Unit 1 Flashcards
Define a monomer
Give examples
4
smaller units from which larger molecules are made
-● monosaccharides (glucose, fructose,
galactose)
● amino acids
● nucleotides
Define a polymer
give examples
4
Molecules made form a large number of monomers joined together
● polysaccharides
● proteins
● DNA / RNA
What happens in a condensation
reaction?
and bonding
Joining of 2 molecules together with the formation of a chemical bond and the elimination of a molecule of water
glycosydic bond
What happens in a hydrolysis reaction?
Breaking of a chemical bond between 2 molecules and involves the use of a water molecule
Name the monosaccharides. 6
● glucose
● fructose
● galactose
all have the molecular formula C6H12O
-ribose 5
deoxyribose 5
Glyceraldehyde 3
Name the type of bond formed when
monosaccharides react 3
And difference between disaccharides and polysaccharides
(1,4 or 1,6) glycosidic bond
2 monomers = 1 chemical bond = disaccharide
multiple monomers = many chemical bonds =
polysaccharide
Name 3 disaccharides. Describe how
they form.
and what is removed 3
condensation reaction forms glycosidic bond
between 2 monosaccharides and removal of H20
● maltose: glucose + glucose
● sucrose: glucose + fructose
● lactose: glucose + galactose
all have molecular formula C12H22O11
Name 3 disaccharides.
reducing and non reducing 4
reducing and non reducing
r= suc, lact
non- malto
Draw the structure of ⍺-glucose.
Draw the structure of
𝛽-glucose.
2
isomers of glucose
alpha down, down, up, down
beta, up, down, up, down
what are polysaccharides and the 3 main ones 4
-polymers formed by many monosaccharides
-joined by glyosidic bonds in a condensation reaction to form chains
alpha- glycogen, starch
beta -cellulose
Describe the structure and functions of
starch 2
Function: storage polymer of 𝛼-glucose in plant cells
all: insoluble = no osmotic effect on cells (water potential)
large = does not diffuse out of cells (lots of glucose in a small space)
Describe the structure and functions of
glycogen.
f=2
s=4
F= main storage polymer of 𝛼-glucose in animal cells
( but also found in plant cells)
s= 1,4 & 1,6 glycosidic bonds
● branched = many terminal ends for hydrolysis, glucose released quickly
● insoluble = no osmotic effect & does not diffuse
out of cells
● compact= store more glucose in a small space
Describe the structure and functions of
cellulose.
f=1
s=4
f= polymer of 𝛽-glucose gives rigidity to plant cell walls
(prevents bursting under turgor pressure, holds stem up
s=● 1,4 glycosidic bonds
● straight-chain, unbranched molecule
● alternate glucose molecules are rotated 180° to bond with OH molecule
● H-bond crosslinks between parallel strands form
microfibrils = high tensile strength
held by weak h bonds, gap between chains so permeable
Describe the Benedict’s test for 3reducing
sugars and 4non reducing
- Add an equal volume of Benedict’s reagent
to a sample. - Heat the mixture in an electric water bath at
100℃ for 5 mins. - Positive result: colour change from blue to
orange & brick-red precipitate forms - Negative result: Benedict’s reagent remains blue
- Hydrolyse non-reducing sugars e.g. sucrose into their
monomers by adding 1cm3
of HCl. Heat in a boiling
water bath for 5 mins. - Neutralise the mixture using sodium carbonate solution.
- Proceed with the Benedict’s test as usual.
Describe the test for starch.
- Add iodine solution.
- Positive result: colour change from
orange to blue-black
Outline how colorimetry could be used to give
qualitative results for the presence of sugars and
starch
- Make standard solutions with known concentrations.
Record absorbance or % transmission values. - Plot calibration curve: absorbance or % transmission
(y-axis), concentration (x-axis). - Record absorbance or % transmission values of unknown
samples. Use calibration curve to read off concentration.
Test for proteins and lipids
p=add biurets solution
positive result lilac from pale blue
lip= add ethanol and mix/ shake sample in solution until clear
add water
positive=emulsion should form and solution go milky
How do triglycerides form?
