Transcript Topic 2 Classroom Notes - General Course Information

Tuesday, August 25th
QUESTION TO PONDER
Read the following.
An astronomer, physicist and mathematician are in a
train headed to Edinburgh. Out the window they see a
lone black sheep. The astronomer says, “That’s
interesting, sheep in Scotland are black.” The physicist
says, “It would be more prudent to say that some of
the sheep in Scotland are black.” The mathematician
says, “To be more precise, we can say that in Scotland
there exists at least one field in which there is at least
one sheep, which is black on at least one side.”
1.
What does this story reveal about scientific observations,
hypotheses and conclusions?
2.
What does it reveal about each of their represented
disciplines?
3.
Is biology less exact than physics or mathematics?
If there were a biologist on board this train,
what would a biologist say about sheep?
THE CHEMISTRY OF LIFE
TOPIC 2
Essential Idea: Living organisms control their composition by
a complex web of chemical reactions.
Today I will…
1.
2.
3.
4.
List various elements crucial to sustaining homeostasis.
State one function of elements crucial to survival.
Describe several properties of water.
Differentiate between organic and inorganic chemistry.
Chemical elements
Some chemistry terms to review…
SI units (time, temperature, electric current, energy, mass, amount),
chemical symbols, atomic number, atomic mass, isotopes.
NAME
FUNCTION
BOND
carbon (C)
organic compounds
forms 4 covalent
hydrogen (H)
organic compounds, water forms 1 covalent
oxygen (O)
organic compounds, water forms 2 covalent
nitrogen (N)
organic compounds
various covalent
The most frequently occurring chemical elements in living things
are carbon, hydrogen, oxygen and nitrogen.
Additionally, a variety of additional
elements include…
Sulfur (sulphur): component enzymes (proteins, amino acids)
– all life forms
Calcium: structural component of bones and teeth, important in
muscle contraction and blood clotting (animals).
Phosphorus: a component of nucleotides (all), important in energy
transfer (ATP) (all), a component of cell membranes (phospholipids)
(all), structural component of bone and teeth (animals).
Iron: component of hemoglobin (animals).
Sodium: important in fluid balance and for conduction of nerve
impulses (action potential) (animals).
A variety of other elements are needed by living
organisms
including
sulfur
(S),
calcium
(Ca),
phosphorus (P), iron (Fe) and sodium (Na).
State one role for each of the elements mentioned above.
Refer to the roles in plants, animals and prokaryotes.
Water (Topic 2.2)
Essential Idea: Water is the medium of life.
Water molecules are polar - that is, they bear a partial (slightly) positive
(H+ side) and a partial negative (O side) charge.
-
Electrons are not shared equally.
-
Attraction of oppositely charged poles of water molecules causes them
to group together.
-
The attractive forces form hydrogen bonds. Although individual
bonds are weak, they collectively form important forces which hold water
molecules together.
Draw and label a diagram showing the structure of water molecules to
show their polarity and hydrogen bond formation.
Wednesday, August 26th
Good afternoon!
QUESTION TO PONDER
How do you diagram a water molecule?
Please draw a diagram and label the
parts of a water molecule.
Today I will…
1. Diagram a water molecule and explain various
properties.
2. Differentiate between adhesion and cohesion.
3. Explain the statement, “Like dissolves like”.
Fig. 3-2
–
Hydrogen
bond
+
H
+
O
–
–
+
H
+
–
Fig. 3-6
Hydrogen
bond
Ice
Hydrogen bonds are stable
Liquid water
Hydrogen bonds break and re-form
Water molecules have a very strong tendency to stick to each other;
that is they are cohesive.
Cohesion: force of attraction between molecules of the same substance.
-
due to the hydrogen bonds between the molecules.
-
The cohesive forces between molecules accounts for the upward pull of
water in xylem (plants). Water molecules stick to many other kinds of
substances (those substances that have charged groups of atoms and molecules
on their surfaces).
-
Adhesion
This is called adhesion.
Water-conducting
cells
Cohesion
Direction
of water
movement
Thermal properties: A high amount of energy is required to heat up water and
change its state (and the reverse) due to the hydrogen bonds; it protects
organisms from damaging changes in temperature.
Water is an excellent solvent for polar substances e.g., salt whose charged ions
dissociate in water when the substance dissolves in water.
•
Non-ionic substances like sugars which contain charged groups within the
molecules also dissolve in water. Water is capable of dissolving many organic
and inorganic particles.
Na+
Cl-
Outline the thermal, cohesive and solvent properties of water.
Water is a coolant medium (high specific heat, evaporation, temperature
control). Evaporation from the surface of a terrestrial plant or animal has a
cooling effect, excess heat is lost to the surroundings and their temperatures are
stabilized (panting, sweating).
Water is a transport medium due to its solvent properties for ions and polar
molecules. It is the transport medium in the blood, excretory and digestive
systems of animals and in the vascular tissues of plants.
Water is a habitat (transparent: light penetrates water so plants can
live in water and still receive light for photosynthesis [food chain],
relatively constant temperature, excellent solvent).
http://www.sumanasinc.com/webcontent/animations/content/propertiesofwater/water.swf
Explain the relationship between the properties of water and its uses in living
organisms as a coolant (thermal properties), medium for metabolic reactions
(solvent and thermal properties) and transport medium (cohesion and
thermal properties).
Thursday, August 27th
Good afternoon. As you come in, grab a number off of
the front counter, please, and take out your TOPIC 2
study guide.
Be prepared to review the questions from Topic 2.1
this afternoon. You are the experts here!
Today I will…
1. Recall various properties of water.
2. Describe vitalism and provide information about the discovery of urea in
falsifying this theory.
3. Provide diagrams of 4 important organic molecules crucial to
homeostasis.
Let’s take some time to
review: WATER
Please take a look at the following
Crash Course: Water video.
QUESTION TO PONDER
Should you be concerned
about the substance known as DHMO (dihydrogen monoxide)?
Visit DHMO.org to find out more information about this substance.
Friday, August 28th
Good afternoon to you.
QUESTION TO PONDER
Recall the molecule glycerol, which you were asked to
diagram. Please provide a diagram of glycerol and state
the functional group associated with this molecule.
Today I will…
1. Complete questions that assess my understanding of basic
chemistry topics.
2. Review properties of water and provide the structural formula
Hydroxyl group
for this molecule.
3. Identify the structural formula of various carbohydrates,
including glucose, ribose and fructose.
Monday, August 31st
QUESTION TO PONDER Discuss this quote with your
table mates. How do Monod’s thoughts coincide with your
own? How do they differ?
Why study Carbon?
 All
of life is built on carbon
Today I will…
 Cells
1.List
the four most commonly used elements by
living
things.
 ~72%
H 2O
2.Differentiate
between molecules and
 ~25% carbon compounds
hydrocarbons.
 carbohydrates
3.Define isomer and provide an example.
 lipids
4.List various
functional groups and provide
 proteins
properties
of these molecules.
 nucleic acids
 ~3%
salts
 Na,
Cl, K…
Chemistry of Life

