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3.1 Biological molecules Flashcards

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1
Q

3.1.1 Monomers & Polymers

What are monomers?

A

Smaller units from which larger molecules are made. (polymers)

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2
Q

3.1.1 Monomers & Polymers

What are polymers?

A

Molecules made from larger number of monomers joined together via a condensation reaction.

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3
Q

3.1.1 Monomers & Polymers

Outline some examples of monomers.

A
  1. Monosaccharides (glucose, fructose, galactose)
  2. Amino acids
  3. Nucleotides.
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4
Q

3.1.1 Monomers & Polymers

Outline some examples of polymers.

A
  1. Polysaccharides
  2. Proteins
  3. DNA / RNA
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5
Q

3.1.1 Monomers & Polymers

What is a condensation reaction?

A

Joins two molecules together with a formation of a **chemical bond **and involves the elimination of a water molecule.

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6
Q

3.1.1 Monomers & Polymers

What is a hydrolysis reaction?

A

Breaks a chemical bond between two molecules and involves the addition of a water molecule.

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7
Q

3.1.2 Carbohydrates

What are monosaccharides?

A

The monomers from which larger carbohydrates are made.

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8
Q

3.1.2 Carbohydrates

Outline some examples of monosaccharides.

A
  1. Glucose
  2. Galactose
  3. Fructose

Formula C6H12O6

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9
Q

3.1.2 Carbohydrates

Outline some features of monosaccharides.

A
  1. Sweet tasting
  2. Soluble substances
  3. General formula : (CH2O)n

n can be any number between 1 and 7

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10
Q

3.1.2 Carbohydrates

Name the type of bond formed when monosaccharides react.

(Including how diasaccharides and polysaccharides are formed and number of GB.)

A

(1,4 or 1,6) glycosidic bond via a condensation reaction.

2 monomers = 1 chemial (glycosidic) bond = diasaccharide.

many monomers = many chemical (glycosidic) bonds = polysaccharide.

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11
Q

3.1.2 Carbohydrates

Name 3 diasaccharides and explain how they are formed.

A
  1. Maltose: glucose + glucose
  2. Sucrose: glucose + fructose
  3. Lactose: glucose + galactose
    all have molecular formula C12H22O11

2 monosaccharides form a glycosidic bond via a condensation reaction.

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12
Q

3.1.2 Carbohydrates

Glucose has two isomers. What are these two isomers called?

A

A-glucose
B-glucose

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13
Q

3.1.2 Carbohydrates

Describe the structure of A-glucose.

A

A H A
(alpha hydrogen above)
-OH bond below

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14
Q

3.1.2 Carbohydrates

Describe the structure of B-glucose.

A

BHB
(beta hydrogen below)
OH bond above.

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15
Q

3.1.2 Carbohydrates

Outline some features of polysaccharides.

A

1.Very large molecules.
2. insoluble
} makes it good for storage.

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16
Q

3.1.2 Carbohydrates

What are 3 examples of polysaccharides?

A
  1. Starch
  2. Glycogen
  3. Cellulose
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17
Q

3.1.2 Carbohydrates

Outline the basic structure of starch and its’ role.

A

BASIC STRUCTURE
1. Found in plants.
2. formed from condensation of A-glucose
3. chains may be branched or unbranched.
4. Amylopectin: 1,4 and 1,6 GB.
5. Amylose: 1,4 GB.

ROLE
1. Insoluble ∴ doesn’t affect ψ, so water not drawn into cells by osmosis.
2. large and insoluble } does not diffuse out of cells.
3. Compact, stored in small space.
4. Branched form has many ends } can be acted upon by enzymes simulatenously } glucose monomers released rapidly.
5. When hydrolysed, forms A-glucose, can be transported and readily used in respiration.

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18
Q

3.1.2 Carbohydrates

What is the difference between amylopectin and amylose?

A

Amylopectin:
1. 1,4 and 1,6 GB
2. A-Glucose
2. Branched form has many ends } can be acted upon by enzymes simulatenously } glucose monomers released rapidly.
3. Increase in SA

Amylose:
1. 1,4 GB
2. Unbranched helix molecule
3. compact so can fit lots of glucose in small space.

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19
Q

3.1.2 Carbohydrates

Outline the basic structure of glycogen and its’ role.