( macromolecules)
condensation reaction between 1 molecule of glycerol C3H8O3
3 fatty acids
(RCOOH)
forms ester bonds
Contrast saturated 4 and unsaturated 4 fatty acids. in the R-group
Saturated:
● carbon single bonds
● Straight-chain molecules
have many contact points
● Higher melting point = solid
at room temperature
● Found in animal fats
Unsaturated:
●C=C double bonds
● ‘Kinked’ molecules have
fewer contact points (bend)
● Lower melting point = liquid
at room temperature
● Found in plant oil
Properties of lipids 4-6
- biological molecules made of carbon, hydrogen and oxygen
-only soluble in
organic solvents such as alcohols (ethanol)
-not polymers (not made of repeating units)
-source of energy, insulation, hormones and protection
(fat around kidneys)
chemical test for unsaturation 2
add to bromine water
postive= orange-brown to colourless
due to double c=c bond
Relate the structure of triglycerides to their functions 3-5
( macromolecules)
● Insoluble hydrocarbon chain = no effect on water potential of cells, hydrophobic insoluble in water
-large molecules, large chemical energy store, protection as molecules in large and lower density (cushioned)
-high ratio of C-H bonds in the fatty acid tails= lots of energy released when respire, water released when hydrolysed (camels)
-low mass to energy ratio- low density, lots of energy stored with a low mass, buoyancy of aquatic
What is the difference between phospholipids and triglycerides 2
Tri- 3 fatty acids and ! glycerol
Phoso- 2 fatty acids, glycerol and 1 phosophate group (PO4) 3- charge
Describe the structure and function of phospholipids
(contains polar and non polar) Amphipathic molecule:
-glycerol backbone
- 2 hydrophobic (water hating) fatty acid cells
-1 hydrophilic(water loving) polar phosphate head
describe phospholipids in water 3
-Forms phospholipid bilayer in water=
component of membranes.
● Tails can splay outwards = waterproofing.
-forms micelle- heads facing water
Compare phospholipids and triglycerides 3-4
● Both have glycerol backbone.
● Both may be attached to a mixture of saturated, monounsaturated &
polyunsaturated fatty acids.
● Both contain the elements C, H, O.
● Both formed by condensation reactions.
( macromolecules) Contrast phospholipids and triglycerides 2 of each
phospholipids:
● 2 fatty acids & 1
phosphate group attached
● Hydrophilic head &
hydrophobic tail
● Used primarily in
membrane formation
triglycerides:
● 3 fatty acids attached
● Entire molecule is
hydrophobic
● Used primarily as a
storage molecule
(oxidation releases
energy)
What is the general structure of an
amino acid? 3
-COOH carboxyl/ carboxylic acid
group
-R variable side group consists of
carbon chain & may include other
functional groups e.g. benzene
ring or -OH (alcohol)
-NH2
amine/ amino group
How many amino acids are there and how do they differ from one another? 2
20
differ only by side ‘R’ group
How do dipeptides and polypeptides form?
3
● Condensation reaction
forms peptide bond= carboxyl of 1 amino acid (OH) and amine of another HNH)
(-CONH-) & eliminates
molecule of water
● Dipeptide: 2 amino acids
● Polypeptide: 3 or more
amino acids
Define ‘primary structure’ of a protein.
p 1*- ● Sequence, number & type of amino acids in the polypeptide.
● Determined by sequence of codons on mRNA.
Define ‘tertiary structure’ of a protein.
Name the bonds present. 4
*3D structure formed by further folding of polypeptide
● disulfide bridges -sulfur
● ionic bonds
● hydrogen bonds -easily broken
Describe each type of bond in the tertiary structure
of proteins.
5
● Disulfide bridges: strong covalent S-S bonds between molecules of the amino acid cysteine
● Ionic bonds: relatively strong bonds between charged R groups (pH changes cause these bonds to break)
● Hydrogen bonds: numerous & easily broken
Define ‘quaternary structure’ of a protein. 3
● Functional proteins may consist of more than one polypeptide.chain
● Precise 3D structure held together by the same types of bond as tertiary structure.
● May involve addition of prosthetic groups e.g metal ions or phosphate groups. (iron haemoglobin)
Describe the structure and function of globular proteins 3
● Spherical & compact.
● Hydrophilic R groups face outwards & hydrophobic
R groups face inwards = usually water-soluble.
● Involved in metabolic processes e.g. enzymes &
haemoglobin
Describe the structure and function of fibrous proteins. 3
● Can form long chains or fibres
● insoluble in water.
● Useful for structure and support e.g.
collagen in skin. elasticity
Structure of amylose and amylopectin
Amylose = compact
-1,4 glycosidic bonds
-linear
-helix with H-bonds = compact
amylopectin:
● 1,4 & 1,6 glycosidic bonds
● branched = many terminal
ends for hydrolysis into
glucose
amylopectin:
● 1,4 & 1,6 glycosidic bonds
● branched = many terminal
ends for hydrolysis into
glucose
Define secondary structure of a protein
s 2*- hydrogen bonds between amino acids so coil or bend
α-helix:
β-pleated sheet:
What are enzymes? 3
● Biological catalysts for intra & extracellular
reactions. (lowers activation energy for reaction it catalyses
● Specific tertiary structure determines shape of active site, complementary to a specific substrate.
