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BIOLOGY, AQA > BIOLOGICAL MOLECULES > Flashcards

BIOLOGICAL MOLECULES Flashcards

1
Q

Define monomer. Give some examples

A 
smaller units that join together to form larger molecules
some examples are : 
- glucose, fructose, galactose
- amino acids
-nucleotides
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2
Q

Define polymer. Give some examples

A 
molecules formed when many monomers join together
some examples are :
- polysaccharides
- proteins
- DNA/RNA
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3
Q

What happens in a condensation reaction?

A

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

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4
Q

What happens in hydroysis reaction?

A

A water molecule is used to break a chemical bond between 2 molecules

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5
Q

Name the 3 hexose monosaccharides?

A

-glucose
- fructose
-galactose
all have the molecular formula C6 H12 06

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6
Q

Name the type of bond formed when monosaccharides react.

A
  • (1,4 or 1,6) glycosidic bond
    2 monomers = 1 chemical bonds= disaccharide
    multiple monomers = many chemical bonds = polysacharide
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7
Q

Name 3 disaccharides. Describe how they form.

A 
condensation reaction forms glycosidic bond between 2 monosaccharides 
- maltose: glucose + glucose
- sucrose: glucose + fructose
-lactose: glucose + galactose
all have molecular formula C12 H22 O11
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8
Q

Draw the structure of a- glucose

A

Oh is down

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9
Q

Draw the structure of B - glucose

A

OH is up

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10
Q

Describe the structure and functions of Starch

A
  • storage polymer a-glucose in plant cells
  • insoluble = no osmotic effect on cells
  • large= does not diffuse out of cells
  • made from amylose and amylopectin:
    Amylose; - 1,4 glycosidic bonds
    - helix with intermolecular
    - hydrogen bonds = compact
    Amylopectin; - 1,4 and 1,6 glycosidic bonds
    - branch = many terminal ends for
    hydrolisis into glucose
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11
Q

Describe the structure and functions of Glycogen

A
  • main storage polymer a-glucose in animal cells (but also found in plant cells)
  • 1.4 & 1,6 glycosidic bonds
  • branched= many terminal ends for hydrolysis
  • insoluble = no osmotic effect & does not diffuse out of cells
  • compact
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12
Q

Describe the structure and functions of cellulose

A
  • polymer of b-glucose gives rigidity to plant cells walls (prevents bursting under turgor pressure, holds stem up)
  • 1,4 glycosidic bonds
  • straight chain, unbranched molecule
  • alternate glucose molecules are rotated 180 degrees
  • H-bonds crosslinks between parallel strands form microfibrils= high tensile strength
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13
Q

Describe the Benedict’s test for reducing sugars

A
  1. Add an equal volume of Benedict’s reagent to a sample
  2. Heat the mixture in an electric water bath at 100 degrees, for 5 minutes
  3. Positive result: colour change from blue to orange & brick- red precipitate forms
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14
Q

Describe the Benedict’s test for non-reducing sugars

A
  1. Negative result: Benedict’s reagent remains blue
  2. Hydrolyse non-reducing sugars e.g sucrose into their monomers by adding 1cm3 of HCl. Heat in a boiling water bath for 5 mins
  3. Neutralise the mixture using sodium carbonate solution
  4. Proceed with the Benedict’s test as usual
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15
Q

Describe the test for starch

A
  1. Add iodine solution

2. Positive result: colour change from orange to blue-black

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16
Q

Outline how colorimetry could be used to give qualitative results for the presence of sugars and starch

A
  1. Make standard solutions with known concentrations.
    Record absorbance or % transmission values
  2. Plot calibration curve: absorbance or % transmission (y-axis), concentration (x-axis)
  3. Record absorbance or % transmission values of unknown samples. Use calibration curve to read off concentration
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17
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
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18
Q

How do triglycerides form?

