Transcript Observation


Solutions
Solution = Solvent + Solute
According to the size of solute particles
there are three types of the solutions:
1- true solution
2- Colloidal solution
3- Suspension and emulsion solution
True solution
Dispersed
particles are
molecules or ions.
Colloidal
solution
Suspension &
Emulsions
Dispersed
particles are
aggregate of
molecules.
Dispersed
particles are
aggregate of
molecules.
Diameter of
Diameter of
Diameter of
particles < 0.001μ particles between particles > 0.1 μ.
0.001μ and 0.1 μ.
Pass through
permeable and
semi-permeable
membrane.
Pass through
permeable
membrane only.
Cannot pass
through any
membranes.
Examples:
NaCl in H2O
Sucrose in H2O
Examples:
Starch in H2O
Protein in H2O
Examples:
Sand in H2O
Oil in H2O
Cellophane sac
5ml NaCl + 5ml starch
Distilled water (dist. H2O)
Dialysis
1
After 20 mins
take 2 ml from
the external
solution then
2
 Observation:
In tube no. 1 white ppt. formed after addition of
AgNO3 solution.
In tube no. 2 there is no change after addition of
I2 solution.
 Comment:
1- Sodium chloride (NaCl) is a true solution
which has small particles size so it can pass
through cellophane membrane (semipermeable
membrane) and after addition of silver nitrate
(AgNO3) this reaction occurs:
AgNO3 + NaCl
NaNO3 + AgCl
white ppt.
2- Starch is a colloidal solution which has
large particle size than the true one so it cannot
pass
through
cellophane
membrane
(semipermeable membrane), so after addition of
iodine solution (I2) no reaction occurs.
 Electrical
adsorption
The particles of colloidal solutions carry
electrical charges on its surface which may be
positive
or
negative
and
every
particles
surrounded by opposite charge to form electrical
double layer. Every solution its particles has the
same charge.
+
+
+
Colloidal
particle
negative
charge
+
+
+
+ +
positive
charge
Filter paper
stripe
Mixture of
M.B. & L.G.
Methylene
Light green
blue dye
dye
leave for
15 mins
 Observation:
1- In case of M.B., the particles of dye
accumulated on the end of the filter paper
stripe, while water move vertically through the
stripe.
2- In case of L.G., the levels of water and dye
are the same, whereas the particles of L.G.
raised through filter paper stripe.
 Comment:
Witted filter paper stripe posses -ve charge, so in
case of M.B. dye, attraction force occur
between the particles and the stripe in contact
region as M.B. particles posses +ve charge.
While in case of L.G. repulsion force occurs
between the particles and the stripe as L.G.
particles posses -ve charge .
I2 solution
Gelatin + Starch
Starch
Gelatin + I2 solution
Leave it
1
for 48 hours
Leave it
for 48 hours
2
 Observation:
In tube no. 1 blue ring formed on the top and
increases gradually downwards.
In tube no. 2 there is no change.
 Comment:
1- Iodine solution (I2) is a true solution which
have small particle size so it can pass through
gelatin membrane so it can react with starch
forming the blue zone:
2- Starch is a colloidal solution which have
large particle size than the true one so it cannot
pass through gelatin membrane and cannot react
with iodine.
FeCl3
NaOH + ph. ph. +
K4[Fe(CN)6] + gelatin
Blue zone
Leave it
for 48 hours
Colourless
zone
 Observation:
Formation of colorless zone followed by blue
zone through gelatin layer.
Comment:
FeCl3
ionization
3 Cl - +
Large in number
and small in
size & charge
React with
NaOH
Fe 3+
small in number
& Large in size
& charge
React with
K4[Fe(CN)6]
3 NaCl
Fe4[Fe(CN)6]3
Colourless zone
Blue zone
Chloride ions are large in number and small
in size so it pass through the gelatin membrane
faster and react with NaOH as follow:
3 Cl- + 3 NaOH
3 NaCl
Sodium chloride (NaCl) is a neutral solution
so in the presence of ph.ph. as indicator it form
the colourless zone
while, ferric ions are small in number and
large in size so it pass through the gelatin
membrane slowly and react with potassium Ferro
cyanide as follow :
4Fe3+ + 3 K4[Fe(CN)6]
Fe4[Fe(CN)6]3
Prussian blue
Ferric Ferro cyanide (Prussian blue) make
the blue zone
 Permeability
It is a property of membranes posses
which mean the capacity of controlling the
exit and entrance of various substances .
Plasma membrane
The plasma membrane consists of two
phospholipids layers contain protein .
Beet discs
pH values
2
4
6
8
Record the time for
appearing the pigment
2
4
6
8
Draw the relation between
pH values and time
2
4
6
8

