Transcript f212 biological molecules

Module 1
Biological Molecules
F212 Molecules, biodiversity,
food and health
Module 1 Topics
• Biological
molecules
– Water
– Intro to biological
molecules
– Proteins
– Carbohydrates
– Lipids
– Practical
biochemistry
• Nucleic acids
• Enzymes
Learning Outcomes
• describe how hydrogen bonding
occurs between water molecules,
and relate this, and other properties
of water, to the roles of water in living
organisms
Definitions
• Covalent bond
– Formed when atoms share electrons
– Strong bonds
• Hydrogen bond
– Weak interaction that occurs when a
negatively charged atom is bonded to a
positively charged hydrogen
Water
• 60 – 70 % of mammals
• About 90% of plants
• Life originated in water
• Good solvent
• What else do you know about little
old dihydrogen monoxide (DHMO)
Water is a liquid
• A polar molecule
• Made up of two positively charged hydrogen
atoms and one negatively charged oxygen
• Covalent bonds form between oxygen and
hydrogen with electrons shared between them.
• Hydrogen bonds form between water molecules
• Up to four may form clusters which break and
reform all the time
Water molecule
Hydrogen Bonds in
water
Hydrogen
bonds
Key features of water
• Key features of water as a
constituent of living organisms
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–
–
–
–
Good solvent
High specific heat capacity
High latent heat of vaporisation
High cohesion
Reactive
Incompressibility
Learning Outcomes
• To be able to
– Define metabolism
– State the functions of biological
molecules
– Name monomers and polymers of
carbohydrates, fats, proteins and
nucleic acids
– Describe general features of
condensation and hydrolysis reaction
Biological Molecules
• Molecular biology
– the study of structure and functioning of
biological molecules.
• Metabolism
– sum total of all biochemical reactions in
the body.
Nutrients and Health
• To maintain a healthy body
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Carbohydrates
Lipids
Proteins
Vitamins and minerals
Nucleic acid
Water
fibre
Key Biological
Molecules
• There are 4 key biological molecules
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–
Carbohydrates
lipids
proteins
nucleic acids
Building blocks of life
• 4 most common elements in the
living organisms
–
–
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–
hydrogen
carbon
oxygen
nitrogen
Biochemicals and bonds
• Covalent bonds join atoms together to
form molecules
• Carbon is able to make 4 covalent bonds
• Carbon can bond to form chains or rings
with other atoms bonded to the chain
• Carbon can also form double bonds
– E.g. C=C or C=O
Polymers
• “poly” means “many” = polymers
• Macromolecules are made up of
repeating subunits that are joined end to
end, they are easy to make as the same
reaction is repeated many times.
• Polymerisation is the making of polymers.
Macromolecules
Macromolecule
Subunit (monomer)
polysaccharide
monosaccharide
proteins
amino acids
nucleic acids
nucleotides
Metabolism
• Metabolism is the sum of all of the
reactions that take place within organisms
– Anabolism
• Build up of larger, more complex molecules from
smaller, simpler ones
• This process requires energy
– Catabolism
• The breakdown of complex molecules into simpler
ones
• This process releases energy
Condensation reactions
• In a condensation reaction
– A water molecule is released
– A new covalent bond is formed
– A larger molecule is formed by bonding
together of smaller molecules
Hydrolysis Reactions
• In hydrolysis reactions
– A water molecule is used
– A covalent bond is broken
– Smaller molecules are formed by the
splitting of a larger molecule
Hydrolysis and condensation
OH
HO
CONDENSATION
HYDROLYSIS
O
Learning Outcomes
• describe, with the aid of diagrams,
the structure of an amino acids
• describe, with the aid of diagrams,
the formation and breakage of
peptide bonds in the synthesis and
hydrolysis of dipeptides and
polypeptides
Introduction to protein
• 50% of the dry mass of cells is protein
• Important functions include
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Cell membranes
Haemoglobin
Anti-bodies
Enzymes
Keratin (hair and skin)
collagen
Structure of proteins
• All proteins are made up of the same
basic components  amino acids
• There are 20 different amino acids,
which alter by having different
residual groups (R groups)
• A single chain of amino acids makes
a polypeptide
Structure of an amino
acid
• Amino acids contain
– Amine group (NH2)
– Carboxylic acid group (COOH)
Joined at the same C atom
Structure of an amino
acid
R group varies in different amino acids
R
H
N
H
Amine
group
C
H
O
C
OH
Carboxyl
group
TEST TIME
• Build an amino acid using the molymod
models
• Glycine is an amino acid where the R
group is hydrogen – change you
molecule into glycine
• Build a dipeptide using the molymod
models
Different Amino Acids
• Glycine
• Alanine
• Valine
R group = H
R group = CH3
R group = C3H7
• You will be expected to learn how to
draw the basic structure of an amino
acid. Remember that each Amino
acid has it’s own specific R group
Learning Outcomes
• explain, with the aid of diagrams, the
term ‘primary structure’
• explain, with the aid of diagrams, the
term ‘secondary structure’ with
reference to ‘hydrogen bonding’
Peptide bond
H
N
H
R
O
H
R
C
C
N
C
H
Peptide
bond
H
O
C
OH
Building a polypeptide
• Peptide bonds are formed in
condensation reactions
• Primary structure
– The primary structure of a polypeptide is
its amino acid sequence
– This is determined by the gene that
codes for the polypeptide Peptide Bond
Amino
acid
Secondary Structure
• Polypeptides become twisted or
coiled
• They fold into one of two structures
– Alpha helix (right handed helix)
– Beta-pleated sheet
• Hydrogen bonds hold coils in place
– Weak but give stability to the parts of a
protein molecule.