Organic chemistry is the study of carbon compounds

C atoms are versatile building blocks

bonding properties

4 stable covalent bonds
Complex molecules assembled like TinkerToys
Hydrocarbons
 Combinations
of C & H
non-polar
not
soluble in H2O
hydrophobic
stable
very
methane
(simplest HC)
little attraction between
molecules
a
gas at room temperature
Hydrocarbons can grow
Isomers

Molecules with same molecular formula but different structures
(shapes)

different chemical properties

different biological functions
6 carbons
6 carbons
6 carbons
Diversity of molecules

Substitute other atoms or groups around the carbon

ethane vs. ethanol
H
replaced by an hydroxyl group (–OH)
nonpolar
gas
vs. polar
vs. liquid
biological
ethane
effects!
ethanol
Tuesday, September 1st
QUESTION TO PONDER Recall the structure
of propane, discussed briefly yesterday. Please
do the following:
•
Diagram this molecule.
•
Describe how it differs from another
common hydrocarbon, ethane.
•
Explain how to convert propane to propanol.
Propane has an additional
methyl group attached to it.
CH3 = methyl
Functional groups
 Parts
of organic molecules that are
involved in chemical reactions
 give
organic molecules distinctive
properties
 hydroxyl
 amino
 carbonyl
 sulfhydryl
 carboxyl
 phosphate
 Affect
reactivity
 makes
hydrocarbons hydrophilic
 increase solubility in water
Hydroxyl

–OH
 organic
 names
compounds with OH = alcohols
typically end in -ol
 ethanol
Carbonyl

C=O

O double bonded to C
if
C=O at end molecule = aldehyde
if
C=O in middle of molecule = ketone
Carboxyl

–COOH

C double bonded to O & single bonded to OH group
compounds
fatty
with COOH = acids
acids
amino
acids
Amino

-NH2

N attached to 2 H
compounds
with NH2 = amines
amino acids
NH2
acts as base
ammonia
picks up H+ from solution
Sulfhydryl

–SH

S bonded to H
compounds
SH
with SH = thiols
groups stabilize the structure of proteins
Phosphate

–PO4

P bound to 4 O
connects
lots
to C through an O
of O = lots of negative charge
 highly
reactive
transfers
 ATP,
energy between organic molecules
GTP, etc.
What about that methyl?
-CH3
•
Methyl groups are non-polar
•
Therefore, molecules are
hydrophobic
•
Lengthen hydrocarbons
•
Serve as an ID badge of sorts
•
Found commonly in lipids
Thursday, September 3rd
QUESTION TO PONDER
Can you identify the functional groups based
upon their structure? Let’s see…
1.
At your tables, you have 7 functional group
names (gold) and 7 structures (white). Please
match the proper structure with the proper
name.
2.
Be ready to share out!
Today I will…
1. Recall the basic structure of 7 functional groups.
2. List the 4 biomolecules necessary for life and describe the structure of
carbohydrates and lipids.
Carbohydrates and Lipids webquest
Please take out your webquest and turn to the second
page. I want to take a moment to complete #3 A with
you as a group. You will need:
•
Colored pencils or markers that are
•
GREEN
•
BLACK
•
RED
•
YELLOW
Friday, September 4th
Good morning! Let’s watch the following
Crash Course: Biomolecules video.
Afterward, you will have time to:
•
Work on Topic 2.3 study guide questions
Have a great weekend!
Viva la difference!

Basic structure of male & female hormones
is identical
 identical
carbon skeleton
 attachment of different functional groups
 interact with different targets in the body

different effects
Macromolecules

Smaller organic molecules join together to
form larger molecules
 macromolecules

4 major classes of
macromolecules:
 carbohydrates
 lipids
 proteins
 nucleic
acids
Polymers

Long molecules built by linking repeating
building blocks in a chain
 monomers
 building
blocks
 repeated
 covalent
small units
H 2O
bonds
HO
H
HO
H
Dehydration synthesis
HO
H
How to build a polymer

Synthesis
 joins
1
monomers by “taking” H2O out
monomer provides OH–
 other
monomer provides H+
 together
 requires
these form H2O
H 2O
energy & enzymes
HO
H
Dehydration synthesis
HO
H
enzyme
Condensation reaction
HO
H
How to break down a polymer
 Digestion
 use
H2 O
to breakdown polymers
 reverse
 cleave
 H2O

