B1 Flashcards

1
Q

the carbon atom’s features that allow organic molecules to be formed

A

carbon atoms readily form bonds with other carbon atoms to make a carbon backbone along which other atoms can attach
in living organisms, there are relatively few other atoms that can attach to carbon
life is therefore based on a small number of chemical elements

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2
Q

monosaccharide

A

basic carbohydrate monomer
sweet tasting
soluble
(CH2O)n - n=3-7
can form crystals
affect water potential
e.g. glucose, fructose, galactose

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3
Q

glucose

A

hexose sugar
C6H12O6
two isomers: a and B gluclose
- they differ in the position of -OH attached to C1
in B glucose, the H is below the plane and the -OH is above the plane on C1

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4
Q

disaccharides

A

two monosaccharides join in s condensation reaction
forms a glycosidic bond
addition of water breaks glycosidic bond- hydrolysis

glucose+glucose = maltose a1-4
glucose+fructose = sucrose
glucose+galactose = lactose B1-4

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5
Q

polysaccharides

A

polymers, made by combining many monosaccharide molecules, joined by glycosidic bonds in condensation

large molecules
insoluble in water
not sweet
does not affect water potential
Cx(H2O)y

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6
Q

polysaccharides differ in:

A

constituent monomers
type of bond
where the glycosidic bond is (polysaccharides)
structure (helical/branched)
function
where they are found

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7
Q

uses of a + B glucose

A

a-glucose: respiratory processes + energy storage
B-glucose: strength, rigidity, support

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8
Q

test for starch

A

place sample in test tube
add drops of iodine in potassium iodide solution
positive result = yellow –> blue/black

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9
Q

condensation

A

joins monomers together
forms a chemical bond
releases water

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10
Q

hydrolysis

A

breaks a chemical bond between monomers
uses water

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11
Q

starch- about + structure

A

polysaccharide found in plants in small grains
esp in seed and storage organs
major energy source in most diets

  • made up of chains of a-glucose, linked by a 1-4 glycosidic bonds, in condensation reactions
  • branched chains- amylopectin
  • unbranched chains- amylose
    amylose = wound in tight coil- compact helical structure
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12
Q

starch structure to function

A
  • insoluble- doesnt affect water potential
  • large- does not diffuse out of cells
  • amylose = compact- lots stored in small space
  • hydrolysed to a-glucose- used in respiration
  • branched amylopectin- many ends increase surface are - which can be acted on by enzymes- releasing a-glucose - monomers readily for respiration
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13
Q

glycogen about + structure

A

found in animals and bacteria, never plant
similar structure to amylopectin, but shorter chains, more highly branched
‘animal starch’- major carbs storage molecule in animals

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14
Q

glycogen structure to function

A
  • insoluble- does not affect water potential
  • large- does not diffuse out of cells
  • compact- lots stored in small space
  • more highly branched than starch- more ends can be acted on simultaneously by enzymes- rapidly hydrolysed to glucose monomers- used in respiration
    this is important to animals which have a higher metabolic rate and therefore respiratory rate than plants because they are more active
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15
Q

suggest how glycogen acts as a source of energy

A

hydrolysed to glucose monomers
used in respiration

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16
Q

cellulose - structure + function

A

made of B-glucose monomers

  • long, straight unbranched chains
  • chains run parallel to each other
  • hydrogen bonds form cross linkages between adjacent chains
  • number of hydrogen bonds adds much strength
  • unlike starch, adjacent glucose molecules are rotated 180 degrees
    this allows hydrogen bonds to form between -OH groups on adjacent parallel chains

cellulose molecules grouped together –> microfibrils –> fibres

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17
Q

how does cellulose support plant cells

A

provides rigidity to plant cell wall that prevents cell bursting when water enters by osmosis
- exerts inwards pressure that stops further influx of water
- living plant cells are turgid and semi-rigid
important in maintaining stems and leaves in a turgid state so they can provide maximum surface area for photosynthesis

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18
Q

explain how the structure of starch and cellulose are different

A

starch made of a-glucose
cellulose made of B-glucose

position of -OH on C1 = inverted in cellulose

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19
Q

characteristics of lipids

A

contain C, H, O
proportion of O to C and H is smaller than in carbohydrates- long fatty acid hydrocarbon chain
insoluble in water
soluble in organic solvents e.g. alcohols and acetone

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20
Q

roles of lipids

A
  • cell surface membrane and membrane surrounding organelles
    phospholipids contribute to the flexibility of membranes and the transfer of lipid soluble substances across them

-source of energy: when oxidised, provide more than twice the energy as same mass of carbs + release water

-waterproofing: insoluble in water- plants and insects have thick waxy cuticles to conserve water, mammals produce oily secretion from sebaceous glands

