SPEC - P1 - M2: BIO MEM Flashcards
(a) how hydrogen bonding occurs between water
molecules, and relate this, and other properties
of water, to the roles of water for living
organisms (9)
- forms hydrogen bonds between H and O
Ice is less dense than water
Ice will float and insulate the water underneath
The whole body of water does not freeze
Water has a high specific heat capacity
Because the hydrogen bonds can absorb a lot of energy
Therefore it is thermally stable
Good habitat, particularly for ecotherms
Water has a high latent heat of evaporation
Because it takes a lot of energy to break the hydrogen bonds
Therefore, can be used to cool organisms via sweating
Water has high cohesion
Because it is a polar molecule
Therefore it flows by mass flow through the xylem and phloem
Water is a good solvent
Because it is a polar molecule
Therefore ions and glucose easily dissolve and can be transported
(b) the concept of monomers and polymers and
A: Describe the difference between a monomer and polymer. Give an example of each. (4)
A: A monomer is a small, basic molecular unit.
For example an amino acid, monosaccharide or nucleotide.
A polymer is a large, complex molecule made up of lots of monomers bonded together in a long chain.
For example proteins, polysaccharides and nucleic acids
(b) the importance of condensation and hydrolysis
reactions in a range of biological molecules: Condensation vs Hydrolysis
Condensation:
Two molecules are bonded together
A covalent bond is formed
Water is formed as a by product
From the OH from one molecules and the H from the other
Hydrolysis:
A larger molecule is split into two smaller ones
A covalent bond is broken
Water is added
To form the OH of one molecules and the H of the other
(c) the chemical elements that make up biological
molecules: carbohydrates, lipids, proteins, nucleic acids
C, H and O for carbohydrates
C, H and O for lipids
C, H, O, N and S for proteins
C, H, O, N and P for nucleic acids
(e) the synthesis and breakdown of a disaccharide
SYNTHESIS:
Condensation reaction
Forms a 1,4 – glycosidic bond
Water is formed
From a hydrogen atom of one sugar and a hydroxyl group (OH) from the other
BREAKDOWNS:
Hydrolysis reaction
Breaks a 1,4 – glycosidic bond
Water is needed for the reaction
Adds a hydrogen atom to one sugar and a hydroxyl group (OH) to the othe
give three disaccharides
sucrose
maltose
lactose
State the monosaccharides present in these disaccharides:
Sucrose
Lactose
Maltose
Sucrose – One alpha glucose molecule, one fructose molecule
Lactose – One beta glucose molecule and one galactose molecule
Maltose – Two alpha glucose molecules
(f) the structure of starch (amylose and
amylopectin)
Polysaccharide of alpha glucose
A mixture of amylose and amylopectin
Amylose is unbranched chain of alpha glucose molecules
All 1,4 – glycosidic bonds
Coiled structure
Amylopectin is branched
Contains 1,4 and 1,6 – glycosidic bonds
(f) the structure of glycogen (3) and cellulose molecules (6)
Glycogen:
Polysaccharide of alpha glucose
Lots of branching
Contains 1,4 and 1,6 – glycosidic bonds
Cellulose:
Polysaccharide of beta glucose
Alternate beta glucose molecules are flipped over
Forms straight cellulose chains
Cellulose chains bonded together with hydrogen bonds
This forms microfibrils
Microfibrils join together to form macrofibrils
(g) how the structures and properties of glucose relate to
their functions in living organisms
Soluble – therefore easily transported
The chemical bonds contain a lot of energy
(g) how the structures and properties of
starch relate to their functions in living organisms
Insoluble – doesn’t affect the water potential of the cell
The coiled structure of amylose makes it compact
Can store a lot of starch in a small space
The branches of amylopectin allow enzymes to hydrolyse it quickly
Glucose can be released more quickly
(g) how the structures and properties of glycogen relate to
their functions in living organisms
Insoluble – doesn’t affect the water potential of the cell
Compact – can store a lot in a small space
Lots of branches allow enzymes to hydrolyse it quickly
Glucose can be released more quickly
(g) how the structures and properties of cellulose molecules relate to their functions in living organisms
Many hydrogen bonds provide structural strength
Necessary for the cell wall
(h) the structure of a triglycerides (5)
One glycerol molecule
Three fatty acids tails
The fatty acids can be saturated or unsaturated
Saturated fatty acids only contain single bonds between the carbon atoms
Unsaturated fatty acids have at least one double bond between the carbon atoms
(h) the structure of a phospholipid
One glycerol molecule
One phosphate group
Two fatty acid tails
The phosphate group is hydrophilic and the fatty acids are hydrophobic
(i) the synthesis of triglycerides by
the formation (esterification)
A condensation reaction occurs
Between the glycerol molecule and the fatty acids
A hydrogen atom is removed from the glycerol
An OH group is removed from the fatty acids
To form three molecules of water
An ester bond is formed
(j) how the properties of triglycerides molecules relate to their functions in living organisms/ Describe the functions of triglycerides and explain how their structure aids the function
Energy storage molecules
They are insoluble
And therefore don’t affect the water potential of the cell
The fatty acid tails contain a lot of chemical energy
This energy is released when the bonds are broken
(j) how the properties of phospholipid
molecules relate to their
functions in living organisms/ Describe the functions of phospholipids and explain how their structure aids the function
Form the bilayer in membranes
The phosphate heads are hydrophilic and the fatty acid tails are hydrophobic
Therefore the phosphate heads face outwards and the tails face inwards
The fatty acid tails act as a barrier to water soluble substances
(j) how the properties of cholesterol
molecules relate to their
functions in living organisms/ Describe the functions of cholesterol and explain how their structure aids the function
Strengthen the cell membrane
Small, flat shape
Therefore, it can fit in between the tails of the fatty acids in the membrane
The cholesterol can bind to the tails of the phospholipids
Therefore making the membrane less fluid and more rigid
(l) the synthesis of dipeptides
SYNTHESIS:
Condensation reaction
The OH of the carboxylic acid group in one amino acid bonds with a H atom from the amino group of the other amino acid
This produces water
Forms a peptide bond
(m) the levels of protein structure
primary, secondary, tertiary and
quaternary structure
Define the term primary structure. (1)
The sequence of amino acids in a polypeptide chain
Describe how the secondary structure of a protein is formed. (2)
The polypeptide chains coil to form an alpha helixes
Or fold to form a beta pleated sheet
Describe how the tertiary structure of a protein is formed (include details of the different bonds that hold the protein in its tertiary structure). (5)
The alpha helixes and beta pleated sheets coil and fold further.
