Topic 2 Molecular Biology

Question 1

Define “molecular biology”.

2.1

Answer 1

explains living processes in terms of the chemical substances involved

Understanding: Molecular biology explains living processes in terms
of the chemical substances involved.

Question 2

Explain how urea flasified the theory of vitalism.

2.1

Answer 2

• In 1828, Frederick Woehler heated an inorganic salt (ammonium cyanate) and produced urea
• Urea is a waste product of nitrogen metabolism and is eliminated by the kidneys in mammals
• The artificial synthesis of urea demonstrates that organic molecules could be produced without a “vital force”

Application: Urea as an example of a compound that is produced by
living organisms but can also be artifcially synthesized.

Question 3

Explain why carbon forms the basis of organic life.

2.1

Answer 3

Due to its ability to form large and complex molecules via covalent bonding
* Carbon atoms can form four covalent bonds
* Forms a wide variety of organic compounds that are chemically stable

Understanding: Carbon atoms can form four bonds allowing a diversity
of compounds to exist.

Question 4

Define “organic compound”.

2.1

Answer 4

An organic compound is a compound that contains carbon and is found in living things

Understanding: Carbon atoms can form four bonds allowing a diversity
of compounds to exist.

Question 5

Define “carbohydrates”.

2.1

Answer 5

Composed of carbon, hydrogen and oxygen, with hydrogen and oxygen in the ratio of two hydrogen atoms to one oxygen, hence the name carbohydrate.

Understanding: Life is based on carbon compounds including carbohydrates, lipids, proteins and nucleic acids.

Question 6

Define “lipids”.

2.1

Answer 6

Broad class of molecules that are insoluble in water, including steroids, waxes, fatty acids and triglycerides.
* In common language, triglycerides are fats if they are solid at room temperature or oils if they are liquid at room temperature.

Understanding: Life is based on carbon compounds including carbohydrates, lipids, proteins and nucleic acids.

Question 7

Define “proteins”.

2.1

Answer 7

Composed of one or more chains of amino acids.
* All of the amino acids in these chains contain the elements carbon, hydrogen, oxygen and nitrogen, but two of the twenty amino acids also contain sulphur.

Understanding: Life is based on carbon compounds including carbohydrates, lipids, proteins and nucleic acids.

Question 8

Define “nucleic acids”.

2.1

Answer 8

Chains of subunits called nucleotides, which contain carbon, hydrogen, oxygen, nitrogen and phosphorus.
* There are two types of nucleic acid: ribonucleic acid (RNA) and deoxyribonucleic acid (DNA).

Understanding: Life is based on carbon compounds including carbohydrates, lipids, proteins and nucleic acids.

Question 9

Define “metabolism”.

2.1

Answer 9

Metabolism is the sum of all reactions that occur in an organism.
* It is the web of all enzyme-catalysed reactions that occur within a cell or organism

Understanding: Metabolism is the web of all the enzyme catalysed reactions in a cell or organism.

Question 10

Outline anabolism.

2.1

Answer 10

Anabolism: builds molecules by making bonds (i.e. proteins, carbohydrates)
* Includes the formation of macromolecules by condensation reactions
* Water is a product of the reaction

Examples:
* Protein synthesis using ribosomes
* DNA synthesis during replication.
* Photosynthesis, including production o glucose rom carbon dioxide and water.
* Synthesis of complex carbohydrates including starch, cellulose and glycogen.

Understanding: Anabolism is the synthesis of complex molecules from simpler molecules including the formation of macromolecules from monomers by condensation reactions.

Question 11

Outline catabolism.

2.1

Answer 11

Catabolism: breaks down molecules by breaking bonds (i.e. digestion)
* Includes the hydrolysis of macromolecules into monomers
* Water is a reactant/added to the reaction

Examples:
* Digestion of food in the mouth, stomach and small intestine.
* Cell respiration in which glucose or lipids are oxidized to carbon dioxide and water.
* Digestion of complex carbon compounds in dead organic matter by decomposers.

Understanding: Anabolism is the synthesis of complex molecules from simpler molecules including the formation of macromolecules from monomers by condensation reactions.

Question 12

Outline a water molecule.

