3.1 Biological Molecules

Question 1

What are monomers/ Name some examples.

Answer 1

Monomers are small basic units that larger molecules are composed of. Monosaccharides are examples of monomers, and these include glucose, amino acids and nucleotides

Question 2

What monomers are carbohydrates composed of?
Name the 3 elements that all carbohydrates contain.

Answer 2

Monosaccharides (aka simple sugars)
Carbon, hydrogen, oxygen

Question 3

Explain how monosaccharides form di and polysaccharides, including the type of bond involved

Answer 3

By condensation reactions, which is where the joining of two molecules with a chemical bond results in the elimination of a water molecule (it’s released). The bond that forms when this happens is a glycosidic bond

Question 4

What are the two isomers of glucose? Draw and describe the difference in both examples

Answer 4

In alpha glucose, the OH is below the plane of the ring, whereas in beta glucose the OH is above the plane of the-ring

Question 5

What type of monomer is glucose?
Why is glucose important?

Answer 5

Hexose sugar (monosaccharide with 6 carbon atoms in every molecule))
Glucose is the main substrate for respiration

Question 6

What are polymers

Answer 6

long chain of monomers that are repeating units

Question 7

What is formed when 2 monosaccharides join together? Give some examples of these

Answer 7

a disaccharide
-maltose is a disaccharide formed by condensation of two alpha glucose molecules
-sucrose is a disaccharide formed by condensation of an alpha glucose molecule and a fructose molecule
-lactose is a disaccharide formed by condensation of an alpha glucose molecule and a galactose molecule.

Question 8

Name the 3 common monosaccharides and give the general formula

Answer 8

glucose, fructose and galactose
(CH2O)n, where n can be any number from 3 to 7

Question 9

What are polysaccharides? Give some examples and describe the general properties of polysaccharides

Answer 9

-Glycogen and starch are formed by the condensation of α-glucose.
-Cellulose is formed by the condensation of β-glucose.

macromolecules that are polymers formed by many monosaccharides joined by glycosidic bonds in a condensation reaction to form chains. These chains may be:
-Branched or unbranched
-Folded (making the molecule compact which is ideal for storage, e.g. starch and glycogen)
-Straight (making the molecules suitable to construct cellular structures, eg. cellulose) or coiled
-Polysaccharides are insoluble in water

Question 10

What is an organic compound and give some examples?

Answer 10

An organic compound contains carbon and hydrogen. Some examples are carbohydrates, lipids, proteins and nucleic acids.

Question 11

State the difference between a monosaccharide and a disaccharide

Answer 11

A monosaccharide is a single monomer whereas a disaccharide is a pair of monomers

Question 12

Explain the importance of carbon atoms to organic compounds

Answer 12

Each carbon atom can form four covalent bonds – this makes the compounds very stable (as covalent bonds are so strong they require a large input of energy to break them)

Carbon atoms can form covalent bonds with oxygen, nitrogen and sulfur

Carbon atoms can bond to form straight chains, branched chains or rings

Question 13

Explain the structure and function of glycogen

Answer 13

-main energy storage molecule in animals and fungi
-formed by many molecules of alpha glucose joined together by 1,4 and 1,6 glycosidic bonds
-present in the liver and skeletal muscles of the human body
-has a large number of side branches, so energy can be released quickly as enzymes can act simultaneously on these branches
-relatively large but compact molecule thus maximising the amount of energy it can store or release of glucose to suit the demands of the cell
-being insoluble means it will not affect the water potential of cells and cannot diffuse out of cells

Question 14

Explain the structure and function of starch

Answer 14

Starch stores energy in plants as granules in plastids (i.e. chloroplasts) It is a mixture of two polysaccharides: amylose (10-30%) and amylopectin (70-90%)
-Amylose is an unbranched helix-shaped chain of glucose molecules joined by 1,4 glycosidic bonds. As a result, amylose is coiled and thus a very compact molecule storing a lot of energy The helix shape also makes it more resistant to digestion
-Amylopectin is branched and is made up of alpha glucose molecules joined by 1, 4 and 1, 6 glycosidic bonds. Due to the presence of many side branches these can be acted upon simultaneously by many enzymes and thus broken down to release its energy. The branches result in many terminal glucose molecules that can be easily hydrolysed for use during cellular respiration or added to for storage

