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A-LEVEL BIO > Biological molecules > Flashcards

Biological molecules Flashcards

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1
Q

Define a monomer

A

Smaller units from which larger molecules are made

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2
Q

Define polymer

A

Molecules made from a large number of monomers joined together

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3
Q

what are most polymers made up of

A

-C
-H
-O2
-N

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4
Q

Examples of monomers

A
  • Monosaccharide
  • Amino acids
  • Nucleotides
  • Fatty acids and glycerol
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5
Q

Describe condensation reaction

A

Reaction that joins 2 molecules together with the formation of a chemical bond and involves the elimination of water molecule

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6
Q

Describe hydrolysis reaction

A

Reaction that breaks a chemical bond between 2 molecules and involved the addition of water

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7
Q

What’s a monosaccharide

A

Monomers from which larger carbohydrates are made e.g. glucose, galactose and fructose

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8
Q

general formula for monosaccharide

A

(CH2O)n

where n can be any number form 3-7

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9
Q

How’s a glycosidic bond formed

A

condensation reaction between 2 monosaccharides

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10
Q

How’s a disaccharide fromed

A

Formed by the condensation reaction between 2 monosaccharides

e.g. sucrose is a disaccharide formed by condensation of a glucose molecule and a fructose molecule

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11
Q

Molecules that make maltose

A

Condensation of 2x glucose molecules

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12
Q

Molecules that make lactose

A

Condensation of glucose and galactose molecules

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13
Q

Molecules that make sucrose

A

Condensation of glucose and fructose molecules

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14
Q

Where’s the OH in an Alpha and beta glucose

A

Alpha = bottom right

beta = top right and bottom left

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15
Q

How are polysaccharides formed

A

Condensation of many glucose units

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16
Q

How is glycogen formed

A

Condensation of alpha glucose

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17
Q

How is starch formed

A

Condensation of alpha glucose

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18
Q

How is cellulose formed

A

Condensation of beta glucose

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19
Q

Describe the test for reducing sugars

A

1) Add 2cm^3 of food sample to test tube. If not liquid form first grind up in water
2) Add equal volume of Benedicts reagent
3) Heat test tube gently for 5 minutes
if solution turns from blue to green, yellow, orange or red then a reducing sugar is present

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20
Q

What does reducing sugar form when heated with benedict’s solution

A

insolubale red ppt of copper (I) oxide

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21
Q

Explain why benedict’s reagent turns red when heated with reducing sugar

A

sugar donates electrons that reduce blue copper (II) sulfate to orange copper (I) oxide

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22
Q

Explain how monosaccharides are linked together to form disaccharides

A

Disaccharides consists of 2 monosaccharides units linked together by a glycosidic bond which is formed by a condensation reaction between the hydroxyl group of one monosaccharide and with the hydrogen of the other monosaccharide

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23
Q

Describe how alpha glucose molecules are linked to form starch

A

Starch is made up of chains of alpha glucose monosaccharides linked by glycosidic bonds that are formed by condensation reactions

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24
Q

Describe the test for non-reducing sugars

A

1) Make sure liquid form if not ground up in water
2) add 2cm^3 food sample being tested to 2cm^3 of Benedict’s reagent in a test tube and filter
3) place tt in a water bath for 5 mins. If Benedict’s reagent doesn’t change colour (remains blue) then a reducing sugar isn’t present
4) Add another 2cm^3 of food sample to 2cm^3 of dilute HCl in a tt and place gently in water bath for 5 mins. Dilute HCl will hydrolyse any disaccharide present into its constituent monosaccharides
5) Slowly add some sodium hydrogencarbonate solution to the tt in order to neutralised the HCl (Benedict’s reagent will not work in acidic conditions). Gets with pH paper to check the solution is alkaline
6) re-test the resulting solution by heating it up with 2cm^3 of Benedict’s reagent in a gently boiling water bath for 5 mins
7) if non-reducing sugar was present in original sample, the Benedict’s reagent will now turn orange-brown (this is due to reducing sugars that were produced from the hydrolysis of the non-reducing sugar)