2.2

Answer 12

• covalent bonds between H and O atoms
• Polar covalent bonds
• Polar molecule: O is slightly negative and H is slightly positive (uneven electron distribution)
• Have two poles and are dipoles (show dipolarity)

Understanding: Water molecules are polar and hydrogen bonds form
between them.

Question 13

Define hydrogen bonding.

2.2

Answer 13

• A hydrogen bond is the intermolecular force that forms when a hydrogen atom in one polar molecule is attracted to a slightly negative atom of another polar covalent molecule.

Understanding: Water molecules are polar and hydrogen bonds form
between them.

Question 14

Outline the cohesive property of water.

2.2

Answer 14

Cohesion: the binding together of two molecules of the same type, for instance two water molecules.
* Water molecules are cohesive: they cohere, which means they stick to each other, due to hydrogen bonding

Examples:
* Water is transported through xylem vessels at low pressure since H bonding keeps the water molecules together.
* Allows organisms (water strider) to move along the suface of water, creates high surface tension

Understanding: Hydrogen bonding and dipolarity explain the cohesive,
adhesive, thermal and solvent properties of water.

Question 15

Outline the adhesive property of water.

2.2

Answer 15

**Adhesion: **form between water and other polar molecules, causing water to stick to them.

Examples:
* Water adheres to cellulose molecules in cell walls, allows for water to cling onto plant vessel
* Water is drawn out from plant vessel during evaporation to keep plant moist to absorb CO2 for photosynthesis

Understanding: Hydrogen bonding and dipolarity explain the cohesive,
adhesive, thermal and solvent properties of water.

Question 16

Outline the thermal propertes of water.

2.2

Answer 16

High specific heat capacity:
* H bonds restrict the motion of water molecules, more energy required to break the bonds to raise the temperatue.
* Importance: Stable temperature for living organisms

High latent heat of vaporization:
* Heat required to evaporate a molecule, more energy is required due to H bonds
* Importance: Energy is taken from environment (endothermic reaction), making surrounding cooled (ex. sweating)

High boiling point:
* Importance: Water is liquid over a broad range of temperatures

Understanding: Hydrogen bonding and dipolarity explain the cohesive,
adhesive, thermal and solvent properties of water.

Question 17

Contrast hydrophobic and hydrophillic molecules.

2.2

Answer 17

Substances that freely associate and readily dissolve in water are characterised as hydrophilic (‘water loving’)
* Hydrophilic substances include all polar molecules and ions

Substances that do not freely associate or dissolve in water are characterised as hydrophobic (‘water-hating’)
* Hydrophobic substances include large, non-polar molecules (such as fats and oils)

Understanding: Substances can be hydrophilic or hydrophobic.

Question 18

Compare and contrast methane and water.

2.2

Answer 18

Compare:
* Comparable size and weight
* Comparable valence structures

Contrast:
* Water has a significantly higher melting and boiling point
* Water has a higher specific heat capacity (energy required to raise the temperature of 1 g of substance by 1ºC)
* Water has a higher heat of vaporisation (energy absorbed per gram as it changes from a liquid to a gas / vapour)
* Water as a higher heat of fusion (energy required to be lost to change 1 g of liquid to 1 g of solid at 0ºC)
* All due to H bonds that methane does not have

Application: Comparison of the thermal properties of water with
those of methane.

Question 19

Outline how water is used as a coolant in sweat.

2.2

Answer 19

1. The change of water from liquid to vapour (evaporation) requires an input of energy
2. This energy comes from the surface of the skin when it is hot, therefore when the sweat evaporates the skin is cooled
3. Because water has a high specific heat capacity, it absorbs a lot of thermal energy before it evaporates
4. Thus water functions as a highly effective coolant, making it the principal component of sweat

Application: Use of water as a coolant in sweat.

Question 20

Outline how glucose, amino acids, cholesterol, fats, oxygen and sodium chloride are transported in blood plasma.

2.2

Answer 20

Water Soluble Substances:
* Sodium chloride (NaCl) is an ionic compound: freely soluble, dissociates in water (NA+ and Cl-)
* Oxygen is non-polar but soluble in water in low amounts due to small size (most oxygen is transported by haemoglobin within red blood cells)
* Glucose is a polar molecule: freely soluble
* Amino acids have both positive and negative charges: solubility degree depends on the R chain

Water Insoluble Substances:
* Lipids (fats and cholesterol) are non-polar and hydrophobic and hence will not dissolve in water
* They form complexes with proteins (lipoproteins) in order to move through the bloodstream
* Hydrophilic portions of proteins, cholesterol and phospholipids will face outwards and shield internal hydrophobic components

Application: Methods of transport of glucose, amino acids, cholesterol, fats, oxygen and sodium chloride in blood in relation to their solubility in water.