Some key properties of starch that make it suitable are that:
-its insoluble so will not affect cell water potential
-it is compact so a lot of energy can be stored in a small space
-when it is hydrolysed the released alpha glucose can be transported easily

Question 15

Explain the structure and function of cellulose

Answer 15

-is a component of cell walls in plants
-is composed of long, unbranched chains of beta glucose which are joined by glycosidic bonds
-as β-glucose is an isomer of α-glucose to form the 1,4 glycosidic bonds consecutive β-glucose molecules must be rotated 180° to each other
-due to the inversion of the β-glucose molecules many hydrogen bonds form between the long chains giving cellulose it’s strength
-microfibrils are strong threads which are made of long cellulose chains running parallel to one another that are joined together by hydrogen bonds forming strong cross linkages
-Cellulose is important in stopping the cell wall from bursting under osmotic pressure, because it exerts inward pressure that stops the influx of water
-this means that cells stay turgid and rigid, helping to maximise the surface area of plants for photosynthesis.
The high tensile strength of cellulose allows it to be stretched without breaking which makes it possible for cell walls to withstand turgor pressure
The cellulose fibres and other molecules (eg. lignin) found in the cell wall form a matrix which increases the strength of the cell walls
Cellulose fibres are freely permeable which allows water and solutes to leave or reach the cell surface membrane
As few organisms have the enzyme (cellulase) to hydrolyse cellulose it is a source of fibre

MT:
every other glucose rotates 180 degrees so that the hydroxides are in the same direction, hence the H2O can be removed and the cellulose can be formed through a condensation reaction
-the alternative glucose rotations cause alternating glycosidic bonds

Question 16

How does cellulose make a plant’s cell wall

Answer 16

-water is a di-polar molecule,as the oxygen has a negative charge and the hydrogen has a positive charge
-many cellulose molecules are cross-linked by hydrogen bonds, made by the hydroxyl groups that are exposed on either ends of the molecule
-many micro fibrils are cross-linked by hydrogen bonds to form cellulose fibres, which strengthen the cell wall and stop it from bursting through osmosis

Question 17

Biochemical tests using Benedict’s solution for reducing sugars

Answer 17

Benedict’s reagent is a blue solution that contains copper (II) sulfate ions (CuSO4 ); in the presence of a reducing sugar copper (I) oxide forms
Copper (I) oxide is not soluble in water, so it forms a precipitate
Method

Add Benedict’s reagent (which is blue as it contains copper (II) sulfate ions) to a sample solution in a test tube
Heat the test tube in a water bath or beaker of water that has been brought to a boil for a few minutes
If a reducing sugar is present, a coloured precipitate will form as copper (II) sulfate is reduced to copper (I) oxide which is insoluble in water
It is important that an excess of Benedict’s solution is used so that there is more than enough copper (II) sulfate present to react with any sugar present
A positive test result is a colour change somewhere along a colour scale from blue (no reducing sugar), through green, yellow and orange (low to medium concentration of reducing sugar) to brown/brick-red (a high concentration of reducing sugar)
This test is semi-quantitative as the degree of the colour change can give an indication of how much (the concentration of) reducing sugar present

Question 18

Biochemical test for non-reducing sugars, using sucrose as an example

Answer 18

To test for non-reducing sugars:
Add dilute hydrochloric acid to the sample and heat in a water bath that has been brought to the boil
Neutralise the solution with sodium hydrogencarbonate
Use a suitable indicator (such as red litmus paper) to identify when the solution has been neutralised, and then add a little more sodium hydrogencarbonate as the conditions need to be slightly alkaline for the Benedict’s test to work
Then carry out the Benedict’s test as normal; add Benedict’s reagent to the sample and heat in a water bath that has been boiled – if a colour change occurs, a reducing sugar is present
Explanation
The addition of acid will hydrolyse any glycosidic bonds present in any carbohydrate molecules
The resulting monosaccharides left will have an aldehyde or ketone functional group that can donate electrons to copper (II) sulfate (reducing the copper), allowing a precipitate to form