-insulation: slow conductors of heat- retain body heat. also electrical insulators in myelin sheath around nerve cells

-protection: stored around delicate organs

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21
Q

hydrogenated fatty acids

A

formed when acids with C=C are bombarded with H so they become saturated

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22
Q

triglyceride structure

A

have three fatty acids combined with glycerol
each fatty acid forms an ester bond with glycerol in a condensation reaction (-OH + -COOH)

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23
Q

simple triglyceride

A

fatty acids all the same

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24
Q

mixed triglyceride

A

different fatty acids

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25
Q

saturated/monounsaturated/polyunsaturated fatty acid

A

no C=C/ 1 C=C/ many C=C

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26
Q

structure of fatty acids related to properties

A

2 saturated, 1 unsaturated
kinky
cannot stack ontop of each other
liquid @ room temp
e.g. oils

3 saturated
can stack ontop of each other
solid @ room temp
e.g. fat

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27
Q

triglyceride structure related to properties

A
  • high ratio of C-H to C (fatty acid) so excellent energy soure
  • low mass to energy ratio- good storage molecule, much energy stored in a small volume- reduces mass animals need to carry around
  • large, non-polar so insoluble in water- does not affect water potential
  • release H2O when oxidised
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28
Q

phospholipid structure

A

differ from triglyceride- one fatty acid molecule replaced by phosophate molecule

hydrophilic head, hydrophobic tail

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29
Q

structure of phospholipids related to function

A
  • polar molecules- in aqueous environment, form hydrophobic bilayer within cell-surface membranes
  • hydrophobic heads hold surface of cell-surface membrane
  • can form glycolipids within cell surface membranes- cell recognition
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30
Q

test for lipids

A

add ethanpl to sample
shake to dissole
add water
shake gently
milky white emulsion = positive

milky colour = due to any lipid being finely dispersed in water to form emulsion
light is refracted as it passes oil-water so appears cloudy

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31
Q

benedicts test
reducing + non reducing

A

reducing
- grind with pestle + mortar
- add water
- filter out solid
- add benedicts
- gently heat with water bath
- blue to brick red

non reducing
- if reducing = negative
- new sample, add HCl
- boil
- add alkali to neutralise
- add benedicts
- heat gently
- blue to brick red

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32
Q

non subjective approach to benedicts

A

filter out precipitate and dry
weigh ppt
the higher the mass, the more sugar present

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33
Q

describe how an ester bond is formed in a phosphodiester molecule

A

condensation reaction
loss of water
between glycerol and fatty acid

34
Q

explain how cellulose molecules are adapted to their function

A

long straight chains
linked by many hydrogen bonds
to form fibrils
grouped to fibres
provide strength to cell walls

35
Q

explain how starch molecules are adapted for their function

A

insoluble- doesnt affect water potential
helical- compact
large molecule- cannot leave cell

36
Q

why do polar molecules dissolve in water?

A

polar molecules form hydrogen bonds with water molecules

37
Q

what is a macromolecule and how is it formed

A

large molecule consisting of thousands of carbon atoms joined together
(polymers = molecules)
formed by condensation reactions

38
Q

why does sucrose produce a positive benedicts test after acid hydrolysis

A

acid hydrolysis of sucrose molecules releases the monomers a-glucose and b-fructose
the monomers are reducing sugars
give a positive result when heated with benedicts reagent

39
Q

plant cell walls structire to function

A

contain cellulose
exerts inward pressure on cell contents
prevents further influx of water
therefore cells do not burst when HYDROSTATIC PRESSURE increases

40
Q

why are starch and glycogen good for storage

A

both virtually insoluble
they DO NOT ADD TO THE SOLUTE CONC in cells
so water potential is not affected

41
Q

what is an emulsion

A

particles of a substance that are not dissolved but dispersed in a volume of water

42
Q

what are amino acids/ polypeptides/ proteins

A

amino acids are the monomer which combine through condensation and formation of peptide bonds to form the polymer polypeptide
polypeptides can combine to form proteins

proteins have a FUNCTION

43
Q

structure of amino acids

A

central (alpha) carbon atom attached to:
- amino group (-NH2)
- carboxyl group (-COOH)
- -H atom
- R group

44
Q

formation of peptide bond

A

-OH of carboxyl combines with -H of amino group of another amino acid

45
Q

why are amino acids zwitterions?

