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The Dynamic Cell > Molecular Machines - Block 4 > Flashcards

Molecular Machines - Block 4 Flashcards

1
Q

What is meant by an enzyme being a “molecular jig”?

A

> It does not need energy

> It aligns things so they are easier to work on

> It protects the components so they are processed easily

> It can be re-used

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2
Q

What is PGK for?

A

It is used in the first ATP-producing step in glycolysis ( converting 1,3-BPG//1,3-PGA to 3PG//3PGA) which is a dephosphorylation.

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3
Q

How may the PGK reaction work?

A

1,3-PGA and ADP join onto different binding sites, and once the phosphate is transferred, then one substrate leaves followed by the other (which one leaves is random).

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4
Q

How does PGK align its components?

A

The in-line mechanism refers to the fact that there are in-line attacks in the phosphorylation reaction that results in an inversion of the O atom configuration. The enzyme aligns the components precisely.

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5
Q

How can we rule out that the PGK mechanism has a phosphoryl-enzyme intermediate?

A
  1. If we saturate the binding site with ADP and add varying concentrations of 1,3-PGA, the enzyme acts as if it were a typical single-site enzyme, and vice versa
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6
Q

How can we rule out that the PGK mechanism has a phosphoryl-enzyme intermediate?

A
  1. If we saturate the binding site with ADP and add varying concentrations of 1,3-PGA, the enzyme acts as if it were a typical single-site enzyme, and vice versa.
  2. If there were a single site, there would have to be a singular binding site, but a defined order in which reactants bind, and both experiments show hyperbolic kinetics at saturating concentrations of the other substrate.

> These observations are inconsistent with the phosphoryl-enzyme mechanism, but consistent with the direct transfer hypothesis (although it does not prove it).

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7
Q

What evidence is there that the leaving PO4 group is attacked directly in line with the bond that is broken in the PGK mechanism?

A

There is an inversion of the configuration of the oxygen groups around the hydrolysed PO4 group.

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8
Q

How does the PGK enzyme stabilise the transition state?

A

> Because of the in-line attack and PO4 group configuration, a pentavalent intermediate forms at the leaving phosphate; this is a transition state.

> The enzyme stabilises this transition state with positively charged groups on its surface.

> A Lys (+) and an Arg (+) residue are within H-bonding distance of the transferring PO4 group.

> The +ve charge of a bound Mg ion also stabilises the transition state

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9
Q

How does the PGK enzyme protect the substrate from hydrolysis? Why is this important?

A

> PGK promotes PO4 transfer to ATP over PO4 transfer to water by meams of a conformational change

> The enzyme closes only when BOTH substrates are present, so the enzyme does not catalyse attack of water, and catalyses the attack of the phosphate on 1,3-PGA instead.

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10
Q

Which step of the PGK enzyme reaction involves a release of energy?

A

> > The release of the products «

The negative substrates bound to the enzyme only do so because of the correct alignment of compensating positive charge.

If the domains flex open at all, even for a short time during which the -ve charges become uncompensates, the products repel each other and rapidly diffuse away.

ΔG comes from the solution of the products in water, entropy increase and the more stable resonance forma available to the free molecules.

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11
Q

Why doesn’t 1,3-PGA get hydrolysed before it finds PGK?

A

Glycolytic enzymes are often present in concentrations greater or equal to that of their substrate, so the intermediates do not freely diffuse in solution; there are no free metabolites in solution.

Glycolytic enzymes are often present in high concentrations, so they may form complexes.

(Enzymes may also form supercomplexes, passing the substrates to each other)

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12
Q

What accompanies the release of the products from PGK?

A

> A conformational change

> ΔG, from the release of products

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13
Q

Why does the release of the products from PGK produce ΔG?

A

> The solution of the products in water

> Increse in entropy

> More stable resonance forms of products

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14
Q

What does Michaelis-Menten kinetics mostly describe?

A

The saturation of the enzyme with substrate

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15
Q

What is Km, in terms of enzymes?

