Prepared by Ashok.V, BiologyTeacher, CHSE.
Introduction
„ Respiration is the break down of glucose with the
use of oxygen and release of energy.
Respiration
Internal
respiration
(Cellular
respiration)
External
respiration
(Breathing)
Aerobic
respiration
Anaerobic
respiration
Internal or Tissue Respiration
(Cellular respiration)
y Internal or tissue respiration is a type of
metabolic pathway in which glucose is
oxidised to produce energy (ATP) needed
for muscle contraction leading to movement,
cell division leading to growth, transmission
of nerve impulses, synthesis of biomolecules, active transport across the cell
surface membrane, bioluminescence in fire
flies and in plants translocation through the
phloem tubes and so on.
ATP- The energy source for the cell
‡ ATP is made up of a nitrogenous base
(Adenine), a ribose sugar and three
phosphate groups.
‡ The covalent bond linking the second and
third phosphate group is unstable, and is
easily broken by hydrolysis.
‡ When this bond is broken a phosphate
group is removed, and ATP becomes ADP
and Pi. This is an exergonic reaction
because energy is released.
‡ The break down of ATP releases energy to
do work.
‡ ATP is not an energy storage molecule, it
is used up within one minute after it is
produced.
y ATP can be resynthesised from ADP and
inorganic phosphates by condensation
reactions.
y The energy for this reaction comes from
respiration.
y The ATP cycle shows the relationship
between ATP, ADP and respiration.
y The condensation of ADP to ATP is
an energy consuming process.
y Energy consuming process are called
endergonic.
AEROBIC RESPIRATION
y Aerobic respiration is the release of
large amount of energy by the break
down of glucose into carbon dioxide
and water in the presence of oxygen.
C6H12O6 + 6O2
6CO2 + 6H2O +
ATP
Glucose + Oxygen Carbon dioxide +
water + ATP
The process of Aerobic Respiration
y Aerobic respiration occurs in three stages1. Glycolysis (takes place without
oxygen) ± occurs in the cytoplasm.
2. The link reaction and Krebs cycle ( Tri
Carboxylic Acid (TCA) cycle)- Occurs in
the mitochondrial matrix.
3. Oxydative phosphorylation- occurs in
the inner mitochondrial membrane.
NAD and FAD are electron carriersNAD (Nicotinamide adenine dinucleotide) is
a coenzyme, which accepts a hydrogen atom
and 2 electrons from a metabolic pathway
and become reduced form(NADH2).
FAD(Flavin Adenine Dinucleotide) is another
coenzyme which accepts hydrogen and
become a reduced form (FADH2).
During oxidative phosphorylation, one
NADH2 ( reduced NAD) produces 3 ATP
molecules and FADH2 ( reduced FAD)
produces 2 ATP .
1. Glycolysis:
y Glycolysis occurs in the cytoplasm of the
cell and it consists of a series of
reactions, each step catalysed by a
specific enzyme, in which the 6C
glucose is broken down into two
molecules of 3 carbon pyruvic acid
(pyruvate) with net gain of 2 ATP
molecules and 2 NADH2.
y This pyruvate is taken into mitochondria
and enters into Krebs cycle.
y Initially glucose is phosphorylated by
ATP to form glucose-6-phosphate.
y Phosphorylation of glucose serves two
process:
a) It prevents glucose from leaving the cell,
as the membrane is impermeable to
sugar phosphates.
b) phosphorylation makes glucose more
reactive, so that it can be readily
converted into phosphorylated threecarbon compounds.
2.The link reaction and
Krebs(TCA) cycle:
y The subsequent steps occurs in the
mitochondria.
y In the link reaction, pyruvate diffuses into
the matrix of the mitochondria and is
metabolised there.
y First, the three carbon pyruvate is
decarboxylated by removal of carbon
dioxide and at the same time, oxidised by
removal of hydrogen.
y NAD is reduced to NADH + H.
y The product of this oxidative
decarboxylation reaction is an acetyl
group (Acetate) - a two carbon compound.
y This Acetate is then combined with a
coenzyme called coenzyme A, forming
Acetyl coenzyme A.
y The production of acetyl coenzyme A from
pyruvate is known as the link reaction
because it connects glycolysis to reactions
of the krebs cycle.
y In the Krebs cycle, acetyle coenzyme A
reacts with a four carbon compound
(oxaloacetate) to form a 6 carbon
compound (citrate). Coenzyme A release
out and reuses.
y Citrate is decarboxylated and oxidised to
form a 5 carbon compound
(હ ketoglutarate), carbon dioxide and
NADH2.
y The 5 carbon compound is then
decarboxylated, oxidised and
dephosphorylated through a series of
reactions, involving many intermediate
compounds and enzymes, to regenerate
oxaloacetate (4C).
y The net result of the 2 krebs cycle for
each glucose molecule is 4 CO2, 6
NADH2, 2 FADH2 and 2 ATP.
3. Oxidative phosphorylation:
y This is the phosphorylation of ADP to ATP
in the presence of oxygen.
y NADH2 from glycolysis, link reaction and
krebs cycle get oxidised to form
electrons (e-) and hydrogen atoms as
protons(H+).
y NAD and FAD molecules then return to
the krebs cycle to collect more hydrogen.
y The electrons pass down the chain of
proteins which undergo a series of redox
reactions, controlled by oxidoreductase or
hydrogenase enzymes.
y As the electrons are passed from one
component to another, hydrogen ions are
pumped from the matrix, into the intermembranal space, by using energy of these
electrons.
y This causes a high concentration of
hydrogen ions in the intermembranal space.
y Finally, the electrons combine with protons
and molecular oxygen to from water, the
final end-product of respiration.
y Cytochrome oxidase is an example of an
oxidoreductase enzyme which catalyses the
transfer of hydrogen to oxygen at the end of
the electron transport chain, to form water.
y ATP is generated from free energy released
when H+ ions move back into mitochondrial
matrix through stalked particles, which
contain the enzyme ATP synthase.
y These stalked particle are called
chemiosmotic channels and the movement
of hydrogen ions into the matrix is called
chemiosmosis.
y Oxygen is used to absorb electrons from
the electron transport chain and combines
these electrons with H+ ions to form water.
This ensures that the electron transport
chain continues to function and NAD can
be regenerated for aerobic respiration to
continue.
ANAEROBIC RESPIRATION
y In human beings anaerobic respiration
takes place in muscles.
y It takes place in oxygen deficient
conditions. The ATP generated only from
glycolysis, which yields 2 ATP molecules
from one glucose molecule.
y The electron transport chain does not
function, as there is no oxygen to
accept electrons at the end of the
chain.
y Electrons cannot leave the respiratory
chain, and so NADH cannot unload any
hydrogen to form NAD.
y Since NAD is not regenerated, krebs
cycle and link reaction do not operate
without sufficient NAD.
y During anaerobic respiration, pyruvate
is converted to lactate. This is a
reduction reaction, so NADH is used
and NAD regenerated.
y The lactate produced during this reaction
change into lactic acid and lowers the pH
in the muscles causing muscle cramps,
due to denaturation of respiratory enzymes
and muscle proteins. The denaturation of
respiratory enzymes leads to lack of ATP.
y When oxygen enters into the muscles, the
lactic acid get oxidised to pyruvate.
y The lactic acid can be directly oxidised to
carbon dioxide and water; the remainder
can be converted into glucose.
RESPIRATORY SUBSTRATES
and RESPIRATORY QUOTIENT
LEARNING OBJECTIVES
Understand what is meant by the term respiratory quotient (RQ).
RESPIRATORY SUBSTRATES
y Different respiratory substrates are used at different
times in the cells of the body. Lipids are an excellent
source of energy and they can be hydrolysed by lipases to
give fatty acids and glycerol.
y The glycerol is phosphorylated and enters the glycolytic
pathway as GALP.
y The fatty acids pass through a series of reactions which
remove 2-carbon sections of acetyl CoA. Hydrogen atoms
are removed in the process and are transferred along the
electron transport chain to form ATP even before the acetyl
CoA enters the Krebs cycle.
RESPIRATORY SUBSTRATES
y A single fatty acid can produce a large amount of
ATP; for example, a single molecule of stearic
acid (an 18-carbon chain) produces about 180
molecules of ATP. Thus, the complete oxidation
of fat produces a large amount of useable energy.
y Protein is not usually used as a respiratory
substrate. It is broken down only if supplies of
both carbohydrates and lipids are very low.
y The amino acids which make up the peptide
chains must be deaminated (the amino groups
must be removed) before the rest of the molecule
can be used for cellular respiration.
RESPIRATORY QUOTIENTS
y The amounts of oxygen used and carbon dioxide
produced during cellular respiration change
depending on the level of activity of the
organism, the type of food being respired and
other factors.
y Respiratory quotient (RQ) is a measure of the
amounts of carbon dioxide produced and oxygen
used by an organism in a given time period.
y The RQ helps us to understand what types of foods are
being oxidised in the body of an organism at a particular
time.
y In theory, carbohydrates give an RQ of 1, fats give an RQ
of 0.7 and protein gives an RQ of 0.9.
y Under normal conditions, protein is not often used to provide
energy. so an RQ of around 1 suggests that a lot of
carbohydrate is being used in cellular respiration.
y An RQ of less than 1 indicates that a combination of
carbohydrate and lipid is being respired.
y If the RQ of an organism is greater than 1, then anaerobic
respiration may be occurring, with relatively little oxygen
being used compared with the carbon dioxide produced.
y You will often observe very low RQ values in photosynthetic
organisms because much of the carbon dioxide produced is
used in making new sugars and so it cannot be measured.
THANK YOU
Core Practical 15:
Use an Artificial Hydrogen Carrier (Redox Indicator) to
Investigate Respiration in Yeast
Objectives:
‡ To learn how temperature affects the rate of respiration
‡ To know how to process and analyse data using appropriate mathematical skills
Background:
‡ Dehydrogenase enzyme involved in aerobic respiration removes hydrogen
‡ Redox indictors such as Methylene Blue is a better H acceptor
Oxidised Methylene Blue (Blue)
Reduced Methylene Blue (colourless)
‡ Increase in temperature increases rate of reaction
Core Practical 15:
Use an Artificial Hydrogen Carrier (Redox Indicator) to
Investigate Respiration in Yeast
Independent variable: Temperature
Dependent variable: Time taken for colour change / Rate of reaction
Confounding variables : Mass of yeast and glucose, volume of yeast extract, volume of
indicator used, same end point
Core Practical 15:
Use an Artificial Hydrogen Carrier (Redox Indicator) to
Investigate Respiration in Yeast
Risk Assessment
Hazard
Risk
Safety Precaution
In emergency
Risk
Level
Methylene
Blue
Irritant to skin
and eyes; may
cause staining
Wear eye protection
Wash from skin/eyes
immediately using cold water
Low
Biohazard
Allergies
Wash hands after use
Seek assistance
Low
Broken glass
Cuts from sharp
object
Take care when handling glass
objects; keep away from edge of
desk
Elevate cuts; apply pressure;
do not remove glass from
wound; seek medical
assistance
Low
Hot liquids
Scalding
Handle with care; use tongs to
remove boiling tubes from water
bath; wear eye protection
Run burn under cold water;
seek medical assistance
Low
Core Practical 15:
Use an Artificial Hydrogen Carrier (Redox Indicator) to
Investigate Respiration in Yeast
Core Practical 15:
Results
Core Practical 15:
Results
Core Practical 15:
Results
Temperature
of water
bath / oC
Time taken for yeast to turn methylene blue colourless / s
T1
T2
T3
Mean
20
641
651
633
642
30
443
441
412
432
40
291
233
306
277
50
136
119
139
131
60
198
208
187
198
Core Practical 15: Result Analysis
1. Create a table to show the rate of reaction for each temperature.
Temperature
of water
bath / oC
Time taken for yeast to turn methylene blue colourless / s
T1
T2
T3
Mean
Mean rate of
reaction / s-1
20
641
651
633
642
0.0016
30
443
441
412
432
0.0023
40
291
233
306
277
0.0036
50
136
119
139
131
0.0076
60
198
208
187
198
0.0051
Core Practical 15: Result Analysis
2. Draw a suitable graph to show how the temperature affected the rate of reaction
0.009
Time taken for yeast to turn Mean rate
methylene blue colourless / s
of
reaction /
T1
T2
T3
Mean
s-1
20
641
651
633
642
0.0016
30
443
441
412
432
0.0023
40
291
233
306
277
0.0036
50
136
119
139
131
0.0076
60
198
208
187
198
0.0051
0.008
0.007
Mean rate of reaction / s-1
Temperature
of water bath
/ oC
0.006
0.005
0.004
0.003
0.002
0.001
0
20
30
40
50
Temperature of water bath / oC
60
70
Core Practical 15: Questions
1. Suggest one source of random error and one possible systematic error in this investigation.
‡ From unpredictable changes during an experiment which
results in one measurement to differ slightly from the next.
‡ Affects Precision
‡ Unavoidable, but cluster around the true value
‡ Always affects measurements the same amount or by the
same proportion, provided that a reading is taken the same
way each time.
‡ Affects Accuracy.
‡ Can be reduced.
Core Practical 15: Questions
1. Suggest one source of random error and one possible systematic error in this investigation.
‡ From unpredictable changes during an experiment which
results in one measurement to differ slightly from the next.
‡ Affects Precision
‡ Unavoidable, but cluster around the true value
‫ٳ‬
‫ٳ‬
‫ٳ‬
Uneven concentration of yeast cells in suspension
Inaccurately measured volumes of solutions
Time taken to start timer once the solutions were
mixed might not be the same determining the endpoint of the reaction.
‡ Always affects measurements the same amount or by the
same proportion, provided that a reading is taken the same
way each time.
‡ Affects Accuracy.
‡ Can be reduced.
‫ٳ‬
‫ٳ‬
‫ٳ‬
Temperature due to inaccurate thermometer
Water bath temperature due to thermostat
Volume of solutions due to mar kings on pipette
Core Practical 15: Questions
Core Practical 15: Questions
5. Explain why the temperature affects the rate of respiration.
Respiration is catalysed by enzymes , which each have an optimum temperature. At lower temperatures,
molecules have less kinetic energy and therefore collide less frequently. At higher temperatures, the enzymes
may become denatured and not work as well, if at all.
Core Practical 16:
Use a simple respirometer to determine the rate of respiration and RQ of a
suitable material (such as germinating seeds or small invertebrates)
Objectives:
To understand how to use a respirometer
To be able to interpret and evaluate respirometer data
Independent variable: Organism
Dependent variable: Volume of O2 consumed / Rate of O2 consumption
Confounding variables : Mass of organism, Temperature
Core Practical 16:
Use a simple respirometer to determine the rate of respiration and RQ of a
suitable material (such as germinating seeds or small invertebrates)
Core Practical 16:
Use a simple respirometer to determine the rate of respiration and RQ of a
suitable material (such as germinating seeds or small invertebrates)
Risk Assessment
Hazard
Risk
Animal and
Potential allergic
plant samples reaction
Safety Precaution
Risk
Level
Wear gloves when handling Wash hands after the Low
practical
Glassware
Cuts from sharp objects Take care when handling glass objects Keep
away from edge of desk
Low
Soda lime or
potassium
hydroxide
Corrosive, potential
allergic reaction and
could irritate skin
Medium
Avoid skin contact. Wear eye protection and
gloves. Be careful when handling (use spatula).
Do not inhale. Wash hands after
Core Practical 16: Results
Distance moved by coloured liquid in 1 min /mm
Organism
Mass / g
T1
T2
T3
T4
T5
Mean
Woodlice
1
5
4
6
4
5
4.8
Peas
5
7
5
9
7
7
7
Core Practical 16: Analysis of Results
1. If your respirometer does not have volumes marked onto it, convert the distance moved by the liquid into the
volume of oxygen used. Use the formula for the volume of a cylinder: ǀŽůƵŵĞсʋƌ2h where r is the radius of the hole
in the glass tube and h is the distance moved.
Diameter of capillary tube = 1 mm
T1
T2
T3
T4
T5
Mean
Mean Volume
of O2
consumed/
mm3
Distance moved by coloured liquid in 1 min /mm
Organism
Mass / g
Woodlice
1
5
4
6
4
5
4.8
3.77
Peas
5
7
5
9
7
7
7
5.5
Core Practical 16: Analysis of Results
2. Calculate the mean rate of oxygen uptake during the 5 minutes, then divide by the mass of organisms used to find the
mean rate per gram of organism (mm3 രŵŝŶоϭ രŐоϭ ).
Diameter of capillary tube = 1 mm
Organism
Mass / g
Distance moved by coloured liquid in 1 min
/mm
T1
T2
T3
T4
T5
Mean
Mean Volume
of O2
consumed/
mm3
Mean rate of O2 consumption
/ mm3 min-1 g-1
Woodlice
1
5
4
6
4
5
4.8
3.77
3.77
Peas
5
7
5
9
7
7
7
5.5
1.10
Core Practical 16: Analysis of Results
3. Collect mean results for the rate of oxygen uptake per gram from other groups in the class. You could collect the
results in a spreadsheet. Calculate the overall mean for each organism. If there are sufficient data, calculate the standard
deviation.
Organism
Mass / g
Distance moved by coloured liquid in 1 min
/mm
T1
T2
T3
T4
T5
Mean
Mean Volume
of O2
consumed/
mm3
Mean rate of O2 consumption
/ mm3 min-1 g-1
Woodlice
1
5
4
6
4
5
4.8
3.77
3.77
Peas
5
7
5
9
7
7
7
5.5
1.10
Woodlic Rate/ mm3
e
min-1 g-1
Where:
භ S is the standard deviation
භ ɇсƐƵŵŽĨ
භ X = each measurement
භ n = the total number of measurements.
5
4
6
4
5
Mean
Standard
3.927
3.142
4.712
3.142
3.927
3.770
0.657
Peas
7
5
9
7
7
Mean
Standard
Rate /
mm3 min-1
g-1
1.100
0.785
1.414
1.100
1.100
1.100
0.22
Core Practical 16: Analysis of Results
3. Collect mean results for the rate of oxygen uptake per gram from other groups in the class. You could collect the
results in a spreadsheet. Calculate the overall mean for each organism. If there are sufficient data, calculate the standard
deviation.
