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Biological molecules Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Define monomer. Give some examples.

A

Smaller units that join together to form larger molecules. E.g: monosaccharides, amino acids, nucelotides

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2
Q

Define polymer. Give some examples.

A

Molecules formed when many monomers join together. E.g: polysaccharides, proteins, DNA/RNA

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3
Q

When happens in a condensation reaction?

A

A chemical bond forms between 2 molecules and a molecule of water is produced and removed.

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4
Q

What happens in a hydrolysis reaction?

A

A water molecule is used to break a chemical bond between 2 molecules.

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5
Q

Name the 3 hexose monosaccharides.

A

Glucose, fructose, galactose (all have molecular formula C6H12O6)

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6
Q

Name the type of bond formed when monosaccharides react.

A

Glycosidic bond(between c1-c4 and c1-c6).
-2 monomers=1 chemical bond=disaccharides
-multiple monomers=many chemical bonds=polysaccharides

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7
Q

Name 3 disaccharides. Describe how they form.

A

Condensation reaction forms glycosidic bond between 2 monosaccharides
-maltose: glucose+glucose
-sucrose: glucose+fructose
-lactose: glucose+galactose

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8
Q

Draw the structure of a-glucose

A
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9
Q

Draw the structure of b-glucose

A
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10
Q

Describe the structure and function of starch.

A

Storage polymer of a-glucose in plant cells.
-insoluble=no osmotic effect on cells
-large=does not diffuse out of cells
Made from amylose(c1-c4 glycosidic bonds, helix with intermolecular H bonds=compact)
Made from amylopectin(c1-c4&c1-c6 glycosidic bonds, branched structure=many terminal ends for hydrolysis into glucose)

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11
Q

Describe the structure and functions of glycogen.

A

Main storage polymer of a-glucose in animal cells (also found in plant cells)
-c1-c4 & c1-c6 glycosidic bonds
-branched=many terminal ends for hydrolysis
-insoluble=no osmotic effect&doesn’t diffuse out of cells
-compact

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12
Q

Describe the structure and functions of cellulose.

A

Polymer of b-glucose gives rigidity to plant cell walls (prevents bursting under pressure and holds stem up)
-c1-c4 glycosidic bonds
-straight chain, unbranched molecule
-alternate glucose molecules are rotated 180
-H bonds crosslinked between parallel strands form microfibrils=high tensile strength

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13
Q

Describe the Benedict’s test for reducing sugars

A
  1. add an equal volume of Benedict’s reagent to sample
  2. heat the mixture in an electric water bath at 100degrees for 5mins
  3. positive result:colour change from blue to orange and brick-red precipitate forms
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14
Q

Describe the Benedict’s test for non-reducing sugars

A
  1. negative result: Benedict’s reagent remains blue
  2. hydrolyse non-reducing sugars e.g. sucrose into their monomers by adding 1cm3 of HCl, heat in a boiling water bath for 5mins
  3. neutralise the mixture using sodium carbonate solution
  4. proceed with the Benedict’s test as usual
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15
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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16
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 values(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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17
Q

Describe how to test for lipids in a sample/ emulsion test

A
  1. dissolve solid samples in ethanol
  2. add an equal volume of water and shake
  3. positive result: milky white emulsion forms
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18
Q

How do triglycerides form? Draw and label one.

A

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

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19
Q

Contrast saturated and unsaturated fatty acids

A

Saturated:
-contain single bonds
-straight-chain molecules
-higher melting point=solid at room temp
-found in animal fats

Unsaturated:
-contain C=C double bonds
-kinked molecules
-lower mp=liquid at room temp
-found in plant oils

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20
Q

Relate the structure of triglycerides to their functions

A

-high energy:mass ratio= high calorific value from oxidation(energy storage)
-insoluble hydrocarbon chain=no effect on water potential of cells&used for waterproofing
-slow conductor of heat=thermal insulation
-less dense than water=buoyancy of aquatic animals

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21
Q

Describe the structure and function of phospholipids

A

Amphipathic molecule: glycerol backbone attached to 2 hydrophobic fatty acid tails and 1 hydrophilic polar phosphate head
-forms phospholipid bilayer in water=component of membranes
-tails can splay outwards=waterproofing

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22
Q

Compare phospholipids and triglycerides

A

-both have glycerol backbone
-both may be attached to a mixture of saturated, monosaturated & polyunsaturated fatty acids
-both contain the elements C, H, O
-both formed by condensation reactions

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23
Q

Contrast phospholipids and triglycerides

A

Phospholipids:
-2 fatty acids&1 phosphate group attached
-hydrophilic head and hydrophobic tail
-used in membrane formation

Triglycerides:
-3 fatty acids attached
-entire molecule is hydrophobic
-used as a storage molecule

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24
Q

Are phospholipids and triglycerides polymers?

A

No; they are NOT made from small repeating units, they are macromolecules

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25
Q

Why is water a polar molecule?