A

The concentration at which the enzyme is at HALF of its maximum velocity (Vmax).

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16
Q

What is Vmax, in terms of enzymes?

A

The maximum velocity of an enzyme.

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17
Q

What us Km usually measured in?

A

μM

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18
Q

How do metabolites find their enzymes if they are present at very low concentrations?

A

It occurs randomly!

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19
Q

What is are the kinds of “track” upon which a motor protein moves?

A
  1. Microtubules

2. Actin filaments

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20
Q

How is directionality determined with microtubule transport?

A

A microtubule is polar, with its minus end ending with an α-tubulin and its plus end ending with a β-tubulin.

The plus ends lead to the cell periphery, and minus ends lead to the organising centre.

Cargo can be carried in either direction.

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21
Q

What is the motor protein that works on microtubules?

A

Tubulin

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22
Q

What is the motor protein that works on actin filaments?

A

Myosin

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23
Q

What are the three functional similarities between Tubulin and Myosin?

A

> They both hydrolyse ATP.

> They have a conf. change coupled to hydrolysis.

> They have a conf. chagne coupled to the polymer

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24
Q

What is the processivity of kinesins vs muscle myosin?

A

Kinesins are always bound by o ne of the two heads almost all of the time

Myosins only bind briefly during the muscle contraction “power stroke” - spends very little time bound and is very explosive

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25
Q

What are some examples of linear motors and rotary motors?

A

Linear motor proteins: kinesins and muscle myosin

Rotary motor protein: F1 ATP Synthase

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26
Q

What can be observed when you sonicate mitochondria?

A

The membrane lipids form vesicles, with “lollipop” structures on the outside. The hexagonal heads of the lollipops are the F1 complex.

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27
Q

What do the F1 “heads” do when they are isolated from the F0 membrane pore?

A

They hydrolise ATP to ADP.

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28
Q

What happens with vesicles containing F0 but not F1?

A

They do not make ATP.

29
Q

What happend when the F1 heads are added back with F0?

A

ATP synthesis is restored (upon addidion of ADP + Pi).

30
Q

What does SDS-PAGE of the F1 particle show?

A

The F1 particle has five subunits (α, β, γ, δ, ε)

31
Q

Describe an experimental proof that shows both ATP synthase and the H+ gradient are necessary.

A

> Make an artificial lipid vesicle with (a) bacteriorhodopsin, a light-driven proton pump, and (b) a mitochondrial ATPase (ATP synthase).

> When the vesicle is illuminated, ATP is produced, and when drugs that permeablise the membrane & destroy the H+ gradient are added, ATP is not produced.

> This proves that the H+ gradient is necessary.

32
Q

Which parts of the complex form the “stator” in the ATP synthase?

A

the a subunit, along with the b and δ subunits, form the stationary part, and the other components rotate relative to it.

33
Q

Which components form the part of the complex that rotates?

A

The α + β head rotates, along with the γ subunit that is connected to the centre of the head, and the ring of c subunits that are imbedded in the membrane.

34
Q

What part of ATP synthase is referred to as F0?

A

The parts in the membrane.

35
Q

What part of ATP synthase is referred to as F1?

A

The “head” that is outside of the membrane.

36
Q

Why do the three β subunits of the F1 head all have different conformations?

A

There are three β subunits: O, L and T (open, loose and tight).

T - Bound to ATP

L - Bound to ADP + Pi

O - Empty, ready to bind new substrate

37
Q

How does the γ subunit affect the three β subunits of the F1 head?

A

The long γ subunit is asymmetric and has a protrusion.

When the γ subunit rotates, it interacts with the β subunits differently, making them go through and adopt all the different conformations (either L, T or O) in turn.

38
Q

What causes the γ subunit to rotate and therefore change the conformation of the three β subunits?

A

The flow of H+ ions.

39
Q

What causes the γ subunit to rotate and therefore change the conformation of the three β subunits?