Organism
Mass / g
Distance moved by coloured liquid in 1 min
/mm
T1
T2
T3
T4
T5
Mean
Mean Volume
of O2
consumed/
mm3
Mean rate of O2 consumption
/ mm3 min-1 g-1
Woodlice
1
5
4
6
4
5
4.8
3.77
3.77
Peas
5
7
5
9
7
7
7
5.5
1.10
Mean rate of O2
Organism consumption / Minimum
mm3 min-1 g-1
Organism
Mean rate of O2
consumption / mm3
min-1 g-1
Standard
Deviation
Woodlice
3.77
0.66
Woodlice
3.77
3.14
4.71
Peas
1.10
0.22
Peas
1.10
0.79
1.41
Maximum
Core Practical 16: Analysis of Results
4. Plot a suitable graph of the class results. Use the range or standard deviation to indicate the precision of the data.
Organism
Mean rate of O2
consumption /
mm3 min-1 g-1
Minimum
Maximum
0.66
Woodlice
3.77
3.14
4.71
0.22
Peas
1.10
0.79
1.41
Mean rate of O2 consumption
/ mm3 min-1 g-1
Standard
Deviation
Woodlice
3.77
Peas
1.10
Organism
5
Mean rate of O2 consumption / mm3
min-1 g-1
Mean rate of O2 consumption / mm3
min-1 g-1
5
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
Woodlice
Peas
Organism
Woodlice
Peas
Organism
Core Practical 16: Analysis of Results
5. Comment on your results and the variability of the data.
5
Mean rate of O2 consumption / mm3
min-1 g-1
Respiration rate in woodlice is significantly higher than in
germinating peas as there is no overlap in the error bars.
Although both plants and animals respire to produce ATP for
cellular processes, animals move around unlike plants,
therefore, require additional ATP for muscle contraction.
Looking at variability of the data, the measurement taken for
woodlice is less reliable as the error bars are longer compared
with peas.
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
Woodlice
Peas
Organism
Core Practical 16: Questions
1. Consider your results and any differences in rates of oxygen uptake between different
types of organism. Animals usually have a higher respiration rate per gram than plants.
Explain this difference.
Both plants and animals respire to produce ATP for cellular processes but animals also
move around and require additional ATP for muscle contraction. Respiration rates are
therefore usually higher in animals.
2. Suggest what factors may have caused any variability seen in class results.
Some variables may not have been effectively controlled, such as temperature, amount of
movement, age or stage of development of organisms.
The low resolution of the manometer scale will also cause some uncertainty in readings.
3. How could this variability be reduced and the precision of the results improved?
Temperature could be controlled using a water bath.
Organisms could be matched between groups for size or stage of development.
Core Practical 16: Questions
4. It would have been better to have used a control respirometer alongside the experimental set-up. In the
control the equipment is the same but the organisms are replaced by non-living material such as glass beads.
(a) Explain what may cause the liquid in the control tube to move towards and away from the respirometer.
A decrease in temperature or an increase in atmospheric pressure would cause movement towards the
respirometer. An increase in temperature or a decrease in atmospheric pressure would cause movement in
the opposite direction.
(b) Explain how you would use the control results to correct your experimental data.
Movement towards the control respirometer would be subtracted from results; movement away from the
control respirometer would be added to results.
5. (a) What is the importance of using soda lime in the respirometer?
Soda lime is used to absorb any carbon dioxide produced by the respiring organisms.
(b) How does this affect the volume of gas in the apparatus?
The gas volume will reduce as oxygen is removed for respiration.
6. How does this influence the movement of the liquid in the capillary tube?
A reduction in gas volume will reduce the pressure inside the tube. As it becomes lower than atmospheric
pressure, the fluid bubble will move towards the respirometer chamber.
Core Practical 16: Different Respirometer
Simple Respirometer
U ʹ Tube Respirometer (DixonʹBarcroft Respirometer)
Advantages- very simple to set up; minimal number
of connections makes a good seal easier to obtain.
Disadvantages - does not allow you to reset; it needs
a control tube used alongside it; no scale so
measurements likely to be less accurate.
Advantages- does not need to have an additional control
as the second tube balances out the effects of changes in
temperature or atmospheric pressure; the syringe allows
you to move the liquid in the U to reset the apparatus.
Disadvantages- tendency for the connections to leak in
elderly school/college models (making the equipment
useless); expensive
Muscles and Movement
Prepared by Ashok.V, BiologyTeacher, CHSE.
Introduction
y Muscles are bundles of elastic fibres
which can be tightened and
loosened to produce movements.
y Muscles are composed of many
elongated cells called muscle fibres
which are all able to contract and
relax.
y The muscles of the vertebrates are of three
types:
1.Skeletal Muscle (also called striated/
striped / voluntary muscle): Muscle
which is attached to bone. It is concerned
with locomotion, contracts very fast and
fatigues very fast.
2.Smooth muscle (also called unstriated/
unstriped / plain / involuntary): Muscles
which is found in the walls of tubular
organs of the body and is concerned with
movement of materials through them.
y It contracts slowly and fatigues slowly.
3.Cardiac Muscle:
y Muscle found only in the wall of the heart.
y These are also striated and they are
involuntary.
y It can contract by its own without nerve
stimulation (myogenic).
y It contracts spontaneously and without
fatigue.
Structure of muscle fibre
y Muscles are made from muscle fibres
(cells) arranged into bundles.
y The plasma membrane of a muscle cell is
called the sarcolemma, the cytoplasm
present called sarcoplasm and the
membrane reticulum is called the
sarcoplasmic reticulum.
y Each fibre is made from bundles of
myofibrils, which are extremely long,
cylindrical fibrils.
Actin and
myosin
(myo
Filaments)
sarcomere
Myo
Muscle
fibril
fibre
Muscle
y Each myofibrils are made up of functional
units called sarcomeres.
y Sarcomere is made up of overlapping
myofilaments called actin and myosin.
y The myosin is often called thick filament
where as actin is a thin filament.
y Under the microscope the muscle fibres
are having striped appearance. Where
actin filaments occur on their own, there is
a light band on the sarcomere (I band).
A sarcomere is a
repeating unit of
the muscle
myofibrils
defined by the
distance between
two Z lines
‰The filaments of actin
and myosin overlap to
give a distinct banding
pattern when seen with
an electron microscope.
‰This model show the
arrangement of the actin
and myosin filaments in
a myofibril
a) Actin only( I band)
b) Myosin only (H band)
c) Myosin attachment region adds stability (M line)
d) Actin and myosin overlap in cross sections (A band)
y Where both actin and myosin filaments
occur, there is a dark band ( A band).
y Where only myosin filaments occur, there
is an intermediate coloured band (H
band).
y Traversing the middle of each I band is a
dark line called the Z line.
y From the Z line actin filaments extend in
both directions.
y The section of a myofibril between two Z
lines is called a sarcomere.
y The H band itself may be bisected by a
dark line, is called the M line.
y The M line joins adjacent myosin filaments
together at a point halfway along their
length.
Structure of Actin filaments:
y They are thin filaments.
y Each actin filament is made up of two
helical strands of globular actin molecules,
which twist round each other.
y Actin consist of myosin binding sites.
y It consist of other two types of protein called
Tropomyosin and troponin.
y The tropomyosin molecules are filamentous
and lie in the grooves of twisted actin
strands.
y Tropomyosin blocks the myosin binding
sites on actin filaments.
y Troponin is a globular protein with three
sub units.
y Troponin combines with calcium ions and
causes tropomyosin to shift. This exposes
the myosin binding sites on the actin.
Structure of myosin filaments:
y Each myosin molecule has rod like tail
and two globular head region at one end.
y The chemical reactions for muscle
contraction occur in the myosin head.
The calcium ions released from the
sarcoplasmic reticulum bind to troponin
and expose the myosin binding sites on
actin filaments and the myosin filaments
bind to actin during contraction.
Muscle Contraction
y This is explained by sliding filament theory.
y Muscle contraction occurs by the pulling of
thin actin filaments towards each other by the
myosin filaments.
y On arrival of a nerve impulse at a
neuromuscular junction, calcium ions are
released by the sarcoplasmic reticulum.
y As Ca++ diffuse through the sarcoplasm,
they initiate the movement of protein
filaments in the following steps.
y First, the Ca++ attach to the troponin
molecule causing it to move.
y This results in the tropomyosin on the actin
filament to change its position exposing the
myosin binding sites on the actin filament.
y The bulbous head of the myosin binds to
actin forming an acto-myosin bridge.
y As ADP and Pi are released from the
myosin head, it changes shape.
y Now the head bends forward moving the
actin filament along the myosin filament,
shortening the sarcomere. This is called
WKHµSRZHUVWURNH¶
y A free ATP binds to the myosin head
causing the myosin head to detach from
the actin filament.
y The enzyme ATPase binds to the myosin
head hydrolyses ATP forming ADP + Pi
activated by Ca++
y This provides the energy for the myosisn
head to return to its original position.This
enables the cycle to start.
Muscle Twitch
y A muscle twitch is the response to a
single action potential (nerve impulse)
from a motor neuron.
y If a single nerve stimulus of sufficient
strength is applied to the muscle, the
muscle responds by giving a quick
contraction called twitch.
y If a train of stimuli are sent to the muscle
the muscles goes into a sustained
contraction called a tetanus.
y There are two types of fibre in human
skeletal muscle, distinguished by how
quickly they contract.
y Muscle fibres that are adapted to undergo
very frequent twitches are called as fast
twitch fibres, where as fibres which are
adapted to undergo less frequent
responses are called as slow twitch
fibres.
y Sprinters need lots of fast twitch muscle.
Therefore, the muscle type of cheetah like
organisms will be predominantly fast
twitch, where as the muscle of a camel or
an elephant will be predominantly slow
twitch.
y Muscle type of humans is predominantly
one or the other due to inherited alleles.
Differences between Slow twitch and Fast twitch
Muscle fibres:
Slow Twitch Fibres
1. Referred to as red muscle
as it contains lots of dark red
pigment called myoglobin Ȃ
which stores oxygen in
muscles.
Fast Twitch fibres
1. Referred to as white
muscle and is light pink in
colour. It has very little
myoglobin and is adapted
for anaerobic respiration.
2. Have many mitochondria,
which facilitate aerobic
respiration.
2. Have few mitochondria as
muscles undergo anaerobic
respiration.
3. Little sarcoplasmic
reticulum, which store and
release calcium ions for
muscle contraction.
3. Have lots of sarcoplasmic
reticulum, which release
plenty of calcium ions for
frequent muscle
contraction.
Slow Twitch Fibres
Fast Twitch fibres
4. Have a low glycogen
4. Have lots of glycogen,
content, as glucose is
which is converted into
supplied continuously by
glucose and used instantly for
numerous capillaries.
respiration.
5. Have numerous
5. Have few capillaries, which
capillaries, which provide a result in reduced oxygen
continuous supply of
supply and anaerobic
oxygen and glucose for
respiration occurs.
aerobic respiration.
6. Fatigue resistant as they 6. Fatigue quickly, due to
undergo aerobic
build up of lactic acid from
respiration.
anaerobic respiration
7. Creatine phosphate is
7. Have lots of creatine
found in low concentration phosphate which reacts with
as these fibres undergo
ADP to form ATP rapidly for
aerobic respiration
muscle contraction.
SUPPORT, MOVEMENT AND
LOCOMOTION
y Support- refers to the skeleton and its role
in holding up the structure and giving it
shape.
y Movement- refers to the change in the
position of one part of the body in relation
to another part.
y Locomotion- refers to the movement of
the entire body from one place to another.
JOINTS
y Joints are the areas where two or more
bones meet. Joints make the movements
possible.
Structure:
Cartilage- Soft, slippery, and slightly elastic
tissue covering the heads of the bones
allows friction free movement, act as a
shock absorber and prevent wearing of the
bones.
y Cartilage is a hard but flexible tissue. It is
made up of cells called chondrocytes within
an organic matrix which consists of varying
amounts of collagen fibrils.
y Cartilage is elastic (stretches and returns to
its original size) and able to withstand
compressive (squeezing ) forces.
y It is a very good shock absorber and is found
between bones such as the vertebrae in your
spine and in the joints.
y There are two main types of cartilage found in
the skeleton.
‰
‰
Hyaline cartilage is found at the ends
of bones (and in the nose, air
passageways and parts of the ear).
White fibrous cartilage has bundles
of densely packed collagen in the
matrix. It has great tensile strength but
is less flexible than the other forms of
cartilage. It forms the discs between
your vertebrae and is found between
the bones in the joints.
Synovial capsule- Encloses joints. Keeps
synovial fluid in place.
Synovial membrane- A tissue lining the joint
and seals it into capsule. Secretes the
synovial fluid.
Synovial fluid- Acts as a lubricant between
joints and also provides nutrients to the
bone surface.
Ligament- Elastic connective tissue which
connects bone to bone. Provide external
support and strengthen bone.
Tendon-Non elastic connective tissue,
connect muscles to bone, allows
movement.
JOINTS
movable
Partly
movable
Gliding
Pivot
Freely
movable
Hinge
Immovable
Eg: Skull
bones
Ball
and
Socket
Types of joints:
1.Ball and socket joints:
y Allows movement in three planes (360ƒ).
Eg: Shoulder and hip.
2. Hinge joint:
y Allows movement in one plane (180ƒ). Eg:
Elbow, knee, fingers and toes .
3. Pivot joint:
y In pivot joint a round process of bone that
fits into a bony socket that allows rotation.
y Seen between axis and atlas vertebra.
y Allows shaking of head from side to side.
4. Gliding (saddle) joint:
y Saddle shaped head of bone permit
movement in two different planes.
Eg: Wrist, ankle, between ribs and between
adjacent vertebra.
How muscle work to move the bones ?
y Muscles can pull but never push.
y Muscles can pull only when they contract.
y When muscles contract, they decrease in
size.
y When returning to their original length they
relax.
y Muscles work in pairs. As one contract the
other in the pair relaxes.
y One muscle produces the opposite
movement from the other muscle, there
fore, the pairs are called antagonistic
pairs.
y Muscles which cause a joint to extend are
called extensors, muscles which cause a
limb to retract, bend or fold are called
flexors.
y Flexion is the movement that reduces the
angle of a joint where as extension
increases the angle of a joint and moves
the bones away from each other.
THANK YOU
Prepared by Ashok.V, BiologyTeacher, CHSE.
Myogenic stimulation of heart:
y The stimulus for the contraction of
cardiac muscles is generated within
the heart itself without electrical
stimulation. So the stimulation is
called myogenic stimulation not
neurogenic. But the rate can be varied
by the nervous system or hormones
like adrenaline.
..
y The impulse for the cardiac cycle
originates at the sino atrial node
(SAN), which is otherwise called the
pacemaker.
y This wave of excitation (impulse)
travels across the cardiac muscle fibres
to the wall of the left atrium and the
atrioventricular node (AVN). This
causes atrial systole. Tricuspid and
bicuspid valves open.
..
..
y Impulses pass to the ventricles via the
AVN.
y From the AVN the impulse passed down to
the apex of the heart by the Bundle of His
which is present in the septum.
y Through the Purkinje fibres (sometimes
called Purkyne fibres) the impulse
spreads to the walls of ventricle and causes
ventricular systole. Tricuspid and bicuspid
valves are closed. Blood enters into
pulmonary artery and aorta.
..
y The non-conducting tissue between
SAN and AVN prevents electrical
impulse being transmitted to the
ventricular valves directly from SAN.
This ensures that ventricular systole is
accompanied only after atrial systole.
..
The Electrocardiogram
y The electrical activity of the heart can be
detected and displayed on an
Electrocardiogram(ECG), a graphic record
of the electrical activity during the cardiac
cycle.
y The machine used for measuring electrical
activity of the heart is Electrocardiograph.
y This is the most common test to check for
problems with the heart.
..
How is an ECG carried out ?
y In an ECG, electrodes are attached to the
SHUVRQ¶VFKHVWDQGOLPEVWRUHFRUGWKH
electrical current produced during the
cardiac cycle.
y Normally, a patient will have 12 electrodes
attached to their body to give 12 views of
the heart.
y An ECG is usually performed while the
patient is at rest, lying down and relaxed,
but it may be used in a stress test.
..
..
What does the ECG trace show us ?
P wave ± It is caused by atrial depolarization.
The p wave is not as elevated as the QRS
complex as the atria have thinner wall
compared to the ventricles.
When the wave of electrical charge reaches
the atrio ventricular node at the base of the
right atrium there is slight delay shown by
the time between the end of the p wave and
the start of the QRS complex.
..
..
..
PR interval- The time taken for impulses to be
conducted from the SAN across the atria to
ventricles, through the AVN.
QRS complex- This is caused by ventricular
depolarization which leads to ventricular
systole.
ST segment- This shows the short period of
time when no further electrical impulse can
be passed through the heart muscle.
T wave ± This is the period when the
ventricles are relaxing and return to their
resting state.
..
Using the ECG trace to measure heart
rate
y Each large square in ECG represents
0.2 seconds.
y Five large squares represent 1 second.
y You can work out the time for one
complete cardiac cycle by multiplying
the number of large squares between 2
QRS complexes by 0.2, and then
dividing this into 60.
..
..
..
SAQ : Work out the time for one cardiac
cycle and the heart rate for the patient
whose ECG trace is shown in the figure.
There are five squares between 2 QRS
complexes.
So, time for one cardiac cycle = 5 x 0.2
= 1 sec.
Heart rate = 60/1 = 60 beats/min.
..
Heart Problems
1. Bradycardia (slow heart beat)y A heart rate less than 60 beats per minute
is known as Bradycardia.
y It is common in fit athletes at rest.
Training increases the amount of blood
the heart can hold and pump out with
each beat.
y It can also be a symptom of heart
problems due to changes in the electrical
activity of the SAN.
..
..
2. Tachycardia (fast heartbeat)y Heart rate greater than 100 beats per minute.
y It is normally the result of anxiety, fear, fever
or exercise.
y It can also be a symptom of coronary heart
disease, heart failure, use of medicines or
drugs.
3. Arrhythmiasy Irregular heart beat caused by electrical
disturbances.
y This is because of the blockage of coronary
arteries.
..