A

O is more electronegative than H, so attracts the electron density in the covalent bond more strongly. This gives O a slight negative charge and H a slights positive charge.

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26
Q

State 4 biologically important properties of water.

A

Due to polarity&intermolecular H-bonds:
-metabolite/solvent for chemical reactions in the body
-high specific heat capacity
-high latent heat of vaporisation
-cohesion between molecules

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27
Q

Explain why water is significant to living organisms.

A

-solvent for polar molecules during metabolic reactions
-enables organisms to avoid fluctuations in core temperature
-cohesion-tension of water molecules in transpiration stream

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28
Q

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

A

They are ions that do not contain carbon atoms
-found in cytoplasm&extracellular fluid
-may be in high or very low concentrations

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29
Q

Explain the role of hydrogen ions in the body.

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

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30
Q

Explain the roles of iron ions in the body.

A

Fe2+ bonds to porphyrin ring to form haem group in haemoglobin. Haem group has binding site to transport 1 molecule of O2 around body in bloodstream. 4 haem groups per haemoglobin molecule.

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31
Q

Explain the role of sodium ions in the body

A

Involved in co-transport for absorption of glucose and amino acids in lumen of gut. Involved in propagation of action potentials in neurons.

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32
Q

Explain the role of phosphate ions in the body

A

component of:
-DNA
-ATP
-NADP
-cAMP

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33
Q

What is the general structure of an amino acid? Draw this.

A

-COOH carboxyl/carboxylic acid group
-R variable side group consists of carbon chain&may include other functional groups
-NH2 amine/amino group

34
Q

Describe how to test for proteins in a sample.

A

Biuret test confirms presence of peptide bonds
1.add equal volume of sodium hydroxide to sample at room temp
2.add drops of dilute copper(II) sulfate solution. Swirl to mix (steps 1 and 2 make Biuret reagent)
-positive result:changes from blue to purple
-negative result:solution remains blue

35
Q

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

A
  1. Differ by R groups
36
Q

How do dipeptides and polypeptides form?

A

-condensation reaction forms peptide bond&eliminates molecule of water
-dipeptide=2 amino acids
-polypeptide=3 or more amino acids

37
Q

How many levels of protein structure are there?

A

4

38
Q

Define ‘primary structure’ of a protein.

A

The order of the amino acids in the polypeptide chain

39
Q

Define ‘secondary structure’ of a protein.

A

The sequence of amino acids causes parts of a protein molecule to bend into a-helix shapes or fold into b-pleated sheets. Hydrogen bonds hold the secondary structure in place( between C=O groups of carboxyl group of 1 amino acid and the H in the amine group of another amino acid)

40
Q

Describe the 2 types of secondary protein structure.

A

A-helix:
-spiral shape, H-bonds parallel to helical axis, all N-H bonds on same side of protein chain

B-pleated sheets:
-N-H and C=O groups alternate from one side to the other

41
Q

Define ‘tertiary structure’ of a protein and name the bonds present.

A

The further folding of the secondary structure to form a unique 3D shape which is held in place by ionic, hydrogen and disulphide bonds(bonds form between R groups)

42
Q

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

A

-disulphide bridges=strong covalent S-S bonds between molecules of the amino acid cysteine.
-ionic bonds=relatively strong bonds between charged R groups
-hydrogen bonds=numerous and easily broken

43
Q

Define ‘quaternary structure’ of a protein.

A

A protein made up of more than one polypeptide chains held together by disulphide, ionic and hydrogen bonds.

44
Q

Describe the structure and functions of globular proteins

A

Spherical and compact. Hydrophilic R groups face outwards&hydrophobic R groups face inwards = water soluble. They are involved in metabolic processes.

45
Q

Describe the structure and functions of fibrous proteins.

A

Can form long chains or fibres, insoluble in water, useful for structure and support

46
Q

Outline how chromatography could be used to identify the amino acids in a mixture.

A

1.use a capillary tube to spot mixture onto pencil origin line & place chromatography paper in solvent
2.allow solvent to run until it almost touches other end of paper. Amino acids move different distances based on relative attraction to paper & solubility in solvent
3.use revealing agent or UV light to see spots
4.calculate R values & match distance to database

47
Q

What are enzymes?

A

Biological catalysts. They have a specific tertiary structures which determines the shape of the active site which is complementary to a specific substrate. Formation of enzyme-substrate complexes lowers activation energy og metabolic reactions.

48
Q

Explain the induced fit model of enzyme action.

A

-Shape of active site is not directly complementary to substrate and is flexible
-Changes to active site enables ES complexes to form
-This puts strain on substrate bonds, lowering activation energy

49
Q

Describe lock and key model of enzyme action.

A

-Enzyme and substrate are complementary
-They collide to form an ES complex
-Once ES complex formed, charged groups within active site distorts substrate
-Activation energy lowered and products released

50
Q

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

A

-enzyme concentration
-substrate concentration
-pH
-temperature
-inhibitors

51
Q

How does substrate concentration affect rate of reaction? Draw this graph.