A

The flow of H+ ions.

40
Q

What happens when the L (loose) β subunit becomes a T (tight) β subunit?

A

> The L subunit is bound to ADP + Pi.

> Once it changes to a T subunit, ATP can be produced.

41
Q

Describe an experiment that could prove that there is rotation of the γ subunit.

A

> α + β subunits were genetically engineered to have hexahistidine tags, so the head could be affixed to a slide coated with Ni.

> An actin filament was attached to the γ subunit, and an actin-binding fluorescent drug was added.

> The filament is seen to move somewhat randomly, but overall in one direction. This is clockwise when viewed from the head towards the membrane.

42
Q

Describe an experiment that could prove that there is rotation of the γ subunit.

A

Short Answer: The α + β head is affixed to a slide, and an actin filament is attached to the γ subunit. The addition of an actin-binding fluorescent drug shows that the filament rotates in mostly one direction.

Long Answer:

> α + β subunits were genetically engineered to have hexahistidine tags, so the head could be affixed to a slide coated with Ni.

> An actin filament was attached to the γ subunit, and an actin-binding fluorescent drug was added.

> The filament is seen to move somewhat randomly, but overall in one direction. This is clockwise when viewed from the head towards the membrane.

43
Q

What is the structure of the (F0) a-subunit?

A

> It is curved to fit the c-subunits

> It has a Cytosolic half-channel…

> …as well as a Matrix half-channel

44
Q

What is the structure of the (F0) c-subunit?

A

> Two hydrophobic helices bend back on each other via hair-pin loops.

> All a.a. side-chains are hydrophobic save for an aspatic acid residue halfway down one helix.

45
Q

Why can the ATPase rotor not turn in any direction without protonation?

A

> The aspartate residue is protonated when in contact with non-polar fatty acid chains

> Protonation//deprotonation can only occur when in contact with the half-channel

> Without addition of more protons, the rotor is stalled because the c-subunits are negatively charged; the opposing energy of negative charge vs non-polar fatty acids is too high

46
Q

What causes the ATPase rotor to turn when it previously could not?

A

> A proton comes down the upper (cytosolic) channel and protonates the aspartate carboxylate group of a c-subunit.

> The ensemble now moves left (anticlockwise) because the two c-subunits to the left part of the a-subunit are protonated

> The new c-subunit moving in from the right to contact the a-subunit is protonated. This means that the aspartate residue can deprotonate and H+ leaves from the lower (matrix) half-channel.

47
Q

In summary, what happens to a proton that enters the top (cytosolic) half-channel of ATPase?

A

> They bind to negative Asp61 on a c-subunit (forming aspartic acid, the protonated form),

> They rotate the whole way round still bound to that subunit,

> When they come to the bottom (matrix) half-channel, aspartic acid deprotonates to become aspartate again and the proton leaves.

48
Q

Why do protons go through the channel in the direction that they do?

A

It is thermodynamically favourable; there is a lower concentration of proteins on the inner (matrix) side of the membrane than the outer (cytosolic) side.

49
Q

What could theoretically make H+ flow in the opposite direction through ATPase, and what would the result of this be?

A

> ATPase could work in the opposite direction if the H+ concentration was higher at the bottom of the complex,

> Then the rotor could run in the opposite direction,

> and ATP could be hydrolysed instead of synthesised.

> A high concentration of ATP would also be required to do this, and could even pump H+ against their concentration gradient.

> Such a mechanism could acidify a membrane component.

50
Q

Where is the F1Fo-ATP synthase complex located in a cell?

A

The inner membrane of the mitochondrion (as well as the thylakoid membrane in a chloroplast).

51
Q

How can the rotation of the γ subunit be described?

A

Its rotation is eccentric.

52
Q

What is the energy-requiring (and rate-limiting) step of ATP synthesis?

A

The release of synthesised ATP; formation of ATP in the enzyme does not require much energy, but releasing it does.

53
Q

Why is there greater flow of water into a cell than out, and what problem does this cause?

