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Biological Molecules Flashcards

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AP® Biology flashcards Biology 101 flashcards
1
Q

Define monomer

A

Smaller units that join together to form larger molecules

Monosaccharides
Amino acids
Nucleotides

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2
Q

Define polymer

A

Molecules formed when many monomers join together

Polysaccharides
Proteins
DNA/RNA

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3
Q

What happens in a condensation reaction

A

A chemical bond forms between 2 molecules and a molecule if water is removed/produced

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4
Q

What happens in a hydrolysis reaction

A

A water molecules is used to break a chemical bind between 2 molecules

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5
Q

Name the three hexose monosaccharides

A

Glucoses
Galactose
Fructose

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6
Q

Name the type of bond formed when monosaccharides react

A

Glycosidic bond

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7
Q

Name three disaccharides

Describe how they form

A

Maltose
Sucrose
Lactose

Condensation reactions forms a glycosidic bond between 2 monosaccharides

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8
Q

Whats different between alpha and beta glucose

A

Alpha has OH at bottom on right side
Beta has OH at top on right side

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9
Q

Describe structure and function of starch

A

Storage polymer of alpha glucose in plant cells
Insoluble- no osmotic effect in cell
Large- does not diffuse out of cell

Made from amylose:
1,4 glycosidic bonds
Helix with intermolecular H-bonds (compact)

And amylopectin:
1,4 and 1,6 glycosidic bonds
Branched- many terminal ends for hydrolysis into glucose

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10
Q

Describe structure and function of glycogen

A

Main storage of alpha glucose in animal cells
1,4 and 1,6 glycosidic bonds
Branched- many terminal ends for hydrolysis
Insoluble- no osmotic effect and does not diffuse out of cells
Compact

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11
Q

Describe the structure and function of cellulose

A

Polymer of beta glucose gives rigidity to plant cell walls

1,4 glycosidic bonds
Straight-chain, unbranched molecule
Alternate glucose molecules are rotated 180
H-bonds crosslinks between parallel strands form microfibrils= high tensile strength

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12
Q

Describe Benedict’s test for reducing sugars

A
  1. Add an equal volume of Benedict’s reagent to a sample
    2.Heat the mixture in an electric water bath at 100•c for 5 minutes
    3.positive result- colour change from blue to brick red
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13
Q

Describe benedicts test for non reducing sugars

A

1.Negative results: Benedicts reagent remains blue
2.Hydrolyse non-reducing sugars by adding 1cm^3 of HCL. Heat in water bath for 5 minutes
3.Neutralise the micture usinh sodium carbonate solution
4.Add an equal volume of Benedict’s reagent to a sample
5.Heat the mixture in an electric water bath at 100•c for 5 minutes
6.positive result- colour change from blue to brick red

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14
Q

Describe test for starch

A

1.Add iodine solution
2.Positive result: colour change from orange to blue/black

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15
Q

Describe how to test for lipids

A

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

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16
Q

How do triglycerides form

A

Condensation reaction between 1 molecule of glycerol and 3 fatty acids form an ester bond

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17
Q

Contrast saturated and unsaturated fatty acids

A

Saturated:
Contain only single bonds
Straight chained molecules
Higher melting point

Unsaturated:
Contain C=C bonds
Have kinks within the molecule
Lower melting point

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18
Q

Relate the structure of triglycerides to their functions

A

Insoluble hydrocarbon chain- no effect on water potential of cells
Less dense then water- buoyancy of aquatic animals

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19
Q

Describe the structure and function of phospholipids

A

Glycerol attached to 2 hydrophobic fatty acids trails and 1 hydrophilic polar phosphate head

Forms phospholipid bilayer in water- component of membraine

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20
Q

Compare phospholipids to triglycerides

A

Both have glycerol backbones
Both may be attached to a mixture of saturated and unsaturated fatty acids
Both contain elements C, H, and O
Both formed by condensation reactions

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21
Q

Contrast phospholipids and triglycerides

A

Phospholipid:
2fatty acids and 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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22
Q

Are phospholipids and triglycerides polymers

A

No
They are not made of small repeating units

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23
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

