different motor proteins. 0 = all the actin is monomers. 100 = no monomers all in the actin form the addition of the salt to make a buffer, the addition of the salt changes the conformation from actin monomers to filaments. Actin length is growing, addition at the plus end is faster. Eventually an equilibrium state will be achieved where the addition will be equaled with the loss.

Related Questions

1 Stathmin and Profilin are similar because ____.

a) Stathmin and Profilin are necessary to form branched actin filaments

b) Stathmin is necessary for nucleation of microtubules as Profilin is necessary for forming straight unbranched actin filaments

c) Stathmin is used to sequester free tubulin subunits as Profilin is used to sequester free actin monomers

d) Stathmin and Profilin are motor proteins

2 Which of the following statements regarding dynamic instability is FALSE?

a) Dynamic instability is required for both actin and microtubule function.

b) The GTP cap helps protect a growing microtubule form depolymerization.

c) GTP hydrolysis by the tubulin dimer promotes microtubule shrinking.

d) Cap proteins help stabilize the microtubules, promoting microtubule growth.

3. Which of the following statements about vesicular membrane fusion is TRUE?

a) Membrane fusion does not always immediately follow vesicle docking.

b) The interactions of the v-SNAREs and t-SNAREs pull the vesicle membrane and the target organelle together so that their lipids can intermix.

c) NSF and SNAP disassemble the SNARE complex so the SNAREs can be recycled.

d) All of the above are true.

Activity: Dynamic Instability of Microtubules

Part A

What causes catastrophe of the microtubule in vitro?

the lack of tubulin heterodimers

non-motor Microtubule Associated Proteins (MAPs)

GTP hydrolysis

mutation of the Î²-tubulin

Part B

What is the role of GTP in microtubule polymerization?

GTP binds the alpha and beta tubulin subunits together to form the tubulin monomer.

GTP hydrolysis provides the energy for the polymerization of the microtubule.

GTP is a second messenger that signals the need for polymerization/de-polymerization.

GTP stabilizes the tip of the microtubule, allowing more monomers to be added.

Part C

What would happen in the treadmilling experiment if a non-hydrolyzable analogue of GTP were used?

The monomers would be unable to add to the plus end, and the microtubules would shrink until they disappeared.

The microtubule would treadmill until the new tubulin, with non-hydrolyzable GTP, reached the minus end, and then it would only extend at the plus end.

The microtubule would add monomers at both the plus and minus ends, growing in both directions.

The non-hydrolyzable GTP would stabilize both ends, causing treadmilling to stop.

Part D

What is the difference between the plus and minus ends of the microtubule in in vitro experiments?

The beta subunit of the tubulin is exposed on the minus end.

Polymerization occurs at the plus end.

The plus end has a lower critical concentration for tubulin heterodimers.

Catastrophe occurs at the minus end.

Part E

How would the drug taxol affect the in vitro dynamic instability and treadmilling experiments?

Taxol would block catastrophe at the plus end in the dynamic instability experiment but not depolymerization at the minus end in the treadmilling experiment.

Taxol would block the addition of tubulin in both experiments, leading to a destruction of the microtubules.

Taxol would stabilize the microtubules in both experiments, leading to polymerization without catastrophe.

Taxol would block depolymerization at the minus end in the treadmilling experiment but not catastrophe at the plus end in the dynamic instability experiment.

1. Kinesin-1 motors can move long distances on microtubule tracks without dissociating. By contrast, myosin II motors in skeletal muscle take only one or a few steps before letting go. What is most likely the difference between these two motors that allow two different types of movement?

A. Differences in length of the stalk of each respective motor allow for differences in step size

B. Differences in ATP versus GTP hydrolysis alter the motor head movement along filaments

C. Actin filaments are smaller in diameter and have less space for movement

D. Binding affinity of the Kinesin-1 motor head to microtubules is high, whereas binding affinity of the myosin II motor head to actin filaments is low

E. Microtubules are larger in diameter and have more space for movement

2. Dynamic instability causes microtubules either to grow or to shrink rapidly. What is the most likely reason for a microtubule to undergo rapid growth?

A. GTP hydrolysis exceeds the rate of heterodimer addition

B. GTP hydrolysis of alpha tubulin causes an outward bend in the heterodimer conformation

C. A low concentration of tubulin heterodimers are in solution

D. Heterodimer addition exceeds the rate of GTP hydrolysis

E. Increased trafficking of microtubule motor proteins favors stability

3. There are no known motor proteins that move on intermediate filaments (IFs). What property of intermediate filaments would most likely explain the lack of motor proteins?

A. Tissue specific expression of IFs prevent a single type of motor protein to evolve over time

B. Lateral bundling of IFs prevent motor movement

C. IFs are less dynamic than microtubules and actin filaments

D. IFs are unpolarized

E. IFs do not bind a nucleotide

4. You discover a mutation that results in less rapid hydrolysis of GTP by the beta-tubulin subunit. What would most likely occur?

A. There would be no change in filament assembly

B. There would be enhanced instability of the microtubule

C. There would be enhanced stability of the microtubule

D. Microtubules would not form

E. There would be enhanced treadmilling

5. Which of the following statements describes common features of cadherins and integrins?

A. They can link to microtubules using specific adapter proteins

B. They mediate heterophillic interactions

C. They achieve high binding affinity through clustering

D. They are responsible for interactions of a cell with the extracellular matrix

E. They are found in the apical plasma membrane of an epithelial cell

6. Intermediate filaments are uniquely non-polar and built by lateral bundling in five steps: 1) synthesis of an alpha-helical monomer, 2) formation of a coiled-coil dimer, 3) formation of a staggered tetramer of two coiled-coil dimers, 4) lateral association of 8 tetramers, and 5) addition of 8 tetramers to a growing filament. At what step is polarity lost?

A. Step 1

B. Step 2

C. Step 3

D. Step 4

E. Step 5

7. Collagenase is an enzyme that breaks peptide bonds in collagen to help destroy extracellular structures. What would be the most likely reason to use collagenase in a cell biology laboratory?

A. Disrupt cadherin interaction with the extracellular matrix

B. Disrupt cell-cell contacts to passage cells in cell culture

C. Dissociate epithelial cells from the underlying basement membrane to separate tissues

D. All of the answers are likely uses for collagenase

E. Lyse cells to expose intracellular proteins and organelles

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BIO130H1 Lecture Notes - Lecture 18: Myosin, Dynein, Hydrolysis

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