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BIOL 2070H Lecture Notes - Lecture 3: Glut4, Antiporter, Resting Potential

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pinksalmon977
24 Aug 2016
School
Trent University
Department
Biology
Course
BIOL 2070H
Professor
William Dew

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Document Summary

Phospholipid bilayer forms a semi-permeable barrier to the passage of many molecules. Many materials don"t need to cross the barrier to perform their function. Movement of these is through the use of channel proteins ad transport proteins. Concentration may be different on the inside and outside of the cell. The differences are important for the operation of enzymes and the generation and use of energy. Permeability of the lipid bilayer is determined by molecule size, polarity, and charge. Small hydrophobic molecules and small uncharged polar molecules generally make it though. Larger molecules and ions usually do not make it through without help. Transport proteins require a molecule to bind in a binding site on the transporter. Conformational change moves the molecule from one side of the bilayer to the other and is then released. Channels will let the molecule through a small place from one side to the other.

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Related Questions

True or False? True False
1. Proteins rapidly pass artificial membrane (liposome).--------------------------------------------------------------------------------------( ) ( )
2. Amphiphatic detergent such as SDS spontaneously forms bilayer inwater. ------------------------------------------------------------( ) ( )
3. Cholesterol does have a hydrophilic head and hydrophobic tail.-------------------------------------------------------------------------( ) ( )
4. Fat molecules are one of the major components of the plasmamembrane.------------------------------------------------------------- ( ) ()
5. Phospholipid molecules such as phosphatidylcholine spontaneouslyform bilayer. ----------------------------------------------------( ) ( )
6. Flippase facilitates lateral movement of phospholipids withinthe plane of the membrane.----------------------------------------- ( ) ( )
7. Glycolipids make bilayer less permeable to small water solublemolecules.----------------------------------------------------------- ( ) ()
8. Membrane inner leaflet is enriched with phosphatidylcholine,Sphingomyelin, and Glycolipid. -----------------------------------( ) ( )
9. Plasma membrane proteins can have an enzyme activity.--------------------------------------------------------------------------------( ) ( )
10. A single pass membrane protein must have an amphiphatic alphahelix.-------------------------------------------------------------- ( )( )
11. Plasma membrane proteins can be linked to a lipid molecule atthe outer leaflet of the plasma membrane. --------------------- ( )( )
12. The lateral motility of membrane proteins can be restricted bytight junctions.------------------------------------------------------ ( ) ()
13. Channel proteins have specific binding sites for ions and thusslower than carrier proteins.--------------------------------------- ( ) ( )
14. An antiport would function as a symport if its orientation inthe membrane were reversed. ---------------------------------------( ) ( )
15. A transmembrane pore can be formed by multiple amphipathicalpha helices. ----------------------------------------- ( ) ()
16. Porin proteins form water-filled channels in the outer membraneof a bacterium and are mostly composed of beta sheets. ---- ( ) ()
17. Glucose uniporters localize at the apical side of theintestinal epithelium.------------------------------------------------------------ ( ) ()
18. GAP junction is a selective pore for anions.-----------------------------------------------------------------------------------------------( ) ( )
19. A voltage-gated Na+ channel adopts only two conformationstates: open and closed.----------------------------------------------- ( ) ( )
20. Voltage gated K+ channel opens faster than the Voltage gatedNa+ channels.-------------------------------------------------------- ( ) ()
21. Opening of chloride channel at the synapse will help to fire anaction potential.------------------------------------------------------ ( ) ()
22. The nuclear pore complex forms a gate through which moleculespass the nucleus uni-directionally. -------------------- ( ) ()
23. A completely folded protein is imported into a mitochondrionthrough a large non-selective pore.--------------------------------- ( ) ( )
24. ER signal sequence contains multiple lysine and arginineresidues.----------------------------------------------------------------------( ) ( )
25. The plasma membrane is a sorting station of inward endocyticpathway.---------------------------------------------------------------- () ( )
26. The trans Golgi functions as a sorting station of outwardsecretory pathway.----------------------------------------------------------- ( ) ()
27. Vesicles budding from ER are coated with Clathrin.----------------------------------------------------------------------------------------( ) ( )
28. RabGTPases and tethering proteins are the major specificitydeterminant of vesicle transport system. --------------------------( ) ( )
29. v-SNARE proteins are one of the tethering proteins.------------------------------------------------------------------------------------( ) ( )
30. In secretory cells, the constitutive and regulatory pathways ofexocytosis diverge at the cis Golgi network. ------------------ ( )( )
31. Many proteins are posttranslationally glycosylated at tyrosineresidues in the ER.------------------------------------------------- ( ) ( )
32. Cells import LDL molecules exclusively through pinocytosis.-------------------------------------------------------------------------( ) ( )
33. Cells recycle LDL receptors from lysosome to the plasmamembrane.----------------------------------------------------------------- () ( )
34. Phagosomes fuse with late endosome.------------------------------------------------------------------------------------------------------( ) ( )
35. Small vesicles from ER near the plasma membrane can fuse withthe plasma membrane upon bypassing Golgi apparatus. -- ( ) ()
36. If a Sec mutant showed an accumulation of newly synthesizedproteins at ER,
then the Sec gene is likely a component of vesicle-buddingmachinery of trans-Golgi network.------------------------------------ ( ) ( )
37. A signal recognition particle is a single pass trans-membraneprotein of the ER membrane.---------------------------------------- ( ) ( )
38. Typical animals cells display higher levels of intracellularsodium than extracellular compartments.------------------------------ ( ) ( )
39. Typical animals cells display higher levels of intracellularchloride than extracellular compartments.----------------------------- ( ) ( )
40. The Na-K Pump is a GTPase that can adopt active and inactiveconformational states.----------------------------------------------- ( ) ( )
1. Phospholipase
C-­-ß
produces
phosphatidylinositol
3,4,5-­-trisphosphate
and
diacylglycerol.
(
)
(
)
2. PDGF
activates
GPCR.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
3. Rho
is
an
example
of
heterotrimeric
G
proteins.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­-
(
)
(
)
4. All
members
of
the
G
protein
coupled
receptors
interact
with
small
GTPase
Ras.
-­--­--­--­--­--­--­--­--­--­-
(
)
(
)
5. Adrenaline
is
an
example
of
the
ligands
for
the
RPTK
family.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
6. Ras
GEF
activates
RPTK.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
7. PH
domain
containing
proteins
bind
to
activated
EGFR.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
8. Ligands
for
GPCR
proteins
induce
dimerization
of
the
receptor.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
9. Grb2
is
an
adaptor
protein
that
binds
to
a
phospho-­-tyrosine
residue
of
activated
RPTK
through
a
PH
domain.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
10. When
a
cell
is
expressing
a
large
number
of
dominant
negative
EGFR
proteins,
an
introduction
of
a
constitutively
active
Ras
mutant
will
allow
EGF
signaling.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
11. PI3K
generates
phosphatidylinositol
3,4,5-­-trisphosphate
from
IP3.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
12. Calcium
inhibits
PKC.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
13. cAMP
activates
PKA.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
14. The
MAP
kinase
ERK
inhibits
MAP
kinase
kinases.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­-
(
)
(
)
15. cAMP
is
an
example
of
a
first
messenger.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­----­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
16. All
7-­-Transmembrane
receptors
are
G
Protein
Coupled
Receptors.
-­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
17. Both
Ga
and
Gß
subunits
of
the
heterotrimeric
G
protein
can
mediate
GPCR
signaling.
-­--­--­-
(
)
(
)
18. Both
Ga
and
Gß
subunits
of
the
heterotrimeric
G
protein
have
GTPase
activity.
-­--­--­--­--­--­--­--­--­--­--­--­-
(
)
(
)
19. The
Gas
subunit
of
the
heterotrimeric
G
protein
suppresses
the
function
of
Adenylyl
cylase.
(
)
(