C O
H N
Learning Outcomes
• explain, with the aid of diagrams, the
term ‘tertiary structure’ with
reference to hydrophobic and
hydrophilic interactions, disulphide
bonds and ionic interactions
Tertiary Structure
• Folding of the polypeptide to give a
more complex 3-D shape, the shape
is specific to the function of the
polypeptide.
• Examples
– Hormone must fit into the hormone
receptor in a target cell
– Enzymes have a complementary active
site to it’s substrate
Tertiary Structure bonds
• Four types of bond help to hold the
folded proteins in their precise
shape.
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Hydrogen Bonds
Disulphide bonds
Ionic bonds
Hydrophobic interactions
Hydrogen Bonds
• Between polar groups
– Electronegative oxygen atoms of the –CO
– Electropositive H atoms on either the –OH
or –NH groups.
Disulphide bonds
• Between sulfur-containing R groups
of the amino acid cysteine.
• Covalent bonds
• Form strong links which make the
tertiary protein structure very stable.
• This bond can be broken by
reducing agents
Ionic Bonds
• Between R groups, which ionise to
form positively and negatively
charged groups that attract each
other.
Hydrophobic Interactions
• These are interactions between the nonpolar side chains of a protein molecule.
• The bond forms between non-polar,
hydrophobic R groups on the amino
acids.
• Once the two hydrophobic molecules are
close together the interaction is reinforced
by Van der Waals attractions (which
provide the weak bond).
Van der Waals attractions
• Electrons are always in motion, and
are not always evenly distributed
about a molecule.
• This results in areas of positive and
negative charge, which are
continuously changing, and enables
molecules to “stick” to one another.
Denaturing Protein
• The Polar R groups of proteins interact with
water forming hydrogen bonds that face
outwards, This creates a hydrophobic core
to the molecule
• When proteins are heated these bonds
break, the tertiary structure changes and
the protein does not function.
• The destruction of shape or loss of function
is denaturation.
Denaturing Proteins
• Frying an egg
Learning Outcomes
• explain, with the aid of diagrams, the
term ‘quaternary structure’, with
reference to the structure of
haemoglobin
Quaternary Structure
• Association of different polypeptide
chains bonded together to form
intricate shapes
• Sometimes contain prosthetic
groups, which are a permanent part
of a protein molecule but not made
of amino acids
Quaternary Structure
• Globular protein
– Molecules curl up into a “ball” shape
– Examples – myoglobin, haemoglobin
– Metabolic roles
• Fibrous Proteins
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Form long strands
Usually insoluble
Have a structural role
Examples – keratin, collagen
Haemoglobin
• Function – oxygen carrying pigment found
in red blood cells
• Structure
– 4 polypeptides
• 2 x α-globin
• 2 x β-globin
– Each polypeptide has a 3o structure stabilised
by hydrophobic interactions in the centre
– In the middle each polypeptide in a haem
group
OK so let’s summarise
proteins
Protein structure and
diversity
• It is difficult to describe in a simple
sentence the role of proteins.
– when there is something to do, it is a protein
that does it.
• Therefore proteins are
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important
numerous
very diverse
very complex,
able to perform actions and reactions under
some circumstances
Some examples of
proteins
• Antibodies:
– they recognise molecules of invading organisms.
• Receptors:
– part of the cell membrane, they recognise other
proteins, or chemicals, and inform the cell...
• Enzymes:
– assemble or digest.
• Neurotransmitters and some hormones:
– Trigger the receptors...
• Channels and pores:
– holes in the cell membrane
Summary of levels of
protein structure
• Primary Structure
– Amino acids linked in a linear sequence
• Secondary Structure
–
folding or coiling of polypeptide
• Tertiary structure
– Folding of polypeptide by disulphide bonds,
ionic bonds, hydrogen bonds or hydrophobic
interactions
• Quaternary structure
– Two or more polypeptides bonded together
Learning Outcomes
• describe, with the aid of diagrams,
the structure of a collagen molecule
• compare the structure and function
of haemoglobin (and example of a
globular protein) and collagen (an
example of a fibrous protein)
Collagen (a fibrous
protein)
• Collagen is found in skin, teeth,
tendons, cartilage, bones and the
walls of blood vessels, making it an
important structural protein.
Structure of collagen
• 3 identical polypeptide chains
wound into a triple helix; this is a lefthanded helix.
• Each polypeptide is about 1000
amino acids long
• Primary structure
• Every 3 amino acids = glycine
Collagen
• Sequences of polypeptide chains are
staggered so that glycine is found at
every position along the triple helix.
• The three polypeptide chains are held
together by hydrogen bonds.
• Adjacent molecules of collagen are held
together by covalent bonds formed
between the carboxyl group of one amino
acid and the amine group of another.
