Download File - Mr Weng`s IB Chemistry

IB CHEMISTRY Option B Biochemistry
Higher level
B.1 Introduction to biochemistry
OBJECTIVES
• The diverse functions of biological molecules depend on their structures and
shapes.
• Metabolic reactions take place in highly controlled aqueous environments.
• Reactions of breakdown are called catabolism and reactions of synthesis are
called anabolism.
• Biopolymers form by condensation reactions and are broken down by
hydrolysis reactions.
• Photosynthesis is the synthesis of energy-rich molecules from carbon dioxide
and water using light energy.
• Respiration is a complex set of metabolic processes providing energy for cells.
• Explanation of the difference between condensation and hydrolysis reactions.
• The use of summary equations of photosynthesis and respiration to explain the
potential balancing of oxygen and carbon dioxide in the atmosphere.
(Intermediates of aerobic respiration and photosynthesis are not required.)
Classification of biomolecules
Anabolic - complex molecules from smaller inorganic
or organic substances
Catabolic - complex molecules broken to smaller
fragments
Metabolic pathways
Most anabolic processes are condensation reactions.
Condensation reaction monomers require 2 function
groups and the release of a small molecule (H2O, HCl, NH3
etc).
Most catabolic processes are hydrolysis reactions.
The most important of these catabolic/anabolic
pairings are respiration and photosynthesis.
Aerobic vs anerobic
120J (lactic acid)
2900J
B.2 Proteins and enzymes
OBJECTIVES
• Proteins are polymers of 2-amino acids, joined by amide links (also known as peptide bonds).
• Amino acids are amphoteric and can exist as zwitterions, cations and anions.
• Protein structures are diverse and are described at the primary, secondary, tertiary and quaternary levels.
• A protein’s three-dimensional shape determines its role in structural components or in metabolic processes.
• Most enzymes are proteins that act as catalysts by binding specifically to a substrate at the active site.
• As enzyme activity depends on the conformation, it is sensitive to changes in temperature and pH and the
presence of heavy metal ions.
• Chromatography separation is based on different physical and chemical principles.
• Deduction of the structural formulas of reactants and products in condensation reactions of amino acids, and
hydrolysis reactions of peptides.
• Explanation of the solubilities and melting points of amino acids in terms of zwitterions.
• Application of the relationships between charge, pH and isoelectric point for amino acids and proteins.
• Description of the four levels of protein structure, including the origin and types of bonds and interactions
involved.
• Deduction and interpretation of graphs of enzyme activity involving changes in substrate concentration, pH and
temperature.
• Explanation of the processes of paper chromatography and gel electrophoresis in amino acid and protein
separation and identification.
(The names and structural formulas of the amino acids are given in the data booklet in section 33. Reference should
be made to alpha helix and beta pleated sheet, and to fibrous and globular proteins with examples of each. In paper
chromatography the use of Rf values and locating agents should be covered. In enzyme kinetics Km and Vmax are
not required.)
Amino acids
• Amino acids are the monomers or building blocks of
proteins (polymers)
• There are 20 amino acids which combine to form proteins
(see data booklet) which vary with the R side chain
• Each amino acid contains an amino group (-NH2) and a
carboxyl group (-COOH)
Amino acid properties
• Amino acids can have both positive and negative charged groups
called a zwitterion, often denoted as the H from the carboxyl group
moving to the amino group
• Amino acids are amphoteric meaning they can act as both acids and
bases
• Amino acids will act as buffers by absorbing excess H+ or OH- ions.
• Each amino acid has a pH in which it is electrically neutral called the
isoelectric point.
• When the pH is below this it will be a positive ion, above this and it
will be a negative ion.
Relationship between pH and ionic character
cation
zwitterion
anion
Problem 1:
ANSWER:
Formation of a Protein
Serine
Peptide Bond
Valine
Tyrosine
3 H2O
Cysteine
A Short Polypeptide
Formation of a Protein
β
Protein structure
Primary structure
• The amino acid sequence of the protein (intramolecular forces).
Covalent (amide peptide) bonding.
Secondary structure
• This is due to regular repeating hydrogen bonding along the same
peptide chain and may consist of turns known as α–helix, or sheets
known as β–pleated sheets.
Tertiary structure
• This is due to the R side groups (note these are called intramolecular
forces)
• These groups not only influence the isoelectric point but also include
hydrophobic/hydophilic groups, hydrogen bonding, ionic bonds
between charged side groups, and disulfide bridges (covalent and
strongest of all these linkages)
• Denaturing of the protein is said to occur at this level
Quaternary structure
• The relationship of the polypeptide to other polypeptides (proteins
such as hemoglobin are actually four interlinking polypeptides)
Secondary structure
Tertiary structure
Fibrous vs Globular proteins
Fibrous proteins – insoluble structural components
that are elongated with a dominant secondary
structure
Globular proteins – soluble functional tools that have
a spherical shape and a dominant tertiary structure
Protein
Type
Role
Function
Keratin
Fibrous
Structure
Protective covering for hair, wool, claws
Collagen
Fibrous
Structure
Connective tissue of skin and tendons
Polymerase
Globular
Enzyme
Catalyzes DNA synthesis
Insulin
Globular
Communication
Hormone for glucose homeostasis
Hemoglobin
Globular
Transport
O2 transport
Antibodies
Globular
Immune system
Protection from pathogens
Protein analysis
• By using hydrolysis, proteins can be broken up into their
constituent amino acids
• Hydrolysis uses enzymes, or heat/acid
• Amino acids can then be detected by either chromatography
or electrophoresis
1. Chromatography
• Amino acids will move at different rates
depending on their solubility with the
solvent and molecular size
• Amino acids are made visible by
spraying with a locating reagent
(ninhydrin).
• Each amino acid will have a special
Rfvalue (Retardation factor).
