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Biology Epic Collection > 3.1 biological molecules > Flashcards

3.1 biological molecules Flashcards

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1
Q

Define monomer and polymer

A

Monomer
Small units from which larger molecules are made.

Polymer
large molecule made up of many identical monomers joined together.

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2
Q

Give 3 examples of a Monomer and a Polymer

A

Monomer
Monosaccharide, Amino acid, Nucleotide

Polymer
Polysaccharide, polypeptide, DNA/RNA

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3
Q

what is a condensation reaction?

A

A condensation reaction:
- joins 2 molecules together
- elimination of a water molecule
- formation of a chemical bond e.g. glycosidic bond

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4
Q

What is a hydrolysis reaction?

A

A hydrolysis reaction:
- separates 2 molecules
- addition of a water molecule
- breakage of a chemical bond e.g. glycosidic bond

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5
Q

Diagram showing the conversion of nucleotide monomers to the polymer DNA (image)

A
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6
Q

how are carbohydrates classified into their 3 groups

A

They are determined by how many units they are made of

monosacharides
- glucose
- fructose
- galactose

disaccharides
- maltose
- sucrose
- lactose

polysaccharides
- starch
- glycogen
- cellulose

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7
Q

what are monosacharides? give 3 examples

A

Monosaccharides are the monomers from which larger carbohydrates are made.
e.g. glucose, galactose and fructose

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8
Q

Draw a molecule of glucose (image)

A
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9
Q

How are disaccharides formed by the condensation of 2 monosaccharides?

A

Disaccharides are formed by the condensation of 2 monosacharides.

e.g. maltose, sucrose and lactose

glucose+glucose=maltose
glucose+fructose=sucrose
glucose+galactose=lactose

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10
Q

Define Isomer

A

when molecules have the same molecular formula but the atoms are arranged differently.

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11
Q

How are polysaccharides formed?

A

they are formed by the condensation of many glucose.

examples include:
starch. glucose and cellulose

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12
Q

How is the structure of glycogen related to its function?

A

Function: energy store in animal cells.

  • Structure:
  • Polysaccharide of alpha-glucose.
  • C1-C4 and C1-C6 glycosidic bonds so branched.
  • Structure related to function:
  • Branched; can be rapidly hydrolysed to release
    glucose for respiration to provide energy.
  • Large polysaccharide molecule; can’t leave cell.
  • Insoluble in water; water potential of cell not affected i.e. no
    osmotic effect.
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13
Q

how is the structure of starch related to its function ?

A

Function: energy store in plant cells.

  • Structure:
  • Polysaccharide of alpha-glucose.
  • Made of amylose and amylopectin.
  • C1-C4 glycosidic bonds in amylose; unbranched.
  • C1-C4 and C1-C6 glycosidic bonds in amylopectin; branched.
  • Structure related to function:
  • Helical; compact for storage in cell.
  • Large polysaccharide molecule; can’t leave cell.
  • Insoluble in water; water potential of cell not affected i.e. no osmotic effect.
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14
Q

which sugars are reducing sugars and which are non reducing sugars? what is the test for them?

A

Reducing sugar
- All monosacharides e.g. glucose
- Some disaccharides e.g. maltose/lactose

Non reducing sugars
- no monosaccharides
- some disaccharides e.g. sucrose

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15
Q

Describe Benedict’s test for reducing sugars

A

Benedict’s test for reducing sugars:
1. Add benedict’s reagent (blue) to food sample.
2. Heat in a boiling water bath.
3. Positive = green / yellow / orange / red precipitate (reducing sugar present).
* If negative result (Benedicts reagent doesn’t change colour) then test for non-reducing sugar.

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16
Q

describe Benedict’s test for non-reducing sugars

A

Benedict’s test for non-reducing sugars:
1. Add an equal volume of sample and dilute hydrochloric acid to hydrolyse the sugar.
2. Heat in a boiling water bath.
3. Neutralise with sodium bicarbonate.
4. Carry out normal Benedict’s test.
5. Non-reducing sugar present = green / yellow / orange / red precipitate.