MT:
-PREPARE A TEST TUBE OF 2CM^3 OF SUCROSE SOLUTION
ADD AN EQUAL VOLUME (2CM^3) OF DILUTE ACID
hEAT/BOIL THE MIXTURE OF ACID AND SUCROSE SOLUTION
ALLOW IT TO COOL DOWN TO ROOM TEMPERATURE
NEUTRALISE THE REACTION MIXTURE BY THE SODIUM HYDROGENCARBONATE (BECAUSE bENEDICT’S SOLUTION REQUIRES AN ALKALINE PH TO WORK)
THEN ADD 2CM^3 OF BENEDICT’S REAGENT
HEAT ABOVE 80 DEGREES FOR ABOUT 2-3 MINUTES
OBSERVATION: BENEDICT’S SOLUTION WLL TURN FROM BLUE TO BRICK RED AS THE cu2+ ions have become a CU+ precipitate

Question 19

Explain what condensation reactions are

Answer 19

A condensation reaction occurs when monomers combine together by covalent bonds to form polymers (polymerisation) or macromolecules (lipids) and water is removed

Question 20

Explain what hydrolysis is and the enzymes that catalyse this in the human body

Answer 20

The glycosidic bond is broken when water is added in a hydrolysis

Disaccharides and polysaccharides are broken down in hydrolysis reactions

Hydrolysis reactions are catalysed by enzymes, these are different to those present in condensation reactions
The enzyme that catalyses the hydrolysis of maltose is maltase.
The enzyme that catalyses the breakdown of lactose is lactase
The enzyme that catalyses the hydrolysis of sucrose is sucrase
These are usually present in the digestion of food in the alimentary tract and the breakdown of stored carbohydrates in muscle and liver cells for use in cellular respiration
Alternatively, the hydrolysis of these disaccharides can be carried out in a test tube in laboratory conditions.. In this case, the hydrolysis would be catalysed by dilute acid (e.g. HCl)

Question 21

What is the importance of disaccharides and polysaccharides, in terms of the glycosidic bond

Answer 21

To make monosaccharides more suitable for transport, storage and to have less influence on a cell’s osmolarity, they are bonded together to form disaccharides and polysaccharides

Disaccharides and polysaccharides are formed when two hydroxyl (-OH) groups (on different saccharides) interact to form a strong covalent bond called the glycosidic bond (the oxygen link that holds the two molecules together)

Every glycosidic bond results in one water molecule being removed, thus glycosidic bonds are formed by condensation

Each glycosidic bond is catalysed by enzymes specific to which OH groups are interacting

As there are many different monosaccharides this results in different types of glycosidic bonds forming (e.g maltose has a α-1,4 glycosidic bond and sucrose has a α-1,2 glycosidic bond)

Question 22

Give the type of glycosidic bond for the following:
Disaccharides:
-maltose
-sucrose

polysaccharides:
-cellulose
-amylose
-amylopectin
-glycogen

Answer 22

-maltose: (alpha) 1,4
-sucrose (alpha) 1,6

-cellulose: (beta) 1,4
-amylose: (alpha) 1,4
-amylopectin: (alpha) 1,4 and (alpha) 1,6
-glycogen: (alpha) 1,4 and (alpha) 1,6

Question 23

Describe what happens in the hydrolysis of sucrose

Answer 23

Sucrose is a non-reducing sugar which gives a negative result in a Benedict’s test. When sucrose is heated with hydrochloric acid this provides the water that hydrolyses the glycosidic bond resulting in two monosaccharides that will produce a positive Benedict’s test

Question 24

work out the molecular formula for disaccharides

Answer 24

2 x α glucose = C12H24O12
DISACCHARIDES ARE MADE BY THE ELIMINATE OF h2o (CONDENSATION REACTION)
Therefore, C12H24O12 –> h2o + c12h22o11
c12h22o11 IS THE NEW SUBSTANCE MADE (MALTOSE IN THIS CASE)

Question 25

biochemical test using iodine/potassium iodide for starch.