of dehydration synthesis
off one monomer at a time
is split into H+ and OH–
H+ & OH– attach to ends
 requires
H2O
enzymes
HO
Hydrolysis
Digestion
HO
enzyme H
H HO
H
Carbohydrates & Lipids
(Topic 2.3)
Essential Idea: Compounds of carbon, hydrogen and oxygen are
used to supply and store energy.
Distinguish between organic and inorganic compounds.
Compounds containing carbon that are found in living
organisms (except hydrogen carbonates, carbonates and
oxides of carbon) are regarded as organic.
Identify amino acids, glucose, ribose and fatty acids
from diagrams showing their structure. Specific names
of amino acids and fatty acids are not expected.
Monosaccharides: ribose, deoxyribose, glucose, fructose,
galactose.
Disaccharides: maltose, sucrose, lactose.
Polysaccharides: starch, glycogen, cellulose.
List
three
examples
each
disaccharides and polysaccharides.
of
monosaccharides,
Functions: cellular fuel (glucose, fructose, lactose, sucrose), energy storage (glycogen
[animals], structural components of plant cell walls (cellulose).
State one function of glucose, lactose and glycogen in animals, and of fructose,
sucrose and cellulose in plants.
* Monomer: a simple, relatively small molecule that can be linked to others to form a
polymer.
* Polymer: a large molecule composed of many similar or identical molecular subunits
called monomers.
Condensation (also called dehydration synthesis): a type of chemical reaction in which
two molecules join to form one larger molecule, simultaneously splitting out a
molecule of water.
Hydrolysis: splitting of one molecule into two by addition of H+ and OH- ions from
water.
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.
Functions for lipids: cellular fuel; energy storage; components of cell membranes
(phospholipids); some are hormones (sex); thermal insulation (lipids conduct heat
only slowly); mechanical protection (lipids serve as a packing material around
delicate organs).
State three functions of lipids.
Both lipids and carbohydrates can be used for energy storage in living organisms.
Lipids: contain more energy per gram than carbohydrates so stores of lipid are lighter
than stores of carbohydrate that contain the same amount of energy; are insoluble in
water, so they do not cause problems with osmosis in cells; for long-term energy
storage.
Carbohydrates: are more easily digested than lipids so the energy stored by them can
be released more rapidly; are soluble in water so are easier to transport to and from the
store; used for energy storage over short periods.
Compare the use of carbohydrates and lipids in energy storage.
3.3 & 7.1 DNA structure
DNA structure: 5-carbon sugar (deoxyribose), organic (nitrogenous) base and
phosphate: nucleotide (building block, monomer).
The bases: adenine A, guanine G, thymine T, cytosine C.
State the names of the four bases in DNA.
The DNA nucleotides are linked together by covalent bonds (phosphodiester 3'-5'
linkage, spatial arrangement, condensation reaction) into a single strand.
Chargaff's rules.
Watson, Crick: a DNA double helix is formed by complementary base pairing and
hydrogen bonds between the bases (A – T, G – C) belonging to the two strands. The
two strands are antiparallel (run in opposite direction) and are not identical.
Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and
phosphate. Simple shapes can be used to represent the component parts. Only the
relative positions are required.
Two DNA nucleotides can be linked together by a covalent
bond between the sugar of one nucleotide and the phosphate
group of the other. More nucleotides can be added in a similar
way to form a strand of nucleotides.
Outline how the DNA nucleotides are linked together by
covalent bonds into a single strand. Only the relative
positions are required.
Explain how a DNA double helix is formed using
complementary base pairing and hydrogen bonds.
Draw and label a simple diagram of the molecular structure of DNA. An extension of
the previous diagram is sufficient to show complementary base pairs of A - T and G C, held together by hydrogen bonds and the sugar-phosphate backbone.
It is possible to assign numbers to the carbon atoms in the deoxyribose. The phosphate
is attached to carbon #5. In a single strand of DNA, the phosphate of the next
nucleotide will attach to carbon #3. This is called a 3’ – 5’ linkage. A DNA strand
therefore ends on one side with a phosphate on the 5’ end of the deoxyribose and a
hydroxyl at the 3’ end. The sequence of the nucleotides is given in the 5’ to 3’ direction.
Describe the structure of DNA including the antiparallel strands, 3‘-5’ linkages and
hydrogen bonding between purines and pyrimidines.
Analysis of chromosomes has shown that they are made of DNA and protein. DNA has
negative charges along the strand and positively charged proteins are bonded to this.
These basic proteins are called histones. The complex of DNA and protein is known as
chromatin.
The total length of DNA in a human is
approximately 2.2 m. To pack all this into a cell means
that the total length is shortened to 276 m = 0.276 mm.
This means that the length has to be reduced by a factor
8000 which needs a good organisation if you want to be able
to unwind it again. The histone proteins form the skeleton
for this. The genetic material looks like beads on a string.
This is caused by the DNA helix combining with 8 small
histone molecules. These structures are called nucleosomes.
Outline the structure of nucleosomes. Limit this to the
fact that a nucleosome consists of DNA wrapped
around eight histone proteins and held together by
another histone protein.
Nucleosomes help to supercoil chromosomes and help
to regulate transcription.
Repetitive DNA: DNA made up of copies of the same
or nearly the same nucleotide sequence. DNA sequences
that are present in many copies per chromosome set.
Repetitive DNA sequences may be closely linked or
dispersed throughout the genome or parts of the
genome.
Unique DNA: a length of DNA with no repetitive DNA
sequences.
Distinguish between unique or single-copy genes and highly repetitive sequences in
nuclear DNA. Highly repetitive sequences (satellite DNA) constitute 5-45% of the
genome. The sequences are typically between 5 and 300 base pairs per repeat, and
may be duplicated as many as 105 times per genome.
Eukaryotic genes can contain exons and introns.
3.4 & 7.2 DNA replication
DNA replication is semi-conservative (each molecule formed
by replication consists of one new strand and one old strand
conserved from the parent DNA molecule).
DNA replication occurs in a 5'→3' direction. The 5’ end of
the free DNA nucleotide is added to the 3’ end of the chain of
nucleotides that is already synthesized.
Explain DNA replication in terms of unwinding of the
double helix and separation of the strands by helicase
(replication
fork,
template
strand),
followed
by
formation of the new complementary strands by DNA
polymerase (it catalyses the linking together of the
nucleotide subunits).
The two daughter DNA molecules are identical in base sequence to each other and to
the parent molecule, because of complementary base pairing. Each of the new strands is
complementary to the template on which it was made and identical to the other
template.
Explain the significance of complementary base pairing in the conservation of the
base sequence of DNA.
In order to synthesise a complete new strand, a
beginning of the new strand must be present. This
beginning in eukaryotic cells is a short strand of RNA
called a primer. RNA primase catalyses the formation
of this primer on the exposed single strand of DNA
(template strand).
DNA replication in one strand will take place
continuously. This is the leading strand and the new
DNA will grow in the 5’→3’ direction.
The other template strand is antiparallel and DNA
synthesis will take place in the opposite direction of the
growing fork.
The short sequences of DNA formed this way are called
Okazaki fragments. DNA ligase will join these
segments together to form the lagging strand. The
lagging strand requires a new primer for each Okazaki
fragment.
DNA
nucleotides
are
in
the
form
of
deoxyribonucleoside triphosphates when they form