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25
Q

Why are disaccharides like sucrose known as non-reducing sugars

A

they do not change the colour of benedict’s reagent when heated

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26
Q

Property of polysaccharide

A

-very large molecules> insoluble > suitable for storage
- some polysaccharides like cellulose are used for structural support

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27
Q

polysaccharide starch does what

A

found in many plants to form of small granules or grains in chloroplast

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28
Q

Describe the test for starch

A

1) add 2cm^3 of sample being tested into a tt (or add 2 drops of sample onto a spotting tile)
2) add 2 drops of iodine solution and shake or stir
3) presence of starch is indicated by colour change of orange-brown to blue-black

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29
Q

What’s starch used for and name the 2 types of starch

A
  • Storage in plants, hydrolysed when glucose is needed in order for respiration
    -Amylose and amylopectin
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30
Q

what is starch made up of

A

-chains of A glucose monosaccharides linked by glycosidic bonds hat are formed by condensation reaction
-chains may be branches/unbranched
-unbranched chain is wound into tight coil that makes molecule very compact

-OH group pointing inwards to form H bonds > hold helix in palce

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31
Q

Amylose and it’s structure

A

-Unbranched alpha glucose polysaccharide (1,4 glycosidic bond)
- Unbranched and will form into a tight coil (helical structure) so it’s compact (takes up little space) > lots of starch in small area
-Insoluble> doesn’t diffuse> doesn’t affect water potential> water isn’t drawn into cells via osmois

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32
Q

Amylopectin and it’s structure

A

-Branches alpha glucose polysaccharide (1,4 and 1,6 glycosidic bond)
-Enzymes work at each end > glycosidic bond is broken at the end
-Branched form has many ends> gets enzymes that work simultaneously at each end > released the alpha glucose rapidly needed for respiration

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33
Q

What’s glycogen used for

A

Excess glucose storage in animals > easily hydrolysed when glucose is needed

34
Q

how is glycogen stored in animals

A

small granules mainly in muscles and liver > mass of carbohydrate stored is relatively small as fat is main storage molecule in animals

35
Q

Structure of glycogen

A
  • Highly branched alpha glucose polysaccharide (1,4 and 1,6 glycosidic bonds)
    -High branched (more than starch) > more compact and shorter chains > more ends so alpha glucose can be released quickly by enzymes > higher metabolic rate (as enzymes can work quickly on glycogen compared to starch)
    -insoluble in water > doesn’t affect water potential of cells> water isn’t drawn in by osmosis
    -Compact> a lot can be stored in a small space
36
Q

What’s cellulose used for

A

Found in plant cell walls to give strength

37
Q

cellulose having unbranched chains that run parallel to one another allows what

A

-H bonds to form cross-linkages between adjacent chains

38
Q

why does cellulose have OH groups occurring on both sides of the molecule

A

so it can form H bonds with other chains on both side

39
Q

how does cellulose cell wall prevent cell from bursting as water enters by osmosis and why this is important in plants

A

exerts inward pressure that stops any further influx of water > living plant cells are turgid nand push against one another > making non-woody parts of plant semi-rigid > important in maintaining stems and leaves in turgid state so that they can provide maximum surface area for photosynthesis

40
Q

Structure of cellulose

A

-Unbranched long and straight beta glucose polymers (1,4 glycosidic bonds)
-Adjacent beta glucose rotated 180 degrees> OH groups both sides of beta glucose molecules> allows hydrogen bonds to form on both sides of the molecule> many hydrogen bonds form> gives cellulose strength and prevents cell bursting from water entering via osmosis> maintains turgidity for plant stem and leaves
-Individual cellulose molecules join together to form microfibrils and microfibrils join together to form fibres which help support the cell wall
-Cross links of hydrogen bonds between layers> adds collective strength to the cell wall> prevents it form bursting by exerting inward pressure which maintains turgidity

41
Q

Draw/describe the basic structure of a triglyceride?