Answer 25

To test for the presence of starch in a sample, add a few drops of orange/brown iodine in potassium iodide solution to the sample
The iodine is in potassium iodide solution as iodine is insoluble in water
If starch is present, iodide ions in the solution interact with the centre of starch molecules, producing a complex with a distinctive blue-black colour
This test is useful in experiments for showing that starch in a sample has been digested by enzymes

Question 26

Name some reducing and non-reducing sugars

Answer 26

Reducing:
fructose
maltose
galactose
glucose

Non-reducing:
sucrose

Question 27

Explain the method to find the concentration of glucose

Question 28

What is a lipid?

Answer 28

Lipids are:
-macromolecules which contain only hydrogen, carbon and oxygen. However, unlike carbohydrates, lipids contain a lower proportion of oxygen
-non-polar (electrons have been shared equally so it is not charged)
-hydrophobic (insoluble in water)

The R-group of a fatty acid may be saturated or unsaturated.

In phospholipids, one of the fatty acids of a triglyceride is substituted by a phosphate-containing group.

Question 29

What are the two types of lipids

Answer 29

triglycerides (the main component of fats and oils) and phospholipids

Question 30

Describe what triglycerides are

Answer 30

non-polar, hydrophobic molecules

monomers are glycerol and fatty acids

Glycerol is an alcohol (an organic molecule that contains a hydroxyl group bonded to a carbon atom)

Fatty acids contain a methyl group at one end of a hydrocarbon chain known as the R group (chains of hydrogens bonded to carbon atoms, typically 4 to 24 carbons long) and at the other is a carboxyl group

The shorthand chemical formula for a fatty acid is RCOOH

Fatty acids can vary in two ways:
-Length of the hydrocarbon chain (R group)
-The fatty acid chain (R group) may be saturated (mainly in animal fat) or unsaturated (mainly vegetable oils, although there are exceptions e.g. coconut and palm oil)

Unsaturated fatty acids can be mono or poly-unsaturated:
If H atoms are on the same side of the double bond they are cis-fatty acids and are metabolised by enzymes
If H atoms are on opposite sides of the double bond they are trans-fatty acids and cannot form enzyme-substrate complexes, therefore, are not metabolised. They are linked with coronary heart disease

MT:
-some are liquid at room temperature (oil), whereas others are solid()fat
-glycerol has 3 hydroxyl groups,hence why it needs 3 fatty acids to form a triglyceride
-each fatty acid has 1 carboxyl group (-COOH)
–the carbon bonds with a hydrocarbon tail to make saturated/unsaturated fatty acids (unsaturated fatty acids can be identified by the presence of a double bond)
the hydrocarbon tail is recognised as the R group

Question 31

How are triglycerides formed?

Answer 31

Triglycerides are formed by esterification
-An ester bond forms when a hydroxyl (-OH) group on glycerol bonds with the carboxyl (-COOH) group of the fatty acid:
-An H from glycerol combines with an OH from the fatty acid to make water
-The formation of an ester bond is a condensation reaction
-For each ester bond formed a water molecule is released
-Three fatty acids join to one glycerol molecule to form a triglyceride
-Therefore for one triglyceride to form, three water molecules are released

Question 32

Describe the structure and function of triglycerides

Answer 32

Energy storage
long hydrocarbon chains contain many carbon-hydrogen bonds with little oxygen (triglycerides are highly reduced)
-when triglycerides are oxidised during cellular respiration this causes these bonds to break releasing energy used to produce ATP
-Triglycerides therefore store more energy per gram than carbohydrates and proteins (37kJ compared to 17kJ)
-As triglycerides are hydrophobic they do not cause osmotic water uptake in cells so more can be stored
-Plants store triglycerides, in the form of oils, in their seeds and fruits. If extracted from seeds and fruits these are generally liquid at room temperature due to the presence of double bonds which add kinks to the fatty acid chains altering their properties
-Mammals store triglycerides as oil droplets in adipose tissue to help them survive when food is scarce (e.g. hibernating bears)
-The oxidation of the carbon-hydrogen bonds releases large numbers of water molecules (metabolic water) during cellular respiration
-Desert animals retain this water if there is no liquid water to drink
Bird and reptile embryos in their shells also use this water

Insulation
Triglycerides are part of the composition of the myelin sheath that surrounds nerve fibres
This provides insulation which increases the speed of transmission of nerve impulses
Triglycerides compose part of the adipose tissue layer below the skin which acts as insulation against heat loss (eg. blubber of whales)