Rf =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑏𝑦 𝑎𝑚𝑖𝑛𝑜 𝑎𝑐𝑖𝑑
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑚𝑜𝑣𝑒𝑑 𝑏𝑦 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
Problem 1:
ANSWER:
2. Electrophoresis
• Amino acids are placed in a pH
buffer solution
• An electric current is then passed
through
• Amino acids will move at different
rates depending on their
isoelectric point
• Detection is made by staining, or
by fluorescing under UV light
Enzymes - action
• Organic molecules (proteins) that increase
reaction rates faster than inorganic catalysts by
binding substrate to a specific active site
• Sensitive to temperature and pH, heavy metal
ions can also denature the proteins
• Can be very specific (and vary in their
specificity) determined by there tertiary and
quarternary shape
• Some proteins require co-factors to bind to
them for activity
• Organic co-factors such as vitamins are also
called co-enzymes
• Inorganic co-factors include metal ions
• Animation of enzyme specificity
• Animation of induced fit model
Enzymes – organic vs inorganic
Enzyme
Inorganic catalyst
• Proteins (except ribosomes)
• Highly specific
• Maximum rate with substrate
saturation
• Homogenous (in aq state)
• Regulated by inhibitors and
activators
• Work within narrow limits
• Increase rates by 1000 to 1000
000
• Metal ions or complex molecules
• Much less specific
• Do not show saturation
• Homogenous and heterogenous
• Not regulated
• Work at high temp and pressure
• Relatively small increase in rate
compared to enzymes
Enzymes – effect of temperature and pH
Temperature:
• Initial inc in rate due to inc in temp as any normal reaction.
• Sudden dec in rate due to the disruption of hydrogen
bonds and other forces holding the tertiary structure
pH:
• Reactions with the R-groups and H ion concentrations
change the tertiary structure
Enzymes – effect of heavy metal ions
Heavy metal ions:
• Pb, Cu, Hg, Ag react with sulfhydryl groups –
SH forming covalent bonds which many
change the tertiary structure
B.3 Lipids
OBJECTIVES
• Fats are more reduced than carbohydrates and so yield more energy when oxidized.
• Triglycerides are produced by condensation of glycerol with three fatty acids and contain ester links.
Fatty acids can be saturated, monounsaturated or polyunsaturated.
• Phospholipids are derivatives of triglycerides.
• Hydrolysis of triglycerides and phospholipids can occur using enzymes or in alkaline or acidic
conditions.
• Steroids have a characteristic fused ring structure, known as a steroidal backbone.
• Lipids act as structural components of cell membranes, in energy storage, thermal and electrical
insulation, as transporters of lipid soluble vitamins and as hormones.
• Deduction of the structural formulas of reactants and products in condensation and hydrolysis
reactions between glycerol and fatty acids and/or phosphate.
• Prediction of the relative melting points of fats and oils from their structures.
• Comparison of the processes of hydrolytic and oxidative rancidity in fats with respect to the site of
reactivity in the molecules and the conditions that favour the reaction.
• Application of the concept of iodine number to determine the unsaturation of a fat.
• Comparison of carbohydrates and lipids as energy storage molecules with respect to their solubility
and energy density.
• Discussion of the impact of lipids on health, including the roles of dietary high-density lipoprotein
(HDL) and low-density lipoprotein (LDL) cholesterol, saturated, unsaturated and trans-fat and the use
and abuse of steroids.
(The structures of some fatty acids are given in the data booklet in section 34. Specific named
examples of fats and oils do not have to be learned. The structural differences between cis- and transfats are not required.)
Lipids
Lipids are characterized as biological molecules that are
mostly non-polar and so insoluble in water. Molecules
interact with each other with weak Van der Waal’s forces.
There are three major lipid groups:
1. Triglycerides (fats and oils)
2. Phospholipid
3. Steroids
1. Triglycerides - formation
• Formation is a condensation reaction to create an ester link
• The reverse – hydrolysis occurs during digestion by lipases
Glycerol
Fatty Acid Tails
3 H2O
Ester Linkage
A Triglyceride
1. Triglycerides - structure
• The length of the carbon chains are usually even numbers between 14-22
carbons long and are of an even number
• No double bonds means it is saturated (with hydrogens). Monounsaturated has one double bond. Polyunsaturated means there are many
double bonds
• Bond angles on saturated carbons is 109.5°
• Omega 3 and omega 6 fatty acids are essential fatty acids in that the body
cannot make these and must consume them
• Omega refers to the last carbon, the 3 (or 6) is the number of carbons from
the end where the first double bond occurs (total 18 carbon chain)
• Most fatty acids are cis. Trans fatty acids are often a by-product of food
processing
1. Triglycerides – saturated vs unsaturated fat
The kink in the chain caused by the double bonds means less
effective Van der Waals forces hence lower melting points –
cf. fat (solid) to oil (lquid) above.
1. Triglycerides – iodine number
• Iodine number is the number of grams of iodine that react with 100g of fat
• It is a measure of the degree of saturation in the fat and expresses the
average number of double bonds pure unit mass of oil/fat.
• Usually fat is reacted with iodine and excess iodine is determined by
titration with sodium thiosulphate (Na2S2O3)
Brown
Colourless
Problem 1: 0.01 moles of linoleic acid reacts with 5.1 g of
iodine. Determine the number of double bonds present in
the acid.
Answer:
n(I2)= m/M = 5.1 g / 254 g/mol
= 0.020 moles I2
This implies a ratio of 0.01 mol linoleic acid : 0.020 mole I2
Therefore,
Every molecule of linoleic acid contains 2 double bonds
Problem 2: Calculate the iodine number of linoloeic acid,
C17H31COOH.
Answer:
Molecular Weight of Acid = 18(12) + 32(1) +2(16) = 280 g/mol
Molecular Weight of Iodine = 2(127) = 254 g/mol
Linoleic Acid has 2 double bonds per molecule (from last problem).
280 grams of fat reacts with 2 x 254 g of iodine = 508 g iodine
100 grams of fat reacts with 508/280 = 181
The iodine number of linoleic acid is 181
RANCIDITY
• Rancidity is the chemical or biological decomposition of fats and oils
that produces unpleasant odours.
Hydrolytic rancidity
• Hydrolytic rancidity is caused by the hydrolysis of ester bonds on
exposure to water creating unpleasant smelling short chain fatty
acids and is caused by enzymes (lipases), high temperatures and
organic acids (autocatalytic effect). Prevention is by sterilizing foods,
keeping at low temperature and adding acidic components at the
last stage of cooking.
Oxidative rancidity
• Oxidative rancidity is caused by oxygen free radical reactions with
double bonds and is accelerated by sunlight creating unpleasant
smelling volatile aldehyde and ketones. Prevention is by blocking
sunlight and adding antioxidants including vitamins A, C and E.
ENERGY
Fats give high energy yields
because:
• Large numbers of reduced
carbons that can be oxidized
• Hydrophobic nature
decreases water content
and compacts molecules
together
• The insolubility of fat makes
it slower to transport and
metabolize in the body
explaining why
carbohydrates provide a
faster source of energy
2. Phospholipids - structure
Phospholipids (or glycerophospholipids) are derivatives of
triglycerides. They have a fatty acid replaced with a phosphate
group.