Question 21

List the examples of monosaccharides.

2.3

Answer 21

Glucose, fructose and ribose

Understanding: Monosaccharide monomers are linked together by condensation reactions to form disaccharides and polysaccharide polymers.

Question 22

List the examples of monosaccharides, disaccharides and polysaccharides.

2.3

Answer 22

Monosaccharides (one sugar unit) are typically sweet-tasting and function as an immediate energy source for cells
* Examples of monosaccharides include glucose, galactose and fructose

Disaccharides (two sugar units) are small enough to be soluble in water and commonly function as a transport form
* Examples of disaccharides include lactose, maltose and sucrose

Polysaccharides (many sugar units) may be used for energy storage or cell structure, and also play a role in cell recognition
* Examples of polysaccharides include cellulose, glycogen and starch

Understanding: Monosaccharide monomers are linked together by condensation reactions to form disaccharides and polysaccharide polymers.

Question 23

Outline the structure and function of cellulose.

2.3

Answer 23

• 1-4 bonds
• from beta-glucose
• alternating orientation
• striaght chain
• unbranched
• High tensile strength that prevents bursting of cell walls

Application: Structure and function of cellulose and starch in plants and glycogen in humans.

Question 24

Outline the structure and function of amylose starch.

2.3

Answer 24

• 1-4 bonds
• from alpha-glucose
• same orientation
• curved chain
• unbranched
• can easily add or take glucose units
• energy storgae of glucose
• helix shape

Application: Structure and function of cellulose and starch in plants and glycogen in humans.

Question 25

Outline the structure and function of amylopectin starch.

2.3

Answer 25

• 1-4 and 1-6 bonds
• from alpha-glucose
• same orientation
• curved chain
• branched
• can easily add or take glucose units
• energy storgae of glucose
• globular shape

Application: Structure and function of cellulose and starch in plants and glycogen in humans.

Question 26

Outline the structure and function of glycogen.

2.3

Answer 26

• 1-4 and 1-6 bonds
• from alpha-glucose
• same orientation
• curved chain
• branched
• energy storage of glucose
• in animals and fungi

Application: Structure and function of cellulose and starch in plants and glycogen in humans.

Question 27

Outline how triglycerides arre formed.

2.3

Answer 27

Triglycerides are formed when condensation reactions occur between one glycerol and three fatty acids
* The hydroxyl groups of glycerol combine with the carboxyl groups of the fatty acids to form an ester linkage
* This condensation reaction results in the formation of three molecules of water

Understanding: Triglycerides are formed by condensation from three fatty
acids and one glycerol.

Question 28

Outline the functions of lipids and carbohydrates for energy storage.

2.3

Answer 28

• Storage (lipids are more suitable for long-term energy storage)
• Osmolality (lipids have less of an effect on the osmotic pressure of a cell)
• Digestion (carbohydrates are easier to digest and utilise)
• ATP Yield (lipids store more energy per gram)
• Solubility (carbohydrates are easier to transport in the bloodstream)
• Mnemonic: SODAS

Application: Lipids are more suitable for long term energy storage in humans than carbohydrates.

Question 29

Outline why lipids are more suitable for long term energy storage.

2.3

Answer 29

• Lipids have secondary roles (heat insulators, protection around tissue, hormone signalling etc)
• The amount of energy released in cell respiration per gram of lipids is double that of carbohydrate’s
• Lipids do not carry associated water, so is six times more efficient than carbs in the amount of energy stored per gram body mass

Application: Lipids are more suitable for long term energy storage in humans than carbohydrates.

Question 30

State the formula for calculating BMI.

2.3

Answer 30

BMI= mass in kg/ (height in metres)^2

Skill: Determination of body mass index by calculation or use of a nomogram.

Question 31

Outline the types of fatty acids.

