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Y12 Biology > 1. Biological Molecules > Flashcards

1. Biological Molecules Flashcards

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1
Q

What are monomers and polymers?

A

Monomers - small/repeating molecules from which larger molecules/polymers are made
Polymers - molecule made up of many identical/similar molecules/monomers

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2
Q

What happens during condensation and hydrolysis reactions?

A

Condensation reaction
- 2 molecules join together
- forming a chemical bond
- releasing a water molecule

Hydrolysis reaction
- 2 molecules separated
- breaking a chemical bond
- using a water molecule

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3
Q

Give examples of polymers and the monomers from which they’re made

A

Nucleotide —> Polynucleotides (DNA/RNA)
Monosaccharide —> Polysaccharide (starch)
Amino acid —> Polypeptide (protein)

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4
Q

Are lipids polymers?

A

No because they’re not made up from repeating monomers

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5
Q

What is a disaccharide also called?

A

A dimer

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6
Q

What are monosaccharides? Give 3 common examples

A
  • Monomers from which larger carbohydrates are made
  • Glucose, fructose and galactose
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7
Q

Describe the difference between the structure of a-glucose and ß-glucose

A
  • OH group is below carbon 1 in a-glucose, but above carbon 1 in ß-glucose
  • Isomers of each other, same molecular formula but differently arranged atoms
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8
Q

What are disaccharides and how are they formed?

A
  • 2 monosaccharides joined together with a glycosidic bond
  • formed by a condensation reaction, releasing a water molecule
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9
Q

List 3 common disaccharides & monosaccharides from which they’re made

A

Maltose = glucose + glucose
Sucrose = glucose + fructose
Lactose = glucose + galactose

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10
Q

Draw a diagram to show how 2 monosaccharides are joined together

A

The 2 OH groups react together, releases a molecule of water and you’re left with a C-O-C bond (glycosidic bond)

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11
Q

What are polysaccharides and how are they formed?

A
  • Many polysaccharides joined together with glycosidic bonds
  • Formed by many condensation reactions, releasing many water molecules
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12
Q

Describe the basic function and structure of starch

A

Starch acts as an energy store in plant cells
- Polysaccharide of a-glucose
- Amylose (1,4 glycosidic bonds), unbranched structure
- Amylopectin (1,4 and 1,6 glycosidic bonds), highly branched structure

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13
Q

Describe the basic function and structure of glycogen

A

Glycogen acts as an energy store in animal cells (mainly found in muscle and liver cells)
- Polysaccharide of a-glucose
- 1,4 and 1,6 glycosidic bonds —> highly branched structure

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14
Q

Explain how the structure of starch relates to its function

A
  • Helical: compact so good for storage in cells
  • Large, insoluble polysaccharide molecule: cannot leave the cell/cross cell membrane
  • Insoluble in water: does not affect the water potential of the cell (no osmotic effect)
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15
Q

Explain how the structure of glycogen relates to its function

A
  • Branched: compact so can fit more molecules in a small area
  • Branched: more terminal ends for faster hydrolysis —> release glucose for respiration to make ATP for energy release
  • Large insoluble polysaccharide molecule: can’t leave the cell/cross cell membrane
  • Insoluble in water: does not affect the water potential of cells/no osmotic effect
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16
Q

Describe the basic function and structure of cellulose

A

Provides strength and structural support to plant/algal cell walls
- Polysaccharide of ß-glucose
- 1,4 glycosidic bond —> straight, unbranched chains
- chains linked in parallel by hydrogen bonds forming many microfibrils

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17
Q

Explain how the structure of cellulose relates to its function

A
  • Every other ß-glucose molecule is inverted in a long, straight, unbranched chain
  • Many hydrogen bonds link parallel strands (cross links) to form microfibrils (strong fibres)
  • Hydrogen bonds are weak individually but strong in high numbers
  • So provides strength to plant cell walls
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18
Q

Describe the test for reducing sugars

A

Reducing sugars = monosaccharides, maltose and lactose
1. Add Benedict’s solution (blue) to the sample
2. Heat in a boiling water bath
3. Positive result = green/yellow/orange/red precipitate