A

One glycerol molecule attached to three fatty acids in a E shape

42
Q

main characteristics of lipids

A

-contains C, H and O
-proportion of O:C:H is smaller than carbohydrates
-insoluble in water
-soluble in organic solvents eg alcohols and acetone

43
Q

Describe the structure of triglycerides and how this relates to their function

A
  • 3 fatty acids attached to a glycerol and each fatty acid forms an ester bond with glycerol in a condensation reaction

-high ratio of energy-storing carbon-hydrogen bonds to carbon atoms > excellent source of energy
-low mass of energy ratio > good storage molecules > much energy can be stored in a small volume > beneficial for animals as it reduces the mass they have to carry when they move
- Being large, non-polar > insoluble in water > doesn’t affect water potential of cells so water isn’t drawn in by osmosis
- High ratio of hydrogen to oxygen atoms > releases water when oxidised > provide water > good for animals in deserts

44
Q

What are lipids made of

A
  • C, H, O. Less O: C+H than in carbohydrates
    -insoluble in water but soluble in organic solvent e.g. alcohol and ethanol
45
Q

Describe the role of lipids

A
  • Energy source > when oxidised, lipids provide more than 2x the energy released compared to carbohydrates
  • waterproofing > lipids are insoluble in water > useful in water proofing e.g. waxy cuticle prevents water loss
  • insulation (electrical and heat) > stored between body surface so help retain body heat.
    Electrical insulators in the myelin sheath around nerve cells
  • protecting organs
  • cell membranes > phospholipids > enables substances to be transferred through the membrane > flexible and can change shape
46
Q

How are triglycerides formed

A

-One glycerol molecule reacts with 3 fatty acids
-condensation reaction (releases water)
-ester bonds are formed

47
Q

where do the differences in properties of different fats and oils come from

A

-variations in fatty acids
-carboxyl group with hydrocarbon chain attached

48
Q

Describe the structure of phospholipids and it relates to their function

A
  • 1 molecule of glycerol bound to 2 fatty acids and a phosphate group

-phospholipids = polar molecules, having a hydrophilic phosphate head and hydrophobic tail of 2 f.a. > when in aq environment, phospholipid form bilayer between the inside and outside of the side
-hydrophilic phosphate heads help hold the surface of cell surface membrane
-phospholipid structure allows them to form glycolipids by combining with carbohydrates within cell surface membrane > these glycolipids are important in cell recognition
-have two ends so when placed in water, polar phospholipid molecules position themselves so hydrophilic heads are as close to the water as possible and hydrophobic tail is as far away as possible

49
Q

What’s saturated, monounsaturated and polyunsaturated

A
  • saturated = no double bonds as c atoms are linked to the max possible number of h atoms (saturated with h atoms)
  • monounsaturated = single double bond

-polyunsaturated = more than 1 double bond present

50
Q

what does a double bond cause

A

-molecule to bend > cannot pack together closely making them liquid at room temp (oils)

51
Q

What gives triglycerides their different properties

A

glycerol form an ester bond with 3 fatty acids

52
Q

how are triglycerides good source of energy

A

have high ratio of c-h atoms which release a large amount of energy and have low mass to energy ratio and so are good storage without increasing mass

53
Q

difference between lipids and phospholipids

A

-one fatty acid molecule is replaced by phosphate molecule

54
Q

Describe the test for lipids

A

1) add 2cm^3 of food sample into a tt and add 5cm^3 of ethanol and shake thoroughly to dissolve
2) add 5cm^3 of water and shake gently
3) white emulsion indicates presence of lipid

control - repeat procedures using water instead of the sample, final sample should remain clear

55
Q

how is the WHITE cloudy emulsion formed

A

cloudy colour is due to any lipid in the sample being finely dispersed in the water to form an emulsion. light passing through this emulsion is refracted as it passes from oil droplets to water droplets making it appear cloudy

