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A- Level Biology > Unit 1 > Flashcards

Unit 1 Flashcards

1
Q

Define a monomer
Give examples
4

A

smaller units from which larger molecules are made

-● monosaccharides (glucose, fructose,
galactose)
● amino acids
● nucleotides

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2
Q

Define a polymer
give examples
4

A

Molecules made form a large number of monomers joined together

● polysaccharides
● proteins
● DNA / RNA

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3
Q

What happens in a condensation
reaction?
and bonding

A

Joining of 2 molecules together with the formation of a chemical bond and the elimination of a molecule of water

glycosydic bond

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4
Q

What happens in a hydrolysis reaction?

A

Breaking of a chemical bond between 2 molecules and involves the use of a water molecule

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5
Q

Name the monosaccharides. 6

A

● glucose
● fructose
● galactose
all have the molecular formula C6H12O

-ribose 5
deoxyribose 5
Glyceraldehyde 3

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6
Q

Name the type of bond formed when
monosaccharides react 3
And difference between disaccharides and polysaccharides

A

(1,4 or 1,6) glycosidic bond
2 monomers = 1 chemical bond = disaccharide
multiple monomers = many chemical bonds =
polysaccharide

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7
Q

Name 3 disaccharides. Describe how
they form.
and what is removed 3

A

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

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7
Q

Name 3 disaccharides.

reducing and non reducing 4

A

reducing and non reducing
r= suc, lact
non- malto

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8
Q

Draw the structure of ⍺-glucose.
Draw the structure of
𝛽-glucose.
2
isomers of glucose

A

alpha down, down, up, down
beta, up, down, up, down

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9
Q

what are polysaccharides and the 3 main ones 4

A

-polymers formed by many monosaccharides
-joined by glyosidic bonds in a condensation reaction to form chains

alpha- glycogen, starch

beta -cellulose

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10
Q

Describe the structure and functions of
starch 2

A

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)

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11
Q

Describe the structure and functions of
glycogen.
f=2
s=4

A

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

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12
Q

Describe the structure and functions of
cellulose.
f=1
s=4

A

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

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13
Q

Describe the Benedict’s test for 3reducing
sugars and 4non reducing

A
  1. Add an equal volume of Benedict’s reagent
    to a sample.
  2. Heat the mixture in an electric water bath at
    100℃ for 5 mins.
  3. Positive result: colour change from blue to
    orange & brick-red precipitate forms
  4. Negative result: Benedict’s reagent remains blue
  5. Hydrolyse non-reducing sugars e.g. sucrose into their
    monomers by adding 1cm3
    of HCl. Heat in a boiling
    water bath for 5 mins.
  6. Neutralise the mixture using sodium carbonate solution.
  7. Proceed with the Benedict’s test as usual.
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14
Q

Describe the test for starch.

A
  1. Add iodine solution.
  2. Positive result: colour change from
    orange to blue-black
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15
Q

Outline how colorimetry could be used to give
qualitative results for the presence of sugars and
starch

A
  1. Make standard solutions with known concentrations.
    Record absorbance or % transmission values.
  2. Plot calibration curve: absorbance or % transmission
    (y-axis), concentration (x-axis).
  3. Record absorbance or % transmission values of unknown
    samples. Use calibration curve to read off concentration.
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16
Q

Test for proteins and lipids

A

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

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17
Q

How do triglycerides form?
( macromolecules)

A

condensation reaction between 1 molecule of glycerol C3H8O3

3 fatty acids
(RCOOH)
forms ester bonds

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18
Q

Contrast saturated 4 and unsaturated 4 fatty acids. in the R-group

A

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

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19
Q

Properties of lipids 4-6

A
  • 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)
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20
Q

chemical test for unsaturation 2

A

add to bromine water
postive= orange-brown to colourless
due to double c=c bond

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21
Q

Relate the structure of triglycerides to their functions 3-5
( macromolecules)

A

● 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

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22
Q

What is the difference between phospholipids and triglycerides 2

A

Tri- 3 fatty acids and ! glycerol

Phoso- 2 fatty acids, glycerol and 1 phosophate group (PO4) 3- charge

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23
Q

Describe the structure and function of phospholipids

A

(contains polar and non polar) Amphipathic molecule:
-glycerol backbone
- 2 hydrophobic (water hating) fatty acid cells
-1 hydrophilic(water loving) polar phosphate head

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24
Q

describe phospholipids in water 3

A

-Forms phospholipid bilayer in water=
component of membranes.
● Tails can splay outwards = waterproofing.
-forms micelle- heads facing water

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25
Q

Compare phospholipids and triglycerides 3-4

A

● 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.

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26
Q

( macromolecules) Contrast phospholipids and triglycerides 2 of each

A

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)

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27
Q

What is the general structure of an
amino acid? 3

A

-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

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28
Q

How many amino acids are there and how do they differ from one another? 2

A

20
differ only by side ‘R’ group

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29
Q

How do dipeptides and polypeptides form?
3

A

● 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

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30
Q

Define ‘primary structure’ of a protein.

A

p 1*- ● Sequence, number & type of amino acids in the polypeptide.
● Determined by sequence of codons on mRNA.