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19
Q

Describe the test for non-reducing sugars

A

Non-reducing sugars = sucrose
1. Do Benedict’s test and stays blue/negative
2. Heat in a boiling water bath with dilute acid (HCl) —> to hydrolyse into reducing sugars
3. Neutralise with alkali (sodium hydrogencarbonate)
4. Heat in a boiling water bath with Benedict’s solution
5. Positive result = green/yellow/orange/red precipitate

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20
Q

Suggest a simple method to measure the quantity of sugar in a solution

A
  • Carry out Benedict’s test, then filter and dry the precipitate
  • Find the mass/weight
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21
Q

Suggest a more complicated method to measure the quantity of sugar in a solution

A
  1. Make sugar solutions of known concentrations (e.g dilutions series)
  2. Heat a set volume of each sample with a set volume of Benedict’s solution for the same amount of time
  3. Use a colorimeter to measure absorbance (of light) of each known concentration
  4. Plot a calibration curve (concentration on x-axis and absorbance on y-axis), draw line of best fit
  5. Repeat Benedict’s test with unknown sample and measure the absorbance
  6. Read off the calibration curve to find concentration associated with unknown sample’s absorbance
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22
Q

Describe the biochemical test for starch

A
  • Add iodine dissolved in potassium iodide (orange/brown) and shake/stir
  • Positive result = blue/black
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23
Q

What is wrong with this sentence?
“Glycogen and starch are energy stores as hydrolysis of glycosidic bonds releases energy”

A

Glycogen and starch are energy stores as they’re made of a-glucose, which is the substrate for respiration in cells. This produces ATP for energy release.

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24
Q

Why must Benedict’s reagent be added in excess?

A

To ensure that all of the sugar reacts

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25
Q

Name 2 groups of lipid

A

Triglycerides and phospholipids

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26
Q

Describe the structure of a fatty acid (RCOOH)

A
  • Variable R-group —> hydrocarbon chain, can be saturated or unsaturated
  • COOH = carboxyl group
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27
Q

Describe the difference between saturated and unsaturated fatty acids

A
  • Saturated: no C=C double bonds in hydrocarbon chain; all carbons are fully saturated with hydrogen
  • Unsaturated: one or more C=C double bonds in hydrocarbon chain (creating bend/kink)
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28
Q

Describe how triglycerides form

A
  • 1 molecule of glycerol and 3 fatty acids
  • Condensation reactions
  • Remove 3 water molecules
  • Forming 3 ester bonds (C-O-C=O)
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29
Q

Explain how the properties of triglycerides are related to their structure

A

Act as energy storage molecules
- High ratio of C-H bonds to carbon atoms in the hydrocarbon chain —> so used in respiration to release more energy than same mass of carbohydrates
- Hydrophobic/non-polar fatty acids so insoluble in water (instead clump together as insoluble droplets) —> no effect on water potential of cells

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30
Q

Describe the difference between the structure of triglycerides and phospholipids

A

One of the fatty acids of a triglyceride is substituted by a phosphate-containing group of

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31
Q

Describe how the properties of phospholipids relate to their structure

A

Form a bilayer in cell membranes, allows diffusion of lipid soluble substances (non-polar) or very small ones and restricts the movement of water-soluble (polar) or larger substances
- Phosphate heads are hydrophilic —> attracted to water either side of membrane
- Fatty acid tails are hydrophobic —> repelled by water so form the interior of the membrane

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32
Q

Describe the test for lipids

A
  1. Add ethanol, shake (to dissolve lipids), then add water
  2. Positive = milky white emulsion
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33
Q

Describe the general structure of an amino acid

A

COOH = carboxyl group
NH2 = amine group
R = variable side chain/group

34
Q

How many amino acids are common in all organisms? How do they vary?

A

The 20 amino acids that are common in all organisms differ only in their side group (R)

35
Q

Describe how amino acids join together

A
  • Condensation reaction
  • Removing a water molecule
  • Between a carboxyl group of one and amine group of another
  • Forming a peptide bond
36
Q

What are dipeptides and polypeptides?