Tachycardia and Arrhythmias
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
LUNG VOLUMES
Inspiratory reserve volume (IRV)y The maximum volume of air that can be forcibly
inhaled after a normal inhalation.
Expiratory reserve volume (ERV)y The maximum volume of air that can be forcibly
exhaled after a normal exhalation.
Tidal Volume (TV)y The volume of air breathed in and out of lungs per
breath, during normal breathing at rest.
y It is about 0.5 dm3 at rest, but varies with each
individual.
y It increases during
exercise.
..
Vital capacity (VC)y The maximum volume of air that can be forcibly
expired after a maximum intake of air.
y It varies between 3 dm3 to 6 dm3 depending on
size and fitness of the person.
VC = IRV + TV + ERV
Residual Volume (RV)y It is the volume of air that remains in the lungs even
after a forced expiration.
y It prevents the lungs from collapsing after
expiration.
Total Lung Capacity (TLC)- It is the sum of the vital
capacity and the residual volume.
..
Inspiratory Capacity (IC)y It is the volume of air that can be inspired from the
end of a normal expiration.
y It is the sum of tidal volume and Inspiratory
Reserve Volume (TV + IRV).
Ventilation rate/minute ventilation
y It is a measure of the volume of air breathed in a
minute.
Ventilation rate = Breathing rate x tidal volume.
Spirometer
y Spirometer is an instrument used for measuring
lung volumes.
..
Spirometer
y A spirometer is the standard equipment
used to measure the human lung
volumes.
y It consists of a chamber (of capacity
approximately 6 dm3) suspended freely
over water and counterbalanced so that
gas passed in or drawn out makes the
chamber rise or fall.
...
y You can make a permanent record of the
movements of the chamber either by
attaching a pen to it and allowing it to
write on a drum revolving slowly
(kymograph) or by attaching a motion
sensor which will convert movement into
electronic signals that are then
interpreted by your data logging software.
...
...
...
...
Preparation - setting up the
spirometer:
y Fill the canister with a chemical for absorbing
carbon dioxide, such as Carbosorb® or selfindicating soda-lime granules.
y Check for leaks by filling the spirometer with
air, closing all the taps and loading the lid of
the chamber with a 200 g mass.
y If the chamber does not move during five
minutes, there are no serious leaks.
...
y If using a kymograph, check that the pen
writes on the drum throughout a complete
revolution of the drum, and that the pen is
towards the top of the recording chart
paper with the chamber full of air/ oxygen.
y If using a motion sensor, make sure that
the sensor and data logging software are
appropriately connected and you know
how to collect the data.
...
...
Working of spirometery When the air tank is filled with oxygen and
the water tank is filled with water, you can
breathe the oxygen through the breathing
tubes.
y The lid of the air tank goes up and down as
you breathe in and out, making a trace on
the chart recorder.
y The speed of the chart recorder can be
set, so the relationship between distance
and time can be calculated on the trace.
...
...
y The volume readings can be calibrated by
making marks on the chart with the pen
before and after a known volume of oxygen
is added to the air tank.
y Breathing rate is calculated by counting the
number of breaths in a given time.
y The volume of air breathed in and out is
calculated from the vertical movements of
the trace.
Noseclipy The nose clip make sure that all breathing is
only through the air tubes of the spirometer.
...
Sodalimey The sodalime absorbs CO2 in the exhaled air.
y This ensures that the CO2 levels in the inhaled air
do not change during the experiment, making
uptake of oxygen easier to determine.
y The removal of CO2 is also necessary for safety
reasons.
safety guidance
y Wear eye protection when handling carbon
dioxide absorbing chemicals; they are
corrosive although safer to use than sodium
hydroxide.
...
y Ensure that the valve in the tubing
connected to the spirometer is positioned so
that air is always exhaled through the soda
lime/ Carbosorb to avoid inhalation of any
soda lime/ Carbosorb dust.
y Oxygen forms explosive compounds with
many oils and greases, so such lubricants
should not be put on the connectors or
regulators of oxygen cylinders or on the
tubing connectors to spirometers.
y Disinfect the mouthpiece after each person
has used it..
...
...
Lung volume trace
...
6000
Lung Volume (mL)
5000
4000
3000
2000
1000
Inspirator
y Reserve
Volume
Vital
Tidal
volume Capacit
Expirato
ry
Reserve
Volume
Residual
Volume
0
Time
y
Interpreting a spirometer reading
Calculate
y Tidal volume
y Breathing rate
y Vital capacity
y Minute volume
y Rate of Oxygen consumption
Interpreting a spirometer reading
y Tidal volume
between 0.5 and 0 .8 dm3
Interpreting a spirometer reading
y Breathing rate
18 to 21 breaths per minute
Interpreting a spirometer reading
y Vital capacity
2.55 dm3 averaged over the two breaths
Interpreting a spirometer reading
y Minute volume = ~10dm3 min-1
Interpreting a spirometer reading
‡ Rate of Oxygen consumption
approximately 1.0 dm3 min-1 for
the first 30 seconds
...
THANK YOU
...
Prepared by Ashok.V, BiologyTeacher, CHSE.
Introduction
y The maintenance of a steady internal
state in the body almost regardless of
changes in either the external or the internal
conditions is known as Homeostasis.
y Homeostasis involves a high level of
coordination and control.
y Changes in the body are detected by a
receptor (also called sensor). This sent
impulses to the central nervous system
through sensory neurone.
..
Negative feed back mechanisms
(Negative feed back loops)y %RG\¶VKRPHRVWDVLVDUHPDLQWDLQHGE\
Negative feed back mechanisms.
y In negative feed back, any change or
deviation from the normal range of
function is opposed.
y The change or deviation in the controlled
value initiates responses that bring the
function of the organ or structure back to
within normal range.
..
..
Positive feed back mechanismsy Positive feed back mechanisms do
not correct change from the set point.
y In fact, deviations from the set point
are made even larger.
y A small change in a factor causes a
further change in the same direction.
..
..
Eg: During birth stretching of the
cervix triggers release of the
hormone oxytocin. This hormone
causes muscles of the uterus wall to
contract, pushing the baby against
the enlarging cervix, stretching it
further.
As stretching increases, so more
oxytocin is released.
..
Nervous control of Heart rate
..
Nervous control of Heart rate
y Heart rate is under the control of the
Cardiovascular control centre located in
the medulla of the brain.
y Nervous control of the heart is by the
Autonomic (Involuntary) nervous system.
y The autonomic nervous system is divided
into two parts- The sympathetic nervous
system and Para sympathetic nervous
system.
..
..
y The sympathetic nervous system is usually
excitatory, that speed up the heart rate.
y Para sympathetic nervous system is inhibitory
which slows down the heart rate.
y These two systems operate through negative
feedback control involving two controlling
centers in the cardio vascular centre. These
are,
(1) cardioacceleratory centre (CAC) which is
concerned with increasing cardiac output.
(2) Cardioinhibitory centre (CIC) which
reduces cardiac output.
..
Stimulation or inhibition of the cardio vascular
centre involves blood pressure receptors
(baroreceptors) and chemoreceptors
situated in
‡Carotid arteries in the neck.
‡$RUWD
Baroreceptors are mechanoreceptors which
detect blood pressure changes.
Chemoreceptors will detect change in carbon
dioxide level and blood pH.
..
..
..
..
When CAC is stimulated, it will sent
impulses through sympathetic nerves to
the sinotrial node, atrioventricular node
and cardiac muscle fibres, where they
release noradrenaline. This stimulates
the sinoatrial node to reduce the delay at
the atrioventricular node and increases the
force of the muscular contractions. Thus
the cardiac output rises.
..
..
When CIC get stimulated, impulses are sent
through parasympathetic nerves (vagus)
pass to the sino atrial node and the
atrioventricular node, where they release
acetylcholine.
This suppresses the activity of the sinoatrial
node and increases the delay at the
atrioventricular node. Thus the cardiac
output falls.
..
Exercise and Cardiac Output:
y Exercise increases the stroke volume.
y Stroke volume is the volume of blood
pumped by the left ventricle in one
contraction.
y Heart rate is the number of heartbeats in
one minute.
y Cardiac output (CO) is the volume of blood
pumped by each side of the heart in one
minute.
CO = (heart rate [HR] x(stroke volume [SV])
..
The mechanism by which cardiac output increases during
exercise is shown below
..
The increased cardiac output ensures,
y Oxygen is supplied rapidly to muscles to
meet the excess demand during
exercise.
y Carbon dioxide and lactate is removed
rapidly from muscles ( otherwise pH
could become too low ± acidic and
denature respiratory enzymes)
y Blood circulates rapidly through the lungs
to ensure rapid oxygenation and removal
of carbon dioxide.
..
Hormonal effects on heart ratey Adrenaline, a hormone released from the
adrenal glands located above the kidneys
have a direct effect on the heart rate
similar to stimulation by sympathetic
nerve.
y Fear, excitement and shock cause the
release of adrenaline.
y Adrenaline has a direct effect on the
sinoatrial node, increasing the heart rate.
..
y Adrenaline also causes dilation of the
arterioles supplying blood to skeletal
muscles, and constriction of arterioles
going to the digestive system and
other non essential organs, this
maximizes the blood flow to the active
muscles.
..
Control of breathing rate:
(Ventilation rate)
y Breathing is controlled by respiratory
centres in the medulla oblongata, which
controls the rate and depth of breathing.
y The centre is referred to as the medullary
rhythmicity centre as it controls the
basic regular rhythm of breathing.
..
y The respiratory centre includes the
inspiratory and expiratory centres, which
are stimulated by the impulses from the
stretch receptors in bronchioles and
chemoreceptors in the aorta and carotid
artery.
y Inspiratory centre increases the rate and
depth of inspiration during exercise, when
carbon dioxide concentration increases or
pH decreases.
y The expiratory centre stimulates the rate
and depth of expiration.
..
Functioning of Respiratory centrey The inspiratory centre send impulses to
external intercostal muscles and
diaphragm to contract, through intercostal
nerves and phrenic nerves respectively.
y Inhalation takes place and air moves into
lungs.
y Lungs inflate and stretch receptors in the
bronchioles are stimulated and send
impulses to expiratory centre through
vagus nerve.
..
..
y Inhalation stops and exhalation begins.
y Internal intercostal muscles contracts and
diaphragm relaxes due to stimulation
from expiratory centre.
y Lungs deflate and stretch receptors in
bronchial tree are deactivated.
y Vagus nerve stimulates the inspiratory
centre to stop exhalation and begin
inhalation.
..
Changes in breathing rate during
exercise
y The chemoreceptors present on the
wall of carotid and aortic bodies sense
changes in CO2 and pH levels and, to
lesser extent, they are also sensitive to
changes in oxygen levels.
y When the chemoreceptors detect a rise in
CO2, or fall in oxygen, they send impulses
to the respiratory centre located in the
medulla oblongata.
..
..
y From the inspiratory centre impulses
travel via the intercostals nerves and
phrenic nerves to the external intercostals
muscles and diaphragm respectively.
y These muscles then contract, initiating
inhalation.
y As air enters the lungs, the lungs inflate
and the stretch receptors in the bronchioles
start firing and send information to the
expiratory centre of the brain which
inactivates the inspiratory centre.
..
y This causes the lungs to deflate and the
elastic recoil of the lungs leads to
expiration.
y Any increase in CO2 or lactate
concentration in the blood lowers its pH.
..
Thermoregulation
y In humans, the temperature is
maintained by a negative feedback
mechanism.
y This system involves receptors that
detect changes in the blood
temperature.
y These receptors are located in the
skin.
..
y If the skin is warm, then impulses are
sent to the hypothalamus initiating
the heat-loss responses and
inhibiting heat-gain responses.
y If the skin is cold, opposite happens.
y These changes help to keep the core
body temperature near its optimum.
..
..
..
..
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
yHormones are organic chemicals produced in endocrine
glands and released into the blood.
yHormones are usually either proteins or peptides (e. g.
insulin, antidiuretic hormone) or steroids (e.g. the sex
hormones oestrogen and testosterone).
yOnce a hormone enters the bloodstream, it is carried
around in the blood until it reaches its target organ or
organs. The cells of the target organs have specific
receptor molecules on the surface of their membranes that
bind to the hormone molecules.
yThis leads to a change in the membrane and produces a
response .
..
..
THE ENDOCRINE GLANDS
yEndocrine glands are found around the body,
often in association with other organ systems.
Several of the glands have more than one
function .
For example , the ovaries produce ova as well as
hormones ; the pancreas is both an exocrine
gland which produces digestive enzymes and an
endocrine gland which produces the hormones
insulin and glucagon.
The glands all have a rich blood supply, with
plenty of capillaries.
..
The exocrine glands that produce the secretions
release along small tubes or ducts.
The endocrine glands do not have ducts and they
release their hormones directly into the blood
stream. They include:
The thyroid glands: thyroid hormones
(Thyroxine) are involved in controlling growth and
metabolism.
The parathyroid glands: parathyroid
hormones are involved in the homeostatic
control of calcium metabolism.
..
The pancreas: the exocrine pancreas produces
digestive enzymes and the endocrine pancreas
produces insulin and glucagon which are in
volved in the homeostatic control of blood
sugar.
The adrenal glands: produce many hormones
including adrenaline, involved in the fight or
flight response, and aldosterone, involved in
the homeostatic control of the osmotic balance
of the body.
The ovaries : produce the female sex hormones
The testes: produce the male sex hormones.
..
HORMONE RELEASE SYSTEMS
The control of hormone release by the nervous system is
relatively simple. If the gland is stimulated, hormone is
released.
If it is not stimulated, no hormone is released. The level of
stimulation determines the level of response.
However, many hormones are released from endocrine
glands in response to another hormone or chemical in the
blood.
For example, the pituitary gland in the brain secretes
several hormones that directly stimulate other endocrine
glands. In addition, changes in chemicals in the blood (e.g.
glucose and salt) can stimulate the release of hormones,
which consequently act to regulate the levels of those
chemicals.
..
Some hormones are released in response to a
chemical stimulus such as another hormone or
glucose.
In this case, a negative feedback loop controls
their secretion. The presence of the appropriate
chemical in the blood stimulates the release of
the hormone. As a result of the rise in the
hormone levels. the amount of stimulating
chemical in the blood drops. Therefore, the
endocrine gland receives less stimulation and so
the hormone levels drop. This gives a very
sensitive level of control that can be adjusted
constantly to the needs of the body.
..
..
THE PITUITARY GLAND: CONTROLLING
HORMONE RELEASE AND HOMEOSTASIS
The pituitary gland in the brain produces and releases
secretions that affect most of the other endocrine glands
in the body The pituitary gland has an anterior lobe and a
posterior lobe. The pituitary itself is mainly under the
control of the hypothalamus.
y The hypothalamus contains neurosecretory cells .
One group of these cells (neurosecretory cells 1) produce
substances that stimulate or inhibit the release of hormones
from the anterior pituitary. They are known as either
releasing factors or release inhibiting factors , depending
on what they do . The other group of cells (neurosecretory
cells 2) produce secretions that are stored in the posterior
..
pituitary and are released
later as hormones.
..
..
y Many hormones can act as signal
molecules. Hormones are of three types,
1.Amines- Eg: Adrenaline and
thyroxine(these are lipid insoluble, so
cannot enter the cell directly).
2.Peptides(proteins)- Eg: Insulin,
glucogon, ADH (They are lipid insolube,
so cannot enter the cell directly).
3.Steroids-Eg: Oestrogen and
testosterone(They are lipid soluble and
can enter the cell).
..
Gene activation
Mode of action of Peptide hormone as
signal molecule: (Eg:ADH)
y The hormone first binds with the receptors
on the cell membrane.
y The receptor releases adenyl cyclase
enzyme into the cytoplasm.
y Adenyl cyclase converts ATP into cyclic
AMP(considered as second messenger).
y This activates transcription factor.
..
..
..
y The enzyme RNA polymerase binds to the
transcription factors and form a
Transcription initiation complex.
y This will switch on genes, transcription
starts and mRNA is synthesized.
..
Mode of action of steroid hormone as
signal molecule:
y Steroid hormones, unlike non-steroid
hormones, can enter directly into the cell
through phospholipid bilayer as they are
fat-soluble.
y Once inside the cell the steroid hormone
binds with a specific receptor found in the
cytoplasm of the target cell.
y The receptor bound steroid hormone act
as transcription factor.
..
..
y The transcription factor binds with the
promoter region.
y The enzyme RNA polymerase binds to the
transcription factors and form a
Transcription initiation complex.
y This will switch on genes, transcription
starts and mRNA is synthesized.
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
Osmoregulation is the maintenance of the
osmotic potential in the tissues of a living
organism within narrow limits by
controlling water and salt concentrations.
It is part of homeostasis and it is vital for life.
In mammals, the main organ involved in the
homeostatic control of the water balance of
the body is the kidney.
The liver is also involved in homeostasis, for
example in the breakdown of excess amino
acids and the removal of toxins.
..
THE LIVER, PROTEIN METABOLISM AND
HOMEOSTASIS
yThe liver plays an important role in the deamination
of excess amino acids in protein metabolism. They
remove the amino group and convert it first into
ammonia, which is very toxic, and then to urea which
is less toxic and can be excreted by the kidneys .
y The ammonia produced in the deamination of
proteins is converted in to urea by a series of enzymecontrolled reactions known as the ornithine cycle.
yThe remainder of the amino acid can then be used in
aerobic respiration, or be converted into lipids for
storage.
..
..
..
OSMOREGULATION IN MAMMALS
yOsmoregulation in mammals is largely brought about by the
kidneys.
yThe kidneys are a pair of organs capable of producing urine, which
can be hypertonic to (more concentrated than) the body fluids. This
makes it possible to conserve water;
yIn humans, as in other mammals , the kidneys are a pair of dark
reddish brown organs attached to the back of the abdominal cavity.
yThey are surrounded by a thick layer of fat, which helps to protect
them from mechanical damage. They control the water potential of
the blood plasma that passes through them. They remove urea and
substances that would affect the water balance. Blood plasma
from the body passes through the kidneys where the urea and
excess salts and water are removed to form urine. The urine is
stored in the bladder and released from the body at intervals.
..
..
..
Structure of nephron
..