Phospholipids in
water
• The formation of
this membrane in
water is called a
phospholipid
bilayer
• Phospholipids are
amphiphilic as the
phosphate group
is hydrophilic and
the fatty acid
group is
hydrophobic.
• The most
common
phospholipid is
lecithin.
Lipids and health
• POSITIVE: Lipids are important
for energy storage, insulation
and protection of organs,
production of components
such as steroid hormones and
the cell membrane, essential
fatty acids reduce risk of heart
disease and lower LDL levels
• NEGATIVE: Lipids can cause
obesity due to their high
energy content, saturated and
trans fats can increase LDL
levels increasing risk of heart
disease and atherosclerosis
Lipoproteins
Lipids are transported around the body bye lipoproteins as
they are insoluble in the blood.
LDL (low density lipoproteins) associated with increasing
cardiovascular disease. It appears to increase the build up of
chlolesterol in the arteries. Levels are raised by consuming
saturated fats and trans fats (hydrogenated fats)
HDL (high density liproproteins) associated with lowering
cardiovascular disease. It appears to decrease the build up
of chlolesterol in the arteries. Levels are raised by
consuming unsaturated fats
(The higher density is due to a
higher protein content in HDL)
3. Steroids
• Steroids are lipids that consist of four fused rings called the
steroidal backbone with three six carbon and one five
carbon rings
• The most important is cholesterol as it is the building block
of other steroids and sex hormones
Production and function of hormones
• A hormone is a chemical messenger
sent from cells in one tissue (endocrine
gland) that act on the metabolism
(formation or destruction of molecules)
of the cells of another tissue (target
cells)
• The control center is the hypothalamus
whose nerve endings release precursor
hormones to the pituitary gland which
release a range of precursor hormones
to target organs around the body.
Hormone producing organs
Hormone summary
Endocrine
gland
Hormone name and
structural type
Target cells
Functions
Pituitary gland
Antidiuretic hormone
(ADH); short polypeptide
Kidneys
Controls retention of water in the
kidneys
Thyroid gland
Thyroxine; modifified amino acid
containing iodine
All cells
Regulates the basal metabolic rate
and affects protein synthesis
Adrenal glands
Aldosterone; steroid
Kidneys
Increase reabsorption of ions and
water in the kidney
Adrenal glands
Adrenaline
(epinephrine); modified amino
acid
Heart, liver and
many
other organs
Fight or flight response—boosts
the supply of oxygen and glucose
to the brain and muscles by range
of actions
Pancreas
Insulin; protein
All cells
Causes glucose to be taken up
from the blood and to be stored in
liver and muscle as glycogen
Ovary
Estrogen and progesterone;
steroids
Uterus, wide
range of cells
Secondary female characteristics,
supporting pregnancy
Testes
Testosterone; steroid
Wide range of
cells
Secondary male characteristics,
increased muscle and bone
strength
Sex hormones - production
Sex hormones - structure
2 hydroxyl
groups - diol
benzene
ring
Extra
ketone
group
hydroxyl
Steroids
Positive
• Can be used medically to
stimulate bone growth and
appetite and induce puberty
• Estrogen supplements after
menopause help older
women from getting
osteoporosis (HRT Hormone
Replacement Therapy)
• Help patients wasting away
from chronic diseases
• Quick recovery from injuries
Negative
•
•
•
•
•
Increases LDL decreases HDL
High blood pressure
Liver damage
Aggressive behavior
Cheating in sport?
(Testosterone is an anabolic
steroid – which means it
builds up tissues including
muscle cf. catabolic steroids
which do the opposite)
B.4 Carbohydrates
OBJECTIVES
• Carbohydrates have the general formula Cx(H2O)y .
• Haworth projections represent the cyclic structures of monosaccharides.
• Monosaccharides contain either an aldehyde group (aldose) or a ketone group
(ketose) and several –OH groups.
• Straight chain forms of sugars cyclize in solution to form ring structures containing
an ether linkage.
• Glycosidic bonds form between monosaccharides forming disaccharides and
polysaccharides.
• Carbohydrates are used as energy sources and energy reserves.
• Deduction of the structural formulas of disaccharides and polysaccharides from
given monosaccharides.
• Relationship of the properties and functions of monosaccharides and
polysaccharides to their chemical structures.
(The straight chain and α-ring forms of glucose and fructose are given in the data
booklet in section 34. The component monosaccharides of specific disaccharides and
the linkage details of polysaccharides are not required. The distinction between αand β- forms and the structure of cellulose are not required.)
Carbohydrates
• Carbohydrates are compounds with the chemical formula
Cx(H2O)y
• There are two types – simple monosaccharides (monomers)
and complex polysaccharides (polymers) which may be as small
as 2 monomers (disaccharides).
1. Monosaccharides
• Contain a carbonyl group (C=O) either in the form of a
aldehyde (aldose) or ketone (ketose).
• Have at least 2 hydroxyl groups (-OH)
• Have the empirical formula CH2O
GLUCOSE
C6H12O6
FRUCTOSE
C6H12O6
Cyclical forms
Straight chain sugars are unstable and become cyclical
by intramolecular nucleophilic addition (AN) forming 5
or 6 membered rings. The reaction occurs between the
carbonyl group and one of the hydroxyl groups.
Haworth projections
Haworth projections are 3D diagrams of cyclical carbohydrates
which omit the carbon atoms in the ring and the attached
hydrogens. They help emphasize the functional groups attached
to the ring. Note also the numbering of the carbons is clockwise
from the furthest right carbon.
Problem 1: See if you can draw the cyclical structures for these:
a.
b.
Reducing sugars
Aldoses (glucose) are called reducing sugars as the can be
oxidized.
Alkaline conditions cause ketoses to isomerize to aldoses and so
cause some cross reactions. Both Fehling’s solution and
Benedict’s reagent use the reduction of copper (II) to (I) creating
a red copper oxide precipitate.
2. Disaccharides
• Disaccharides have the formula C12H22O11
Fructose
Glucose
H2O
Sucrose
Glycosidic
Bond
This condensation
reaction can be
reversed:
hydrolysis by the
use of enzymes or
acid
Lactose (disaccharide)
• Glycosidic link: β-glucose and β-galactose
Maltose (disaccharide)
• Glycosidic link: α-glucose and α-glucose
Sucrose (disaccharide)
• Glycosidic link: α-glucose and α-fructose
Problem 1:
ANSWER:
3. Polysaccharides
Common polysaccharides are:
Starch: Mixture of amylose and amylopectin, main
energy storage molecule of plants. Forms 80% of wheat,
rice, corn etc. Can be broken down by amylase enzyme in
saliva or inorganic acids into glucose.