A

Dipeptide = 2 amino acids joined together
Polypeptide = many amino acids joined together

37
Q

Describe the primary structure of a protein

A

Sequence of amino acids in a polypeptide chain, joined by peptide bonds

38
Q

Describe the secondary structure of a protein

A
  • Folding of the polypeptide chain, e.g a-helix or ß-pleated sheet
  • Due to hydrogen bonding between amino acids
  • Between NH2 group of one and C=O group of another
39
Q

Describe the tertiary structure of a protein

A
  • 3D folding of polypeptide chain
  • Due to interactions between amino acid R groups (dependent on sequence of amino acids)
  • Forming hydrogen bonds, ionic bonds and disulphide bridges
40
Q

Describe the quaternary structure of a protein

A
  • More than 1 polypeptide chain
  • Formed by interactions between polypeptides (hydrogen bonds, ionic bonds and disulphide bridges)
41
Q

Describe the test for proteins

A
  1. Add Buiret reagent (sodium hydroxide + copper (II) sulfate)
  2. Positive result = purple/lilac colour (negative stays blue) —> indicates presence of peptide bonds
42
Q

How do enzymes act as biological catalysts?

A
  • Each enzyme lowers the activation energy of the reaction it catalyses
  • To speed up the rate of reaction
43
Q

Describe the induced-fit model of enzyme action

A
  1. Substrate binds to (not completely complementary) active site of enzyme
  2. Causing active site to change shape (slightly) so it is complementary to substrate
  3. So enzyme-substrate complex forms
  4. Causing bonds in substrate to bend/distort, lowering the activation energy
44
Q

Describe how models of enzyme action have changed over time

A
  • Initially lock & key model (now outdated) —> active site a fixed shape, complementary to 1 substrate
  • Now induced-fit model
45
Q

Explain the specificity of enzymes

A
  • Specific tertiary structure determines shape of active site —> dependent on sequence of amino acids (primary structure)
  • Active site is complementary to a specific substrate
  • Only this substrate can bind to the active site, inducing fit and forming an enzyme-substrate complex
46
Q

Describe and explain the effect of enzyme concentration on the rate of enzyme-controlled reactions

A
  • As enzyme conc increases, rate of reaction increases
    > enzyme conc = limiting factor (excess substrate)
    > more enzymes so more available active sites
    > so more enzyme-substrate complexes form
  • At a certain point, rate of reaction stops increasing/levels off
    > substrate conc = limiting factor (all of substrates in use)
47
Q

Describe and explain the effect of substrate concentration on the rate of enzyme-controlled reactions

A
  • As the substrate conc increases, rate of reaction increases
    > substrate conc = limiting factor (too few enzyme molecules to occupy all active sites)
    > more enzyme-substrate complexes form
  • At a certain point, rate of reaction stops increasing/levels off
    > enzyme conc = limiting factor
    > As all active sites are saturated/occupied (at a given time)
48
Q

Describe and explain the effect of temperature on the rate of enzyme-controlled reactions

A
  • As temperature increases up to optimum, rate of reaction increases
    > more kinetic energy so more E/S complexes form
  • As temp increases above optimum, rate of reaction decreases
    > enzymes denature - tertiary structure and active site change shape
    > as hydrogen/ionic bonds break
    > so active site no longer complementary in shape
    > so fewer E/S complexes form
49
Q

Describe and explain the effect of pH on the rate of enzyme-controlled reactions

A
  • As pH increases/decreases above/below and optimum, rate of reaction decreases
    > enzymes denature - tertiary structure and active site change shape
    > as hydrogen/ionic bonds break
    > so active site no longer complementary
    > so fewer E/S complexes form
50
Q

Describe and explain the effect of concentration of competitive inhibitors on the rate of enzyme-controlled reactions

A
  • As concentration of comp inhibitor increases, rate of reaction decreases
    > similar shape to substrate
    > competes for/binds to/blocks active site
    > so substrates can’t bind and fewer E/S complexes form
  • Increasing substrate conc reduces the effect of inhibitors (dependent on relative conc of substrate and inhibitors)
51
Q

Describe and explain the effect of concentration non-competitive inhibitors on the rate of enzyme-controlled reactions

A
  • As concentration of non-comp inhibitors increases, rate of reaction decreases
    > binds to a site other than the active site (allosteric site)
    > changes enzyme tertiary structure/active site shape
    > so active site no longer complementary to substrate
    > so substrates can’t bind so fewer E/S complexes form
  • Increasing substrate conc has no effect on rate of reaction as change to active site is permanent
52
Q

