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Biological Molecules (Topic 1) Flashcards

1
Q

Define monomer. Give some examples

A

Smaller units that join together to form larger molecules
•monosaccharides (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
•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 a hydrolysis 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
Same formula- C6H12O6

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6
Q

Name the bond formed when monosaccharides react.

A

(1,4 or 1,6) glycosidic bonds
2 monomers = 1 bond = disaccharide
Multiple monomers = many bonds = polysaccharide

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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 C12H22O11

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8
Q

Draw the structure of ⍺-glucose.

A

CH2OH
|_________O
H / | \ H
| / H |
| \ OH H /|
OH |________|/ OH
| |
H OH

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9
Q

Draw the structure of 𝛽-glucose.

A

CH2OH
|_________O
H / | \ OH
| / H |
| \ OH H /|
OH |________|/ H
| |
H OH

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10
Q

Describe the structure and functions of
starch.

A

Storage polymer of 𝛼-glucose in plant cells
• insoluble = no osmotic effect on cells
• large = does not diffuse out of cells
Made from amylopectin:
• 1,4 & 1,6 glycosidic bonds
• branched = many terminal
ends for hydrolysis into
glucose
And amylose:
• 1,4 glycosidic bonds
• helix with intermolecular
H-bonds = compact

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11
Q

Describe the structure and functions of
glycogen.

A

main storage polymer of 𝛼-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 𝛽-glucose gives rigidity to plant cell walls
(prevents bursting under turgor pressure, holds stem up)
• 1,4 glycosidic bonds
• straight-chain, unbranched molecule
• alternate glucose molecules are rotated 180°
• H-bond 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℃ for 5 mins.
  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 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

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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 storage).
• Insoluble hydrocarbon chain = no effect on water
potential of cells & used for waterproofing.
• 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 &
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)

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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
Why is water a polar molecule?
O is more electronegative than H, so attracts the electron density in the covalent bond more strongly. forms O 𝛿- (slight negative charge) & H 𝛿+ (slight positive charge).
26
State 4 biologically important properties of water.
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.
27
Explain why water is significant to living organisms.
• Solvent for polar molecules during metabolic reactions. • Enables organisms to avoid fluctuations in core temperature. • Cohesion-tension of water molecules in transpiration stream.
28
What are inorganic ions and where are they found in the body?
• Ions that do not contain carbon atoms. • Found in cytoplasm & extracellular fluid. • May be in high or very low concentrations.
29
Explain the role of hydrogen ions in the body.
• 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
Explain the role of iron ions in the body.
**Fe2+** 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
Explain the role of sodium ions in the body.
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
Explain the role of phosphate ions in the body.
component of: • DNA • ATP • NADP (Topic 5.1) • cAMP (Topic 6.4)
33
What is the general structure of an 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
Describe how to test for proteins in a sample.
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.
35
How many amino acids are there and how do they differ from one another?
20 differ only by side ‘R’ group
36
How do dipeptides and polypeptides form?
• Condensation reaction forms peptide bond (-CONH-) & eliminates molecule of water • Dipeptide: 2 amino acids • Polypeptide: 3 or more amino acids
37
How many levels of protein structure are there?
4; primary, secondary, tertiary and quaternary
38
Define ‘primary structure’ of a protein.
• Sequence, number & type of amino acids in the polypeptide. • Determined by sequence of codons on mRNA.
39
Define ‘secondary structure’ of a protein.
Hydrogen bonds form between O 𝛿- (slightly negative) attached to ‒C=O & H 𝛿+ (slightly positive) attached to ‒NH.
40
Describe the 2 types of secondary protein structure.
α-helix: • all N-H bonds on same side of protein chain • spiral shape • H-bonds parallel to helical axis β-pleated sheet: • N-H & C=O groups alternate from one side to the other
41
Define ‘tertiary structure’ of a protein. Name the bonds present.
3D structure formed by further folding of polypeptide • disulfide bridges • ionic bonds • hydrogen bonds
42
Describe each type of bond in the tertiary structure of proteins.
• 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
43
Define ‘quaternary structure’ of a protein.
• 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
Describe the structure and function of globular proteins.
• Spherical & compact. • Hydrophilic R groups face outwards & hydrophobic R groups face inwards = usually water-soluble. • Involved in metabolic processes e.g. enzymes & haemoglobin.
45
Describe the structure and function of fibrous proteins.
• Can form long chains or fibres • insoluble in water. • Useful for structure and support e.g. collagen in skin.
46
Outline how chromatography could be used to identify the amino acids in a mixture.
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
What are enzymes?
• 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
Explain the induced fit model of enzyme action.
• 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
How have models of enzyme action changed?
• 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
How could a student identify the activation energy of a metabolic reaction from an energy level diagram?
Difference between free energy of substrate & peak of curve.
51
Name 5 factors that affect the rate of enzyme-controlled reactions.
• enzyme concentration • substrate concentration • concentration of inhibitors • pH • temperature
52
How does substrate concentration affect rate of reaction?
Given 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
How does enzyme concentration affect rate of reaction?
Given that substrate is in excess, rate increases proportionally to enzyme concentration Rate levels off when maximum number of ES complexes form at any given time.
54
How does temperature affect rate of reaction?
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
How does pH affect rate of reaction?
Enzymes have a narrow optimum pH range. Outside range, H+ / OH- ions interact with H-bonds & ionic bonds in 3° structure = denaturation.
56
Contrast competitive and non-competitive inhibitors.
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 Non- competitive inhibitors: • bind at allosteric binding site • may permanently stop reaction; triggers active site to denature • increasing substrate concentration has no effect on their effect
57
Outline how to calculate rate of reaction from a graph.
• calculate gradient of line or gradient of tangent to a point. • initial rate: draw tangent at t = 0.
58
Outline how to calculate rate of reaction from raw data.
Change in concentration of product or reactant / time.
59
Why is it advantageous to calculate initial rate?
Represents maximum rate of reaction before concentration of reactants decreases & ‘end product inhibition’.
60
State the formula for pH.
pH = -log10[H+]
61
Name the pentose sugars in DNA & RNA.
DNA: deoxyribose RNA: ribose
62
State the role of DNA in living cells.
Base sequence of genes codes for functional RNA & amino acid sequence of polypeptides. Genetic information determines inherited characteristics = influences structure & function of organisms.
63
State the role of RNA in living cells.
**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
64
How do polynucleotides form?
Condensation reactions between nucleotides form strong phosphodiester bonds (sugar-phosphate backbone).
65
Describe the structure of DNA.
**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)
66
Which bases are purine and which are pyrimidine?
A & G = **2-ring purine bases** T & C & U = **1-ring pyrimidine bases**
67
Name the complementary base pairs in DNA.
2 H-bonds between adenine (A) + thymine (T) 3 H-bonds between guanine (G) + cytosine (C)
68
Name the complementary base pairs in RNA.
2 H-bonds between adenine (A) + uracil (U) 3 H-bonds between guanine (G) + cytosine (C)
69
Relate the structure of DNA to its functions.
• 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 • weak H-bonds break so strands separate for replication
70
Describe the structure of messenger RNA (mRNA).
•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.
71
Relate the structure of messenger RNA (mRNA) to its functions.
• Breaks down quickly so no excess polypeptide forms. • Ribosome can move along strand & tRNA can bind to exposed bases. • Can be translated into a specific polypeptide by ribosomes.
72
Describe the structure of transfer RNA (tRNA).
• Single strand of about 80 nucleotides. • Folded into clover shape (some paired bases). • Anticodon on one end, amino acid binding site on the other: a) anticodon binds to complementary mRNA codon b) amino acid corresponds to anticodon
73
Order DNA, mRNA and tRNA according to increasing length.
tRNA mRNA DNA
74
Why did scientists initially doubt that DNA carried the genetic code?
Chemically simple molecule with few components.
75
Why is DNA replication described as ‘semiconservative’?
• Strands from original DNA molecule act as a template. • New DNA molecule contains 1 old strand & 1 new strand.
76
Outline the process of semiconservative DNA replication.
1. DNA helicase breaks H-bonds between base pairs. 2. Each strand acts as a template. 3. Free nucleotides from nuclear sap attach to exposed bases by complementary base pairing. 4. DNA polymerase catalyses condensation reactions that join adjacent nucleotides on new strand. 5. H-bonds reform.
77
Describe the Meselson-Stahl experiment.
1. Bacteria were grown in a medium containing heavy isotope 15N for many generations. 2. Some bacteria were moved to a medium containing light isotope 14N. Samples were extracted after 1 & 2 cycles of DNA replication. 3. Centrifugation formed a pellet. Heavier DNA (bases made from 15N) settled closer to bottom of tube.
78
Explain how the Meselson-Stahl experiment validated semiconservative replication.
Gen 0: all 15N Gen 1: band moved to in between where 15N and 14N would be, as they have one of each strand Gen 2: a new band is seen at 14N, 50% have 2 14N strands, 50% have one 14N strand and one 15N strand
79
Describe the structure of adenosine triphosphate (ATP).
nucleotide derivative of adenine with 3 phosphate groups
80
Explain the role of ATP in cells.
**ATP hydrolase** catalyses **ATP → ADP + Pi** • Energy released is **coupled** to metabolic reactions. • Phosphate group **phosphorylates** compounds to make them more reactive.
81
How is ATP resynthesised in cells?
• **ATP synthase** catalyses condensation reaction between ADP & Pi • during photosynthesis & respiration
82
Explain why ATP is suitable as the ‘energy currency’ of cells.
• High energy bonds between phosphate groups. • Small amounts of energy released at a time = less energy wasted as heat. • Single-step hydrolysis = energy available quickly. • Readily resynthesised.

Biology Revision (9 decks)

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