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AS Biology > Chapter 1 - Biological Molecules > Flashcards

Chapter 1 - Biological Molecules Flashcards

1
Q

What are monomers?

A
  • smaller/repeating unit from which larger molecules/polymers are made
  • examples: monosaccharides like glucose, amino acids and nucleotides
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2
Q

What are polymers?

A

Molecules made from many (repeating) monomers joined together
- examples: polysaccharides like starch, proteins and DNA/RNA (nucleic acids)

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3
Q

what is polymerisation?

A
  • process by which polymers are formed
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4
Q

what is a condensation reaction?

A
  • joins two molecules together with formation of a chemical bond and involves elimination of water molecule
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5
Q

what is a hydrolysis reaction?

A
  • breaks a chemical bond between two molecules and involves the use of a water molecule
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6
Q

what are monosaccharides? Give an example

A
  • monomers from which larger carbohydrates are made

- glucose, galactose and fructose

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7
Q

what are disaccharides?

A

molecules formed by condensation reaction of two monosaccharides
(held together by glycosidic bond)

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8
Q

how is a glycosidic bond formed?

A

a condensation reaction between two monosaccharides

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9
Q

formation of disacchardies:
formation of a maltose
formation of sucrose
formation of lactose

A

via condensation reactions
maltose = 2x alpha glucose
sucrose = glucose + fructose
lactose = glucose + galactose

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10
Q

what is an isomer? draw the two isomers of glucose

A
  • molceules w/ same molecular formula, but have a different arrangement of the atoms in space

alpha glucose: DUDD
beta glucose: DUDU

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11
Q

what are polysacchardies?

A

larger molecules formed by the condensation of many monosacchardies

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12
Q

how is glycogen, starch and cellulose formed?

A
  • glycogen and starch: condensation of alpha glucose

cellulose: condensation of beta glucose

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13
Q

what’s a reducing sugar? give examples

A
  • any sugar capable of acting as a reducing agent (b/c having free aldehyde or ketone group)
  • all monosacchardies and some disacchardies (e.g. lactose and maltose, NOT sucrose, sucrose is a non-reducing sugar)
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14
Q

Describe the test for reducing sugars

A
  • Add benedict’s reagent (blue) to sample (add excess to make sure all sugar reacts)
  • heat in water bath that’s been brought to boil
  • if test positive, coloured precipitate will be formed
  • blue (none), green (v low), yellow (low), orange (medium), brick red (high)
    higher conc of reducing sugar= further colour change

more accurate: filter solution and weigh precipitate

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15
Q

Describe the test for non-reducing sugars

A
  • boil in dilute HCl (to hydrolyse non-reducing sugar)
  • neutralise solution by adding sodium hydrogen carbonate
  • check pH with pH strip
  • repeat benedict’s test:
    results will now be positive (brick red) if non-reducing sugar present so if solution remains blue, there’s NO sugar present
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16
Q

principle of non-reducing sugars test

A
  • all monosaccharides are reducing sugars
  • so non-reducing sugar hydrolysed to monosaccharide (by dilute HCl)
  • which will turn solution brick red
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17
Q

how to test for presence of starch

A
  • add potassium iodide to food sample (yellow)
  • if starch present turns blue-black
  • starch not present - remains yellow
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18
Q

how is a polysaccharide formed?

A

by condensation of many monosaccharides

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19
Q

what is starch?

A
  • alpha glucose polysaccharide held together by glycosidic bonds
  • major energy source in plants
  • mixture of amylopectin and amylose
  • found in chloroplast in starch granules
  • mostly found in plant cells with high energy demand e.g. photosynthesising leaves in cells
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20
Q

describe the structure of amylose

A
  • long alpha glucose polysaccharide joined by 1-4 glycosidic bonds
  • so it coils into helix shape which makes it more compact
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21
Q

describe the structure of amylopectin

A
  • long alpha glucose polysaccharide joined by 1-4 glycosidic bonds and occasional 1-6 glycosidic bonds
  • more branched ends = more side branches w/ more accessible ends for amylase to hydrolyse starch
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22
Q

name and describe the properties that make starch suitable for energy storage

A
  • insoluble therefore water doesn’t affect WP and so water isn’t drawn into cells by osmosis
  • large and insoluble so doesn’t diffuse out of cells
  • compact so a lot can be stored in a small space
  • when hydrolysed forms alpha glucose monomers which is easily transported and readily used in respiration
  • branched so enzymes can act on branches simultaneously and glucose monomers can be released quickly
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23
Q

what is glycogen?

