Unit 3
3.1 Chemistry of life
3.1.1
State that the most frequently occurring chemical elements in living things are carbon, hydrogen, oxygen and nitrogen.
3.1 Chemistry of life
3.1.1
State that the most frequently occurring chemical elements in living things are carbon, hydrogen, oxygen and nitrogen.
3.1.2
State that a variety of other elements are needed
by living organisms, including sulfur, calcium, phosphorus, iron and sodium.
3.1.3
State one role for each of the elements
mentioned in 3.1.2.
Nitrogen:
Nitrogen:
- Required by proteins (included in amino acid
structure).
- Essential to enzymes required for plant function
- Essential to enzymes required for plant function
Calcium:
- Strengthens bone and teeth
- Regulates cell wall construction in plants
Phosphorus:
- Part of the phosphate groups in ATP and DNA molecules
- Make up part of the ‘backbone’ of DNA
- Formation of the phospholipid bilayer
(Hydrophilic phosphate heads ATTRACTED + Hydrophobic hydrocarbon tails)
Iron:
- Found in the structure of hemoglobin and is essential for the production of red blood cells.
- Involved in light energy transferring compounds involved in photosynthesis in plants.
Sodium:
- major ion associated with the propagation of a nerve impulse.
- Can replace potassium in some plants
Sulfur:
- Regulates cell wall construction in plants
Phosphorus:
- Part of the phosphate groups in ATP and DNA molecules
- Make up part of the ‘backbone’ of DNA
- Formation of the phospholipid bilayer
(Hydrophilic phosphate heads ATTRACTED + Hydrophobic hydrocarbon tails)
Iron:
- Found in the structure of hemoglobin and is essential for the production of red blood cells.
- Involved in light energy transferring compounds involved in photosynthesis in plants.
Sodium:
- major ion associated with the propagation of a nerve impulse.
- Can replace potassium in some plants
Sulfur:
- Found in some amino acids (proteins)
3.1.5
Outline the thermal, cohesive and solvent properties of water.
Thermal properties
- Hydrogen
bonds between polar water molecules cause water to resist change
- High specific
heat (energy required to change water temperature)
- High heat of
vaporization (energy required to boil water)
- High heat of
fusion (loss of energy required to freeze water)
- Thus, water
produces a stable environment for aquatic organisms
Cohesion
- Hydrogen
bonds between polar water molecules cause them to cohere
- Allowing for
transpiration in plants moving water against gravity
- Surface
tension between cohering water molecules
- Allowing for
animals such as water striders to walk over the surface of ponds even
though they are denser than water
Solvent properties
- The polarity of water attracts, or dissolves, any other
polar or charged particles by forming hydrogen bonds with them
- Proteins,
glucose, or ions, such as sodium or calcium are all soluble
3.1.6. Explain the relationship between
the properties of water and its uses in living organisms as a coolant, medium
for metabolic reactions and transport medium.
Coolant
- Hydrogen
bonds between polar water molecules cause water to resist change
- High heat of vaporization (energy required to
change liquid water to vapor) because hydrogen bonds must be broken
- Thus, evaporation of water from plant leaves
(transpiration) or from human skin (sweat) removes heat, acting as a
coolant
Medium
for metabolic reactions
- Cytoplasm is
primarily water, providing a polar medium in which other polar or charged molecules dissolve
- Many enzymes
are globular proteins that are water soluble so they dissolve in
cytoplasm where they control
metabolic reactions
Transport medium
- Hydrogen
bonds between polar water molecules cause them to cohere
- as water is lost by transpiration/evaporation
from plant leaves, hydrogen bonds
between adjacent water molecules pull water up columns of xylem
- Thus, plants move water against gravity,
by as much as 100 meters
3.2 Carbohydrates, lipids and proteins
3.2.1 Distinguish between organic and inorganic compounds.
Organic: Compounds containing carbon that are found in living organisms (except
hydrogencarbonates, carbonates and oxides of carbon) are regarded as
organic.
- exceptions: carbonates (e.g.
CaCO3), hydrogen carbonates (e.g. HCO3), and oxides of carbon (e.g. CO,
CO2)
- all other compounds are regarded
as inorganic
3.2.2
Identify amino acids, glucose, ribose and fatty
acids from diagrams showing their structure
Amino Acid.