Forms O delta neg (slight neg charge)
And H delta pos (slight pos charge)

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24
Q

State 4 biologically important properties of water

A

.Metabolite- solvent for chemical reactions in the body
.High specific heat capacity
.High lateny heat of vapourisation
.Cohesion between molecules

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25
Q

Explain why water is significant to living organisms

A

Solvent for polar molecules during metabolite reactions

Enables organisms to avoid fluctatiobs in core temperatures

Cohesion-tension of water molecules in transpiration stream

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26
Q

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

A

Ions that do not contain carbon atoms
Found in cytoplasm and extracellular fluid
May be high or very low concentrations

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27
Q

Explain the role of hydrogen ions in the body

A

High concentrations of H+ = loe (acidic) pH
H+ ions interact with H-bonds and ionic bonds in tertiary structure of proteins which cab cause them to denature

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28
Q

Explain the role of ions in the body

A

Fe2+ bonds to porphyrin ring to form haem group in haemoglobin

Haem group has binding site to transport 1 O2 molecule around the body in bloodstream

4 haem groups per haemoglobin molecule

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29
Q

Explosion 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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30
Q

Explain the role of phosphate ions in the body

A

Component of:
DNA
ATP
NADP
cAMP

31
Q

What is the general structure of an amino acid

A

COOH-carboxyl group

R- variable group

NH2- amino group

32
Q

Describe the test for proteins

A

1.add equal volume of sodium hydroxide to sample at room temperature
2.Add drops of dilute copper sulfate solution then swirl to mix
3. POS results: blue to purple color change
NEG results: remains blue

33
Q

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

A

20 and they only change by their R group

34
Q

How do dipeptides and polypeptides form

A

Condensation reaction forms peptide bonds between amino acids and removes a molecule of water

35
Q

How many levels of protein structure are there

A

4

36
Q

Define the primary structure of a protein

A

Sequence number and type of amino acid in the polypeptide
Determined by the sequence of codons on mRNA

37
Q

Define secondary structure of a protein

A

Hydrogen bonds form between O delta neg attached to -C=O and H delta pos attached to -NH

38
Q

Describe the 2 types of secondary structures

A

Alpha helix:
All N-H bonds on same side of protein chain
Spiral shape
H-bonds parallel to helical axis

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

39
Q

Define tertiary structure of a protein and name all the bonds present

A

3D structure formed by further folding of polypeptide

Disulphide bridges
Ionic bonds
Hydrogen bonds

40
Q

Describe each bond in the tertiary structure of proteins

A

Disulphide bridges- strong covalent S-S bibs between molecules of the amino acid cysteine

Ionic bonds- relatively strong bonds between charged R groups

Hydrogen bonds- numerous and easily broken

41
Q

Define quaternary structure of a protein

A

Functional proteins may consist of more than one polypeptide

Precise 3D structure held together by the same toes of bonds as tertiary structure

May involve addition of prosthetic groups e.g. metal ions or phosphate groups

42
Q

Describe structure and function of globular proteins

A

Spherical and compact

Hydrophilic R groups face outwards and hydrophobic R groups face inwards= usually water soluble

Involved in metabolic processes e.g. enzymes and haemoglobin

43
Q

Describe the structure and function of fibrous proteins

A

Can form long chains or fibres

Insoluble in water

Useful for support and structure

44
Q

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

A

1.Use capillary tube to spot mixture onto pencil origin line and place chromatography paper in solvent

  1. Allow solvent to run until it almost touches the other end of paper. Amino acids move different distances based on relative attraction to paper and solubility in solvent
  2. Use revealing agent or UV light to see spots
  3. Calculate R value and match to database
45
Q

What are enzymes

A

Biological catalysts for intra and extracellular reactions

Specific tertiary structure determines shape of active site, complimentary to a specific substrate

Formation of enzyme-substrate complex lowers activation energy of metabolic reactions