hydrogen bonds with the complementary nucleotide of
the DNA strand. The deoxyribonucleoside triphosphates
are sometimes known as dATP, dTTP, dCTP and dGTP.
The second and third phosphate group are removed as
the nucleotide is attached to the growing DNA strand by
DNA polymerase III. This provides the energy to drive
the reaction.
DNA polymerase III adds deoxynucleoside triphosphate
to a free 3’ OH group. DNA polymerase I will erase the
primers and fill the gaps thus created. DNA ligase seals
up this gap by making another sugar-phosphate bond.
DNA polymerase III can only catalyse DNA synthesis in a 5’→3’
direction (of the new chain). This means that the lagging strand
cannot simply add on new nucleotides as the helicase unzips the
double strand. To do so would mean adding them on to the 5’ end
of the growing strand (which would be 3’→5’ direction of
replication). This is not possible as DNA polymerase III can only
work in a 5’→3’ direction.
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. The explanation of Okazaki fragments in relation to
the direction of action of DNA polymerase III action is required.
DNA polymerase III adds nucleotides in the 5’→3’ direction. DNA
polymerase I excises the RNA primers and replaces them with
DNA.
DNA replication is initiated at many points in eukaryotic
chromosomes.
http://ed.ted.com/on/qk28a2q9
3.5
&
7.3,
7.4
Transcription
and
translation
DNA
RNA
strand
double
single
sugar
deoxyribose
ribose
bases
adenine, guanine
adenine,
guanine
cytosine, thymine
cytosine,
uracil
Compare the structure of DNA and RNA. Limit this to the names of sugars, bases
and the number of strands.
Three kinds of RNA: messenger RNA (mRNA), transfer RNA (tRNA), ribosomal RNA
(rRNA).
mRNA: RNA transcribed from DNA that specifies the amino acid sequence of a
protein, carries the genetic information from the nucleus to the ribosomes, where it is
translated into protein.
tRNA: accepts a specific amino acid and transfers it to a growing polypeptide chain as
specified by the nucleotide sequence of the messenger RNA being translated.
rRNA: is manufactured by the DNA of the nucleus, it is found in the cytoplasm where
it makes up more than half of the mass of the ribosomes.
Transcription is the process by which RNA is produced from a DNA template.
Transcription is similar to DNA replication in that it takes place in the nucleus and
involves a section of DNA which needs to unzip. Then only one of the two strands of
DNA is transcribed and a RNA complementary strand to this strand is made. This is
called mRNA. After transcription, the mRNA leaves the nucleus through the pores in
the nuclear envelope and goes into the cytoplasm.
Outline DNA transcription in terms of the formation of an RNA strand
complementary to the DNA strand by RNA polymerase.
Transcription is carried out in a 5’→3’ direction. The 5’ end of the free RNA
nucleotide is added to the 3’ end of the RNA molecule that is already synthesized.
DNA is split into two strands by RNA polymerase. One of these
strands forms the template for transcription (is transcribed and
called the antisense strand). The base sequence of the mRNA is
complementary to it. The other strand has the same base sequence of
the mRNA (except for T instead of U) and called the sense strand.
Distinguish between the sense and antisense strands of DNA. The
sense strand (coding strand) has the same base sequence as mRNA
(with uracil instead of thymine). The antisense (template) strand is
transcribed.
Transcription involves a promoter region (a base
sequence on the sense strand that causes RNA
polymerase to bind and start transcribing the antisense
strand). The terminator is a base sequence on the sense
strand
that
transcription.
causes
RNA
polymerase
to
stop
RNA polymerase is similar to DNA polymerase and
works in almost the same way. RNA nucleotides in the
form of ribonucleoside triphosphates form hydrogen
bonds with the complementary nucleotide of the DNA
strand.
As in DNA replication, the ribonucleoside triphosphate will attach
to the 3’ hydroxyl group of the growing strand. The second and third
phosphate groups will be removed, providing the energy required
driving this reaction.
Explain the process of transcription in prokaryotes, including the
role of the promoter region, RNA polymerase, nucleoside
triphosphates and the terminator.
Many genes in eukaryotes contain introns. These are non-coding sequences that are
transcribed but not translated. They are found in newly transcribed mRNA but later are
removed. Mature mRNA does not contain introns. The sequences that are not removed
are called exons. Prokaryotes do not usually have introns in their genes.
Eukaryotic RNA needs the removal of introns to form mature mRNA.
Translation: protein synthesis, mRNA codon, tRNA anticodon, rRNA ribosome. The
genetic code: composed of codons (triplets of bases); a base triplet (why 3?) to code for
one amino acid; non-overlapping, redundant, universal.
Describe the genetic code in terms of codons composed of triplets of bases.
Explain the process of translation, leading to polypeptide formation. Include the
roles of messenger RNA (mRNA), transfer RNA (tRNA), codons, anticodons,
ribosomes and amino acids.
All tRNA molecules have:
a triplet of bases called the anticodon in a loop;
two other loops;
the base sequence CCA at the 3’ terminal, which forms a site for attaching an amino
acid;
sections that become double stranded by base pairing.
The distinctive three-dimensional shape allows the
correct amino acid to be attached to the 3’ terminal by
an enzyme called a tRNA activating enzyme.
Explain that each tRNA molecule is recognized by a
tRNA-activating enzyme that binds a specific amino
acid to the tRNA, using ATP for energy. Each amino
acid has a specific tRNA-activating enzyme. The shape
of tRNA and CCA at the 3’end should be included.
Ribosomes measure approximately 25 nm and are found in both prokaryotic and
eukaryotic cells and are larger in eukaryotic cells than in prokaryotic cells. Ribosomes
are measured by their size and density in units called S. Prokaryotic ribosomes and
ribosomes of chloroplasts and mitochondria are 70S, whereas eukaryotic ribosomes are
80S.
All ribosomes are made of protein and rRNA. They
consist of two subunits: a smaller and a larger unit.
Outline the structure of ribosomes, including protein
and RNA composition, large and small subunits, three
tRNA binding sites and mRNA binding sites.
Draw and label a diagram showing the structure of a
peptide bond between two amino acids.
Translation
consists
of
initiation,
elongation,
translocation and termination.
Translation occurs in a 5’→3’ direction. During
translation, the ribosome moves along the mRNA
towards the 3’ end. The start codon is nearer to the 5’
end.
A polysome (or polyribosome) is a group of ribosomes
moving along the same mRNA, as they simultaneously
translate it.
Initiation. The smaller subunit of the ribosome will attach
to the mRNA 5’ end. The first codon is the start codon
(AUG), also coding for the amino acid methionine (met).
The tRNA with the anticodon UAC, carrying methionine
will attach to the mRNA (hydrogen bonds). The tRNA fits
into one of the two binding sites (the P site).
Elongation. At the end of initiation, the second codon
of the mRNA is lined up with the second binding site of
the larger subunit of the ribosome (A site). A tRNA
(carrying
an
amino
acid)
with
an
anticodon
complementary to the mRNA codon will fit into this
slot. An enzyme will form a bond between the two
amino acids.
The ribosome moves on the next codon, releasing the
first tRNA. The second tRNA now occupies the P site
and the A site is available for another tRNA. This
process continues until the ribosome reaches a stop
codon on the mRNA.
Termination. When the ribosome reaches a stop codon (UAG, UAA or UGA), there is
no tRNA available which as the correct anticodon. The A site will not be occupied and
translation stops.
Explain the process of translation including ribosomes, polysomes, start codon and
stop codons.
Free ribosomes synthesize proteins for use primarily
within the cell, and that bound ribosomes synthesize