THE STRUCTURE AND FUNCTIONS OF THE KIDNEY
y The mammalian kidney has two main roles in the body. One is excretion: the
removal of urea from the body. The other is osmoregulation.
..
STRUCTURAL & FUNCTIONAL UNITS OF
KIDNEY, THE NEPHRONS
y Each kidney is made up of microscopic
tubules called nephrons.
y There are two main types of nephron.
y Cortical nephrons are found mainly in the
renal cortex. They have a loop of Henle (a
U-shaped tubule ) that only just reaches into
the medulla.
y Juxtamedullary nephrons have long loops
of Henle that penetrate right through the
medulla. They are particularly efficient at
producing concentrated urine.
..
The kidney performs three main functions in its
osmoregulatory role. These are ultrafiltration, selective
reabsorption and tubular secretion .
1. ULTRAFILTRATION
Ultrafiltration in the kidney tubules occurs because of a
combination of very high blood pressure in the glomerular
capillaries and the structure of the Bowman 's capsule and
glomerulus. The glomerulus and the Bowman's capsule
together make up the Malpighian body.
High blood pressure develops in the capillaries of the
glomerulus, because the diameter of the blood vessel
coming into the glomerulus is greater than that of the blood
vessel leaving.
..
Any protein
molecule with a
relative MM of
69 000 or more
cannot pass
through the
basement
membrane, and so
cannot escape
from the
glomerular
capillaries. The
basement
membrane
acts as a molecular
filter.
y The high pressure squeezes the blood out
through the pores in the capillary wall; The
size of the pores means that almost all the
contents of the plasma can pass out of the
capillary except the blood cells and the largest
plasma proteins that cannot pass through the
pores.
y The cells of the Bowman's capsule next to the
capillaries act as an additional filter. The filtrate
that enters the capsule contains glucose, salt,
urea and many other substances in the same
concentrations as they are in the blood plasma.
..
2. SELECTIVE REABSORPTION
Ultrafiltration is passive and not selective. It
removes urea from the blood , but it also
removes a lot of water with glucose, salt and
other substances that are present in the
plasma are also needed by the body. After the
ultrafiltrate has entered the nephron, the main
function of the kidney tubule is to return
most of these substances back into the blood.
..
The kidney tubule consists of different parts:
1. THE PROXIMAL TUBULE
yThe proximal tubule reabsorbs over 80% of the
glomerular filtrate into the blood. The cells lining this
tubule are covered with microvilli; these greatly increase
the surface area through which substances can be
absorbed. The cells also have large numbers of
mitochondria for providing energy for active transport.
yActive transport in the proximal tubule results in all the
glucose, amino acids, vitamins and most hormones
being returned to the blood . About 85% of the sodium
chloride and water is reabsorbed as well.
..
THE SODIUM
CO-TRANSPORT
yThe sodium ions are actively transported, and the
chloride ions and water follow passively down
concentration gradients.
yOnce these substances are removed from the
tubule cells into the intracellular spaces, they then
pass by diffusion in to the extensive capillary
network that surrounds the tubules.
yThe blood is constantly moving through the
capillaries, maintaining a concentration gradient
for diffusion. By the time the filtrate reaches the
loop of Henle, it is isotonic with the tissue fluid
that surrounds the tubule.
..
2. THE LOOP OF HENLE
yEach loop of Henle is found in the medulla of the
kidney, in close contact with a network of capillaries.
Together they create a water potential gradient between
the filtrate and the medullary tissue fluid. This makes it
possible for water to be reabsorbed from the distal
tubule and collecting duct. It is this water potential
gradient that allows mammals to produce urine that is
more concentrated than their own blood.
yThe creation of the high concentration of sodium
and chloride ions in the tissue fluid of the medulla
happens as a result of the flow of fluid in opposite
directions in the adjacent limbs of the loop of Henle.
..
..
o Having the two limbs of the loop running side by
side like this, with the fluid flowing down in one
and up in the other, enables the maximum
concentration of solutes to be built up both inside
and outside the tube at the bottom of the loop.
This mechanism is called a counter-current
multiplier.
..
The changes that occur in the loop of Henle:
(a) In the descending limb:
The descending limb is freely permeable to water but
is not very permeable to sodium and chloride ions.
No active transport occurs here. The fluid entering this
limb is isotonic with the blood. As the fluid travels
down the limb deeper into the medulla, the external
concentration of sodium and chloride ions in the tissue
fluid of the medulla and the blood in the capillaries
becomes higher and higher. As a result, water moves
out of the descending limb into the tissue fluid by
osmosis down a concentration gradient. It then moves
into the blood in the capillaries, again down a water
potential gradient.
By the time the fluid reaches the U-shaped bend at the
bottom of the loop it is very concentrated and
hypertonic to the arterial blood.
(b)In the ascending limb:
The first section of the ascending limb is very permeable
to sodium and chloride ions but not permeable to water. No
active transport occurs in this section.
Sodium and chloride ions move down concentration
gradients out of the very concentrated fluid in the loop of
Henle into the tissue fluid of the medulla.
The second, thicker section of the ascending limb is also
impermeable to water, but sodium and chloride ions are
actively pumped out of the tubule into the tissue fluid of
the medulla and the blood of the capillary network.
The tissues of the medulla, therefore, have very
high sodium and chloride ion concentrations that
cause the water to pass out of the descending limb
by osmosis. However, the ascending limb is
impermeable to water, so water cannot follow the
chloride and sodium ions out down the
concentration gradient.
The fluid left in the ascending limb, therefore,
becomes less concentrated.
..
(c)THE DISTAL TUBULE:
yThe distal tubule is permeable to water,
but the permeability varies with the levels
of antidiuretic hormone (ADH). If there is
not enough salt in the body, sodium can be
actively pumped out of the tubule, with
chloride ions, down an electrochemical
gradient. Water leaves by diffusion, if the
walls of the tubule are permeable at that
time.
..
THE COLLECTING DUCT
yThe permeability of the collecting duct is strongly affected by
the hormone ADH.
yWater moves out of the collecting duct down a water potential
gradient as it passes through the medulla, with its high levels of
sodium and chloride ions.
yAs water leaves the collecting duct, the urine becomes steadily
more concentrated. The concentration of sodium ions in the
surrounding fluid increases through the medulla towards the
pelvis of the kidney (where the urine leaves the collecting duct),
so water can be removed along the whole length of the
collecting duct. This makes it possible for the kidney to produce
very hypertonic urine (high osmotic pressure) when it is
necessary to conserve water for the cells of the body.
..
THE URINE
The urine is the fluid that the kidney tubules produce.
First, it is collected in the central chamber (the pelvis )
of each kidney. It then passes along the ureters to the
bladder and is stored there until the bladder is
sufficiently stretched to stimulate urination. The urine
passes out of the body along a tube called the urethra.
The urine contains varying amounts of water and salts,
depending on the diet and the demands of the body. It
also contains relatively large quantities of urea. The
colour of the urine varies from almost colourless to
deep yellow/brown, depending on its concentration.
Substances such as glucose or protein should never
appear in the urine .
..
CONTROL OF THE KIDNEY
AND HOMEOSTASIS
..
OSMOREGULATION
yThe osmotic potential of the blood is maintained
within narrow boundaries by balancing the water and
salts taken in by eating and drinking with the water and
salts lost by sweating, defaecation and in the urine.
yThe concentration of the urine is controlled by a
negative feed back system involving antidiuretic
hormone (ADH ).
yADH is produced by the hypothalamus and secreted
into the posterior lobe of the pituitary and released out
from posterior lobe. ADH increases the permeability to
water of the distal tubule and the collecting duct.
..
..
..
MECHANISM OF ADH ACTION
yThe mechanism by which ADH increases the
permeability to water of the walls of the distal tubule
and the collecting duct is very elegant.
yADH does not cross the membrane of the tubule cells.
It binds to specific receptors, which triggers reactions
that result in the formation of cAMP as the second
messenger.
yThe cAMP starts a series of reactions that cause
vesicles within the cells lining the tubules to move to
and fuse with the cell membranes.
yThe vesicles contain water channels (aquaporins)
which are inserted into the membrane and make it
permeable to ..water.
..
Water then moves through the channels out of the
tubules and into the surrounding blood capillaries by
osmosis.
yThe amount of ADH released controls the number of
channels that are inserted. This means the permeability
of the tubules can be very closely controlled to match
the water demands of the body.
yWhen ADH levels fall, levels of cAMP also drop and
the water channels are taken out of the membranes and
repackaged in vesicles. This makes the tubule
impermeable to water again. The channels are stored in
vesicles ready for reuse when they are needed again.
..
ADH AND NEGATIVE FEEDBACK CONTROL
yIf water is in short supply, the osmotic balance of the
tissue fluids would become disturbed, causing cell
damage. This is prevented by a negative feed back
system involving ADH.
yOsmoreceptors in the hypothalamus detect an
increasing plasma concentration of inorganic ions.
yThey send nerve impulses to the posterior pituitary,
which consequently releases stored ADH in to the
blood.
yThe ADH is accepted by receptors in the cells of the
kidney tubules. ADH increases the permeability of the
distal tubule and the collecting duct to water.
..
..
..
yAs a result, water leaves the tubules by osmosis into the
surrounding capillary network. This means blood plasma
receives more water from the filtrate, and a small volume
of concentrated urine is produced.
y If you drink large volumes of water, the blood plasma
becomes more dilute . The same osmoreceptors of the
hypothalamus detect the change.
y The fall in the concentration of the blood plasma inhibits
the release of ADH by the pituitary gland. The walls of the
distal tubule and the collecting duct remain impermeable to
water and so little or no reabsorption occurs.
y The kidneys therefore produce large amounts of very dilute
urine and the concentration of the blood is maintained.
..
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
Nervous System
Central Nervous
System (CNS)
Peripheral Nervous
System (PNS)
Autonomic Nervous System
(ANS)(involuntary)
Somatic nervous
System (voluntary)
Sympathetic Nervous
System
Parasympathetic
Nervous System
2
CNS
PNS
1. Autonomic
(involuntary)
nervous system
2. Somatic
(voluntary)
nervous system
3
Nerves:
y Nerves are thread like structures, which
emerge from brain and spinal cord and
branch out to almost all parts of the body.
y The nerves are made up of neurons or
nerve fibres (nerve cells) bundled
together.
..
..
..
Structure of Neuron:
Neuron mainly consist of three parts1. The Cell Body or Soma (also called
Perikaryon):
y Contains nucleus, organelles plus other
structures.
2. Dendrites
y Extensively branching from the cell body
y Transmit electrical signals toward the cell
body.
..
Basic nerve cell (Neurone) structure
3.Axons (nerve fibers)
y Neuron has only one axon, but it can
branch to form axonites.
y Transmits action potentials away from the
cell body.
Myelin Sheath:
y Membranous wrapping around axons
made by Schwann cells.
y These are made up of 70% lipid and 30%
proteins.
..
Node of Ranvier:
y Space between two myelin-sheaths where
axon is exposed.
y These regions are not insulated and can
transmit nerve impulses.
y This speed up the transmission of nerve
impulses, as the impulse jumps from one
node of Ranvier to the next- This is called
the Saltatory effect.
..
..
Classification of Neurons by Structure:
1. Multipolar neurons:
y One axon and several dendrites. Eg. Motor
neuron.
2.Bipolar neurons:
y Bipolar neurons have one axon and one
dendron.. Eg. Relay neuron.
3.Unipolar neurons:
y One process extending from the cell body are
termed unipolar neurons.
y The one process divides with one part acting as
an axon and the other part functioning as a
dendrite. Eg: Sensory neurons
..
3 main types of nerve cells
sensory
neurone
relay
neurone
motor
neurone
Unipolar: Sensory neurons
Carries impulses from receptors e.g pain
receptors in skin to the CNS( brain or spinal
cord)
Bipolar -Relay neuron
Carries impulses from sensory nerves to
motor nerves.
Multipolar- Motor neuron
Carries impulses from CNS to effector e.g. muscle
to bring about movement or gland to bring about
secretion of hormone e.g ADH
Conduction of Nerve impulses
(Action potential)
Factors affecting speed of nerve impulse
y Diameter of the axon- Speed is
proportional to diameter.
y Myelination- In myelinated axons the
action potential travels from one node of
Ranvier to the next- this is called the
Saltatory effect, which speeds up
transmission of impulses.
..
Components of the neuron cell
membrane:
1.Sodium channels:
y These are permeable to sodium ions.
y These are also referred to as voltage
gated channels as they open and close
based upon the potential in the cell.
y These gates close at +40 mV.
2. Potassium channels:
y These are permeable to potassium ions.
..
..
y These are also referred to as voltage
gated channels as they open and close
based upon the potential in the cell.
y These gates open at +40 mV and close at
-80 mV.
3. Sodium potassium pumps:
y These are carrier proteins which pump 3
Na+ ions out and 2 K+ ions into the cell, by
active transport.
..
The cytoplasm of the neuron(Axoplasm)
contains negatively charged proteins.
This will remain in the cytoplasm.
The extra cellular fluid contains Cl- ions,
which will not move into the cytoplasm.
Changes in the potential of the membrane
(neurilemma) are brought by movement of
sodium and potassium ions across the
membrane.
..
TRANSMISSION OF NERVE IMPULSES:
1. When a nerve fibre is not conducting
impulses, it is said to be at its resting
potential (-70mV). The membrane is said
to be polarized. At resting potential,
there is a high concentration of sodium
ions outside and a high concentration of
potassium inside the neurone.
Inner, cytoplasmic side is negatively
charged as negatively charged proteins
are more inside than the K+ ions.
..
..
2. When the neurone is stimulated there is a
momentary influx of sodium ions through
the sodium channels along a
concentration gradient. This is called
depolarisation.
The change in potential difference of the
axoplasm from an inside
-70mV to +40mV causes the action
potential (+40 mV).
..
3. Sodium ions move to the adjacent resting
region causing a change in the electrical
charge (potential difference).
The change in potential difference in the
membrane adjacent to first action potential
initiates a second action potential.
4. At the site of the first action potential the
sodium ion channel close and potassium
ion channel open and potassium ions
begin to leave the axon along a
concentration gradient.
..
As the potassium ions leave the axon
through channels the membrane become
repolarised behind the impulse.
An overshoot of potassium ions occurs as
channel proteins allow movement of more
potassium ions cause hyperpolarisation.
This causes the potential to fall lower than
-70 mV; that is it will fall to -80 mV.
5. A third action potential is initiated by the
second. In this way nerve impulses move
along the axon.
..
6. After the impulse has passed and the
neurone is repolarised, sodium is actively
expelled by the Na-K pump in order to
increase the external concentration and so
allow the passage of another impulse.
For every 3Na+ ions pumped out 2K+ ions
are pumped in restoring resting potential.
..
Threshold potential:
y It is the minimum depolarization that is
needed to generate an action potential.
y If the membrane depolarizes upto the
threshold potential, then more Na+
channels open ( an example of positive
feed back) and an action potential is
generated.
y If the depolarization does not reach the
threshold potential, then no action
potential is generated . This is often
referred to as the all or nothing principle
of nerve impulse transmission.
..
Refractory period:
y This is the time period required for the
membrane to regain the resting potential,
after firing an action potential.
..
..
..
The Synapse
y Neurons are not connected to each other.
The gap between the end of one neuron
and the beginning of another neuron is
called synapse.
y The axon of one neurone meets the
dendrites or the cell body of another
neuron in the synapse.
..
..
..
y Synapse consist of
Presynaptic ending (knob)-where
neurotransmitters are made.
Post synaptic ending- Where
neuroreceptors are present.
Synaptic clefty Action potentials cannot cross the synaptic
cleft.
y Across the synapse the nerve impulses are
carried by neurotransmitters.
..
TRANSMISSION OF NERVE IMPULSES
ACROSS A SYNAPSE
y Arrival of an action potential causes
depolarisation of the presynaptic
membrane.
y This depolarisation of the pre-synaptic
membrane increases the permeability
of the membrane to calcium ions.
y Calcium ions diffuses into the pre-synaptic
knob from the synaptic cleft through the
calcium channels.
..
..
y This causes the movement
neurotransmitter (Eg: Acetylecholine)
containing vesicles towards the
presynaptic membrane. These vesicles
fuse with the presynaptic membrane and
release neurotransmitter substances into
synaptic cleft by exocytosis.
y The neurotransmitter diffuse across the
synaptic cleft and binds with receptors on
the postsynaptic membrane.
..
y The binding of neurotransmitter with the
receptors on the postsynaptic membrane
cause the membrane to become
permeable to sodium ions.
y The rapid influx of sodium ions causes the
generation of an action potential in the
post synaptic neuron.
y The neurotransmitter substance is then
removed from the receptors on the post
synaptic membrane by enzymes.
..
y The enzyme cholinesterase hydrolyses
acetylecholine to choline and acetate,
which are inactive.
y This enables the postsynaptic
membrane to repolarise, so that next
action potential can be generated.
y Choline and acetate are reabsorbed into
the presynaptic neuron and are used to
resynthesise acetylecholine.
y Mitochondria are used to release energy
for the synthesis of neurotransmitters.
..
Functions of Synapse:
y Control of nerve pathways.
y Synapses make sure that the flow of
impulses is in one direction only, as the
neurotransmitter vesicles are only in the
presynaptic membrane.
y They allow integration- impulses can be
sent to or received from several neurons.
..
Types of Synapse:
1. Inhibitory synapses:
y The neurotransmitters released from the
presynaptic neuron will not cause depolarization of
the postsynaptic neuron.
y Instead they cause the postsynaptic membrane to
become more negative than usual.
y This makes the neuron less likely to trigger an
action potential. This is called the inhibitory
postsynaptic potential.
Eg: Glutamate is an inhibitory neurotransmitter
released from the rod cells of the eye.
..
2. Excitatory synapses:
y These makes the post synaptic membrane
more permeable to sodium ions and cause
depolarization of the postsynaptic membrane.
Summation- A single nerve impulse typically
does not depolarise the post synaptic
membrane enough to produce an action
potential, but several impulses arriving within
a short time leads to the production of more
vesicles and more release of
neurotransmitters. This addition effect is
called summation.
..
Two types of summation1. Spatial summation:
y In spatial summation, several presynaptic neurons
converge at a synapse with a single post synaptic
neuron.
y More neurotransmitters released into the synaptic
cleft to propagate an action potential.