Observation:
Red color formed in tubes having pH values 2,4
and 6.
And yellow color formed in tube which have pH
value 8.
 Comment:
The
plasma
membrane
consists
of
two
phospholipids layers bordered by protein layer.
Variation in hydrogen ion concentration (pH)
affect on protein layers of membrane as protein a
polypeptide which consists of amino acids.
Amino acids are amphoteric , so it may react on
alkaline or acidic medium as follow
in acidic
medium
R
R
NH3+
CH
COOH
NH2
CH
COOH
Amino acid
in alkaline
medium
R
NH2
CH
COO
In both cases denaturation occurs to protein
which cause plasma membrane destruction. So,
plasmamembrane lose the control on exit and
entrance
of
substances
which
lead
to
exit
anthocyanin pigment easily.
Anthocyanin pigment act as indicator which
have red color in acidic medium and yellow color in
alkaline medium
Beet discs
Alcohol
concentrations
30%
50%
70%
Record the time for
appearing the pigment
30%
50%
70%
Draw the relation between
alcohol concentrations and time
30%
50%
70%

Observation:
Red color formed in all tubes 30%, 50% and
70%.
 Comment:
The plasma membrane consists of two
phospholipids layers contain protein .
Alcohol is an organic solvent which effect
on the phospholipid layer (fats) this cause
plasma- membrane destruction.
So, plasmamembrane lose the control on
exit and entrance of substances (selective
permeability) which lead to exit anthocyanine
pigment easily in the medium. The time needed
for anthocyanine appearance is reversely
proportional with alcohol concentration.
 Enzymes
Enzymes are biomolecules that catalyze
(increase the rate of) chemical reactions. Almost
all enzymes are proteins. In enzymatic reactions,
the molecules at the beginning of the process
are called substrates, and the enzyme converts
them into different molecules, the products.
Almost all processes in a biological cell need
enzymes to occur at significant rates. Enzymes
are selective for their substrates.
Enzyme activity can be affected by other
molecules.
decrease
Inhibitors
enzyme
are
activity;
molecules
activators
molecules that increase enzyme activity.
that
are
Activity is also affected by temperature, chemical
environment (e.g. pH), and the concentration of
substrate.
IEC Classification of Enzymes
Group Name
Type of Reaction Catalyzed
Oxidases or
Dehydrogenases
Oxidation-reduction
reactions
Transfer of functional
groups
Transferases
Hydrolases
Hydrolysis reactions
Lyases
Addition to double bonds or
its reverse
Isomerases
Isomerization reactions
Ligases or
Synthetases
Formation of bonds with
ATP cleavage
2 ml sucrase
(enzyme)
3 ml sucrose
(substrate)
Add 5ml fehling
(A:B) soln.
Water bath at
Direct flame
37± 2° C for
15 mins
Reddish
brown ppt.

 Observation:
Reddish brown ppt. was formed.
 Comment:
Invertase is a hydrolytic enzyme, at
suitable condition it converts sucrose
(disaccharide) into glucose and fructose
(monosaccharides).
C12H22O11 + H2O
Invertase
37 ± 2°C
C6H12O6 + C6H12O6
Glucose
fructose
After addition of fehling (A:B) solution to the
monosaccharides, formed before, reddish brown
ppt. is formed as a result of the following
equation:
COOH
CHO
(CHOH)4 or
CH2OH
(CHOH)4
CH2OH
C
O
+ CuSO4
(CHOH)3
CH2OH
CH2OH
Gluconic
acid
+
Cu2O
cuprous oxide
Red brown ppt.
Paraffin oil
Put 2 ml of glucose
then add 1 ml of
methylene blue dye
5 ml non- 5 ml boiled
boiled yeast
yeast
Water bath at
37± 2° C for
10 mins
Non-boiled
yeast
Boiled
yeast
 Observation:
The blue colour disappeared in the
tube which contain non-boiled yeast,
while the blue colour remain in the
tube which contain boiled yeast.
 Comment:
In
non-boiled
yeast,
glucose
dehydrogenase enzyme is in active
form. So, in suitable conditions this
reaction occur :
C6H12O6 + H2O + M.B.
37±2°C
C6H12O7 + M.B. H2
Glucose
dehydrogenase
The boiled yeast contains enzyme
in inactive form. So, the oxidation –
reduction between M.B. and glucose
not occur.
Catechol
atmospheric
oxygen
Potato disc
Brown colour
 Observation:
A deep brown colour formed on
the surface of potato disc.
 Comment:
Due to the presence of oxidase
enzyme which work as an oxidizing
agent,
reaction
it
catalyze
in
the
atmospheric oxygen:
the
following
presence
of
O
OH
2
+ O2
OH
Catechol
“ Phenol form”
Colourless
Oxidase
2
+ 2 H2O
O
Catechol
“ Quinone form”
Brown colour
Catechol
H 2O 2
Radish disc
Brown colour
 Observation:
A deep brown colour formed on
the surface of radish disc after
addition of hydrogen peroxide H2O2 .
 Comment:
Radish tissues contain peroxidase
enzyme which oxidize the catechol in
the presence of H2O2 (as oxygen
donor) as follow:
2 H2O2
2 H2O + O2
O
OH
2
+ O2
OH
Catechol
“ Phenol form”
Colourless
peroxidase
2
+ 2 H2O
O
Catechol
“ Quinone form”
Brown colour
1
H 2O 2
2
5 ml non5 ml boiled
boiled yeast
yeast
1
2