A

BASIC STRUCTURE
1. found in animals only.
2. formed from condensation of A-glucose
3. 1,4 and 1,6 GB.
4. Has shorter chains and is more branched.
5 . stored in small granules mainly in muscle and liver.

ROLE
1. Insoluble ∴ doesn’t affect ψ, so water not drawn into cells by osmosis.
2. Insoluble } does not diffuse out of cells.
3. Compact } lots of it can be stored in small space.
4. More branched than starch } has more ends that can be acted upon simultaneously by enzymes } more rapidly broken down to form glucose monomers used in respiration } important as animals have a higher metabolic rate than plants as they are more active.

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20
Q

3.1.2 Carbohydrates

Outline the basic structure of cellulose and its’ role.

A

Basic structure
1. Plants only
2. formed from condensation of B-glucose
3. 1,4 GB.
4. straight chain & unbranched molecule.
5. Chains run paralell to each other } allows H bonds to form cross linkages between adjacent chains.
6. Has adjacent glucose molecules that can rotate 180
7. Celluose molecules are grouped together to form microfibrils.

ROLE
1. Many H bonds provide collective strength.
2. Provides ridgity to plant cell.
3. Cellulose cell wall prevents cell from bursting as water enters it by osmosis.
4. B-glucose } forms long, straight unbranched chains
5. Cellulose molecules grouped together to form microfibrils which are grouped again to form fibres providing more strength.

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21
Q

3.1.2 Carbohydrates

Desribe the steps for the biochemical test for reducing sugars.

A
  1. Add 2cm3 of food sample to a test tube. If sample is not in a liquid form first grind it up.
  2. Add equal volume of Benedict’s reagent.
  3. Heat mixture in gently boiling water bath for 5 minuites.
  4. If reducing sugar present solution turns from blue to orange / brown.
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22
Q

3.1.2 Carbohydrates

What is a reducing sugar?

A

Sugar that can donate electrons to reduce another chemical in this case Benedict’s reagent.

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23
Q

3.1.2 Carbohydrates

Describe the steps for the biochemical test for non-reducing sugars.

A
  1. (Using a new sample) if sample is not in liquid form, grind it up.
  2. Add 2cm3 of the food sample and add 2cm3 of HCL in a test tube and place in a gently boiling water bath for 5 minuites. Dilute HCL will hydrolyse any diasaccharides into its’ consitutent monosaccharides.
  3. Slowly add NAHCO3 to test tube in order to neutralise HCL.
  4. Re-test solution by adding 2cm3 of benedict’s reagent into gently boiling water bath for 5 minuites.
  5. Blue to orange / brown (if non-reducing sugar was present in the original sample)
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24
Q

3.1.3 Lipids

Describe the characteristics of lipids.

A

Contain C,H,O

proportions of O: C and H is smaller than in carbohydrates

insoluble in water

Soluble in organic solvents such as alchohol and acetone.

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25
Q

3.1.3 Lipids

What are the roles of lipids?

A
  1. Source of energy : When [O] lipids provide >X2 the energy as the same mass of carbohydrate and release valuble water.
  2. Waterproofing : Lipids are insoluble in H2O
  3. Insulation : Fats = slow conductors of heat, stored beneath body surface to retain body heat. Act as electrical insulators in myelin sheath around nerve cells.
  4. Protection : Fats stored around delicate organs such as kidneys.
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26
Q

3.1.3 Lipids

What are the two main groups of lipids called?

A
  1. Triglycerides
  2. Phospholipids
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27
Q

3.1.3 Lipids

How are triglycerides formed?

A

By the condensation of one molecule of a glycerol and three molecules of fatty acid forming an ester bond.

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28
Q

3.1.3 Lipids

What is the general formula of a fatty acid?

A

RCOOH

R group may be saturated or unsaturated.

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29
Q

3.1.3 Lipids

What is a saturated fatty acid?

A

Contains no double bond. (Carbons are linked to max n.o of hydrogen atoms. i.e. saturated with hydrogens.)

Straight chain molecules } have many contact points.

Higher mp } solid at room temp

Found in animal fats.

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30
Q

3.1.3 Lipids

What is an unsaturated fatty acid?

A

Contains double bond.

Mono-unsaturated } single double bond between carbon- carbon bond.

Poly-unsaturated } +1 double bond between carbon-carbon bonds.