2. Temporal summation:In temporal summation,
only one presynaptic and one post synaptic
neuron forms a synapse.
Several impulses arrive at a synapse having
travelled along a single neuron one after the
other.
..
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
Introduction
y Receptors allow us to perceive and
respond to a wide variety of stimuli.
y Many receptors spread throughout the
body, and they group in some regions form
sense organs.
y Human beings have a multitude of senses.
y Traditionally recognized senses are sight (ophthalmoception), hearing
(audioception), taste (gustaoception),
smell (olfacoception or olfacception),
and touch (tactioception).
..
..
y Other senses include temperature
(thermoception), pain (nociception),
balance (equilibrioception) etc...
..
PHOTORECEPTION IN
ANIMALS:
..
PHOTORECEPTION:
y The eyes are the most important sense
organs, as we receive 80% of the
information through the eye.
y This occurs with the help of photoreceptors
present in the eye.
y The photosensitive pigments in
photoreceptors convert light energy
into electrical signals, in the form of
nerve impulses.
..
..
..
Structure of retina
y The retina consists of three layers of cells
± the outermost layer of photoreceptor
cells (towards the choroid), the layer with
bipolar neurons and the inner most
layer with ganglion (mass of nerve
cell bodies) cells, towards the vitreous
humour).
y Retina contains two types of photoreceptor
cells sensitive to light: Rods and Cones.
..
y Cones allow colour vision in bright
light. Rods only give black and white
vision but, unlike cones, this will work in
dim light as well as in bright light.
y Rods contain the photosensitive pigment
Rhodopsin, which is sensitive to light of
low intensity/ dim light.
y Cone cells contain the photosensitive
pigment Iodopsin, which is sensitive to the
wavelength of light/colour vision.
..
..
..
..
..
Structure of rod cells
y Rod cells consists of three parts ± Outer
segment with vesicles containing
rhodopsin, inner segment with nucleus
mitochondria and other organelles; and
synaptic region.
y The outer segment is rod shaped.
y Rhodopsin consist of membrane bound
protein called opsin and a covalently
bound prosthetic group (a cofactor,required
for the protein's biological activity) called
Retinal.
..
..
y Retinal is made from vitamin A (Deficiency
of vitamin A causes night-blindness -poor
vision in dim light).
y Retinal is the light sensitive part, and it can
exists in 2 forms: cis retinal and trans
retinal.
y Light causes cis-retinal to be converted
into trans-retinal, which splits away from
opsin.
y Trans- retinal is reconverted into cis-retinal
by an enzyme retinal isomerase.
..
..
y The cis-retinal recombines with opsin to form
rhodopsin. The reformation of rhodopsin is
called Dark adaptation. ATP is used for this
process.
y In the dark retinal is in the cis form, but when
it absorbs a photon of light it quickly
switches to the trans form. This causes the
trans-retinal to split from the opsin. This
process is called bleaching.
y Conversion of trans to cis form is catalysed
by an enzyme (retinal isomerase) and is very
slow process taking few minutes.
..
..
y This is why we are initially blind when you
walk from sunlight to a dark room. Some
time is needed to resynthesise rhodopsin.
This is called as the dark reaction.
y Rod cell membranes contain a special
sodium channel that is controlled by
rhodopsin.
y Rhodopsin with cis-retinal opens the
sodium channel and rhodopsin with transretinal closes it.
..
In dark,
y Rhodopsin ( with cis retinal) accumulates
in rod cells (called dark adaptation).
y sodium channel is open, allowing sodium
ions to flow in. These sodium ions moves
down the concentration gradient into the
inner segment where sodium pumps
continuously transport them back out of
the cell.
y The influx of sodium ions produces a
slight depolarization of the cell.
..
..
y The potential difference across the
membrane is about -40mV, compared to
the -70mV resting potential.
y This slight depolarization leads to the
release of neurotransmitter glutamate.
y The synapse with the bipolar neuron is an
inhibitory synapse, so the glutamate binds
to bipolar neuron and stopping it
depolarizing and vision not possible in
darkness.
..
In light
y Rhodopsin is bleached into opsin and
trans-retinal.
y This causes sodium channels to close. The
outer membrane become less permeable
to sodium ions.
y However, Sodium ions are continuously
pumped out of the inner segment.
y The inside of the rod cell become more
negative than usual. This is called
hyperpolarisation.
..
..
y Hyperpolarisation stops the release of
neurotransmitter substance.
y Lack of glutamate in the tissue fluid
causes the sodium channels in the
bipolar neuron to open and generates
an action potential. The action potential is
transmitted to the brain through optic
nerve, and vision possible.
..
Structure of cone cells:
y Cones are adapted to give colour vision.
y The outer segment is cone shaped.
y There are fewer membranous vesicles
and they are formed from infoldings of the
outer membrane.
y They contain iodopsin, which is thought to
occur in three different forms (blue, green
or red) each responding to light in a
narrow range of wavelengths.
..
..
y Each form of iodopsin occur in a
different cone and the relative stimulation
of each type is interpreted by the brain as
a particular colour.
y Depending on the intensity of light and the
wavelength that strikes the cone, we see
different colours.
y Since a single bipolar neuron arises from
each cone (temporal summation) more
information is sent to brain and give
greater visual acuity .
..
..
..
..
y In the case of rod cells, many rod cells
synapse with a single bipolar neuron
(spatial summation)- called synaptic
convergence- leads to low visual
acuity.
..
Response of pupil towards light:
y To operate well in different light conditions
the eye must be able to control the amount
of light entering the eye.
y It does this by changing the diameter of the
pupil.
y The pupils dilation and constriction
according to the intensity of light is
regulated by autonomic nervous system.
..
..
Pupil reflex towards light
In dim light
In bright light
1. Light sensitive cells in the
retina detect the light intensity.
1. Light sensitive cells in the
retina detect the light intensity.
2. Impulses are sent along the
optic nerve to the brain.
2. Impulses are sent along the
optic nerve to the brain.
3. The brain returns the
impulses along occulomotor
nerve to the radial muscles of
the iris.
3. The brain returns the
impulses along occulomotor
nerve to the circular muscles of
the iris.
4. The radial iris muscles
contract, while the circular iris
muscles relax.
4. The circular iris muscles
contract, while the radial iris
muscles relax.
5. The diameter of the pupil
5. The diameter of the pupil
increases, allowing more light to decreases, allowing less light to
enter.
enter.
THANK YOU
..
INVESTIGATING HUMAN
BRAIN
Prepared by Ashok.V, BiologyTeacher, CHSE.
MEDICAL IMAGING TECHNOLOGY
1. Computerised Tomography (CT Scan):
y While doing CT scan thousands of narrowbeam X-rays pass through the patients
head from a rotating source.
y The X-rays are collected on the other side
of the head and their strength measured.
y The density of the tissue the X-ray passes
through decreases the strength of the
signal, and therefore, can identify what type
of tissue is in the brain.
..
..
..
..
CT scan shows brain structures, not brain
activity.
y They only give still images.
y X rays cause damage to tissues.
y They are very useful for picking up
diseases, such as cancer, stroke and
oedema.
..
2. Magnetic Resonance Imaging (MRI Scan):
y MRI scans produce images that show much
finer detail than CT scan.
y They are produced using magnetic fields and
radio waves to image the soft tissues.
y Radio waves removes even the small risk of
damage from the X-rays used in CT scans.
y Hydrogen atoms in water present in tissues
are monitored in MRI.
y The hydrogen atoms take energy from the
radio waves when it spins in two magnetic
fields.
..
..
..
y When the radio waves are turned off, the
hydrogen nuclei return to their original
alignment and release the energy they
absorbed. This energy is detected.
y MRI scan gives 3D image of the brain.
y MRI produces very detailed images which
give doctors a great deal of information
about the living brain.
y It is widely used to diagnose brain injuries,
strokes, tumours and infections of the
brain or the spine.
..
3.Functional Magnetic Resonance
Imaging(fMRI):
y fMRI shows which part of the brain are
active during a particular task.
y Deoxyhaemoglobin absorbs the radiowave
signal and later re-emits it, while oxy
haemoglobin does not. When an area of
the brain is active, the blood flow to that
area increases and more oxyhaemoglobin
is delivered to supply the active cells with
the oxygen they need for aerobic
respiration.
..
y Less of the signal is absorbed as a result.
So an active area of the brain absorb less
energy than a less active area.
y This is used to rapidly produce a three
dimensional (3D) image of the brain.
y In addition to benefits of MRI scan, fMRI
shows brain activity.
..
..
4. Positive emission tomography (PET) scans :
Positive emission tomography (PET) scans
give another way of forming detailed, 3D images
of the inside of the body , including the brain.
PET scans reveal abnormal areas in the body
and can also show how well different areas are
working.
PET scans can be combined with CT scans and
MRI scans to produce very detailed images to
help with diagnosis.
The great advantage of PET scans is that they
can show how parts of your brain are actually
working.
..
They can be used to plan surgery by giving surgeons a 3D
image of the areas of the brain that are affected.
To produce a PET scan, the patient is injected with a
radiotracer (radio active isotope) which is similar to
glucose. This means the body treats it in a similar way and
it is carried to all the cells.
The scanner works by detecting the radiation which the
radiotracer gives off and the computer system analyses
where it accumulates and where it does not. So, for
example, cancer cells absorb much more of the radio
tracer than normal cells, so a PET scan shows cancerous
cells clearly.
Areas of the brain that are less active than they should be
(for example, areas that have died as a result of diseases
such as Alzheimer's, which causes dementia) absorb less
of the radiotracer than expected.
..
..
THANK YOU
..
CHEMICAL IMBALANCE
OF BRAIN
Prepared by Ashok.V, BiologyTeacher, CHSE.
..
Chemical imbalance of brain:
y Chemical imbalance of neurotransmitters can
contribute to ill health.
3DUNLQVRQ¶V'LVHDVHV 3' y 3'LVD³PXOWLIDFWRULDO´GLVHDVHVWKDWLVFDXVHG
by a mixture of environmental and genetic
factors.
y This is due to lack of the neurotransmitter
Dopamine.
y Dopamine is crucial neurotransmitter that helps
to transmit messages that both initiate and
control movement and balance.
..
y These Dopamine molecules ensure that
muscles work smoothly, under precise
control and without unwanted movement.
y In PD neurons which secretes dopamine in
the brain die.
Main symptoms¾Stiffness of muscles.
¾Tremor of the muscles.
¾Slowness of movement.
¾Poor balance
¾Difficulty in speaking and breathing.
..
..
Treatment of Symptoms:
1. Levodopa (L-dopa)y Dopamine cannot cross the blood-brain
barrier, so it cannot be used to treat PD.
y L-dopa is a precursor of dopamine and small
enough to cross the barrier and enter the
neurons.
y It is then converted to dopamine and
secreted, which relieves the symptoms.
..
2. Dopamine Agonists:
y These drugs bind with dopamine receptors
and mimic the action of dopamine.
3. Monoamine Oxidase B (MAOB)
inhibiters:
y Monoamine oxidase B is an enzyme, which
break down dopamine in the brain.
y MAO B inhibitors are drugs which inhibit the
action of MAO B.
y The level of dopamine remains high and the
symptoms are relieved.
..
Depression:
y Depression may be caused by external factors
like, relationship stress, bereavement, loss of
jobs and internal factors like, low levels of
serotonin and dopamin.
y In depression neurons in the brain that secrete
serotonin neurotransmitter stop working
properly and serotonin levels fall.
Common signs and symptoms of
depression:
¾Feelings of helplessness and hopelessness
¾Loss of interest in daily activities.
..
¾Appetite or weight changes.
¾Sleep changes.
¾Irritability or restlessness.
¾Concentration problems.
¾Unexplained aches and pains.
Treatment
Antidepressant medications:
1. Selective serotonin reuptake inhibitors(SSRIs)
y These are medications that increase the amount
of serotonin in the brain.
y SSRIs selectively inhibits serotonin reuptake in
the brain.
..
..
y This keeps high concentration of serotonin in
synapses.
y This helps to activate the neurons that have
been deactivated by depression , thereby
UHOLHYLQJWKHGHSUHVVHGSHUVRQ¶VV\PSWRPV
2. Monoamine Oxidase Inhibitors (MAOIs):
y MAOIs elevate the levels of neurotransmitters
in the brain by inhibiting the enzyme
Monoamine Oxidase.
y When these enzymes are broken down, the
amount of neurotransmitter in the brain is
increased.
..
THE EFFECT OF DRUGS ON
THE NERVOUS SYSTEM
..
ILLEGAL DRUGS AND THE BRAIN
Action of MDMA (ecstasy):
y Ecstasy is one of the most dangerous drugs
threatening young people today called MDMA (3-4
Methylenedioxymethylamphetamine) by scientists.
y It is a synthetic chemical that can be derived from
an essential oil of the sassafras tree.
y MDMA blocks the transporter proteins in the
presynaptic membrane, preventing the reuptake of
serotonin.
y This increases the concentration of serotonin in the
synaptic cleft.
..
..
..
y Moreover, MDMA molecule have similar
structure to serotonin molecules,
enabling them to bind to serotonin
receptor sites on the transporter protein
in the presynaptic membrane.
y High concentrations of serotonin in the
synaptic cleft will stimulate repeated
impulses in the postsynaptic neuron.
y This stimulate pleasure pathways in the
brain.
..
yUsing ecstasy causes physical changes, such as
increased heart rate, and can cause problems
in the body's thermoregulatory system. There
may be no desire to drink, which can lead to
hyperthermia (overheating) with raised blood
pressure and irregular heartbeat. In a small
number of cases, this can lead to death.
yEcstasy can also affect the hypothalamus so
that it secretes more antidiuretic hormone.
This effectively stops the kidneys from producing
urine and can lead to problems if someone keeps
drinking water in an attempt to stay hydrated and
cool down.
..
THE EFFECTS OF OTHER DRUGS ON
THE NERVOUS SYSTEM
..
1. Nicotine:
y Nicotine is the addictive drug found in tobacco. It
mimics the effect of acetylcholine and binds to
specific acetylcholine receptors in post-synaptic
membranes known as nicotinic receptors.
y It triggers an action potential in the post-synaptic
neuron, but then the receptor remains unresponsive
to more stimulation for some time. Nicotine causes
raised heart rate and blood pressure. It also
triggers the release of another type of neuro
transmitter in the brain called dopamine. This is
associated with pleasure sensations. At low doses,
nicotine has a stimulating effect, but at high doses it
blocks the acetylcholine receptors.
..
..
2. Lidocaine :
It is a drug used as a local anaesthetic. It is commonly
used by dentists before drilling or removing a tooth.
Lidocaine molecules block voltage-gated sodium
channels, preventing the production of an action
potential in sensory nerves and so it prevents from
feeling pain.
It is also used to prevent some heart arrhythmias
(irregular heartbeat). In this situation, it works in the
same way (i.e. blocking sodium channels). This
reduces or prevents early or extra action potentials
from the SAN region that can cause arrhythmias.
..
..
3. Cobra venom (alpha-cobratoxin):
It is a substance made by several species of cobra. It is toxic
and often fatal in snake bites.
It binds reversibly to acetylcholine receptors in postsynaptic membranes and neuromuscular junctions.
Thus, it prevents the transmission of impulses across
synapses, including the neuromuscular junctions between
motor neuron and muscles.
Consequently, muscles are not stimulated to contract and
gradually the person affected becomes paralysed.
When the toxin reaches the muscles involved in breathing it
causes death.
However, in very low concentrations, it can relax the muscles
of the trachea and bronchi in severe asthma attacks, and so
save lives.
..
..
THANK YOU
..
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www.pmt.education
Prepared by Ashok.V, BiologyTeacher, CHSE.
Nervous System
Central Nervous
System (CNS)
Peripheral Nervous
System (PNS)
Autonomic Nervous System
(ANS)(involuntary)
Somatic nervous
System (voluntary)
Sympathetic Nervous
System
Parasympathetic
Nervous System
2
CNS
PNS
1. Autonomic
(involuntary)
nervous system
2. Somatic
(voluntary)
nervous system
3
..
STRUCTURE
OF BRAIN
Introduction
THE FORMATION OF THE BRAIN
yIn vertebrates (including humans) the
brain forms as a swelling (a larger area) in
the hollow neural tube at the anterior
(front) end of a vertebrate embryo that
folds back on itself .
yThe basic brain pattern has three distinct
areas: the forebrain, midbrain and
hindbrain.
..
..
..
y The human brain, the mind, has a consistency of
semisoft cheese.
y The human brain weighs about 1.5 kg and contains
thousands of millions of neurones.
y Brain protected by cranium.
y The brain is surrounded by three membranes
together known as meninges.
y In space between these membranes Cerebro
spinal fluid present.
y Cerebro-spinal fluid supplies oxygen &
nutrients; absorbs shock; maintain shape &
prevent collapse.
y The meninges protect the delicate brain and carry
the blood vessels
to it.
..
..
Brain structure
Cerebrum
cerebellum
hypothalamus
Pituitary gland
medulla
..
1. cerebrum
‡ This is the largest part of the human brain.
‡ It is made up of two halves called
cerebral hemispheres.
‡ It has an outer layer of neuron cell bodies
(greymatter) and an inner layer of nerve
fibers with myelin sheath (white matter) .
‡ It has much folded and wrinkled
surface called cortex. The inner part is
not folded and called medulla.
..
yThe cerebrum is divided into four
lobes- Frontal lobe.
- Temporal lobe.
- Parietal lobe.
- Occipital lobe.
..
..
2. Hypothalamus:
y It is a reflex centre concerned with a
number of homeostatic mechanisms such
as temperature control, water balance
and CO2 concentration in the blood.
y Controls hunger, thirst and sexual activity.
y Controls the autonomic nervous system.
y Controls feeding, sleeping and aggression.
y Act as an endocrine gland.
y Monitors the composition of blood.
..
..
3. Corpus callosum:
y Connects left and right cerebral
hemispheres.
y Allows information to move between
hemispheres.
..
..
4.Medulla Oblongata:
y Control involuntary actions like
swallowing, salivation, sneezing,
vomiting and coughing.
y It plays an important role in
maintaining homeostasis by
controlling autonomic activities like
heart rate, blood pressure and
ventilation.
y Contains chemoreceptors.
