Biochemistry (Option B)

Question 1

What are biological polymers? How do they form? Provide an example

Answer 1

Biological polymers (biopolymers) are formed by condensation
reactions. These involve the reaction between two smaller
molecules to form one larger molecule with the evolution
o a small molecule such as water. For condensation
polymerization to occur each reacting molecule must possess
at least two reactive unctional groups. Classic examples
include the condensation o amino acids to form proteins and
the condensation o sugars to form starch

Question 2

Describe and explain hydrolysis

Answer 2

Hydrolysis is the reverse of condensation. A molecule is hydrolysed when a water molecule (often in the presence
of acid or alkali) reacts with a larger molecule to break a
bond and form two smaller molecules.

Question 3

State and explain the state of amino acids at room temperature

Answer 3

They are solids at room temperature and have quite high melting points. This solid state is because they
can exist as zwitterions in which the hydrogen atom from the carboxylic acid group protonates the amine group to form a carboxylate anion and a substituted ammonium cation within the same compound

Question 4

What is a secondary structure of a protein?

Answer 4

The secondary structure describes the way in which the chain of amino acids folds itself due to intramolecular hydrogen bonding.

Question 5

What is the tertiary structure and what forms it?

Answer 5

The tertiary structure describes the overall folding of the chain by interactions between distant amino acids (intra molecular) to give the protein its three-dimensional shape. These interactions may be due to
-hydrogen bonds,
-London dispersion forces between non- polar side groups
-ionic attractions between polar groups
-disulfide bridges
-hydrophobic interactions

Question 6

What is a quaternary structure of a protein? what forms it?

Answer 6

Separate polypeptide chains can interact together (inter molecular bonding) to give a more complex structure of this is known as the quaternary structure.

These interactions may be due to
-hydrogen bonds,
-London dispersion forces between non- polar side groups
-ionic attractions between polar groups
-disulfide bridges
-hydrophobic interactions

Question 7

State and explain the structure, function and properties of fibrous and globular protein providing one example for each

Answer 7

Haemoglobin is an example of a globular protein. Globular proteins have complex tertiary and sometimes quaternary structures (e.g. haemoglobin) folded into spherical (globular) shapes. They are usually soluble to some extent in water as the hydrophobic side chains tend to be in the centre of the structure. Fibrous proteins, such as collagen, have little or no tertiary structure and form long parallel polypeptide chains. Fibrous proteins have cross-linking
at intervals to form long fibres or sheets and have mainly structural roles such as keratin in hair and collagen, which is found in skin and the walls of blood vessels and acts as connective tissue.

Question 8

Describe how you would carry out paper chromatography on amino acids

Answer 8

A small spot of the unknown amino acid sample is placed near the bottom of a piece of chromatographic paper. The paper is placed in a solvent, which then rises up the paper due to capillary action. As it meets the sample spots the different amino acids partition themselves between the solvent and
the paper to different extents, and so move up the paper at different rates. When the solvent has nearly reached the top, the paper is removed from the tank, dried, and then sprayed with a location agent (ninhydrin) and placed in a heated oven to develop the chromatogram by colouring the acids. The positions of all the spots can then be compared using Rf values.

Question 9

Formula for Rf

Answer 9

Rf value = distance travelled by sample /
distance travelled by solvent

Rf value < 1 (ALWAYS)

Question 10

What is the isoelectric point of proteins

Answer 10

For each amino acid there is a unique pH value (known as the isoelectric point) where the acid will exist as the zwitterion.

Question 11

Predict with an explanation whether an amino acid will move towards the cathode or anode during electrophoresis when:

pH< isoelectric point

pH = isoelectric point

ph> isoelectric point

Answer 11

The structure of amino acids alters at different pH values. At low pH (acid medium) the amine group will be protonated, thus a general positive charge on the molecule and will move towards the negative anode. At high pH (alkaline medium) the carboxylic acid group will lose a proton and thus the molecule will have a general negative charge and move towards the positive cathode.