Cellulose: Forms the cell wall in plants and is a major
form of fibre in the diet as indigestible (except for some
animal species)
Glycogen: Main energy storage molecule in humans.
Concentrated in muscle and liver.
Starch (polysaccharide) 2 types
Amylose, a straight chain with 1-4 α-glucose glycosidic linkages
Amylopectin, a branched chain with 1-4, and 1-6 α-glucose glycosidic linkages
Iodine test for starch
Iodide ions and iodine form triiodide ions:
These react with amylose to make a blue-black complex:
Glycogen (polysaccharide)
• Energy storage molecule for the body
• Like amylopectin, branched chain with 1-4, and 1-6 αglucose glycosidic linkages, but with more 1-6 branches
Cellulose (polysaccharide)
• a straight chain with 1-4 β-glucose glycosidic linkages
• humans lack the cellulase enzyme to break up these β
linkages
• good source of fibre (molecule forms rigid microfibers with
H-bonding)
• Multiple intermolecular H-bonding make it insoluble
Problem 1:
ANSWER:
Dietary fibre
Undigestable carbohydrates form dietary
fibre eg. cellulose, lignin, pectin
This is needed to:
1. Provide bulk to move food through the
digestive tract
2. Prevent constipation and hemorrhoids
3. Reduce sugar intake preventing
diabetes and obesity
4. Reduce the risk of colon cancer,
diverticulosis and irritable bowel
syndrome
B.5 Vitamins
OBJECTIVES
• Vitamins are organic micronutrients which (mostly) cannot be synthesized by
the body but must be obtained from suitable food sources.
• The solubility (water or fat) of a vitamin can be predicted from its structure.
• Most vitamins are sensitive to heat.
• Vitamin deficiencies in the diet cause particular diseases and affect millions of
people worldwide.
• Comparison of the structures of vitamins A, C and D.
• Discussion of the causes and effects of vitamin deficiencies in different
countries and suggestion of solutions.
(The structures of vitamins A, C and D are provided in the data booklet section
35. Specific food sources of vitamins or names of deficiency diseases do not have
to be learned.)
Nutrients
Macronutrients – substances that make up more than
0.005% of our body weight. These include proteins, fats,
carbohydrates and minerals (Na, Mg, K, Ca, P, S, Cl).
Micronutrients – substances that make up less than 0.005%
of our body weight. These include the vitamins (water
soluable – B, C, fat soluable – A, D, E, K) and minerals (Fe,
Cu, Zn, I, Se, Mn, Mo, Cr, Co, B).
Co-factors – substances that help enzymes work
Vitamins
Vitamins are organic micronutrients that cannot be
synthesized by the organism in sufficient amounts and must
either be obtained from suitable foods or taken as food
supplements.
Water soluble vitamins:
Need daily supply as can be eliminated from 30min to several
weeks.
Fat soluble vitamins:
Accumulate in liver and fat tissue. Can be stored for up to
several months. Excess can cause vitamin poisoning.
Vitamin A
(retinol)
• Group of organic compounds including an alcohol (retinol),
and aldehyde (retinal) and several polyunsaturated
hydrocarbons (carotenes)
• Fat soluble as only has one polar hydroxyl group and long
non-polar hydrocarbon chain and ring
• Important for vision at low intensity light levels, (disease
xerophthalmia – also red and dry eyes)
• Fried food can lose up to 50% of the vitamin content
Vitamin D
• A group of four compounds
chemically similar to
cholesterol
• Fat soluble as only contains
one polar hydroxyl group
• Stimulates calcium uptake
so is needed for healthy
bones and teeth (disease
rickets)
• Sunscreen protects from UV
but prevents the skin
producing vitamin D!
• Fried food can lose up to
50% of the vitamin content
(calciferol D3)
Vitamin C
• A single compound – ascorbic acid
• Water soluble due to several
hydroxyl groups to bind to water
• Cofactor for some enzymes
• Important for tissue regeneration
(disease scurvy) and the
biosynthesis of collagen
• Helps gives resistance in some
diseases
• Cooking in water can leech out
Vitamin C, which is also sensitive
to heat and atmospheric oxygen
Pale skin, sunken eyes, loss of teeth
and gum problems – common with pirates
(Other nutritional diseases)
Nutrient
deficiency
Name of
disease
Notes
Iron
Anaemia
Tiredness
Iodine
Goitre
In adults
Mental
retardation
Lack of iodine
is the world’s
largest cause
of mental
retardation in
children
(Other nutritional diseases)
Nutrient
deficiency
Name of disease Notes
Niacin
(Vit B3)
Pellagra
Dermatitis, diarrhea, dementia
Thiamin
(Vit B1)
Beriberi
Weight loss, fatigue, swelling
Protein
Marasmus
Weight loss in infants after
weaning from mother’s milk
Kwashiorkor
Weight loss in infants with high
starch low protein diets
World health
Malnutrition is the lack of regular, balanced supply of nutrients. The
term hence can also be applied to obesity and diabetes due to the
consumption of energy-dense, micronutrient-poor foods in
industrialized countries.
Solutions include:
• providing food rations that are composed of fresh and vitamin- and
mineral-rich foods
• fortification - adding nutrients missing in commonly consumed
foods (eg I2 to salt, Vit A to margarine, Vit B, D and C to cereals)
• genetic modification of food
• providing nutritional supplements
• providing selenium supplements to people eating foods grown in
selenium-poor soil
B.6 Biochemistry and the environment
OBJECTIVES
• Xenobiotics refer to chemicals that are found in an organism that are not normally present there.
• Biodegradable/compostable plastics can be consumed or broken down by bacteria or other living
organisms.
• Host–guest chemistry involves the creation of synthetic host molecules that mimic some of the
actions performed by enzymes in cells, by selectively binding to specific guest species, such as toxic
materials in the environment.
• Enzymes have been developed to help in the breakdown of oil spills and other industrial wastes.
• Enzymes in biological detergents can improve energy efficiency by enabling effective cleaning at
lower temperatures.
• Biomagnification is the increase in concentration of a substance in a food chain.
• Green chemistry, also called sustainable chemistry, is an approach to chemical research and
engineering that seeks to minimize the production and release to the environment of hazardous
substances.