proteins primarily for secretion or for lysosomes.
A gene is a segment of DNA. Polypeptide is a chain of
amino acids. One gene codes for making one
polypeptide. DNA is transcribed to mRNA then mRNA
is translated to polypeptide.
Exception: immunoglobulin lambda chain structure: two
genes, one polypeptide chain.
The lambda chain seems to be encoded by two separate
genes which are expressed as a single, continuous
polypeptide chain. Antibody light chains appear to be an
exception to the rule of "one gene, one polypeptide chain".
Explain the relationship between one gene and one
polypeptide. Originally, it was assumed that one gene
would invariably code for one polypeptide, but many
exceptions have been discovered.
7.5 Proteins
PRIMARY
Linear sequence of amino acids with peptide linkages.
This sequence determines proteins' properties and
shape. Changes in the sequence may have several
effects on the overall structure and activity. The number
of sequence is almost infinite (length, composition).
SECONDARY
-helix : hydrogen bonds form between the N-H of one
peptide bond and the C=O of another peptide bond; the
R groups are held to the outside. (e.g., keratin - hair,
wool, horn, feathers, nail etc.). Polypeptide is coiled
into a helix.
-pleated sheet: hydrogen bonds form between the NH of a peptide bond and the C=O of another peptide
bond in another part of the chain; the interacting strands
of the chain are anti-parallel (e.g. silk), chains are held
together by hydrogen bonds between the CO and NH2
groups of the polypeptide chain (same , different ).
TERTIARY
This level describes the effects of all the interactions
between the R groups of the amino acids. The
possibility of creating active sites:
added strength due to ionic bonds (between positively
and negatively charged R groups),
disulphide linkages (bridges, strong covalent bonds
between pairs of cysteines),
hydrogen bonds (between some R groups),
hydrophobic interactions (weak bonds, between nonpolar R groups),
the possibility of prosthetic groups and coenzymes.
A prosthetic group: a non-protein compound that
associates with a polypeptide, like a haem in
haemoglobin. Haem is the actual oxygen carrier in
blood. The haem molecule is held in the folded globin
molecule by electrostatic attractions between two
electrically charged side chains and the Fe2+ ion in the
haem.
Cofactor: a non-protein substance needed by the
enzyme for normal activity; some cofactors are
inorganic; others are organic cofactors, known as
coenzymes (e.g., NAD+, NADP+).
QUATERNARY
The polypeptide can interact with other polypeptide chains. Most large, non-structural
proteins (haemoglobin) have more than one polypeptide chain and it leads to greater
range of biological activity. The forces holding chains together are the forces that hold
tertiary structure in place.
This structure may involve the binding of a prosthetic group (a non-polypeptide
structure) to form a conjugated protein.
Explain the four levels of protein structure, indicating the significance of each level.
In fibrous proteins (collagen, elastin, fibrin, keratin) the polypeptide chains are
arranged in long chains which may run parallel to one another, being linked by cross
bridges (only secondary structure exists: -helix or -pleated sheet). They are very
stable molecules, insoluble in water and have structural roles within organisms.
In globular proteins (tertiary structure: albumin, gamma
globulin, enzymes, membrane proteins), the polypeptide
chains are tightly folded into a compact spherical shape. These
molecules are relatively unstable and soluble in water.
Outline the difference between fibrous and globular proteins,
with reference to two examples of each protein type.
Polar amino acids
on the surface of proteins make them water soluble;
create channels through which hydrophilic substances can diffuse, positively charged R
groups allow negatively charged ions through and vice versa;
cause parts of membrane proteins to project from the membrane, transmembrane
proteins have two such regions.
Non-polar amino acids
in the centre of water-soluble proteins stabilise their structure;
cause proteins remain embedded in membranes.
Explain the significance of polar and non-polar amino acids. Limit this to
controlling the position of proteins in membranes, creating hydrophilic channels
through membranes and the specificity of active sites in enzymes.
Proteins perform a wide variety of functions in living organisms:
1.
enzymes (catalase),
2.
contractile proteins (myosin, actin),
3.
transport of oxygen (haemoglobin),
4.
immune protection (antibodies or immunoglobulins),
5.
receptors for light stimuli (rhodopsin),
6.
components of skin and bones (collagen),
7.
hormones (insulin).
State four functions of proteins, giving a named example of each.
3.6 & 7.6 Enzymes
Enzyme: a globular protein functioning as a
biological catalyst of a chemical reaction.
Catalyst: speeds up (catalyze) the rate of a chemical
reaction. The enzyme affects the speed of a chemical
reaction without being consumed by the reaction.
Name: (most enzymes) derives from the name of the
substrate + function
-ase (lipase, protease). Some have
a traditional or trivial name: pepsin, trypsin.
Active site: the region on the surface of an enzyme to
which substrate or substrates bind and which catalyzes
a chemical reaction involving the substrates.
Substrate: a reactant in a reaction catalyzed by an enzyme.
Metabolism: anabolism, catabolism.
Metabolic reactions take place in cells. A substrate is converted to an end product via a
series of intermediates. Example: glycolysis (glucose is converted to pyruvate). Some
are cycles. Example: Krebs cycle. End product also is a starting point.
Metabolic pathways consist of chains and cycles of enzyme-catalysed reactions.
Specificity
an enzyme is specific to its substrate because of the shape of its active site. This makes
up a small part of the total protein molecule and is a three-dimensional arrangement
which fits the shape of the substrate. The rest of the protein is a framework which holds
the amino acids of the active site in the correct position.
This earliest hypothesis to explain the specificity of
enzymes was the "lock-and-key" hypothesis (model).
The “lock-and-key” model: enzymes are thought to
operate on a lock (active site) and key (substrate)
mechanism. The two molecules form a temporary
structure called the enzyme-substrate complex.
The products have different shape from the substrate and
so, once formed, they escape from the active site, leaving
it free to become attached to another substrate molecule.
Enzymes are specific one or a few substrates. Change in
shape of active site by denaturation inactivates enzymes.
Explain enzyme-substrate specificity. The lock-and-key
model can be used as a basis for the explanation. Refer
to the three-dimensional structure.
The “induced fit model”: enzyme is not a rigid shape. The active site of certain
enzymes becomes comforted to the shape of the substrate molecule. It accounts for the
broad specificity of some enzymes (the ability to bind several substances).
Describe the induced fit model. This is an extension of the lock-and-key model. Its
importance in accounting for the ability of some enzymes to bind to several substrates
should be mentioned.
Activation energy: the energy is required to initiate a
chemical reaction or the energy required to break the
existing bonds and to begin the reaction.
Explain that enzymes lower the activation energy of
the chemical reactions that they catalyse. Only
exothermic reactions should be considered.
Temperature
most enzymes have an optimal temperature at which the rate of reaction is the fastest.
The rates of most enzyme-controlled reactions increase with increasing temperature,
within limits. High temperatures rapidly inactivate most enzymes by denaturing the
protein. This inactivation is usually not reversible; that is, activity is not regained when
the enzyme is cooled.
pH
most enzymes have an optimal pH and are active only
over a narrow pH range. Many enzymes become
inactive and usually irreversibly denatured when the
medium is made very acidic or very basic.
Substrate concentration