1. The plasma membrane of liver cells contains a glucose uniportthat transports glucose by facilitated diffusion. When the[glucose] is higher inside the liver cell than the [glucose]outside the liver cell, the glucose uniport will

A. couple the transport of glucose out of the cell to thehydrolysis of ATP

B. couple the transport of glucose into the cell to thehydrolysis of ATP

C. switch to a closed conformation, allowing no transport ofglucose

D. transport glucose out of the cell by passive transport

E. transport glucose in of the cell by passive transport

2. Unlike the positive charged ion shown in the middle diagramof figure 11-4B, K+ is higher in concentration inside the cell. Howwill the membrane potential of a cell at rest (shown in both themiddle and right-hand diagrams of Figure 11-4B) affect theelectrochemical gradient for K+?

A. The membrane potential will increase the electrochemicalgradient

B. The membrane potential will decrease the electrochemicalgradient

C. The membrane potential will have no effect on theelectrochemical gradient

3. Which of the following proteins carries out activetransport?

A. gluclose symport

B. Bacteriorhodopsin

C. ABC transporter

D. NA K pump

E. All of the above

4. Aquaporin will not transport Na+ ions because

A. Two asparagine residues in the channel prevent the passage ofNa+ ions by binding the oxygen of each water molecule as it passesby

B. Na+ ions are repelled by the positive charges at the entranceto the channel

C. Na+ ions cannot make contact with the carbonyl (C=O) groupson all four sides of the channel

D. The passage of Na+ ions is energetically unfavorable becauseone side of the channel is hydrophobic in nature

5. The acetylcholine receptor is different from other channelproteins we have discussed, such as the K+ leak channel,because

A. The acetylcholine channel is composed of a single polypeptiderather than multiple subunits

B. The acetylcholine channel transports Na+ ions against theconcentration gradient for sodium

C. The acetylcholine channel has a much smaller diameter thanother channel proteins

D. The acetylcholine channel transports ions with theirassociated shell of water molecules

E. The acetylcholine channel is always open, and never switchesto a closed conformation

6. After opening briefly along the axon of a neuron,voltage-gated ion channels switch to an inactivated conformationthat slowly returns to the original closed conformation of theprotein. This slow change in conformation, requiring about amillisecond, is important because

A. all of the ions that were transported across the membranewhen the ion channels were open must be actively transported backto the other side of the membrane, restoring the originalconcentration gradient, before the ion channels can open again

B. ATP hydrolysis is necessary to drive this energeticallyunfavorable change in conformation

C. it prevents the action potential from reversing direction andtraveling back toward the cell body of the neuron

D. the channel protein must be dephosphorylated in order tocomplete the transport process and return to the originalnonphosphorylated conformation

Question 1


Which of the following provides the best description of adifference between prokaryotic and eukaryotic cells?.l.14.3>

The cell membrane of prokaryotic cells contains proteins, whilethe cell membrane of eukaryotic cells contains only lipids andcarbohydrates.