Buoyancy
The low density of fat tissue increases the ability of animals to float more easily

Protection
The adipose tissue in mammals contains stored triglycerides and this tissue helps protect organs from the risk of damage

Question 33

Describe the structure and function of phospholipids

Answer 33

Structure:
-Unlike triglycerides, there are only two fatty acids bonded to a glycerol molecule in a phospholipid as one has been replaced by a phosphate ion (PO43-)
-As the phosphate is polar it is soluble in water (hydrophilic)
-The fatty acid ‘tails’ are non-polar and therefore insoluble in water (hydrophobic)
-They have fatty acid tails that are hydrophobic and a phosphate head, that is hydrophilic, attached to a glycerol molecule.

Function:
-The main component (building block) of cell membranes
-Due to the presence of hydrophobic fatty acid tails, a hydrophobic core is created when a phospholipid bilayer forms
-This acts as a barrier to water-soluble molecules
-The hydrophilic phosphate heads form H-bonds with water allowing the cell membrane to be used to compartmentalise
-This enables the cells to organise specific roles into organelles helping with efficiency
-Composition of phospholipids contributes to the fluidity of the cell membrane
-If there are mainly saturated fatty acid tails then the membrane will be less fluid
-If there are mainly unsaturated fatty acid tails then the membrane will be more fluid
-Phospholipids control membrane protein orientation
-Weak hydrophobic interactions between the phospholipids and membrane proteins hold the proteins within the membrane but still allow movement within the layer
-In the presence of water due to the hydrophobic and hydrophilic parts phospholipids will form monolayers or bilayers.

Question 34

Name the monomers that make proteins

Answer 34

Amino acids

Question 35

What do amino acids consist of?

Answer 35

Carbon in the centre
A hydrogen atom
NH2 to represent an amine group
COOH t represent a carboxyl group
R group represents a side chain

Question 36

How are proteins formed

Answer 36

-condensation reaction occurs between amino acids to form a peptide bond
-1 water molecule released in this process
-reverse reaction occurs during digestion

-dipeptides are formed by the condensation of two amino acids
-polypeptides are formed by the condensation of many amino acids

-sequence of amino acids in a polypeptide is determined by the sequence of the genetic code on mRNA being translated in the ribosomes

Question 37

Explain the role of hydrogen bonds, ionic bonds and disulfide bridges in the different protein structures

Answer 37

Primary structure:
-unique sequence of amino acids in polypeptide chain
Primary structure proteins vary depending on 3 factors:
-number of amino acids
-sequence of amino acids
-different R groups

Secondary structure:
-primary structure folds
-held together by hydrogen bonds between the carboxyl groups and amino groups in the polypeptide backbone
-can folded in two ways:
-alpha-helix, a tight, twisted strand
-beta pleated sheet, a zig-zag line of amino acids bonds with the next to form a ribbon shape

Tertiary structure:
-further folding of the secondary structure
-held together by bonds between the R groups of the amino acids in the protein, so depends on the sequence of amino acids
-3 types of bonds involved:
-weak hydrogen bonds
-ionic bonds between R-groups with positive/negative charges
-disulfide bridges - covalent S-S bonds between two cysteine amino acids which are the strongest type of bond, allowing it to withstand higher temperatures and extreme pH
(Note: disulfide bonds are not present in every tertiary structure: only when R groups contain sulfur)

Quaternary structure:
-formed from several polypeptides aka subunits
-common examples include:
-collagen - fibrous protein of three polypeptide chains (trimetric)
-haemoglobin - globular protein with four polypeptide chains (tetrametric)
-insulin - two polypeptide chains (dimetric) held together by disulfide bond

Tertiary and quaternary structure proteins can be globular or fibrous:
Globular proteins:
-spherical
-hydrophilic amino acids on the outside, allowing them to interact with aqueous environment
-hydrophobic amino acids shielded on the inside of the protein, as they are non-polar
-hence globular proteins are soluble in water
Examples of functional proteins:
-plasma proteins in bloodstream
-enzymes are able to catalyse reactions due to their solubility
-haemoglobin in RBCs transport oxygen and contains iron as a prosthetic group (inorganic compound)