● Formation of enzyme-substrate (ES) complexes
lowers activation energy of metabolic reactions
Explain the induced fit model of enzyme
action. 3
● Shape of active site is not directly complementary to substrate & is flexible.
● Conformational change enables ES complexes to form.
● This puts strain on substrate bonds, lowering
activation energy.
How have models of enzyme action
changed?
● Initially lock & key model: rigid shape of
active site complementary to only 1
substrate.
● Currently induced fit model: also explains
why binding at allosteric sites can change
shape of active site
-so enzyme is flexible and can change shape slightly when substrate binds
How have models of enzyme action
changed?
● Initially lock & key model: rigid shape of
active site complementary to only 1
substrate.
● Currently induced fit model: also explains
why binding at allosteric sites can change
shape of active site
-so enzyme is flexible and can change shape slightly when substrate binds
How have models of enzyme action
changed?
● Initially lock & key model: rigid shape of
active site complementary to only 1
substrate.
● Currently induced fit model: also explains
why binding at allosteric sites can change
shape of active site
-so enzyme is flexible and can change shape slightly when substrate binds
Name 5 factors that affect the rate of
enzyme-controlled reactions.
● enzyme concentration
● substrate concentration
● concentration of inhibitors
● pH
● temperature
How does substrate concentration affect rate of reaction? 3
the concentration of enzyme is fixed so-
*more enzyme molecules= more substrates molecules coliding= more ESCs
- increasing enzyme concentration = increases the rate of reaction
- amount of substrate limiting= enzyme concentration no further effect
all active sites occupied at any 1 time
How does enzyme concentration affect
rate of reaction? 3
higher enzyme concentration= greater number of active sites availabel= more ESCs formed
substrate is in excess,rate increases proportionally to enzyme concentration
Rate levels off when maximum
number of ES complexes form at
any given time.
How does temperature affect the rate of reaction? 4
*rate increases= kinetic energy increases= more collisons= more ESCs made
*Till reaches optimum, peak
*kinetic energy increases= vibrations causing H/Ionic bonds to break in tertiary structure
*active site changes shape= no longer complementary , denatured
How does pH affect rate of reaction? 3
*increase or decrease in pH can alter the charges of R groups of amino acids in the active site
*H/Ionic bonds of tertiary structure break, altering the tertiary structure of enzyme
*enzyme denatures= no more ESCs
Contrast competitive 3 and
non-competitive inhibitors 3
Competitive inhibitors
-similar shape to substrate = bind
to active site *stopes substrate binding to as
-do not stop reaction; ES complex
forms when inhibitor is released
-increasing substrate concentration
decreases their effect
Non-competitive inhibitors
-bind at allosteric binding site
-may permanently stop reaction;
triggers active site to change
shape
-increasing substrate concentration
has no impact on their effect
Contrast competitive 3 and
non-competitive inhibitors 3
Competitive inhibitors
-similar shape to substrate = bind
to active site *stopes substrate binding to as
-do not stop reaction; ES complex
forms when inhibitor is released
-increasing substrate concentration
decreases their effect
Non-competitive inhibitors
-bind at allosteric binding site
-may permanently stop reaction;
triggers active site to change
shape
-increasing substrate concentration
has no impact on their effect
Why is it advantageous to calculate initial rate?
Represents maximum rate of reaction
before concentration of reactants
decreases & ‘end product inhibition’.
what are ribosomes formed from
RNA and proteins
what does a nucleotide contain and pentose sugars in DNA/ RNA
phosphate group
pentose sugar
*DNA: deoxyribose
RNA: ribose
nitrogenous base
State the role of DNA and RNA in living cells
holds genetic information
RNA transfers genetic information
from DNA to the ribosomes.
How do polynucleotides form?
Condensation reactions between
nucleotides form strong phosphodiester
bonds (sugar-phosphate backbone).
Describe the structure of DNA 3
purine a g
pyrimadine c, t, uracil (RNA)
double helix of 2 polynucleotide strands
(deoxyribose)
H-bonds between complementary purine &
pyrimidine base pairs on opposite strands:
adenine (A) + thymine (T) 2 h bonds
guanine (G) + cytosine (C) 3 h bonds
Name the complementary base pairs in RNA. 3
2 H-bonds between
adenine (A) + uracil (U)
3 H-bonds between
guanine (G) + cytosine (C)
Relate the structure of DNA to its functions
3-6
● sugar-phosphate backbone & many H-bonds provide stability
● Polynucleotide=long molecule stores lots of
genetic information
● weak H-bonds break so strands separate for replication
- lots of H bonds stabilise helix
● helix is compact for storage in nucleus
● double-stranded for semi-conservative replication
● complementary base pairing for accurate replication= 2 strands parallel
Why did scientists initially doubt that
DNA carried the genetic code?
