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AQA A Level Biology > Biological Molecules > Flashcards

Biological Molecules Flashcards

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1
Q

Define monomer. Give some examples.

A

small repeating units that join together to form larger molecules

  • monosaccharides (glucose, fructose, galactose)
  • amino acids
  • nucleotides
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2
Q

What happens in a condensation reaction?

A

joins monomers together and forms a
chemical bond, releasing water

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3
Q

What happens in a hydrolysis reaction?

A

breaks a chemical bond between
joined molecules and uses water

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4
Q

Name the three hexose monosaccharides.

A

glucose, fructose, galactose

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5
Q

Name the type of bond formed when monosaccharides react.

A

(1,4 or 1,6) glycosidic bond

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6
Q

Describe the structure of amylopectin (starch)

A

1,4 and 1,6 glycosidic bonds

Branched- many terminal ends for hydrolysis to break it down into glucose

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7
Q

Describe the structure of amylose (starch)

A

only 1,4 glycosidic bond

Helix/coiled with intermolecular H bonds

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8
Q

Describe the function of starch.

A

storage of alpha glucose in plants

-insoluble= doesn’t affect water potential

-large= doesn’t diffuse out of the cell

-compact- allows lots of glucose to be stored

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9
Q

Describe the structure and functions of glycogen.

A

Main storage polymer of 𝛼-glucose in animal cells

● 1,4 & 1,6 glycosidic bonds.

● Branched = many terminal ends for hydrolysis.

● Insoluble = no osmotic effect & does not diffuse out of cells.

● Compact.

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10
Q

Describe the structure and functions of cellulose.

A

polymer of 𝛽-glucose gives rigidity to plant cell walls

● 1,4 glycosidic bonds

● straight-chain, unbranched molecule

● H-bond crosslinks between parallel strands form microfibrils = high tensile strength

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11
Q

Benedict’s test for reducing sugars

A
  1. Add benedict’s reagent (blue) to sample
  2. Heat in a boiling water bath
  3. Positive = colour change from blue to orange & brick red precipitate (reducing sugar present)
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12
Q

Benedict’s test for non-reducing sugars

A
  1. Negative result as benedict’s remains blue
  2. Hydrolyse solution by adding HCl
  3. Heat in water bath for 5 minutes
  4. Neutralise with Sodium Carbonate Solution
  5. Proceed with Benedict’s test
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13
Q

How do triglycerides form?

A

Condensation reaction between 1 molecule of glycerol & 3 fatty acids forms ester bonds.

Entirely hydrophobic

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14
Q

Properties of saturated fatty acids

A

-solid at room temperature

-single covalent bonds only

  • straight chain molecules

-mostly found in animal fats.

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15
Q

properties of unsaturated fatty acids

A

-have “kink” in the fatty acid chains

-liquid at room temp

  • Found in plant cells

-Contain a C=C double bond

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16
Q

Relate the structure of triglycerides to their functions

A

High energy: mass ratioo= good energy storage

Has an Insoluble hydrocarbon chain= no effect on water

Slow conductor of heat, making it a great Thermal insulator

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17
Q

Describe the structure and function of phospholipids.

A
  • glycerol backbone
  • Attached to 2 hydrophobic fatty acid tails and 1 hydrophilic polar phosphate head
  • Forms phospholipid bilayer in water ( a component of membranes)

-Tails can splay outwards= waterproofing

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18
Q

Similarities between triglycerides and phospholipids

A
  • Both have glycerol backbone

-Both formed by condensation reactions

  • Both contain C, H and O
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19
Q

Are phospholipids and triglycerides polymers?

A

No; they are not made from a small repeating unit. They are macromolecules.

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20
Q

Why is water a polar molecule?

A

water has positively charged hydrogen atoms and negatively charged oxygen atoms

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21
Q

What are inorganic ions and where are they found in the body?

A

-Ions that don’t contain carbon atoms

  • Found in the cytoplasm and extracellular fluid
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22
Q

Explain the role of hydrogen ions in the body.

A

lower the pH of solutions and impact enzyme and haemoglobin function

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23
Q

Explain the roles of iron ions in the body

A

A component of haemoglobin in the transport of oxygen

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24
Q

Explain the role of sodium ions in the body.

A

Involved in co-transport for absorption of glucose & amino acids

25
Q

Explain the role of phosphate ions in the body

A

Component of:

  • DNA (forms phosphodiester bonds with deoxyribose)
  • ATP (makes ADP more reactive)
26
Q

State the role of DNA in living cells.

A

Base sequence of genes codes for functional RNA & amino acid sequence of polypeptides.

Genetic information determines inherited characteristics = influences structure & function of organisms.

27
Q

State the role of RNA in living cells.

A

to transfer the genetic code from DNA in the nucleus to the ribosomes.

28
Q

Which bases are purines?

A

Adenine and Guanine

29
Q

Which bases are pyrimidines?