 Pupil Activity 
• Using your brains and what you have
been taught
– compare the structure and function of
haemoglobin and collagen
• Try to make a bullet point list of at least 10
things
Collagen vs Haemoglobin
• Collagen
– Repeating sequence of
amino acids
– Most of molecule has
left handed helix
structures
– Does not contain
prosthetic group
– Insoluble in water
– Metabolically
unreactive
– Structural role
• Haemoglobin
– Precise 1o structure
– 2o structure wound into
alpha helix
– Contains prosthetic
group
– Soluble in water
– Metabolically reactive
Learning Outcomes
• describe, with the aid of diagrams,
the molecular structure of alphaglucose as an example of a
monosaccharide carbohydrate
• state the structural difference
between alpha and beta glucose
Carbohydrates
• contain carbon, hydrogen & oxygen
• organic compounds
• general formula Cx(H2O)y
– glucose C6H12O6
• 3 main groups
– monosaccharides
– disaccharides
– polysaccharides
Monosaccharides
• dissolve easily in water to form sweet
solution
• general formula (CH2O)n, where n is
the number of carbons
• 3 main types
– Trioses
– Pentoses
– Hexoses
(3C)
(5C)
(6C)
Glucose - a hexose
• Glucose is made of a chain of atoms
long enough to close up upon itself
and form a stable ring structure.
• Carbon atom 1 (1C) joins to the O on
5C.
• The six sided structure formed is
known as a pyranose ring.
Chain for a glucose
H
1C
O
H
2C
OH
OH
3C
H
H
4C
H
5C
OH
OH
6CH OH
2
α-glucose ring form
6CH OH
2
5C
H
4C
OH
H
OH
3C
H
O
H
2C
OH
H
1C
OH
Making the drawing
easier
O
H
OH
Glucose – a hexose
• Isomers
– possess the same molecular formula
but differ in arrangement of atoms.
• α-glucose and β-glucose are
isomers of glucose.
– Depending on whether the OH of 1C is
above or below the plane of the ring.
The Isomers
• α-glucose
O
• β-glucose
H
OH
O
OH
H
Learning Outcomes
• describe, with the aid of diagrams,
the formation and breakage of
glycosidic bonds in the synthesis and
hydrolysis of a disaccharide
(maltose) and a polysaccharide
(amylose)
Disaccharides and the
Glycosidic Bond
• Monosaccharides combine in pairs to
give a disaccharide, this involves the loss
of a single water molecule
• This reaction is called condensation
• The bond formed is known as a glycosidic
bond.
• To break a disaccharide the addition of
water is needed, this reaction is called
hydrolysis.
Formation and breakage
of the glycosidic bond
Polysaccharides
• Final molecules maybe 1000’s of
monosaccharides, the size of these
molecules make them insoluble.
• Polysaccharides are NOT sugars
• The most important polysaccharides are
built up entirely of glucose molecules.
• These are starch, glycogen and cellulose.
Learning Outcomes
• describe, with the aid of diagrams,
the structure of starch
• describe, with the aid of diagrams,
the structure of glycogen
Starch
• A mixture of two substances amylose and
amylopectin.
• Starch granules are insoluble in water.
• The form of carbohydrate used for storage
in plants.
• Starch grains build up in chloroplasts, or in
storage organs such as potato tubers.
Amylose
• Long unbranching chains
• 1-4 glycosidic bonds
• formed by condensation reactions.
• The chains curve and coil into helical
structures.
Amylopectin
• 1,4 linked α-glucose molecules form
chains
• shorter
• branch out to the sides.
– The branches form by 1-6 linkages
Comparison of the structure of
amylose and amylopectin molecules
Glycogen
• The form in which carbohydrate is stored in the
animal body.
• Glucose is converted to glycogen in the liver and
muscles,
– it is kept until required
– then it is broken down again into glucose.
• Formed by α-glucose molecules joining in 1-4
and 1-6 links
• There are more branches containing a smaller
number of glucose molecules than amylopectin
Structure of glycogen
Starch and glycogen
• Starch and Glycogen are energy
storage molecules
• which take up little space due to
their compact shapes
• They help to prevent too high
concentrations of glucose in cells.
Learning outcomes
• describe, with the aid of diagrams,
the structure of cellulose
Cellulose
• Most abundant organic molecule on the planet
due to its presence in cell walls.
• Slow rate of breakdown in nature.
• Polymer of about 10,000 β-glucose molecules in
a long unbranched chain.
• Many chains run parallel to each other and have
cross linkages between them, giving increased
stability.
• hydrogen bonds form these links between chains,
which collectively give the structure increased
strength.
Structure of cellulose
Cellulose
• To join together one β-glucose molecule
must be rotated at 1800 relative to the
other.
• Successive glucose molecules are linked
at 1800 to each other.
• Cellulose molecules become tightly crosslinked with each other to form bundles
called micro fibrils.
• Micro fibrils form cellulose fibres by
hydrogen bonding giving a high tensile
strength similar to steel.