• Discussion of the increasing problem of xenobiotics such as antibiotics in sewage treatment plants.
• Description of the role of starch in biodegradable plastics.
• Application of host–guest chemistry to the removal of a specific pollutant in the environment.
• Description of an example of biomagnification, including the chemical source of the substance.
Examples could include heavy metals or pesticides.
• Discussion of the challenges and criteria in assessing the “greenness” of a substance used in
biochemical research, including the atom economy.
(Specific names of “green chemicals” such as solvents are not expected. The emphasis in explanations
of host–guest chemistry should be on non-covalent bonding within the supramolecule.)
Xenobiotics
Xenobiotics are chemical compounds that are found in a
living organism, but which are foreign to that organism. This
includes the build up of excess levels of a compound in an
organism. Xenobiotics that remain the in environment for
long periods of time without being metabolized are called
persistent organic pollutants (POPs).
Xenobiotics can include:
• Antibiotics
• Insecticides (DDT)
• Heavy metals (Hg, Pb etc)
• Hormones
Examples xenobiotic problems
1. Bioaccumulation (eg. DDT, Hg)
2. Pharmaceutically active compounds – PACs (eg.
xenoestrogens )
3. Non-biodegradable plastics
1a. Bioaccumulation of DDT
An insecticide created to
prevent malaria, DDT
accumulated and increased
in concentration up the food
chain (increasing trophic
levels) causing egg shells to
be thin and so break which
caused a large reduction in
the numbers of birds of prey
in the 1960s.
1b. Bioaccumulation of Hg
Another example of
bioaccumulation
increasing up through the
trophic levels is Minimata
disease in which the
heavy metal mercury (Hg)
affects the neurological
system causing death.
2. PAC (pharmaceutically active compounds) - xenoestrogen
Estrogen
Estrogen type compounds can
be released into water sources.
One source are the
unmetabolized products
released in urine from those
taking the pill. Other sources
are polychlorinated biphenols
(PCBs).
These ‘feminize’ the fish
making them unable to breed.
PCB
3. Non-biodegradable materials
Substances such as
non-biodegradable
plastics can
accumulate in the
stomachs of birds and
marine animals
decreasing their
motility and digestive
systems.
Amelioration of xenobiotics
1. Host-guest complexes
2. Biodegradable materials
3. Enzymes and microorganisms
1. Host-guest complexes
The guest molecule (pollutant) is bound to a host molecule
via non-covalent interactions such as van der Waal’s forces,
ionic bonds, hydrogen bonds and hydrophobic interactions.
The pollutant can hence be immobilized on a host material
and mechanically separated where it can be further
processed or incinerated. The host-guest complex is also
known as a supramolecule.
Example host-guest complex. Radioactive cesium-137 ion is
removed from radioactive waste using this supramolecule.
2. Biodegradable materials
Biodegradable plastics consist of over 50% starch based
polymers. By creating polymers from starch or cellulose it
not only decreases the use of fossil fuels, but allows the
use of microorganisms in the environment to break it
down.
3. Enzymes and microorganisms
• Strains of microorganisms have
been found that can break down
hydrocarbons and have been used
to help clean up oil spills – also
called bioremediation.
• Enzymes used in detergents allow
for fast and effective breakdown
of fats, carbohydrates and lipids
even in cold water
Green chemistry
Green chemistry is an approach that works to minimize the
use and generation of hazardous chemicals. The 12
principles are:
1. Prevention It is better to prevent waste than to treat or clean up waste after it has been
created.
2. Atom Economy Synthetic methods should be designed to maximize the incorporation of
all materials used in the process into the final product.
3. Less Hazardous Chemical Syntheses Wherever practicable, synthetic methods should be
designed to use and generate substances that possess little or no toxicity to human health
and the environment.
4. Designing Safer Chemicals Chemical products should be designed to affect their desired
function while minimizing their toxicity.
5. Safer Solvents and Auxiliaries The use of auxiliary substances (e.g., solvents, separation
agents, etc.) should be made unnecessary wherever possible and innocuous when used.
6. Design for Energy Efficiency Energy requirements of chemical processes should be
recognized for their environmental and economic impacts and should be minimized. If
possible, synthetic methods should be conducted at ambient temperature and pressure.
7. Use of Renewable Feedstocks A raw material or feedstock should be renewable rather
than depleting whenever technically and economically practicable.
8. Reduce Derivatives Unnecessary derivatization (use of blocking groups, protection/
deprotection, temporary modification of physical/chemical processes) should be minimized
or avoided if possible, because such steps require additional reagents and can generate
waste.
9. Catalysis Catalytic reagents (as selective as possible) are superior to stoichiometric
reagents.
10. Design for Degradation Chemical products should be designed so that at the end of their
function they break down into innocuous degradation products and do not persist in the
environment.
11. Real-time analysis for Pollution Prevention Analytical methodologies need to be further
developed to allow for real-time, in-process monitoring and control prior to the formation of
hazardous substances.
12 Inherently Safer Chemistry for Accident Prevention Substances and the form of a
substance used in a chemical process should be chosen to minimize the potential for
chemical accidents, including releases, explosions, and fires.
Atom economy
Atom economy uses the ratio of masses to determine the
practical reaction yield.
A reaction with a target product of C will be much better
produced by A + B  C (=100% atom economy), than A +
B  C + D, as D ends up being waste (<100% atom
economy), .
OBJECTIVES
• Inhibitors play an important role in regulating the activities of enzymes.
• Amino acids and proteins can act as buffers in solution.
• Protein assays commonly use UV-vis spectroscopy and a calibration curve based on
known standards.
• Determination of the maximum rate of reaction (V max) and the value of the
Michaelis constant (Km) for an enzyme by graphical means, and explanation of its
significance.
• Comparison of competitive and non-competitive inhibition of enzymes with
reference to protein structure, the active site and allosteric site.
• Explanation of the concept of product inhibition in metabolic pathways.
• Calculation of the pH of buffer solutions, such as those used in protein analysis and
in reactions involving amino acids in solution.
• Determination of the concentration of a protein in solution from a calibration curve
using the Beer–Lambert law.
(The effects of competitive and non-competitive inhibitors on K m and V max values
should be covered. The Henderson–Hasselbalch equation is given in the data booklet
in section 1. For UV-vis spectroscopy, knowledge of particular reagents and
wavelengths is not required.)