A
  • major source of energy storage in animals (found in bacteria too)
  • stored in muscles and liver
  • alpha glucose polysaccharide with 1-4 and 1-6 glycosidic bonds (more 1-6 glycosidic bonds than starch)
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24
Q

compare the structure of glycogen to that of starch

A

similar to starch but:

  • has shorter chains
  • more highly branched
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25
Q

describe the properties of glycogen that make it suitable for energy storage

A
  • insoluble therefore doesn’t tend to draw water into cells by osmosis
  • insoluble so doesn’t diffuse out of cells
  • compact so it can be stored in a small space
  • highly branched so more ends can be acted on simultaneously (than starch) by enzymes. So it’s more rapidly hydrolysed (broken down) to glucose monomers which are used in respiration. This is important to animals which have a higher metabolic rate and their respiratory rate is higher than plants b/c they’re more active
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26
Q

function of cellulose

A
  • major component of cell wall of plant cell

- strengthens cell wall and prevents cell from bursting when too much water enters cell by osmosis

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27
Q

describe the structure of cellulose

A
  • composed of beta glucose monomers joined by 1-4 glycosidic monomers, to be able to form 1-4 glycosidic bonds, each beta glucose molecule is inverted 180 degrees from previous molecule
  • beta glucose monosaccharide
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28
Q

explain how H bonds are formed in cellulose molecule

A
  • cellulose chain, unlike starch, has adjacent glucose molecules rotated by 180
  • this allows H bonds to be formed between hydroxyl groups (-OH) on adjacent chains that help give cellulose structural stability
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29
Q

what are microfibrils?

A

microfibrls: multiple cellulose chains connected via H bonds
- provide structural support

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30
Q

how is the structure of cellulose suited to its function?

A
  • made up beta glucose monomers that form straight, unbranched chains
  • cellulose molecule chains run parallel to each other and cross-linked by H bonds which adds collective strength
  • forms microfibrils which provide more strength
  • bonds difficult to break (glycosidic and H bonds)
    ….making cell wall strong/resists osmotic pressure
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31
Q

the monomers of proteins are called what?

A

amino acids

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32
Q

what’s a dipeptide?

A

two amino acids joined together by a peptide bond, formed via a condensation reaction

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33
Q

what’s a polypeptide?

A

formed when more than two amino acids are joined together via peptide bonds, through condensation reactioons

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34
Q

what is the general structure of an amino acid?

A
  • google search *

- Central carbon, R variable/side group, H atom, COOH group and amine group (NH2)

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35
Q

How many amino acids are there and

how do they differ from one another?

A

20

differ only by side ‘R’ group

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36
Q

what is a protein?

A

polymers of amino acids (made up of one or more polypeptides), formed via condensation reactions
- held together by peptide bonds

37
Q

a functional protein may contain what?

A
  • one or more polypeptides
38
Q

what is the primary structure of a protein?

A

the sequence of AAs in the polypeptide chain

39
Q

what is the secondary structure of a protein?

A
  • polypeptide chain doesn’t remain flat/straight
  • H bonds form between AAs in chain, makes it coil into alpha helix or fold into beta pleated sheets
    ( H bonds form between positive NH group on one end to negative O of C=O of other end of another AA; causes long polypeptide chain to be twisted into a 3D shape)
40
Q

what is the tertiary structure of a protein?

A
  • secondary structure twisted/folded further to form final 3D shape of proteins made up of 1 polypeptide chain
  • tertiary structure maintained by disulfide bridges, ionic bonds and hydrogen bonds
  • where bonds occur depends on primary structure of protein
  • tertiary structure makes protein distinctive and allows it to interact with other molecules in a specific way
41
Q

Describe each type of bond in the tertiary structure of proteins.

A

● Disulfide bridges: strong covalent S-S bonds between molecules of the amino acid cysteine
● Ionic bonds: relatively strong bonds between charged R groups (pH changes cause these bonds to break)
● Hydrogen bonds: numerous & easily broken

42
Q

what is the quarternary structure of proteins?