Learning Outcomes
• compare and contrast the structure
and functions of starch (amylose)
and cellulose
• explain how the structures of
glucose, starch (amylose), glycogen
and cellulose molecules relate to
their functions in living organisms
Comparing polysaccharides
Characteristic amylose amylopectin glycogen cellulose
Found in
Found as
Function
Monomer
Bonds
chain
Homework Question
• Discuss the structures of glucose,
starch, glycogen and cellulose in
relation to their functions; include
diagrams to illustrate your answer
Learning outcomes
• compare, with the aid of diagrams,
the structure of a triglyceride and a
phospholipids
• explain how the structure of a
triglyceride, phospholipids and
cholesterol molecules relate to their
functions in living organisms
Lipids are not polymers
• Large molecules
• few oxygen atoms
• many carbon and hydrogen atoms
• hydrophobic
• Less dense than water
Lipids
• Two important groups
– Triglycerides
• Fats – solid at room temperature
• Oils – liquid at room temperature
– phospholipids
Lipids - functions
•
•
•
•
•
•
A source of energy
Store of energy (adipose tissues)
Biological membranes
Thermal insulators / insulation
Buoyancy
Protection
– Cuticle of a leaf
– Internal organs
• Metabolic source of water
• hormones
Glycerol and fatty acids
• glycerol
H
H
C
OH
H
C
OH
H
C
OH
H
• Fatty acid
O
H H H H H
C C C C C C H
HO
O
OH
C
H H H H H
Fatty Acids
• Fatty acids have
– an acid group at one end (COOH)
– Hydrocarbon chain (2  20 carbons
long)
• Fatty acids can be
– Saturated
– Unsaturated
Saturated fatty acid
• All possible bonds are made with
hydrogen
O
C
HO
H
C
H
C
H
C
H
C
H
C
H
H
H
H
H
H
Unsaturated fatty acid
• One or more double bond between
carbon atoms
O
C
HO
H
C
C
H
H
C
H
C
H
C
H
H
H
H
Saturated and
unsaturated fatty acids
• Polyunsaturated
– more than one double bond
• Monounsaturated
– only one double bond
• Animal lipids are often saturated and
occur as fats
• plant lipids are often unsaturated
and occur as oils
Triglycerides
• Most common form of lipid
• Combination of 3 fatty acid
molecules and one glycerol
molecule.
– Glycerol is a type of alcohol
– Fatty acids are organic molecules with
a COOH group attached to a
hydrocarbon tail.
Triglycerides
• Each of the glycerol molecules 3 OH groups reacts with the carboxyl
group of a fatty acid.
• This is a condensation reaction, and
an ester bond is established.
Structure of a
triglyceride
• Glycerol
+
3 fatty acids
O
H
H
C
OH
HO
C
O
H
C
OH
HO
C
O
H
C
OH
HO
C
H
Condensation reaction and
formation of an ester bond
O
H
H
C
O
C
O
H
C
O
C
O
H
C
O
C
H
Ester bond
Triglycerides
• Triglycerides are
– insoluble in water,
– soluble in some organic solvents, e.g.
ether or ethanol.
– non-polar
– hydrophobic.
Roles of triglycerides
• Energy reserve
• Insulator against heat loss
• Buoyancy
• Protection (vital organs)
• Metabolic source of water.
Phospholipids
• Special type of lipid
• one of the fatty acid groups is
replaced by phosphoric acid.
• phosphoric acid is hydrophilic
(attracts water)
• Biological significance of this
molecule is its role in the cell
membrane.
Simplified structure of
phospholipid
Structure of a phopholipid
O
H
O
P
OH
O
H
C
H
C
O
C
O
H
C
O
C
H
Phosphate group
Structure of a
phospholipid
Cholesterol - structure
• Small molecule
• -OH group is polar
• 4 carbon rings and hydrocarbon tail
are non polar
Cholesterol - Structure
Cholesterol - function
• Found in biological membranes
• Steroids e.g. testosterone, oestrogen
and progesterone are made from
cholesterol
• Excess cholesterol
– Form gallstones in bile
– Cause atherosclerosis in blood vessels
Learning Outcomes
• describe how to carry out chemical
tests to identify the presence of the
following molecules: protein (Biuret
test), reducing and non-reducing
sugars (Benedict’s test), Starch
(iodine solution) and lipids (emulsion
test)
Chemical Tests
• Chemical tests can be done to
confirm the presence of various
biological molecules within a
sample
• These tests are qualitative tests
– They indicate presence of a molecule
not how much is present
Testing for presence
of a carbohydrate
• Starch
• Reducing sugar
• Non reducing sugar
starch
• Iodine solution
– iodine in potassium iodide
– Add to solution will turn blue-black
quickly if comes into contact with
starch.
Starch
• Starch molecules curl up into long
spirals, with a hole down the middle
of the spiral, just the right size for an
iodine molecule.
• The starch-iodine complex forms a
strong blue-black colour.
Reducing sugar
• Benedict’s Reagent (copper II
sulphate in alkaline solution)
– Add benedict’s reagent to the solution
testing
– Heat in a water bath (80oC) for 3
minutes
Reducing sugars
• If added to a reducing agent Cu2+ ions are
reduced to Cu+, and the change in colour to red
of Copper (I) sulphate.
• All monosaccharides are reducing sugars;
• Reducing sugars have an aldehyde group (HC=0) somewhere in their molecule, which
contribute an electron to the copper.
• Reducing sugars become oxidised.