Question 12

Explain why amino acids can function as buffers

Answer 12

The structure of amino acids alters at different pH values. At low pH (acid medium) the amine group will be protonated. At high pH (alkaline medium) the carboxylic acid group will lose a proton. This explains why amino acids can function as buffers. If H+ ions are added they are removed as NH4 + and if OH- ions are added the COOH loses a proton to remove the OH- ions as water

Question 13

State and explain the effect of temperature, pH and heavy metals on enzyme activity

Answer 13

The action o an enzyme depends on its specific shape. Increasing the temperature will initially increase the rate
of enzyme-catalysed reactions, as more of the reactants
will possess the minimum activation energy. Additionally the collision frequency will increase according to kinetic theory. The optimum temperature for most enzymes is about 40ºC. Above this temperature enzymes rapidly become denatured as the weak bonds holding the tertiary structure together break.

At different pH values the charges on the amino acid change affecting the bonds between them, and so altering the tertiary structure and making the enzyme ineffective.

Heavy metals can poison enzymes by reacting with -SH groups replacing the hydrogen atom with a heavy metal atom or ion so that the tertiary structure is altered.

Question 14

List three types of lipids

Answer 14

Three important types of lipids are:
-triglycerides (fats and oils)
-phospholipids (lecithin)
-steroids (cholesterol).

Question 15

What are fats and oils and how do they form?

Answer 15

Fats and oils are triesters (triglycerides) formed from the condensation reaction of propane- 1,2,3-triol (glycerol) with long chain carboxylic acids (fatty acids).

Question 16

What are the four components of a phospholipid?

Answer 16

A backbone such as propane-1,2,3-triol (glycerol), linked by esterifcation to two fatty acids and a phosphate group which is itself condensed to a nitrogen-containing alcohol

Question 17

What is characteristic of all steroid structures. Provide an example

Answer 17

Cholesterol has the characteristic our-ring structure possessed by all steroids

Question 18

Explain why unsaturated fats exist as oils at room temp and saturated fats are solid at room temp

Answer 18

The regular tetrahedral arrangement of saturated fatty acids means that they can pack together closely, so the London dispersion forces holding molecules together are stronger as the surface area between them is greater. As the bond angle at the C=C double bonds changes from 109.5º to 120º they are unable to pack so closely and the London dispersion forces between the molecules become weaker, which results in lower melting points

Question 19

Describe how hydrogenation of unsaturated fats occurs and discuss the possible concerns

Answer 19

Unsaturated fats can be hydrogenated to saturated fats by adding hydrogen under pressure in the presence of a heated nickel catalyst.

However during the hydrogenation process, partial hydrogenation can occur and the trans-isomers may be formed. Unlike natural mono- and poly-unsaturated oils, trans-unsaturated fats increase the formation of LDL cholesterol (bad cholesterol) and thus increase the risk of heart disease.

Question 20

When has a lipid gone rancid (common terms)

Answer 20

Lipids (fats and oils) in food become rancid when our senses perceive them to have gone off due to a disagreeable smell, texture or appearance.

Question 21

Describe the two ways in which a lipid may go rancid

Answer 21

-Hydrolysis o the triesters (hydrolytic rancidity) as shown above to produce disagreeable smelling fatty acids

-Oxidation of the fatty acid chains due to the addition of oxygen across the C=C double bonds in unsaturated fatty acids. The process proceeds by a free radical mechanism catalysed by light in the presence of enzymes. Volatile aldehydes and ketones are the products that make the fat rancid.

Question 22

What does 1/2 Vmax mean?

Answer 22

This is the substrate concentration at which the reaction rate is equal to one half its
maximum value. In other words, [S] = Km when the rate is Vmax/2.

Question 23

What is the effect of adding excess substrate to enzyme solution which contains;

A competitive inhibitor

Non-competitive inhibitor

Answer 23

In the comp. situation adding excess substrate will allow Vmax too be reached as the probability of substrate-enzyme complex forming increases

In non-comp situation excess reactant will have no effect as inhibitor is still able to affect the enzymes allosteric site irreversibly

Question 24

What is end product inhibition?

Answer 24

The product of a reaction sometimes acts as an inhibitor of the enzyme for its synthesis, thereby setting up a feedback loop regulating its own concentration. This is known as product inhibition.