Glucose
Ribose
Fatty acids
3.2.3 List three examples each
of monosaccharide, disaccharides and polysaccharides
Monosaccharide:
- Glucose
- Fructose
- Galactose
Monosaccharide:
- Glucose
- Fructose
- Galactose
(ribose)
Disaccharides:
- Maltose
- Lactose
- Lactose
- sucrose
Polysaccharides:
- starch
- glycogen
- cellulose
- glycogen
- cellulose
3.2.4 State one function of glucose, lactose and glycogen in animals, and of fructose, sucrose and cellulose in plants.
glucose, lactose and glycogen in animals
- glucose =
monosaccharide: major energy source for
fueling cellular respiration
- lactose =
disaccharide: energy source found in mammalian
milk
- glycogen =
polysaccharide: energy storage in
liver
and fructose, sucrose and cellulose in
plants
- fructose =
monosaccharide: energy component in flower
nectar
- sucrose =
disaccharide: energy molecule transported
via phloem
- cellulose =
polysaccharide: major structural
component of plant cell walls
3.2.5
Outline the role of condensation and hydrolysis in the relationships between
monosaccharides, disaccharides and polysaccharides; between fatty acids,
glycerol and triglycerides; and between amino acids and polypeptides.
condensation synthesis
- condensation
synthesis reactions link two monosaccharide
monomers
- forming one disaccharide molecule and one H2O molecule
- repeated additions of monosaccharides
produces a polysaccharide
hydrolysis
- The addition of water to split
polysaccharides into smaller fragments. –OH and –H are added to the
fragments.
- H2O molecules
used as a sources of -H and a -OH groups
- catalyzed by enzymes
fatty acids, glycerol and triglycerides:
condensation synthesis
- three separate condensation synthesis
reactions
- link three fatty acid monomers
to a single glycerol monomer
- forming one triglyceride molecule and three H2O molecules
hydrolysis
- a
triglyceride is broken down into one glycerol and three fatty acid molecules
- with three H2O molecules used as
a sources of -H and a -OH groups
- catalyzed by enzymes
amino acids and polypeptides:
condensation synthesis
- two amino acid monomers are linked to form a
dipeptide
- releasing one H2O molecule
- repeated condensation synthesis
reactions produce
polypeptides (=proteins)
hydrolysis
- a polypeptide is broken down into separate
amino acid molecules
- with H2O molecules used as a
sources of -H and a -OH groups
- catalyzed by enzymes
3.2.6 State three
functions of lipids.
- Energy storage
- Energy storage
- Thermal insulation
- Hormones – used in messenger molecules
(used as insulators to speed up the transfer of nerve
impulses from nerves) 3.2.7 Compare the use of carbohydrates in energy storage:
A. carbohydrates
- Glucose, when
in excess, linked together by condensation
synthesis to form polysaccharides
- Such as
starch, glycogen, and chitin
starch
- the primary
energy storage compound in plants, with å-glucose
- forming
either as amylose (unbranched chains)
- amylopectin
(branched chains)
glycogen
- a branched
polysaccharide formed from å-glucose
- primary
energy storage compound in animals
- found mainly
in liver and muscle cells
B. lipids
triglycerides
- a primary long-term energy storage compound
- containing
twice the amount of energy per unit mass or volume as do either
carbohydrates or proteins (more energy compared to
proteins/carbohydrates)
C-H bonds
- along the length of fatty acids are very
stable
- because the
electrons in the C-H bond are equally shared
- thus, when
these bonds are broken, they yield large sums of energy
3.3 DNA structure
3.3.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate.
- A DNA is composed of a deoxyribose, a phosphate group and a nitrogen base (Adenine, Cytosine, Thymine, and Guanine).
3.3.1 Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate.
- A DNA is composed of a deoxyribose, a phosphate group and a nitrogen base (Adenine, Cytosine, Thymine, and Guanine).
- The phosphate group is covalently bonded to the carbon of
the deoxyribose and then nitrogenous base is attached to the deoxyribose.
3.3.2
State the names of the four bases in DNA.
Cytosine
Guanine
Thymine
Adenine
Cytosine
Guanine
Thymine
Adenine
3.3. 3 Outline how DNA nucleotides are linked together by
covalent bonds into a single strand.