..
5.Cerebellum:
y Made of 2 hemispheres.
y Right half control left side and left half
control right side.
y Coordinate (but not initiate) muscle
contraction (voluntary) and maintain
balance & posture (involuntary).
y Well developed in fast moving
animals. (e.g birds & fishes).
..
6. Pituitary gland:
yIt is a protrusion of the bottom of
the hypothalamus at the base of
the brain.
yIt is the master gland, which
control all other endocrine gland.
..
THE STRUCTURE AND FUNCTIONS OF THE
SPINAL CORD
yThe spinal cord is a tube made up of a core of
grey matter surrounded by white matter, which
runs out from the base of the brain (the medulla
oblongata) through the vertebrae. Impulses from
receptors travel along sensory nerves in to the
spinal cord through the dorsal roots, and then
travel in sensory nerves up the spinal cord to the
brain. Instructions from the brain travel as
impulses down motor nerves in the spinal cord
and out in motor nerves through the ventral roots
to the effector organs.
..
..
Reflex action
y It is a rapid and automatic response to a
stimulus by an organ or organs without
conscious thought.
Two types :
1. Spinal reflex
2. Cranial reflex
y Reflex actions controlled by spinal cord are
called as spinal reflex actions.
y Reflex actions controlled by Brain are called
cranial reflex actions.
y The nervous pathway of a reflex action is called
Reflex arc.
27
(1) Stimulus
(2) Receptor
sensory neurone
(3) CNS (spinal cord)
motor neurone
(4) Effector
(5) Response
28
The sequence of events in a
spinal reflex
y Heat or pain receptors in the skin are
stimulated and fire off impulses which
travel along the sensory neuron.
y The sensory neuron enters the spinal cord
via the dorsal root.
y In the grey matter of the spinal cord, the
impulse pass from the sensory neuron to a
relay neuron across a synapse.
29
y The relay neuron in turn makes a synapse
with one or more of the motor neuron.
y The impulses are transmitted to the motor
neuron which leave the spinal cord through
the ventral root and pass to a muscle.
y The impulse causes the bicep muscle to
contract and triceps muscle to relax.
y The antagonistic movement of the
muscles help to remove the hand from the
painful stimulus and prevent damaging the
tissue.
30
..
Cranial reflexes
..
CNS
PNS
1. Autonomic
(involuntary)
nervous system
2. Somatic
(voluntary)
nervous system
34
THE PERIPHERAL NERVOUS SYSTEM are of
two main types.
(a) SOMATIC NERVOUS SYSTEM:
This is the voluntary nervous system which
involves motor neurones that are under
voluntary or conscious control involving the
cerebrum.
Voluntary motor neurones function as a result of
conscious thought.
For example, when you consider an action, such
as picking up a drink or switching on the computer,
the instructions that need to be issued to the
muscles will be carried along voluntary nerve
fibres.
‡
‡
..
(B)THE AUTONOMIC NERVOUS SYSTEM
The autonomic nervous system (the involuntary
nervous system ) involves motor neurones that
control bodily functions that are normally involuntary.
Examples include control of the heart and breathing
rate, sweating, the dilation or constriction of the iris of
the eye in response to changing light levels etc
The autonomic nervous system is sub-divided in to
the sympathetic nervous system and the
parasympathetic nervous system .
The differences between them are both anatomical
and functional.
..
STRUCTURAL DIFFERENCES BETWEEN THE
PARASYMPATHETIC AND SYMPATHETIC SYSTEMS
Both the parasympathetic and the sympathetic nervous
systems have myelinated preganglionic neurons that
leave the CNS and synapse in a ganglion (a collection
of cell bodies outside the central nervous system ) with
unmyelinated postganglionic neurons.
In the sympathetic system, the ganglia are very close
to the CNS, so the preganglionic fibres are short and
the postganglionic fibres are long.
In the parasympathetic system, the situation is
reversed. The ganglia are near to or in the effector
organ, so the preganglionic fibres are very long and
the postganglionic fibres are very short.
..
..
..
FUNCTIONAL DIFFERENCES BETWEEN THE
PARASYMPATHETIC AND SYMPATHETIC NERVOUS
SYSTEMS
There are two basic differences between these two nervous
systems.
The sympathetic nervous system produces noradrenaline at
the synapses and usually produces a rapid response in the
target organ system.
The parasympathetic nervous system often has a slower, or
inhibitory effect on organ systems and it produces the
neurotransmitter acetylcholine at the synapses.
The parasympathetic system maintains normal functioning of
the body and restores calm after a stressful situation
..
..
Comparison Between Nervous and Hormonal
control.
Nervous Control
Hormonal control
Mode of transmission is
Transmission is purely chemical
electrochemical
Transmission and responses Transmission and responses are
usually slow.
are rapid.
Often short term changes.
Often long term changes.
Pathway is specific through Pathway through blood, so not specific
neurones.
but target specific.
Response is often localised. Response is widespread. e.g. growth
e.g. muscle
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
Growth in Plants
y The growth in plants is coordinated by plant
growth substances (PGS).
y They are chemicals produced in the plant in
very low concentrations and transported to
other regions where they cause a response.
y Plant growth substances exert their effects
alone or may interact with each other either
synergistically or antagonistically.
..
Plant growth
substances
(PGS)
Growth
promoters
Auxins
GIBBERELLINS
..
Growth
inhibitors
CYTOKININS
ABSCISIC
ACID
ETHENE
1. Auxins:
y Auxins are group of chemical substances of which
Indole Acetic Acid-IAA is the common.
y Auxins produced at the shoot tips. It helps for
elongation of cells. It will stretch the cell walls. This
begins as auxins [IAA- Indole Acetic Acid]
produced by the meristematic cells descend from
the shoot tip towards the cells in the zone of
elongation.
y The response of plants to light in which the plants
bend towards the light is called positive
phototropism.
..
..
y The auxins move away from the
illuminated side of the shoot and
accumulate on the darker side, where it
stimulates cell elongation.
y It binds to the receptors on the cell
membrane of the peripheral cells and
activates the proton pump.
y As H+ ions are actively pumped towards
cellwall of these cells, the H+ ion
concentration rises and pH become acidic.
..
..
..
y This provide an optimum pH, which
activates enzymes which break down
hydrogen bonds in the cellulose microfibrils
and cell wall become flexible.
y As the cell absorb water by osmosis, the
very flexible cell wall stretch, leading to cell
wall elongation and allow cells to expand.
y When the cells mature, enzymatic
degradation of auxin causes the pH to rise
and hydrogen bonds to be formed between
cellulose microfibrils.
..
..
..
..
GIBBERELLINS AND SEED GERMINATION
y Gibberellins are other important group of plant
hormones. These compounds act in several ways,
including as growth regulators. They also promote
the growth of fruit.
y They are involved in breaking dormancy (inactive
period) in seeds and in germination, because they
stimulate the formation of enzymes in seeds. For
example, they stimulate the production of amylase,
which breaks down starch stores in cereal plants.
This makes glucose available for respiration in the
embryo plant as it develops during germination.
y Gibberellins also stimulate bolting, a period of
sudden rapid growth and flowering, in biennial plants.
..
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
Photoreception in Flowering Plants.
y Plants detect the quantity, direction and
wavelength of light using photoreceptors
and respond to the changes in light
conditions.
y Unlike animals, all messages in plants
are chemical, and so all their responses
are slower.
y Plants contain several families of
photoreceptors.
y Most important one is Phytochromes.
..
Phytochromes- Plant photoreceptors:
y Phytochrome is a photoreceptor pigment
found in leaves, seeds, stems and buds of
plants.
y It is a conjugated protein consists of a
protein molecule bonded with a
pigment molecule.
y It exists in two inter-convertible formsPhytochrome red (PR or P660 ) absorbs red
light and Phytochrome far red (PFR or
P730 ) absorbs far-red light and the other.
..
y Pfr is the active form of phytochrome. The
presence of Pfr in a tissue stimulates
events. Eg: initiation of flowering, seed
germination etc..
..
y Plants synthesize phytochromes in the Pr
form (it is a blue pigment). Absorption of
red light converts Pr into Pfr (it is a blue
green form).
y Absorption of far-red light converts Pfr
back into Pr.
y The conversion of Pr into Pfr dominates
in sunlight because more red absorbed.
y Pfr accumulates in the light. In the dark,
any Pfr present is slowly converted to
Pr.
..
..
PHYTOCHROME AND TRANSCRIPTION
..
Phytochromes and gene
regulation:
When Pr is converted in to Pfr in the presence of
light, it moves in to the nucleus through the pores
in the nuclear membrane.
In the nucleus, it binds to a nuclear protein
known as the phytochrome- interacting factor 3
(PIF3).
PIF 3 will act as a transcription factor and it can
only binds to Pfr. It does not bind to Pr.
PIF 3 only activates gene transcription and the
formation of mRNA if it is bound to Pfr.
..
..
Effect of Phytochromes:
1. Phytochromes and germination of seeds:
y Pfr stimulates germination where as Pr
inhibits germination. This ensures that seeds
germinate only when long durations of light
are available.
y This increases the chances of survival of the
seedling.
y Only seeds just below the surface (which
receive some light) will germinate, ensuring
that they emerge from the soil and start to
photosynthesise.
..
y Seeds falling on forest floor, below the
canopy, will not germinate until a large tree
falls down.
y This mechanism is used by seeds to start
germinating in the spring or summer.
y During summer the days are longer than in
winter. More Pfr accumulates in the seed
and stimulates germination.
y On the other hand, seeds do not germinate
in winter because the days are shorter and
there would be a higher concentration of Pr in
the seeds.
..
..
3.Phytochromes and Greening:
y Greening refers to the changes that a
young seedling undergoes as soon as it
emerges from the soil.
y The exposure to light stimulates the
production of Pfr, which promotes the
formation of chlorophyll, leaf development
and unrolling of primary leaves.
y It also inhibits the elongation of internodes.
..
..
y Etiolation is the opposite of greening. It
occurs when plants are kept in complete
darkness.
y Pr is produced in darkness and brings
about the elongation of internodes and
yellowing of leaves.
y The elongation of stem helps the plant to
grow rapidly upward in search of light.
..
2. Phytochromes and flowering:
y The ability of certain plants to sense the
relative amounts of light and dark in a 24hour period is called photoperiodism.
y Photoperiodism controls flowering in many
plants.
y Phytochromes provide a mechanism to
detect photoperiodism.
..
Based on Photoperiodism plants are
classified into 3 types
a. Long Day Plants(LDP) :
y Flower buds are formed only in summer,
when the days are longer than nights.
y In long day plants, flowering is stimulated
by Pfr. Eg: strawberries, oats, poppies and
lettuce, cabbage
b. Short Day Plants (SDP):
y Flower only in winter, when the days are
shorter than nights.
..
..
y In short day plants flowering
stimulated by Pr. Eg: Rice, cotton.
c. Day neutral plants (DNP):
y These are unaffected by the length of
the day.
Eg: Cucumbers, tomatoes and pea
plants.
..
How photoperiodism occurs ?
In SDPs, Pfr inhibits flowering, and a lack of Pfr
allows flowering to occur. During long periods of
darkness, the levels of Pfr fall, as it is almost all
converted to Pr. This allows flowering to take place.
y In LDPs, high levels of Pfr, stimulate flowering. The
nights are short so relatively little Pfr is converted back
to Pr. As a result, relatively high Pfr levels are
maintained all the time, stimulating flowering.
yDNPs irrespective of the levels of Pr, and Pfr
flowering occurs. Other factors trigger their
flowering.
..
..
..
yIn 1930s, scientists first hypothesized the presence of a
‡
‡
‡
plant hormone known as florigen . They thought that
plants produce florigen in response to the changing
levels of phytochromes and the plant transport system
carried it to the flower buds.
If the whole plant is kept in the dark, apart from one leaf
which is exposed to the appropriate periods of light and
dark, flowering occurs as normal. A plant kept in total
darkness does not flower.
Using the same experimental set-up, if the photo
periodically exposed leaf is removed immediately after the
stimulus, the plant does not flower. If the leaf is left in
place for a few hours, it does flower.
If two or more plants are grafted together and only one
is exposed to appropriate light patterns, all the plants
will flower.
..
For years no one could isolate the theoretical hormone and so
the florigen theory lost support.
However, recently scientists have shown that, when a leaf is
exposed to a given amount of light and dark, a particular form
of mRNA is produced in the leaf, linked with a gene associated
with flowering(the FT gene or Flowering Locus T).
It is known as FTmRNA . It was thought that a large molecule
like FTmRNA could not be florigen, as it would not be able to
leave the cell. Now scientists have shown that FTmRNA can
move from cell to cell to the transport tissues through the
plasmodesmata.
They have also shown that FTmRN A travels from the leaves in
which it is formed to the apex of the shoot, where other
genes associated with flowering are activated.
So at the moment, it looks as though FTmRNA is the chemical
known as florigen.
..
THANK YOU
..
Prepared by Ashok.V, BiologyTeacher, CHSE.
Use of Genetically Modified
Organisms ( GMO- plants, animals
and microorganisms) for producing
drugs:
y Genetically Modified Organisms
(transgenic organisms/transformed)
are produced by the artificial
introduction of genes from another
organism, through genetic engineering.
..
..
..
This genetic engineering (gene technology)
includes different stages1. Isolation of a desired gene by using
DNA or mRNA
y The mRNA is extracted from the cell.
y A complementary DNA (cDNA) will be
made from the mRNA using the enzyme
reverse transcriptase.
y A double stranded DNA is produced from
the cDNA using the enzyme DNA
polymerase.
..
y The DNA can be cut into small fragments
by the help of enzymes named restriction
endonuclease, which cut the DNA in
specific regions.
y This allows the isolation of individual
genes.
y Each gene is cut by specific restriction
endonuclease.
Example -ECORI and HIND III .
..
y The specific nucleotide sequence between
which the endonuclease cut the DNA is
called recognition sites.
Some restriction endonucleases can cut the
DNA strands in a way that leaves a few
base pairs longer on one strand than the
other, forming a sticky end.
Sticky ends attach to other compatible sticky
ends.
2. Inserting the gene in to a plasmid (recipient
DNA) to make recombinant DNAy Plasmids are circular loops of extra chromosomal
DNA in bacteria- They are smaller in size and can
replicate with in the cells.
y Before insertion of gene( Donor gene), the
plasmid is cut by the same restriction
endonuclease.
y After insertion, the sticky ends are joined by the
enzyme DNA ligase.
y The donor DNA incorporated into a recipient
DNA is called recombinant DNA.
3. Inserting Recombinant DNA into host
cellsy The host cell with an inserted
recombinant DNA are called transformed
cells or transgenic cells.
y The DNA and genes are large molecule
which do not readily cross cell
membranes. So, it is done by using
vectors.
VECTORS
Vectors play a key role in the formation of recombinant
DNA .
They transfer the required gene, with any marker genes,
into the new cells. A successful vector targets the right
cells, ensures that the desired gene is incorporated in to
the host genetic material so it can be activated
(transcribed and translated) and does not have any
adverse side-effects.
Plasmids are particularly useful as vectors in the
formation of GM bacteria and in the formation of GM
plants. However, other vectors are needed to carry new
DNA into animal cells, especially human cells, and into
some types of plant cell.
..
The recombinant DNA or new genes, can be
introduced into host cells by different methods:
1. Gene guns:
These are used to shoot DNA carried on very
small gold or tungsten pellets (balls) into the cell
at high speed. Some cells survive this treatment
and accept the DNA as part of the genetic
material.
2. By using virus:
Harmless viruses can be engineered to carry a
desirable gene and then used to infect an
animal's cells thus introducing the desirable
DNA.
..
..
3. Liposome wrapping:
It is a technique in which the gene to be inserted
is wrapped in liposomes (spheres formed from a
lipid bilayer). The liposomes fuse with the target
cell membrane and can pass through it to deliver
the DNA into the cytoplasm.
4.Microinjection (DNA injection):
It is a way of introducing DNA by injecting it into a
cell through a very fine micropipette. The method
is not very efficient because many cells have to
be injected before one accepts the DNA
successfully. However, it is the method that has
resulted in most successful transgenic animals.
..
..
Transgenic plants(GM plants)
Many scientists hope that transgenic plants will
become an important weapon in the worldwide fight
against disease. Trials are already in place in the USA
and China. Bananas are emerging as a prime
candidate for genetic modification to carry plant
vaccines, particularly against hepatitis B.
fig : Around 350 million people globally are
already infected with hepatitis B. If
genetically modified bananas could
be grown carrying the hepatitis B vaccine,
it would save many lives and be a great
alternative to injections
MAKING TRANSGENIC PLANTS
The bacterium Agrobacterium tumefaciens is
usually used to introduce genes from one type of
plant into another or even from an animal into a
plant.
A. tumefaciens causes tumours in plants which
are known as crown galls. It contains a plasmid
called the Ti plasmid which transfers bacterial
genetic information directly into the plant DNA.
Refer picture given to identify the process.
..
Transgenic animals(GM animals)
yThe production of proteins using transgenic animals
involves introducing a copy of a human gene which
codes for the desired protein into the genetic material of an
egg of a different animal species. As well as the gene for
the specific protein there is also a promoter sequence
which makes sure the gene will be expressed in the
mammary gland of the lactating female.
yThe fertilised egg, now a developing transgenic embryo,
is then replaced inside a surrogate mother, is born and
grows to maturity. When the animal is mature and
produces milk that milk is harvested, purified and the
human protein extracted.
Transgenesis in animals
..
Microorganisms and human insulin:
..
IDENTIFYING TRANSFORMED ORGANISMS
Usually scientists transfer special marker genes with
the desired DNA so they can identify the
microorganisms in which transformation has taken
place. These marker genes are usually genes that
make a bacterium dependent on a particular nutrient,
or which cause the organism to fluoresce in UV light.
Replica plating method is commonly used to identify
recombinant cells. This involves growing identical
patterns of bacterial colonies on agar plates with
different media. It allows to identify colonies that
cannot survive without a particular nutrient. These
are the bacteria which have been genetically
modified (GM).
..
..
Benefits and risks associated with the
use of genetically modified organisms:
BENEFITS
1. GENETIC MODIFICATION OF CROPS like
devolpment of Flood-resistant rice, Pest
resistance in crops as well as changing the
nutrient values of plants.