Question 25

Beer–Lambert law

Answer 25

I0 = the intensity of light before passing through the sample
I = the intensity of light after passing through the sample

The absorbance depends on:
* the molar absorptivity, ε, defined as the absorbance of a 1.00 mol dm–3 solution in a 1.00 cm cell at a specified wavelength
* the concentration of the solution, c * the path length, l.

log10(I0/I) = εlc

Question 26

Are Saturated or Unsaturated fats more stable and why>

Answer 26

As they cannot undergo auto-oxidation, saturated fats are more stable than unsaturated fats.

Question 27

4 roles of lipids in the body

Answer 27

-Energy storage. Because they contain proportionally less oxygen than carbohydrates they release more energy when oxidized.

-Insulation and protection of organs. Fats are stored in adipose tissue, which provides both insulation and protection to parts of the body.

-Steroid hormones. Examples include female and male sex hormones such as progesterone and testosterone and the contraceptive pill. Sometimes steroids are abused. Anabolic steroids have similar structures to testosterone and are taken to build up muscle.

-Cell membranes. Lipids provide the structural component of cell membranes.

Question 28

What is the iodine number and how is it calculated?

Answer 28

Since one mole of iodine will react quantitatively with one mole of C=C double bonds. Iodine is coloured. As the iodine is added to the unsaturated fat the purple colour of
the iodine will disappear as the addition reaction takes place. Once the colour remains the amount o iodine needed to react with all the C=C double bonds can be determined. Often fats are described by their iodine number, which is the number of grams o iodine that add to 100 g o the at.

Question 29

What are the features of carbohydrates and how may they be subdivided?

Answer 29

they contain a carbonyl group (C=O) and at least two -OH groups. I the carbonyl group is an aldehyde they are known as an aldose, if the carbonyl
group is a ketone they are known as a ketose.

Question 30

What are the most common forms of amino acids and carbohydrates found i nature?

Answer 30

D - Sacharides
L - Amino acids

Question 31

What are the most common forms of amino acids and carbohydrates found i nature?

Answer 31

D - Sacharides
L - Amino acids

Question 32

Roles of Carbohydrates in the body

Answer 32

-to provide energy: foods such as bread, biscuits, cakes, potatoes and cereals are all high in carbohydrates.

-to store energy: starch is stored in the livers of animals in the form of glycogen, also known as animal starch. Glycogen has almost the same chemical structure as amylopectin.

-as precursors for other important biological molecules, e.g. they are components of nucleic acids and thus play an important role in the biosynthesis of proteins.

Question 33

What type of link forms between two sugar monomers?

Answer 33

The link between the two sugars is known as a glycosidic link.

Question 34

What are the different forms of starch. Describe their structure and purpose

Answer 34

Starch exists in two forms: amylose, which is water soluble, and amylopectin, which is insoluble in water.

Amylose is a straight chain polymer of d-glucose units with 1,4 glycosidic bonds

Amylopectin also consists of d-glucose units but it has a branched structure with both 1,4 and 1,6 glycosidic bonds:

Question 35

What are micronutrients and what are their functions?

Answer 35

Micro-nutrients are substances required in very small amounts.

They mainly function as a co-factor of enzymes and include not only vitamins but also trace minerals

Question 36

Features and uses of Vitamin A (retinol)

Answer 36

vitamin A is at soluble due to the long non-polar hydrocarbon chain. Unlike most other vitamins it is not broken down readily by cooking. Vitamin A is an aid to night vision.

Question 37

Features and uses of Vitamin C (ascorbic acid)

Answer 37

Due to the large number of polar -OH groups vitamin C is soluble in water so is not retained for long by the body.

Vitamin C oxidises readily

The most famous disease associated with a lack of vitamin C is (scurvy).

Question 38

Features and uses of Vitamin D (calciferol)

Answer 38

A large hydrocarbon with one OH group and is fat soluble.

A defciency of vitamin D leads to bone softening and malformation condition known as rickets.

Question 39

What are most vitamins subject to>

Answer 39

Most vitamins are sensitive to heat.

Question 40

What are xenobiotics

Answer 40

Xenobiotics refer to chemicals that are found in an organism that are not normally present there (at all or at high concentration).

Question 41

What are biodegradable plastics?

Answer 41

Biodegradable plastics are plastics capable of being broken down by bacteria or other organisms, ultimately to carbon dioxide and water. They are based on natural renewable polymers containing ester or glycosidic links, such as starch, that can be hydrolysed

Question 42

Why are plastics made from starch not a fix?