Reducing sugar + Cu2+ = oxidised sugar + Cu+
Non reducing sugar
• Heat sugar solution with acid to hydrolyse
any glycosidic bonds present
• Neutralise solution by adding sodium
hydroxide
• Add benedict’s reagent
• Heat in a water bath
• If it goes orange/red a non-reducing
sugar is present.
Non-reducing sugars
• Not all disaccharides are reducing sugars.
• To check for the presence of a reducing
sugar, the disaccharide needs to be
broken down into its constituent
monosaccharides,
• monosaccharides are reducing sugars
and will react with benedict’s solution.
Testing for the
presence of proteins
Proteins
• Biuret reagent
– copper sulphate and potassium or
sodium hydroxide
– Add Biuret solution to the substance
– If protein present get a purple colour
proteins
• All proteins have several amine, NH2,
groups within their molecules.
• These groups react with copper ions
to form a complex that has a strong
purple colour.
Testing for the
presence of lipids
lipids
• “Emulsion test”
– Shake substance (lipid) with absolute
ethanol
– Pour ethanol into a tube containing
water
– If no lipid is present mixture looks
transparent
– If lipids are present – looks white and
cloudy.
lipids
• Lipids are insoluble in water, but soluble in
ethanol.
• As the ethanol mixture is poured into
water, lipid molecules cannot remain
mixed in water and clump together to
form little groups.
• The lipid molecules impede light and we
see an emulsion (white cloudiness).
Learning Outcomes
• describe how the concentration of
glucose in a solution may be
determined by using colorimetry
Banana Qualitative
• Bananas, at each of five different stages of
ripeness.
– The stages must range from very green (inedible) to
very ripe (brown skin).
– Each student will require an approximately 5 cm length
of each banana.
– The bananas must be labelled or presented on labelled
watch glasses.
• 50cm3 fresh iodine in potassium iodide solution in
a beaker labelled iodine solution.
• 50cm3 fresh Benedict’s solution in a beaker
labelled Benedict’s solution.
Nucleic Acids
Module 1 Biological Molecules
Unit 2 Molecules, Biodiversity, food and health
Learning Outcomes
• state that deoxyribonucleic acid (DNA) is
a polynucleotide, usually double stranded
and made up of the nucleotides adenine
(A), thymine (T), cytosine (C) and guanine
(G)
• state that ribonucleic acid (RNA) is a
polynucleotide usually single-stranded
and made up of the nucleotides adenine
(A), uracil (U), cytosine (C) and guanine
(G)
Nucleic Acids – DNA
and RNA
• The nucleic acids have
– The ability to carry instructions
– The ability to be copied
• DNA and RNA are polymers; the individual
nucleotides are the monomers that build
up the polynucleotides.
– DNA = deoxyribonucleic acid
– RNA = ribonucleic acid
Nucleotides
• Nucleotides are made up of three smaller
components
– Nitrogen containing base
– Pentose sugar (5 carbon atoms)
– Phosphate group
Phosphate
sugar
base
Bases
• There are 5 different nitrogen-containing
bases:
–
–
–
–
–
A
T
U
G
C
• DNA
• RNA
Adenine
Thymine (DNA only)
Uracil (RNA only)
Guanine
Cytosine
– A, G, C and T
- A, G, C and U
Bases
• Purines (larger)
– These have double rings of carbon and nitrogen atoms
– adenine
– Guanine
• Pyrimidines (smaller)
–
–
–
–
These have a single ring of carbon and nitrogen atoms
Thymine
uracil
cytosine
Polynucleotides
• Polynucleotides strands are
formed of alternating sugars
and phosphates
DNA
• Cut and paste activity
– Cut out the nucleotides and stick them
down to form a double stranded DNA
molecule
Learning Outcomes
• describe, with the aid of diagrams,
– how hydrogen bonding between
complementary base pairs (A-T, G-C)
on two anti-parallel DNA
polynucleotide leads to the formation of
a DNA molecule,
– how the twisting of DNA produces it’s
‘double-helix’ shape outline, with the
aid of diagrams,
DNA
• 2 strands
side-by-side
running in
opposite
directions
(antiparallel)
• The two
strands are
held together
by hydrogen
bonds.
Complementary base
pairs
• A purine in one strand is always opposite
a pyramidine in the other strand.
– Adenine – thymine
– Guanine - cytosine
• DNA forms a double helix, the strands are
held in place by hydrogen bonds.
• These bonds can be broken relatively
easily, this is important for protein
synthesis and DNA replication.
Pupil Activity
• Build your own DNA molecule
• Equipment needed:
–
–
–
–
–
–
2 purple pipe cleaners
2 white pipe cleaners
6 red beads
6 yellow beads
12 aqua beads
12 purple beads
• Follow the instructions on the handout
DNA – a double helix
• Two polynucleotides held together by
hydrogen bonds
• Complementary base pairs
– AT (2 hydrogen bonds)
– GC (3 hydrogen bonds)
• Polynucleotides are anti-parallel
– Parallel but with chains running in opposite
directions
• 3’ to 5’direction
• 5’ to 3’direction
Structure to function
• Information storage
– Long molecules
– replication
• Base-paring rules
• Hydrogen bonds
– Stable
Learning Outcomes
• how DNA replicates semiconservatively, with reference to the
role of DNA polymerase
DNA Replication
• Each polynucleotide acts as a
template for making a new
polynucleotide
• This is known as semi-conservative
replication
Experimental Evidence for the semiconservative replication of DNA
• Three ways were suggested for DNA
replication
– Conservative replication
– Semi-conservative replication
– Dispersive replication
• Scientists thought that semi-conservative
replication was most likely but there was
no evidence to support this theory.
