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A-Level Biology AQA > Biological Molecules > Flashcards

Biological Molecules Flashcards

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1
Q

What’s a polymer?

A

Long molecule made from smaller molecules(monomers)

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2
Q

What’s a monomer?

A

Small molecules which are the building blocks of polymers

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3
Q

Bonds in a carbohydrate?
fats/lipids?
protein?
nucleic acids?

A

Glycosidic
Ester
Peptide
Phosphodiester

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4
Q

Monomer/polymer of a carbohydrate?

A

Monosaccharide

Polysaccharide

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5
Q

Monomer/polymer of a protein

A

Amino Acid

Polypeptide/protein

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6
Q

Monomer/polymer of fats/lipids?

A

Monomer: Triglyceride

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7
Q

Monomer/polymer of nucleic acids?

A

Nucleotide

DNA/RNA

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8
Q

Examples of monosaccharides

A

Glucose
Fructose
Galactose

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9
Q

Formula for monosaccharides?

A

C6H12O6

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10
Q

Example of disaccharides?

A

glucose + glucose = maltose
glucose + galactose = lactose
glucose + fructose = sucrose

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11
Q

Formula for disaccharides?

A

C12H22O11

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12
Q

What’s condensation?

A

A chemical bond forms between 2
molecules & a molecule of water is
produced.

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13
Q

What’s hydrolysis?

A

A water molecule is used to break a

chemical bond between 2 molecules.

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14
Q

What’s an organic molecule?

A

Contains carbon

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15
Q

Difference between alpha and beta glucose?

A

Hydroxyl group is flipped
on alpha glucose OH is on the bottom
on beta glucose the OH is on the top

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16
Q

How is glucose used as energy storage?

A

Glucose units contain a lot of bonds that can be broken down to release energy during respiration to create ATP.

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17
Q

Functions of carbohydrates?

A

–respiratory substrates

  • provides energy
  • stores energy
  • Glycoproteins/ receptors
  • builds macromolecules
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18
Q

What’s cellulose?

A

polymer of 𝛽-glucose gives rigidity to plant cell walls

prevents bursting under turgid pressure, holds stem up

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19
Q

What’s glycogen?

A

main storage polymer of 𝛼-glucose in animal cells

but also found in plant cells

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20
Q

What’s starch?

A

storage polymer of 𝛼-glucose in plant cells

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21
Q

Structure and functions of

cellulose

A

-1,4 glycosidic bonds-straight-chain, unbranched molecule
-alternate glucose molecules are rotated 180°
-H-bonds between parallel strands form
microfibrils -provides strength and structure for cell wall

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22
Q

Structure and functions of

glycogen

A

-1,4 & 1,6 glycosidic bonds
-branched = many terminal ends for faster hydrolysis,increases SA
-insoluble = no osmotic effect & does not diffuse
out of cells
-compact

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23
Q

Structure and functions of

starch

A

-insoluble = no osmotic effect on cells
-large = does not diffuse out of cells
Amylopectin:
-1,4 & 1,6 glycosidic bonds
-branched = many terminal
-ends for hydrolysis into glucose
Amylose:
-1,4 glycosidic bonds
-helical = compact

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24
Q

Where is glucose stored in the body?

A

As glycogen in the muscles/liver

25
Q

Similarities between starch and glycogen

A 
- both made of α-glucose
monomers
-both contain branched
chains
-insoluble, do not affect
osmotic potential
-Compact – maximum
storage
26
Q

Compare Cellulose and Amylose

A

-Glycosidic Bonds
-Insoluble in water
However, Cellulose is composed of β Glucose whereas Amylose is composed of α Glucose
Cellulose forms long straight chains, Amylose forms a coiled chain and tends to be found in granules.
Amylose is used as an energy storage in starch, whereas Cellulose plays a structural role.

27
Q

Describe the test for reducing

sugars

A

-Add an equal volume of Benedict’s reagent
to a sample.
-Heat the mixture in an electric water bath at
100℃ for 5 mins.
-Positive result: colour change from blue to
orange & brick-red precipitate forms.

28
Q

Describe the test for

non-reducing sugars

A
  • Negative result: Benedict’s reagent remains blue
  • Hydrolyse non-reducing sugars e.g. sucrose into their monomers by adding 1cm3 of HCl. Heat in a boiling water bath for 5 mins.
  • Neutralise the mixture using sodium carbonate solution.
  • Proceed with the Benedict’s test as usual
29
Q

Describe the test for starch

A

Add iodine, turns blue/black

30
Q

Describe the test for lipids

A

Add ethanol, then add water;

White (emulsion shows lipid)

31
Q

Describe the test for proteins

A
  • Sample is mixed with an equal volume of sodium hydroxide at room temperature
  • Add a few drops of dilute copper sulphate solution then mix
  • Purple colour change if not then stays blue
32
Q

Uses of lipids in animals

A
  • Storage as fat
  • Hormones – steroids
  • Cell membranes
  • Thermal Insulation
  • Electrical insulation (myelin)
  • Waterproofing
33
Q

Describe the structure of a triglyceride

A
  • One glycerol molecule
  • Attached to three fatty acid chains
  • Glycerol head is hydrophilic –interacts with water
  • Fatty acid tails are hydrophobic –repel water
34
Q

Contrast saturated and unsaturated fatty

acids

A

Saturated-Contain only single bonds-straight-chain
Higher melting point = solid at room temperature
Unsaturated -Contain C=C double bonds -‘Kinked’ molecules ,Lower melting point = liquid
at room temperature

35
Q

structure and functions of triglycerides

A

-Energy storage molecule
-Insoluble hydrocarbon chain = no effect on water
potential of cells & used for waterproofing.
-Slow conductor of heat = thermal insulation

36
Q

structure and function of

phospholipids

A

Glycerol backbone attached to 2 hydrophobic fatty acid tails & 1 hydrophilic polar phosphate head
Forms phospholipid bilayer in water = membranes.
Tails = waterproofing.