A

Given that enzyme concentration is fixed, rate increases proportionally to substrate concentration. Rate levels off when maximum number of ES complexes form at any given time.

52
Q

How does enzyme concentration affect rate of reaction? Draw this graph.

A

Given that substrate is in excess, rate increases proportionally to enzyme concentration. Rate levels off when maximum number of ES complexes form.

53
Q

How temperature affect rat of reaction? Draw this graph.

A

Rate increases as kinetic energy/temp increases & peaks at optimum temperature. Above optimum, ionic and H bonds in tertiary structure break=active site no longer complementary to substrate.

54
Q

How does pH affect rate of reaction?

A

Enzymes have a narrow optimum pH range. Outside range: H+ and OH- ions interact with H and ionic bonds in tertiary structure= denaturing.

55
Q

Describe competitive inhibitors.

A

Same shape as substrate and can bind to the active site. This prevents the substrate from binding and the reaction occurring. If you add more substrate this will flood out the inhibitor, knocking them out of the active site.

56
Q

Describe non-competitive inhibitors.

A

Bind to the allosteric site(away from active site). This causes the bonds in the 3D tertiary structure to change shape, and therefore the active site changes shape so substrates can no longer bind, regardless of how much substrate is added.

57
Q

Outline how to calculate rate of reaction from a graph.

A

-calculate gradient of line or gradient of tangent to a point
-____________/time

58
Q

Why is it good to calculate initial rate of reaction?

A

Represents maximum rate of reaction before concentration of reactants decreases.

59
Q

State the formula for pH.

A

pH= -log10[H+]

60
Q

Draw the structure of a nucelotide.

A
61
Q

Name the pentose sugars in DNA and RNA.

A

-DNA = deoxyribose
-RNA = ribose

62
Q

State the role of DNA in living cells.

A

Codes for the sequence of amino acids in the primary structure of a protein which determines the final 3D structure and function of a protein.

63
Q

State the role of RNA in living cells.

A

To copy and transfer the genetic code from DNA in the nucleus to the ribosomes, some RNA is also combined with proteins to create ribosomes.

64
Q

How do polynucelotides form?

A

Condensation reactions between nucelotides form strong phosphodiester bonds (sugar phosphate backbone)

65
Q

Describe the structure of DNA.

A

Double-helix of 2 polynucleotide strands. H bonds between complementary base pairs on opposite strands. (A-T, G-C)

66
Q

Name the complementary base pairs in DNA.

A

-2 H bonds between Adenine and Thymine.
-3 H bonds between Guanine and Cytosine.

67
Q

Name the complementary base pairs in RNA.

A

-2 H bonds between Adenine and Uracil.
-3 H bonds between Guanine and Cytosine

68
Q

Relate the structure of DNA to its functions.

A

-Sugar-phosphate backbone and 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

69
Q

Describe the structure of messenger RNA (mRNA)

A

-Long ribose polynucleotide
-Contains uracil instead of thymine
-Single-stranded and linear
-No complementary base pairing

70
Q

Relate the structure of messenger RNA (mRNA) to its functions.

A

-Breaks down quickly so no excess polypeptide forms
-Ribosomes can move along strand and tRNA can bind to exposed bases
-Can be translated into specific polypeptide by ribosomes

71
Q

Describe the structure of transfer RNA (tRNA)

A

-Single strand of about 80 nucleotides
-Folded into clover shape
-Anticodon on one end, amino acid binding site on the other
a)anticodon binds to complementary mRNA codon
b)amino acid corresponds to anticodon

72
Q

Order DNA, mRNA, and tRNA according to increasing length.

A

tRNA, mRNA, DNA

73
Q

Why did scientists doubt DNA carried the genetic code

A

Chemically simple molecule with few components

74
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 and 1 new strand

75
Q

Outline the process of semiconservative replication.

A

1.Dna helicase breaks H bonds between base pairs
2.Each strand acts as a template
3.Free nucleotides from nuclear sap attach to exposed bases by complementary base pairing
4.DNA polymerase catalyses condensation reactions that join adjacent nucleotides on new strand
5.H bonds reform

76
Q

Describe the Meselson-Stahl experiment.

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 extracted after 1&2 cycles of DNA replication
3.Centrifugation formed a pellet. Heavier DNA settled closer to bottom of tube

77
Q

Explain how the Meselson-Stahl experiment validated semiconservative replication.

A

-Grown in 15N
-1 division=all molecules have 1 strand 15N and 1 strand 14N
-2 divisions=50% have 2 strands 15N and 50% have 1 strand 15N and 1 strand 14N

78
Q

Draw the structure of adenosine triphosphate (ATP)

A
79
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

80
Q

How is ATP resynthesised in cells?

A

-ATP synthase catalyses condensation reaction between ADP and Pi
-during photosynthesis and respiration

81
Q

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

A

-high energy bonds between phosphate groups
-small amounts of energy released at a time=less energy wasted as heat
-Single step hydrolysis=energy available quickly
-Readily synthesised

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