Quiz 6 (L19, 20, 21(beginning))

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20 Terms

1
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Define zymogen

Zymogen: inactive form (“unfinished” form) of an enzyme to protect cells from self-digestion/self-degradation

i.e if enzyme was released in active form it would destory the body

2
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Describe how a zymogen is activated

  • Activation of Trypsinogen & Chymotrpsinogen


Zymogen is activated via Proteolytic cleavage

  • Trypsinogen (inactive) → Trypsin (active) through enteropeptidase

  • Chymotrypsinogen → Chymotrypsin through trypsin


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1) Describe where most zymogens are synthesized

2) Where and how they are activated

3) Why they cannot be synthesized in their active form.

1) Most zymogens are synthesized in the pancreas

2) Zymogens are activated by proteolytic cleavage to become its active form from unfinished form at the duodenum.

3) If synthesized in active form, cells will self-digest/self-degradation


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Describe specific details of what happens within the chymotrypsin structure when the Arg15-Ile16 peptide bond is cleaved by trypsin in the duodenum, and how this impacts the chymotrypsin active site. (AKA Details of how Chymotrpsin is activated)

1) The Arg15-Ile16 peptide bond is cleaved by trypsin in the duodenum.

2) Ile16 forms a new amino terminus, where the positively charged amino group (NH3+) forms a new salt bridge/ion pair with Aspartate 194 (Asp) by moving 17 Angstroms from its original position in the zymogen. Trypsin catalyzes the hydrolysis of key peptide bond to form π-chymotrypsin (fully active), which then catalyzes more hydrolysis to form ɑ-chymotrypsin.

3) Aspartate194 turns 170 degrees to form the ion pair, neighboring backbone around Gly193 changes position & correctly arranges the backbone N–H groups of Gly193 and Ser195 to form the oxyanion hole. The amino acid 188-192 position becomes fixed in place to turn into one side of the substrate binding pocket, which accommodates bulky hydrophobic side chains.

<p>1) <span style="background-color: transparent;">The <strong>Arg15-Ile16 peptide bond </strong>is cleaved by trypsin in the duodenum.</span></p><p><span style="background-color: transparent;">2) Ile16 forms a new amino terminus, where the positively charged amino group (NH3+) forms a new salt bridge/ion pair with Aspartate 194 (Asp) by moving 17 Angstroms from its original position in the zymogen. Trypsin catalyzes the hydrolysis of key peptide bond to form π-chymotrypsin (fully active), which then catalyzes more hydrolysis to form ɑ-chymotrypsin.</span></p><p>3) <span style="background-color: transparent;">Aspartate194 turns 170 degrees to form the ion pair, neighboring backbone around Gly193 changes position &amp; correctly arranges the backbone N–H groups of Gly193 and Ser195 to form the oxyanion hole. The amino acid 188-192 position&nbsp;becomes fixed in place to turn into one side of the substrate binding pocket,  which accommodates bulky hydrophobic side chains.</span></p>
5
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In the activation of chymotrpsin, why is the 17 anstrong movement significant

Since ion pair electrostatic energy doesn’t reach as far as 17 anstrong, large movements indicates that strain is encoded into protein structure,when 15-16 bond is cleave, is released by structural relaxtion - process allows ile16 to be close to active site

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<p>Describe a biomembrane</p><p>1) general aspects of phospholipid, protein and carbohydrate composition</p><p>2) integral vs peripheral membrane proteins</p><p>3) asymmetry of inner and outer leaflets</p>

Describe a biomembrane

1) general aspects of phospholipid, protein and carbohydrate composition

2) integral vs peripheral membrane proteins

3) asymmetry of inner and outer leaflets

1) Phospholipid: 30-70% by mass, Protein: 30-70% by mass, Carbohydrate: 5% by mass

2)

Integral membrane protein: embedded into membrane completely 

  • Function: to transport/to provide structural support

Peripheral membrane protein: partially embedded into membrane and partially exposed to extracellular matrix 