Chemically simple molecule with few
components.
Why is DNA replication described as
‘semiconservative’
● Strands from original DNA molecule
act as a template.
● New DNA molecule contains 1 old
strand & 1 new strand
Outline the process of semiconservative DNA
replication 6
- Unwinding of double helix
2.DNA helicase breaks H-bonds between base pairs in polynucleotide strands
- Each strand acts as a template.
4.attraction of new DNA nucleotides to exposed bases on template strands and base pairing so H bonds form
- DNA polymerase catalyses condensation reactions that join adjacent nucleotides on new strand.
- H-bonds reform
Describe the Meselson-Stahl
experiment.
understand the theory
- Bacteria were grown in a medium containing heavy
isotope 15N for many generations. - Some bacteria were moved to a medium containing
light isotope 14N. Samples were extracted after 1 & 2
cycles of DNA replication. - Centrifugation formed a pellet. Heavier DNA (bases
made from 15N) settled closer to bottom of tub
Explain how the
Meselson-Stahl
experiment
validated
semiconservative
replication. 2
in the first division
-1 strand 15N and 1 strand 14N
2nd division
-50% have two strands 14N
1 strand of 15N and 1 strand 14N
Describe the structure of adenosine triphosphate
(ATP).
nucleotide
derivative of
adenine with 3
phosphate groups and ribose sugar
How does ATP store energy
The bonds between the 3 inorganic phosphate groups are unstable and when broken they release energy
in hydrolysis reactions it is catlaysed by the enzyme ATP hyrolase
Explain the role of ATP in cells3
ATP + H2O → ADP + Pi, is the hydrolysis & exothermic chemical reaction
● Energy released is coupled to metabolic
reactions.
● Phosphate group phosphorylates
compounds to make them more reactive.
How is ATP resynthesised in cells? 3
● ATP synthase catalyses condensation endothermic
reaction between ADP & Pi
● during photosynthesis & respiration
Explain why ATP is suitable as the ‘energy currency’ of cells
= useful in many biological processes.
-ATP releases in small manageable amounts so energy isn’t wasted
-Quickly broken down so energy is released quickly
-ATP can’t pass out of cell so the cell always has an immediate supply of energy
-ATP small and manageable so can be quickly transported around the cell
State 2 biologically important properties of water.4
shc Slhv
high specific heat capacity= H-bonds= acts as a buffer so the water in living organisms can stay at a sable temperature
High specific latent heat of vapourisation= provides a cooling effect= little loss of water through evaporation
State 3 biologically important properties of water.
coh,solvent, meabolite
-good solvent= excellent transport medium= metabolite reacions
-strong cohesion forces (lots of H bonds in water)= surface tension (water-air) so surface of water acts as habitat
-metabolite- hydrolysis, condensation7u7
Explain the role of hydrogen ions in the body.2
● High concentration of H+= low (acidic)
pH.
● H+
ions interact with H-bonds & ionic
bonds in tertiary structure of proteins,
which can cause them to denature.
Explain the role of iron ions in the body
-Fe2+ binds to oxygen in haemoglobin and temporarily becomes Fe3+ till oxygen is released
-4 polypeptide chains, 4 haem groups
haemoglobin transports oxygen
role of sodium ion 2
co-transport for for absorption of
glucose & amino acids in lumen of gut
affect the osmosis and water potential
role of phosphate ions 2
-holds sugar-phosphate backbone in DNA+RNA
-head of phospholipid= hydrophobic
component of:
● DNA
● ATP
● NADP (Topic 5.1)
● cAMP (Topic 6.4)
describe the structure of mRNA 2-4
● Long ribose polynucleotide (but shorter than
DNA).
● Contains uracil instead of thymine.
● Single-stranded & linear (no complementary
base pairing).
● Codon sequence is complementary to exons of 1 gene from 1 DNA strand
Describe the structure of transfer RNA
(tRNA). 3
● Single strand of about 80 nucleotides.
● Folded into clover shape (some paired bases).
● Anticodon on one end, amino acid binding site on the other:
a) anticodon binds to complementary mRNA codon
b) amino acid corresponds to anticodon