Higher level
B.7 Proteins and enzymes
= pI
This ‘pH of isoelectric point’ for specific amino
acids is found in the Data Booklet. This pH is
known as the pI (at the isoelectric point).
Higher level
Amino acids as buffers
= pI
Higher level
The pH of a buffer solution is calculated using the
Henderson-Hasselbalch equation from your Data
Booklet:
The buffer capacity of an amino acid is best when the pH
is equal to the pKa and so it is important to set the pH to
within ±1 pH value of the pKa to make a buffer.
Higher level
Henderson-Hasselbalch equation
Solution contains anions therefore high pH due to H+ removed.
pKa = 9.1 zwitterion
anion + H+
Therefore zwitterion is the acid and anion is the conj base.
From data booklet:
pH = pKa + log ([A-]/[HA])
= pKa + log ([anion]/[zwitterion])
= 9.1 + log (0.2/0.8)
= 9.1 + -0.6
= 8.5
Higher level
Problem 1: Calculate the pH of an aqueous solution that contains
0.8M zwitterionic and 0.2M anionic forms of serine.
Higher level
Problem 2: Calculate the pH changes after the addition of 1.0g of
solid NaOH to
a. 1.00dm3 of pure water
b. 1.00dm3 of a buffer solution containing 0.40mol of
zwitterionic and 0.16mol of cationic forms of glycine.
Assume the densities of the solution are 1.0 kg/dm3 and the
solution volumes do not change with the addition of NaOH.
Higher level
a. NaOH  Na+ + OHn(NaOH) = m/M = 1g/40g/mol = 0.025mol NaOH
1 mol NaOH : 1 mol OHTherefore [OH] = 0.025mol
pOH = -log [OH] = -log0.025 = 1.059
pKw = pOH + pH
pH = 14 – pOH = 14-1.6 = 12.4
pH water is 7 so change in pH = 12.4 – 7.0 = 5.4
(C=n/V with volumes all 1dm3)
R
I
cation
1
0.16
OH1
0.025
zwitterion
1
0.4
C
-0.025
-0.025
+0.025
E
0.135
0
0.425
Higher level
b. Solution contains cations therefore low pH due to H+ added.
pKa = 2.3 zwitterion + H+
cation
Therefore zwitterion is the base and anion is the conj acid.
NaOH  Na+ + OHcation + OH-  zwitterion + H2O
pH(after) = pKa + log ([A-]/[HA])
= pKa + log ([zwitterion]/[cation])
= 2.3 + log (0.425/0.135)
= 2.3 + 0.5
= 2.8
Therefore the change in pH = 2.8-2.7 = 0.1
*The comparison of parts a and b show the effectiveness of a
buffer to resist pH changes.
Higher level
From data booklet:
pH(before) = pKa + log ([A-]/[HA])
= pKa + log ([zwitterion]/[cation])
= 2.3 + log (0.4/0.16)
= 2.3 + 0.4
= 2.7
pH of solution is below the isoelectric point so it contains
cations.
pKa = 2.3 zwitterion + H+
cation
Therefore zwitterion is the base and cation is the conj acid.
Higher level
Problem 3: Identify the conjugate acid and the conjugate base in
a 0.500M solution of glycine at pH = 5.0. Calculate the
concentrations of both glycine species.
If [cation] = x, then [zwitterion] = 501x
Total concentration is x + 501x = 0.500M
502x = 0.500M
x = 0.0001M
(We already know this is not an effective buffer because the pH is more
than 1 pH value away from the pKa)
Higher level
From data booklet:
pH = pKa + log ([A-]/[HA])
5.0 = 2.3 + log ([zwitterion]/[cation])
log ([zwitterion]/[cation]) = 2.7
[zwitterion]/[cation] = 501
Induced fit model:
• Active site is flexible as a stable enzyme-substrate complex can
be achieved.
• The slight change of shape weakens some bonds lowering the
activation energy of the transition state allowing the chemical
reaction to take place.
Higher level
Enzymes
Lock and key model:
• Explained the specificity of substrate to enzyme
• X-ray crystallography and computer modelling did not match
Substances can bind to an
enzyme outside of the
active site to inhibit its
effectiveness. This site is
called an allosteric site
and causes noncompetitive inhibition.
Competitive inhibition is
when a substance that is
not the target substrate
binds directly to the active
site.
Higher level
Enzymes
Higher level
Product inhibition is
the most common
form of competitive
inhibition and is also a
form of negative
feedback or biological
form of Le Chatelier’s
principle helping to
prevent excess product
and so maintain
homeostasis in the
body.
Enzymes – rates of reaction
The maximum velocity Vmax, is the maximum rate of an enzyme under
particular pH and temperature conditions.
The Michaelis constant Km, is the substrate concentration at half of
the Vmax
Enzymes – inhibitors
Competitive inhibitors – attach to the active site of the enzyme
preventing the substrate from doing so
Non-competitive inhibitors – attach to some part of the enzyme
thereby reducing its efficiency
Higher level
Problem 1:
Higher level
ANSWER:
Protein assays determine the concentration of protein
in a sample. A UV-vis spectrophotometer in the UV
range of 280nm absorbs light due to the aromatic
rings in the amino acids of the protein.
Higher level
Protein assay
The Beer-Lambert law (in the data booklet) determines
concentration from absorbance as follows:
A = ɛcL
Where:
ɛ = is the molar absorptivity or extinction coefficient (depends
on nature of solvent and temperature)
c = concentration of the protein
L = cuvette length
Absorbance is calculated as follows:
A = log I0/I where I0 = light intensity before sample
I = light intensity after sample
Higher level
Beer-Lambert law
To identify a particular protein, it first needs to be separated out
by high-performace liquid chromatography (HPLC) and then
compared to a database with information from UV-vis
spectra, gel electrophoresis, HNMR, and mass spectrometry.
Higher level
Qualitative determination of proteins
Higher level
Problem 1:
Higher level
ANSWER:
Higher level
ANSWER:
OBJECTIVES
• Nucleotides are the condensation products of a pentose sugar, phosphoric acid and a nitrogenous base—
adenine (A), guanine (G), cytosine (C), thymine (T) or uracil (U).
• Polynucleotides form by condensation reactions.
• DNA is a double helix of two polynucleotide strands held together by hydrogen bonds.
• RNA is usually a single polynucleotide chain that contains uracil in place of thymine, and a sugar ribose in
place of deoxyribose.
• The sequence of bases in DNA determines the primary structure of proteins synthesized by the cell using a
triplet code, known as the genetic code, which is universal.
