biolmol1

Question 1

Monomers.

Answer 1

The smaller units from which larger molecules are made (via condensation reactions).

Question 2

Polymers.

Answer 2

Molecules made from large number of monomers joined together (via condensation reactions).

Question 3

Condensation.

Answer 3

A reaction joining two molecules together with the formation of a chemical bond. Eliminates a molecule of water.

Question 4

Hydrolysis.

Answer 4

A reaction breaking a chemical bond between two molecules upon the addition of a water molecule, under suitable conditions.

Question 5

Molar solution.

Answer 5

A solution containing one mole of solute in each litre of solution.

Question 6

Metabolism.

Answer 6

Collective name for all the chemical processes taking place in living organisms.

Question 7

Monosaccharides.

Answer 7

The monomers from which larger carbohydrates are made - condensation reactions form glycosidic bonds between monosaccharides. SOLUBLE, reducing sugars.

Question 8

Disaccharides.

Answer 8

Two monosaccharides held together by a glycosidic bond after a condensation reaction.

Question 9

Polysaccharides.

Answer 9

Polymers formed by combining MANY monosaccharide molecules via condensation reactions. Large, insoluble molecules good for storage.

Question 10

Maltose (constituent monosaccharides?)

Answer 10

Glucose and glucose.

Question 11

Sucrose (constituent monosaccharides?)

Answer 11

Glucose and fructose.

Question 12

Lactose (constituent monosaccharides?)

Answer 12

Glucose and galactose.

Question 13

Difference between alpha and beta glucose?

Answer 13

The -H and -OH groups on the 1st carbon atom (only, NOT 4th aswell) reversed.

Question 14

What is Starch?

Answer 14

Alpha glucose polysaccharide used as a major energy source.

Question 15

Starch found where, in what form?

Answer 15

In plants only, in the seeds and storage organs such as potato tubers. In the form of small grains.

Question 16

Amylose’s structure-function relationship?

Answer 16

Long, unbranched alpha glucose chain, with a coiled structure.

Compact, so good for storage as more energy-holding bonds packed in a small space.

Question 17

Amylopectin’s structure-function relationship?

Answer 17

Long, branched alpha glucose chain.

Side branches give easy access to MANY glycosidic bonds which can be acted on simultaneously by enzymes - rapid release of alpha glucose monomers.

Question 18

Other benefits of starch’s structure? (Other than coiled, branched etc.)

Answer 18

Insoluble in water, so doesn’t affect wp of cells, so water not drawn in by osmosis.

Large and insoluble, does not diffuse out of cells.

When hydrolysed, alpha glucose produced which easily transported and readily used in respiration.

Question 19

What is Glycogen?

Answer 19

Main carbohydrate storage molecule.

Question 20

Where is Glycogen found?

Answer 20

In animals and bacteria.

Found as small granules in muscles and liver, where there is a high metabolic activity.

Question 21

Glycogen’s structure-function relationship?

Answer 21

Very highly branched. Insoluble. Compact.

Many branches mean that enzymes can act simultaneously on the many glycosidic bonds they have easy access to - rapidly broken down to alpha glucose monomers.

Insolubility means that it can’t diffuse out of cells or affect cell wp, so no water drawn into cells by osmosis.

Compact, so lots can be stored in a small space.

Question 22

Why is glycogen’s higher levels of branching than amylopectin beneficial for animals?

Answer 22

Animals have a higher metabolic rate than plants so also have a higher respiratory rate.

Question 23

What is cellulose?

Answer 23

A beta glucose polymer found in plants, more specifically their cell walls.

Question 24

What is cellulose’s structure-function relationship?

Answer 24

Straight, unbranched parallel chains, held together by MANY hydrogen bonds. Grouped together to form microfibrils which are then grouped together to form fibres.

Provides strength and rigidity in cell walls - can do this because of microfibrils etc.- gives rigidity.

Also prevents cells from bursting by exerting an inward pressure stopping further influx of water into cells BY OSMOSIS.

Maintains stem and leaves in a turgid and therefore semi-rigid state so they can provide the maximum surface area for photosynthesis.