37
Q

Compare phospholipids and triglycerides

Similarities

A

Both have :

  • glycerol backbone.
  • elements C, H, O.
  • been formed by condensation reactions.
38
Q

Contrast phospholipids and triglycerides

differences

A 
phospholipids:
-2 fatty acids & 1 phosphate group attached
-Hydrophilic head &hydrophobic tail 
-Use:in membrane formation
triglycerides:
-3 fatty acids attached
-Entire molecule is hydrophobic
-Use:storage molecule (oxidation releases energy)
39
Q

Are phospholipids and triglycerides polymers?

A

No.They are macromolecules.

40
Q

How does the presence of a double bond in a lipid affect its boiling point?

A

Lowers the boiling point as double bonds in unsaturated fats cause kinks in fatty acid chain which prevent the molecules from packing closely together

41
Q

How is the structure of a phospholipid different from the structure of a triglyceride?

A

One of the fatty acids tails is replaced with a phosphate molecule

42
Q

What are the 3 types of Lipids?

A
  • Triglycerides (fat for energy store, insulation, protection of organs)
  • Phopholipids (to make membranes)
  • Cholesterol (for membrane stability and make hormones)
43
Q

What are globular proteins?

A

soluble proteins with a specific 3D shape e.g. enzymes, hormones,antibodies, haemoglobin

44
Q

What are fibrous proteins?

A

strong/insoluble/inflexible material e.g. collagen and keratin

45
Q

How do amino acids differ?

A

have different R groups,there are 20 different amino acids

46
Q

Define primary, secondary, tertiary, quaternary structure?

A

-Primary = sequence of AA, polypeptide chain (held by peptide bonds)
-Secondary = the primary structure (polypeptide chain) coils to form a helix, held by hydrogen bonds
-Tertiary = secondary structure folds again to form final 3d shape, held together by
hydrogen/ionic/disulfide bonds
-Quaternary = made of more then one polypeptide chain

47
Q

Outline how chromatography could be used to

identify the amino acids in a mixture

A

-Use capillary tube to spot mixture onto pencil origin line &
place chromatography paper in solvent.
-Allow solvent to run until it almost touches other end of paper. Amino acids move different distances based on relative attraction to paper & solubility in solvent.
-Use UV light to see spots.
-Calculate Rf values & match to database

48
Q

What are enzymes?

A

-Biological catalysts for cellular reactions.
-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.

49
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

50
Q

Explain the induced fit model of enzyme

action

A

-Shape of active site is not directly complementary
to substrate is flexible-enables ES complexes to
form.

51
Q

What’s the lock and key model?

A

rigid shape of active site complementary to only 1 substrate.

52
Q

Name 5 factors that affect the rate of

enzyme-controlled reactions

A
  • enzyme concentration
  • substrate concentration
  • concentration of inhibitors
  • pH
  • temperature
53
Q

How does substrate concentration affect rate of

reaction?

A
  • increase substrate concentration, increases chance of successful collisions, increase chance of forming an
    ES complex, increase rate of reaction
  • this continues until all the enzyme’s active sites are full/saturated = maximum rate of reaction-rate then levels off
54
Q

Effect of enzyme concentration on enzyme activity?

A
  • increase enzyme concentration, increases chance of successful collisions, increase chance of forming an ES complex, increase rate of reaction
  • this continues until all the substrates are used up = maximum rate of reaction
55
Q

Effect of temperature on enzyme activity?

A
  • as temperature increases
  • the KE increases as the molecules move faster
  • increase chance of successful collisions
  • increase chance of forming ES complex
  • increase rate of reaction
  • carries on till optimum
  • after optimum-bonds in tertiary structure break(hydrogen and ionic bonds)
  • lose active site shape ,substrate no longer complementary
  • cant form ES complexes
  • enzyme denatured
56
Q

Effect of pH on enzyme activity?

A
  • change pH away from optimum, bonds in tertiary structure break, lose active site shape, can no longer form ES complex,
  • enzyme denatured
57
Q

Competitive vs Non-Competitive Inhibitors?

A

-Competitive = a substance with a similar shape to the substrate and a complementary shape to the
enzyme’s active site, binds to the active site, blocking it, preventing ES complexes from forming
-Non-Competitive = a substance that binds to another site on the enzyme other then the active site, causes
the active site to change shape, so less ES complexes can form

58
Q

How to calculate rate of reaction from raw

data/graph

A

Change in concentration of product or
reactant / time.
Graph:Draw a tangent

A-Level Biology AQA (1 decks)

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