Fibrous proteins:
-strands
-both hydrophobic and hydrophilic are exposed to the outside, hence making it insoluble and ideal for structural purposes
Examples of structural proteins:
-collagen (connective tissue)
-elastin (found in alveoli and blood vessels)

Question 38

Describe the test for proteins

Answer 38

Method:
1. Take 2 ml of the solution to be tested in a test tube
2. Add 2 ml of 5% sodium hydroxide solution to make the sample alkaline
3. Mix the solutions
4. Add two drops of 1% copper sulfate solution
5. A colour change from blue to violet indicates the presence of proteins

How it works:
The Cu2+ ions react with the nitrogen of the peptide bond to form a purple colour

Question 39

What is an enzyme

Answer 39

-biological catalyst (speeds up reaction without being used up)
-globular protein
-soluble in water, so can move freely in organelles and cytoplasm

Question 40

How does enzyme specificity relate to tertiary structure of its active site and its ability to combine with complementary substrate(s) to form E-S complexes

Answer 40

-due to complementary nature between the shape of the active site on the enzyme and its substrate(s)

-shape of active site (and hence the specificity of the enzyme) is determined by the complex tertiary structure of the protein that makes up the enzyme:

-proteins are formed from chains of amino acids held together by peptide bonds

-order of amino acids determines the shape of an enzyme

-if the order is altered, the resulting three-dimensional shape changes

-enzyme-substrate complex forms when an enzyme and its substrate join together, but only temporarily, before the enzyme catalyses the reaction and the product(s) are released

Question 41

How do enzymes work in terms of activation energy

Answer 41

-activation energy = amount of energy needed by substrate to become just unstable enough for a reaction to occur and for products to form
-enzymes speed up chemical reactions because they influence the stability of bonds in the reactants
-the destabilisation of bonds in the substrate makes it more reactive
-enzymes work by lowering the activation energy of a reaction and in doing so they provide an alternative energy pathway

Question 42

catabolic/anabolic reactions

Answer 42

Catabolic reactions
-involve the breakdown of complex molecules into simpler products, which happens when a single substrate is drawn into the active site and broken apart into two or more distinct molecules
e.g. cellular respiration and hydrolysis reactions

Anabolic reactions
-involve the building of more complex molecules from simpler ones by drawing two or more substrates into the active site, forming bonds between them and releasing a single product
e.g. protein synthesis and photosynthesis

Question 43

lock and key hypothesis for enzymes

Answer 43

-enzymes are globular proteins, so their shape (as well as the shape of the active site) is determined by the complex tertiary structure of the protein that makes up the enzyme

-therefore highly specific

-first model of enzyme activity

-suggested that both enzymes and substrates were rigid structures that locked into each other very precisely

-later modified and adapted to our current understanding of enzyme activity, permitted by advances in techniques in the molecular sciences

Question 44

induced fit model

Answer 44

-scientists discovered that proteins are not rigid structures
-found that multiple regions of an enzyme molecule moved in response to the environment
-larger movements occurred when the substrate bound to the enzyme

-prior to binding, the substrate and active site and not completely complementary in shape
-when the substrate binds the active site alters shape and moulds around the substrate (known as conformational changes)
-maximises the ability of the enzyme to catalyse the reaction

Question 45

effect of enzyme concentration on enzyme activity

Answer 45

-the higher the enzyme concentration in a reaction mixture, the greater the number of active sites available and the greater the likelihood of enzyme-substrate complex formation

-as long as there is sufficient substrate available, the initial rate of reaction increases linearly with enzyme concentration

-if the amount of substrate is limited, at a certain point any further increase in enzyme concentration will not increase the reaction rate as the amount of substrate becomes a limiting factor

Question 46

effect of substrate concentration on enzyme activity

Answer 46

-the greater the substrate concentration, the higher the rate of reaction

-as the number of substrate molecules increases, the likelihood of enzyme-substrate complex formation increases

-if the enzyme concentration remains fixed but the amount of substrate is increased past a certain point, however, all available active sites eventually become saturated and any further increase in substrate concentration will not increase the reaction rate

-when the active sites are all full, any substrate molecules that are added have nowhere to bind to form an enzyme-substrate complex