2.3

Answer 31

Fatty acids that possess no double bonds are saturated (have maximum number of H atoms)
* Linear in structure, originate from animal sources (i.e. fats) and are typically solid at room temperatures

Fatty acids with double bonds are unsaturated – either monounsaturated (1 double bond) or polyunsaturated (>1 double bond)
* Bent in structure, originate from plant sources (i.e. oils) and are typically liquid at room temperatures

Understanding: Fatty acids can be saturated, monounsaturated or polyunsaturated.

Question 32

Distunguish between cis and trans isomers of unsaturated fatty acids.

2.3

Answer 32

Cis: The hydrogen atoms attached to the carbon double bond are on the same side
* bend in the hydrocarbon chain at the double bond
* liquid at room temperature (they are oils)

Trans: The hydrogen atoms attached to the carbon double bond are on different sides
* no bend in the hydrocarbon chain at the double bond,
* higher melting point
* solid at room temperature
* produced artifcially

Understanding: Unsaturated fatty acids can be cis or trans isomers.

Question 33

Outline the evidence for health risks of trans-fats and saturated fatty acids.

2,3

Answer 33

Positive correlation between saturated fatty acid intake and rate of coronary heart disease (CHD)
* but does not prove saturated fats cause the disease

Positive correlation between trans0fats and CHD
* patients who died from CHD had fatty deposits in diseased arteries with high concentrations of trans-fats
* more evidence than a casual link

Application: Scientifc evidence for health risks of trans-fats and
saturated fats.

Question 34

Evaulate the evidence and the methods used to obtain evidence for health claims made about lipids.

2.3

Answer 34

1. Implications - do the results of the research support the health claim strongly, moderately or not at all?
2. Limitations - were the research methods used rigorous, or are there uncertainties about
   the conclusions because o weaknesses in methodology?

Application: Evaluation of evidence and the methods used to obtain the evidence for health claims made about lipids.

Question 35

Define “amino acids”.

2.4

Answer 35

Amino acids are joined together on the ribosome to form long chains called polypeptides, which make up proteins
* Each type of amino acid differs in the composition of the variable side chain
* Hydrogen group
* amine group (-NH2)
* carboxyl group (-COOH)
* R group

Understanding: Amino acids are linked together by condensation to form
polypeptides.

Question 36

State the number of amino acids there are in polypeptides synthesized on ribosomes.

2.4

Answer 36

There are 20 different amino acids which are universal to all living organisms

Understanding: There are twenty diferent amino acids in polypeptides
synthesized on ribosomes.

Question 37

Outline why there is a huge range of possible polypeptides.

2.4

Answer 37

1. can be any length
2. can be any of the 20 amino acids
3. can be arranged in any order or combination

Understanding: Amino acids can be linked together in any sequence
giving a huge range of possible polypeptides.

Question 38

Define “open reading frame”.

2.4

Answer 38

The base sequence that actually codes for a polypeptide

Understanding: The amino acid sequence of polypeptides is coded for
by genes.

Question 39

Outline the protein lysozome in relation to its number of polypeptides.

2.4

Answer 39

1 polypeptide
* Enzyme in secretions such as nasal mucus and tears; it kills some bacteria by digesting the peptidoglycan in their cell walls.

Understanding: A protein may consist of a single polypeptide or more than
one polypeptide linked together.

Question 40

Outline the protein integrin in relation to its number of polypeptides.

2.4

Answer 40

2 polypeptides
* Membrane protein used to make connections between structures inside and outside a cell.

Understanding: A protein may consist of a single polypeptide or more than
one polypeptide linked together.

Question 41

Outline the protein collagen in relation to its number of polypeptides.

2.4

Answer 41

3 polypeptides
* Structural protein in tendons, ligaments, skin and blood vessel walls; it provides high tensile strength, with limited stretching.

Understanding: A protein may consist of a single polypeptide or more than
one polypeptide linked together.

Question 42

Outline the protein hemoglobin in relation to its number of polypeptides.

2.4

Answer 42

4 polypeptides
* Transport protein in red blood cells; it binds oxygen in the lungs and releases it in tissues with a reduced oxygen concentration.

Understanding: A protein may consist of a single polypeptide or more than
one polypeptide linked together.

Question 43

Contrast fibrous and globular proteins

2.4

Answer 43

fibrous
* long and narrow
* structural role
* insoluble
* repetitive AA sequence
* ex. collagen, fibrin

globular protein
* rounded and spherical
* functional role
* soluble
* irregular AA sequence
* ex. catalse, hemogloblin

Understanding: The amino acid sequence determines the three-dimensional
conormation of a protein.

Question 44

Define “denaturation”.

2.4

Answer 44

Denaturation is a structural change in a protein that results in the loss (usually permanent) of its biological properties
* Because the way a protein folds determines its function, any change or abrogation of the tertiary structure will alter its activity
* Denaturation of proteins can usually be caused by two key conditions – temperature and pH

Application: Denaturation of proteins by heat or pH extremes.

Question 45

Outline how temperature can cause denaturation.

2.4

Answer 45

• High levels of thermal energy may disrupt the hydrogen bonds that hold the protein together
• As these bonds are broken, the protein will begin to unfold and lose its capacity to function as intended

Application: Denaturation of proteins by heat or pH extremes.

Question 46

Outline how pH can cause denaturation.

2.4

Answer 46

• Changing the pH will alter the charge of the protein, which in turn will alter protein solubility and overall shape
• All proteins have an optimal pH which is dependent on the environment in which it functions
• Denatiration can be reversed if returned to optimal pH.

Application: Denaturation of proteins by heat or pH extremes.

Question 47

Outline the range of functions of proteins.

2.4

Answer 47

• Catalysis: catalyse specifc chemical reactions
• Muscle contraction: actin and myosin together used in locomotion
• Cytoskeletons: tubulin pull on chromosomes during mitosis.
• Tensile strengthening: fibrous proteins give tensile strength
• Blood clotting: plasma proteins act as clotting factors
• Transport of nutrients and gases: transport oxygen, carbon dioxide, iron and lipids
• Cell adhesion: membrane proteins cause adjacent animal cells to stick to each other within tissues.
• Membrane transport: membrane proteins used for facilitated diffusion and active transport etc.
• Hormones: insulin, FSH and LH are proteins
• Receptors: binding sites for hormones, neurotransmitters etc
• Packing of DNA: histones help chromosomes to condense during mitosis.
• Immunity: cells can make huge numbers of different antibodies.

Understanding: Living organisms synthesize many diferent proteins with
a wide range of functions.

Question 48

Outline the protein collagen

2.4

Answer 48

• A component of the connective tissue of animals
• prevents tearing in skin, prevents fractures in bones, gives tensile strength
• structural protein

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 49

Outline the protein spidersilk

2.4

Answer 49

• used to make spider webs
• resists breakage, tensile strength
• structural protein

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 50

Outline the protein insulin

2.4

Answer 50

• Protein produced by the pancreas and triggers a reduction in blood glucose levels
• a hormone

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 51

Outline the protein immunoglobin

2.4

Answer 51

• Are antibodies produced by plasma cells that are capable of targeting specific antigens
• is an antibody

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 52

Outline the protein hemoglobin.

2.4

Answer 52

• A protein found in red blood cells that is responsible for the transport of oxygen

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 53

Outline the protein rhodopsin.

2.4

Answer 53

• A pigment in the photoreceptor cells of the retina that is responsible for the detection of light
• is a pigment

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 54

Outline the protein rubisco.

2.4

Answer 54

• Catalyses photosynthesis
• An enzyme involved in the light independent stage of photosynthesis

Application: Rubisco, insulin, immunoglobulins, rhodopsin, collagen and spider silk as examples of the range of protein functions.

Question 55

Contrast proteome and genome.

2.4

Answer 55

**A proteome is all of the proteins produced by a cell, a tissue or an organism. **
* By contrast, the genome is all of the genes of a cell, a tissue or an organism.

Understanding: Every individual has a unique proteome.

Question 56

Explain why every individual has a unique proteome.

2.4

Answer 56

Whereas the genome of an organism is fixed, the proteome is variable because different cells in an organism make different proteins.
* because of differences of activity but also because of small differences in the amino acid sequence of proteins

Understanding: Every individual has a unique proteome.

Question 57

Define “enzyme”.

2.5

Answer 57

a globular protein which acts as a biological catalyst by speeding up the rate of a chemical reaction

Understanding: Enzymes have an active site to which specic substrates bind.

Question 58

Define “substrate”.

2.5

Answer 58

the reactant in a biochemical reaction

Understanding: Enzymes have an active site to which specic substrates bind.

Question 59

Define “active site”.

