BIO 320 Lecture Notes - Lecture 26: Profilin, Thymosin, Gelsolin

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28 Jan 2017

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Many (>60) other proteins bind to and affect the dynamic properties, organization and functions of actin filaments in a living cell. In most non-muscle cells, the concentration of unpolymerized actin is very high (~50-200 mm), well above the cc (~0. 2 mm) for actin assembly in vitro. However, much of the unpolymerized actin molecules in a cell are bound to the protein thymosin (thymosin-b4), which prevents such actin molecules from assembling onto either end of actin filaments. Profilin competes with thymosin for binding to g-actin. In the actin-profilin complex, the end of the actin monomer that normally binds to the filament minus end is blocked. Thus, this actin monomer can only be added to filament plus ends. When profilin binds adp-actin, it opens the cleft and greatly enchances the release of adp and subsequent binding of atp. Thus, it also helps to regenerate g-actin that is suitable for assembly. The function of profilin can be regulated (e. g. , by phosphorylation).

Related Questions

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

yellowherring775

Question 1. Increased peripheral edema could be caused by:

a. blocked lymph vessels

b. decreased capillary blood pressure

c. increased concentration of albumin in the blood

d. slowed heart rate

Question 2. The pulse pressure of a healthy student is most likely to be:

a. 5 mm Hg

b. 30 mm Hg

c. 95 mm Hg

d. 120 mm Hg

Question 3. beta-blockers inhibit the activity of beta-adrenergic receptors and change the function of the heart by:

a. increasing heart rate and increasing contractility

b. increasing heart rate and decreasing contractility

c. decreasing heart rate and increasing contractility

d. decreasing heart rate and decreasing contractility

Question 4. In cardiac pacemaker cells, the L-type calcium channels:

a. open during diastole

b. close when the membrane potential rises above threshold

c. are located in the cell membrane

d. pump Ca2+ ions into the sarcoplasmic reticulum

Question 5. Dr. Campbell used an elliptical trainer for 40 minutes at the gym. Although the workload remained constant, Dr. Campbellâs heart-rate rose to 150 beats per minute during the first 5 minutes, stayed at that level for the next 25 minutes and then started to rise. It was 180 beats per minute at the end of the exercise period.

What was the time interval between QRS complexes when Dr. Campbellâs heart rate was 180 beats per minute?

a. 3.3 ms

b. 33 ms

c. 333 ms

d. 3333 ms

Question 6. In the question 5 above, Dr. Campbellâs heart rate is most likely to have risen during the last 10 minutes of exercise because his:

a. lungs were becoming congested

b. atrial cells were fatigued

c. extracellular Ca2+ concentration had increased

d. hematocrit increased due to sweating

Question 7. A Physiology student is sitting still in a cold igloo. After 30 minutes, she stands up, walks slowly around the igloo, and suddenly starts to shiver.

The student preserved their core temperature while they were sitting by:

a. depressing their Frank-Starling mechanism

b. decreasing the extracellular Ca2+ concentration

c. relaxing smooth muscles in their peripheral circulation

d. increasing the resistance of arterioles in their arms and legs

Question 8. In the question 7 above, the student started to shiver soon after walking because:

a. their atria started to fibrillate

b. their ventricles started to fibrillate

c. blood moved to the periphery, became cold, and chilled the core when it returned to the chest

d. the ATP concentration in their skeletal muscles had dropped to zero

Please use the following scenario for answering the questions 9 and 10.

A newly discovered drug binds to myosin molecules in cardiac muscle (and nowhere else) and increases the probability that myosin molecules in ventricular cells will bind to actin filaments when the intracellular Ca2+ concentration rises. The drug is being tested as a potential therapy for patients with heart failure.

Question 9. This drug might help people who have low cardiac output by:

a. increasing peripheral resistance

b. increasing the force developed by ventricular cells

c. increasing the isovolumic contraction time

d. increasing heart rate

Question 10. The increased binding of cardiac myosin molecules to actin might prevent the drug from being a useful treatment by:

a. slowing ventricular relaxation

b. decreasing the duration of systole

c. decreasing resistance of capillaries

d. slowing heart rate

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BIO 320 Lecture Notes - Lecture 26: Profilin, Thymosin, Gelsolin

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