“Kinked” molecules } fewer contact points.

Lower mp } liquid at room temp.

Found in plant oils.

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31
Q

3.1.3 Lipids

Describe the relation between triglyceride’s structure and their properties.

A
  1. High ratio of energy storing C-H bonds : C atoms } good source of energy.
  2. Low mass : energy } good storage molecules as much energy can be stored in small vol. Beneficial to animals as it reduces mass they have to carry when they move around.
  3. Large & non-polar } insoluble in water } their storage does not affect osmosis in cells or ψ of cells.
  4. Have high H:O } release water when [O] } provide important source of H2O especially for organisms living in dry deserts.
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32
Q

3.1.3 Lipids

Outline the structure of a phospholipid.

A

Glycerol, 2 fatty acids (hydrophobic) and 1 phosphate head (hydrophillic).

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33
Q

3.1.3 Lipids

What is a phospholipid made up of?

A

Hydrophobic tail: Orients itself away from water but moves readily towards fats.

Hydrophillic (polar) head: Interacts with water but not fat.

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34
Q

3.1.3 Lipids

Describe the relation between phosphlipids structure and their properties.

A
  1. Polar molecules } has a hydrophillic phosphate head and hydrophobic tail of two fatty acids } forms a bilayer within CSM } hydrophobic barrier formed between in and out of cell.
  2. Hydrophillic phosphate heads } help to hold at the surface of CSM.
  3. Phospholipid structure } allows them to form glycolipids by combining with carbohydrates within CSM. } important in cell recognition.
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35
Q

3.1.3 Lipids

Compare phospholipids and triglycerides.

A
  1. Both have glycerol backbone.
  2. Both may be attached to a mixture of saturated, mono-unsaturated and polyunsaturated fatty acids.
  3. Both contain element C,H,O
  4. Both formed via condensation reactions.
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36
Q

3.1.3 Lipids

Contrast phosphlipids and triglycerides.

A

Phospholipids:
2 fatty acids and 1 phosphate group attached.
Hydrophillic head and hydrophobic tail.
Used mainly in membrane formation.

Triglycerides:
3 fatty acids attached.
Entire molecule hydrophobic.
Used mainly as a storage molecule } [O] releases energy.

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37
Q

3.1.3 Lipids

Are phospholipids and triglycerides polymers?

A

No, they are not made from small repeating units. They are
macromolecules.

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38
Q

3.1.3 Lipids

Describe how to test for lipids in a sample.

A
  1. Add 2cm3 of sample being tested into a test tube (that’s dry and grease free) and add 5cm3 of ethanol.
  2. Shake the tube thoroughly to dissolve lipids in sample.
  3. Then add 5cm3 of water and shake gently.
  4. A milky white emulsion } positive for lipid prescence.

As control, repeat procedures using water instead of sample, the final solution should remain clear.

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39
Q

3.1.4 Proteins

What is an amino acid?

A
  • monomers from which proteins are made from.
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40
Q

3.1.4 Proteins

What is the general structure of an amino acid?

A

R
H2N - C - COOH
H
* NH2 = amine group
* COOH = carboxyl group
* R = side chain

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41
Q

3.1.4 Proteins

How many amino acid groups are common in all organisms and what do they only differ in?

A
  • 20 amino acid groups
  • differ in their side group
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42
Q

3.1.4 Proteins

What is a dipeptide?

A
  • formed by the condensation reaction between two amino acid groups
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43
Q

3.1.4 Proteins

What is a polypeptide?

A
  • formed by the condensation reaction of many amino acids.
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44
Q

3.1.4 Proteins

How is a peptide bond formed?

A
  • condensation reaction between two amino acids
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45
Q

3.1.4 Proteins

Functional protein?

A
  • may contain one or more polypeptides.
46
Q

3.1.4 Proteins

What is the role of hydrogen bonds in the protein structure?

A

Secondary structure
* hydrogen bonds form between amino acids in polypeptide chain.
* Makes chain coiled (a helix)
* Makes chain folded (B pleated sheets)
* There are many but are easily broken

47
Q

3.1.4 Proteins

What is the role of ionic bonds in the protein structure?