A

can internally transfer ions
H from COOH –> NH2
net charge of 0

46
Q

primary protein structure

A

the amino acid sequence in its polypeptide chain
this sequence determines its shape, properties, function

47
Q

secondary protein structure

A

the shape which the polypeptide chain forms as a result of hydrogen bonding

the linked aa possess -NH and -C=O on either side of peptide bond
H of -NH has 8+, O of C=O has 8- (due to high electronegativity of O and N)

these groups form hydrogen bonds
shape : a-helix or B pleated sheet

48
Q

tertiary protein structure

A

due to the bending and twisting of the polypeptide helix into a compact structure
disulfide, ionic, hydrogen bonds present, as well as hydrophobic interactions

it is the 3D shape of a protein that is important in terms of its function
it makes each protein distinctive and allows it to recogise/ be recognised by other molecules

49
Q

quaternary protein structure

A

combination of many different polypeptide chains and associated non protein prosthetic groups into a large, complex protein molecule (conjugated)

50
Q

test for proteins

A

biruret test- detects peptide bonds
- place sample in test tube
- add equal volume NaOH
- add few drops CuSO4 and mix
blue to purple is positive

51
Q

fiborous proteins structure + functions

A

unbranched
tightly wound
tertiary structure twisted in second helix to make quaternary

water insoluble
physically tough
parallel polypeptide chains in long fibres

functions
structural role + contractile

52
Q

globular protein structure + functions

A

roughy spherical
water soluble
tertiary structure critical for function
polypeptide chains folded into a spherical shape

functions
metabolic
catalytic
regulatory
transport
protective

53
Q

what are enzymes

A

enzymes are globular proteins that act as catalysts
they alter the rate of a chemical reaction without undergoing a permanent change themelves

54
Q

what conditions must be satisfied for a reaction to take place

A
  • the reactants must collide with sufficient energy to alter the arrangement of their atoms
  • the free energy of the products must be less than that of the substrates
  • activation energy must be reached
55
Q

how do enzymes lower the activation energy

A

breaking bonds
bringing molecules together

56
Q

enzyme structure

A

globular proteins
specific 3D structure as a result of primary protein structure

functional region called active site
- the specific region of the enzyme where the substrate binds and catalysis takes place

57
Q

substrate

A

the molecule on which the enzyme acts
active site + enzyme are complementary shape
forms enzyme-substrate complex

substrate is held in active site by bonds that temporarily form between certain a.a in the active site and the groups on the substrate molecule

58
Q

how is substrate held in active site

A

R-groups of amino acids face inside the active site
- made up of +ve and -ve charges

  • bonds temporarily form between a.a. and substrate
  • helps substrate settle in active site
59
Q

what is enzyme specificity

A

enzyme only catalyses reaction with substrates with complementary shapes

60
Q

induced fit model

A

proximity of substrate
enzyme changes shape
substrate in active site
enzyme puts strain on substrate
distorts bonds in substrate
lowers activation energy needed to break bonds

61
Q

lock and key model

A

an enzyme only fits the shape of 1 specific substrate - the enzyme is highly specific and rigid

62
Q

limitation of lock and key

A

enzymes are not rigid structures
- other molecules can bind to the enzyme, NOT on the active site (e.g. non competitive inhibitors) and alter the active site shape
- enzymes are therefore flexible structures

so:
if an inhibitor binds to enzyme (not act. site), the active site shape distorts and the substrate cannot fit
- therefore, the active site changes depending on bonding molecules to the enzyme

63
Q

measuring enzyme catalysed reactions

A

use its time course:

rate = amount/time

amount = mass or volume
disappearance of substrate or formation of product

64
Q

the rate of an enzyme catalysed reaction (stages)

A
  • lots of substrate, 0 product

-easy for substrate to meet empty active sites

-all active sites are filled at one time

-amount of substrate decreases as its broken down

-amount of product increases

-becomes more difficult for substrate to come into contact with enzyme- fewer substrate molecules as broken down, and product molecules get in the way.

-substrate molecules take longer to be broken down and rate of reaction slows

-rate continues to slow as substrate concentration decreases

-until substrate concentration is so low so that any more change its concentration cannot be measured.
-until no substrate left so rate stops

65
Q

how to measure the rate of an enzymes catalysed reaction

A

tangent to curve

66
Q

effect of temperature on enzyme action

A
  • increasing temperature increases kinetic energy of molecules
    they move around more rapidly and collide more often
    more enzyme substrate complexes formed so rate increases
  • increasing temperature further begins to cause hydrogen bonds to break- change in tertiary structure
    active site changes shape- substrate fits less easily
  • increasing temperature further (around 60 degrees) the enzyme is so disrupted that it completely denatures and stops working.

denaturation is a permanent change- enzyme is no longer functional

67
Q

why is our body temperature not higher for increased enzyme action

A

although higher body temperature would increase the metabolic rate slightly
advantages offset by additional energy that would be needed to maintain the higher temperatures

other proteins, apart from enzymes, may be denatured at higher temperatures

at higher temperatures, any further rise in temperatures e.g. during illness, may denature the enzymes