 Observation:
In tube no. 1, effervescence occur
and gaseous bubbles evolved.
In tube no. 2, there is no change
 Comment:
Any living tissue contain catalase
enzyme
which
catalyze
the
breakdown of toxic H2O2 into H2O and
O2 as follow:
2 H2O2
2 H2O + O2
This reaction did not occur in nonliving tissues due to absence of
catalase enzyme in active form.
Photosynthesis
is
a
metabolic
pathway that converts light energy
into
chemical
energy.
Its
initial
substrates are carbon dioxide and
water; the energy source is sunlight
(electromagnetic radiation).
The end-products are oxygen and
carbohydrates,
such
as
sucrose,
glucose or starch.
This process is one of the most
important biochemical pathways, since
nearly all life on earth either directly or
indirectly depends on it as a source of
energy.
It is a complex process occurring
in plants, algae, as well as bacteria
such as cyanobacteria.
Photosynthetic organisms are
also referred to as photoautotrophs.
Equation for
photosynthesis
water
6CO2 + 6H2O + Energy
Carbon
dioxide
Chlorophyll
Oxygen
C6H12O6 + 6O2
Glucose
CO2
Phenol red
In dark
Leave one of them
in light and the
other one in dark
In light
Spirogyra
Observation:
The red color reappeared in tube
which present in light, while the
yellow colour remained in tube which
present in dark.
Comment:
Phenol red has a red color in
neutral medium and yellow color in
acidic medium.
So, in the first step, phenol red
change to yellow color when the
medium became acidic due to reaction
between CO2 and water as follow:
CO2 + H2O
In
tube
H2CO3
present
in
light
CO2
consumed in photosynthesis due to
presence of all suitable conditions.
So, the solution became neutral again
and
red
color
following equation:
reappear
as
the
6CO2 + 2 H2O
In
Chlorophyll
light
tube
C6H12O6 + 6H2O + 6O2
present
in
dark
photosynthesis not occur due to the
absence
of
light.
So,
CO2
not
consumed and the medium remain
acidic with yellow color.
Non green
region
Put in alcohol and
transfer it to water
bath (70° C)
Green region
Duranta leaf
Add I2
solution
 Observation:
The green region of leaf became
blue while non green region remain
as it is.
 Comment:
In the green region, all conditions
necessary for photosynthesis are
available . So, the process occurs
and starch formed.
6CO2 + 2 H2O
Chlorophyll
light
C6H12O6 + 6H2O + 6O2
After addition of I2 solution a blue
colour
was
formed
due
to
the
reaction with starch.
In non green region, there is no
chlorophyll
present,
so,
the
photosynthesis process not occur
and there is no starch formed.
Light
Leave 24 hour
Elodea
(hydrophyte)
Water+ NaHCO3
 Observation:
Formation
of
air
bubbles
and
decreasing in water level in test tube
 Comment:
all
condition
photosynthesis
are
necessary
available.
for
So,
photosynthesis process occurs and
O2 evolved causing formation of
gaseous
bubbles
which
lead
to
decrease in the level of water at the
base of the test tube.
6CO2 + 2 H2O
Chlorophyll
light
C6H12O6 + 6H2O + 6O2