• 1958 Matthew Meselsohn and Franklin
Stahl demonstrated that DNA replication
was semi-conservative following
experiments with E. Coli.
Stage 1
• E. Coli were grown in a medium
containing a heavy isotope nitrogen
(15N).
• The bacteria used 15N to make the
purine and pyrimidine bases in its
DNA.
Stage 2
• After many generations, they were
then transferred to light isotope
nitrogen (14N)
Stage 3
• Bacteria were taken from the new
medium after one generation, two
generations and later generations.
• DNA was extracted from each group
of bacteria,
• samples were placed in a solution of
caesium chloride and spun in a
centrifuge.
Results
Generation
1
2
3
Conclusions
1.
2.
3.
4.
5.
Explain why the band of DNA in the first
generation is higher than that in the parental
generation.
If replication were conservative what results
would you expect in the first generation?
If the DNA had replicated dispersively what
results would you expect in the first generation?
Explain how the second generation provides
evidence that the DNA has reproduced semiconservatively and not dispersively
What results would you expect to see from a third
generation, draw a diagram of the results?
Explanation of results
• Parental generation - both strands
made with 15N
• First generation – DNA made of one
strand 15N and one strand 14N
• Second generation – some DNA made
of 2 strands of 14N and some made of
15N and 14N.
DNA Replication
• Double helix unwinds and the DNA “unzips” as
hydrogen bonds break
• Existing polynucleotides acts as a template for
assembly of nucleotides
• Free nucleotides move towards exposed bases of
DNA
• Base pairing occurs between free nucleotides
and exposed bases
• Enzyme DNA polymerase forms covalent bonds
between free nucleotides
• Two daughter DNA molecules form separate
double helices.
Learning Outcomes
• state that a gene is a sequence of
DNA nucleotides that codes for a
polypeptide
• outline the roles of DNA and RNA in
living organisms (the concept of
protein synthesis must be considered
in outline only)
RNA
• single strand, containing
– uracil not thymine
– Ribose sugar
• There are 3 forms of RNA
– Messenger RNA
– Transfer RNA
– Ribosomal RNA
mRNA
tRNA
rRNA
DNA and Protein Synthesis
• All chemical reactions are controlled by
enzymes, all enzymes are proteins, DNA
codes for proteins, therefore DNA controls
all the activities of a cell.
• The shape and behaviour of a protein
depends on the exact sequence of amino
acids in the primary structure
(polypeptide).
The Genetic Code
• DNA determines the exact order in which
amino acids join together.
• The genetic code
–
–
–
–
sequence of bases along the DNA molecule,
There are 20 different amino acids, only 4 bases,
a sequence of 3 bases codes for an amino acid.
This is called the triplet code.
• A gene is the part of a DNA molecule,
which codes for just one polypeptide.
Protein Synthesis
• The process of protein synthesis
occurs in four stages:
– transcription of DNA to make
messenger RNA (mRNA)
– movement of mRNA from the nucleus to
the cytoplasm
– amino acid activation
– translation of mRNA to make a
polypeptide
Transcription
• This is the process by which mRNA is built
up against one side of an opened up
piece of DNA.
• The relevant section of DNA unwinds, the
hydrogen bonds between base pairs are
broken and the two strands split apart.
• Free nucleotides then assemble against
one strand of DNA.
• The enzyme RNA polymerase moves
along the DNA adding on RNA nucleotide
at a time.
Movement of mRNA to
ribosomes
• mRNA leaves the nucleus through a
nuclear pore into the cytoplasm, and
attaches to a ribosome.
Amino Acid Activation
• Enzymes attach amino acids to their
specific tRNA molecule.
• This needs energy supplied by ATP.
• An anti-codon is a triplet of bases
forming part of a tRNA molecule and
it is complementary to a codon.
Translation
• Amino acid attaches to the ribosome
• Adjacent amino acids are joined together
by peptide bonds and a polypeptide
chain is built up.
• This carries on until the ribosome reaches
a stop codon, the polypeptide breaks
loose from the ribosome and translation is
complete.
Enzymes
Learning Outcomes
• state that enzymes are globular
proteins, with a specific tertiary
structure, which catalyse metabolic
reactions in living organisms;
Recap
• What is metabolism?
– sum total of all biochemical reactions in
the body.
Enzymes
• All enzymes are
–
–
–
–
globular proteins
catalysts
Specific
affected by temperature and pH
More about enzymes
• Two basic functions within cells:
– Act as biological catalysts
– Provide a mechanism whereby
individual chemical reactions can be
controlled
• Enzyme molecules have a specific
3D shape and all possess an active
site.
Learning Outcomes
• Follow the progress of an enzymecatalysed reaction;
Catalase
• The enzyme catalase breaks down hydrogen
peroxide into water and oxygen.
2H2O2 => 2H2O + O2
• Hydrogen peroxide is formed continually as a biproduct of various chemical reactions in living
cells.