A

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

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19
Q

Contrast saturated and unsaturated fatty acids

A

SATURATED:

  • Contain only single bonds
  • Straight - chain molecules have many contacts 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
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20
Q

Relate the structure of triglycerides to their functions

A
  • High energy : mass ratio = high calorific value from oxidation (energy store)
  • Insoluble hydrocarbon chain = no effect on water
  • Slow conductor of heat = thermal insulation e.g adipose tissue
  • Less dense than water = buoyancy of aquatic animals
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21
Q

Describe the structure and function of phospholipids

A

Amphipathic molecule: 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
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22
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
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23
Q

Contrast phospholipids and triglycerides

A

Phospholipids:

  • 2 fatty acids & 1 phosphate group attached
  • Hydrophilic head & hydrophilic tail
  • Used primarily in membrane formation

Triglycerides:

  • 3 fatty acids attached
  • Entire molecule is hydrophobic
  • Used primarly as a storage molecule (oxidation releases energy)
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24
Q

Are phospholipids and triglycerides polymers?

A

No, they are not made from a small repeating unit. They are macromolecules

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25
Q

Why is water a polar molecule?

A
  • O is more electronegative than H, so attracts the electron density in the covalent bond more strongly forms O &- (slight negative charge) and H &+ (slight positive charge)
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26
Q

State 4 biologically important properties of water

A

Due to polarity & intermolecular H-bonds:

  • Metabolite/ solvent for chemical reactions in the body
  • high specific heat capacity
  • high latent heat of vapourisation
  • cohesion between molecules
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27
Q

Explain why water is significant to living organisms

A
  • Solvent for polar molecules during metabolic reactions
  • Enables organisms to avoid fluctuations in core temperature
  • Cohesion- tension of water molecules in transpiration stream
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28
Q

What are inorganic ions and where are they found in the body?

A
  • ions that do not contain carbon atoms
  • found in cytoplasm & extracellular fluid
  • may be in high or very low concentrations
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29
Q

Explain the role of hydrogen ions in the body

A
  • High concentration of H+ = low (acidic) pH

- H+ ions interact with H-bonds & ionic bonds in tertiary structure of proteins, which can cause them to denature

30
Q

explain the role of iron ions in the body

A
  • Fe 2+ bonds to porphyrin ring to form haem group in haemoglobin
  • Haem group has binding site to transport 1 molecule of O2 around body in bloodstream
  • 4 haem groups per haemoglobin molecule
31
Q

Explain the role of sodium ions in the body

A
  • Involved in co-transport for absorption of glucose & amino acids in lumen of gut (Topic 2.3)
  • Involved in propagation of action potentials in neurons (Topic 6.2)
32
Q

Explain the role of phosphate ions in the body

A

component of:

  • DNA
  • ATP
  • NADP( Topic 5.1)
  • cAMP ( Topic 6.4)
33
Q

What is the general structure of an amino acid?

A 
structure of amino acid 
- COOH carboxyl/ carboxylic acid group
- R variable side group consists of carbon chain & may include other functional groups e.g benzene 
ring or -OH (alcohol) 
- NH2 amine/ amino group
34
Q

Describe how to test for proteins in a sample

A

Biuret test confirms presence for a peptide bond
1. Add equal volume of sodium hydroxide to sample at room temperature
2. Add drops of dilute copper (ll) 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

35
Q

How many amino acids are there and how do they differ from one another?

A
  • 20

- differ only by side ‘R’ group

36
Q

How do dipeptides and polypeptides form?

A
  • Condensation reaction forms peptide bond (-CONH_) & eliminates molecule of water
  • Dipeptide: 2 amino acids
  • Polypeptides: 3 or more amino acids
37
Q

How many levels of protein structure are there?

A

4

38
Q

Define ‘primary structure’ of a protein

A
  • Sequence, number & type of amino acids in the polypeptide

- Determined by sequence of codons on mRNA

39
Q

Define ‘ secondary structure’ of a protein

A
  • Hydrogen bonds form between 0&- (slightly negative) attached to -C= O & H
  • &+ ( slightly positive) attached to -NH
40
Q

Describe the 2 types of secondary protein structure

A

a-helix
- all N-H bonds on same side of protein chain
- spiral shade
- H-bonds parallel to helical axis
b-pleated sheet
- N-H & C=O groups alternate from one side to the other

41
Q

Define ‘tertiary structure’ of a protein.

Name the bonds present

A

3D structure formed by further folding of polypeptide

  • disulfide bridges
  • ionic bonds
  • hydrogen bonds
42
Q

Describe each type of bond in the tertiary structure of proteins

A
  • Disulfide bridges = strong covalent bonds 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
43
Q

Define ‘quaternary structure’ of a protein

A
  • Functional proteins may consist of more than one polypeptide
  • Precise 3D structure held together by the same types of bond as tertiary structure
  • May involve addition of prosthetic groups e.g metal ions or phosphate groups
44
Q

Define 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 & haemoglobin
45
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
46
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 & 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 & solubility in solvent
  3. Use revealing agent or UV light to see spots
  4. Calculate Rf values & match to database
47
Q

What are enzymes

A
  • Biological catalysts for intra & extracellular 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.
48
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
49
Q

How have models of enzyme action changed?