A

cytosine, thymine, uracil

30
Q

Name the complementary base pairs on DNA

A

2 H-bonds between adenine + thymine

3 H-bonds between guanine + cytosine

31
Q

Name the complementary base pairs in RNA

A

2 H-bonds between adenine + uracil

3 H-bonds between guanine + cytosine

32
Q

Relate the structure of DNA to its functions

A
  • sugar-phosphate backbone & many H-bonds provide stability
  • long molecule stores lots of information
  • helix is compact for storage in nucleus
  • base sequence of triplets codes for amino acids
  • double-stranded for semi-conservative replication
  • complementary base pairing for accurate replication
  • weak H-bonds break so strands separate for replication
33
Q

Describe the structure of messenger RNA

A
  • short polynucleotide strand
  • Contains uracil instead of thymine.
  • Single-stranded & linear (no complementary base pairing).
34
Q

Why is DNA replication called semi-conservative?

A

-Strands from original DNA molecule act as templates.

  • New DNA molecule contains 1 old strand & 1 new strand
35
Q

Outline the process of semi conservative replication

A
  1. DNA Helicase breaks H-bonds between base pairs
  2. Each strand acts a template
  3. Free nucleotides form nuclear sap attach to exposed bases via complementary base pairing
  4. DNA Polymerase catalyses condensation reactions attaches to adjacent nucleotides on new strand
  5. H Bonds reform
  6. Sugar phosphate backbone reforms
36
Q

Describe the Messelson-Stahl Experiment

A
  1. Bacteria were grown in a medium containing heavy 15N for many generations
  2. Some bacteria were moved to a medium containing a light isotope of 14N. Samples were extracted after 1&2 cycles of DNA Replication
  3. Centrifugation formed a pellet. Heavier DNA settled closer to bottom of tube
37
Q

Explain the role of ATP in cells.

A

ATP hydrolase catalyses ATP → ADP + Pi

  • Energy released is coupled to metabolic reactions.
  • Phosphate group phosphorylates compounds to make them more reactive.
38
Q

How is ATP resynthesised in cells?

A
  • ATP synthase catalyses condensation reaction between ADP & Pi
  • during photosynthesis & respiration
39
Q

Explain why ATP is suitable as the ‘energy currency’ of cells.

A
  • High energy bonds between phosphate groups.
  • Small amounts of energy at a time= less energy wasted
  • Single step hydrolysis= energy avaliable

-Readily resynthesised

40
Q

How many amino acids are there? and how do they differ?

A

20

differ only by R group

41
Q

How do dipeptides and polypeptides form?

A
  • Condensation reaction forms peptide bond (-CO and NH-) & eliminates molecule of water

● Dipeptide: 2 amino acids

● Polypeptide: 3 or more amino acids

42
Q

Define primary structure

A

the order of the amino acids in a protein as determined by DNA

43
Q

Define secondary structure

A

The twisting of a polypeptide into either an alpha helix or beta-pleated sheets due to hydrogen bonding between the amino acids

44
Q

Define tertiary structure

A

further folding of the secondary structure to form a unique 3D shape. Disulphide bridges, Ionic bonds and Hydrogen bonds

45
Q

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

A

-Disulfide bridges: strong covalent S-S bonds

-Ionic bonds: relatively strong bonds between charged R groups

-Hydrogen bonds: numerous and easily broken

46
Q

Define quaternary structure

A
  • Functional proteins that may consist of more than one polypeptide

-Precise 3D Structure held together by same bonds in tertiary structure

  • May involve addition of prosthetic groups e.g. metal ions
47
Q

Explain the induced fit model of enzyme action.

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.
48
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.
49
Q

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

A
  • enzyme concentration
  • substrate concentration
  • concentration of inhibitors

-pH

-temperature

50
Q

How does substrate concentration affect the rate of reaction?

A

Given that enzyme concentration is fixed, rate increases proportionally to substrate concentration

Rate levels off when max number of E-S complexes form

51
Q

How does enzyme concentration affect the rate of reaction?

A

Given that substrate is in excess, rate increases proportionally to enzyme concentration

Rate levels off when max number of E-S complexes form

52
Q

How does temperature affect the rate of reaction?

A

Rate increases as kinetic energy increases & peaks at optimum temperature.

Above optimum, ionic & H-bonds in tertiary(3°) structure break = active site no longer complementary to substrate (denaturation).

53
Q

How does pH affect rate of reaction?

A

Enzymes have a narrow optimum pH range.

Too high or too low of a pH interferes with amino acid charges causing bonds in the tertiary structure to break

54
Q

What are the properties of competitive inhibitors?

A

-similar shape to substrate
- binds to active site
-prevents E-S complexes
- adding more substrates will knock out inhibitor resuming E-S complex formation

55
Q

What are the properties of non-competitive inhibitors?

A
  • binds to the allosteric site
  • causes active site to change shape
  • no more E-S complexes regardless how much more substrate is added
56
Q

what is phosphorylation?

A

the process of ATP transferring energy to other compounds through the inorganic phosphate released during hydrolysis binding to different compounds making them reactive.

57
Q

what are the 5 properties of water

A

metabolite - in condensation and hydrolysis reactions

important solvent in reactions- can dissolve and easily transport solutes in cytoplasm or plasma

a high specific heat capacity- buffers temperature

a large latent heat of vaporisation- provides a cooling effect with loss of water through evaporation (e.g. transpiration)

strong cohesion between water molecules- due yo hydrogen bonds between water molecules supports water columns(e.g. moving up xylem ) and provides surface tension

58
Q
A

AQA A Level Biology (8 decks)

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