if the enzyme concentration, pH, and temperature are kept constant, the rate of an
enzymatic reaction is proportional to the concentration of substrate present at low
substrate concentrations because random collisions between substrate and active site
happen more frequently with higher substrate concentrations.
At high substrate concentrations, all the active site sites
of the enzyme are fully occupied, so raising the
substrate concentration has no effect.
Explain the effects of temperature, pH and substrate
concentration on enzyme activity.
Competitive - an inhibiting molecule structurally
similar to the substrate molecule binds to the active
site, preventing substrate binding.
The same quantity of product is formed, because the
substrate continues to use any enzyme molecules which
are unaffected by the inhibitor. It takes longer to make
the products. If the concentration of the substrate is
increased, less inhibition occurs.
This is because, as the substrate and inhibitor are in
direct competition, the greater the proportion of
substrate molecules the greater their chance of finding
the active site, leaving fewer to be occupied by the
inhibitor.
1. inhibition of succinate dehydrogenase by malonate in
the Krebs cycle;
2. sulfonamide Prontosil™ (an antibiotic) inhibits folic
acid synthesis in bacteria (cell wall).
Non-competitive- limited to an inhibitor molecule
binding to an enzyme (not to its active site) that causes
a conformational change in its active site, resulting in
a decrease in activity.
As the substrate and the inhibitor molecules attach to
different parts of the enzyme they are not competing for
the same sites. An increase in substrate concentration
will not therefore reduce the effect on the inhibitor.
1. nerve gases like Sarin inhibit acetyl cholinesterase,
2. Hg2+, Ag+, Cu2+ and CN-inhibit many enzymes by
binding to –SH groups, therefore breaking –S-Slinkages.
COMPETITIVE
1.
The substrate and inhibitor are chemically very similar.
2.
The inhibitor binds to the active site of the enzyme.
3.
While the inhibitor occupies the active site, it prevents
the substrate from binding and so the activity of the
enzyme is prevented until the inhibitor dissociates.
4.
With a fixed low concentration of inhibitor, increases in
the substrate concentration gradually reduce the effect of
the inhibitor.
When the substrate binds to the active site, the inhibitor
cannot bind, so the proportion of enzyme molecules that
are inhibited becomes less and less. When there are
more substrate molecules than inhibitor molecules, the
substrate always wins the competition and binds to the
active site. The same maximum enzyme activity rate is
then reached as when there is no inhibitor.
NON-COMPETITIVE
1.The
substrate and active site are not similar.
2.The
inhibitor binds to the enzyme at a different site
from the active site.
3.The
inhibitor changes the conformation of the
enzyme. The active site does not catalyse the reaction
or catalyses it at a slower rate.
4.With
a fixed low concentration of inhibitor, increases
in the substrate concentration increase enzyme activity.
The substrate cannot prevent the binding of the inhibitor.
Some of the enzyme molecules therefore remain inhibited
and the maximum enzyme activity rate reached is lower than
when there is no inhibitor.
Explain the difference between competitive and noncompetitive inhibition, with reference to one example of
each.
Metabolic pathways have the following features:
1.
they consist of many chemical reactions that are
carried out in a particular sequence;
2.
an enzyme catalyses each reaction;
3.
all the reactions occur inside cells;
4.
some pathways build up organic compounds
(anabolic pathways) and some break them down
(catabolic pathways);
5.
some metabolic pathways consist of chains of
reactions (for example glycolysis);
6.
some metabolic pathways consist of cycles of
reactions, where a substrate of the cycle continually
regenerated by the cycle (for example Krebs cycle).
In many metabolic pathways, the product of the last
reaction in the pathway inhibits the enzyme that
catalyses the first reaction. This is called end-product
inhibition.
The enzyme that is inhibited by the end products is an
example of an allosteric enzyme. Allosteric enzymes
have two binding sites. One of these is the active site.
The other is the allosteric site. It is a binding site for the
end product. When it binds, the structure of the enzyme
is altered so that the substrate is less likely to bind to the
active site. This is how the end-product acts as an
inhibitor.
Binding of the inhibitor is reversible and if it detaches,
the enzyme returns to its original conformation, so the
active site can bind the substrate easily again.
The advantage of this method of controlling metabolic
pathways is that if there is an excess of the end-product
the whole pathway is switched off and intermediates do
not build up.
Conversely, as the level of the end-product falls, more
and more of the enzymes that catalyse the first reaction
will start to work and the whole pathway will become
activated. End-product inhibition is an example of
negative feedback.
Explain the control of metabolic pathways by end-
product inhibition, including the role of allosteric
sites.
Allostery is a form of non-competitive inhibition.
Metabolites can act as allosteric inhibitors of enzymes
earlier in a metabolic pathway and regulate metabolism
according to the requirements of organisms; a form of
negative feedback.
ATP inhibition of phosphofructokinase (glycolysis)
Denaturation: a structural change in a protein that
results in the loss (usually permanent) of its biological
properties. Refer only to heat and pH as agents.
Lactose-free milk can be produced by passing milk over
lactase enzyme bound to an inert carrier: once the
molecule is cleaved, there are no lactose ill-effects.
Explain the use of lactase in the production of lactosefree milk.
3.7 & 8.1 Cell respiration
Oxidation and reduction are chemical processes that
must occur together.
Oxidation involves the loss of electrons from an
element and frequently involves gaining oxygen or
losing hydrogen.
Reduction involves a gain of electrons and frequently
involves losing of oxygen or gaining hydrogen.
A generalised redox reaction is shown below:
↑---------- oxidation ---------------↓
Compound Ae- + Compound B  Compound A + Compound Be↓-------------- reduction -----------↑
In the cell, oxidation and reduction usually involve the movement of an entire hydrogen
atom with its single electron.
When hydrogen atoms are removed from an organic
compound, they take along some of their chemical bond
energy. This energy is transferred to a hydrogen
acceptor molecule.
Electron acceptors are frequently arranged in chains
(ETC). As electrons are transferred from one acceptor to
the next, energy is slowly released. This mechanism
allows the cell to capture energy slowly and efficiently
from the fuel molecule (compare burning and cellular
respiration).
In the complete process of cellular respiration, hydrogen
is transferred from glucose to oxygen. Glucose is
oxidised and oxygen is reduced.
The reaction is shown below:
--------- oxidation----------
C6H12O6 + 6 O2 + 6 H2O  6 CO2 + 12 H2O + Energy
---------- reduction ----------
During this process, the potential energy of the electrons
is decreased and that chemical energy is released.
Cell respiration: controlled release of energy from
organic compounds in cells to form ATP.
In cell respiration, glucose in the cytoplasm is broken
down by glycolysis into pyruvate, with a small yield of
ATP.
In yeast in the absence of oxygen the pyruvate is
converted to ethanol (the reaction is irreversible) with
the release of carbon dioxide and a small yield of ATP.
The CO2 production is important in the baking industry,
while alcohol production is important in producing beer,
wine, and other alcoholic beverages.
In animal cells (humans) the pyruvate is usually
converted to lactate in the absence of oxygen (the
reaction is reversible). Lactate is produced when
bacteria cause milk to sour. Lactic acid fermentation
occurs in strenuous muscle activity, when there is a
shortage of oxygen.