46
Q

Explain the induced fit model of enzyme action

A

Share of active site is not directly complementary to substrate and is flexible

Conformational change enabled enzyme-substrate complexes to form

This puts strain on substrate bonds, lowering activation energy

47
Q

How have models of enzyme action changed

A

Initially lock and 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

48
Q

Name 5 factors that affect the tastes of enzyme controlled reactions

A

.Enzyme concentration
. Substrate concentration
. Concentration of inhibitors
.pH
. Temperature

49
Q

How does substrate concentration affect rate of reaction

A

Rate increases proportionally to substrate concentraction

50
Q

How does enzyme concentration affect the rate of reaction

A

Rate increases proportionally to enzyme concentration

51
Q

How does temperature affect rate of reaction

A

Rate increases as kinetic energy increases and peaks at optimum temperature

52
Q

How does pH affect rate of reaction

A

Enzymes have a narrow optimum pH range

Outside range enzymes denature

53
Q

Constant competitive and non competitive inhibitors

A

Competitive inhibitors:
.Similar shape to substrate- bind to active site
.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

54
Q

Outline how to calculate rate of reaction from a graph

A

Calculate gradient of line or gradient of tangent to a point
Initial rate: draw tangent at t= 0

55
Q

Outline how to calculate rate of reaction from raw data

A

Change in concentration of product or reactant/time

56
Q

Why is it advantageous to calculate initial rate

A

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

57
Q

Name the pentose sugar in DNA and RNA

A

DNA-deoxyribose

RNA-oxyribose

58
Q

State the role of DNA in living cells

A

Base sequence of genes codes for functional RNA and amino acid sequence of polypeptides

Genetic information determines inherited characteristics = influences structure and function of organisms

59
Q

State the role of RNA in living cells

A

mRNA-complementary sequence to 1 gene from DNA with introns spliced out. Codons can be translated into a polypeptide by ribosomes.

rRNA-component of ribosomes

tRNA-supplies complimentary amino acid to mRNA codons during translation

60
Q

How do polypeptides form

A

Condensation reactions between nucleotides form strong phosphodiester bonds

61
Q

Describe the structure of DNA

A

double helix of 2 polynucleotide strands

H-bonds between complementary base pairs on opposite strands

62
Q

Name the complementary base pairs in DNA

A

Adenine + thymine

Guanine + cytosine

63
Q

Name the complimentary base pairs in RNA

A

Adeline + Uracil

Cytosine + Guanine

64
Q

Name the complimentary base pairs in RNA

A

Adeline + Uracil

Cytosine + Guanine

65
Q

Relate the structure of DNA to its functions

A

.Sugar phosphate backbone and many H-binds provide stability
. Long molecules stores lot of information
. Helix is contact for storage in nucleus
. Base sequence of triplets code die amino acids
. Double stranded for semi conservative replication
. Complimentary base pairing for accurate replication
. Weak H-bonds break so strands separate for replication

66
Q

Describe the structure of mRNA

A

.Long ribose polynucleotide(shorter then DNA)
. Contains uracil not thymine
. Single stranded and linear
. Codon sequence is complimentary to exons of 1 gene from 1 DNA strand

67
Q

Relate the structure of mRNA to its function

A

. Breaks does quickly so no excess polypeptide forms
.ribosome can move along stand and tRNA can bind to exposed bases
. Can be translated into a specific polypeptide by ribosomes

68
Q

Describe the structure of tRNA

A

.Single stranded of about 80 nucleotides
.Filed into clover shape
.Anticodon on one end, amino acid binding site on the other

Anticodon binds to complementary mRNA codon
Amino acid corresponds to anticodon

69
Q

Why is DNA replication described as semi conservative

A

Sienna from original DNA molecule act as a temple
New DNA molecule contains 1 old strand and 1 new strand

70
Q

Outline the process of semi conservative DNA replication

A

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

71
Q

Describe the structure of adenosine triphosphate (ATP)

A

Nucleotide derivate of adenine with 3 phosphate groups

72
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

73
Q

How is ATP re synthesised in cells

A

ATP synthase catalyses condensation reaction between ADP + Pi

During photosynthesis and respiration

74
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 resynthesised

Biology (8 decks)

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