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16
Q

explain how the concentration of glucose in a sample can be determined

A
  1. Produce a dilution series of glucose solutions of known concentrations.
  2. Perform a Benedict’s test on each sample (use same amount of solution for each test) and remove any precipitate (e.g. by centrifuging).
  3. Using a colorimeter, measure the absorbance of each sample to establish a calibration curve.
  4. Repeat with unknown sample and compare the absorbance to the calibration curve to determine glucose concentration.
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17
Q

Describe the test for Starch

A

1) add iodine dissolved in potassium iodide to the solution and shake it
2) blue-black colour = starch present

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18
Q

Name 2 groups of lipid

A

Triglycerides and phospholipids

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19
Q

describe the differences between satutated and unsaturated fatty acids

A
  • The R-group of a fatty acid may be saturated or unsaturated.
  • Saturated: no C=C double bonds in hydrocarbon chain; all carbons fully saturated with hydrogen.
  • Unsaturated: one or more C=C double bonds in hydrocarbon chain.
20
Q

Describe the structure of a phospholipid

A

In phospholipids, one of the fatty acids of a triglyceride is substituted by a phosphate containing group.

21
Q

Describe the emulsion test for lipids

A

1)Add ethanol and shake to dissolve lipids
2)Add water

positive - cloudy white emulsion

22
Q

Describe how the porperties of triglycerides relate to their structure

A
  • Triglycerides: energy storage molecules / energy source.
  • High ratio of C-H bonds to C atoms in hydrocarbon tail.
  • Release more energy than same mass of carbohydrates.
  • Insoluble in water (clump together as droplets)
  • No effect on water potential of cell i.e. no osmotic effect.
  • Triglycerides: energy storage molecules / energy source.
  • High ratio of C-H bonds to C atoms in hydrocarbon tail.
  • Release more energy than same mass of carbohydrates.
  • Insoluble in water (clump together as droplets)
  • No effect on water potential of cell i.e. no osmotic effect.
23
Q

Describe what amino acids are. Give the structure as well.

A

Amino acids are the monomers from which proteins are made.

The general structure of an amino acid
- NH2 is an amine group
- COOH represents a carboxyl group
- R represents a variable side chain

24
Q

Describe the formation and product(s) of a peptide bond.

A

A condensation reaction between 2 amino acids forms a peptide bond.
* Dipeptides are formed by the condensation of two amino acids.
* Polypeptides are formed by the condensation of many amino acids.
* A functional protein may contain one or more polypeptides.

25
Q

Describe the different structural levels of proteins and the bonds present at each level.

A

Primary (1°) structure: Sequence of amino acids in a polypeptide chain (joined by peptide bonds).
Secondary (2°) structure: Hydrogen bonding between amino acids (between the carbonyl O of one amino acid and the amino H of another).
This causes the polypeptide chain to fold into a repeating pattern e.g. alpha helix or beta pleated sheet
Tertiary (3°) structure: Overall 3D structure of a polypeptide.
Held together by interactions between the amino acid side chains (R groups): Ionic bonds; Disulfide bridges; Hydrogen bonds
Quaternary (4°) structure: Some proteins are made of 2+ polypeptide chains.
Also held together by more hydrogen, ionic and disulfide bonds.

26
Q

Describe the test for proteins

A

The biuret test for proteins:
- Add biuret solution: sodium hydroxide + copper (II) sulfate
- protein present: purple. If there is no protein it stays blue
- detects presence of peptide bonds

27
Q

Describe the nature and function of enzymes. Draw and label the free energy graph for reactions with and without enzymes.

A

Enzymes are biological catalysts; they catalyse a wide range of intracellular (within cells) and extracellular (outside cells) reactions that determine structures and functions from cellular to whole-organism level.
Each enzyme lowers the activation energy of the reaction it catalyses (see diagram) → speed up rate of reaction

28
Q

describe the models for enzyme action

A

Lock and key
-active site is a fixed shape. It is complimentary to one substrate

Induced Fit Model
1)before reaction, enzyme active site does not fit substrate
2)active site shape changes as substrate binds and an enzyme substrate complex forms
3)This stresses/distorts bonds in substrate leading to a reaction

29
Q

Describe the specificity of an enzyme with regards to its tertiary structure

A

The properties of an enzyme relate to the tertiary structure of its active site and its ability to combine with complementary substrate(s) to form an enzyme-substrate complex.
Enzymes have a specific shaped tertiary structure and active site - Sequence of amino acids (primary structure) determines tertiary structure.
Active site is complementary to a specific substrate.
Enabling only this substrate to bind to the active site (induce fit) → enzyme-substrate complex.