• Genetically modified organisms have genetic material that has been altered by genetic engineering
techniques, involving transferring DNA between species.
• Explanation of the stability of DNA in terms of the interactions between its hydrophilic and hydrophobic
components.
• Explanation of the origin of the negative charge on DNA and its association with basic proteins (histones) in
chromosomes.
• Deduction of the nucleotide sequence in a complementary strand of DNA or a molecule of RNA from a
given polynucleotide sequence.
• Explanation of how the complementary pairing between bases enables DNA to replicate itself exactly.
• Discussion of the benefits and concerns of using genetically modified foods.
(Structures of the nitrogenous bases and ribose and deoxyribose sugars are
given in the data booklet in section 34. Knowledge of the different forms of RNA is not required. Details of
the process of DNA replication are not required. Limit expression of DNA to the concept of a four-unit base
code determining a twenty-unit amino acid sequence. Details of transcription and translation are not
required.)
Higher level
B.8 Nucleic acids
Nucleic acid
Nucleic acids refer to the acidic molecules DNA
(deoxyribonucleic acid) and RNA (ribonucleic acid).
The difference beween the two is located in the
pentose sugars for the two types of DNA. The hydroxyl
group is missing from C2 of DNA sugars.
Pentose: Deoxyribose sugar
Pentose: Ribose sugar
Nucleic acid backbone structure
The backbone
consists of a sugar
phosphate polymer.
The linkages with
phosphate are on
carbons 3’ and 5’ of
the pentose.
Nucleic acid – backbone structure
DNA is double
stranded,
where RNA is
only single
stranded in
humans.
The DNA twists
in a structure
known as an
α–helix.
Anionic nucleic acids wrap
around basic positively charged
proteins called histones.
At physiological pH (7.4), outer
phosphate groups are ionized
as a large anion. It is
hydrophilic forming hydrogen
bonds with water.
Hydrophobic and hydrophilic
parts of the polynucleotide
changes stabilize the DNA helix
and make it resistant to
chemical cleavage.
Nitrogenous bases
Bases (purines and pyrimidines) that are found in DNA are A=T, G≡C. Bases
found in RNA are A=U, G≡C
• PURINES
1. Adenine (A)
• PYRIMIDINES
1. Thymine (T)
2. Guanine (G)
2. Cytosine (C)
3. Uracil (U)
Nucleotides
Nucleotides form by making linkages from the nitrogen
of a base to the ribose sugar group. A nucleotide
consists of a phosphate group, a pentose, and a base. All
these linkages form through condensation reactions.
These nucleotides make up the monomers that through
condensation reactions form the nucleic acid polymer.
Nucleotides link up by
hydrogen bonding,
between oxygen from
carboxyl groups and
hydrogen from amine
groups.
Summary:
DNA
vs
RNA
polymers
DNA – synthesis
DNA replicates by complementary
pairing. The order of bases is known
as the primary structure.
Genetic code
All organisms use the same genetic code consisting of
three nucleotide sequences (triplets or codons) that
determine an amino acid of a polypeptide chain.
Genetically modified organisms (GMOs)
Genetically modified organisms are created when the
DNA of one species is inserted into the DNA of another
species.
An example is Bt corn which contains a gene from a
bacterium that produces a toxin to destroy a common
pest to the corn plant.
Bt corn (GMOs)
Most corn is now Bt corn!
Benefits of GMOs
•
•
•
•
•
•
Longer shelf life
Improved flavour, texture, nutritional value
Increased resistance to diseases and pests
Producing a supply of vitamins and vaccines
Increased crop yeilds
Tolerance for wider growing conditions eg. drought
(Taken directly from Pearson 2014)
Concerns of GMOs
•
•
•
•
•
•
Lack of information about long-term effects
Changes to the natural ecosystem from crosspollination
Possible links to allergies
Altering the natural composition of food
Concerns of breeding species that are resistant to
control
Lack of information on food labeling
(Taken directly from Pearson 2014)
OBJECTIVES
• Biological pigments are coloured compounds produced by metabolism.
• The colour of pigments is due to highly conjugated systems with delocalized electrons, which have intense absorption
bands in the visible region.
• Porphyrin compounds, such as hemoglobin, myoglobin, chlorophyll and many cytochromes are chelates of metals with
large nitrogen-containing macrocyclic ligands.
• Hemoglobin and myoglobin contain heme groups with the porphyrin group bound to an iron(II) ion.
• Cytochromes contain heme groups in which the iron ion interconverts between iron(II) and iron(III) during redox reactions.
• Anthocyanins are aromatic, water-soluble pigments widely distributed in plants. Their specific colour depends on metal
ions and pH.
• Carotenoids are lipid-soluble pigments, and are involved in harvesting light in photosynthesis. They are susceptible to
oxidation, catalysed by light.
• Explanation of the sigmoidal shape of hemoglobin’s oxygen dissociation curve in terms of the cooperative binding of
hemoglobin to oxygen.
• Discussion of the factors that influence oxygen saturation of hemoglobin, including temperature, pH and carbon dioxide.
• Description of the greater affinity of oxygen for foetal hemoglobin.
• Explanation of the action of carbon monoxide as a competitive inhibitor of oxygen binding.
• Outline of the factors that affect the stabilities of anthocyanins, carotenoids and chlorophyll in relation to their structures.
• Explanation of the ability of anthocyanins to act as indicators based on their sensitivity to pH.
• Description of the function of photosynthetic pigments in trapping light energy during photosynthesis.
• Investigation of pigments through paper and thin layer chromatography.
(The structures of chlorophyll, heme B and specific examples of anthocyanins and carotenoids are given in the data booklet
in section 35; details of other pigment names and structures are not required. Explanation of cooperative binding in
hemoglobin should be limited to conformational changes occurring in one polypeptide when it becomes oxygenated.
Knowledge of specific colour changes with changing conditions is not required.)
Higher level
B.9 Biological pigments
Biological pigments are coloured compounds produced
by metabolism. They absorb light in the visible
spectrum due to the lower energy transitions of
delocalized electrons in the p orbitals. Hence these
compounds include many alternating single and double
bonds called conjugation. The part of the molecule
absorbing light is called the chromophore.
Biological pigments include:
1. Cartotenes
2. Porphyrins
3. Anthocyanins
Higher level
Biological pigments
Higher level
The major porphyrins are:
a. Hemoglobin (and myoglobin)
b. Cytochromes
c. Chlorophyll
Higher level
The transition metal ion in porphyrins further adds to
the ability of the molecule to interact with visible light.