  • Function: to signal reactions to occur/to perform enzymatic activity 

3) “Asymmetry” of inner and outer leaflet consist of different lipid and protein compositions in different compartments of the biomembrane

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Describe a biomembrane

1) meaning of “amphiphilic”

2) What is a biomembrane permeable to

3) What is a biomembrane not permeable to

1) Ampiphilic: has both hydrophobic and hydrophilic component(Same meaning as amphipathic)

2) Permeable to small, uncharged molecules such as O2, CO2, H2O, N2, EtOH, uncharged steroids

3) Not permeable to large, charged molecules such as sugars, ions, proteins/AAs, nucleotides

<p>1) Ampiphilic: has both hydrophobic and hydrophilic component(Same meaning as amphipathic)</p><p>2) Permeable to small, uncharged molecules such as O2, CO2, H2O, N2, EtOH, uncharged steroids</p><p>3) Not permeable to large, charged molecules such as sugars, ions, proteins/AAs, nucleotides</p>
8
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Describe the three bilayer phases (solid, liquid, liquid ordered) in terms of

  • Mobility of components (relative level of rotational and translational mobility, relative rates)

  • Typical composition (saturated vs. unsaturated fatty acid chains, cholesterol)


Solid Phase: no double bonds, saturated, slow diffusion

Liquid Phase: double bonds, unsaturated, presence of kinks, fast diffusion

Liquid-Ordered Phase: high levels of cholesterol but saturated so it is highly ordered with liquid properties

<p>Solid Phase: no double bonds, saturated, slow diffusion </p><p>Liquid Phase: double bonds, unsaturated, presence of kinks, fast diffusion</p><p>Liquid-Ordered Phase: high levels of cholesterol but saturated so it is highly ordered with liquid properties </p>
9
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<p>Draw (including correct numbering of the carbons) a fatty acid (FA), given its “delta system” code (e.g., C 18:2 Δ9,12)</p>

Draw (including correct numbering of the carbons) a fatty acid (FA), given its “delta system” code (e.g., C 18:2 Δ9,12)

Lipid Nomenclature (General Form): C # of carbons in the chain : # of double bonds

Δposition of double bonds

<p>Lipid Nomenclature (General Form): C # of carbons in the chain : # of double bonds </p><p>Δposition of double bonds</p>
10
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Describe the three parts of a glycerophospholipid (glycerol backbone, polar/charged headgroup, two hydrocarbon chains) and how they are connected (FA’s O attached to glycerol’s C1 and C2 via acyl linkage, headgroup phosphate’s O attached to C3)

Glycerophospholipid (bilayer-forming lipid) components

Three parts: Glycerol backbone consists of a 3-carbon chain with two hydrocarbon chains and one polar, charged “headgroup”

Connection: Two hydrocarbon chains connected through acyl linkage + “headgroup” connected through the 3rd carbon in the chain.


<p>Glycerophospholipid (bilayer-forming lipid) components </p><p>Three parts: Glycerol backbone consists of a 3-carbon chain with two hydrocarbon chains and one polar, charged “headgroup”</p><p>Connection: Two hydrocarbon chains connected through acyl linkage + “headgroup” connected through the 3rd carbon in the chain.</p><p></p>
11
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Distinguish between biologically important glycerophospholipids in mammals: PA, PE, PC, PS, PG, and PIP2 (i.e., recognize the different head groups and where they are attached to the glycerol backbone).

PA: acidic so we have a hydrogen to donate (hanging hydrogen)

PE: contains ethane with amine group (Ethanolamine)

PC: ethane connected to 3 methyl groups attached to the amine group at the end (Choline) | same head group as sphingomylein

PS: COO- and NH3+ connected to -CH which is connected to CH2 (serine)

PG: glycerol so two OH and three CHn[CH2-CH-CH2] (glycerol)

PIP2: two phosphate groups hanging off the sugar (the only ring structure)

<p>PA: acidic so we have a hydrogen to donate (hanging hydrogen)</p><p>PE: contains ethane with amine group (Ethanolamine)</p><p>PC: ethane connected to 3 methyl groups attached to the amine group at the end (Choline) | same head group as <strong>sphingomylein</strong></p><p>PS: COO- and NH3+ connected to -CH which is connected to CH2 (serine)</p><p>PG: glycerol so two OH and three CHn[CH2-CH-CH2] (glycerol)</p><p>PIP2: two phosphate groups hanging off the sugar (the only ring structure)</p>
12
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Explain why the curvature of a membrane depends on the relative cross-sectional area of headgroup and FA chains.

The way a membrane curves depends on whether each lipid looks more like a cone, a cylinder, or an inverted cone which depends on how big the head is compared to the tails. (Essentially, the shape of the polar head group changes the relative cross-sectional area of the membrane)

→ asymmetry between the polar head group and fatty acid tails creates a cone or inverted-cone shape, which favors membrane curvature.


pentagon shape, which creates a spherical shape for the membrane.


Cell’s Membrane Bilayer doesn’t require a big cross sectiona difference between head and tail

13
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Explain in general terms how the biomembrane serves as a “chemical factory”

The biomembrane is not “merely” the outer barrier of the cell: It also serves as a source of small molecules that are used by the cell for further reactions.

i.e a “chemical factory”


Essentially, it is a site of chemical reactions not just a physical barrier.


14
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Identify the site of PL cleavage and the resulting products for Phospholipase A2, Phospholipase C, and Phospholipase D.

1) Phospholipase A2

  • Site of cleavage: acyl chain attached at C2 glycerol carbon.

  • Product: fatty acid and lyso-phospholipid

  • Result: generates products involved in inflammation and pain

2) Phospholipase C

  • Site of cleavage: phosphodiester bond between glycerol backbone and phosphate

  • Product: choline phosphate and diacylglycerol

  • Result: acting as signal in many pathways

3) Phospholipase D

  • Site of cleavage: phosphodiester bond between phosphate and “headgroup”

  • Product: choline and phosphatidic acid

  • Result: membrane trafficking, possibly cancer



<p>1) Phospholipase A2</p><ul><li><p>Site of cleavage: acyl chain attached at C2 glycerol carbon. </p></li><li><p>Product: fatty acid and lyso-phospholipid</p></li><li><p>Result: generates products involved in inflammation and pain</p></li></ul><p>2) Phospholipase C</p><ul><li><p>Site of cleavage: phosphodiester bond between glycerol backbone and phosphate</p></li><li><p>Product: choline phosphate and diacylglycerol</p></li><li><p>Result: acting as signal in many pathways </p></li></ul><p>3) Phospholipase D </p><ul><li><p>Site of cleavage: phosphodiester bond between phosphate and “headgroup”</p></li><li><p>Product: choline and phosphatidic acid </p></li><li><p>Result: membrane trafficking, possibly cancer </p></li></ul><p></p><p></p>
15
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Explain the concept of “second messenger” and describe the role Phospholipases play in this signaling strategy

“Second messenger”: Certain membrane lipids can act as “second messenger” to carry information to another part of the cell.

Role Phospholipases play: The “phospholipid fragments” carry information to another part of the cell. The inositol triphosphate (IP3) and diacylglycerol (DAG) are the second messengers made from lipids by phospholipase C cleaving IP3


16
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Describe the 6 ways/modes membrane proteins bind to the lipid bilayer.

  1. A single alpha helix consisting of nearly all non-polar residues

  2. Multiple alpha helices (up to 14 helices for one protein), consisting of 7 or 12 helix super-families → interior of multi-helix proteins can be relatively polar/charged

  3. Two or more long acyl chains: Many g-proteins have one acyl chain

  4. GPI (carbohydrate linker: glycosylphosphatidylinositol): Membrane anchor is PI, but unknown why carbohydrate linkers are needed. Prion protein from nerves cells has GPI anchor.

