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You have discovered a single cellular eukaryotic organism that has very little vesicle trafficking. You suspect that this may be due to the relatively few formed microtubules that are observed in this species. Upon close examination, you find the microtubules that are present are unusually stable. When looking at an amino acid comparison with human alpha and beta tubulin, you notice that the alpha tubulin is identical, but the beta tubulin is different in the region responsible for GTP hydrolysis. You isolated the heterodimer subunits from this new species and put them in a cell free system in a GTP bound form of the heterodimer along with centrosome nucleation sites. How would the dynamics be different from normal heterodimer subunits in the same situation?

a) The new species’ microtubules would treadmill at a maximum concentration of heterodimers

b)The new species’ microtubules would be less susceptible to catastrophe

c)The new species’ microtubules would loose more from the (-) end than (+) end

d)The new species’ microtubules would grow to a longer length than normal cellular microtubules

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Patrina Schowalter
Patrina SchowalterLv2
28 Sep 2019

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Related questions

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 β-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

7 questions could use some help on for Molecular Biology. Pleasehelp. Just answers will do (I'll figure out), but wouldn't mindexplanations. If answered correctly 5 star guarantee & goodattempt definite 'best answer'. Thanks!

Question1

Which of the following cytoskeletal components is thelargest?

  1. Microtubules

  2. Microfilaments (actin)

  3. Intermediate filaments

Question2

During mitosis, all microtubules emanate from the centrosome.Therefore, the centrosome is an example of:

  1. a microtubule organizing center (MTOC)

  2. A +TIP protein

  3. A catastrophin

  4. A monomer sequestering protein

Question3

Tubulin is a G-protein. It can stabilize microtubules when boundto which of the following?

  1. GDP

  2. GTP

  3. Either GTP or GDP

  4. Neither GTP nor GDP

Question4

In the formation of microtubules, dimers are built when analpha-tubulin protein binds to a beta-tubulin protein. This is anexample of what level of protein structure?

  1. Primary structure

  2. Secondary structure

  3. Tertiary structure

  4. Quartenary structure

Question5

Different tissues in the body express slightly differentversions of alpha tubulin. Which of the following could lead toslightly different alpha-tubulin molecules with a modified proteinsequence?

  1. One tissue has more DNA methylation upstream of thealpha-tubulin gene.

  2. One tissue has more miRNAs produced than the other.

  3. One tissue uses different splice sites than the other.

  4. One tissue has different trans-elements than the other.

Question6

  1. How is confocal microscopy different from traditionalfluorescent microscopy?

  2. The molecules are labelled red instead of green.

  3. The molecules are observed over time.

  4. The molecules are observed in many thin planes, which are thenstacked together to create a crisper image.

  5. The confocal microscope has higher objective power than regularfluorescent microscopes.

Question7

Taxol is a chemotherapy drug that permanently stabilizesmicrotubules and prevents their degradation. Describe howmicrotubules are typically destroyed and why permanentstabilization would kill dividing cells. Note: you may need toreview the basics of cell mitosis from a previous course.

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