30
Q

Explain how enzyme concentration affects the rate of reaction

A

Increasing enzyme conc. → rate of reaction increases.
Enzyme conc. = limiting factor (substrate in excess).
More enzymes → more available active sites.
More successful E-S collisions and E-S complexes.

At a certain point, rate of reaction plateaus.
Substrate conc. = limiting factor (all substrates in use).

31
Q

Explain how substrate concentration affects the rate of an enzyme-catalysed reaction.

A

Increasing substrate conc. → rate of reaction increases.
Substrate concentration = limiting factor (too few enzyme molecules to occupy all active sites).
More successful E-S collisions and E-S complexes.

At a certain point, rate of reaction plateaus.
Enzyme conc. = limiting factor (all active sites saturated; excess substrate).

32
Q

Explain how temperature affects the rate of an enzyme-catalysed reaction.

A

Increasing temp. up to optimum → rate of reaction increases.
Increase in kinetic energy.
More successful E-S collisions and E-S complexes.
Increasing temp. above optimum → rate of reaction falls.
Enzymes denature; tertiary structure and active site change shape (hydrogen and ionic bonds break).
Fewer E-S collisions and E-S complexes (substrate no longer binds to active site).
Rate of reaction 0 when all enzymes denatured.

33
Q

Explain how pH affects the rate of an enzyme-catalysed reaction.

A

pH above/below optimum pH → rate of reaction decreases.
Enzymes denature; tertiary structure and active site change shape (hydrogen and ionic bonds break).
Complementary substrate no longer binds/fits to active site.
Fewer E-S collisions and E-S complexes.

34
Q

Explain how concentration of competitive and non-competitive inhibitors affects the rate of an enzyme-catalysed reaction.

A

Competitive inhibitors decrease rate of reaction:
-Similar shape to substrate.
-Competes for / binds to / blocks active site so substrates can’t bind.
-Fewer E-S complexes.
-Increasing substrate conc. reduces effect of inhibitor (level of inhibition dependent on relative concs. of substrate and inhibitor).

Non-competitive inhibitors decrease rate of reaction:
-Binds to site away from the active site (allosteric site).
-Enzyme tertiary structure / active site change shape so substrate can’t bind to active site.
-Fewer E-S complexes.
-Increasing substrate concentration has no effect on rate of reaction as causes permanent change to active site.

35
Q

What are the functions of DNA and RNA in living cells?

A

Deoxyribonucleic acid (DNA) holds genetic information.
Ribonucleic acid (RNA) transfers genetic information from DNA to ribosomes.

36
Q

What are the differences between DNA and RNA structure?

A

DNA nucleotides have pentose sugar deoxyribose whereas RNA nucleotides have ribose.
DNA nucleotides can have thymine whereas RNA nucleotides have uracil instead.
DNA molecules are double stranded whereas RNA molecules are single stranded.
DNA is longer whereas RNA is shorter.

37
Q

Compare the structure and bonding between DNA & RNA polymers?

A

DNA: 2 strands joined in anti-parallel held together by hydrogen bonds between specific complementary base pairs – AT (2 H bonds) and CG (3 H bonds).
A condensation reaction between 2 nucleotides forms a phosphodiester bond.

38
Q

Describe how the structure of DNA is related to its function.