Answer 42

In theory, starch-based bioplastics produced as biomass could be almost carbon neutral but there are problems such as using land that could otherwise be used for growing food and the release of the greenhouse gas methane if the plastics are decomposed anaerobically in landfill sites.

Question 43

Where may enzymes be used for biodegradbility?

Answer 43

Enzymes are used to aid the breakdown and dispersal of oil spills.

The use of enzymes in biological detergents. This has the advantage to the environment of lowering the temperature at which clothes need to be washed so making the process more effcient.

Question 44

Describe and explain Host-Guest chemistry, and provide an example of its use.

Answer 44

Host-guest complexes are made
of two or more molecules or ions bonded together through non- covalent bonding, which is critical in maintaining the 3-D structure o the molecule. Non-covalent interactions include hydrogen bonds, ionic bonds and van der Waals forces. These
forces, which are weaker than covalent bonding, allow large molecules to
bind specifcally but transiently to one another to form supramolecules. They work by mimicking some of the actions performed by enzymes by selectively binding to guest species.

They can be used used to remove toxic materials (xenobiotics) from the environment. For example, radioactive 137-caesium from nuclear waste and carcinogenic amines from polluted water.

Question 45

What is biomagnification?

Answer 45

Biomagnification is the increase in concentration of a substance in a food chain.

Question 46

What is the aim of green chemistry?

Answer 46

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.

Question 47

Discuss xenobiotics in the context of antibiotics and its biomagnification in humans

Answer 47

Antibiotics, for example, are not produced by animals nor are they part of a normal diet. The use of antibiotics in animal feed and in sewage plants has meant that they pass through into the human food chain and increase resistant strains of bacteria

Question 48

Discuss xenobiotics in the context of Dioxins and its biomagnification in humans

Answer 48

Dioxins can be formed when polymers are combusted. They do not decompose in the environment and can be passed on in the food chain. Many dioxins, are highly carcinogenic as they can disrupt the endocrine system (hormone action) and lead to cellular and genetic damage.

Question 49

Discuss xenobiotics in the context of DDT and its biomagnification in animals

Answer 49

DDT is an effective insecticide particularly against the malaria mosquito but its use is now banned as it accumulates at high levels in birds of prey, which threatens their survival.

Question 50

What is the importance of buffers in the human body ?

Answer 50

Enzymes only function efficiently within a narrow pH region. Outside of this region the structure is altered and the enzyme becomes denatured, hence the need for buffering.

Question 51

How can an acidic buffer be created?

Answer 51

An acidic buffer solution can be made by mixing a weak acid together with a salt (produced from the weak acid and a strong base)

Question 52

Explain how an acidic buffer opposes a decrease or increase in pH

Answer 52

The weak acid is only slightly dissociated in solution, but the salt is fully dissociated into its ions, so the concentration of the acids/salts anions (neutral in terms of its affect on pH) ions is high.

If an acid is added the extra H+ ions coming from the acid are removed as they combine with the acids/salts anions to form undissociated weak acid (which doesn’t dissociate much), so
the concentration of H+ ions remains unaltered.

If an alkali is added the hydroxide ions from the alkali are removed by their reaction with the H+ ions from the weak acid to form water, so again the H+ ion concentration stays constant.

Question 53

Explain how Alkali buffers made up of ammonia and ammonium chloride works

Answer 53

NH4Cl(aq) –> NH4 + (aq) + Cl- (aq)

NH3(aq) + H2O (l) <–>NH4+( a q ) + OH-( aq)

If H+ ions are added they will combine with OH- ions to form water and more of the ammonia will dissociate to replace them.

If more OH- ions are added they will combine with ammonium ions to form undissociated ammonia.

In both cases the hydroxide ion concentration and the hydrogen ion concentration remain constant.

Question 54

Calculate the pH of a buffer containing 0.200mol of sodium ethanoate in 0.5dm3 of 0.1mol/dm3 ethanoic acid.