  5. Indirect association by binding to another protein or membrane surface (“peripheral” or “extrinsic”)  

  6. Beta barrel (typically found in bacteria)


<ol><li><p><span style="background-color: transparent;">A single alpha helix consisting of <strong>nearly all non-polar residues</strong></span></p></li><li><p><span style="background-color: transparent;">Multiple alpha helices (up to 14 helices for one protein), consisting of 7 or 12 helix super-families → interior of multi-helix proteins can be relatively polar/charged</span></p></li><li><p><span style="background-color: transparent;"><strong>Two or more long acyl chains: </strong>Many g-proteins have one acyl chain</span></p></li><li><p><span style="background-color: transparent;"><strong>GPI (carbohydrate linker: </strong></span>glycosylphosphatidylinositol<span style="background-color: transparent;"><strong>): </strong>Membrane anchor is PI, but unknown why carbohydrate linkers are needed. Prion protein from nerves cells has GPI anchor.</span></p></li><li><p><span style="background-color: transparent;">Indirect association by binding to another protein or membrane surface (“peripheral” or “extrinsic”)&nbsp;&nbsp;</span></p></li><li><p><span style="background-color: transparent;">Beta barrel (typically found in bacteria)</span></p></li></ol><p></p>
17
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In a Red Blood Cell membrane what are the roles of Glycophorin A and Anion channel

Glycophorin - A single alpha helix; Function unknown! But malaria parasite & flu virus binds here!


Anion Chennl is a 12 alpha helices that allows exchange of CO2 & Cl-

<p>Glycophorin - A single alpha helix; Function unknown! But malaria parasite &amp; flu virus binds here!</p><p></p><p>Anion Chennl is a 12 alpha helices that allows exchange of CO2 &amp; Cl-</p>
18
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Identify the structural biology advancement that has increased the number of high-resolution 3D structures of membrane proteins

New methods of cryo-electron microscopy have made it possible to determine the 3D structure of membrane proteins without requiring protein crystals, which are hard to grow.

  • Although membrane proteins make up ⅓ of all proteins, only about 1 membrane protein structure has been solved by x-ray diffraction for every 100 structures of water-soluble proteins.


19
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Define the pyranose vs. furanose forms of saccharides

“Pyranose”: 6-membered ring

“Furanose”: 5-membered ring

<p>“Pyranose”: 6-membered ring </p><p>“Furanose”: 5-membered ring</p>
20
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Draw the pyranose form of D-Glucose in both a and b forms, number the Cs, and

identify the anomeric C.

Five-membered ring (ribofuranose): the OH on carbon 4 reacts with carbon 1. The ring contains four carbons and one oxygen; carbon 5 stays outside.

Six-membered ring (ribopyranose): the OH on carbon 5 reacts with carbon 1. The ring contains five carbons and one oxygen.


Notes:

D vs L: find where the last chrial carbons OH group is either on the left(L) or right(D) on fischer projection. For Haworth Position, CH2OH below(L) above(D)

a vs b: alpha = trans(OH down) and beta(OH up) = cis on the haworth projection(ring form)

Anomeric carbon is the first carbon with either up or down OH

Haworth Projection is drawn clockwise carbon 1 - carbon 4 or 5 then to O

Fischer Projection: Right side points down on haworth and Left side points up on haworth

<p>Five-membered ring (ribofuranose): the OH on carbon 4 reacts with carbon 1. The ring contains four carbons and one oxygen; carbon 5 stays outside.</p><p>Six-membered ring (ribopyranose): the OH on carbon 5 reacts with carbon 1. The ring contains five carbons and one oxygen.</p><p></p><p>Notes:</p><p>D vs L: find where the last chrial carbons OH group is either on the left(L) or right(D) on fischer projection. For Haworth Position, CH2OH below(L) above(D)</p><p>a vs b: alpha = trans(OH down) and beta(OH up) = cis on the haworth projection(ring form)</p><p>Anomeric carbon is the first carbon with either up or down OH</p><p>Haworth Projection is drawn clockwise carbon 1 - carbon 4 or 5 then to O</p><p>Fischer Projection: Right side points down on haworth and Left side points up on haworth</p>