A

Tertiary structure
* more bonds form between different parts of the polypeptide chain, including hydrogen and ionic bonds further coiling or folding the chain.
* formed between any carboxyl and amino groups that are not involved in forming peptide bonds.
* weaker than disulfide bridges.
* easily broken by changes in pH

48
Q

3.1.4 Proteins

What is the role of disulfide bridges in the protein structure?

A
  • form whenever 2 molecules of amino acid cystenine come close together.
  • Sulfur atom in one cystenine binds to the sulfur atom in the other.
  • Fairly strong and are not easily broken.
49
Q

3.1.4 Proteins

Can you describe the relationship between the primary, secondary, tertiary and quaternary structure and protein function?

A

PRIMARY:
* sequence of amino acids in its polypeptide chains.
* function: sequence determines its properties and shapes.

SECONDARY:
* Shape the polypeptide chain forms is due to hydrogen bonding.
* forms alpha helix or beta pleated sheets.

TERTIARY:
* Compact: due to bending and twisting of the polypeptide helix
* All 3 bonds (H, I, DB) contribute to maintenance of tetiary structure.

QUATERNARY:
* Arises from the combination of many different polypeptides chains and associated non-protein groups (prosthetic) into large, complex protein molecule e.g: haemoglobin.

50
Q

3.1.4 Proteins

Biuret test for proteins:

A
  1. Add equal volume of NaOH to sample at room temp.
  2. Add drops of dilute copper (II) sulfate solution. Swirl to mix. (steps 1 & 2 make biuret reagent)
  3. Positive result = colour change from blue to purple.
    Negative result = solution remains blue.
51
Q

3.1.4 Proteins

Describe the structure and function of globular proteins.

A
  • spherical & compact.
  • Hydrophillic R group faces outwards.
  • Hydrophobic R group faces inwards = usually water-soluble.
  • involved in metabolic processes e.g: enzymes and haemoglobin.
52
Q

3.1.4 Proteins

Describe the structure and function of fibrous proteins.

A
  • can form long chains or fibres
  • insoluble in water.
  • useful for structure and support e.g: collagen in skin.
53
Q

3.1.4 Proteins

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

A
  1. Use capillary tube to spot mixture onto pencil orgin line and place chromatography paper in solvent
  2. Allow solvent to run until it almost touches the other end of paper. Amino acids move different distances based on the relative attraction to paper and solubility in solvents.
  3. Use UV or revealing agent to see spots.
  4. Calculate Rf value and match to database.
54
Q

3.1.4 Proteins

What are enzymes?

A
  • Biological catalysts for intra and extracellular reactions.
  • Specific tetiary structure determines shape of the active site complementary to a specific substrate
  • Formation of E-S complexes lowers the Ea of metabolic reactions.
55
Q

3.1.4 Proteins

Explain the induced fit model of enzyme action.

A
  • Shape of active site is not complementary to substrate and is flexible
  • conformational change enables ES complexes to form.
  • Puts a strain on substrate bonds, lowering Ea.
56
Q

3.1.4 Proteins

Explain the lock and key model.

A
  • Enzymes are complementary to the shape of the substrate
  • Subsrate will fit into the active site and an E-S complex will form.
  • Enzyme will catalyse the reaction and products will form E-S complexes.
57
Q

3.1.4 Proteins

Specificity of enzymes:

A
  • is due to the tetiary structure of its active site, allowing complementary binding to substrates.
  • Enzymes catalyse intra and extracellular reactions that determine structure and functions from cellular to whole organism.
58
Q

3.1.4 Proteins

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

A
  1. Enzyme concentration
  2. Susbtrate concentration
  3. Concentration of inhibitors
  4. pH
  5. Temperature
59
Q

3.1.4 Proteins

How does enzyme concentration affect rate of reaction?

A
  • As concentration of enzyme is increased, enzyme activity also increases.
  • More substrate will be broken down if more enzyme is added.
  • Once all the substrate have been broken down, the enzyme will no longer have anything to break down therefore the increase in enzyme activity does not occur forever.
60
Q

3.1.4 Proteins

How does substrate concentration affect rate of reaction?

A
  • Once all enzymes have bound, any susbtrate increase will have no effect on rate of reaction.
  • Available enzymes will be saturated and working at their maximum rate.
61
Q

3.1.4 Proteins

How does concentration of inhibitors affect rate of reaction?