Two DNA nucleotides can be linked together by a covalent
bond between the sugar of one nucleotide and the phosphate of another. More
nucleotides can be added to form a single strand.
3.3.4
Explain how a DNA double helix is formed using complementary base pairing and
hydrogen bonds.
DNA molecules consist of two strands of nucleotides which are then wound together to form a double helix. These are formed between the bases of two strands. However, it is formed by complementary base pairing because Adenine only forms hydrogen bonds with Thymine and Cytosine only forms hydrogen bonds with Guanine.
DNA molecules consist of two strands of nucleotides which are then wound together to form a double helix. These are formed between the bases of two strands. However, it is formed by complementary base pairing because Adenine only forms hydrogen bonds with Thymine and Cytosine only forms hydrogen bonds with Guanine.
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.
3.4.1 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.
1) Production of new DNA à semi conservative à one old and one new strand
2) Helicase unwinds the DNA/double helix to allow the
strands to be separated
by breaking hydrogen bonds between bases.
3) Single strand act as a template.
4) Free bases form H-bonds with complementary bases.
5) Nucleotides are linked by DNA polymerase.
6) Daughter DNA molecules rewind.
7) Daughter strand identical to parent strand.
by breaking hydrogen bonds between bases.
3) Single strand act as a template.
4) Free bases form H-bonds with complementary bases.
5) Nucleotides are linked by DNA polymerase.
6) Daughter DNA molecules rewind.
7) Daughter strand identical to parent strand.
3.4.2 Explain the significance of complementary base pairing in the conservation of the base sequence of DNA.
Because the nitrogenous bases that compose DNA can only pair with complementary bases, any two linked strands of DNA are complementary. This ensures that the old base sequence is conserved.
A bonds with T and G bonds with C. Complementary base pairing ensures proper base incorporated into DNA strand thus making identical copies of the DNA strand (ensuring conservation not the base sequence).
How the base sequence of DNA is conserved during replication.
DNA replication is semi-conservative.
DNA is split into two single strands.
Nucleotides are assembled on to each single strand by complementary base pairing
AT, CG
Strand newly formed on each single strand is identical to other single strand.
DNA is split into two single strands.
Nucleotides are assembled on to each single strand by complementary base pairing
AT, CG
Strand newly formed on each single strand is identical to other single strand.
DNA polymerase used.
3.4.3 State that DNA replication is semi-conservative.
DNA replication copies DNA to produce
new molecules with the same base sequence. It is semi-conservative because each
new molecule formed by replication uses one new strand and one old strand which
is conserved from the parent DNA molecule.
3.5 Transcirption and translation
3.5.1 Compare the structure of RNA and DNA.
1. The number of strands.
·
DNA has two strands forming a double
helix.
·
RNA has one strand.
2. The type of sugar.
·
DNA has a deoxyribose.
·
RNA has a Ribose.
3. The Nucleotides.
·
DNA has A,C,G,T
·
RNA has A,C,G,U, (Uracil replaces Thymine)
3.5.2 Outline DNA transcription in terms of the formation of an RNA strand complementary to the DNA strand by RNA polymerase.
Transcription:
The copying of the base sequence of a gene by making an RNA molecule.
Initiation:
- The DNA double helix uncoils and the
two strands separate.
- RNA Polymerase attaches to promoter
regions of the DNA strand (starting point)
Elongation:
- RNA polymerase makes a mRNA molecule that is complementary to the DNA.
- (Uses DNA anti-sense strand as a template, synthesizes a complementary RNA strand using base pairing rules)
Termination:
- RNA polymerase makes a mRNA molecule that is complementary to the DNA.
- (Uses DNA anti-sense strand as a template, synthesizes a complementary RNA strand using base pairing rules)
Termination:
- The mRNA separates from the DNA.
- The DNA strands reforms into a double
helix.
1.
3.5.3
Describe the genetic code in terms of codons composed of triplets of bases
triplet code = 3 nucleotide bases code for one
amino acid
codon = a group of 3 nucleotide bases à
codon
there are 64 different codons (4 x 4 x4 = 64)
3.5.4 Explain the process of translation, leading to polypeptide formation
1) mRNA attaches
to ribosome (small unit)
2) Many ribosome
bind to the same mRNA,
which carries codons.
