BMSC 200 Final no pics Flashcards

1
Q

difference between coenzymes and cofactors

A

coenzymes = organic (vitamins)

cofactors = inorganic (metalions)

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2
Q

describe the difference between the rate and the equilibrium of a reaction

A

the equilibrium is determined by the difference in free energy between the substrate and the product

the rate of the reaction is determined by the difference between the energy of the transition state and the substrate, enzymes can lower this influencing the rate of the reaction

however, an enzyme has no effect on the equilibrium of a reaction

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3
Q

what are the binding effects of an enzyme catalyst

A

substrate binding –> reduces entropy
aligns functional groups
desolves substrate ( removes water)
distorts substrate
induced fit

transition state stabilization
increased interaction between enzyme and substrate during transition state
active site is complimentary to TS
must be similar enough to substrate to ensure specificity but different enough to promote change
enzymes have higher affinities to TS than S
provides microenvironment

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4
Q

what are transition state analogs

A

stable compound resembling unstable transition state
type of competitive inhibitor
bind to active site w high affinity

Antibodies against transition state analogues may have catalytic activity.

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5
Q

what are the chemical effects of enzyme catalyst

A

acid base catalysis –>
uses his
catalytic proton transfer

covalent catalysis –>
covalent linkage to substrate
regenerates free enzyme
ex) sucrose phosphorylase

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6
Q

describe Michalis menton kinetics

A

describe relationship of substrate concentration and initial velocity
allosteric enzymes do not obey this

V0= vmax(S) / km+(S)

km = 1/2vmax

When [S] < Km, enzymes are highly sensitive to changes in substrate
concentration but have very little activity.

When [S] > Km, enzymes have high activity but are insensitive to
changes in substrate concentration.

When [S] = Km, enzyme has significant activity and is responsive to
changes in substrate concentration.

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7
Q

describe lineweaver burke plots

A

can find vmax and km

Comp –> Y axis same but X axis closer
increase Km
bind only free enzyme E
Vmax is same

Non Comp –> Y axis higher X axis same
decrease Vmax
bind to E and ES
Km same

Uncomp –> y and x both farther
decrease vmax
decrease Km
bind to only ES

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8
Q

discuss serine proteases

A

cleave polypep chains (peptide bonds)
diff proteases have diff specificity
example of both acid base and covalent catalysis

trypsin –> cleave pos (lys arg)
chymotrypsin –> cleave aromatics (phe met)
elastase –> cleaves small hydrophobic (gly ala)

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9
Q

what is the catalytic triad

A

His, H, acid base catalysis

aspartate, Asp, D, stabilization

Ser, S covalent catalysis

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10
Q

enzyme regulation

A

through covalent modification (phosphorylation)
non covalent –> allosteric regulation

uses negative feedback loops
one at end can regulate one at beginning, or first one of branch
if two join to form one, can regulate first of each branch

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11
Q

allosteric enzymes

A

usually quaternary structure
usually rate limiting step
do not obey Michalis mention kinetics instead have S curve
bind non covalently

have active R and inactive T (R ready to go)
activators bind to R, so can substrates but not T

more sensitive to changes in substrate concentration near Km then MM enzymes
this sensitivity is called the threshold effect (think back to r t states of hemoglobin, once one switches they all do

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12
Q

Describe phosphofructokinase 1

A

in glycolysis
PEP is heterotropic allosteric inhibitor
ADP is heterotropic allosteric activator
noncovalent modification

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13
Q

describe glycogen metabolism

A

covalent modification
glucose –> glycogen
glycogen synthase to form glycogen anabolic, when non phosphorylated/ insulin
glycogen phosphorylase to form glucose catabolic when phosphorylated/ hungry

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14
Q

how do you find the number of stereoisomers in a monosaccharide

A

2^n where n is number of chiral carbons

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15
Q

what are epimers

A

sugars that differ at only a single chiral center

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16
Q

what are the two ring structures cyclized sugars can form

A

pyran of pyranose 6 c

furan or furanose 5c (in the ring)

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17
Q

what is the anomeric carbon

A

carbon that becomes chiral as a result of cyclization
cyclization always involves either 1-5 or 2-5c

on 6c or pyran ring forms the anomeric carbon is C1 (ALDOSES)
on 5c or furan ring forms the anomeric carbon is on c2 (KETOSES)

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18
Q

how can you tell the alpha or beta forms of cyclized carbons

what is switching between the two called

A

beta on top
alpha on bottom

switching between the two is called mutarotation

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19
Q

describe the nomenclature of disaccharides

A

glycosidic bonds are the structural linkage
0-glycosisic occurs through oxygen
N glycosidic is on nitrogen or amide

monosaccharides involved
ring type
configurations
‘linkages

end chain with free anomeric carbon is reducing end
always name non reducing end first (osyl then ose)

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20
Q

describe the energy storage polysaccharides

A

in plants
starch (amylose, unbranched; amylopectin, branched)
has a1-6 branch every 25 residues
many non reducing ends

in animals
glycogen
has a a1-6 branch every 9 residues
even more non reducing ends
means animals can mobilize their energy faster than plants bc more branches

all use a1-4 linkages
all homopolysaccharides

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21
Q

describe the structural polysaccharides

A

cellulose (fiber in plant cells)
chitin

use B1-4 linkages instead
linear
fibrils formed from parallel long linear chains linked through hbonds
means they are rigid and cannot be broken easily by amylase

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22
Q

glycolipids

A

used in blood group antigens

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23
Q

difference between glycoproteins and proteoglycans

A

glycoproteins
more protein
variety of roles
either 0 or N linked
ex) EPO

Proteoglycans
more sugar (carb)
structural and lubricating
ex) glycosaminoglycans –> introduces negatively charged fibrous strands

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24
Q

what is a fatty acid, what are the different types of FA

A

hydrocarbon tail with a carboxylic head group, associate thru hphobic interactions

saturated = no double bonds –> solid, hard to melt
unsaturated = one double bond –> less solid, easier to melt
polyunsaturated = multiple double bonds –> liquid, easiest to melt

when naming carbon of carboxy group is carbon one

25
Q

what are the main energy storage molecules in the body and why

A

triacylglycerols which are 3 fatty acids linked thru ester linkages
have glycerol backbone
hydrophobic

works well because it has a low oxidation state and low hydration state

26
Q

how do you release the fatty acids from the triacylglycerol?

A

saponification

27
Q

what are structural lipids in membranes

A

have two hydrophobic tails and one head group
can be classified based on backbone
either glycerol of sphingosine

28
Q

glycerophospholipids

A

most abundant in membranes
glycerol backbone with 2 FA and 1 phosphate
can have diff head groups
ex) phosphatidylinositol –> signal transduction with both head group and backbone, would be inside cell

29
Q

sphingolipids

A

single long chain fatty acid linked by an amide bond
sphingosine backbone
can have a variety of polar head groups
sphingomyelins
cerebrosides
gangliosides

30
Q

what do glycosphingolipids do

A

determine blood type, reflects sugar patterns in head group

31
Q

explain steroids

A

sterols are structural membrane lipids
contain 4 fused ring nucleus (3 6c, 1 5c)
rings are rigid

steroid hormones also exist
sterol derivative
can pass thru plasma membrane and bind receptors in nucleus
alters gene expression and metabolism

32
Q

explain cholesterol

A

mediates membrane fluidity

precursor of steroids

33
Q

what are the three types of eicosanoids

A

paracrine hormones (act near point of production)

prostaglandins –> constriction of blood vessels
Thromboxanes –> blood clot formation
Leukotrines –> smooth muscle contraction

34
Q

Lipid vitamins

A

A eyes, from egg
D bones, from light
E no free radicals
K blood coagulation

35
Q

explain the membrane lipid bilayer

A

amphipathic
self assembling through hydrophobic effect
lipids with two hydrocarbon tails form bilayers
one hydrocarbon tail forms micells\

only small or non polar molecules can cross the membrane without help

the more active the membrane the more proteins than lipids it has

follows the fluid mosaic model, meaning that shit moves around in there within the membrane, but movement across the membrane is restricted

36
Q

explain lipid rafts within a membrane

A

frosm from spontaneous association of lipids with tails of a similar length
usually sphingolipids cuz they are longer
often serves signalling functions
docking points for lipid anchored proteins

37
Q

explain peripheral membrane proteins

A

associated with one side of the membrane
easiest to remove from membrane–> changes in pH
associate through either hydrogen bonding or electrostatic interactions
non covalent

38
Q

lipid anchored membrane proteins

A

covalently linked on (harder to remove)
GPI anchored proteins on outside (lipid rafts)
proteins with only one hydrocarbon tail on inside

39
Q

Integral membrane proteins, 3 broad types

A

span the membrane
either single pass ahelical, helical bundles or b-barrels
residues with non polar side chains dominate middle
usually can tell if there’s 24 hydrophobic residues in a row

