Biological Molecules Flashcards

1
Q

Define monomer. Give examples.

A

Smaller units that join together to form larger molecules
•monosaccharides
•amino acids
•nucleotides

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2
Q

Define a polymer. Give some examples.

A

Molecules formed when many monomers join together
•polysaccharides
•proteins
•DNA/RNA

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3
Q

Condensation reaction

A

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

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4
Q

Hydrolysis reaction

A

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

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5
Q

Name the three hexose monosaccharides

A

•glucose
•fructose
•galactose
All have the molecular formula C6H12O6

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6
Q

Bond formed between monosaccharides

A

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

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7
Q

Three disaccharides and 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

Structure of alpha glucose

A

Hydroxyl group is on the bottom

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9
Q

Structure of beta glucose

A

Hydroxyl is on the bottom

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10
Q

Structure and function of starch

A

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

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11
Q

Structure and function of glycogen

A

Main storage polymer of alpha glucose in animal cells (but also found in plant cell)
•1,4 and 1,6 glycosidic bonds
•branched=many terminal ends for hydrolysis
•insoluble= no osmotic effect and does not diffuse out of cells
•compact=lots of storage in a small space

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12
Q

Structure and function of cellulose

A

Polymer of beta glucose that gives rigidity to plant cells (prevents bursting under turgor pressure and holds stem up)
•1,4 glycosidic bonds
•straight-chain unbranched molecule
•alternate glucose molecules are rotated 180°
•H-bonds cross link between parallel strands to form microfibrils= high tensile strength

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13
Q

Benedict’s test for reducing sugars

A
  1. Add equal volume of Benedict’s reagent to the sample
  2. Heat the mixture in an electric water bath at 100°c for 5 minutes
    3.Positive result= colour change from blue to orange and brick red precipitation forms
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14
Q

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 the monomers by adding 1cm^3 of HCL. Heat in a boiling water bath for 5 mins
  3. Neutralise the mixture using sodium carbonate solution
  4. Proceed with Benedict’s test as usual
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15
Q

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 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

Test for lipids

A
  1. Dissolve solid samples in ethanol
  2. Add an equal volume of water and shake
  3. Positive result- milky/cloudy white emulsion forms
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18
Q

How do triglycerides form

A

Condensation reaction between 2 molecules of glycerol and 3 fatty acids forms an ester bond

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19
Q

Contrast saturated and unsaturated fatty acids

A

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

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 and used for waterproofing
-Slow conductor of heat= thermal insulation e.g. adipose tissue
-Less dense than water = buoyancy of aquatic animals

21
Q

Describe the structure and function of phospholipids

A

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

22
Q

Compare phospholipids and triglycerides

A

-Both have a glycerol backbone
-Both may be attached to a mixture of saturated, monounsaturated and polyunsaturated fatty acids
-Both contain the elements C, H, O
-Both formed by condensation reactions

23
Q

Contrast phospholipids and triglycerides

A

Phospholipid:
-2 fatty acids and 1 phosphate group attached
-Hydrophilic head and hydrophobic tail
-Used primarily in membrane formation
Triglyceride:
-3 fatty acids attached
-Entire molecule is hydrophobic
-Used primarily as a storage molecule (oxidation releases energy)

24
Q

Why are phospholipids and triglycerides, not polymers

A

They are not made from small repeating units, they are macromolecules

25
Q

Why is water a polar molecule

A

O is more electronegative than H, so attracts the electron density in the covenant bond more strongly
Forms O - (slightly negative charge) and H+ (slightly positive charge)

26
Q

State 4 biologically important properties of water

A

Due to polarity and intermolecular H-bonds
-Metabolite/solvent for chemical reactions in the body
-High specific heat capacity
-High latent heat of vapourisation
-Cohesion between molecules

27
Q

Explain why water is significant to live organisms

A

-Solventmfor polar molecules during metabolic reactions
-Enables organisms to avoid fluctuations in core temperature
-Cohesion-tensions of water molecules in transpiration

28
Q

Inorganic ions and where they are found in the body

A

-Ions that do not contain carbon atoms
-Found in cytoplasm and extracellular fluid
-May be in high or very low concentrations

29
Q

Role of hydrogen ions in the body

A

-High concentration of H+=low (acidic) pH
-H+ ions interact with H-bonds and ionic bonds in tertiary structure of proteins, which can cause them to denature

30
Q

Role of iron ions in the body

A

FE2+ bonds to the porphyrin ring to form a haem group in haemoglobin
Haem group has a binding site to transport 1 molecule of O2 around the body in bloodstream
4 haem groups per haemoglobin molecule

31
Q

Role of sodium ions in the body

A

Involved in the co-transport for absorption of glucose and amino acids in the lumen of the gut.
Involved in propagation of action of potentials in neurons

32
Q

Role of phosphate ions in the body

A

Component of:
-DNA
-ATP
-NADP
-cAMP

33
Q

General structure of an amino acid

A

•COOH- carboxyl/ carboxylic acid group
•R variable side group consists of carbon chain and may include other functional groups
•NH2- amino group

34
Q

Test for protein

A

Biuret test confirms the presence of a peptide bond
1. Add equal volumes of sodium hydroxide to a sample at room temperature
2. Add drops of dilute copper (II) sulfate solution. Swirl to mix
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-) and eliminated molecule of water
•Dipeptide: 2 amino acids
•Polypeptide: 3 or more amino acids

37
Q

How many levels of protein structure are there?

A

4

38
Q

Primary structure of a protein

A

•Sequence, number and type of amino acids in the polypeptide
•Determined by sequence of codons on mRNA

39
Q

Secondary structure of a protein

A

Hydrogen bonds formed between atoms of the polypeptide backbone, forming alpha helices and beta sheets

40
Q

Describe the 2 types of secondary protein structure

A

Alpha helix:
•all N-H bonds on same side of protein chain
•spiral shape
•H-bonds parallel to helical axis
Beta pleated sheet:
•N-H and C=O groups alternate from one side to the other

41
Q

Tertiary structure of a protein. Name the bonds present

A

3D structure formed by further folding of polypeptide
•disulphide bridges
•ionic bonds
•hydrogen bonds

42
Q

Describe each type of bond in the tertiary structure of proteins

A

•Disulphide bridges: strong covenant 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 and easily broken

43
Q

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 bonds as tertiary structure
•May involve addition of prosthetic groups e.g. metal ions or phosphate groups

44
Q

Structure and function of globular proteins

A

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

45
Q

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

What are enzyme?

A

•Biological catalysts for intra and extra 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

47
Q

Induced fit model of enzyme action

A

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