2.5

Answer 59

the region on the surface of the enzyme which binds to the substrate

Understanding: Enzymes have an active site to which specic substrates bind.

Question 60

Outline the process of enzyme catalysis.

2.5

Answer 60

1. The substrate binds to the active site of the enzyme.
2. While the substrates are bound to the active site they change into different chemical substances, which are the products of the reaction.
3. The products separate from the active site, leaving it vacant for substrates to bind again.

Understanding: Enzyme catalysis involves molecular motion and the
collision of substrates with the active site.

Question 61

Outline how substrates can bind to the active site.

2.5

Answer 61

• substrate and enzyme are usually moving randomly within the solution
• only reacts if there is a collision between actuve sites and substrates

Understanding: Enzyme catalysis involves molecular motion and the
collision of substrates with the active site.

Question 62

Outline how temperature affects enzyme activity.

2.5

Answer 62

• Low temperatures result in insufficient thermal energy for the activation of an enzyme-catalysed reaction to proceed
• Increasing the temperature will increase the speed and motion of both enzyme and substrate, resulting in higher enzyme activity
• This is because a higher kinetic energy will result in more frequent collisions between the enzymes and substrates
• At an optimal temperature (may vary for different enzymes), the rate of enzyme activity will be at its peak
• Higher temperatures will cause enzyme stability to decrease, as the thermal energy disrupts the enzyme’s hydrogen bonds
• This causes the enzyme (particularly the active site) to lose its shape, resulting in the loss of activity (denaturation)

Understanding: Temperature, pH and substrate concentration afect the rate of activity of enzymes.

Question 63

Outline how pH affects enzyme activity.

2.5

Answer 63

• Changing the pH will alter the charge of the enzyme, which in turn will alter protein solubility and overall shape
• Changing the shape or charge of the active site will diminish its ability to bind the substrate, abrogating enzyme function
• Enzymes have an optimal pH (may differ between enzymes) and moving outside this range diminishes enzyme activity

Understanding: Temperature, pH and substrate concentration afect the rate of activity of enzymes.

Question 64

Outline how substrate concentration affects enzyme activity.

2.5

Answer 64

• Increasing substrate concentration will increase the activity of a corresponding enzyme
• More substrates mean there is an increased chance of enzyme and substrate colliding and reacting within a given period
• After a certain point, the rate of activity will cease to rise regardless of any further increases in substrate levels
• This is because the environment is saturated with substrate and all enzymes are bound and reacting (Vmax)

Understanding: Temperature, pH and substrate concentration afect the rate of activity of enzymes.

Question 65

Outline the denaturation of enzymes.

2.5

Answer 65

• Enzymes are proteins, and like other proteins their structure can be irreversibly altered by certain conditions.
• This process is denaturation and both high temperatures and either high or low pH can cause it.
• When an enzyme has been denatured, the active site is altered so the substrate can no longer bind.

Understanding: Enzymes can be denatured.

Question 66

Outline the advantages of immobilized enzymes in industry.

2.5

Answer 66

• The enzyme can easily be separated from the products of the reaction.
• May be recycled, giving useful cost savings, especially as many enzymes are very expensive.
• Immobilization increases the stability of enzymes to changes in temperature and pH
• Substrates can be exposed to higher enzyme concentrations than with dissolved enzymes, speeding up reaction rates.

Understanding: Immobilized enzymes are widely used in industry.

Question 67

Outline the advantages of lactose-free milk.

2.5

Answer 67

• As a source of dairy for lactose-intolerant individuals
• As a means of increasing sweetness in the absence of artificial sweeteners (monosaccharides are sweeter tasting)
• As a way of reducing the crystallisation of ice-creams (monosaccharides are more soluble, less likely to crystalise)
• As a means of reducing production time for cheeses and yogurts (bacteria ferment monosaccharides more readily)

Application: Methods of production of lactose-free milk and its advantages.

Question 68

Outline the process of creating lactose-free milk.

2.5

Answer 68

• Lactose-free milk can be produced by treating the milk with the enzyme lactase
• The lactase is purified from yeast or bacteria and then bound to an inert substance (such as alginate beads)
• Milk is then repeatedly passed over this immobilised enzyme, becoming lactose-free
• lactose —> glucose + galactose.

Application: Methods of production of lactose-free milk and its advantages.