Question 47

effect of concentration of competitive and non-competitive inhibitors on enzyme activity

Answer 47

two types of reversible inhibitors:
-competitive inhibitors have a similar shape to that of the substrate molecules and therefore compete with the substrate for the active site
-non-competitive inhibitors bind to the enzyme at an alternative site, which alters the shape of the active site and therefore prevents the substrate from binding to it
-slow down reactions (opposite of catalyst)

Competitive inhibitors:
-active site, substrate and inhibitors all have similar shape
-means that inhibitors bind to active site, which blocks the substrate from binding to active site
-hence less E-S complexes are made, so less products are formed per unit time, so rate of reaction decreases
-tend to be reversible, so bonds are temporary and active site is not changed, meaning the enzyme can still function as a catalyst, just not as effectively

Non-competitive inhibitors:
-do not have the same shape as active site
-therefore, inhibitor binds to any part of the enzyme, away from active site (known as the allosteric site)
-this alters shape of protein, due to enzyme flexibility (proven by induced fit model, but not lock and key)
-therefore, change in active site, so it is no longer complementary to substrate
-hence less E-S complexes are made, so less products are formed per unit time, so rate of reaction decreases
-tend to be irreversible, due to permanent bond between allosteric site and inhibitor

Question 48

effect of temperature on enzyme activity

Answer 48

-enzymes have a specific optimum temperature
-lower temperatures either prevent reactions from proceeding or slow them down:
-molecules move relatively slow
-lower frequency of successful collisions between substrate molecules and active site of enzyme
-frequent enzyme-substrate complex formation
-substrate and enzyme collide with less energy, making it less likely for bonds to be formed or broken (stopping the reaction from occurring)

-higher temperatures speed up reactions:
-molecules move more quickly
-higher frequency successful collisions between substrate molecules and active site of enzyme
-more frequent enzyme-substrate complex formation
-substrate and enzyme collide with more energy, making it more likely for bonds to be formed or broken (allowing the reaction to occur)

However, as temperatures continue to increase, the rate at which an enzyme catalyses a reaction drops sharply, as the enzyme begins to denature:
-bonds (eg. hydrogen bonds) holding the enzyme molecule in its precise shape start to break
-causes the tertiary structure of the protein (ie. the enzyme) to change
-permanently damages the active site, preventing the substrate from binding
-denaturation has occurred if the substrate can no longer bind
-few human enzymes can function at temperatures above 50°C
-due to humans maintaining a body temperature of about 37°C, therefore even temperatures exceeding 40°C will cause the denaturation of enzymes
-high temperatures causes hydrogen bonds between amino acids to break, changing the conformation of the enzyme

Question 49

effect of pH on enzyme activity

Answer 49

-enzymes are denatured at extremes of pH
-hydrogen and ionic bonds hold the tertiary structure of the protein (ie. the enzyme) together
-below and above the optimum pH of an enzyme, solutions with an excess of H+ ions (acidic solutions) and OH- ions (alkaline solutions) can cause these bonds to break
-alters the shape of the active site, which means enzyme-substrate complexes form less easily
-eventually, enzyme-substrate complexes can no longer form at all, so complete denaturation of the enzyme has occurred

Where an enzyme functions can be an indicator of its optimal environment:
-Eg. pepsin is found in the stomach, an acidic environment at pH 2 (due to the presence of hydrochloric acid in the stomach’s gastric juice)
-Pepsin’s optimum pH, not surprisingly, is pH 2

Question 50

nucleotide structure in DNA and RNA

Answer 50

Generally:
A pentose sugar (a sugar with 5 carbon atoms)
A nitrogen-containing organic base
A phosphate group

DNA
-1 deoxyribose sugar with hydrogen at the 2’ position
-a phosphate group
-a nitrogenous base either adenine (A), cytosine(C), guanine(G) or thymine(T)
-by complementary base pairing, hydrogen bonds form between bases
-A and T have 2 hydrogen bonds
-C and G have 3 hydrogen bonds

RNA
-a ribose sugar with a hydroxyl (OH) group at the 2’ position
-a phosphate group
-a nitrogenous base either adenine (A), cytosine(C), guanine(G) or uracil(U)

Question 51

Why are DNA and RNA important in terms of their functions?