A

COMPETITIVE:
* binds to the active site and prevents subtrate from binding there

NON-COMPETITVE:
* binds to a different site on the enzyme
* doesn’t block substrate binding
* BUT causes other changes in the enzyme so that it can no longer catalyse the reaction efficiently.

62
Q

3.1.4 Proteins

How does pH affect the rate of reaction?

A
  • Each enzyme has an optimum enzyme pH range.
  • Changing the pH outside of this range will result in slow enzyme activity
  • Extreme pH values will cause enzymes to denature
63
Q

3.1.4 Proteins

How does temperature affect the rate of reaction?

A
  • Increasing temp, speeds up rate of reaction.
  • Above optimum temperatures will cause hydrogen and ionic bonds in the tertiary structure to break = active site is no longer complementary to substrate = denaturation
64
Q

3.1.4 Proteins

Contrast competitive and non-competitive inhibitors.

A

COMPETITIVE
* similar shape to substrate = bind to AS
* do not stop reaction = ES complex formed when inhibitor is released.
* Increasing substrate concentration = decreases their effect.

NON-COMPETITIVE
* bind at a different site on the enzyme but the active site.
* may permanently stop the reaction = triggers AS to change shape.
* Increasing substrate concentration = has no impact on their no effect

65
Q

3.1.4 Proteins

State the formula for pH.

A

pH = -log₁₀ [H⁺]

66
Q

3.1.4 Proteins

Outline how to calculate rate of reaction from a graph.

A
  • calculate gradient of line
  • intial rate = draw tangent at t = 0
67
Q

3.1.4 Proteins

Outline how to calculate rate of reaction from raw data.

A
  • change in concentration of product or reactant / time
68
Q

3.1.4 Proteins

Why is advantageous to calculate inital rate?

A
  • represent maximum rate of reaction before concentration of reactants decreases and “end product inhibition”.
69
Q

3.1.4 Proteins

RP 01 Rate of an Enzyme Controlled Reaction:

A
  1. Make 2 control samples:
    * take 2 flat bottom tubes
    * Add 5cm3 of milk suspension to each tube.
    * Add 5cm3 of distilled water to one tube = will indicate abscence of enzyme activity
    * Add 5cm3 of HCl to the other = indicates colour of completely hydrolysed sample
  2. Take 3 test tubes and measure 5cm3 of milk into each. Place in the water bath at 10°C for 5 mins to equilibriate.
  3. Add 5cm3 of trypsin to each test tube and start the timer.
  4. Record how long it takes for the milk sample to completely hydrolyse and become colourless
  5. Repeat step 2 & 3 at temperatures of 20°C, 30°C, 40°C and 50°C.
  6. Find mean time for the milk to be hydrolysed at each temp and find out rate of reaction.
  7. Rate of reaction = 1 / mean time.
70
Q

3.1.4 Proteins

What is the conclusion of RP 01 Rate of an Enzyme Controlled Reaction?

A
  • Milk contains protein called casein, which when broken down causes milk to become colourless
    Trypsin is protease enzyme which hydrolyses the casein protein
  • Temp increases from 10°C, kinetic energy increases = more E-S complexes form = rate of reaction increases up to optimum temperature.
  • above optimum temperature = bonds in tertiary structure break, which changes the shape of active site = enzyme and substrate are no longer complementary.
71
Q

3.1.5 Nucleic acids are important information-carrying molecules

What is the importance of DNA and RNA?

A
  • important information-caryying molecules.
  • In all living cells, DNA holds genetic info and RNA transfers genetic info from DNA to ribosomes.
  • Both are polymers of nucleotides.
72
Q

3.1.5 Nucleic acids are important information-carrying molecules

Describe the structure of a nucleotide.

A
  • Phosphate group
  • Pentose sugar
  • Nitrogenous base.
73
Q

3.1.5 Nucleic acids are important information-carrying molecules

Name the pentose sugars in DNA and RNA.

A

DNA: deoxyribose
RNA: ribose.

74
Q

3.1.5 Nucleic acids are important information-carrying molecules

How are ribosomes formed?

A
  • RNA and proteins
75
Q

3.1.5 Nucleic acids are important information-carrying molecules

State the role of DNA in living cells.

A
  • Base sequence of genetic code for functional RNA and amino acid sequence of polypeptides.
  • Genetic information determines inherited characteristics = influences structure and function of organism.
76
Q

3.1.5 Nucleic acids are important information-carrying molecules

State the role of RNA in living cells.

including mRNA, rRNA, and tRNA.

A
  • mRNA: complementary sequence of 1 gene from DNA with introns (non-coding regions) which are spliced out. Codons translated into polypeptide by ribosomes.
  • rRNA: component of ribosomes (with proteins).
  • tRNA: supplies complementary amino acid to mRNA codons during translation.
77
Q

3.1.5 Nucleic acids are important information-carrying molecules

How do polynucleotides form?

A
  • Condensation reaction between two nucleotide forming phosphodiester bond.
78
Q

3.1.5 Nucleic acids are important information-carrying molecules

Describe the structure of DNA.

A
  • double helix of two polynucleotide strands (deoxyribose) held together by hydrogen bonds between specific complementary base pairs.
  • (A - T)
  • (C - G)

Apples grow on Trees. Cars are parked in Garages.

79
Q

3.1.5 Nucleic acids are important information-carrying molecules

Name the components of a DNA nucleotide.

A
  • Deoxyribose
  • A phosphate group
  • One of the organic bases adenine, cytosine, guanine or thymine.
80
Q

3.1.5 Nucleic acids are important information-carrying molecules

Name the components of an RNA nucleotide.

A
  • Ribose
  • A phosphate group
  • One of the organic bases adenine, cytosine, guanine or uracil
81
Q

3.1.5 Nucleic acids are important information-carrying molecules

Which bases are purine and which are pyrimidine?

A

A , G = 2-ring purine bases
T , C , U = 1-ring pyrimidine bases

82
Q

3.1.5 Nucleic acids are important information-carrying molecules

Name the complementary bases in DNA.

A
  • A + T = 2 H bonds.
  • G + C = 3 H bonds.
83
Q

3.1.5 Nucleic acids are important information-carrying molecules

Name the complementary bases in RNA.

A
  • A + U = 2 H bonds
  • G + C = 3 H bonds
84
Q

3.1.5 Nucleic acids are important information-carrying molecules

Relate the structure of DNA to its function.

A
  • sugar phosphate backbone and many H bonds = provides stability.
  • long molecule = stores lots of information
  • helix is compact = good storage in nucleus.
  • base sequence of triplets codes for amino acids.
  • double stranded = for semi-conservative replication.
  • complementary base pairing = accurate replication.
  • weak H bonds break = strands seperate for replication.
85
Q

3.1.5 Nucleic acids are important information-carrying molecules

Describe the structure of mRNA (messenger RNA) relative to its function.

A
  • Long ribose polynucleotide (shorter than DNA) = breaks down quickly so no excess polypeptide forms.
  • Contains uracil instead of thymine.
  • Single-stranded + linear (no complementary base pairing) = ribosomes can move along strand, tRNA can bind to exposed bases
  • Codon sequence complementary to exons of 1 gene from 1 DNA strand = can be translated into specific polypeptide by ribosomes.
86
Q

3.1.5 Nucleic acids are important information-carrying molecules

Describe the structure of tRNA (transfer).

A
  • Single strand of about 80 nucelotides.
  • folded into a clover shape (H bonds between some complementary bases)
  • Anticodon on one end , amino acid binding site on the other.
  • Anticodon binds to the complementary mRNA codon.
  • Amino acid corresponds to anticodon.
87
Q

3.1.5 Nucleic acids are important information-carrying molecules

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

A
  • tRNA
  • mRNA
  • DNA
88
Q

3.1.5 Nucleic acids are important information-carrying molecules

Describe the structure of RNA molecule.

A
  • relatively short polynucelotide.
89
Q

3.1.5 Nucleic acids are important information-carrying molecules

Why did scientists doubt that DNA carried genetic code?

A
  • DNA is a chemically simple molecule with few components = only has four bases so how could it code for 20 amino acids? They reasoned that identical molecules could not carry different instructions across all organisms.
90
Q

3.1.5 Nucleic acids are important information-carrying molecules

What method is used to replicate DNA?

A
  • semi-conservative replication
91
Q

3.1.5 Nucleic acids are important information-carrying molecules

What does semi-conservative replication of DNA ensure?

A
  • Genetic continuity between generations of cells.
92
Q

3.1.5 Nucleic acids are important information-carrying molecules

Why is DNA replication described as semi-conservative?

