Unit 1 Topic 1 Biological Molecules Flashcards
(48 cards)
1
Q
Explain how the properties of water make it an ideal transport medium
A
- Water is a solvent / dissolves substances
- Because water molecules surround polar molecules by forming hydrogen bonds between water molecules and solute molecules
- Water is a liquid so there is cohesion between water molecules
2
Q
How are primary proteins formed
A
- Primary: sequence of amino acids joined by peptide bonds in a polypeptide chain
- Peptide bonds between the carboxyl group of one amino acid and the amino group of another amino acid
3
Q
How are secondary proteins formed
A
- Secondary: coiling of polypeptide chain into an α-helix / folding of polypeptide chain into a β-pleated sheet
- Hydrogen bonds between oxygen in the carboxyl groups and hydrogen in the amino acids at different locations in polypeptide chain
4
Q
How are tertiary proteins formed
A
- Tertiary: 3D conformation of polypeptide resulted from folding of α-helix or β-pleated sheet due to hydrogen bonds, disulfide bonds, ionic bonds and hydrophobic interactions between side chains of amino acids
5
Q
How are quaternary proteins formed
A
- Quaternary: 3D conformation of protein consists of 2 or more polypeptide chains binding together
6
Q
What are the properties of starch?
A
- Many α-glucose joined by formation of glycosidic bonds by condensation
7
Q
Compare the properties of amylose and amylopectin
A
- Amylose: α-glucose joined by 1,4-glycosidic bond, unbranched and coils into spiral shape, less readily hydrolysed: stable energy storage molecule
- Amylopectin: α-glucose joined by 1,4-glycosidic bonds and some 1,6-glycosidic bonds, branched (1,6 connects amylose), more readily hydrolyzed (more enzymes catalyze hydrolysis) - more rapid supply of glucose as fuel for cellular respiration
8
Q
Properties of glycogen
A
- Energy storage molecule in animals
- Many α-glucose joined by 1,4-glycosidic bonds and many 1,6-glycosidic bonds
- readily hydrolyzed: allow more rapid supply of glucose as fuel for cellular respiration
9
Q
Properties of cellulose
A
- Constituent of plant cell walls -> strong building materials
- Unbranched, straight polymer of β-glucose joined by 1,4-glycosidic bonds by condensation
- Hydrogen bonds hold parallel cellulose molecules
10
Q
Compare and contrast the molecular structures of globular and fibrous proteins.
A
- Both are chains of amino acids joined by peptide bonds
- Both contain named bonds (i.e. hydrogen bonds, disulfide bonds, ionic bonds)
- Globular proteins have hydrophilic groups on the outside whereas fibrous proteins have hydrophobic groups on the outside
- Globular have tertiary or quaternary structures whereas fibrous have little or no tertiary structure
- Globular are folded into compact shapes whereas fibrous have long chains
11
Q
Explain the relationship between the composition of starch and the rate of hydrolysis by enzymes (hint: amylose content increases, percentage of starch hydrolyzed decreases)
A
- As amylose content increases, the percentage of starch hydrolyzed decreases
- The less amylose present, the greater the proportion of amylopectin
- Amylose is unbranched / amylopectin is branched
- Amylopectin contains (1,4- and) 1,6-glycosidic bond
- {Branches} increases the rate of hydrolysis
12
Q
Define ionic bond
A
strong attraction between oppositely charged ions
13
Q
Define covalent bond
A
- Strong attraction between atoms that share a pair of electrons in their outermost electron shell
- More electronegative atom pulls shared electrons in covalent bond more strongly towards itself
14
Q
Define polar bond
A
- Covalent bond with unequal share of electrons
- Polar molecules:
a) shared electrons in covalent bonds are not equally shared by atoms
b) Leads to separation of charges in the molecules
c) Dipole (molecule with partially positive and partially negative charges)
15
Q
Define hydrogen bonds
A
- the weak electrostatic force of attraction between a partially positive hydrogen atom in a polar molecule and an electronegative atom of another molecule
- Oxygen atom of one water molecule is joined to the hydrogen atom of another water molecule by a hydrogen bond
- Formed between intermolecular molecules, not within intramolecular
16
Q
Properties of water
A
- High specific heat capacity (absorb large amounts of energy to raise temperature)
- High latent heat of vapourization (energy needed for water to evaporate)
- High surface tension of water
- Lower density of ice than liquid water -> floating ice forms insulating layer to prevent remaining water from freezing
***5. Powerful solvent -> polar -> dissolve substances in water
- Incompressible (e.g. turgor pressure, hydrostatic pressure)
- Cohesion: water molecules attach with each other
- Adhesion: water molecules attach to other molecules (e.g. xylem)
17
Q
Name organic compounds and inorganic compounds
A
Organic:
- Nucleic acids
- Proteins
- Lipids
- Carbohydrates
Inorganic:
- Water
- Mineral ions / salts
18
Q
Organic compounds: examples of functional groups of organic compounds
A
HYDROXYL - polar -> more electronegative -> soluble
CARBOXYL -> polar -> more electronegative -> soluble
CARBONYL -> releases H+ -> acidic (pH drop)
AMINO -> accepts H+ -> basic (pH rise)
19
Q
Functions of monosaccharides
A
- Energy source for metabolism
- Major fuel for cellular respiration -> oxidized respiration: release energy stored in ATP (provide energy for metabolic need)
- Carbon skeletons for synthesis of other organic molecules
20
Q
alpha-glucose vs beta-glucose
A
2 Ring forms of glucose
- Alpha: hydroxyl group on carbon 1 is below the plane of the ring
- Beta: hydroxyl group on carbon 1 is above the plane of the ring
21
Q
Disaccharides
A
2 monosaccharides joined by formation of a glycosidic bond by condensation
22
Q
Why are polysaccharides good energy storage molecules?