3) Transfer tRNA each have specific anticodon that binds to ribosomes to corresponding triplet base/codon.
4) A second tRNA binds to next codon.
5) Two amino acids bind together in a peptide linkage.
6) First tRNA detaches, ribosomes move along mRNA.
7) Another tRNA binds to next codon and it continues until polypeptide/protein is formed to stop codon (UAG, UAA and UGA.)
8) Stop codon has no corresponding tRNA, causes release of polypeptide
3) Transfer tRNA each have specific anticodon that binds to ribosomes to corresponding triplet base/codon.
4) A second tRNA binds to next codon.
5) Two amino acids bind together in a peptide linkage.
6) First tRNA detaches, ribosomes move along mRNA.
7) Another tRNA binds to next codon and it continues until polypeptide/protein is formed to stop codon (UAG, UAA and UGA.)
8) Stop codon has no corresponding tRNA, causes release of polypeptide
3.5.5 Discuss the relationship between
one gene and one polypeptide.
Amino
acids must be linked up in a precise sequence to make a polypeptide.
Genes store the information needed
to make a polypeptide in a coded form.
The
sequence of bases in a gene codes
for the sequence of amino acids in a
polypeptide.
The
information in the gene is decoded during the making of the polypeptide.
This
process is conducted in two stages known as transcription and translation.
DNA
(sequence) is copied to (m)RNA.DNA is separated.
RNA polymerase separate the strands, only one strand is copied.
RNA polymerase binds to promoter of template strand.
ATP provides energy for attachment.
RNA polymerase catalyzes the formation of the polymer
RNA lengthens in 5’ to 3’ direction.
RNA polymerase reaches terminator on DNA to stop transcription process.
mRNA is separated from DNA.
3.6 Enzymes
3.6.1 Define enzymes and active site.
Enzyme: A globular protein functioning as a biological catalyst, speeding up reaction rates by lowering activation energy
Active site: Site on surface of the enzyme to which the substrate binds.
3.6.2 Explain enzyme–substrate specificity.
Enzymes fit together with substrates similar to a lock and key.
Active site has shape that gives specificity.
Enzymes catalyze a reaction with a specific substrate.
3.6.3 Explain the effects of temperature, pH and
substrate concentration on enzyme activity.
pH:
Enzymes have an optimum pH, active site works best at this pH.
Activity decreases above and below the optimum by interfering with active site structure, denaturing by extremes of pH so enzyme activity stops.
Temperature:
pH:
Enzymes have an optimum pH, active site works best at this pH.
Activity decreases above and below the optimum by interfering with active site structure, denaturing by extremes of pH so enzyme activity stops.
Temperature:
Each
enzyme has an optimal temperature for it’s maximum activity
Small temperature increases result in increased enzyme activity to a point.
Increase activity due to increased movement of molecules (more KE).
Temperature increases above the optimum causes loss of activity due to denaturation.
Small temperature increases result in increased enzyme activity to a point.
Increase activity due to increased movement of molecules (more KE).
Temperature increases above the optimum causes loss of activity due to denaturation.
Substrate concentration:
Low – As increase concentration get increase in rate of reaction.
More chance of collision between substrate and enzyme/active site.
At high substrate concentration- have no change in rate as increase concentration; all active sites occupied.
3.6.4 Define denaturation
Denaturation is a structural change in a protein that results in the loss of it’s biological properties.
3.6.5 Explain the use of lactase in the production of lactose-free milk.
Low – As increase concentration get increase in rate of reaction.
More chance of collision between substrate and enzyme/active site.
At high substrate concentration- have no change in rate as increase concentration; all active sites occupied.
3.6.4 Define denaturation
Denaturation is a structural change in a protein that results in the loss of it’s biological properties.
3.6.5 Explain the use of lactase in the production of lactose-free milk.
- Used to improve the digestion of milk by some people.
-
Breaks down lactose into glucose and galactose.
- Increase sweetness of milk (glucose and galactose are sweeter)
- No need to add extra sugar in manufacture of flavoured milk drinks/frozen desserts,
- Increase sweetness of milk (glucose and galactose are sweeter)
- No need to add extra sugar in manufacture of flavoured milk drinks/frozen desserts,
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