• It is toxic and if the cells did not immediately
break it down it would kill them.
Investigation 1
• Catalase is the fastest enzyme known.
• In this investigation you will be able to watch the
action of catalase and compare it with an
inorganic catalyst that catalyses the same
reaction.
1. Pour hydrogen peroxide into two test tubes to a depth
of about 2cm.
2. Into one test tube sprinkle about 0.1g of manganese
dioxide.
3. Into the 2nd test tube put in a 1cm2 piece of potato.
4. Observe the two test tubes and record what happens.
Results
• Describe the difference in reaction
with the inorganic catalyst and the
organic catalyst
Investigation 2
Graduated
measuring
cylinder
15ml
Hydrogen
peroxide
water
Method
•
•
•
Design a results table to record the
oxygen produced every 10 seconds.
cut up 4cm3 piece of potato into this
slices into the conical flask, and start
recording results immediately.
Take a reading for the amount of oxygen
produced every 10 seconds, until the
oxygen is no longer being produced.
Extension
• If you have time, you could repeat
the above experiment, but this time
grind up the 4cm3 of potato with
some fine sand. How do the results
compare?
Results
•
•
•
–
–
–
–
Draw a graph of oxygen produced against
time.
Describe the graph in terms of interaction
between the molecules of catalase and
hydrogen peroxide.
How could you adapt this experiment to
investigate the effect of the following on the
rate of the reaction.
temperature
pH
substrate concentration
enzyme concentration
Learning Outcomes
• state that enzyme action may be intracellular or
extra cellular;
• describe, with the aid of diagrams, the
mechanism of action of enzyme molecules, with
reference to
–
–
–
–
–
–
–
specificity,
active site,
lock and key hypothesis,
induced-fit hypothesis,
enzyme-substrate complex,
enzyme-product complex
lowering of activation energy
Active Site
• The Active site is the region to which
another molecule or molecules can bind.
This molecule is the substrate of the
enzyme.
• The enzyme and substrate form an
enzyme-substrate complex.
• When enzyme and substrate collide in the
correct orientation, the substrate becomes
attached and held temporarily in position
at the active site.
Substrate  end products
• Enzyme and substrate molecules
then interact so that a chemical
reaction involving the substrates
takes place and the appropriate
products are formed.
• When the reaction is complete, the
product or products leave the active
site.
Enzyme Specificity
• Active sites are specific for one type of
molecule
• Examples of specificity
– Amylase breaks down glycosidic bonds in
starch to form maltose
– Catalase breaks down hydrogen peroxide into
water and oxygen
– Trypsin is a protease that only breaks peptide
bonds next to the amino acids arginine and
lysine
Lock and Key Theory
• Some part of the enzyme has an
active site, which is exactly the
correct shape to fit the substrate.
– Active site = lock
– Substrate = key
Induced fit Theory
• Active site is a cavity of a particular shape
• initially the active site is not the correct
shape in which to fit the substrate.
• As the substrate approaches the active
site, the site changes and results in being
a perfect fit.
• After the reaction has taken place and
the products have gone.
• The active site returns to its normal shape.
Metabolism
• A catabolic reaction
– substrate has been broken down
• An anabolic reaction
– substrate used to build a new molecule
Lowering of Activation
Energy
• Activation energy is the energy given
temporarily to a substrate to convert it into
a product.
• The higher the activation energy the
slower the reaction.
• Enzymes help to decrease activation
energy by providing an active site where
reactions can occur more easily than
elsewhere.
Lowering Activation
Energy
Activation
energy
without
enzyme
Activation
energy
with
enzyme
Learning Outcomes
• To follow the progress of an enzymecatalysed reaction;
Experiments with
enzymes
• Follow the time course of an enzyme-catalysed
reaction by measuring
– rates of formation of products (for example using
catalase),
– rate of disappearance of substrate (for example using
amylase).
• When an enzyme and a substrate are mixed
together, a reaction begins. Substrate molecules
collide with the enzyme and bind to its active
site; product molecules are formed.
Experiments with
enzymes
• As the reaction proceeds the number of
substrate molecules decreases and the
number of product molecules increase.
The number of enzyme molecules remains
constant.
• We can measure the rate of a reaction by
measuring either:
– Increasing product
– Decreasing substrate
Increasing Product
Example: catalase breaks down hydrogen
peroxide into water and oxygen
Decreasing Substrate
Example: amylase breaks down starch into
maltose
Explanations for the
course of reaction
• As the reaction proceeds there is less
substrate available, therefore less product
gets released.
• Rate of reaction is quickest at the
beginning when there is a high
concentration of substrate.
• Later the substrate becomes the limiting
factor and the reaction slows down.
• Eventually all substrate is used up, so the
reaction stops
Learning Outcomes
• describe and explain the effects of pH,
temperature, enzyme concentration and
substrate concentration on enzyme
activity;
• describe how the effects of pH,
temperature, enzyme concentration and
substrate concentration on enzyme
activity can be investigated
experimentally
Factors Affecting
enzyme Activity
• Enzyme Concentration
• Substrate concentration
• Temperature
• pH
Enzyme Concentration
• The rate of reaction is directly
proportional to the enzyme
concentration
• assuming that there are plenty of
substrate molecules and enzymes
are the only limiting factors.