A

● Functional proteins may consist of more than one polypeptide i.e. final 3D structure of proteins w/ more than 1 polypeptide chain
● Precise 3D structure held together by the same types of bond as tertiary structure.
● May involve addition of prosthetic groups e.g iron containing haem group in haemoglobin

43
Q

Describe how to test for proteins in a sample

A

Biuret test confirms presence of peptide bond
1. Add equal volume of sodium hydroxide to sample at room temperature.
2. Add drops of dilute copper (II) sulfate solution. Swirl to mix.
(steps 1 & 2 make Biuret reagent)
3. Positive result: colour changes from blue to purple
Negative result: solution remains blue.

44
Q

where do H bonds form in secondary structure of proteins?

A

Hydrogen bonds form between O 𝛿-
(slightly negative) attached to ‒C=O & H
𝛿+ (slightly positive) attached to ‒NH.

45
Q

what is the differnece between fibrous and globular proteins?

A
  • fibrous proteins generally composed of long and narrow strands and have a structural role (they are something) e.g. α-keratin found in hair and nails
  • globular proteins generally have a more compact and rounded shape and have functional roles (they do something) e.g. many enzymes and haemoglobin (theyre most water-soluble)
46
Q

what is the difference between fibrous and globular proteins?

A
  • fibrous proteins generally composed of long and narrow strands and have a structural role (they are something) e.g. α-keratin found in hair and nails (they have structural functions)
  • globular proteins generally have a more compact and rounded shape and have functional roles (they do something) e.g. many enzymes and haemoglobin (theyre most water-soluble), (they carry out metabolic functions)
47
Q

Describe the structure and function of globular proteins

A

● Spherical & compact.
● Hydrophilic R groups face outwards & hydrophobic R groups face inwards = usually water-soluble.
● Involved in metabolic processes e.g. enzymes and haemoglobin.

48
Q

Describe the structure and function of fibrous proteins

A

● Can form long chains or fibres
● insoluble in water.
● Useful for structure and support e.g. collagen in skin

49
Q

Outline how chromatography could be used to identify the amino acids in a mixture

A
  1. Use capillary tube to spot mixture onto pencil origin line and place chromatography paper in solvent.
  2. Allow solvent to run until it almost touches other end of paper. Amino acids move different distances based on
    relative attraction to paper and solubility in solvent.
  3. Use revealing agent or UV light to see spots.
  4. Calculate Rf values and match to database.
50
Q

what are enzymes?

A

● Biological catalysts (i.e. don’t get used up) for intra & extracellular
reactions (e.g. respiration and digestion)
● Specific tertiary structure determines shape of active site, complementary to a specific substrate.
● Formation of enzyme-substrate (ES) complexes lowers activation energy of metabolic reactions

51
Q

Explain the induced fit model of enzyme action

A

● Shape of active site is not directly complementary to substrate & is flexible.
● Conformational change enables ES complexes to form.
● This puts strain on substrate bonds, lowering activation energy

52
Q

How have models of enzyme action changed?

A

● Initially lock and key model: rigid shape of active site complementary to only 1
substrate.
● Currently induced fit model: also explains why binding at allosteric sites can change shape of active site (non-competitive inhibitor)

53
Q

How could a student identify the activation energy of a metabolic
reaction from an energy level diagram?

A

Difference between free energy of substrate & peak of curve

54
Q

describe the shape of a normal enzyme-controlled reaction

A
  • as its heated, energy level of substrate increases, until it meets Ea
  • at this point, reaction occurs
  • substrates start getting used up and products get in way of collisions (so substrate energy level decreases as there are fewer substrates)
55
Q

state two difference between a reaction w/o and enzyme and an enzyme-controlled reactions (in terms of graph)

A
  • same amount of product formed but it takes longer w/o enzyme
  • more energy required in reaction w/o enzyme
56
Q

Name 5 factors that affect the rate of

enzyme-controlled reactions.

A 
● enzyme concentration
● substrate concentration
● concentration of inhibitors
● pH
● temperature
57
Q

How does substrate concentration affect rate of reaction?

A
  • increases the rate of reaction to a certain point (more likely to collide and react if more substrates)
  • once all of the enzymes active sites have bound to all available substrates, any substrate increase will have no effect on the rate of reaction
  • b/c available enzymes will be saturated and working at their maximum rate.
58
Q

How does enzyme concentration affect rate of reaction?

A
  • will speed up the reaction, as long as there is substrate available to bind to.
  • once all of the substrate is bound, the reaction will no longer speed up (substrates have become limiting factor)
59
Q

How does temperature affect rate of reaction?