™Help to feed the developing world- As GM
crops are resistant to diseases and pests
would increase the yield.
..
™Could helps to preserve natural habitat-
Higher yielding crops reduce the destruction
of more habitats.
™Could produce edible drugs or medicines
in milk, eggs, or fruit.
..
2. GENE THERAPY:
A benefit of gene technology as by this
method it is possible to introduce normal
genes into gametes, zygote, early embryo or
somatic cells to replace faulty genes and
cure genetic disorders.
For example cystic fibrosis might be cured by
inserting healthy CFTR gene into the cells of
the lungs.
..
Risks:
Health risks:
a) Unexpected gene interactions may produce
different proteins that affect normal
metabolism.
b) Possibility of creating new allergens by the new
proteins or by their interaction with usual proteins
Potential threat to the welfare of farmers who wish
to produce non-GM products.
Mainly benefits big biotech companies, as they are
taking patent for the plants or animals they made
by gene technology.
..
MICROARRAYS AND
BIOINFORMATICS
..
MICROARRAYS
Microarray is a laboratory technique used to detect thousands
of active genes at the same time.
USING MICROARRAYS
y When genes are active, they are expressed. Messenger
RNA (mRNA) is produced from active genes and used as a
template for the production of amino acid chains.
y If there is a mutation in a gene, a different form of mRNA
will be produced. However, a gene contains a very large
number of DNA bases, and there are many places where a
mutation can occur.
y A microarray is a tool which scientists use widely to show if
a DNA sample from an individual contains any mutations.
These samples are also known as 'DNA chips'.
..
Basically, a DNA microarray is a slide on which there are
thousands of spots. Each spot is in a specific position and
contains a known DNA sequence. The mRNA samples are
then collected. Scientists usually use a reference sample with
a known gene sequence and an experimental sample.
Typically, this will be from an individual with a particular
disease which may have a genetic element, for example a
form of cancer.
Reverse transcriptase enzymes convert the mRNA into cDNA.
Each sample is given a fluorescent label. Usually, the known
sample is given a green fluorescent label and the
experimental sample is given a red fluorescent label.
The labelled DNA samples are mixed together and applied to
the microarray slide, where they bind to the matching DNA
probes. This process is called hybridisation . After
hybridisation, the microarray is scanned to measure the
fluorescent light produced by the different spots.
..
..
..
..
..
..
If both samples are expressing a gene equally, the
light will appear yellow (a mixture of red and green
Light). If the experimental sample is expressing
more than the control, the spot will appear red.
If the sample is expressing less than the control,
the spot will appear green. We can collect very
large amounts of data using this technology .
The analysis of the data provides detailed
information about gene profiles, the causes of
many diseases and the effectiveness of some
treatments.
..
BIOINFORMATICS
DNA technologies such as microarrays, DNA sequencing and
DNA profiling generate huge amounts of data, far more than
was ever produced in the past during scientific investigations.
Bioinformatics is the development of the software and
computing tools needed to organise and analyse raw
biological data. This includes the development of
algorithms, mathematical models and statistical tests
which can help us interpret the enormous quantities of
data that are generated.
Using bioinformatics, we can process and use the information
generated using rnicroarrays and other forms of DNA
analysis. Bioinformatics enable scientists and doctors to use
the information from these DNA techniques to adapt the care
of individual patients and learn more about the biology of
many different organisms.
..
7KDQN\RX««
..
Observation - Crab burrow diameter increases up shore
(as distance from water line increases)
‡ Collect data by systematic sampling method and record in a
table
Ocypode cursor
‡ Collect data by systematic sampling method and record in a table
‡ Draw a scatter chart to graphically present the data
burrow
diameter
(mm)
13
16
19
18
27
25
29
33
23
29
27
31
34
29
40
Burrow diameter (mm)
Distance
from
waterline
(m)
1.2
2.8
3.1
4.7
6.1
7.2
8.6
9.9
10.1
12.0
14.3
15.1
17.2
17.8
35
30
25
20
15
10
5
0
0
5
10
15
Distance from waterline (m)
20
Prepared by ,
Ashok.V ,Biology Teacher, CHSE.
DATA ANALYSIS
Mean- It is defined as a sum of
measurements divided by the total
number of measurements.
Q1. In an investigation, the number of
aphids found on 14 bean plants where
counted and the results are84, 58, 129, 94, 106, 80, 114, 105, 119, 97,
86. 59, 152 and 141.
Calculate the mean ?
84 58 129 94 106 80 114 105 119 97 86 59 152 141
14
=102 aphids
Median-The median is the middle number
or value, when all the values have been
ranked.
16,22, 24, 6, 9, 15, 12, 14, 19, 22,
27,18
y In the example above, the median is hours.
y When there is an even number of values,
the median is calculated as the mean of
the two middle values
ANS : 17
Mode-The mode is the most common
number or value in a set of observations.
In the example above, --- hours is the mode.
ANS : 22
The Hypothesis(H) and Null Hypothesis
( Ho )
y In an investigation, there will be a
hypothesis (i.e. an intelligent guess or your
prediction) and a corresponding null
hypothesis.
The hypothesis could be stated as:
y ³There will be a significant correlation
between the distance from waterline and
crab burrow diameter´
The null hypothesis ( H0) is:
³There will be no significant
correlation between the distance
from waterline and crab burrow
diameter´
Levels of Significance
y In biology, the 5% (<1 in 20 or p = 0.05) level
of significance is most commonly used.
y This means that a particular event is
predicted to happen by chance less than
once in twenty times.
y By referring to statistical tables, we can
compare our calculated or computed value
with the tabulated value at 5% (or p = 0.05),
known as the critical value or the table
value.
Statistical tests
Correlation
between 2
variables
- Spearman
rank
correlation
coefficient
Comparison
(difference
between 2 data
sets)
- t-test
Comparison of
difference
between observed
and expected
results
- Chi-squared
test (ɝ2)
y In biology, you are likely to encounter three
types of statistical test, as follows.
1. Correlation between two variables
[Spearman¶s rank correlation coefficient]
2. Comparison between two sets of data
[students t-test)
3. Comparison of differences between
observed and expected results [Chisquared test (F2)]
FOR,
SPEARMAN¶S RANK CORRELATION
COEFFICIENT, STUDENTS t-TEST AND CHISQUARED TEST (F2), ± IF THE CALCULATED
VALUE IS LESS THAN THE CRITICAL VALUE,
ACCEPT NULL HYPOTHESIS.
(SSC ± Ca.V<Cr.V, ACCEPT H0)
1. Spearman¶s Rank Correlation coefficient:
y This test is used to determine or check
whether the two variables are correlated or
not.
y If the value of rs (ignoring the +; and -;ve
values) is equal to or greater than the
critical value , we can say that there is a
significant correlation between the two
variables and the null hypothesis is rejected
at 5% significance level.
y If rs value is positive, there is positive
correlation and if it is negative there is a
negative correlation.
y If rs value is zero,no correlation between the
variables.
Degree of freedom = n-1
Where, n= number of measurements.
Working out the meaning of rs value
y If rs value is positive, there is positive
correlation and if it is negative there is a
negative correlation.
y No correlation if the variables do not
correlate.
SAQ:
A student has formed the
hypothesis that mayfly nymphs
has found to be abundant in the
fast flowing waters of streams
when compared to the slow
flowing section of the stream.
He concluded that the greater the speed of
flow of water the greater is the number of
nymphs.
(a) Suggest a null hypothesis for this
investigation.
(b) Complete the table.
(c) Calculate rs value using the formula
(d) What conclusion can you draw from
these investigation?
Site
Flow
rate/m/s
1
0.3
86
2
0.6
46
3
0.5
39
4
0.1
15
5
0.1
41
6
0.4
52
7
0.9
100
8
0.8
63
9
0.2
60
10
0.5
30
11
0.7
72
12
1.1
71
™
Rank
(R1)
Animals/
quadrat
Rank
(R2)
(R1-R2)=
D
D2
Answers:
(a) There will be no significant
correlation between the flow rate
of water and the number of may
fly nymphs found.
(b)
Site
Flow
rate/m/s
Rank
(R1)
Animals/
quadrat
Rank
(R2)
(R1-R2)=
D
1
0.3
4
86
11
-7
49
2
0.6
8
46
5
+3
9
3
0.5
6.5
39
3
+3.5
12.25
4
0.1
1.5
15
1
+0.5
0.25
5
0.1
1.5
41
4
-2.5
6.25
6
0.4
5
52
6
-1
1
7
0.9
11
100
12
-1
1
8
0.8
10
63
8
+2
4
9
0.2
3
60
7
-4
16
10
0.5
6.5
30
2
+4.5
20.25
11
0.7
9
72
10
-1
1
12
1.1
12
71
9
+3
9
0
129
™
D2
Spearmans rank correlation
(c)
rs = 1 = 1
6¦ D
n(n
2
2
1)
6 x129
2
12(12 1)
= 1-0.45 = 0.55.
(d) Since the calculated value is less than the critical
value (0.59) for 12 number of pairs of measurements
at 5% significance level we can accept null
hypothesis.
We can conclude that there is no correlation between
the flow rate of water and the may fly nymphs found.
Correlation coefficient- Practice
question 2
1. An investigation was carried out into the relationship
between length and mass in a species of tuna, caught in
the Pacific Ocean. Twelve specimens were caught and the
length and mass of each fish was recorded. The table
below shows the results of this investigation.
(a) Plot a scatter diagram of the data and describe the
relationship.
(b) Calculate the Spearmanǯs rank correlation
coefficient for the data.
(b) Calculate the Spearmanǯs rank
correlation coefficient for the data.
1
6¦ D
n(n
2
2
1)
ߑd2
=૛૞+૙+૝+૛૞+૝+૝+૝+૚+૚+૙+ૢ+૚=ૠૡ
࢙࢘= ૚ χ૟(ૠૡ)/ ૚૛(૚44χ૚)
= 0.73
(c) What does your calculated value indicate
about the relationship between these two
variables?
Answer
The critical value for rs, at p = 0.05, for 12
pairs of values, is 0.59.
The calculated value for rs (0.73) is greater
than 0.59. We can therefore conclude that
there is a significant positive correlation
between length and mass in this sample of
fish.
Comparison between two sets
of data.
1. Students t-test.
(Ca. V < Cr. V, ACCEPT H0)
2. Students T test:
y This test is carried out to see whether the
means of two samples differ significantly or not.
t=
Where,
= Mean of the sample x.
= Mean of the sample y.
Sx= Standard deviation of sample x,
Sy = Standard deviation of sample y.
nx= No. of individual measurements in sample x.
ny= No. of individual measurements in sample y.
y If the calculated t value is less than the
critical value for appropriate degrees of
freedom accept the null hypothesis.
y Degrees of freedom = (na+nb- 2)
na and nb are the number of samples.
Graph : Bar graph to represent mean
values.
PRACTICE QUESTION- t test
y Dogǯs mercury is a plant that grows in shady areas in
woodland and in open clearings. An investigation was
carried out to determine whether there was a significant
difference between the surface area of the leaves from
dogǯs mercury plants growing in the shaded areas and
open clearings of a wood.
y Seventeen leaves of dogǯs mercury were collected from
plants growing in a shaded area (Site A) and seventeen
leaves were also collected from plants growing in an open
clearing (Site B).
y The surface area of each leaf was measured. The table
shows the results of this investigation.
(a) Calculate the mean surface area of the leaves from Site
B.
Answer
The mean surface area for the leaves from Site B is 15.71
cm2.
(b )State the number of degrees of
freedom for this investigation.
Answer
The number of degrees of
freedom = (17 + 17 Ȃ 2)
= 32
c) (c) A statistical table showed that the critical value
at p = 0.05 is 2.04. Use Studentǯs t-test to determine
whether the difference in mean surface area is
significant.
Answer
y The calculated value for t is 2.96. This value is
greater than the critical value at p = 0.05 (2.96 >
2.04). We can therefore say that there is a
significant difference between the mean surface
area of leaves in shaded area and open clearing.
3. The Chi-squared Test (F2):
y This test is used to compare observed (O)
results with expected (E) results, to
determine whether the difference between
the two is significant.
x
2
¦
(o E )
2
E
y In general, the smaller the value of F2, the
closer the observed results are to the
expected results.
y It is used in the field of genetics.
Degrees of freedom
y In the Chi-squared test, the number
of degrees of freedom is n ± 1, where
n is the number of pairs of observed
and expected values.
Graph : Bar graph
Phenotype
Observed numbers (O)
Expected numbers (E)
OȂE
(O Ȃ E)2
(O Ȃ E)2 ÷ E
ϝ
Chi square test- Practice question
A test cross was carried out between two maize plants, one
with yellow, smooth cobs and the other with white,
wrinkled cobs. The table shows the results of this cross.
Use a Chi-squared test to investigate whether these results
fit the expected 1 : 1 : 1 : 1 ratio.
Phenotype
yellow
and
smooth
Observed numbers (O) 28
Expected numbers (E)
OȂE
(O Ȃ E)2
(O Ȃ E)2 ÷ E
ϝ
yellow
and
wrinkled
white
and
smooth
white and
wrinkled
33
30
34
¼ x 125
The calculated value for Chi-squared
(0.73) is less than the critical value
(7.82) at p = 0.05.
We can therefore say that these results
do fit the expected 1 : 1 : 1 : 1 ratio.
Further questions
Question 1
Ultraviolet light (UV) can be used to kill microorganisms. A
student carried out an investigation into the effect of UV
light on the survival of bacteria.
The student spread bacteria evenly on three agar plates.
Each plate was then exposed to UV light for one minute.
The student then placed the agar plates in an incubator.
The procedure was repeated for exposure times of 2, 3, 4
and 5 minutes.
After 48 hours, the number of bacterial colonies on each
plate were recorded.
The results are shown below.
(a) Write a suitable null hypothesis for this
investigation.
There is no significant correlation
between time of exposure to UV light
and the mean number of bacterial
colonies.
b Complete the following table, by calculating the
mean number of bacterial colonies for each UV
exposure time.
c Plot a scatter diagram to show the relationship
between the mean number of bacterial colonies and
the exposure time.
(d) The student used a statistical test to investigate the
significance of the correlation between the number
of bacterial colonies and exposure time to UV light.
y A correlation coefficient of - 0.99 was found.
y The following table shows the critical values, for p = 0.05,
for this statistical test.
What conclusion can be drawn from the results of this
investigation?
y The calculated value of the correlation coefficient
(0.99) is greater than the critical value (0.90) at p =
0.05 . So, reject the null hypothesis.
y The negative sign indicates an inverse
relationship. We can therefore conclude that there
is a significant inverse relationship between the
time of exposure and the mean number of
bacterial colonies.
y The graph shows an inverse relationship (or
negative correlation) between the time of
exposure to UV light and the mean number of
bacterial colonies.
Question 2
A group of nine athletes (A to I) wanted to
see if training for two weeks at a
mountain camp, 2000 m above sea level,
had an effect on the number of red blood
cells in their blood. Samples of blood
were taken from each of the athletes at
their normal training camp at sea level.
Blood samples were taken again after two
weeks of training at the mountain camp.
The following table shows the results.
a Complete the table by calculating the mean numbers
of red blood cells × 1012 per dm3 of blood for the
athletes before and after mountain training.
b Calculate the percentage increase in the
number of red blood cells after mountain
training.
Answer
c Write a null hypothesis for this investigation.
There is no significant difference between
the mean number of red blood cells before
and after mountain training.
ȋ†Ȍ–—†‡–ǯ•–-test was applied to the data to
test the null hypothesis.
(i) State the number of degrees of freedom for
these results.
degrees of freedom = N1 + N2 Ȃ 2
(9 + 9) Ȃ 2 = 16.
ii The calculated value of t was found to be 2.24.
The following table gives critical values of t, for the
appropriate number of degrees of freedom.
What conclusion can be drawn from the results this
investigation?
The calculated value of t (2.24) is greater than the
critical value of t at p = 0.05 (2.12). We can therefore
conclude that there is a significant difference
between the mean numbers of red blood cells
before and after mountain training and reject the
null hypothesis.
Question 3
Global warming can affect abiotic factors that determine
the distribution of organisms.
The presence of sodium chloride in soil is an abiotic factor
that affects the germination of seeds.
The effects of sodium chloride solution and gibberellin on
the germination of rice seeds have been investigated.
Gibberellin regulates developmental processes in plants.
Fifty seeds were placed in each of three Petri dishes
containing different solutions.
The seeds were incubated for 96 hours and the number that
germinated in each
Petri dish was counted.
(a) Give a null hypothesis for this
experiment. (1)
There will be no significant
difference between treatment of
seeds with sodium chloride or
sodium chloride and gibberellin on
the number of seeds that
germinate (1)
(b) (i) Calculate the chi-squared (࣑૛) value for these results,
using the formula provided.
Treatment
Control
Sodium
chloride
Sodium
chloride and
gibberellin
™
OBSERVED
NUMBER
(O)
EXPECTED
NUMBER
(E)
(O ² E)2
(O ² E)2
/ E
(b) (i) Calculate the chi-squared (࣑૛) value for these results,
using the formula provided.
Treatment
OBSERVED
NUMBER
(O)
48
EXPECTED
NUMBER
(E)
50
(O ² E)2
4
0.08
Sodium
chloride
33
50
289
5.78
Sodium
chloride and
gibberellin
45
50
25
0.5
Control
(O ² E)2
/ E
6.36
™
(ii) In a second experiment, using the same three
treatments, the chi-squared (߯2) value was found to be
6.635.
The table gives the critical values for the chi-squared (߯2)
test at different probability levels.
Deduce the statistical significance of
the results of the second experiment.
(2)
The calculated value 6.635 is greater
than critical value 5.991 at p = 0.05,
at 2 degrees of freedom ;
Ȉ‘ǡ ƒŽ —Žƒ–‡†˜ƒŽ—‡‹••‹‰‹ˆ‹ ƒ–Ǣ
THANK YOU
Unit 6: Practical Skills in Biology II (WBI16/01)
Duration: 1 hour and 20 minutes
Total marks: 50
3 or 4 questions
‡ Practical procedure related
‡ Statistical analysis
‡ Planning an investigation
Questions may include ones that assess you on:
‡ What are you controlling? INDEPENDENT VARIABLE
‡ How are you controlling it? Do you have a control to compare it against?