Enzyme Concentration
Substrate
concentration
• For a given amount of enzyme, the
rate of an enzyme controlled
reaction increases with substrate
concentration, up to a certain point.
• This point is Vmax, which is the
maximum rate of reaction; the
amount of enzyme becomes the
limiting factor.
Substrate
concentration
Temperature
• An increase in temperature affects the
rate of reaction in two ways
• Factor 1
– As the temperature increase the kinetic
energy of the substrate and enzyme
molecules increases and they move faster.
– The faster the molecules move the more often
they collide and the greater the rate of
reaction.
Temperature
• Factor 2
– As temperature increases, more atoms
which make up the enzyme molecules
vibrate.
– This breaks down the bonds which hold
the molecules in the precise shape.
– The enzyme becomes denatured and
loses catalytic properties.
Temperature
• OPTIMUM TEMPERATURE
– temperature at which an enzyme
catalyses a reaction at a maximum
rate.
Temperature
pH
• The precise 3-D shape of an enzyme is
partly a result of hydrogen bonding.
• These bonds maybe broken down by high
concentrations of H+ ions.
• When pH changes from the optimum
– shape of enzyme changes
– affinity of substrate for the active site
decreases
pH
Online resources
• Online simulation of practical available at
– http://mvhs.mbhs.edu/coresims/enzyme/inde
x.php
• Good simulation of the theory of temp/pH
available at AS guru
– www.bbc.co.uk
• Chemistry for biologists
– www.chemsoc.org/networks/learnnet/cfb/
Learning Outcomes
• explain the effects of competitive
and non-competitive inhibitors on
the rate of enzyme-controlled
reactions,
– with reference to both reversible and
non-reversible inhibitors;
Enzyme Inhibitors
• Inhibitors prevent enzymes from
working
• There are two types of inhibitor
– competitive
– non-competitive.
Competitive Inhibitors
• Have a similar shape to the normal
substrate and are able to bind to the
active site.
• Do not react with the active site but leave
after a time without any product forming.
• The rate of reaction decreases because
the substrate molecules have to compete
with the inhibitor for the active site.
• It is possible to reduce the effect of the
inhibitor by adding more substrate
Competitive inhibitor
Effect of concentrations of inhibitor
and substrate on the rate of an
enzyme controlled reaction
Rate of reaction
No inhibitor
With fixed
concentration
of competitive
inhibitor
Substrate concentration
Examples
• Competitive inhibitor
– Reversible
• Statins compete with a liver enzyme which
helps to make cholesterol
– Non-reversible
• Penicillin inhibits an enzyme that makes cell
walls in some bacteria
Non-competitive
inhibitors
• Molecules bind to some part of an
enzyme other than the active site.
• This changes the active site so that the
substrate can no longer fit.
• If the concentration of this type of inhibitor
is high enough, all enzymes maybe
inhibited and the reaction slows to
nothing.
• Increasing the concentration of the
substrate has no effect on this type of
inhibition.
Non competitive
inhibitor
Rate of an enzyme controlled reaction
with and without a non-competitive
inhibitor
Rate of reaction
No inhibitor
With non-competitive
inhibitor
Substrate concentration
Examples
• Non-competitive inhibitor
– Potassium cyanide bind to haem, which
is part of cytochrome oxidase
– This is non-reversible
End product inhibition
• Metabolic reactions must be finely
controlled and balanced;
• end product inhibition regulates
certain enzyme-catalysed processes
in organisms.
End product inhibition
End product inhibition
• This is an example of noncompetitive inhibition
– product 3 binds to another part of the
enzyme other than the active site.
• It is also an example of a feedback
mechanism.
Learning Outcomes
• explain the importance of cofactors and
coenzymes in enzyme-controlled
reactions;
• state that metabolic poisons may be
enzyme inhibitors, and describe the
action of one named poison;
• state that some medicinal drugs work by
inhibiting the activity of enzyme
Co-factor
• A non-protein component
• Required by enzymes to carry out
reactions
• Examples
– Metal ions in carbonic anhydrase
– Haem in catalase
– Chloride ions and amylase
Co-enzyme
• Organic, non protein molecules
• Role is to carry chemical groups
between enzymes, linking together
enzyme controlled reactions
• Examples
– NAD, FAD and coenzyme A – involved
in respiration
– NADP – involved in photosythesis
Prosthetic groups
• A coenzyme that is a permanent
part of the enzyme
• Example
– Carbonic anhydrase contains a zincbased prosthetic group
Metabolic poisons
• Metabolic poisons can be enzyme
inhibitors
• Example
– Potassium cyanide
• inhibits cell respiration
• Non-competitive inhibitor for the enzyme
cytochrome oxidase
• Decreases the use of oxygen so that ATP can not be
made
• The organism respires anaerobically and lactic acid
builds up in the blood
Medicines and enzymes
• Infection by viruses are treated by
using chemicals that act as protease
inhibitors which the virus needs to
build new viral coats.
• Antibiotics
– Penicillin inhibits a bacterial enzyme
which makes bacterial cell walls
Learning Outcomes
• Measure the effect of different
independent variables and
independent variable ranges on an
enzyme-catalysed reaction;
• Measure the effect of an inhibitor on
an enzyme-catalysed reaction.