Anaerobic metabolism is inefficient because the fuel is only
partly oxidised. Alcohol is still contains a great deal of energy.
Lactate has three carbons and retains even more of the energy
of glucose than alcohol. The net ATP production is only 2
ATPs (in glycolysis) compared to 38 ATPs produced by
aerobic respiration
Explain that, during anaerobic respiration, pyruvate can be
converted into lactate (in humans) or ethanol and carbon
dioxide (in yeast) in the cytoplasm, with no further yield of
ATP.
If oxygen is available, the pyruvate is absorbed by the mitochondrion. Inside the
mitochondrion, the pyruvate is broken down into carbon dioxide and water. Aerobic
cell respiration has a much higher yield of ATP per gram of glucose than anaerobic cell
respiration.
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.
The process of glycolysis, taking place in the
cytoplasm, does not require oxygen and includes
phosphorylation, lysis, oxidation and ATP formation
(by substrate level phosphorylation: a process when a
phosphate group is transferred to ADP from a
phosphorylated intermediate).
The four stages in glycolysis
1. Two phosphate groups are added to a molecule of glucose to form hexose
biphosphate. Adding a phosphate group is called phosphorylation. Two molecules of
ATP provide the phosphate groups. The energy level of the hexose is raised by
phosphorylation and this makes the subsequent reactions possible.
2. The hexose biphosphate is split to form two
molecules of triose phosphate. Splitting molecules is
called lysis.
3. Two atoms of hydrogen are removed from each triose phosphate molecule. This is an
oxidation. The energy released by this oxidation is used to link on another phosphate
group, producing a 3-carbon compound carrying two phosphate groups. NAD+ is the
hydrogen carrier that accepts the hydrogen atoms.
4. Pyruvate is formed by removing the two phosphate
groups and by passing them to ADP. This results in ATP
formation.
In the cytoplasm, one hexose sugar is converted into
two three-carbon atom compounds (pyruvate) with a
net gain of two ATPs and two NADH + H+.
Outline
the
process
of
glycolysis
including
phosphorylation, lysis, oxidation and ATP formation.
The pyruvate, the end product of glycolysis, contains
most of the energy present in the original glucose. When
oxygen is present, pyruvate molecules move into the
mitochondrion, where all subsequent reactions occur.
The importance of the mitochondria in the generation of
ATP for a cell is reflected in the numbers found in a
cell. Lengths range from 1-10 m and widths from
0.25-1 m. The folds increase the surface area for ATP
generation.
Outer membrane: a regular membrane, separating the
mitochondrion from the cytoplasm. It is permeable to
H+ ions.
Inner membrane: is folded into cristae to provide
maximum space for the electron carriers and ATP
synthetase (ATP-ase). It is impermeable to H+ ions. The
ATP synthetase molecules can be seen on the cristae.
Intermembrane space: has a higher concentration of
H+ ions (protons) because of the electron transport
chain. Its pH is lower (more acidic).
Matrix: contains the enzymes which enable the link
reaction and the Krebs cycle to proceed.
Draw and label a diagram showing the structure of a
mitochondrion as seen in electron micrographs.
Explain the relationship between the structure of the
mitochondrion and its function. Cristae form a large
surface area for the electron transport chain, the small
space between inner and outer membranes for
accumulation of protons and the fluid matrix contains
enzymes of the Krebs cycle.
Pyruvate
from
mitochondrion.
glycolysis
Enzymes
is
in
absorbed
the
matrix
by
the
of
the
mitochondrion remove hydrogen and carbon dioxide
from the pyruvate.
Removal of hydrogen is oxidation. Removal of carbon
dioxide is decarboxylation. The whole conversion is
therefore oxidative decarboxylation.
In aerobic respiration (in mitochondria in eukaryotes)
each pyruvate is decarboxylated (CO2 removed).
The remaining two-carbon molecule (acetyl group)
reacts with reduced coenzyme A, and at the same time
one NADH + H+ is formed. This is known as the link
reaction:
CH3COCOOH + CoA-S-H + NAD+  CO2 + NADH +
H+ + CH3CO-S-CoA
In the Krebs cycle, each acetyl group (CH3CO) formed
in the link reaction yields two CO2. The names of the
intermediate compounds in the cycle are not required
(C2 + C4 = C6  C5  C4).
The cycle must run twice for each glucose. One turn of
Krebs cycle yields: 2 CO2, 3 NADH + 3 H+, 1 FADH2
and 1 ATP.
Summary of the Krebs cycle
1.
Carbon dioxide is removed in two of the reactions.
These reactions are decarboxylations. The carbon
dioxide is a metabolic waste product.
2.
Hydrogen is removed in four of the reactions. These
reactions are oxidations. In three of the oxidations the
hydrogen is accepted by NAD+. In other oxidation
FAD accepts it.
3.
ATP is produced directly in one of the reactions.
Oxidative phosphorylation
the chemiosmotic theory (Mitchell, 1961). The synthesis of
ATP is coupled to electron transport and the movement of
protons (H+ ions). As hydrogens pass along the system,
protons become separated from electrons. The electrons are
passed along a chain of acceptors. Entering electrons have
relatively high energy content.
As they pass through the system, the electrons lose much of
their energy. Some of the energy lost by electrons in their
passage is used up to pump the protons across the inner
mitochondrial
membrane
from
the
matrix
to
the
intermembrane space. The proton pumps produce a much
greater concentration of protons in the intermembrane space
than in the matrix.
This produces both an electrical gradient and a chemical
gradient, which provides the energy for ATP synthesis.
The ET carriers strategically arranged over the inner
membrane of the mitochondrion.
They oxidise NADH + H+ and FADH2, energy from this
process
forces
protons
to
move,
against
the
concentration gradient, from the mitochondrial matrix to
the space between the membranes (using proton
pumps).
Eventually the H+ ions flow back into the matrix
through special gates (protein channels) in the ATP
synthetase molecules in the membrane.
As the ions flow down the gradient, energy is released
and ATP is made. The final electron acceptor is
molecular oxygen. As the low-energy electrons are
passed to oxygen, they simultaneously reunite with
protons from the surrounding medium.
A chemical reaction between the rejoined hydrogen
atoms and oxygen produces water. If no oxygen is
present to act as a final electron acceptor, the system
stops working all the way back to the hydrogen-carrying
coenzymes.
Explain oxidative phosphorylation
in terms of
chemiosmosis.
Explain aerobic respiration including
the link
reaction, the Krebs cycle, the role of NADH + H+, the
electron transport chain and the role of oxygen.
3.8 & 8.2 Photosynthesis
Photosynthesis is the process used by plants to produce
all their organic substances (food) using only light
energy and simple inorganic substances.
Photosynthesis involves an energy conversion. Light
energy (sunlight) is converted into chemical energy.
Sunlight is called white light, but it is actually made up
of a wide range of wavelengths, including red, green
and blue. Some chemical substances called pigments
can absorb light.
The
main
pigment
photosynthesis
is
used
to
chlorophyll.
absorb
The
light
in
structure
of
chlorophyll allows it to absorb some colors or
wavelengths of light better than others.
Red and blue light are absorbed. The green light that
chlorophyll cannot absorb is reflected. This makes
chlorophyll and therefore chloroplasts and plant leaves
look green.
Some of the energy absorbed by chlorophyll is used to produce ATP.
Some of the energy absorbed by chlorophyll is used to split water molecules. This is
called photolysis of water. Photolysis of water results in the formation of oxygen and
hydrogen. The oxygen is released as a waste product.
Carbon dioxide is absorbed to make a wide range of
organic substances. The conversion of carbon in a gas to
carbon in solid compounds is called carbon fixation.
Carbon fixation involves the use of hydrogen from
photolysis of water and energy from ATP.