A
  • Initially lock & 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
50
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
51
Q

Name 5 factors that affect the rate of enzyme-controlled reactions

A
  • enzyme concentration
  • substrate concentration
  • concentration of inhibitors
  • pH
  • temperature
52
Q

How does substrate concentration affect rate of reaction?

A
  • Give, that enzyme concentration is fixed, rate increases proportionally to substrate concentration
  • Rate levels off when maximum number of ES complexes form at any given time
53
Q

How does enzyme concentration effect rate of reaction ?

A

Give that substrate is in excess, rate increases proportioanlly to enzyme concentration
Rate levels off when maximum number of ES complexes form at any given time

54
Q

How does temperature affect rate of reaction?

A
  • Rate increases as kinetic energy increases & peaks at optimum temperature
  • Above optimum, ionic & H-bonds in 3* structure break= active site no longer complementary to substrate (denaturation)
55
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

56
Q

Contrast competitive and non-competitive inhibitors

A

Competitive inhibitors

  • similar shape to substrate = bind to active site
  • do not stop reaction ; ES complex forms when inhibitor is released
  • increasing substrate concentration decreases their effect
57
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 t=0
58
Q

Outline how to calculate rate of reaction from raw data

A

Change in concentration of product or reactant/ time

59
Q

Why is it advantageous to calculate initial rate?

A
  • Represents maximum rate o reaction before concentration of reactants decreases & ‘ end product inhibition’
60
Q

State the formula of pH

A

pH= -log10 [H=]

61
Q

Draw the structure of a nucleotide

A
  • phosphte group (cercle)
  • pentose sugar (pentose)
  • nitrogen containing base ( rectangle)
62
Q

Name the pentose sugars in DNA & RNA

A

DNA: deoxyribose
RNA: ribose

63
Q

State the role DNA in living cells

A
  • Base sequence of genes codes for fucntional RNA & amino acid sequence of polypeptides
  • Genetic information determines inherited characteristics = influences structure & function of organisms
64
Q

State the role of RNA in living cells

A

mRNA: Complementary sequence to 1 gene from DNA with introns ( non-coding regions) spliced out. Codons can be translated into a polypeptide by ribosomes
rRNA: Component of ribosomes (along with proteins)
tRNA: supplies complementary amino acid to mRNA codons during translation

65
Q

How do polynucleotides form?

A

Condensation reaction between nucleotides form strong phosphodiester bonds ( sugar-phosphate backbone)

66
Q

Describe the structure of DNA

A
  • double helix of 2 polynucleotide strands (deoxyribose)
  • H- bonds between complementary purine & pyrimidine base pairs on opposite strands
    Adenine ( A) + Thymine (T)
    Guanine (G) + cytosine (C)
67
Q

Which bases are purine and which bases are pyrimidine?

A

A & G = 2-ring purine bases

T & C & U = 1-ring pyrimidine bases

68
Q

Name the complementary base in pairs

A

2 H-bonds between adenine (A) + thymine (T)

3 H- bonds between guanine (G) + cytosine (C)

69
Q

Name the complementary base in pairs in RNA

A

2 H-bonds between adenine (A) + uracil (U)

3 H-bonds between guanine (G) + cytosine (C)

70
Q

Relate the structure of DNA to its functions

A
  • sugar-phosphate backbone & many H-bonds provide stability
  • long molecule stores lots of information
  • helix is compact for storage in nucleus
  • base sequence of triplets codes for amino acids
  • double-stranded for semi-conservative replication
  • complementary base pairing for accurate replication
  • week H-bonds break so strands separate for replication
71
Q

Describe the structure of messenger RNA (mRNA)

A
  • Long ribose polynucleotide (but shorter than DNA)
  • Contains uracil instead of thymine
  • Single-stranded & linear ( no complementary base pairing)
  • Codon sequence is complementary to exons of 1 gene from 1 DNA strand

BIOLOGY, AQA (4 decks)

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