Unit 1
1. Discuss the following unique properties of water molecules:
● configuration ( structure)
Water consists of an oxygen atom covalently bonded to two hydrogen atoms
● Polarity
The oxygen atom is strongly electronegative, which means that it has a
tendency to attract electrons.the oxygen atom carries a partial negative
charge, and a corresponding partial positive charge is shared between the
two hydrogen atoms.
● Hydrogen bonding
the separation of
partial negative and positive charges generates a strong mutual (electrical)
attraction between adjacent water molecules or between water and other
polar molecules. This attraction is called hydrogen bonding
● Covalent bond
2. Discuss the following unique properties of water, with specific reference to the role of
hydrogen bonds
● Liquid at physiological temperature
The boiling and melting depend on molecule size, which means that changes
of state of smaller molecules occur at lower temperature than of the larger
molecules, so hydrogen bonds in water need higher temperature to break
down. This is because the presence of oxygen creates strong hydrogen
bonds between water molecules
● Specific heat
The specific heat of water is 4.184 J g−1 ◦C−1
● Latent heat of vaporisation and heat of fusion
Just as hydrogen bonding increases the amount of energy required to melt
ice, it also increases the energy required to evaporate water. The heat of
vaporization of water, or the energy required to convert one mole of liquid
water to one mole of water vapor, is about 44 kJmol−1 at 25◦C.
The heat of fusion for water is 335 J g−1, which means that 335 J of energy
are required to convert 1 gram of ice to 1 gram of liquid water at 0◦C (
● Water as a solvent
The excellent solvent properties of water are due to the highly polar character
of the water molecule. Water hasthe ability to partially neutralize electrical
attractions between charged solute molecules or ions by surrounding the ion
, or molecule with one or more layers of oriented water molecules, called a
hydration shell.
3. Define ficks law which describes the process of diffusion and also provide the formula
Diffusion can be interpreted as a directed movement from high concentration to a
region of lower concentration, but it is accomplished through the random thermal
motion of individual molecules. Fick’s law tells us that the rate of diffusion is directly
proportional to the cross-sectional area of the diffusion path and to the concentration
or vapor pressure gradient, and it is inversely proportional to the length of the
diffusion path. Ficks law is a quantitative description of the process of diffusion.
4. Distinguish between bulk flow and diffusion and also provide a complete definition of
bulk flow and diffusion.
Bulk flow occurs when an external force, such as gravity or pressure, is applied.
Diffusion can be interpreted as a directed movement from a region of a high
concentration to a region of lower concentration, but it is accomplished through the
random thermal motion of individual molecules
while bulk flow is pressure-driven, diffusion is driven principally by concentration
differences
5. With aid of graphic representation, explain how the system with which osmosis can
be measured in the laboratory works. What is this device called?
The device is called a osmometer
A selectively permeable membrane is stretched across the end of a thistle tube
containing a sucrose solution and the tube is inverted in a container of pure water.
Initially, water will diffuse across the membrane in response to a chemical potential
gradient. Diffusion will continue until the force tending to drive water into the tube is
balanced by (A) the force generated by the hydrostatic head (h) in the tube
6. Distinguish between semi permeable and differentially permeable membranes
, A membrane which permits the passage of pure solvent molecules to pass through it
and not the solute particles is called semi-permeable.
A membrane which allows some substances to pass through it more readily than
others is known as selectively/differentially permeable.
7. Water potential has various components. Discuss how these components influence
the water potential and how they relate to each other.
ΨP- Pressure potential represents the hydrostatic pressure in excess of ambient
atmospheric pressure and contributes to water potential
ΨS- Osmotic potential is the negative value for osmotic pressure and contributes to
water potential
Matric potential is a result of the adsorption of water to solid surfaces.
Ψ=ΨP+ΨS
8. Analyse and evaluate the relationship between the energy and pressure units of
water potential.
9. Discuss the phenomenon of dilution in plant cells during osmosis and support your
discussion with a labelled drawing .
When a cell is surrounded in a hypotonic solution such as water with a water
potential of 0, water will enter the cell as it moves down the water potential gradient.
This causes simultaneously small dilution of the vascular content with a
corresponding increase in osmotic potential and the generation of a turgor pressure.
Net movement of water into the cell will cease when the osmotic potential of the cell
is balanced by its turgor pressure and, by equation 1.21, the water potential of the
cell is therefore also zero.
10. Define incipient plasmolysis and plasmolysis and explain how these
phenomena relate to the osmotic and pressure potential inside cells.
Incipient plasmolysis is the condition in which the protoplast just fills the cell
volume.At incipient plasmolysis, the protoplast exerts no pressure against the wall
but neither is it withdrawn from the wall. Consequently, turgor pressure (P) is zero
and the water potential of the cell (cell) is equal to its osmotic potential (S).
The protoplast then shrinks away from the cell wall, a condition known as
plasmolysis.. Continued removal of water concentrates the vacuolar contents,
further lowering the osmotic potential.Turgor pressure remains at zero and the water
potential of the cell is determined solely by its osmotic potential.
, 11. define matrix potential, citing come examples
Matric potential is a result of the adsorption of water to solid surfaces. It is
particularly important in the early stages of water uptake by dry seeds (called
imbibition) and when considering water held in soils
UNIT 2
1. Briefly explain the various methods by which the rate of transpiration can be
determined.
A potometer can be used to measure the rate of transpiration. A potometer
measures how factors such as light, temperature, humidity, light intensity and wind
will affect the rate of transpiration. The 'bubble' potometer is the main potometer. The
potometer measures the amount of water lost from a leafy shoot by observing the
rate at which an air bubble moves along the narrow tube as the leafy shoot sucks up
water to replace the water lost by the transpiration of the plant.
The Heat Pulse Technique (HPV) approach is a method for the direct measurement
of transpiration using a continuous supply of heat as a tracer. It is based on the
temperature difference between two thermocouples inserted into the stem to
estimate sap velocity, and the mass sap flow is calculated from heat balance.
Time domain reflectometry (TOR) method is based on the propagation velocity of an
electromagnetic pulse at microwave frequencies. Electromagnetic pulses travel down
a transmission line that terminates in a couple of parallel stainless steel rods inserted
into the soil. The propagation velocity is determined by probe length (the two rods)
and pulse return time, which is induced by a high speed oscilloscope. This velocity is
influenced by the dielectric constant of soil which depends mainly on soil water
content.
Portable infrared gas exchange units are used for the determination of
transpiration of single leaves. These units allow to estimate the gas exchanges of a
unique leaf within 60 - 120 s.
Another method approved to evaluate the whole plant transpiration consists of
placing the plant in a chamber to measure gas exchanges. It can be used together
with the portable infrared gas analyzer units to obtain affordable measurements of
the whole canopy transpiration.
2. Evaluate the possible advantages of transpiration for plants.
Transpiration develops a pressure gradient to move water from the roots to
the leaves. Transpiration plays a significant role in cooling down of leaves,
transpiration can thus account for dissipation of approximately one half of the
net radiation balance.
3. Discuss previous efforts( excluding cohesion theory) to explain the
mechanisms by which water is conveyed upwards in trees.
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller claughtonpm. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $8.73. You're not tied to anything after your purchase.