Held in place by lots of bonds:
Ionic bonds between amino acids of opposite charge
Hydrogen bonds between amino acids with slight charges
Disulphide bonds that form between sulphur atoms in two cysteine amino acids
Hydrophilic interactions – hydrophilic amino acids are found on the outside of the protein.
Define the term quaternary structure. (1)
A protein made up of different polypeptide chains
the structure and function of globular proteins
including a conjugated protein/ Describe the general structure of a globular protein + Define Conjugated protein
Glob;
Spherical
Amino acids with hydrophilic R groups are on the outside
Soluble
Conj:
A protein with a non-protein group attached.
The non-protein group is called a prosthetic group
3 examples of globular proteins
haemoglobin, insulin, amylase
Structure and function of 3 globular proteins:
Haemoglobin Four polypeptide chains Each polypeptide chain has a haem group as a prosthetic group. Each haem group contains iron. To carry oxygen in erythrocytes.
Insulin Two polypeptide chains Held together with disulphide bonds Hormone Reduces blood glucose concentration
Amylase A single chain of amino acids Enzyme Catalyses the hydrolysis of starch
Describe the general features of fibrous proteins. (3)
Insoluble
Strong
Structural proteins
Give three fiborous proteins
keratin, elastin, collagen
(o) the properties and functions of fibrous proteins
Collagen Very strong Minerals can bind to increase rigidity Found in connective tissue, such as skin, bone and muscle.
Keratin Can be flexible (if fewer disulphide bonds) Or can be hard and tough (if more disulphide bonds) Found in nails, skin, hair, feathers and horns
Elastin Allows tissues to return to their original shape after they have been stretched Found in elastic connective tissue, such as skin, large blood vessels and ligaments.
(p) the key inorganic ions that are involved in
biological processes: 5 cations
cations: calcium ions (Ca2+), sodium ions (Na+),
potassium ions (K+), hydrogen ions (H+), ammonium
ions (NH4+)
(p) the key inorganic ions that are involved in
biological processes: 5 anions
anions: nitrate (NO3–), hydrogencarbonate (HCO3–),chloride (Cl –), phosphate (PO43–), hydroxide, (OH–).
Describe a biological function of these cations: (5)
Ca2+
Na+
K+
H+
NH4+
Ca2+ - Stimulates the vesicles with neurotransmitter to move to the presynaptic membrane
Na+ - Involved in transmission of action potentials along a neurone
K+ - Involved in transmission of action potentials along a neurone
H+ - Used in the active loading of sucrose into the companion cells
NH4+ - Absorbed from the soil by plants – used a source of nitrogen
Describe a biological function of these anions: (5)
NO3-
HCO3-
Cl-
PO43-
OH-
NO3- - Absorbed from the soil by plants – used a source of nitrogen
HCO3- - How the majority of carbon dioxide is carried in the blood
Cl- - Acts a cofactor for amylase
PO43- - Needed for the synthesis of nucleotides
OH- - pH determination – makes solutions alkaline
Describe how to carry out a qualitative test for reducing sugars. (4)
Add Benedict’s solution to the sugar solution
Place in a water bath and heat to 80°C
If solution stays blue is a negative result – no reducing sugar
A green, yellow, orange or red precipitate is a positive result
Can be a semi quantitative tes
Describe how to carry out a test for a non-reducing sugar. (7)
Carry out reducing sugar test first
If negative, obtain a new sample of the sugar solution
Add dilute hydrochloric acid and boil
Neutralise with sodium hydrogen carbonate
Repeat the Benedict’s test for a reducing sugar
If blue, no sugar present (reducing or non-reducing)
A green, yellow, orange or red precipitate indicates non-reducing sugar is present
Describe how to carry out a quantitative test for a reducing sugar. (9)
Complete serial dilutions to obtain different concentrations of reducing sugar
Measure out the same volume of each reducing sugar solution
Add the same volume of Benedict’s solution to each reducing sugar solution
Heat in a water bath at 80°C for 10 minutes
Filter each solution to remove the precipitate
Pour each solution into a cuvette
Turn on the colourimeter and put on the red filter
Zero with a cuvette of distilled water
Measure the transmission of light of each solution
Plot a calibration curve
Test the unknown solution of reducing sugar and use the calibration curve to estimate the concentration of reducing suga
Describe the biological test for lipids. (4)
Shake the sample with ethanol
Pour the solution into water
If lipid is present the solution turns white
If no lipid is present the solution stays clear