Ka for ethanoic acid = 1.8 x 10^(-5) mol/dm3

Answer 54

Total volume of solution = 0.5dm3

[CH3COO-] = 0.200/0.5 = 0.400mol/dm3
[CH3COOH] = 0.100mol/dm3 (given)

Ka = 1.8 x 10^(-5) mol/dm3 (given)
Ka = ( [H+] x 0.400 ) / 0.100

1.8 x 10^(-5) mol/dm3 = ( [H+] x 0.400 ) / 0.100
{solve for [H+}

[H+] = 4.5 x 10^(-6) mol/dm3

pH= 5.35

Question 55

Calculate what mass of sodium propanonate must be dissolved in 1.00dm3 of 1.00mol/dm3 propanoic acid (pKa = 4.87) to give a buffer solution with pH 4.5

Answer 55

Ka = ( [H+] x [C2H5COO-] ) / [C2H5COOH]
{rearrange}

[C2H5COO-] = ( Ka x [C2H5COOH] ) / [H+]
= ( 10^(-4.87) x 1.00 ) / 10^(-4.5)

= 0.427 mol/dm3

Mr (SODIUM propanoate) = 96.07

0.427 x 96.07 = 41.0g

Question 56

On what two relationships is UV-VIS spectroscopy based?

Answer 56

-Protein is added to dye and the intensity of the colour depends upon the concentration of the protein in the solution. The coloured complex with the dye absorbs light at a particular wavelength

-Beer-Lmabert equation

Question 57

What are nucleic acids made of?

Answer 57

Nucleic acids are made up of repeating base-sugar- phosphate units called nucleotides.

Question 58

Describe the structure of DNA with reference to its stability.

Answer 58

In DNA, the polynucleotide units are wound into a helical shape with about 10 nucleotide units per complete turn. Two helices are then held together by hydrogen bonds between the bases to give the characteristic double helix structure. The helices run anti-parallel to each other The stability of this double helix structure is due to the base-stacking interactions between the hydrophilic and hydrophobic components as well as hydrogen bonding between the nucleotides. The hydrogen bonds are very specific. Cytosine can only hydrogen bond with guanine and adenine can only hydrogen bond with thymine (uracil in RNA)

Question 59

How is DNA compacted in a eukaryotic cell?

Answer 59

The DNA is compacted efficiently in the eukaryotic nucleus by forming DNA-protein complexes with histones. Histones are positively charged proteins that bond tightly to the negatively charged phosphate groups in the DNAs phosphate-sugar backbone.

Question 60

Benefits and Concerns of GM foods

Answer 60

Benefits:

-enhance the taste, favour, texture and nutritional value and also increase the maturation time.

-Plants can be made more resistant to disease, herbicides and insect attack.

• GM foods can increase resistance to disease, increase productivity and feed efficiency to give higher yields of milk and eggs.

-Increased amounts of vitamins (such as vitamin A in rice) could be incorporated and exposure to less healthy fats reduced.

-Environmentally friendly bio-herbicides and bio- insecticides can be formed. GM foods can lead to soil, water and energy conservation and improve natural waste management.

Concerns:

-The outcome of alterations is uncertain as not enough is known about how genes operate.

-They may cause disease as antibiotic-resistant genes could be passed to harmful microorganisms.

-Genetically engineered genes may escape to contaminate normal crops with unknown effects.

-They may alter the balance of delicate ecosystems as food chains become damaged.

-There are possible links to an increase in allergic reactions (particularly with those involved in food processing) .

Question 61

Which compounds can absorb light?

Answer 61

Organic compounds containing unsaturated groups such as C=C, C=O, N=N, NO and the benzene ring can absorb in the
ultraviolet or visible part of the spectrum. Such groups are known as chromophores and the precise energy of absorption is affected
by the other groups attached to the chromophore. The absorption is due to electrons in the bond being excited to an empty orbital of higher energy, usually an anti-bonding orbital. The energy involved in this process is relatively high and most organic compounds absorb in the ultraviolet region and thus appear colourless.

Question 62

What causes anthocyanins to be colorful?

Answer 62

The conjugation of the pi electrons contained in this structure that accounts for the colour of anthocyanins

In acidic solution anthocyanins form a positive ion and there is less conjugation than in alkaline solution where the pi electrons in the extra double bond between the carbon and oxygen atom are also delocalized.

Difference in groups (e.g. hydroxyls) affect the wavelength absorbed and hence color.

Question 63

What are anthocyanins?

Answer 63

Anthocyanins are aromatic, water-soluble pigments widely distributed in plants. They contain the favonoid C6C3C6 skeleton

Question 64

What are Carotenoids?