68
Q

why do different species have different body temperatures

A

some animals e.g. birds have a normal body temperature of around 40 degrees because they have a high metabolic rate for the high energy requirement of flight

69
Q

when describing a rate / temp graph

A
  1. where it starts
  2. where it peaks- optimum temp
  3. where it goes down
  4. where it ends
70
Q

effect of pH on enzyme action

A

pH is a measure of H= conc
each enzyme has optimum pH

at a different pH than the optimum, the H bonds in the active site are broken- enzyme denatires

change in pH also alters the charges on amino acids in the active site- so it is no longer complementary

can also break ionic+hydrogen bonds in the tertiary structure so the active site changes shape

71
Q

temp + pH- pattern of marks

A
  1. identify bonds
  2. state effect on tertiary structure of active site
  3. decrease in the number of enzyme substrate complexes forming
    loss of complimentary shapes
  4. denaturation
    reduction of rate of reaction
72
Q

effect of enzyme concentration on rate

A

as long as excess substrate, increase in enzyme conc leads to proportionate increase in rate of reaction

graph initially shows proportionate increase
- because there is more substrate than the enzyme’s active sites can cope with
- increasing enzyme concentration, some excess substrate can be acted upon so rate increases

if substrate is limiting (i.e. not sufficient substrate to supply all active sites) any increase in enzyme conc will have no effect on rate
rate will stabilize at constant level
because available substrate is already being used as rapidly as it can be by existing enzyme molecules

73
Q

effect of substrate concentration on rate

A

if enzyme conc = fixed and substrate conc increased, rate increases in proportion

  • at low substrate conc, enzyme molecules have limited number of substrate molecules to collide with so active sites are not working at full capacity
  • as more substrate added, active sites gradually become filled, until max rate Vmax

when substrate is in excess, rate levels off

74
Q

inhibitor

A

a substance which reduces the activity of an enzyme, catalyst or reactant

75
Q

enzyme inhibitor

A

substances that directly or indirectly interfere with the functioning of the active site of an enzyme and so reduce its activity

76
Q

types of enzyme inhibitors

A

competitive- bind to active site

non competitive- bind to enzyme at a position other than the active site

77
Q

competitive inhibitor properties

A
  • similar molecular shape to substrate
    allows them to occupy active site

it is the difference between the concentrations of the inhibitor and substrate that determines the effect on enzyme activity

if the substrate conc is increased, effect of inhibitor is reduced

the inhibitor is NOT PERMANENTLY BOUND to the active site so when it leaves another molecule can take its place

sooner or later, all substrate molecules will occupy an active site, but the greater the concentration of inhibitor, the longer this will take

78
Q

non competitive inhibitor

A

attach to enzyme at binding site which isnt active site

upon attachment, alters shape of enzyme and thus its active site so that it is no longer complementary to substrate

a PERMANENT change to shape of enzyme

as the substrate and inhibitor are not competing, increase in substrate concentration does not decrease effect of inhibitor

eventually no enzyme substrate complexes formed

79
Q

what is a metabolic pathway

A

a series of reactions in which each step is catalysed by an enzyme
each reaction is connected by their intermediates i.e. the product of one is the reactant of the next

80
Q

how are metabolic pathways structured

A

the enzymes that control a pathway are often attached to the membrane of a cell organelle in a precise sequence.

  • in a metabolic pathway, the product of one reaction acts as a substrate for the next
  • by having the enzymes in an appropriate sequence, there is a greater change of each enzyme coming into contact with its substrate than if the enzymes were free in the organelle

this is a more efficient means of producing the end product

to keep a steady concentration of a particular chemical in a cell, the same chemical often acts as an inhibitor of an enzyme at the start of a reaction,

81
Q

end product inhibition

A

end product inhibits 1st enzyme
if the concentration of end product increases above normal, there will be greater inhibition of 1st enzyme
as a result, less product will be produced and its concentration will return to normal
if the concentration of the end product falls below normal, there will be less to inhibit 1st enzyme
so more product produced

end product inhibition is normally non competitive

82
Q

suggest one advantage of end product inhibition being non competitive
relate your answer to how the two types of inhibition take place

A
  • the level of end product does not fluctuate with substrate

non- competitive inhibitors occur at a site on the enzyme other than the active site-

hence isnt affected by substrate concentration

therefore in non-competitive inhibition, changes in the level of substrate do not affect the level of inhibitor, nor the level of end product

competitive inhibition involves competition for active sites
a change in the level of substrate would therefore affect how many end products molecules combine with active sites

therefore the degree of inhibition would fluctuate and so would the level of end product.