A
  • Large number of (individually) weak hydrogen bonds between complementary bases on different strands → stable / strong molecule and can be unzipped for replication.
  • Double helix (with sugar phosphate backbone) → protects bases / hydrogen bonds.
  • Long molecule → store lots of (genetic) information (can code for sequence of amino acids in the primary structure of a protein).
    Double stranded → semi-conservative replication can occur as both strands can act as templates.
  • Complementary base pairing (A-T, C-G) → accurate replication / identical copies can be made
  • Double helix (coiled) → compact.
39
Q

Describe the process of DNA replication

A

1) DNA Helicase breaks hydrogen bonds between bases, unwinds double helix.
🡪 Two strands which both act as templates.
2) Free floating DNA nucleotides attracted to exposed bases via specific complementary base pairing, hydrogen bonds form (adenine-guanine; guanine-cytosine).
3) DNA polymerase joins adjacent nucleotides on new strand by condensation, forming phosphodiester bonds (🡪 sugar phosphate backbone).

🡪 Replication is semi-conservative – each new strand formed contains one original / template strand and one new strand.
🡪 Ensures genetic continuity between generations of cells.

40
Q

Describe the experiment carried out by Meselson & Stahl and how it was used to provide evidence for semi-conservative replication.

A

Bacteria grown in a nutrient solution containing heavy nitrogen (15N) for several generations.
Nitrogen incorporated into bacterial DNA bases.
Bacteria then transferred to a nutrient solution containing light nitrogen (14N) and allowed to grow and divide twice.
During this process, DNA from different samples of bacteria was extracted, suspended in a solution in separate tubes and spun in a centrifuge.

41
Q

Draw and label ATP

A

Ribose, a molecule of adenine, 3 phosphate groups (Pi).
Nucleotide derivative (modified form of nucleotide).
The structure of ADP (adenosine diphosphate) is the same as ATP minus a phosphate.

42
Q

Give the basic equation for ATP hydrolysis and name the enzyme involved in this process.

What occurs during ATP hydrolysis and what does this result in?

A

ATP → ADP + Pi
Catalysed by the enzyme ATP hydrolase.

Bonds between inorganic phosphate groups are high energy bonds so by breaking one of these bonds a small amount of energy is released.
- Can be coupled to energy requiring reactions within cells, to provide energy for e.g. active transport, protein synthesis.
- The inorganic phosphate released can be used to phosphorylate other compounds e.g. glucose, often making them more reactive (i.e. lowers activation energy).

43
Q

Give the basic equation for ATP condensation and name the enzyme involved in this process.

When does ATP condensation occur?

A

ADP + Pi 🡪ATP
Catalysed by the enzyme ATP synthase.

Happens during respiration or photosynthesis

44
Q

Describe the properties of ATP that make it a suitable immediate energy source.

A
  • ATP releases energy in small, manageable amounts (so no energy wasted).
  • Only one bond is hydrolysed (single reaction) to release energy (which is why energy release is immediate).
45
Q

Describe the structure and bonding present in water.

A
  • Water is a covalent compound made up of 2 hydrogen atoms + 1 oxygen.
  • Oxygen has a slight negative charge and hydrogen atoms have a slight positive charge = (di)polar.
  • Hydrogen bonds (attractive force between opposite charges) form between water molecules causing them to ‘stick together’.
46
Q

State where inorganic ions can be found in the body.

A

Occur in solution in the cytoplasm and body fluids of organisms.
Some in high concentrations and others in very low concentrations.

47
Q

What are the 4 key inorganic ions and what role do they have in the body?

A

PHOSPHATE IONS – PO43-
Attached to other molecules as a phosphate group.
Phosphate ions form the phosphate groups of DNA/RNA nucleotides.
Enables nucleotides to join together – phosphodiester bonds form.
Phosphate ions form the phosphate groups of ATP - Breaking the bonds between the phosphate groups in ATP releases energy.

SODIUM IONS – Na+
Co transport of glucose and amino acids across cell membranes.

HYDROGEN IONS – H+
Maintain pH levels in the body. * Too much H+ = acidic (low pH).
Too little H+ = alkaline (high pH).
Affects rate of enzyme controlled reactions as can cause enzymes to denature.

IRON IONS – Fe2+
Component of (haem group of) haemoglobin which is contained in red blood cells.
Transports oxygen around the body – oxygen temporarily binds to it so it becomes Fe3+.

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