Question 25

Why do hydrogen bonds form between adjacent cellulose chains?

Answer 25

Because adjacent beta glucose monomers are rotated 180 degrees due to -H and -OH reversal on 1st Carbon. Formed between -OH groups.

Question 26

Describe how to test for REDUCING sugars.

Answer 26

Benedict’s Test

1. Add 2cm3 of sample solution or sample ground up in water to test tube with 2cm3 of Benedict’s reagent (alkaline CuSO4 solution).
2. Heat in a gently boiling water bath for 5 minutes.
3. REDUCING sugar present in increasing amounts by following colours
   blue - green - yellow - orange - brick red.
   (test is therefore semi-quantitative as we can use the colour change to estimate the approximate amount of REDUCING sugar present. A better, more accurate, method would be to filter the resulting solution and to weigh the CuO precipitate).

Question 27

Describe how to test for NON-REDUCING sugars.

Answer 27

First carry out REDUCING sugar test to ensure no REDUCING sugar present already.

1. Add 2cm3 of food sample solution to 2cm3 of dilute HCl. HCl hydrolyses any disaccharides etc. into its constituent, REDUCING monosaccharides.
2. Add sodium hydrogencarbonate to neutralise any excess acid as Benedict’s will not work in an acidic environment.
3. Check with pH paper to ensure ensure solution is alkaline (Benedict’s works in an alkaline environment).
4. Re-test resulting solution by heating with 2cm3 of Benedict’s in a gently boiling water bath for 5 minutes.
5. If a NON-REDUCING sugar was present in the original sample, colour change observed like with REDUCING sugar test.

NB: Test is the same, but includes hydrolysis of any NON-REDUCING sugars to constituent REDUCING sugars.

Question 28

Describe the test for starch.

Answer 28

1. Add Iodine DISSOLVED IN KI to the sample.

2. Sample will change colour from orange-brown to dark blue-black if starch present.

Question 29

Solubility of Lipids?

Answer 29

Insoluble in water, but soluble in organic solvents like alcohols and acetone.

Question 30

What are the two types of lipids? Describe their structures.

Answer 30

Triglycerides - glycerol molecule ester bonded to three fatty acid molecules, formed via condensation reactions.

Phospholipids - glycerol molecule ester bonded to two fatty acid molecules. Also bonded to a polar phosphate group, instead of a third fatty acid group.

Question 31

Describe a saturated fatty acid.

Answer 31

Saturated fatty acid = a fatty acid with no C=C double bonds (i.e. max saturation with H atoms).

Question 32

Describe an unsaturated fatty acid.

Answer 32

Unsaturated fatty acid = a fatty acid WITH C=C bonds present (i.e. not max saturation with H atoms).

Question 33

4 functions of triglycerides?

Answer 33

1. Source of energy - oxidation releases water and twice more energy per gram than carbohydrates.
2. Waterproofing - insoluble in water. Plants and insects have waxy, lipid cuticles to prevent water loss. Animals produce an oily secretion from sebaceous glands in skin.
3. Insulation - fats help to retain body heat and act as electrical insulators around nerve cells (in myelin sheaths).
4. Protection - fat is often stored around delicate organs, such as the kidney.

Question 34

Triglycerides’ structure-function relationship?

Answer 34

Long hydrocarbon tails of fatty acids have high C-H (high-energy bonds) : C ratio, so good source of energy.

Low mass : energy ratio - beneficial to animals as less to carry around.

Large, non-polar molecules therefore insoluble so don’t affect cell wp, and don’t draw in water BY OSMOSIS into cells - good for storage.

High ratio of H : O atoms, so water released when oxidised - important source of water for organisms living in arid climates.

Question 35

Phospholipids’ structure-function relationship?

Answer 35

In aqueous environments, phospholipid molecules form a bilayer within cell-surface membranes - form a hydrophobic barrier between inside and outside of cells.

= They do this because the hydrophilic phosphate head orients itself towards water, while the hydrophobic fatty acid tail orients itself away from water.

The phospholipid structure allows them to combine with carbohydrates to form glycolipids - important in cell recognition.

Question 36

Describe the test for lipids.

Answer 36

Emulsion test

1. Take a dry, grease-free test tube.
2. 2cm3 sample and 5cm3 ethanol.
3. Shake tube thoroughly to dissolve sample’s lipids.
4. Add 5cm3 of water and shake gently.
5. Cloudy-White colour indicates the presence of a lipid.
6. Repeat 1-5 as a control, using water instead of sample. Solution should remain clear.

Question 37

Amino acids.

Answer 37

Basic monomer units which combine to form a polypeptide polymer via a series of condensation reactions.

Over 20 naturally occurring acids common to all organisms.

Acids only differ in their R side group.

Question 38

Draw the structures of alpha and beta glucose, triglycerides, phospholipids and an amino acid.

Answer 38

Check if correct.

Question 39

Primary structure of a protein.

Answer 39

Sequence of amino acids in a polypeptide chain, held together by peptide bonds.

Mutation causing change in only 1 amino acid - could prevent protein function as proteins are highly specific.

Question 40

Secondary structure of a protein.

Answer 40

Hydrogen bonds form between -NH (d+) and -C=O (d-), causing the long polypeptide chain to twist into a 3D shape: either the alpha helix or the beta pleated shape.

Question 41

Tertiary structure of a protein.

Answer 41

Further folding and twisting of the secondary structure caused by disulphide bridges (if S in R groups) which are strong covalent bonds, ionic bonds (which are easily broken by pH changes - think about zwitterions) and hydrogen bonds.

=> giving the specific 3D structure of a protein - crucial in recognition of protein and its function.

Question 42

Quaternary structure of a protein.

Answer 42

Number of different polypeptide chains linked together in various ways (by vdw etc.).

Can contain prosthetic groups such as Fe2+ in haemoglobin.

3D structure crucial to function, but primary structure decides 3D structure.

Question 43

Give 4 main function areas of proteins.

Answer 43

1. Transport - channel, carrier etc.
2. Enzymes - soluble, biological metabolic catalysts.
3. Structural proteins - keratin and collagen.
4. Antibodies - have variable regions specific to antigens. Involved in the immune response.

Question 44

Describe the test for proteins.

Answer 44

Biuret Test - detects peptide bonds.

1. Add sample of solution to test tube with an equal volume of NaOH solution at room temperature.
2. Add a few drops of very dilute (0.05%) CuSO4 solution and mix gently.
3. If peptide bonds present in sample, blue to purple colour change observed.

Question 45

What are enzymes?

Answer 45

Biological catalysts, which speed up the rate of metabolic reactions by lowering their activation energy.

They work at both a cellular level (respiration) and for the organism as a whole (digestion).

Can be intra-/extra-cellular.

Question 46

Comment on the specificity of enzymes.

Answer 46

Enzymes are highly specific due to their tertiary structure, due to their primary structure which is coded for by DNA.

Specific functional region of enzyme = active site - usually only catalyses one reaction.

Question 47

What is an enzyme-substrate complex?

Answer 47

The complex of the substrate bound to the active site of an enzyme - temporary bonds form between certain amino acids of the active site and groups on the substrate molecule.

Question 48

Describe the ‘induced fit’ model of enzyme action.

Answer 48

M1 - Active site is not initially complementary.

M2 - Active site is flexible and changes shape to wrap around the substrate.

M3 - …allowing the substrate to fit into the active site and the E-S complex to form.

Question 49

Describe the ‘lock and key’ model and describe the main difference from the ‘induced fit’ model.

Answer 49

M1 - active site does not change shape and is fixed in lock and key model.

Similar to induced fit, but was changed when new evidenced showed that E-S complex changed shape to complete the fit.

Question 50

How do enzyme properties relate to their tertiary structure?

Answer 50

1. Very specific - will usually only catalyse one reaction, as only one complementary substrate will fit into the active site.
2. Active site’s shape determined by tertiary structure, determined by primary structure, determined by a gene - a mutation could therefore prevent properly-functioning enzymes.
3. Different enzymes have different tertiary structures and therefore different-shaped active site - so no E-S complex.
4. Changes in pH/temperature can alter tertiary structure and therefore the active site - can therefore prevent E-S complexes from forming.