Describe the basic functions of DNA and RNA in all living cells

A

DNA: holds genetic information which codes for polypeptides (proteins)
RNA: transfers genetic information from DNA to ribosomes

53
Q

Name the 2 types of molecule from which a ribosome is made

A

RNA and proteins

54
Q

Draw and label a DNA nucleotide and an RNA nucleotide

A

Both have a phosphate group, pentose sugar (either deoxyribose or ribose) and a nitrogen-containing organic base (A,T,C,G in DNA/A,U,C,G in RNA)

55
Q

Describe the differences between a DNA nucleotide and an RNA nucleotide

A

DNA - pentose sugar is deoxyribose - base can be thymine
RNA - pentose sugar is ribose - base can be uracil

56
Q

Describe how nucleotides join together to form polynucleotides

A
  • Condensation reactions, removing water molecules
  • Between phosphate group of 1 nucleotide and pentose sugar of another
  • Forming phosphodiester bonds
57
Q

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

A

The relative simplicity of DNA - chemically simple molecule with few components

58
Q

Describe the structure of DNA

A
  • polymer of nucleotides (polynucleotide)
  • each nucleotide formed from deoxyribose, a phosphate group and a nitrogen-containing base
  • phosphodiester bonds join adjacent nucleotides
  • 2 polynucleotide chains held together by hydrogen bonds
  • between specific complementary base pairs —> 2 between A&T and 3 between C&G
  • double helix
59
Q

Describe the structure of messenger RNA

A
  • polymer of nucleotides (polynucleotide)
  • each nucleotide formed from ribose, a phosphate group and a nitrogen-containing base
  • bases = A,U,C&G
  • phosphodiester bonds join adjacent nucleotides
  • single helix
60
Q

Compare and contrast the structure of DNA and mRNA

A

DNA:
- pentose sugar is deoxyribose
- has the base thymine
- double stranded/double helix
- long (many nucleotides)
- has hydrogen bonds/base pairing

mRNA:
- pentose sugar is ribose
- has the base uracil
- single stranded/single helix
- shorter (fewer nucleotides)
- does not have hydrogen bonds

61
Q

Suggest how the structure of DNA relates to its functions (7)

A
  • 2 strands —> both can act as a template for semi-conservative replication
  • hydrogen bonds between bases are weak —> strands can be separated for replication
  • complementary base pairing —> accurate replication
  • many hydrogen bonds between bases —> stable/strong molecule
  • double helix with sugar-phosphate backbone —> protects bases/hydrogen bonds
  • long molecule —> stores lots of genetic information (that codes for polypeptides)
  • double helix (coiled) —> compact
62
Q

Suggest how you can use incomplete information about the frequency of bases on DNA strands to find the frequency of other bases

A
  1. % of adenine in strand 1 = % of thymine in strand 2 (and vice versa)
  2. % of cytosine in strand 1 = % of guanine in strand 2 (and vice versa)
    - because of complementary base pairing between 2 strands
63
Q

Why is semi-conservative replication important?

A

Ensures genetic continuity between generations of cells

64
Q

What is semi-conservative replication?

A

Each new DNA molecule consists of one original/template strand and one new strand

65
Q

Describe the process of semi-conservative DNA replication

A
  1. DNA helicase breaks hydrogen bonds between complementary bases, unwinding the double helix
  2. Both strands act as templates
  3. Free floating DNA nucleotides are attracted to exposed bases and join by specific complementary base pairing
  4. Hydrogen bonds form between A-T and C-G
  5. DNA polymerase joins adjacent nucleotides on new strand by condensation reactions
  6. Forming phosphodiester bonds
66
Q

Use your knowledge of enzyme action to suggest why DNA polymerase moves in opposite directions along DNA strands

A
  • DNA has antiparallel strands
  • So shapes/arrangements of nucleotides on 2 ends are different
  • DNA polymerase is an enzyme with a specific shaped active site
  • So can only bind to substrate with complementary shape (3’ end of lagging/developing strand)
  • New strand is made in a 5’ - 3’ direction
67
Q