40
Q

simple diffusion

A

small nonpolar molecules dont need no man
determined by concentration gradient
can only move down gradient
no carrier
not saturable
down gradient
no energy

41
Q

Facilitated diffusion

A

removes hydration shell
lower activation energy
no energy

channels
fast, do not saturate

carriers
slow, do saturate
ex) uptake of glucose into RBC/ erythrocytes

42
Q

Primary active transport

A

driven by atp
can move against gradient

ptype –> phosphorylated intermediate –> NaKATPase
3 Na out/ 2 K in

vtype –> pumps protons in Vesicles

ABC type–> removes toxins from cells

43
Q

Secondary active transport

A

uses concentration gradient as a source of energy
couples the movement of one molecule down its concentration gradient with another against

ex) glucose uptake into epithelial cells
uses Na and glucose , Na down, glucose up

44
Q

ion channels

A

enable rapid movement of ions across the membrane
very selective and regulated
faster and no saturation limits

K+ channel

even though Na is smaller than K it still cannot get through based on ability to shed water molecules

45
Q

structural features of nucleotides

A

building blocks for nucleic acids
all have ribose sugar ( ribose or deoxyribose)
nitrogenous bases ( purine or pyrimidine)
phosphate(s)

46
Q

which has ribose DNA or RNA

A

DNA has deoxyribose, RNA has ribose

47
Q

Nitrogenous bases

A

either purine (A,G) or pyrimidine (C,T,U)
planar and non polar
link to ribose through N-glycosidic bonds
all nitrogenous bases link to C1 of sugar

In purines, N-glycosidic bond is to N9 of nitrogenous base
in pyrimidines N-glycosidic bond is to N1 of nitrogenous base

48
Q

difference between nucleotides and nucleosides

A

differ in wether they are phosphorylated at the C5 position
Nucleotides have 1 to 3 phosphates on the 5’ position\
Nucleotides are phosphorylated nucleosides

49
Q

forming and compostion of nucleic acids

A

nucleotides form linear strands through 3’-5’ phosphodiester linkages
identical in DNA and RNA, independent of the nucleotides that are being joined

strand of sugars linked by phosphodiester bonds is the backbone of nucleic acids

50
Q

how is RNA different from DNA

A

rna contains ribose instead of deoxyribose
contains uracil rather than thymine

is single stranded but can adopt complex 3d structures (don’t care lol)

less stable because of 2’ hydroxyl group, intentional ad DNA is needed for long term information storage not RNA

51
Q

explain the structure and formation of the double helix of DNa

A

`right handed helix
complimentary and antiparallel
sugars outside, n bases inside
purine matched with pyrimidine

A-T
G-C
(determined by hydrogen bonding)

chargaffs rule implies that A+G= T+C in DNA

52
Q

what are the four weak forces that stabilize a double helix

A

hydrophobic effects –> burying n bases in center
stacking interactions –> stacked base pairs form VDW interactions
hydrogen bonds –> hydrogen bonding between base pairs
Charge-charge interactions –> electrostatic repulsion of neg charged phosphate groups

53
Q

restriction endonucleases

A

cleave specific DNA sequences
bacterial defense mechanism against viral invasion
restriction enzymes cut at palindrome sequences (flipped below not across)
restriction enzymes used as molecular scissors

54
Q

DNA as a carrier of genetic information

A

complimentary nature is important for replication and repair
since one sequence determines the other, each strand can be used as a template
the resulting DNA duplexes will be identical

55
Q

Information about destroying and rebuilding DNA`

A

can be denatured and separate the strands via heat
will reform –> annealing

Melting point Tm is when half the DNA has become single stranded
reflects composition –> more GC = harder to melt

56
Q

Packaging of eukaryotic DNA

A

involves formation of nucleosomes –> DNA histone complexes
5 Histone proteins –> H1, H2A, H2B, H3, H4
nucleosome has 8 histone proteins –> 2 of H2A, H2B, H3, H4
H1 binds the region of linker DNA
histones can be reversibly modified

57
Q

Bacterial Genome

A

closed and circular
no introns
may have additional info in the form of plasmids

58
Q

Eukaryotic genome

A

have multiple chromosomes
chromosomes are linear –> some gets lost each replication –> telomeres help this
have introns
some may contain extra DNA in mitochondria and chloroplasts

introns are removed from mRNA prior to translation
introns mean multiple mRNas of diff sequences can be generated from single gene