Carotenes are long hydrocarbon chains containing
many double bonds. Carotenes are fat soluble and
sensitive to photo-oxidation. They are also involved in
increasing the efficiency of photosynthesis.
Retinol is a carotene that absorbs at around 400-420nm
and so appears yellow. This is one of the pigments that
give fat tissue its yellow colour.
Higher level
1. Carotenes
Higher level
Β-Carotenes is a carotene that absorbs
at longer wavelengths of around 430480nm due to a longer hydrocarbon
chain as so more conjugation. It makes
carrots orange and flamingos pink.
Porphyrins are biological pigments that consist of a
complex of metal ions and a large cyclic ligand. This
ligand is called porphin and contains four nitrogen
atoms in a conjugated aromatic (cyclical with resonance)
heterocycle (a cyclical compound of more than one type
of atom).
Porphin
Higher level
2. Porphyrins
Heme group
Higher level
Hemoglobin and myoglobin are responsible for
transport and release of oxygen. Both contain a heme
group with the iron ion able to form further
coordination bonds with histidine residue in protein
and free electron pairs in oxygen or water.
Higher level
The iron (II) in myoglobin makes meat red and the
oxidation to iron (III) with cooking changes it to
brown.
Higher level
The hemoglobin molecule consists of four
subunits. Each subunit contains a heme
prosthetic group. So each molecule can hold
four oxygen molecules.
Higher level
Bound oxygen changes the shape
of the entire molecule including
the subunits, which increases the
ability of the molecule to further
bind oxygen.
This is called cooperative binding.
This is why the oxygen affinity
curve is sigmoidal.
This is important to help
aid the diffusion of
oxygen across the
placenta.
Higher level
Fetal hemoglobin (HbF)
has a higher affinity for
oxygen than adult
hemoglobin (HbA).
Higher level
Other factors influencing hemoglobin's affinity for O2:
• decreased with increased temperature due to
conformational changes at active sites and positive
entropy of dissociation
• decreased pH (slightly) and increased pCO2 act as noncompetitive inhibitors allowing more oxygen release in
venous blood where tissues may be oxygen deprived
Higher level
In hemoglobin,
carbon monoxide
acts as a
competitive
inhibitor to oxygen.
Levels above 0.5%
in the air can be
fatal. Hemoglobin’s
affinity for CO2 is
about 200 times
greater than that for
O2.
Active site of cytochrome c oxidase
Higher level
Cytochromes are enzymes contain heme groups with
metal ions (iron, copper, magnesium, zinc) that get
oxidised as oxygen is reduced to water through
aerobic respiration.
Higher level
Chlorophyll is the primary pigment in photosynthesis.
The heme group always contains magnesium.
Higher level
Besides chlorophyll other accessory pigments also
absorb light in different parts of the spectrum and
pass this energy to chlorophyll.
Higher level
Captured energy is processed down the electron
transport chain which is a series of energy rich
molecules undergoing redox reactions to oxidise water
and synthesis ATP.
Anthocyanins are biological pigments that are
water soluble and concentrated in the vacuoles of
plants producing bright colours.
They are used to attract animals for pollination and
seed dispersal, as antioxidants and protect against
UV light.
Higher level
3. Anthocyanins
Higher level
Anthocyanins can act as pH indicators because the
transfer of H+ ions from hydroxyl groups in acidbase reactions alter their conjugation and so alter
the colours they absorb. Hence their colour
depends on their metal ions and pH.
OBJECTIVES
• With one exception, amino acids are chiral, and only the L-configuration is found in proteins.
• Naturally occurring unsaturated fat is mostly in the cis form, but food processing can convert it
into the trans form.
• D and L stereoisomers of sugars refer to the configuration of the chiral carbon atom furthest
from the aldehyde or ketone group, and D forms occur most frequently in nature.
• Ring forms of sugars have isomers, known as α and β, depending on whether the position of
the hydroxyl group at carbon 1 (glucose) or carbon 2 (fructose) lies below the plane of the ring
(α) or above the plane of the ring (β).
• Vision chemistry involves the light activated interconversion of cis- and trans-isomers of
retinal.
• Description of the hydrogenation and partial hydrogenation of unsaturated fats, including the
production of trans-fats, and a discussion of the advantages and disadvantages of these
processes.
• Explanation of the structure and properties of cellulose, and comparison with starch.
• Discussion of the importance of cellulose as a structural material and in the diet.
• Outline of the role of vitamin A in vision, including the roles of opsin, rhodopsin and cis- and
trans-retinal.
(Names of the enzymes involved in the visual cycle are not required. Relative melting points of
saturated and cis-/trans-unsaturated fats should be covered.)
Higher level
B.10 Stereochemistry in biomolecules
S
C
Higher level
Isomers (organic review)
Amino acids (except glycine) are optical isomers
(enantiomers) because they have a chiral carbon.
Except for some bacterial cell walls and antibiotics,
amino acids occur as L-isomers.
Higher level
Amino acids
Higher level
The CORN rule:
1. Place the H facing away from you
2. If COOH, R, NH2, groups are spelt clockwise it is a
D-enantiomer, if anti-clockwise it is an L-enantiomer.
Most fatty acids occur in the cis form:
Higher level
Fatty acids and triglycerides
Trans fats have been linked to heart disease.
Higher level
High temperature and hydrogenation in food processing
can turn fatty acids to the trans form:
This turns them into saturated fatty acids. This makes oils
spreadable, more resistant to heat, and have a longer shelf
life.
Higher level
Triglycerides can be hydrogenated with high temperatures
and a Ni or Pd catalyst:
L or D enantiomers are
determined from the chiral carbon
furtherest away from the most
senior functional group.
Only D forms are found in nature.
Higher level
Carbohydrates
Higher level
Cyclic sugars also have α and β isomers which are
determined by the hydroxyl group on C1 in glucose. α
isomers have the hydroxyl group on the opposite side to
the C6 group. β isomers have the hydroxyl group on the
same side as the C6 group.
Higher level
Similarly for fructose but the focus is on the C2.
In the rods of your eyes, cisretinal (derived from
Vitamin A) combines with
the opsin protein to
produce rhodopsin in which
trans-retinal is created with
the interaction of light.
Trans-retinal dissociates
from opsin which sends an
electrical impulse. Enzymes
convert the trans back to cis
to complete the vision
cycle.
Higher level
Retinal and vision chemistry
Higher level
The cis to trans conversion occurs as follows: