Bio Chapter 3, 7 and 8 (Respiration)

7/29/2019 Bio Chapter 3, 7 and 8 (Respiration)

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TOPIC 3: CHEMISTRY OF LIFE

3.1 CHEMICAL ELEMENTS AND WATER

3.1.1STATE THE MOST FREQUENT OCCURRING CHEMICAL ELEMENTS IN LIVING THINGS

ARE CARBON, HYDROGEN, OXYGEN, AND NITROGEN.

3.1.2STATE THE VARIETY OF OTHER ELEMENTS ARE NEEDED BY LIVING ORGANISM

INCLUDING SULFUR, CALCIUM , PHOSPHORUS, IRON AND SODIUM.

3.1.3STATE ONE ROLE FOR EACH OF THE ELEMENTS MENTIONED IN3.1.2

Element Role in plant Role in animal Role in prokaryote

Sulfur In some amino acid In some amino acid In some amino acid

Phosphorus Phosphate group in

ATP molecules

Phosphate group in

ATP molecules

Phosphate group in

ATP molecules

Iron In cytochrome In cytochrome and

hemoglobin

In cytochrome

Calcium Co-factor in some

enzyme

Co-factor in some

enzyme and as

component of bones

Co-factor of enzyme

Sodium In membrane

function

In membrane

function and sending

nerve impulse

In membrane

function

3.1.4DRAW AND LABEL WATER MOLECULES TO SHOW THEIR POLARITY AND HYDROGEN

BOND FORMATION

Hydrogen bond

Water molecule


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3.1.5OUTLINE THE THERMAL, COHESIVE AND SOLVENT PROPERTIES OF WATER.

water has high specific heat capacity which

means that water can absorb or give off agreat deal of heat without changing the

temperature greatly.

It has high heat of vaporization which means

the water absorb a great deal of heat when

it evaporate to act as cooling mechanism

Thermal Properties

cohession is when molecules of the same type

are attracted to each other. Water has this

properties because of the hydrogen bond

formed between the water molecules

This properties enable the water to move as a

column in the vascular tissue of plant

Cohesive Properties

Many different substances dissolve in water

because of its polarity

inorganic molecules with positive or negative

charge dissolve in water

organic substances with polar molecules

dissolve.

water is the medium for metabolic reaction.

Solvent Properties


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3.1.6EXPLAIN THE RELATIONSHIP BETWEEN THE PROPERTIES OF WATER AND ITS USES

IN LIVING ORGANISMS AS A COOLANT, MEDIUM FOR METABOLIC REACTION AND

TRANSPORT MEDIUM.

Properties of water Relationship between the properties of

water and its uses in living organism

Thermal properties Blood (mainly compose of water) can

carry heat from warmer parts of the

body to cooler parts

Evaporation of water from plant

(transpiration) and human skin

(sweat) has useful cooling effect.

Water act as a coolantSolvent properties Water is the medium for metabolic

reaction

Many substances to be carried

dissolve in water in the blood of

animals and the sap of plants.

Cohesive properties Strong pulling forces can be exerted

to suck columns of water up to the

tops of the tallest trees in the

transport systems.

Water is used as transport medium inthe xylem of plants.

3.2 CARBOHYDRATES, LIPIDS, AND PROTEINS

3.2.1DISTINGUISH BETWEEN ORGANIC AND INORGANIC COMPOUND

Organic compound Inorganic Compound

Are produced by living things and includeall compound containing carbon that are

found in living organism except for

hydrogen carbonate(HCO3-), carbonate

(CO32-) and oxide of carbon (CO2, CO)

All compound that contain no carbon areinorganic


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3.2.2IDENTIFY AMINO ACID, GLUCOSE, RIBOSE AND FATTY ACID FROM DIAGRAMS

SHOWING THEIR STRUCTURE.

3.2.3LIST THREE EXAMPLES EACH OF THE MONOSACCHARIDE, DISACCHARIDES AND

POLYSACCHARIDES.

Subcategory Examples

Monosaccharide Glucose, Galactose, Fructose

Disaccharide Maltose, Lactose, Sucrose

Polysaccharide Starch, Glycogen and cellulose

Amino Acid Glucose

Ribose Fatty Acid


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3.2.4STATE ONE FUNCTION OF GLUCOSE, LACTOSE AND GLYCOGEN IN ANIMALS AND OF

FRUCTOSE, SUCROSE AND CELLULOSE IN PLANTS.

Compound Function in living thingGlucose Chemical fuel for cell respiration

Lactose Make up some of the solutes in milk

Glycogen Store glucose in liver and muscle

Fructose Found in many fruits (make them sweet)

Sucrose Often transported from leaves of plants to

other locations in plants by vascular tissues.

Cellulose One of the primary components ofplants

cell wall.

3.2.5OUTLINE THE ROLE OF CONDENSATION AND HYDROLYSIS IN THE RELATIONSHIPS

BETWEEN MONOSACCHARIDES, DISACCHARIDES AND POLYSACCHARIDES, BETWEEN

FATTY ACIDS, GLYCEROL AND TRIGLYCERIDES AND BETWEEN AMINO ACID AND

POLYPEPTIDE.


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3.2.6STATE THREE FUNCTIONS OF LIPIDS

Role of lipids:-

As energy storage

Thermal insulation

Make up double layer of cell membrane

3.2.7COMPARE THE USE OF CARBOHYDRATES AND LIPIDS IN ENERGY STORAGE.

Carbohydrates Lipids

Carbohydrates are more easilydigested than lipid so the energy

stored by them can be release

rapidly.

Carbohydrates are soluble in water,so are easier to transport to and from

the store

Contain more energy per gramthan carbohydrates. Therefore,

store of lipids are lighter than the

store of carbohydrates that

contain the same amount of

energy.

Lipids are insoluble in water, sothey do not cause problem with

osmosis in cells.


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3.3 DNA structure

3.3.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and

phosphate. (2)

A nucleotide is made of;

Deoxyribose sugar (differs from ribose in having one less oxygen on carbon 2) A base (which can be either adenine, guanine, cytosine or thymine) and A phosphate group (PO43-)

3.3.2 State the names of the four bases in DNA. (1)

Adenine (A), Guanine (G), Cytosine (C) and Thymine (T)

A and GPurines (big 2-ring structure) C T and UPyrimidines (small 1-ring structure)

3.3.3 Outline how DNA nucleotides are linked together by covalent bonds into a single

strand. (2)

DNA is composed of two strands of nucleotides. Nucleotides are linked into a single strand via condensation reaction;

phosphate + deoxyribose sugar + organic base nucleotides + 2 H2O

The phosphate group of one nucleotide (attached to the 5-end) joins to the hydroxylgroup on sugar of the second nucleotide (3-end)

The phosphate group creates a bridge connecting C5 on one pentose with the C3 on thenext pentose.

This results in a covalent bond, called phosphodiester bond.


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3.3.4 Explain how a DNA double helix is formed using complementary base pairing and

hydrogen bonds (3)

i. DNA is made up of two nucleotide strands.ii.

The nucleotides are connected together by covalent bonds within each strand.iii. The sugar of one nucleotide forms a covalent bond (phosphodiester bond) with the

phosphate group of another.

iv. The two strands themselves are connected by hydrogen bonds. The hydrogen bondsare found between the bases of the two strands of nucleotide;

Adenine pairs with thymine (A = T) Guanine pairs with cytosine (G C)

v. This is called complementary base pairing. Below is a digram showing the molecularstructure and bonds within DNA

3.3.5 Draw and label a simple diagram of the molecular structure of DNA (1)

Two polynucleotide chains. Sugar and phosphate backbone run

antiparallelforming the sides of the double

helix with the organic bases pairs rung in

between.

Two chains are in opposite directions and arewound round each other to form a double

helix They joined together by hydrogen bonds

between the bases.

Adenine hydrogen bonds to Thymine

(2 Hydrogen bonds)

Cytosine hydrogen bonds to Guanine.

(3 hydrogen bonds)

Bonds between the components of thenucleotides are covalently bonded, by

condensation. However, the H bond strength < covalent

bond strength. But, there abundant amount of

H bonds that keep the DNA securely.

Phosphate

group

Complementarybase pairing

Pentose

sugar

Covalent bond

(phosphodiester

bond)

Hydrogen

bond

Nitrogenous

base


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7.1.1 Describe the structure of DNA, including the antiparallel strands, 35 linkages

and hydrogen bonding between purines and pyrimidines. (2)

- Double helix shaped

-Consists of nucleotides

- Nucleotide have one base, one

deoxyribose sugar and one

phosphate

- Consists of 4 different bases

Adenine, Guanine, Thymine and

Cytosine

- Complementary base pairing:

A bonds to T, G bonds to C

- Bases bond with hydrogen bonds

- Nucleotides linked up with

covalent bond/sugar-phosphate

bonds;

- links between nucleotides;

- Strands are anti-parallel

7.1.2 Outline the structure of nucleosomes. (2)

-Nucleosome core consisting of 8 histone protein molecules

-DNA wrapped twice around nucleosome core

-Another histone protein holds the nucleosome together

-Has DNA linker continuing to the next nucleosome

53

5

5

3

3


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7.1.3 State that nucleosomes help to supercoil chromosomes and help to regulate

transcription. (1)

Supercoiling will condense the chromosome because only certain areas of supercoiled DNA

are accessible to transcription enzymes.

7.1.4 Distinguish between unique or single-copy genes and highly repetitive

sequences in nuclear DNA. (2)

Single copy Highly repetitive

One locatable region on DNA molecule Can repeat up to 100 000 times on various

location on DNA

Long base sequence Short sequence/ 5-300 bases

Have coding function Have no know function

May be translated Not translated

7.1.5 State that eukaryotic genes can contain exons and introns. (1)

Exons are sequences of bases that are transcribed and translated, and introns are non-coding

fragments that are transcribed but not translated.

3.4 DNA Replication

3.4.1 Explain DNA replication in terms of unwinding the double helix and separation of

the strands by helicase, followed by formation of the new complementary strands by

DNA polymerase (3)

7.2.2 Explain the process of DNA replication in prokaryotes, including the role of

enzymes (helicase, DNA polymerase, RNA primase and DNA ligase), Okazaki

fragments and deoxynucleoside triphosphates. (3)

The cell produces many free nucleotides for DNA replication. Each nucleotide has 3

phosphate groups (deoxyribonucleoside triophosphate). During replication 2 phosphate

groups are removed to release energy.

1.Helicase uncoils the DNA double helix and splits them into 2 template strands called

lagging strandand leading strandby breaking the hydrogen bonds between the bases.

2. The leading and lagging strands of DNA are now templates for the new strands to form.


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3. On the leading strand, theRNA primase synthesizes a short RNA primer on the DNA to

begin replication. This acts as a primer, allowing the enzyme DNA polymerase III to bind.

(The RNA primer will later be removed by DNA polymerase I)

4.DNA polymerase IIIthen starts adding deoxynucleoside triphosphates to the strand in a 5

to 3 direction by complementary base pairing.

5. On the lagging strand,RNA primase synthesizes a short RNA primercomplementary to the

exposed DNA for replication.

6. DNA polymerase III then starts adding deoxynucleoside triphosphates to the strand in a 5

to 3 direction by complementary base pairing, moving away from the replication fork.

7. DNA polymerase I removes the RNA primer and replaces it with deoxynucleoside

triphosphates. Short lengths of DNA are formed between RNA primers, called Okazaki

Fragments.

8. There is a nick where two nucleotides are still unconnected and DNA ligase seals the nick

by making another sugar-phosphate bond.

10. Two identical DNA double helices are formed with each consisting of one old and one

new strand.

11. The DNA strands then rewind to form a double helix.

The replication process has produced a new DNA molecule which is identical to the initial

one

Leading strand; The new strand that is synthesised continuously and follows the replication fork

Lagging strand; The new strand that is synthesised in short fragments in the opposite direction to the

movement of the replication fork

3.4.2 Explain the significance of complementary base pairing in the conservation of the

base sequence of DNA. (3)

i. During DNA replication, base pairing occurs (A = T) and (G C)ii. Thus, the sequence of bases in one strand exactly determines the sequence of bases in

the other strand

iii. This complementary base pairing allows the two DNA molecules to be identical toeach other as they have the same base sequence. It ensures proper base/correct base

incorporated into DNA strands. It also ensures the conservation of base sequence.

iv. The new strands formed are complementary to the template strands but also identicalto the other template.

v. However, mistakes do occur, and these are called mutations.


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3.4.3 State that DNA replication is semi-conservative (1)

One strand will be from the original molecule and another strand will be newly synthesized.

7.2.1 State that DNA replication occurs in a direction. (1)

The end of the free DNA nucleotide is added to the end of the chain of nucleotides that

is already synthesized. This DNA replication occurs in a 5 to the 3 direction.

7.2.3 State that DNA replication is initiated at many points in eukaryotic chromosomes.

(1)

There are as many as 80 million bases to replicate in eukaryotic chromosome. That is why

there are many replication forks on the chromosome.

3.5 Transcription and Translation

3.5.1 Compare the structure of RNA and DNA (3)

Feature DNA RNA

Number of strands 2 strands - Forming a double helix 1 strand

Type of sugar Deoxyribose (lack of one oxygen) Ribose

Bases A, G, C, T A, G, C, U

3.5.2 Outline DNA transcription in terms of the formation of an RNA strand

complementary to the DNA strand by RNA polymerase. (2)

Transcription is a process ofsynthesizing RNA from a DNA template and it happens inthe nucleus

Enzyme involved: RNA polymerase Involves apromoterand a terminatorregion Subunit used:RNA nucleotides Only one of the two DNA strands will be transcribed;Antisense strand: Transcribed Sense strand: Not transcribed

The RNA polymerase attaches to the DNA; unwinds and separates the double strands. RNA

polymerase covalently joins the complementary RNA nucleotides together to form a single

strand (U instead of T). The RNA polymerase detaches from the DNA, so the double helix

reforms.

5 3


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Example of RNA made by transcription: Messenger RNA (mRNA), Transfer RNA (tRNA)

and Ribosomal RNA (rRNA)

3.5.3 Describe the genetic code in terms of codons composed of triplets of bases (2)

1. Genetic information in DNA controls the manufacture of specific proteins by cell.2. The codes require for protein synthesis are known as thegenetic code.3. The genetic codes are in the form of a series of triplets of bases in DNA, from which

is transcribed into a complementary sequence of codons in messenger RNA.

4. Codons composed of triplets of bases.5. The sequence of these codons determines the sequence of amino acids during protein

synthesis.

6. There are 4 bases in DNA/RNA (A, G, C, T/U).7. The 4 bases are based in sets of 3 called triplets.8. Therefore, there are 43 possible triplets of DNA = 64 triplets (These are codons).9. There are only 20 amino acids.10.64 triplets/codons are mapped to 20 amino acids.11.Genetic code is degenerate.12.Degenerate means more than one triplet or codon can code/map for one amino acid.13.Genetic code is universalall living organisms on earth share the same genetic code.

3.5.4 Explain the process of translation, leading to polypeptide formation. (3)

1. Translation involves initiation, elongation/ translocation & termination.2. Translation takes place in the cytoplasm, ribosomes attach to the mRNA.3. mRNA binds to the small subunit of the ribosome.4. The ribosome covers an area of three codons on the mRNA.5. Slides along the mRNA tostart codon.6. The ribosome read the mRNA in triplets of bases called codon (starting at AUG)7. The first tRNA carrying an amino acidwill come in and the anti-codon exposed on

the tRNA will have complementary binding with the start codon of the mRNA in the

P site of the ribosome

8. The ribosome will move and the first tRNA will be in the P site on the ribosome.9. The second tRNA with its own anti-codon, and carrying specific amino acid, will

complementary bind to the second codon on the mRNA, filling the A site in the

ribosome.

10.The second amino acid will attach to the first (formation of a peptide bond by acondensation reaction) and the first amino acid will be released from the first tRNA.

11.The ribosome will move by one codon in relation to the mRNA (5 3)12.The first tRNA is now found in the E site of the ribosome.13.The second tRNA will be in the P site and the A site is empty.14.The anti-codon of the third tRNA will bind to the third codon of the mRNA in the A

site of the ribosome.


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15.The anti-codon of thefirst tRNA will dissociate from the first codon of the mRNA inthe E site of the ribosome.

16.The second amino acid will form a peptide bondwith the third amino acid and thesecond amino acid will occupy the first site of the ribosome.

17.The second site is occupied by the third tRNA. The third site is free and the nexttRNA will come in, carrying its own amino acid.

18.The process continues until a STOPcodon is reached. The STOP codon doesnt codefor an amino acid but terminates translation.

3.5.5 Discuss the relationship between one gene and one polypeptide. (3)

i. A polypeptide is formed by amino acids liking together throughpeptide bonds.ii. There are 20 different amino acids so a wide range of polypeptides are possible.

iii. Genes store the information required for making polypeptides.iv. The information is stored in a coded form by the use of triplets of bases which form

codons.

v. The sequence of bases in a gene codes for the sequence of amino acids in a polypeptide.vi. The information in the genes is decoded during transcription and translation leading to

protein synthesis.

7.3.1 State that transcription is carried out in a direction. (1)

Transcription is carried out in a 5 to 3 direction.


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7.3.2 Distinguish between the sense and antisense strands of DNA. (2)

Sense strand has the same base sequence as the mRNA, except for T instead of U.

Anti-sense strand is the strand that is transcribed

7.3.3 Explain the process of transcription in prokaryotes, including the role of the

promoter region, RNA polymerase, nucleoside triphosphates and the terminator. (3)

RNA Polymerase binds to the promoter region on the DNA. Then it unwinds the DNA strand by breaking the hydrogen bonds between the DNA. This forms the sense strand and anti sense strand. The antisense strand is used for transcription Direction for transcription is 5 to 3 Promoter region is where nucleoside triphosphates are added to extend the growth of

the mRNA.

When the RNA polymerase reaches the terminator, the RNA polymerase stops andtranscription stops.

mRNA detaches from the template, DNA rewinds RNA polymerase detaches from the DNA Introns removed in eukaryotes to form mature mRNA

7.3.4 State that eukaryotic RNA needs the removal of introns to form mature mRNA (1)

The non- coding introns are spliced out of the mRNA. The remaining mRNA is called mature

mRNA and is exported from the nucleus to the cytoplasm for translation into the polypeptide


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7.4.1 Explain that each tRNA molecule is recognized by a tRNA-activating enzymethat

binds a specific amino acid to the tRNA, using ATP for energy. (3)

1. Amino acid is specific to each tRNA.2. The amino acid will react with ATP and become activated. ATP loses energy in this

process.

3. Activated amino acid will then bind to the acceptor stem of its own tRNA with thehelp of activating enzyme.

4. tRNA is composed of one chain of RNA nucleotideso 3 loops and is clover shapedo has double stranded sections formed by base pairingo has a site where amino acids attach too has anti codon which bind to mRNA codono 3 end terminal of ACC/CCA

7.4.2 Outline the structure of ribosomes, including protein and RNA composition, large

and small subunits, three tRNA binding sites and mRNA binding sites. (2)

Ribosomes consist of 2 subunits, one large and one small made up of protein andrRNA.

There is a binding site for mRNA on the small unit of ribosome. There are 3 binding sites for tRNA on the large unit of ribosome.

7.4.3 State that translation consists of initiation, elongation, translocation and

termination. (1)

Translation consists of initiation, elongation, translocation and termination.

7.4.4 State that translation occurs in a direction. (1)

Translation occurs in a 5 to 3 direction


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7.4.5 Draw and label a diagram showing the structure of a peptide bond between two

amino acids. (1)

- During translation amino acids will bind together with a peptide bond

7.4.6 Explain the process of translation, including ribosomes, polysomes, start

codons and stop codons.(3)

1. The small unit of the ribosome binds to 5 end of mRNA.2. Small subunit slides along mRNA until it reaches the START codon AUG.3. An activated tRNA with the anticodon UAC carrying amino acid: Methionine, binds

to the small subunit of the ribosome.

4. Then, the large subunit of the ribosome binds to the smaller unit5. There are three binding sites for tRNA on the large sub unit.(A,P,E)6. Another tRNA with the anticodon complementary to the next mRNA binds to the

ribosome. Elongation of polypeptides now start.

7. The large subunit of the ribosome advances over the small subunit and detaches thepolypeptide from the tRNA.

8. The small subunit slides across the large subunit and moves three nucleotides alongthe mRNA in a 5 to 3 direction.

9. A polypeptide chain is formed10.When the ribosomes reach the STOP codon, UGA no tRNA has a molecule

complementary to the anticodon.

11.The large subunit advances over the small subunit. The polypeptide is released fromthe tRNA.

12.The tRNA detaches and the large subunit, small subunit and mRNA all separate.


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7.4.7 State that free ribosomes synthesize proteins for use primarily within the cell,and

that bound ribosomes synthesize proteins primarily for secretion or for lysosomes. (1)

Free ribosomes in the cytoplasm are associated with the synthesis of proteins for internal use

in the cell.

Ribosomes which are attached to the wall of the endoplasmic reticulum are associated with

proteins which will be placed into vesicles and secreted form the cell.

7.5.1 Explain the four levels of protein structure, indicating the significance of each

level. (3)

Primary Structure

The number and sequence of amino acids in a polypeptide

Linked by peptide bonds

Determines the 2 and 3 structures and function

Reflects genetic information of the protein

Secondary structure:

Folding of polypeptides to form beta - pleated sheets

Coiling of polypeptides to form alpha-helix;

Held together by hydrogen bonds

Contributes to the strength of fibrous proteins

Eg: Keratin

Tertiary Structure:

A 3 dimensional conformation of a polypeptide

Due to intramolecular bonds between amino R-groups:

Hydrogen bonds, Ionic Bonds, Disulphide Bridges, Sulphur bonds

Determines overall shape of the protein

Eg: Lysozyme


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Quaternary Structure:

Linking together of two or more polypeptides to form a single protein

Same type of bonding as in tertiary structure

Linking of non-polypeptide (prosthetic group)

4 polypeptides are linked to heme group

Eg: Hemoglobin and insulin

7.5.2 Outline the difference between fibrous and globular proteins, with reference to

two examples of each protein type. (2)

Type Fibrous Globular

Shape Long, narrow Rounded shape

Solubility in water Insoluble Soluble

Functions Providing strength and

support to tissue

Act as pigments and transport

proteins

Example Myosin: Contraction in

muscle fibers for movement

in animals.

Collagen: Strengthen tendons,

bone and skin.

Hemoglobin: Bind to oxygen

in lungs to transport to tissues

Immunoglobulin: Act as

antibodies.

7.5.3 Explain the significance of polar and non-polar amino acids (3)

Polar amino acids are hydrophilic, water soluble, become channels for transport of ions/polar

substances

Non polar acids are hydrophobic, non-soluble, are embedded in the within the lipid

membrane

7.5.4 State four functions of proteins, giving a named example of each (1)

Hormones- Insulin

Transport- Haemoglobin

Movement- Myosin

Enzymes- Amylase


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3.6 Enzymes

3.6.1 Define enzyme and active site (1)

Enzymes: Globular proteins which act as catalysts of chemical reaction.Active site: The site on the surface of an enzyme to which substrates bind / the site on the

enzyme where it catalyzes a chemical reaction.

3.6.2 Explain enzymesubstrate specificity (3)

Enzyme has a specific shape. Active site of enzyme binds to specific substrate. Shape of the active site and substrate fit/complement each other. Active site works as a lock and substrate as a key. Active site fits substrate molecule. Enzyme-substrate complex formed. Weakens the bonds in substrate to lower activation energy.

3.6.3 Explain the effects of temperature, pH and substrate concentration on enzyme

activity (3)

Temperature

Rate of reaction increases as temperature increases (or vice versa).. This is because molecules

have more kinetic energy [or faster in movement of molecules] that result in more collisions

between active site of enzyme and substrate. Optimum temperature rate of enzyme-

catalyzed reaction is fastest; At a very high temperatures enzymes are denatured and stop

working. Denatured means change of structure resulting in loss of its biological properties

which makes it no longer can carry out its function;


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pH

Optimum pH is rate of enzyme-catalyzed reaction is fastest. Rate of reaction reduced as

increase or decrease pH (from optimum). Strong acids and alkalis can denature enzymes.

Affect (weak, ionic, hydrogen) bonds that hold enzyme in specific shape alter the

intermolecular interactions within the protein.

Example;PepsinpH of 2 (active in acidic stomach)

TrypsinpH of 8 (active in alkaline duodenum & small intestine)

Most enzymes in human cellspH of 7

Substrate concentration

At low substrate concentrations, as increase concentration get increase in rate of reaction.More chance of collision between substrate and active site of enzyme (more enzyme-

substrate complex forms). At high substrate concentration, have no change in rate as increase

concentration as all active sites occupied. Additional substrate will not lead to a greater rate

of product formation at this point

3.6.4 Define denaturation (1)

Denaturation: The changing of the structure of an enzyme (or other protein) so it can no

longer carry out its function. It is usually permanent.


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3.6.5 Explain the use of lactase in the production of lactose-free milk. (3)

Lactose is the sugar found in milk. Lactase is the enzyme and is obtained fromKluveromyces

lactis.

Lactose-free milk can be made in two ways;

a) Adding the enzyme lactase to the milk so that the milk contains the enzyme.b) Immobilizing the enzyme on a surface or in beads of a porous material. The milk is

then allowed to flow past the beads or surface with the immobilized lactase. Avoids

having lactase in the milk.

Reasons for using lactase in food processing;

a) Lactose intolerance high in some human population lactase is used to producelactose-free / low-lactose milk.

b)

Galactose and glucose are sweeter than lactose

no need to add extra sugar inmanufacture of flavoured milk drinks / frozen desserts.

c) Lactose tends to crystallize during production ofice cream, giving gritty structure.Glucose and galactose are more soluble and thus remain dissolved smoother

texture of ice cream.

d) Bacteria ferment glucose and galactose more quickly than lactose, results in fasterproduction of cottage cheese and yogurt.

7.6.1 State that metabolic pathways consist of chains and cycles of enzyme catalysed

reactions (1)

Metabolic pathways consist of chains and cycles of enzyme catalysed reactions.

7.6.2 Describe the induced-fit model (1)

Substrate approaches active site.

Shape of the active site will change to fit the substrate

As this occurs the substrates bonds are weakened and lowers its activation energy

This permits some enzymes to bond with several substances

Eg: Protease


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7.6.3 Explain that enzymes lower the activation energy of the chemical reactions that

they catalyse. (2)

1. Activation energy is the energy needed for a reaction to occur.2.

Enzymes lower the activation energy of the chemical reaction that they catalyse3. In the activated complex, energy is put into the substrate and weakens the structure.This allows the reaction to occur with a minimal amount of additional energy

required.

4. Normal activation energy would cause damage to the proteins of the cell. Thusreduced activation energy makes these reactions possible in a cell.

5. After the product is formed energy is released.7.6.4 Explain the difference between competitive and non-competitive inhibition, with

reference to one example of each (3)

Type Competitive Non Competitive

Binding Inhibitor binds to the same active site of

enzyme

Inhibitor binds to enzyme at

different site from active site

(allosteric site)

Similarity Substrate and inhibitor are chemically

similar

Substrate and inhibitor are not

similar

Activity When inhibitor occupies the active site,

substrate cannot bind to it. Activity of

enzyme is decreased.

When inhibitor binds to the

allosteric site, this can cause the

enzymes active site to changeshape (conformation change). Thus

the substrate will not be able to bind

to the active site.

Example Malonate- inhibitor

Succinate- substrate

Opioid- inhibitor

Nitric Oxide- substrate


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7.6.5 Explain the control of metabolic pathways by end-product inhibition, including

the role of allosteric sites. (3)

1. Allostery is a form of non-competitive inhibition.2. The allosteric site is an area of the enzyme separate from the active site.3. The end product in the pathway inhibits the enzyme that catalyses the first reaction of

the pathway. This is called end-product inhibition and shape of allosteric enzymes and

the active sites are altered by this process. So the substrate is less likely to bind to the

enzyme

4. Once the inhibitor is released from the allosteric site, the active site returns to itsoriginal conformation and the substrate is able to bind again.

5. End product inhibition is an example of negative feedback.6. When there is an excess of end-product, the whole metabolic pathway is shut down.

Therefore less of the end product gets produced and by inhibiting the first enzyme it

also prevents the formation of intermediates.

7. When the levels of the end product decrease, the enzymes start to work again and themetabolic pathway is switched on.

3.7 Cell respiration

3.7.1 Define cell respiration (1)

Cell respiration is the controlled release of energy from organic compounds (glucose) in cells

to form ATP. It occurs in every living cell.

3.7.2 State that, in cell respiration, glucose in the cytoplasm is broken down by

glycolysis into pyruvate, with a small yield of ATP (1)

Location: Cytoplasm Process: Glycolysis (does not require O2) Substrate: Glucose Products: 2 pyruvates and small amount of ATP


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3.7.3 Explain that, during anaerobic cell respiration, pyruvate can be converted in the

cytoplasm into lactate, or ethanol and carbon dioxide, with no further yield of ATP. (3)

Humans:

Yeast (Fermentation process without O2):

No ATP produced in this reactions Location: cytoplasm These reactions start with glycolysis.

3.7.4 Explain that, during aerobic cell respiration, pyruvate can be broken down in the

mitochondrion into carbon dioxide and water with a large yield of ATP. (3)

With the presence of O2 Location: Mitochondria (Pyruvate and O2 diffuses into mitochondria) Substrate: Pyruvate Products: CO2, H2O, large amount of ATP and heat

The overall equation of glycolysis and aerobic respiration:

The overall equation of glycolysis and anaerobic respiration:

(The ATP comes from glycolysis)

Topic 8 : Cell Respiration And Photosynthesis

Cell respiration

8.1.1 State that oxidation involves the loss of electrons from an element, whereas reduction

involves a gain of electrons; and that oxidation frequently involves gaining oxygen or losing

hydrogen, whereas reduction frequently involves losing oxygen or gaining hydrogen.

Comparison of oxidation and reduction

oxidation reduction loss of electron gain of electron loss of hydrogen atoms gain of hydrogen atoms gain of oxygen atoms loss of oxygen atoms


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REMEMBER OIL RIG = OXIDATION IS LOSS (in terms of electron and hydrogen)

, REDUCTION IS GAIN (in terms of electron and hydrogen)

8.1.2 Outline the process of glycolysis, including phosphorylation, lysis, oxidation and ATP

formation.

4 Main Stages in Glycolysis

1. PhosphorylationTwo phosphate groups are

added to a molecule of

glucose to form hexose

biphosphate . These two

phosphate groups are

provided by two molecules

of ATP.

2.

LysisHexose biphosphate is split to

form 2 moleculesoftriose

phosphate

3. OxidationTwo atoms of hydrogen are

removed from each triose

phosphate. The energy

released by the oxidation is used to add another phosphate group to each molecule. This

will result in two 3-carbon compounds, each carrying two phosphate groups. NAD+

is the

hydrogen carrier that accepts the hydrogen atoms lost from each triose phosphate

molecule.


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4. ATP FormationTwo pyruvate molecules are formed by removing two phosphate groups from each

molecule. These phosphate groups are given to ADP molecules to form ATP.Four ATP

will be produced.

Glycolysis occurs in the cytoplasm of cells. Two ATP molecules are used and 4 ATP

molecules are produced. Therefore there is a net yield of two ATP molecules. Also, two

NAD+

are converted into NADH + H+ during glycolysis.

8.1.3 Draw and label a diagram showing the structure of a mitochondrion as seen in

electron micrographs.

8.1.4 Explain aerobic respiration, including the link reaction, the Krebs cycle, the role of

NADH +H+, the electron transport chain and the role of oxygen.

Anaerobic respiration

Glycolysis can take place without oxygen.

Pyruvate produced from glycolysis cannot be oxidised further without thepresence of oxygen


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Aerobic respiration

occurs in the mitochondria of cells. consists of three stages ;Link reaction, The Krebs Cycle, The Electron Transport

Chain

The Link Reaction

Pyruvate from glycolysis is absorbed by the mitochondria Enzymes within the matrix of the mitochondrion remove hydrogen #(oxidation) and

carbon dioxide *(decarboxylation) from the pyruvate.

Therefore, the process is called oxidative decarboxylation The hydrogen removed is accepted by NAD+ results in the formation of an acetyl group which then accepted by CoA and forms

acetyl CoA.

# = removal of hydrogen or addition of oxygen

* = removal of carbon dioxide


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The Krebs Cycle

Step 1 - In the first stage of the Krebs cycle, the acetyl group from acetyl CoA is transferred

to a four carbon compound. This forms a six carbon compound.

Step 2 - This six carbon compound then undergoes decarboxylation (CO2 is removed) and

oxidation (hydrogen is removed) to form a five carbon compound. The hydrogen is

accepted by NAD+and forms NADH + H+.

Step 3 - The five carbon compound undergoes decarboxylation and oxidation (hydrogen is

removed) again to form a four carbon compound. The hydrogen is accepted by NAD+ and

forms NADH + H+.

Step 4 - The four carbon compound then undergoes substrate-level phosphorylation and

during this reaction it produces ATP. Oxidation also occurs twice (2 hydrogens are

removed). One hydrogen is accepted by NAD+ and forms NADH + H+. The other is accepted

by FAD and forms FADH2. The four carbon compound is then ready to accept a new acetyl

group and the cycle is repeated.

1

2

3

4


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Summary:

Carbon dioxide is removed in two reactions Hydrogen is removed in 4 reactions NAD+ accepts the hydrogen in 3 reactions FAD accepts the hydrogen in 1 reaction ATP is produced in one of the reactions

The carbon dioxide that is removed in these reactions is a waste product and is excreted from

the body. The oxidations release energy which is then stored by the carriers when they accept

the hydrogen. This energy is then later on used by the electron transport chain to produce

ATP.

The Electron Transport Chain

Series of electron carriers,located in the inner membrane of the mitochondrion NADH + H+supplies 2 electrons to the first carrier in the chain. FADH2 also donates electrons but at a later stage than NADH. The electrons come from oxidation reaction in earlier stages of cell respiration As the electrons are passed from one carriers to another,energy is released . The energy released is used to synthesize ATP via ATP synthase

ATP synthase is an enzyme that is also found in the inner mitochondrial membrane.


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The oxygen-dependent synthesis of ATP within mitochondria using energy releasedfrom redox reaction is called oxidative phosphorylation

The final electron acceptor is oxygen where it will combine with hydrogen ions toform water.

The Role of Oxygen = as the terminal electron acceptor

Oxygen is important for cell respiration as at the end of the electron transport chain, the

electrons are donated to oxygen. This occurs in the matrix at the surface of the inner

membrane. At the same time oxygen binds with hydrogen ions and forms water. This is the

only stage that oxygen is used in cell respiration.

If there is no oxygen then electron flow along the electron transport chain stops and NADH +

H+ can no longer be reconverted into NAD+. Eventually supplies of NAD+ in the

mitochondrion runs out and therefore the link reaction and Krebs cycle no longer take place.

8.1.5 Explain oxidative phosphorylation in terms of chemiosmosis


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Chemiosmosis = is the coupling of ATP synthesis to electron transport via a concentration

gradient of protons

There is a link between electrons being passed down the electron transport chain andthe production of ATP.

NADH + H + and FADH2 deposit their electrons to the electron transport chain in theinner membrane

As the high energy electrons pass through the electron transport chain,they releaseenergy.

The energy released is used to pump H+ from the matrix across the innermitochondrial membrane into the intermembrane spaces

A concentration gradient of H+ is formed ,which is a store of chemical potentialenergy (high concentration of protons in the intermembrane spaces and a low

concentration of protons in the matrix.)

ATP synthase (enzyme) located in the inner mitochondrial membrane transport theH

+backacross the membrane down the concentration gradient

As the protons pass across the membrane ,they release energy and this is used byATP synthase to convert ADP to ATP.

Since the electrons come from previous oxidation reactions of cell respiration and theATP synthase catalyses the phosphorylation of ADP into ATP, this process is called

oxidative phosphorylation.

8.1.6 Explain the relationship between the structure of the mitochondrion and its function.

Matrix: Watery substance that contains ribosomes and many enzymes. These enzymes are

vital for the link reaction and the Krebs cycle.

Inner membrane: The electron transport chain and ATP synthase are found in this

membrane. These are vital for oxidative phosphorylation.

Space between inner and outer membranes: Small volume space into which protons are

pumped into. Due to its small volume, a high concentration gradient can be reached very

quickly. This is vital for chemiosmosis.

Outer membrane: This membrane separates the contents of the mitochondrion from the rest

of the cell. It creates a good environment for cell respiration.

Cristae: These tubular projections of the inner membrane increase the surface area for

oxidative phosphorylation.


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3.8 Photosynthesis

3.8.1 State that photosynthesis involves the conversion of light energy into chemical

energy (1)

Photosynthesis is the process used by plants & some other organisms to produce their own

organic substances.

Reaction: Traps light energy (photons) and converts it into chemical energy Substrates: CO2, H2O Products: Organic compounds (sugar), CO2

3.8.2 State that light from the Sun is composed of a range of wavelengths (colours) (1)

Sunlight is called white light, but it is actually made up of a wide range of wavelengths

(colours) including red, green and blue.

3.8.3 State that chlorophyll is the main photosynthetic pigment (1)

Some substances called pigments can absorb light Chlorophyll is the main photosynthetic pigment. This is where light energy is trapped

and turned into chemical energy.

The structure of chlorophyll allows it to absorb some colours or wavelengths of lightbetter than others.

Red and blue light are absorbed more than green Green light is reflected chlorophyll, chloroplast and plant leaves look green

3.8.4 Outline the differences in absorption of red, blue and green light by chlorophyll.

(2)

The 'peaks' show which wavelength of light is being absorbed. The x-axis shows the colours of light that is being absorbed at the 'peaks'. The main colour of light absorbed by chlorophyll is red and blue. The main colour reflected (not absorbed) is green.


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3.8.5 State that light energy is used to produce ATP, and to split water molecules

(photolysis) to form oxygen and hydrogen. (1)

Light energy is used to;i. Produced ATPii. To split water molecules (photolysis) to form oxygen and hydrogen

3.8.6 State that ATP and hydrogen (derived from the photolysis of water) are used to fix

carbon dioxide to make organic molecules. (1)

ATP and hydrogen derived from photolysis of water are used to combine with carbondioxide to form organic compounds like sugar.

Bonds are formed between the carbon, hydrogen and oxygen using the energy fromATP (which came from the sun). C, H, O are enough to form lipids and

carbohydrates

3.8.7 Explain that the rate of photosynthesis can be measured directly by the production

of oxygen or the uptake of carbon dioxide, or indirectly by an increase in biomass (3)

Measuring the rate of photosynthesis;

Production of oxygen

Aquatic plants release oxygen bubbles during photosynthesis and so the volume can be

collected and measured.

The uptake of carbon dioxide

Difficult to measure so it is usually done indirectly. When carbon dioxide is absorbed from

water the pH of the water rises and so this can be measured with pH indicators or pH meters.

Increase in biomass

If batches of plants are harvested at a series of times and the biomass of these batches is

calculated, the rate increase in biomass gives an indirect measure of the rate of

photosynthesis in the plants.


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3.8.8 Outline the effects of temperature, light intensity and carbon dioxide

concentration on the rate of photosynthesis. (2)

Temperature

As temperature increases, the rate of photosynthesis increases more and more steeply(as the kinetic energy of the reactants increase) until the optimum temperature is

reached

If temperature keeps increasing above the optimum temperature then photosynthesisstarts to decrease very rapidly (denaturation of enzymes)

Light intensity

As light intensity increases so does photosynthesis until a certain point. At high light intensities photosynthesis reaches aplateau and so does not increase any

more

At low and medium light intensity the rate of photosynthesis is directly proportionalto the light intensity.

CO2 concentration

As the carbon dioxide concentration increases so does the rate of photosynthesis. There is no photosynthesis at very low levels of carbon dioxide At high concentration of CO2, the rate reaches aplateaubecause other factors become

the limiting factor.

Documents

Bio Chapter 3, 7 and 8 (Respiration)