A
1.Compact molecules -> large amount of energy can be stored in a small space
2.Insoluble
- No osmotic effect, doesn’t affect water potential
- When water potential of the cell is less than the water potential of the surrounding, there is a net movement of water molecules into cells by osmosis, causing the cell to burst
- As they are insoluble, they will not diffuse down the concentration gradient
3.Physically and chemically inactive -> will not affect other metabolic activities
23
Q
Principle of Benedict’s test
A
- To test for the presence of reducing sugars
- Cu2 + e- -> Cu+
- Reducing sugars can reduce blue copper 2+ ions in Benedict’s solution into C+ ions, which forms brick-red precipitate (Copper I oxide) -> insoluble
- Benedict’s solution should be in excess, and should be heated in boiling water bath
24
Q
Quantitative vs semi-quantitative vs qualitative
A
Quantitative: results in numerical values
- use calibrated instruments
- objective data
Semi-quantitative: provides approximate measurements (e.g. high, low)
- results are scaled or ranked
Qualitative: non-numerical observations
- descriptive results
25
Benedict's test result analysis
a) Comparison of blue colour intensity of filtrate
- Blue colour comes from Cu2+ ions in the Benedict's solution which are not reduced
b) Lower reducing sugar content in sample
c) Less Cu2+ ions in the Benedict's solution reduced by reducing sugars
d) More Cu2+ ions in Benedict's solution not reduced
e) Intensity of blue colour of the filtrate is higher
f) Absorbance of orange light (complementary to the blue colour)
g) Deeper colour, absorbs more light
- Higher absorbance, less reducing sugar content
26
Polyunsaturated properties
- 2/2+ C-C double bonds
a) Can cause the hydrocarbon chain in unsaturated fatty acids to bend / create kinks
b) Prevents the molecules from packing closely together
c) Smaller intermolecular force among the molecules
d) Less energy is needed to separate them
- The greater the number of C-C double bond, the lower the melting point of the fatty acid
- Fat has more saturated fatty acids, higher melting point than unsaturated
27
Solubility of globular protein
a) Hydrophilic, polar side chains of amino acids are usually on the surface of protein while
b) The hydrophobic, non-polar side chains of amino acids in globular proteins are usually at the centre of the protein
28
Collagen
- A fibrous protein consisting of 3 polypeptide A-chains held by hydrogen bonds
- Three polypeptides supercoil to form a triple helical structure of great tensile strength, which has resistance to breaking under tension when being stretched
- Each polypeptide A-chain in collagen already consists of hundreds of amino acids in repeating sequences of amino acids
29
DNA properties
1. Double stranded: consists of 2 polynucleotide chains formed with deoxyribonucleotides, where two chains twist to form a double helix
2. Anti parallel: 2 polynucleotide chains run in opposite directions, held by hydrogen bonds between nitrogenous base pairs
30
Describe how an unsaturated triglyceride is synthesized
- 1 glycerol and 3 fatty acids
- Joined by ester bonds by condensation
- More than 1 fatty acid is {unsaturated / has a c-c double bond}
31
Which polypeptide does the liver store
glycogen
32
Which carbohydrates can be digested to release glucose
Lactose and sucrose
33
Explain the role of primary structure in determining the properties of protein of cell membrane
* because the sequence of amino acids determine the {tertiary / quaternary} structure of the protein (1)
* by determining the {position / type} of bonds that form between the R groups (1)
* {hydrophobic / non-polar} {(R) groups / amino acids on (the outside of) the part of the protein that is embedded in the fatty acid tails (1)
* {hydrophilic / polar} {(R) groups / amino acids} (on the outside of) the part of the protein that is {amongst the phosphate heads / facing the cytoplasm / facing the aqueous environment} (1)
34
Explain how the structure of glycogen relates to its role as an energy storage molecule.