A

Proteins within the cell have charged surfaces, attracting counter-ions, so the water concentration inside the cell is lower than outside the cell. Osmosis means that water flows in along the concentration gradient, and so the cell needs to take counter measures or else it could burst.

54
Q

Why does a cell need to pump Na+ ions out, and what problem does this cause?

A

This is in order to acheive osmotic balance, since the extracellular concentration of Na+ becomes larger than the intracellular concentration.

The problem is that this leaves the inside of the cell is -ve with respect to the outside, so it gets harder to pump Na+ out.

55
Q

How is membrane polarisation due to Na+ being pumped out solved?

A

K+ ions are taken into the cell as counter-ions to balance this.

56
Q

What does the Na+//K+ pump do in terms of ion exchange?

A

> Pumps out 3 Na+ ions

> Takes in 2 K+ ions

57
Q

How is K+ buildup in the cell avoided?

A

K+ ions leak out via K+ ion channels, so internal [K+] reaches a steady state.

58
Q

Briefly describe the architecture of a channel, in terms of (1) motif and (2) repeats of these motifs.

A
  1. A single channel has a SIX-helix motif
  2. This six-helix motif has FOUR repeats

> The loop between helices 5 and 6 determines which ion is selected

59
Q

How does the K+ channel prevent smaller ions like Na+ from slipping through? (L.O.)

A

> At the narrow opening of the pore (leading to outside the cell) the selectivity loop can be found

> The carbonyl groups of the selectivity loop are in the exact correct orientation to bind unhydrated K+ ions

> As Na+ ions are smaller, the distance between the carbonyl groups and the Na+ ion are too large to mimic the water molucles in hydrated Na+

> Hydrated Na+ ions cannot pass through as they are too large.

60
Q

Why can Na+ not pass through the K+ channel, in thermodynamic terms?

A

The free energy needed to dehydrate Na+ is too large, as the dehydrated state is not sufficiently supported by the protein.

The free energy needed to dehydrate K+ is smaller because the protein compensates perfectly for the stripped-off water molecules.

61
Q

K+ channel gotta go fast; why??

A
  1. Mutual Repulsion
    K+ ions hop between the top and bottom binding sites in the selectivity pore, until another ion binds to the bottom site. Mutual repulsion means that the ion in the top site leaves.
  2. Potassium Resevoir (thermodynamics)
    The reason that the ions keep going out rather than in is becasue there are far more K+ inside than out, so it is quite likely that one will bind to the bottom site & force the next one out.
62
Q

What is the general role of MFS transporters?

A

> Transport small molecules

> Always in consort with a Na+ or H+ ion

63
Q

What are the two kinds of MFS transporters?

A

> Symporters - Transport the ion and molecule in the SAME direction.

> Antiporters - Transport the ion and molecule in OPPOSITE directions.

64
Q

What is the MFS?

A

The Major Facilitator Family (of transporters).

65
Q

What does the Maltose Symporter do?

A

> Accumulates maltose.

> Uses high extracellular H+ concentration to do so.

66
Q

What does Lactose Permease (LacY gene) do?

A

> Imports lactose (transcribed in the absence of glucose).

> Uses H+ ions to do so.

67
Q

What other gene is transcribed when LacY is transcribed?

A

> LacZ.

> This gene is for the β-galactosidase enzyme which catalyses Lactose&raquo_space;> Galactose + Glucose.

> This is only transcribed when LacY is transcribed to activate Lactose Permease.

68
Q

Describe the 6 steps in the Lactose Permease mechanism.

A
  1. Protonation at Glu (extracellular side)
  2. Sugar binds at the (extracellular side)
  3. Preinversion - H+ released
  4. Inversion - H+ binds to lower Glu, open configuration stabilised
  5. Sugar leaves
  6. H+ leaves
69
Q

What would make Lactose Permease run in the opposite direction?

A

An alkaline pH

The Dynamic Cell (8 decks)

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