Name the 2 scientists who proposed models of the chemical structure of DNA and of DNA replication

A

Watson & Crick

68
Q

Describe the work of Meselsohn & Stahl in validating the Watson-Crick model of semi-conservative DNA replication

A
  1. Bacteria grown in medium containing heavy 15N and light 14N
  2. Nitrogen is incorporated into DNA bases
    > DNA extracted and centrifuged —> settles near the bottom for heavy 15N and near the top for light 14N
  3. 15N bacteria transferred to medium containing light 14N and allowed to divide once
    > DNA extracted and centrifuged —> settles in the middle as all DNA molecules contain 1 heavy and 1 new light strand
  4. Bacteria in light 14N is allowed to divide again
    > DNA extracted and centrifuged —> half settles in the middle, and half settles near the top (contains 2 light strands)
69
Q

What is ATP?

A

Adenosine triphosphate

70
Q

Describe the structure of ATP

A
  • Ribose bound to a molecule of adenine (base) and 3 phosphate groups
  • Nuclotide derivative (modified nucelotide)
71
Q

Describe how ATP is broken down

A
  • ATP (+ water) —> ADP (adenosine diphosphate) + Pi (inorganic phosphate)
  • Hydrolysis reaction, using a water molecule
  • Catalysed by ATP hydrolase (enzyme)
72
Q

Give 2 ways in which the hydrolysis of ATP is used in cells

A
  1. Coupled to energy requiring reactions within cells (releases/provides energy) —> e.g active transport, protein synthesis
  2. Inorganic phosphate released can be used to phosphorylate (add phosphate to other compounds), making them more reactive
73
Q

Describe how ATP is resynthesised in cells

A
  • ADP + Pi —> ATP (+ water)
  • Condensation reaction, removing a water molecule
  • Catalysed by ATP synthase (enzyme)
  • During respiration and photosynthesis
74
Q

Suggest how the properties of ATP make it a suitable immediate source of energy for cells

A
  • Releases energy in (relatively) small amounts/little energy lost as heat
  • Single reaction/one bond hydrolysed to release energy (so immediate release)
  • Cannot pass out of cell
75
Q

Explain how hydrogen bonds occur between water molecules

A
  • Water is a polar molecule
  • Slightly negatively charged oxygen atoms attract slightly positively charged hydrogen atoms of other water molecules
  • 180º bond angle
76
Q

Explain 5 properties of water that are important in biology

A
  1. Metabolite - used in condensation/hydrolysis/respiration/photosynthesis
  2. Solvent - allows metabolic reactions to occur (faster in solution), allows transport of substances, e.g nitrates in xylem, urea in blood
  3. High specific heat capacity - buffers changes in temperature as can gain/lose a lot of heat energy without changing temperature —> good habitat for aquatic organisms as temp more stable than land, helps organisms maintain a constant internal body temperature
  4. High latent heat of vaporisation - allows effective cooling via evaporation of a small volume (e.g sweat) so helps organisms maintain a constant internal body temperature
  5. Strong cohesion between water molecules - supports columns of water, e.g transpiration pulls water through xylem in plants, produces surface tension supporting small organisms (walk on water)
77
Q

Where are inorganic ions found in the body?

A

In solution in cytoplasm and body fluid, some in high concentrations and others in very low concentrations

78
Q

Describe the role of hydrogen ions

A
  • Maintain pH levels in the body —> high conc = low pH/acidic
  • Affects enzyme rate of reaction as can cause enzymes to denature
79
Q

Describe the role of iron ions

A
  • Component of haem group of haemoglobin
  • Allowing oxygen to bind/associate for transport as oxyhaemoglobin
80
Q

Describe the role of sodium ions

A
  1. Involved in co-transport of glucose/amino acids into cells
  2. Involved in action potentials in neurons
  3. Affects water potential of cells/osmosis
81
Q

Describe the role of phosphate ions

A
  1. Component of nucleotides, allowing phosphodiester bonds to form in DNA & RNA
  2. Component of ATP, allowing energy release
  3. Phosphorylate other compounds making them more reactive
  4. Hydrophilic part of phospholipids, allowing a bilayer to form

Y12 Biology (4 decks)

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