56
Q

lipids complete sentence

A

fats and oils make up a group of lipids called triglyceride which then hydrolysed form glycerol and fatty acids. a fatty acid with more than one c-c double bond is describe as polyunsaturated. In a phospholipid the number of fatty acids is 2 , these are describe as hydrophobic as they repel water

57
Q

List the differences between triglyceride molecules and a phospholipid molecule

A

Triglyceride:
- 3 fatty acids
- no phosphate group
-non-polar

phospholipid:
- 2 fatty acids
- 1 phosphate group
- hydrophilic head and hydrophobic tail

58
Q

Organisms that move use lipids rather than carbohydrates as an energy store suggest one reason and why so

A
  • lipids provide more than 2x as much as energy as carbohydrates when they’re oxidised
  • if fat is stored, the same amount of energy can be provided for less than half the mass
  • it is therefore a lighter storage product - adv is that organism is motile
59
Q

Explain how amino acids are linked to form polypeptides

A

Amino acids are joined in condensation reactions to form polypeptide chains, where water molecules are eliminated (-OH from carboxyl and H from amine). This forms a bond between the two monomers.

60
Q

Proteins

A
  • amino acids are basic monomer units which combined to make up a polymer called a polypeptide
  • contain C, H, N, O and sometimes S
61
Q

Amine group

A

NH2

62
Q

Carboxyl group

A

COOH

63
Q

What’s a peptide bond

A

condensation reaction that joins 2 amino acids with a peptide bond, producing a dipeptide and a molecule of water

64
Q

What’s a polypeptide

A

polymer of amino acids (long chain of amino acids joined with peptide bond)

65
Q

Where does peptide bond form

A

between amino and carboxyl group

OH and H

peptide bond = C-N

66
Q

Explain how primary structures are formed

A

A condensation reaction joins many amino acid monomers through the process of polymerisation, resulting a long chain of amino acids joined with a peptide bond (polypeptide). The sequence of amino acids in a polypeptide chain

  • This is important as it determines the shape that the protein is going to end up in.
67
Q

Explain how polypeptides are arranged to form secondary structures

A

-The H on the -NH group has an overall positive charge while the other O of the -C=O group has an overall negative charge. These two groups therefore really form weak hydrogen bonds causing the long polypeptide chain to be twisted into 3d shape such as a coil (alpha glucose)

Important as hydrogen bonds determine shape

68
Q

Explain how polypeptides are arranged to form tertiary structures A

A

Alpha helices of secondary protein structure can be twisted and folded even more to be given the complex and often specific 3D structure of each protein.
Structures are maintained by 3 bonds, where the bonds occur depend on primary structure:
- disulphide bridges - fairly strong and so not easily broken
- ionic bonds - formed between any carboxyl and amino groups that aren’t involved in forming peptide bonds. Weaker than disulphide bond and easily broken by changes in pH
-hydrogen bonds - numerous and easily broken

69
Q

Explain how the quaternary structure of a protein is formed

A

-formed from different tertiary structures linking together in various ways.
There may also be non-proteins (prosthetic groups) associated with the molecules
-more than 1 polypeptide
-might end up being an enzyme or functional protein e.g. haemoglobin

70
Q

All 4 proteins

A

PRIMARY - specific sequence of amino acids found in its polypeptide chains the specific sequernce of a.a coded for by dna . This sequence determines its properties and shape.
SECONDARY - shape which the polypeptide chain forms as a result of hydrogen bonding. This is most often a spiral known as the alpha helix, although other configurations occurs
TERTIARY - due to bending and twisting of polypeptide helix into a compact structure to create specific 3d shape. All 3 types of bond, disulfide, ionic and hydrogen, contribute to the maintenance of the tertiary structure
QUATERNARY - arises from the combination of a number of different polypeptide chains and associated non-protein (prosthetic) groups into a large, complex protein molecule e.g. haemoglobin