Answer 64

Carotenoids are lipid-soluble pigments, and are involved in harvesting light in photosynthesis.

Question 65

Wat causes electron conjugation in caratenoids?

Answer 65

The conjugation in carotenoids is mainly due to a long hydrocarbon chain consisting of alternate single and double carbon to carbon bonds.The conjugation in carotenoids is mainly due to a long hydrocarbon chain consisting of alternate single and double carbon to carbon bonds.

Question 66

Why and ow does vitamin A go rancid?

Answer 66

Because of the unsaturation in the double bond carotenoids are susceptible to oxidation. This oxidation process can
be catalysed by light, metals and hydroperoxides. It results in a change of
colour, loss ofactivity in vitamin A and is the cause of bad smells.

Question 67

What are porphyrin compounds?

Answer 67

Porphyrin compounds, are chelates of metals with large
nitrogen-containing macrocyclic ligands. Porphyrins contain a cyclic system in which all the carbon atoms are sp2 hybridized.
This results in a planar structure with extensive pi conjugation. The non-bonding pairs of electrons on the four nitrogen atoms
enable the porphyrin to form coordinate bonds with metal ions.

Question 68

Describe the difference between Chlorophyll A and B

Answer 68

Chlorophyll contains a magnesium ion.
In chlorophyll a, the -R group is a methyl group,
-CH3 , and in chlorophyll b the -R group is an aldehyde group, -CHO

Question 69

Describe the structure of hemoglobin and describe how it reacts with oxygen in a cooperative process.

Answer 69

Porphyrin group bound to an iron(II) ion. When oxygen binds to one of the iron atoms in the complex to form HbO2 it causes the iron atom to move towards the centre of the porphyrin ring and at the same time the imidazole side-chain of a histidine residue is pulled
towards the porphyrin ring. This produces a strain that is transmitted to the remaining three monomers in the quaternary haemoglobin. This brings about a similar conformational change in the other haem sites so that it is easier for a second oxygen molecule to bind to a second iron atom. Each time the haem groups affinity to attract oxygen increases as the remaining sites become filled. Haem becomes fully saturated with oxygen when all four iron atoms have been utilized forming HbO* . The binding of oxygen is, thus, a cooperative process.

Question 70

Why is carbon monoxide dangerous?

Answer 70

Carbon monoxide is a dangerous poison as carbon monoxide is a stronger ligand than oxygen and forms an irreversible complex with the iron in haemoglobin. It thus acts as a competitive inhibitor and prevents the haemoglobin from binding with oxygen. As a result the tissues within the body are not receiving oxygen and can not respire

Question 71

Describe thin layer chromatgraphy

Answer 71

Similar to paper chromatography but uses a thin layer o a solid, such as alumina, Al2O3 , or silica, SiO2 , on an inert support such as glass. When absolutely dry it works by adsorption but, like paper, silica and alumina have a high affinity for water, therefore the separation occurs more by partition with water as the stationary phase. The choice of a suitable solvent depends on the polarity or otherwise of the particular pigment

Question 72

What is the advantage of thin layer chromatagraphy?

Answer 72

One real advantage of thin layer chromatography over paper chromatography is that each of the separated components can be recovered pure. The section containing the component is scraped off the glass and then dissolved in a suitable solvent. The solution is then filtered to remove the solid support and the solvent can then be evaporated to leave just the pure component.

Question 73

What dictates whether a sugar is an D or L isomer?

Answer 73

The d and l
stereoisomers of sugars refer to the configuration on the chiral carbon atom furthest from the aldehyde or ketone group.

Question 74

What is the visual cycle?

Answer 74

The process whereby photons of light are converted into
electrical signals in the retina at the back of the eye is called the
visual cycle.

Question 75

Describe the visual cycle

Answer 75

Rhodopsin in the retina consists of a protein, opsin and a covalently bonded co-factor retinal, which is produced which is produced in the retina from vitamin A. When light falls on the retina it converts the carotenoid retinal from
the cis- to the trans- form and as this happens a nerve impulse is transmitted via an interaction, which causes a conformational change in the structure of opsin to send a signal along the optic nerve to the brain.

Question 76

What must we look at when deciding wether a stereoisomer of sugar is D or L?

Answer 76

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.