* {polymer of glucose / polysaccharide} therefore has a high energy content (1)
* {large molecule / polymer / polysaccharide} therefore {insoluble / has no osmotic effect} (1)
* branched structure therefore {broken down / energy released / hydrolysis is} faster (1)
* compact therefore has a high energy density (1)
35
Explain the properties of amino acids located on the outer surface of proteins.
* must have R groups that are {polar / hydrophilic} (1)
* so that the {haemoglobin / protein} {can dissolve in (red blood cell) cytoplasm / is soluble in water} (1)
36
Explain how dietary antioxidants reduce the risk of CVD. (3)
* because antioxidants reduce free radicals (1)
* free radicals cause {cell damage /tissue damage / oxidative stress / damage to endothelial lining} (1)
* (antioxidants) reduce {plaque / atheroma} formation (1)
37
(past paper: spec) Carbohydrates used as energy storage molecules include glycogen and starch. Compare and contrast the structures of glycogen and starch. (4)
Similarities
- glycosidic bonds
- both polysaccharides / alpha glucose monomers
Differences
- starch: 2 polysaccharides, glycogen: 1
- amylose is straight / helical shape, while amylopectin and glycogen are branched
38
What is protein
- amino acid monomer joined by peptide bond
primary: amino acid sequence
secondary: folding
teriary: 3D folding
quarternary: involves more than 1 polypeptide chain
39
Why is glucose more soluble in water
- has more polar OH groups so can form more H-bonds with water
40
Explain how the properties of water molecules result in surface tension (3)
- dipolar
- form hydrogen bonds with each other
- cohesion between them
- inward force at its surface
41
What is the purpose of water's high specific heat capacity
- ensures aquatic life as temp is constant for enzymatic activity
42
Maltose
alpha glucose + alpha glucose
43
Sucrose
alpha glucose + fructose
44
Lactose
alpha glucose + galatose
45
Why do fatty acids have different melting points
number of double bonds increase
melting point decreases as fatty acid has more kinks
fatty acid have weaker intermolecular forces
packed less closely
46
Explain how the primary structure of an enzyme determines its three-dimensional structure and its properties
- primary structure of the polypeptide involves the **linear sequence of amino acids** in the chain
- amino acids joined together by **peptide bond** between carboxyl group of one amino acid and amine group of another amino acid
- determine the **arrangement and type** of R groups of amino acids
- determine the **type of bonds and interactions** formed between R groups
- hold / determine **overall folding and coiling** of the polypeptide chain into tertiary structure with a specific shape of active site
47
(past paper) Explain how this primary strucutre produce soluble protein
in primary structure
- involves **linear sequence of amino acids** joined by **peptide bond** (between carboxyl group of one amino acid and amine gropu of another amino acid)
- formed by **condensation** reactions, removal of water
in secondary structure
- folding of the **polypeptide** forming secondary structure
- forming **beta pleated sheets and alpha helix**
- held by **hydrogen bonds** (between O of carbonyl group of one amino acid and H of amine group of another amino acid) with no R group involvement
in tertiary structure
- overall folding and coiling of polypeptide chain into specific 3D shape to form tertiary structure
- maintained by **R group interaction and bonding** such as hydrogen bonds between polar R groups, ionic bond between ionised R groups, disulfied between cysteine amino acids
48
(past paper: Oct24) Compare and contrast the structures of an unsaturated fatty acid and the structure of a saturated fatty acid. (3)
Similarities
- Both contain carbon and hydrogen and oxygen
- Both contain covalent bonds only
Differences
- Unsaturated fatty acids contain carbon-carbon double bonds but saturated fatty acids have only single carbon-carbon bonds
- ACCEPT saturated fatty acids do not contain carbon-carbon double bond C=C
- DO NOT ACCEPT saturated fatty acids have no double bonds
- Unsaturated fatty acids have fewer hydrogens for the same number of carbon / lower hydrogen to carbon ratio
- Unsaturated fatty acids are kinked / bent, and saturated fatty acids are straight / linear