71
Q

Describe the test for proteins

A

1) Place sample of solution to be tested into a tt and add equal volume of sodium hydroxide solution at room temp
2) add a few drops of very dilute (0.05%) copper(II) sulphate solution and mix gently
3) if peptide bonds are present solution will change from blue to purple

72
Q

Explain how enzymes speed up chemical reactions

A

-decreasing the activation energy
-allowing reactions to take place at a lower temperature than normal
-reactions can only occur when activation energy level is overcome

73
Q

Describe how the structure of enzyme molecules relates to their function

A

-The active site is the specific functional part of an enzyme. It’s shape is determined by the primary (and therefore tertiary) structure of the protein.
-The substrate and active site are
complementary to each other and form enzyme substrate complexes.

74
Q

Explain the lock and key model of enzyme action

A

-The shape of the substrate (key) exactly fits the shape of the enzyme (lock), this explains the observation that enzymes are specific to the reaction that they catalyse.

75
Q

Explain the induced fit model

A

-shape of the active site is changed by the proximity of the substrate to the enzyme
-H bond in the enzyme are changed to accomadate the substrate
-this puts strain on the substrate which distorts bonds and lowers activation energy

The active site of an enzyme forms as the enzyme and substrate interact. In the presence of a substrate the shape of the active site changes, this places a strain on the bonds in the substrate which lowers the activation energy. The active site resumes its original shape when products leave.

76
Q

Describe how the rate of an enzyme
controlled reaction is measured.

A

Measuring the formation of products or disappearance of reactants in relation to time

77
Q

Explain how temperature affects the
rate of an enzyme controlled
reaction.

A

-As temperature increases enzymes
and substrates have more kinetic
energy, so are moving more quickly
and with more energy, so are more
likely to collide with the result of a
successful reaction.
-Above the optimum temperature
the bonds in the tertiary structure
of the protein breakdown, so the
active site changes shape and can
no longer form an enzyme
substrate complex.

78
Q

Explain how pH affects the rate of
an enzyme controlled reaction.

A

-Each enzyme has an optimum pH
in which it works fastest.
-At a lower or higher pH than the
optimum the rate of the reaction
is reduced. The charges in the
amino acids making up the active
site change preventing enzyme
substrate complexes forming.
-Extreme changes in pH cause H
bonds to break and the shape of
the active site to change
significantly (denatured enzymes)
-Therefore an enzyme substrate
complex cannot form

79
Q

Explain how substrate and enzyme
concentration affect the rate of
reaction.

A

-Reaction begins- starts off fast,
concentration of product increases
quickly. Because there is a high
concentration of the enzyme and the
substrate, which means lots of
collisions can happen and reaction
take place.
-Reaction rate is decreasing,
because there is less substrate as it
has been broken down already, so it
is less likely for the substrate to
collide with the enzymes.
-Rate of reaction slows due to less
and less substrate being available
and eventually the reaction will be
complete, no new product can be
produced.

80
Q

Describe the nature of
enzyme inhibition.

A

-Enzyme inhibitors are substances that directly or indirectly interfere with the functioning of an enzymes active site
-therefore reduce its activity

81
Q

Explain how
competitive inhibitors
and non-competitive
inhibitor affect the
active site.

A

-Competitive inhibitors: have a molecular shape similar to thesubstrate,
-they can compete for the active site and occupy it instead of / blocking a.s.
-reducing rate of reaction.
-Fewer E-S complexes
-The effect of a competitive inhibitor depends on the relative concentration of substrate and inhibitor
-Increasing substrate conc. reduces effect of inhibitor(level of inhibition dependent on relative concs. of substrate and inhibitor)

Non-competitive inhibitors:
-molecules that bind to an enzyme at a binding site away from active site (allosteric site)
-Binding of the inhibitor alters the 3 structure of enzyme and therefore shape of the active site
-so substrate can no longer bind to the enzyme preventing a reaction.
-fewer e-s complexes formed
-Increasing substrate concentration has no effect on inhibitor as they are not competing for the same binding site.

A-LEVEL BIO (20 decks)

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