A
  • Rate increases as kinetic energy
    increases (molecules move faster, more likely to collide and react), peaks at optimum temp
  • Above optimum, ionic & H-bonds in 3°
    structure break (that hold active site together) = active site no longer
    complementary to substrate
    (denaturation).
60
Q

How does pH affect rate of reaction?

A
  • Enzymes have a narrow optimum
    pH range
  • Outside range, H+/ OH- ions
    interact with H-bonds & ionic bonds in 3° structure = denaturation of active site
61
Q

Contrast competitive and non-competitive inhibitors

A 
Competitive inhibitors:
1. similar shape to substrate = bind
to active site
2. do not stop reaction; ES complex
forms when inhibitor is released
3. increasing substrate concentration
decreases their effect
Non-competitive inhibitors:
1. bind at allosteric binding site
2. may permanently stop reaction;
triggers active site to change
shape
3. increasing substrate concentration
has no impact on their effect
62
Q

Outline how to calculate rate of reaction from a graph

A

● calculate gradient of line or gradient of
tangent to a point.
● initial rate: draw tangent at time = 0

63
Q

Outline how to calculate rate of reaction from raw data

A

Change in concentration of product or

reactant / time

64
Q

Why is it advantageous to calculate initial

rate?

A

Represents maximum rate of reaction
before concentration of reactants
decreases & ‘end product inhibition’

65
Q

State the formula for pH

A

pH = -log10[H+]

66
Q

describe the importance of an enzyme’s tertiary structure as related to its function

A
  • specific AA sequence determines bond placement in tertiary structure (disulfide and ionic)
  • so different attractions between AAs form which affect the way polypeptide chain twist and folds, so different tertiary structure therefore different active site
  • this is why enzymes have high specificity, so different active site no longer complementary to substrate
  • therefore no formation of E-S complex
67
Q

describe two ways the impact of an enzyme on a reaction can be measured

A
  1. measure formation of product

2. disappearance of substrate

68
Q

Describe how to test for lipids in a sample

A
  1. Dissolve solid samples in ethanol.
  2. Add an equal volume of water and
    shake.
  3. Positive result: milky white emulsion
    forms
69
Q

How do triglycerides form?

A

condensation reaction between 1 molecule of glycerol & 3 fatty acids forms ester bonds

70
Q

Contrast saturated and unsaturated fatty acids

A 
Saturated:
● Contain only single bonds
● Straight-chain molecules
have many contact points
● Higher melting point = solid
at room temperature
● Found in animal fats
Unsaturated:
● Contain C=C double bonds
● ‘Kinked’ molecules have
fewer contact points
● Lower melting point = liquid
at room temperature
● Found in plant oils
71
Q

Relate the structure of triglycerides to their functions

A

● High energy:mass ratio = high calorific value from
oxidation (energy storage).
● Insoluble hydrocarbon chain = no effect on water
potential of cells & used for waterproofing.
● Slow conductor of heat = thermal insulation e.g.
adipose tissue (stores energy in form of fat)
● Less dense than water = buoyancy of aquatic
animals (i.e. they won’t sink)

72
Q

Describe the structure and function of phospholipids

A

Amphipathic molecule (has polar and non-polar parts):
glycerol backbone attached to 2 hydrophobic fatty acid tails & 1 hydrophilic polar phosphate head.
● Forms phospholipid bilayer in water =
component of membranes.
● Tails can splay outwards = waterproofing

73
Q

Compare phospholipids and triglycerides

A 
● Both have glycerol backbone.
● Both may be attached to a mixture of
saturated, monounsaturated &
polyunsaturated fatty acids.
● Both contain the elements C, H, O.
● Both formed by condensation reactions
74
Q

Contrast phospholipids and triglycerides

A

phospholipids:
● 2 fatty acids & 1 phosphate group attached
● Hydrophilic head & hydrophobic tail
● Used primarily in membrane formation

triglycerides:
● 3 fatty acids attached
● Entire molecule is hydrophobic
● Used primarily as a storage molecule (oxidation releases energy)

75
Q

Are phospholipids and triglycerides polymers?

A

No; they are not made from a small
repeating unit (monomers). They are
macromolecules.

76
Q

A condensation reaction between glycerol and a fatty acid (RCOOH) forms what?

A

an ester bond

77
Q

how many ester bonds in one triglyceride?

A

3

78
Q

The R-group of a fatty acid may be what two things?