Photosynthesis involves the conversion of light energy into chemical energy.
Light from the Sun is composed of a range of wavelengths (colors).
Chlorophyll is the main photosynthetic pigment.
Outline the differences in absorption of red, blue and
green light by chlorophyll. Pigments absorb certain
colors of light (blue, red). The remaining colors
(green) of light are reflected.
Light energy is used to split water molecules
(photolysis) to form oxygen and hydrogen, and to
produce ATP.
ATP and hydrogen, (derived from the photolysis of
water) are used to fix carbon dioxide to make organic
molecules.
1. Since photosynthesis utilizes carbon dioxide, it is
theoretically possible to place a plant in an enclosed
space, measure the available carbon dioxide before and
after the experiment (by using radioactive CO2). This
will tell you how much carbon dioxide was used for
photosynthesis.
It is also possible to allow carbon dioxide to interact
with water, producing bicarbonate and hydrogen. Hence
the acidity of the resulting solution will indicate the
amount of carbon dioxide present (measured with a pH
meter).
2. Photosynthesis produces oxygen and glucose. It is
possible to measure how much oxygen a plant produces
over time (counting bubbles, volume produced).
3. It is also possible to measure how much heavier a plant is
after photosynthesis. We need to make sure that we measure
the change in organic matter and not, for example, the
change in water content. Therefore, we need to determine
the biomass by completely dehydrating (drying) the plant
before weighing it.
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.
The name chloroplast means green shape in Greek. They
are between 4-10 m in length and 2-3 m in width.
Draw and label a diagram showing the structure of a
chloroplast (1-10 m) as seen in electron micrographs
(include the following structures: outer membrane, inner
membrane, stroma, fat droplet, circular DNA, granum,
thylakoid membrane, ribosomes, starch grain.)
Photosynthesis consists of light-dependent and lightindependent reactions.
Chlorophyll a molecules in PS II absorb light at 680
nm, electrons are excited and removed by a chain of
electron carriers. This process is coupled (aided by Chl
a+ molecules) with the photolysis of water so that
oxygen, protons and electrons are released. The latter
reduce Chl a+ back to Chl a.
Light energy is transmitted to PSI reaction centre;
resulting excited state chlorophyll a. It gives up an
electron, becoming oxidised to chlorophyll a+.
Chlorophyll a+ gains electrons from H2O and is reduced
to chlorophyll a, H2O is oxidised to O2.
Light energy is transmitted to PSII reaction centre.
PSII pigment is oxidised to chlorophyll a+ as electron is
ejected.
PSII pigment gains electrons from electron carriers and
is reduced to Chl a.
Electrons pass from the reaction centre to an electron
carrier then to NADP+.
Electrons pass from Chl a to electron carriers and then
to PS I.
(1) 12 H2O + 12 NADP+ + 18 ADP + 18 Pi + light 
6 O2 + 12 NADPH + 12 H+ + 18 ATP
Photophosphorylation: chemiosmosis. Each step in the
Z-scheme is a coupled oxidation-reduction reaction.
One electron carrier becomes oxidised by giving up
electron(s); the coupled electron carrier becomes
reduced by gaining electron(s).
Non-cyclic photophosphorylation
energy from light is used to make ATP;
there is a one-way flow of electrons from water to
NADP+;
there is an energy yield of one ATP molecule and one
NADPH + H+;
proton gradient forms across the thylakoid membrane.
Cyclic photophosphorylation
only photosystem I is involved in this pathway;
the electrons that are passed through the transport chain
come from photosystem I and ultimately return
photosystem I;
no NADP+ is reduced.
Explain the light-dependent reactions: include the
photoactivation of photosystem II, photolysis of water,
electron
transport,
cyclic
and
non-cyclic
photophosphorylation, photoactivation of photosystem
I and reduction of NADP+.
Electron transport causes the pumping of protons to the
inside of the thylakoids. They accumulate and
eventually move out to the stroma through protein
channels in the ATP synthetase enzymes. This provides
energy for ATP synthesis.
Explain
photophosphorylation
chemiosmosis.
in
terms
of
In the stroma of the chloroplasts the ATP provides
energy, and the NADPH provides energy and reducing
power for biosynthesis using carbon dioxide.
The Calvin cycle in which the five-carbon molecule
RuBP acts as CO2 acceptor (catalysed by Rubisco) is
important, then forming two three-carbon molecules
with the initials GP and TP.
TP can be converted to glucose, sucrose, starch, fatty acids,
amino acids and other products. Some of the TP is used to
regenerate RuBP.
(2) 12 NADPH + 12 H+ + 18 ATP + 6 CO2  C6H12O6 + 12
NADP+ + 18 ADP + 18 Pi + 6H2O
Summary of reactions (1) and (2):
12 H2O + 6 CO2 + light  C6H12O6 + 6 O2 + 6 H2O
Explain the light-independent reactions: occur in the chloroplast stroma and include
the roles of ribulose biphosphate (RuBP) carboxylase (Rubisco), reduction of
glycerate 3-phosphate (GP) to triose phosphate (glyceraldehyde-3-phosphate) (TP),
NADPH + H+, ATP, regeneration of RuBP and subsequent synthesis of more
complex carbohydrates.
Explain the relationship between the structure of the chloroplast and its function.
Limit this to the large surface area of thylakoids for light absorption, the small space
inside thylakoids for accumulation of protons and the fluid stroma for the enzymes of
the Calvin cycle.
The absorption and action spectrum for chlorophyll
pigments showed that the wavelengths of light absorbed
by chlorophyll pigments are very similar to the
wavelengths that cause photosynthesis. The absorption
and action spectrum match quite well.
So the wavelengths optimally absorbed are the ones that
provide most energy for photosynthesis; both blue and
red light are used by the green plant.
Explain the relationship between the action spectrum
and the absorption spectrum of photosynthetic
pigments in green plants.
The factor the furthest away from its optimum value will limit the amount of
photosynthesis is called a limiting factor. If you improve this factor, the rate of
photosynthesis will increase until another factor becomes the limiting factor. If you plot
a graph of the rate of photosynthesis versus light intensity, the graph will go up until
light is no longer the limiting factor.
Then, the amount of photosynthesis will remain
constant.
Outline the effects of temperature, light intensity and
carbon
dioxide
concentration
on
the
rate
photosynthesis. The shape of the graphs is required.
of
The amount of light, the carbon dioxide supply, the
temperature, the water supply, and the availability of
minerals are the most important environmental factors
that directly affect the rate of photosynthesis in land
plants.
The complex mechanism of photosynthesis includes
photochemical (light-dependent) stage and an enzymatic
(light-independent)
reactions.
stage
that
involves
chemical
These stages can be distinguished by studying the rates
of photosynthesis at various degrees of light intensity
and at different temperatures. Over a range of moderate
temperatures and at low to medium light intensities, the
rate of photosynthesis increases as the light intensity
increases and is independent of temperature.
As the light intensity increases to higher levels,
however, the rate becomes increasingly dependent on
temperature and less dependent on intensity; light
saturation is achieved at a specific light intensity, and
the rate then is dependent only on temperature if all
other factors are constant.
At high light intensities, some of the chemical reactions
of the light-independent stage become rate limiting. At
light saturation, rate increases with temperature until a
point is reached beyond which no further rate increase
can occur.
Included among the rate-limiting steps of light-independent
stage of photosynthesis are the chemical reactions by which
organic compounds are formed using carbon dioxide as a
carbon source.
Explain the concept of limiting factors in photosynthesis,
with reference to light intensity, temperature and
concentration of carbon dioxide.