31
Q

Define ‘tertiary structure’ of a protein.
Name the bonds present. 4

A

*3D structure formed by further folding of polypeptide
● disulfide bridges -sulfur
● ionic bonds
● hydrogen bonds -easily broken

32
Q

Describe each type of bond in the tertiary structure
of proteins.
5

A

● 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

33
Q

Define ‘quaternary structure’ of a protein. 3

A

● 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)

34
Q

Describe the structure and function of globular proteins 3

A

● Spherical & compact.
● Hydrophilic R groups face outwards & hydrophobic
R groups face inwards = usually water-soluble.
● Involved in metabolic processes e.g. enzymes &
haemoglobin

35
Q

Describe the structure and function of fibrous proteins. 3

A

● Can form long chains or fibres
● insoluble in water.
● Useful for structure and support e.g.
collagen in skin. elasticity

36
Q

Structure of amylose and amylopectin

A

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

37
Q

Define secondary structure of a protein

A

s 2*- hydrogen bonds between amino acids so coil or bend

α-helix:
β-pleated sheet:

38
Q

What are enzymes? 3

A

● 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

39
Q

Explain the induced fit model of enzyme
action. 3

A

● 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.

40
Q

How have models of enzyme action
changed?

A

● 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

40
Q

How have models of enzyme action
changed?

A

● 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

41
Q

How have models of enzyme action
changed?

A

● 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

42
Q

Name 5 factors that affect the rate of
enzyme-controlled reactions.

A

● enzyme concentration
● substrate concentration
● concentration of inhibitors
● pH
● temperature

43
Q

How does substrate concentration affect rate of reaction? 3

A

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
44
Q

How does enzyme concentration affect
rate of reaction? 3

A

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.

45
Q

How does temperature affect the rate of reaction? 4

A

*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

46
Q

How does pH affect rate of reaction? 3

A

*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

47
Q

Contrast competitive 3 and
non-competitive inhibitors 3

A

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

48
Q

Contrast competitive 3 and
non-competitive inhibitors 3

A

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

49
Q

Why is it advantageous to calculate initial rate?

A

Represents maximum rate of reaction
before concentration of reactants
decreases & ‘end product inhibition’.

50
Q

what are ribosomes formed from

A

RNA and proteins

51
Q

what does a nucleotide contain and pentose sugars in DNA/ RNA

A

phosphate group
pentose sugar
*DNA: deoxyribose
RNA: ribose
nitrogenous base

52
Q

State the role of DNA and RNA in living cells

A

holds genetic information

RNA transfers genetic information
from DNA to the ribosomes.

53
Q

How do polynucleotides form?

A

Condensation reactions between
nucleotides form strong phosphodiester
bonds (sugar-phosphate backbone).

54
Q

Describe the structure of DNA 3

purine a g
pyrimadine c, t, uracil (RNA)

A

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

55
Q

Name the complementary base pairs in RNA. 3

A

2 H-bonds between
adenine (A) + uracil (U)

3 H-bonds between
guanine (G) + cytosine (C)

56
Q

Relate the structure of DNA to its functions

3-6

A

● 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

57
Q

Why did scientists initially doubt that
DNA carried the genetic code?

A

Chemically simple molecule with few
components.

58
Q

Why is DNA replication described as
‘semiconservative’

A

● Strands from original DNA molecule
act as a template.
● New DNA molecule contains 1 old
strand & 1 new strand

59
Q

Outline the process of semiconservative DNA
replication 6

A
  1. Unwinding of double helix

2.DNA helicase breaks H-bonds between base pairs in polynucleotide strands

  1. 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

  1. DNA polymerase catalyses condensation reactions that join adjacent nucleotides on new strand.
  2. H-bonds reform
60
Q

Describe the Meselson-Stahl
experiment.

understand the theory

A
  1. Bacteria were grown in a medium containing heavy
    isotope 15N for many generations.
  2. Some bacteria were moved to a medium containing
    light isotope 14N. Samples were extracted after 1 & 2
    cycles of DNA replication.
  3. Centrifugation formed a pellet. Heavier DNA (bases
    made from 15N) settled closer to bottom of tub
61
Q

Explain how the
Meselson-Stahl
experiment
validated
semiconservative
replication. 2

A

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

62
Q

Describe the structure of adenosine triphosphate
(ATP).

A

nucleotide
derivative of
adenine with 3
phosphate groups and ribose sugar

63
Q

How does ATP store energy

A

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

64
Q

Explain the role of ATP in cells3

A

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.

65
Q

How is ATP resynthesised in cells? 3

A

● ATP synthase catalyses condensation endothermic
reaction between ADP & Pi
● during photosynthesis & respiration

66
Q

Explain why ATP is suitable as the ‘energy currency’ of cells
= useful in many biological processes.

A

-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

67
Q

State 2 biologically important properties of water.4
shc Slhv

A

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

68
Q

State 3 biologically important properties of water.

coh,solvent, meabolite

A

-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

69
Q

Explain the role of hydrogen ions in the body.2

A

● 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.

70
Q

Explain the role of iron ions in the body

A

-Fe2+ binds to oxygen in haemoglobin and temporarily becomes Fe3+ till oxygen is released

-4 polypeptide chains, 4 haem groups

haemoglobin transports oxygen

71
Q

role of sodium ion 2

A

co-transport for for absorption of
glucose & amino acids in lumen of gut

affect the osmosis and water potential

72
Q

role of phosphate ions 2

A

-holds sugar-phosphate backbone in DNA+RNA
-head of phospholipid= hydrophobic

component of:
● DNA
● ATP
● NADP (Topic 5.1)
● cAMP (Topic 6.4)

73
Q

describe the structure of mRNA 2-4

A

● 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

74
Q

Describe the structure of transfer RNA
(tRNA). 3

A

● 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

A- Level Biology (9 decks)

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