‡ What are you measuring? DEPENDENT VARIABLE
‡ How are you measuring it?
‡ Some factors to control
‡ Most obvious factors to control.
‡ How to control and effect of not controlling factor on results
‡ Any issues to procedure
‡ Safety considerations
‡ Ethical considerations
‡ Data organisation, presentation, analysis and interpretation
‡ Draw tables & graphs
‡ Calculate mean, median, mode, SD, range
‡ Formulate H0, calculate test statistic (rs, t, ʖ2), derive conclusion using
critical value table
‡ Testing biological knowledge of a concept
AS ʹ year 11 (U1 & U2)
A2 ʹ year 12 (U4 & U5)
CP1: Test for reducing sugars and starch
CP2: Vitamin C content of foods
CP3: Membrane permeability
CP4: Factors affecting rate of enzymatic reactions
RP: Dissection of the heart
RP: Test for proteins
RP: Investigating tissue water potential
CP10: Rate of photosynthesis using
photosynthometer
CP11: Ecological sampling
CP12: Brine shrimp hatching
CP13: Measuring microbial growth
CP14: Effect of antibiotics
CP5: Animal cell observation, use of graticule
CP6: Root tip squash - mitosis
CP7: Plant tissue (root, stem, leaves) observation
CP8: Tensile strength of plant fibres
CP9: Antimicrobial properties of plant extracts
RP: Pollen tube growth
RP: Plant mineral deficiency
CP15: Yeast respiration rate using a redox indicator
CP16: Respirometer
CP17: Spirometry
CP18: Seed germination using gibberellin
RP: Habituation
Type 1
Similar to question 1 in unit 3 paper
™ Experimental procedure (5 ʹ 6 marks)
™ Some factors to control (2 marks)
™ How to control and effect of not
controlling factor on results (2 marks)
™ Testing biological knowledge of
concept (3 ʹ 4 marks)
™ Calculations (3 ʹ 4 marks)
Type 2 ʹ Statistical Analysis
™ State null hypothesis (1 mark)
™ Present raw data in the form of tables
after calculating means (3 marks)
™ Plot a graph (3 marks)
™ Calculation on the given statistical
test (3 marks)
™ Draw conclusions based on statistical
analysis (2 marks)
™ Suggest improvements (2 marks)
Age of
Leaves
(days)
Time taken for
solution to become Rate of Reaction
colourless (mins)
(min-1)
T1 T2 T3 Mean
Type 3 ʹ Plan an Investigation
(a) Safety and/ ethical issues
™Question specific content
™Description of sampling technique
™Why a protein solution becomes clear when protease is added
™Why the colour of solution change
b) Preliminary work
c) Detailed description of method
d) Results recording, presentation and analysing
e) Limitations of proposed method
USING CORMS TO IDENTIFY ALL POINTS
Change/Compare
What factor are you investigating? Will you have a range of
/Control
values? (always a minimum of 5 values) Or will you have
(independent variable) two groups, one with the factor and one without? What is
your control? How to change it?
Organism
What species/size/age/gender will you use? Note:planning
(other biological
an investigation into enzyme activity, identify the enzyme
material)
and the substrate.
Reliability
You MUST take more than one reading- you should take a
Repetition/Replication minimum of 3 readings and calculate a mean for each group
of the independent variable.
Measurement
What will you be measuring, how often and what are the
(of dependent
units? describe how you will take the measurement, and
variable)
with what equipment.
Same/Standardisation what factors to keep the same and how to do this, to
ensure you have carried out a fair test? (identify
.
at least 2)
e.g. temperature/light intensity/volume of water etc.
Marking using CORMS
A student observed that vegetables are often cooked and then kept warm
for a period of time before being eaten. The student formulated the
following hypothesis: The longer the vegetables are kept warm after
cooking the lower their vitamin C concentration.
‡ Plan an investigation to test this hypothesis, using one named vegetable.
‡ A detailed method, including an explanation of how important variables
are to be controlled or monitored.
1. correct experimental sequence involving cooking vegetables, keeping them warm
then blending or crushing ; ACCEPT extracting juice
2. clear statement of independent variable as length of time food is stored (after
cooking) ; ACCEPT time left after cooking
3. at least 5 times stated ;
4. clear statement of dependent variable as vitamin C concentration ; . IGNORE
amount of vitamin C
5. clear description of method of measuring dependent variable as titration using
redox indicator including correct colour change ; . NOT use of cooking water alone
6. idea of recording volume of { DCPIP / vegetable extract } needed (to reach the
end point) ; ACCEPT number of drops
7. identification of two variables that could affect the result obtained ; IGNORE room
temperature e.g. DCPIP concentration, { cooking / storage } conditions, { source /
preparation } of vegetable
8. and 9. description of methods by which each of the variables can be controlled ;;
10. clear reference to need for repeats ;
11. idea of calibration of vitamin C assay using solution of known vitamin C
concentration ;
Response 1
Response 2
Practice question
The photograph shows one species of the genus Cabomba, an aquatic plant.
Cabomba grows in streams and ponds in many parts of the world.
It adds oxygen to the water.
A student observed that there were fewer Cabomba plants growing in
the shaded parts of a pond.
The student formed the following hypothesis.
The greater the light intensity the faster the rate of photosynthesis in
Cabomba plants.
Plan an investigation to test this hypothesis.
(a) State one safety issue you would need to take into
account. (1)
Practice question
(a) State one safety issue you would need to take into
account. (1)
ͻŽŶĞƐĞŶƐŝďůĞƌŝƐŬŝĚĞŶƚŝĨŝĞĚ
eg cuts/burns/allergies/infections/electrocution
‡Avoid skin contact with sodium hydrogen carbonate and be
careful not to raise or inhale any dust from it as it is an irritant or
can cause allergic reactions
‡Take care with use of water near electrical equipment. Ensure
hands are dry when using lamps to minimize risk of
electrocution.
Also, keep all containers of water away from sockets and from
areas where spills could drip into sockets. Boiling tube racks
could be placed into trays to catch any spills and prevent water
running onto electrical equipment.
Practice question
The photograph shows one species of the genus Cabomba, an aquatic plant.
Cabomba grows in streams and ponds in many parts of the world.
It adds oxygen to the water.
A student observed that there were fewer Cabomba plants growing in
the shaded parts of a pond.
The student formed the following hypothesis.
The greater the light intensity the faster the rate of photosynthesis in
Cabomba plants.
Plan an investigation to test this hypothesis.
(b) Describe the preliminary practical work that you
might undertake to ensure your proposed method
would provide meaningful data. (3)
Preliminary Work
‡ A pilot experiment
‡ Can establish the effectiveness of a particular protocol
‡ Determine what needs to be measured as the dependent variable and how to
measure
‡ Range of concentrations / Temperatures
‡ Determine other variables to control and how to control
Practice question
(b) Describe the preliminary practical work that you might undertake to
ensure your proposed method would provide meaningful data. (3)
A description that includes the following points:
ͻĨŝŶĚĂƐƵŝƚĂďůĞŵĞƚŚŽĚĨŽƌŵĞĂƐƵƌŝŶŐƚŚĞƉƌŽĚƵĐƚŝŽŶŽĨŽdžLJŐĞŶ;ϭͿ
IGNORE practise the method. IGNORE rate of photosynthesis / counting
oxygen bubbles
ͻŽǀĞƌĂƐƵŝƚĂďůĞƚŝŵĞƉĞƌŝŽĚ;ϭͿ>>KtƐƚĂƚĞĚŶƵŵďĞƌŽĨŵŝŶƵƚĞƐͬ
hours e.g. 10 minutes to 8 hours
ͻĨŝŶĚĂƐƵŝƚĂďůĞŵĞƚŚŽĚĨŽƌǀĂƌLJŝŶŐƚŚĞůŝŐŚƚŝŶƚĞŶƐŝƚLJ;ϭͿ
Determine a suitable method for measuring the production of oxygen
Determine a suitable time period for collection of oxygen gas to
calculate the rate of photosynthesis
Find a suitable method for varying the light intensities, e.g. different
distances of light source or use different light bulbs in order to subject
the plant to different light intensities
Practice question
The photograph shows one species of the genus Cabomba, an aquatic plant.
Cabomba grows in streams and ponds in many parts of the world.
It adds oxygen to the water.
A student observed that there were fewer Cabomba plants growing in
the shaded parts of a pond.
The student formed the following hypothesis.
The greater the light intensity the faster the rate of photosynthesis in
Cabomba plants.
Plan an investigation to test this hypothesis.
(c) Devise a detailed method, including an
explanation of how you would control and monitor
important variables. (10)
Detailed Investigation
‡ Identify proper dependent & independent variable ; ;
‡ Method to measure dependent variable ;
‡ Control ;
‡ Range (if applicable minimum 5 values needed with units) ;
‡ Identify 2 confounding variables to be controlled ;
‡ Ways to control these variables ; ;
‡ Details of how you are going to carry out the investigation ;
‡ Repetition and calculating mean ;
Detailed Investigation
(using ruler)
(using bulbs of different brightness/more bulbs)
Luminous (light) intensity/lux
Brightness of bulb/ lumens
IV? Distance from beaker /cm
DV? Volume of oxygen produced per minute (using photosynthometer and stopwatch, then calculate rate)
Other variables to control? Temperature ʹ using a thermostatic water bath set at 30oC
Length of Cabomba plant ʹ cut 10cm pondweed using ruler and scalpel/scissors
How to control?
CO2 concentration ʹ use same stock solution of NaHCO3, or measure same mass (1g) of NaHCO3 and
dissolve in same volume (1L) of water to use every time
Practice question
(c) Devise a detailed method, including an explanation of how you would control
and monitor important variables. (10)
An answer that includes ten of the following points:
ͻĐůĞĂƌƐƚĂƚĞŵĞŶƚŽĨƚŚĞĚĞƉĞŶĚĞŶƚǀĂƌŝĂďůĞ;ϭͿ>>Kt΂ǀŽůƵŵĞͬĐŽŶĐĞŶƚƌĂƚŝŽŶ΃
of oxygen. IGNORE counting oxygen bubbles
ͻĚĞƐĐƌŝƉƚŝŽŶŽĨŵĞƚŚŽĚƚŽƉƌŽǀŝĚĞϱĚŝĨĨĞƌĞŶƚŝŶƚĞŶƐŝƚŝĞƐŽĨůŝŐŚƚ;ϭͿ
ͻƉůĂŶƚŐŝǀĞŶƚŝŵĞƚŽacclimatise (1) ALLOW equilibrate
ͻĚĞƐĐƌŝƉƚŝŽŶŽĨŵĞƚŚŽĚĨŽƌŵĞĂƐƵƌŝŶŐŐĂƐǀŽůƵŵĞ;ϭͿ/'EKZĐŽƵŶƚŝŶŐďƵďďůĞƐ
ͻŵĞƚŚŽĚŽĨĐĂůĐƵůĂƚŝŶŐƌĂƚĞŽĨŐĂƐƌĞůĞĂƐĞĚ;ϭͿ>>KtŶƵŵďĞƌŽĨďƵďďůĞƐƉĞƌ
unit time
ͻƚŚĞĐĂƌďŽŶĚŝŽdžŝĚĞŵƵƐƚďĞŝŶĞdžĐĞƐƐ;ϭͿ>>KtƵƐĞŽĨƐŽĚŝƵŵŚLJĚƌŽŐĞŶ
carbonate
ͻǀĂƌŝĂďůĞƐƚŚĂƚŶĞĞĚƚŽďĞƚĂŬĞŶŝŶƚŽĂĐĐŽƵŶƚ;ϮͿĞ͘Ő͘ƚĞŵƉĞƌĂƚƵƌĞ͕Ɖ,͕
background light
ͻĚĞƐĐƌŝƉƚŝŽŶŽĨŚŽǁƚŚĞƐĞǀĂƌŝĂďůĞƐĂƌĞ΂ĐŽŶƚƌŽůůĞĚͬŵĞĂƐƵƌĞĚ΃;ϮͿĞ͘Ő͘ĐŽŶƚƌŽůůĞĚ
room temperature or waterbath. buffer or pH meter, blackout other light sources
ͻƌĞƉĞĂƚƐĨŽƌĞĂĐŚůŝŐŚƚŝŶƚĞŶƐŝƚLJ;ϭͿ
Practice question
(c) Devise a detailed method, including an explanation of how you would control and monitor important variables. (10)
Set up apparatus as shown.
1.Independent variable is the distance of light source from beaker containing plant. Place the light source at 10, 20, 30, 40 and
50 cm from beaker;
2. Dependent variable is the volume of oxygen gas produced;
3. How to measure dependent variable ;
‡ As bubbles of oxygen begin to form and pass into the capillary tubing, start the stop clock. After a suitable time (15
minutes), use the syringe to draw any oxygen produced into the capillary tubing. Record the volume of gas produced.
4. Allow the pondweed to adjust for 5 minutes before starting ;
5 and 6. Confounding variables;;
‡ Temperature, background light, length of plant, CO2 concentration
7 and 8. How to control these variables ;;
‡ Temperature ʹ using a thermostatic water bath set at 30oC
‡ Length of pondweed ʹ cut 10cm pondweed using ruler and scalpel/scissors
‡ CO2 concentration ʹ use same stock solution of NaHCO3, or measure same mass (1g) of NaHCO3 and dissolve in same
volume (1L) of water to use every time
‡ Background light ʹ cover one side of the beaker with aluminium foil, so that light can only enter the beaker from the other
side
9. Repeating the investigation 3 times at each distance from the beaker and calculate mean volume of oxygen released at each
distance.
10. Determine mean rate of photosynthesis by dividing volume of gas collected by time;
Practice question
The photograph shows one species of the genus Cabomba, an aquatic plant.
Cabomba grows in streams and ponds in many parts of the world.
It adds oxygen to the water.
A student observed that there were fewer Cabomba plants growing in
the shaded parts of a pond.
The student formed the following hypothesis.
The greater the light intensity the faster the rate of photosynthesis in
Cabomba plants.
Plan an investigation to test this hypothesis.
(d) Describe how your results should be recorded,
presented and analysed in order to draw conclusions
from your investigation. (4)
Record, Present & Analyse
1. Table with headings ;
2. Means calculated from repeats ;
3. Appropriate graph format with labelled axes ;
4. Use of an appropriate statistical test ;
Practice question
(d) Describe how your results should be recorded, presented and analysed in order
to draw conclusions from your investigation. (4)
An answer that includes the following points:ALLOW
concentration of oxygen
ͻƚĂďůĞǁŝƚŚŚĞĂĚŝŶŐƐ;ϭͿhŶŝƚƐŽŶůLJŶĞĞĚĞĚŽŶĐĞ
from table or graph
ͻŵĞĂŶƐĐĂůĐƵůĂƚĞĚĨƌŽŵƌĞƉĞĂƚƐ;ϭͿ
ͻůŝŶĞŐƌĂƉŚĨŽƌŵĂƚǁŝƚŚůĂďĞůůĞĚĂdžĞƐ;ϭͿ>>Kt
sketch graph either rising or falling depending on axis
labels
ͻƵƐĞŽĨĂŶĂƉƉƌŽƉƌŝĂƚĞĐŽƌƌĞůĂƚŝŽŶƐƚĂƚŝƐƚŝĐĂůƚĞƐƚ;ϭͿ
e.g. Pearsons͕^ƉĞĂƌŵĂŶ͛ƐZĂŶŬ
Use Spearman rank correlation test to see if there is a
significant relationship between the distance of lamp
from the beaker and the mean rate of photosynthesis.
Distance from lamp
to beaker/ cm
10
20
30
40
50
Rate of photosynthesis/ mm3min-1
T1
T2
T3
mean
Practice question
The photograph shows one species of the genus Cabomba, an aquatic plant.
Cabomba grows in streams and ponds in many parts of the world.
It adds oxygen to the water.
A student observed that there were fewer Cabomba plants growing in
the shaded parts of a pond.
The student formed the following hypothesis.
The greater the light intensity the faster the rate of photosynthesis in
Cabomba plants.
Plan an investigation to test this hypothesis.
(e) Suggest three limitations of your proposed
method. (3)
Practice question
(e) Suggest three limitations of your proposed method. (3)
An answer that includes three of the following points:
ͻĚŝĨĨŝĐƵůƚƚŽĐŽŶƚƌŽůĂůůǀĂƌŝĂďůĞƐĂĨĨĞĐƚŝŶŐ΂ŐƌŽǁƚŚͬƉŚŽƚŽƐLJŶƚŚĞƐŝƐ΃ŽĨƉůĂŶƚƐ;ϭͿ
ALLOW biotic/ abiotic variables
ͻŝĚĞĂŽĨĚŝĨĨŝĐƵůƚLJŽĨĐŽŶƚƌŽůůŝŶŐĂŶĂŵĞĚǀĂƌŝĂďůĞ;ϭͿĞ͘Ő͘ƚĞŵƉĞƌĂƚƵƌĞŽƌƉ,
ͻŝĚĞĂƚŚĂƚƐƵƌĨĂĐĞĂƌĞĂĐĂŶŶŽƚďĞĐŽŶƚƌŽůůĞĚͬŵĞĂƐƵƌĞĚ;ϭͿ>>KtƐŝnjĞŽĨůĞĂǀĞƐ
ͻƉŽƐƐŝďůĞĞƌƌŽƌƐŝŶŵĞĂƐƵƌŝŶŐǀŽůƵŵĞŽĨŐĂƐƌĞůĞĂƐĞĚ;ϭͿ>>KtƐŝnjĞŽĨďƵďďůĞ
may vary
Not counting bubbles unqualified
1. It is difficult to control all the variables affecting photosynthesis of plants.
2. It is difficult to control temperature of the water bath as light source can
increase the temperature.
3. Although the same length of the plant is used, the size of the leaves may vary
which can affect the rate of photosynthesis.
4. We assume that the gas bubbles released contains only oxygen gas. But there
could be a small amount of carbon dioxide in the bubbles from respiration.
Nitrogen may also come out of the solution in the water. At the same time,
some of the oxygen produced will be used internally in respiration.
7+$1.<28«