A

saturated or unsaturated

79
Q

draw the condensation reaction between a glycerol molecule and three fatty acids to produce a triglyceride

A
  • search google images for triglyceride formation *
80
Q

Describe how amino acids join to form a polypeptide so there is always NH2 at one end and COOH at the other end

A
  • NH2 group joins to COOH group to form a peptide bond

- so in chain there’s a free NH2 group at one end and a free COOH group at the other

81
Q

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

Give two named examples of polymers and their associated monomers to illustrate your answer (5 marks)

A
  1. A condensation reaction joins monomers together and forms a (chemical) bond and releases water
  2. A hydrolysis reaction breaks the (chemical) bond between monomers and uses water
  3. A suitable example of polymers and monomers from which they are made
  4. A second suitable example of polymers and the monomers from which they are made
  5. Reference to a correct bond within a named polymer
Suitable Example =
Amino Acids --> Polypeptide/Protein/Enzyme/Antibody
Nucleotides --> Polynucleotide/DNA/RNA
Alpha Glucose --> Starch/Glycogen
Beta Glucose --> Cellulose
82
Q

Describe how the structures of starch and cellulose molecules are related to their functions. (5 marks)

A

Starch (max 3)

  1. Helical/spiral shape so compact;
  2. Molecule is insoluble so osmotically inactive (does not affect water potential)
  3. Branched so glucose is easily accessible by enzymes to break down for respiration;
  4. Large molecule so cannot leave cell/cross cell-surface membrane

Cellulose (max 3)

  1. Long, straight & unbranched chains of β glucose;
  2. Joined by hydrogen bonding, to form (micro/macro) fibrils
  3. These provide rigidity/strength;
83
Q 
Describe the structure of a cellulose molecule and explain how cellulose is adapted for its function
in cells (6 marks)
A
  1. made from β-glucose;
  2. joined by condensation to form glycosidic bond;
  3. 1 : 4 link described;
  4. “flipping over” of alternate molecules;
  5. hydrogen bonds linking long straight chains;
  6. cellulose makes cell walls strong;
  7. can resist turgor pressure/osmotic pressure;
  8. bond difficult to break;
  9. resists action of enzymes
84
Q

Describe the structure of proteins (5 marks)

A
  1. Polymer of amino acids;
  2. Joined by peptide bonds;
  3. That are formed by condensation;
  4. Primary structure is the order of amino acids;
  5. Secondary structure is folding of polypeptide chain due to hydrogen bonding;
  6. Tertiary structure is 3-D folding due to hydrogen bonding and ionic / disulfide
    bonds;
  7. Quaternary structure is two or more polypeptide chains
85
Q

Describe competitive and non-competitive inhibition of an enzyme (5 marks)

A
  1. Inhibitors reduce binding of enzyme to substrate & prevent the formation of E-S complexes

(Competitive inhibition),

  1. Inhibitor has a similar shape to substrate;
  2. It binds to the active site (of enzyme);
  3. Inhibition can be overcome by adding more substrate;

(Non-competitive inhibition),

  1. Inhibitor binds to a site on enzyme other than active site;
  2. This changes the shape of the active site
  3. Inhibition cannot be overcome by adding more substrate
86
Q

Describe the role of the enzymes of the digestive system in the complete breakdown of starch
(5 marks)

A

Amylase (mouth);
Breaks down starch to maltose

Maltase (Small Intestine);
Breaks down maltose to glucose
Hydrolysis of starch involves breaking of glycosidic bond

87
Q

The epithelial cells that line the small intestine are adapted for the absorption of glucose.
Explain how. (6 marks)

A
  1. Microvilli provide a large surface area;
  2. Many mitochondria produce ATP for active transport;
  3. Carrier proteins present for active transport;
  4. Channel / carrier proteins for facilitated diffusion;
  5. Co-transport of sodium and glucose achieved through carrier protein for sodium (ions) and glucose;
  6. Membrane-bound enzymes digest disaccharides to produce glucose;
88
Q

Compare and contrast the structure and properties of triglycerides and phospholipids [5 Marks]

A

compare:

  • both contain ester bonds
  • both contain glycerol
  • fatty acids on both could be saturated or unsaturated
  • both are insoluble
  • both contain C, H and O but phospholipids also contain P

contrast:

  • triglyceride has 3 fatty acids and phospholipid has 2 fatty acids plus phosphate group
  • triglycerides are hydrophobic and phospholipids have hydrophilic and hydrophobic region
  • phospholipids form a monolayer (on surface)/bilayer (in water) but triglycerides don’t

AS Biology (9 decks)

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