The DNA in eukaryotic cells is enclosed in the nucleus, whilethe DNA in prokaryotic cells is not separated from the rest of thecell’s contents.

Eukaryotic cells have cell walls that consist mainly of sugars,while prokaryotic cells do not.

Prokaryotic cells contain chloroplasts and mitochondria, whileeukaryotic cells do not contain specialized organelles.

Question 2

Cellular respiration is a process involving the breakdown ofglucose molecules in order to generate energy for cellularactivities. Given the general reaction below, which of thefollowing best describes the products labeled x andy?

C6H12O6 + O2 → _x___+ __y___ + energy.l.18.8>

The products are CO (carbon monoxide) andH2O2 (hydrogen peroxide), which thendecompose into carbon dioxide (CO2) and water(H2O).

The products are H2O (water) and CO2 (carbondioxide), which then undergo anaerobic respiration.

The products are CO2 (carbon dioxide) andH2O (water), which then become the reactants forphotosynthesis.

The products are O2 (oxygen) and H2O(water), which are then used for growth and to keep the organismhydrated.

Question 3

If a researcher added a quantity of phospholipid molecules to abeaker of water, what would be most likely to happen to theorientation of the phospholipid molecules?.l.14.2>

They would break apart, and the remaining lipid tails wouldfloat up and form a layer on the surface of the water.

Their lipid tails would align within a spherical double layerwith their phosphate heads in the water inside and outside thesphere.

They would form a single-layered sphere with the phosphate headspointing outward and the lipid tails pointing inward.

Their lipid tails would adhere to the glass of the beaker andtheir phosphate heads would float in the water.

Question 4

While looking at a cell under a microscope, a scientist is ableto see a biological molecule. This molecule is a nucleic acid witha ladder structure and contains the nitrogenous bases adenine,thymine, cytosine, and guanine. Based on this information, whatconclusion can the scientist draw regarding this nucleicacid?.l.18.1>

The nucleic acid is RNA and would be found in the cell’snucleus.

The nucleic acid is RNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cytoplasm.

The nucleic acid is DNA and would be found in the cell’snucleus.

Question 5


Which of the following is needed for cellular respiration tooccur?.l.18.8>

Sodium

Oxygen

Water

Nitrogen

Question 6


What would be the most likely effect if the side chain of an aminoacid group was unable to vary its structure?.l.18.4>

The organism would not be able to make and store starches.

The organism would not be able to process glucose moleculeseffectively.

The organism would not be able to regulate the creation ofspecialized cells.

The organism would not be able to make more than one type ofprotein.

Question 7


During photosynthesis, unused oxygen atoms form oxygen gas(O2) which is released as a waste product. How did theseoxygen atoms originally enter the plant?.l.18.7>

They were part of the ATP molecules gathered by thechloroplasts.

They were part of the sugar molecules(C6H12O6) the plant brought in forfood.

They were part of the carbon dioxide (CO2) moleculesthat entered through the leaves.

They were part of the water molecules (H2O) pulled inthrough the roots.

Question 8


Which of the following organelles houses DNA in eukaryoticcells?.l.14.3>

Endoplasmic reticulum

Golgi apparatus

Nucleus

Mitochondrium

Question 9


M. J. Schleiden shifted the scientific community's attention tocellular processes with his.l.14.1>

attention to disproving spontaneous generation

publication of ideas on cell generation

theories on the division of cells

claim that cells were the basic unit of all plant matter

Question 10


While the average human is able to hold his or her breath forapproximately one minute, a whale can dive for over 30 minuteswithout returning to the surface. Which of the following correctlydescribes this difference?.l.18.10>

Whales need less energy than humans.

Whales gather energy from their environment better thanhumans.

Whales are able to tolerate the products of cellular respirationbetter than humans.

Whales have a slower cellular respiration metabolism rate thanhumans.

Question 11


What can you conclude about the advantage of the cell membranehaving hydrophobic ends in the phospholipid bilayer?.l.18.3>

The cell requires the ability to digest and eliminateenzymes.

The cell requires a barrier between itself and itsenvironment.

The cell requires a means to store starches and glucose.

The cell requires a structure for eliminating waste.

Question 12


What would be the most likely outcome if the ribosomes of a cellwere damaged and stopped functioning?.l.14.3>

The cell would lose structure and shape.

Cell growth and repair would slow or stop.

Cell waste would no longer be processed.

The cell would produce more mitochondria.