BIO 3153 Lecture Notes - Lecture 4: Atp Hydrolysis, Treadmilling, Thymosin

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15 Dec 2017

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Lecture 4 cytoskeleton i: filament proteins. 3 types: actin filaments, microtubules, intermediate filaments. Tubulin heterodimer, each binds gtp, hydrolyzes at 1 site, 13 protofilaments, plus and minus ends. Actin monomer, atp, plus and minus end: treadmilling. Nucleation drawback of treadmilling and destruction of filaments, lag time where oligomers of actin must form, then growth phase causing fewer subunits in monomer pool, occurs at minus end by y-tubulin protein complex. Growth (leading up to treadmilling: actin subunits are monomers with atp binding site orient with positive/negative ends, actin subunits polymerize to form filaments, atp hydrolysis, if actin atp high = subunits added to both ends, rapid growth. Gtp cap lost because hydrolysis occurs slower, tubulin concentration decreases within critical values, hydrolysis affects conformation: intermediate filaments. Tetramers packed into array of 16 dimers in cross-section, rope like appearance, formation by spontaneous interaction, disassembly regulated by phosphorylation.

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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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1. ___ Which of thefollowing statements about the cytoskeleton is false?

(a) The cytoskeletonis made up of three types of protein filament.

(b) The bacterialcytoskeleton is important for cell division and DNAsegregation.

(d) The cytoskeleton of acell can change in response to the environment.

2. ___ Which of thestatements below about intermediate filaments is false?

(a) They can stayintact in cells treated with concentrated salt solutions.

(b) They can befound in the cytoplasm and the nucleus.

(d) Each filament is about10 ?m in diameter.

3. ___ Keratins,neurofilaments, and vimentins are all categories of intermediatefilaments. Which of the following properties below is nottrue of these types of intermediate filaments?

(a) They strengthencells against mechanical stress.

(b) Dimers associateby noncovalent bonding to form a tetramer.

(d) Phosphorylation causesdisassembly during every mitotic cycle.

4. ___ Which of thefollowing statements about microtubules is true?

(a) Motor proteinsmove in a directional fashion along microtubules by using theinherent structural polarity of a protofilament.

(b) The centromerenucleates the microtubules of the mitotic spindle.

(d) ATP hydrolysis by atubulin heterodimer is important for controlling the growth of amicrotubule.

5. ___ Thehydrolysis of GTP to GDP carried out by tubulin molecules________________.

(a) provides theenergy needed for tubulin to polymerize

(b) occurs becausethe pool of free GDP has run out

(d) allows the behavior ofmicrotubules called dynamic instability

6. ___ The microtubulesin a cell form a structural framework that can have all thefollowing functions except which of the following?

(a) holding internalorganelles such as the Golgi apparatus in particular positions inthe cell

(b) creating longthin cytoplasmic extensions that protrude from one side of thecell

(d) moving materials fromone place to another inside a cell

7. ___ Which of thefollowing statements about organellar movement in the cell isfalse?

(a) Organellesundergo saltatory movement in the cell.

(b) Only themicrotubule cytoskeleton is involved in organellar movement.

(d) Organelles areattached to the tail domain of motor proteins.

8. ___ Which of thefollowing statements is correct?

Kinesins and dyneins____________________.

(a) have tails thatbind to the filaments

(b) move along bothmicrotubules and actin filaments

(d) derive their energyfrom GTP hydrolysis

9. ___ Cell movementinvolves the coordination of many events in the cell. Which of thefollowing phenomena are not required for cell motility?

(a) Myosin-mediatedcontraction at the rear of the moving cell.

(b) Integrinassociation with the extracellular environment.

(d) Release ofCa²⁺ from the sarcoplasmic reticulum.

10. ___Which of the followingstructures shorten during muscle contraction?

(a) myosinfilaments

(b) flagella

(d) actin filaments

7. ___ Referring to Figure Q16-29,which of the following statements below is false?

(a) A constitutivelyactive mutant form of PKA in skeletal muscle cells would lead to adecrease in the amount of unphosphorylated phosphorylasekinase.

(b) A constitutivelyactive mutant form of PKA in skeletal muscle cells would notincrease the affinity of adrenaline for the adrenergicreceptor.

(c) A constitutivelyactive mutant form of PKA in skeletal muscle cells would lead to anexcess in the amount of glucose available.

(d) A constitutivelyactive mutant form of PKA in skeletal muscle cells would lead to anexcess in the amount of glycogen available.

8. ___ Match the targetof the G protein with the appropriate signaling outcome.

adenylyl cyclase________ A. cleavageof inositol phospholipids

ion channels_________ B.increase in cAMP levels

phospholipase C_________ C. changes in membranepotential

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BIO 3153 Lecture Notes - Lecture 4: Atp Hydrolysis, Treadmilling, Thymosin

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