A
  • Strands from orginigal DNA molecule act as a template strand.
  • New DNA molecule contains 1 old strand and 1 new strand.
93
Q

3.1.5 Nucleic acids are important information-carrying molecules

Outline the process of semi-conservative replication of DNA.

A
  1. DNA helicase unwindes the double helix and breaks the hydrogen bonds between the complementary base pairs in the polynucelotide strands.
  2. Free nucleotides attach to the exposed bases by complementary base pairing on the template strand.
  3. DNA polymerase catalyses condensation reactions that join adjacent nucleotides on new strand.
  4. H bonds reform and each new DNA molecule contains one strand of the orignal and one strand of the new DNA.
94
Q

3.1.6 ATP

Describe the structure of ATP.

A
  • Nucleotide derivative.
  • formed from a molecule of ribose (pentose sugar), a molecule of adenine (base) and three phosphate groups.
95
Q

3.1.6 ATP

Explain the role of ATP in cells.

A
  • ATP HYDROLASE catalyses ATP -> ADP + Pi.
  • energy released is coupled to metabolic reactions
  • phosphate groups phosphorylate compounds to make them more reactive.
96
Q

3.1.6 ATP

How is ATP resynthesised in cells?

A
  • ATP SYNTHASE catalyses condensation reaction of ADP + Pi -> ATP during photosynthesis or during respiration.
97
Q

3.1.6 ATP

Explain why ATP is suitable as the main energy source in cells.

A
  • High energy bonds between phosphate groups.
  • Small amount of energy is released at a time = less energy wasted as heat.
  • Single-step hydrolysis = energy is available quickly.
  • Readily resythesised.
98
Q

3.1.7 Water

What percentage of cells is made up of water?

A
  • 65%.
99
Q

3.1.7 Water

Water as a metabolite:

A
  • in many metabolic reactions such as condensation and hydroylsis reactions which are used in forming and breaking of chemical bonds.
100
Q

3.1.7 Water

Water as a solvent:

A
  • readily dissolves other substances (O2 and CO2), wastes (ammonia and urea), inorganic ions and small hydrohillic molecules (amino acids, monosaccharides etc) and enzymes whose reaction takes place in solution.
101
Q

3.1.7 Water

Water having a high specific heat capacity:

A
  • water molecules stick together via hydrogen bonds = lots of energy required to break these bonds.
  • Acts as a buffer = minimising temperature fluctuations.
102
Q

3.1.7 Water

Water having a large latent heat of vapourisation:

A
  • Hydrogen bonding makes evapouration more difficult to break
  • Allows organisms to control their body temperatures i.e. providing a cooling effect with little water loss - sweating.
103
Q

3.1.7 Water

Water having strong cohesion between molecules:

A
  • enables effective transport of water in tube like the xylem.
  • supports columns of water in the tube-like transport cells of plants
  • Produces surface tension = when the molecules meet air they tend to pull back into the body of water than escape,
    acting like a skin that is strong enough to support small organisms
104
Q

3.1.7 Water

What is water?

A
  • major component of cells.
  • made up from slightly delta + H atoms and a slightly delta - O atom.
105
Q

3.1.8 Inorganic ions

Where do inorganic ions occur?

A
  • In the solution of the cytoplasm and body fluids of organisms some that are high in concentrations and other that is very low.
106
Q

3.1.8 Inorganic ions

Outline some essential ions.

A
  • Hydrogen ions
  • Iron ions
  • Sodium ions
  • Phosphate ions
107
Q

3.1.8 Inorganic ions

Outline the location and function of hydrogen ions.

A
  1. Solutions
  2. Determines the pH of substances such as blood and functioning of enzymes.
    The higher the concentration of hydrogen ions, lower concentration of pH.
108
Q

3.1.8 Inorganic ions

Outline the location and function of iron ions.

A
  1. Component of haemoglobin
  2. Oxygen carrying molecule in red blood cells.
109
Q

3.1.8 Inorganic ions

Outline the location and function of sodium ions.

A
  1. Cells
  2. Co-transport of glucose and amino acids across membranes.
110
Q

3.1.8 Inorganic ions

Outline the location and function of phosphate ions.

A
  • component of DNA and ATP.
  • structural and store of energy.

AS Biology (5 decks)

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