Question 69

List the two types of nucleic acids.

2.6

Answer 69

• DNA (deoxyribonucleic acid) is a more stable double stranded form that stores the genetic blueprint for cells
• RNA (ribonucleic acid) is a more versatile single stranded form that transfers the genetic information for decoding

Understanding: The nucleic acids DNA and RNA are polymers of nucleotides.

Question 70

Outline the components of a nucleotide.

2.6

Answer 70

• 5-carbon pentose sugar (pentagon)
• Phosphate group (circle)
• Nitrogenous base (rectangle)

Understanding: The nucleic acids DNA and RNA are polymers of nucleotides.

Question 71

Contrast DNA with RNA

2.6

Answer 71

1. The sugar within DNA is deoxyribose and the sugar in RNA is ribose.
2. Two polymers of nucleotides in DNA (double stranded) but only one in RNA (single stranded).
3. The four bases in DNA are adenine, cytosine, guanine and thymine. The four bases in RNA are adenine, cytosine, guanine and uracil.

Understanding: DNA difers from RNA in the number of strands normally
present, the base composition and the type of pentose.

Question 72

Outline the structure of DNA.

2.6

Answer 72

Each strand consists of a chain of nucleotides linked by covalent bonds.

The two strands are parallel but run in opposite directions so they are said to be antiparallel. One strand is oriented in the direction 5’ to 3’ and the other is oriented in the direction 3’ to 5’.

The two strands are wound together to form a double helix.

Two polynucleotide chains of DNA are held together via hydrogen bonding between complementary nitrogenous bases
* Adenine (A) pairs with Thymine (T) via two hydrogen bonds
* Guanine (G) pairs with Cytosine (C) via three hydrogen bonds

Understanding: DNA is a double helix molecule made of two antiparallel strands of nucleotides linked by hydrogen bonding between complementary base pairs

Question 73

Outline how model making elucidated the structure of DNA.

2.6

Answer 73

Guided by an understanding of molecular distances and bond angles developed by Linus Pauling, and were based upon some key experimental discoveries:
* DNA is composed of nucleotides made up of a sugar, phosphate and base – Phoebus Levene, 1919
* DNA is composed of an equal number of purines (A + G) and pyrimidines (C + T) – Erwin Chargaff, 1950
* DNA is organised into a helical structure – Rosalind Franklin, 1953 (data shared without permission)

Application: Crick and Watson’s elucidation of the structure of DNA using model making

Question 74

Outline how DNA is semi-conservative.

2.7

Answer 74

DNA replication is a semi-conservative process, because when a new double-stranded DNA molecule is formed:
* One strand will be from the original template molecule
* One strand will be newly synthesised
* This occurs because each nitrogenous base can only pair with its complementary partner

Understanding: The replication of DNA is semi-conservative and depends on complementary base pairing.

Question 75

Outline how DNA is semi-conservative.

2.7

Answer 75

DNA replication is a semi-conservative process, because when a new double-stranded DNA molecule is formed:
* One strand will be from the original template molecule
* One strand will be newly synthesised
* This occurs because each nitrogenous base can only pair with its complementary partner

Understanding: The replication of DNA is semi-conservative and depends on complementary base pairing.

Question 76

Outline Meselson and Stahl’s experiment.

2.7

Answer 76

1. Grew E. coli bacteria in medium containing N15
2. Because it N15 is heavier, it is denser than N14
3. After 15 generations, they transferred the bacteria to a medium containing only N14
4. Collected samples over several hours, then extracted the DNA
5. Spun DNA it in a centrifuge, DNA created a layer based on its density compared to cesium chloride’s diffusion gradient

Skill: Analysis of Meselson and Stahl’s results to obtain support for the theory of semi-conservative replication of DNA

Question 77

Outline Meselson and Stahl’s experiment.

2.7

Answer 77

1. Grew E. coli bacteria in medium containing N15
2. Because it N15 is heavier, it is denser than N14
3. After 15 generations, they transferred the bacteria to a medium containing only N14
4. Collected samples over several hours, then extracted the DNA
5. Spun DNA it in a centrifuge, DNA created a layer based on its density compared to cesium chloride’s diffusion gradient

Skill: Analysis of Meselson and Stahl’s results to obtain support for the theory of semi-conservative replication of DNA