Answer 51

-DNA holds genetic information
-contains the instructions for the growth and development of all organisms

-RNA transfers genetic information from DNA to the ribosomes, which read RNA to synthesise proteins in translation

-both needed to build proteins, which are essential for the proper functioning of cells

Question 52

What are purines and pyramidines

Answer 52

-nitrogenous base molecules that are found in DNA (A, T, C, G) and RNA (A, U, C, G) occur in two structural forms: purines and pyrimidines

-adenine and guanine are purines – they have a double ring structure
-cytosine, thymine and uracil are pyrimidines – they have a single ring structure

Question 53

Shape of a DNA and RNA molecule

Answer 53

DNA
-double helix with two polynucleotide chains held together by hydrogen bonds between specific complementary base pairs
-antiparallel
-each DNA polynucleotide strand has a 3’ end and a 5’ end (these numbers relate to which carbon on the pentose sugar could be bonded with another nucleotide)
-as the strands run in opposite directions, one is known as the 5’ to 3’ strand and the other is known as the 3’ to 5’ strand

RNA
-relatively short polynucleotide chain
-only made of one strand
-consists of alternating ribose sugars and phosphate groups linked together
The sugar-phosphate bonds (between different nucleotides in the same strand) are covalent / phosphodiester bonds
-form sugar-phosphate backbone of the RNA polynucleotide strand
-phosphodiester bonds link the 5-carbon of one ribose sugar molecule to phosphate group from same nucleotide, which is linked by another phosphodiester bond to the 3-carbon of the ribose sugar molecule of the next nucleotide in the strand

-both are polynucleotides
-nucleotides are joined via condensation reactions between phosphate group of one nucleotide and pentose sugar of the next nucleotide
-forms phosphodiester bond as it consists of a phosphate group and two ester bonds
-sugar phosphate backbone = chain of alternating phosphate groups and pentose sugars produced as a result of many phosphodiester bonds

Question 54

What is the importance of semi-conservative replication?

Answer 54

ensures genetic continuity between generations of cells

Question 55

evaluate the work of scientists in validating the Watson–Crick model of DNA replication.

Question 56

Explain the process of semi-conservative replication of DNA

Answer 56

-in each new DNA molecule produced, one of the polynucleotide DNA strands is from the original DNA molecule being copied
-the other polynucleotide DNA strand has to be newly created by the cell

-DNA Helicase breaks hydrogen bonds between the complementary bases
-double strand separates, leaves 2 template stands
-free complementary nucleotides are attracted to and bind to exposed bases on template strands (A to T, C to G)
-DNA Polymerase joins the sugar-phosphate backbone of the new strand

Question 57

Evidence for semi-conservative replicaiton

Answer 57

-Replicating Bacterial DNA in 2 types of Nitrogen Isotopes, 15N and 14N
-15N = heavy isotope
-14N = light isotope
-Nitrogen found in nitrogenous bases of DNA
-Bacterial DNA made from 15N will have a Heavy Density
-Bacterial DNA made from 14N will have a Light Density
-Experiment = Bacterial DNA made of 15N is replicated in an environment of 14N – produces DNA molecules with half 15/half 14 (semi-conservative replication, original strand = 15N & new strand = 14N), therefore, DNA molecule has medium density

Question 58

Properties of water

Answer 58

-metabolite in many metabolic reactions, including condensation/hydrolysis reactions
-solvent in which metabolic reactions occur
-has relatively high heat capacity, buffers changes in temperature
-has relatively large latent heat of vaporisation, providing a cooling effect with little loss of water through evaporation
-has strong cohesion between water molecules, which supports columns of water in the xylem and produces surface tension where water meets air

Question 59

ATP components

What enzyme catalyses its hydrolysis and give equation

What enzyme catalyses its resynthesis and give equation

Answer 59

a molecule of ribose, a molecule of adenine and three phosphate groups

Hydrolysis:
ATP —> ADP + Pi (catalysed by ATP hydrolase)
-inorganic phosphate released during the hydrolysis of ATP can be used to phosphorylate other compounds
-often makes them more reactive

Resynthesis
ADP + Pi —> ATP (catalysed by ATP synthase)