Biology A Level - Biological Molecules summary questions

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Last updated 4:56 PM on 8/17/26
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1
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How does hydrogen bonding occur between water molecules? Include diagram.

Water molecule are polar (meaning charge is unevenly distributed), hence contain d+ hydrogen atoms and d- oxygen atoms. Between water molecules, there is a force of attraction between the d+ H and d- O, so a weak hydrogen bond forms.

<p>Water molecule are polar (meaning charge is unevenly distributed), hence contain d+ hydrogen atoms and d- oxygen atoms. Between water molecules, there is a force of attraction between the d+ H and d- O, so a weak hydrogen bond forms. </p>
2
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Name the 6 properties of water that are important for life processes. Relate each one to living organisms. 

Cohesion/ surface tension - supports small insects eg. pond skaters.

Adhesion - Capillary action: movement of H2O through blood vessels and xylem.

Latent heat of evaporation - Acts as a coolant eg. sweat

High specific heat capacity - Maintains constant temperatures in cellular environments.

Density of ice - Insulating layer above lake water so fish can survive in winter.

Universal solvent - Medium for chemical reactions and transport eg. blood.

3
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Explain hydrolysis and condensation reactions and provide examples in ALL biological molecules

Condensation: joining of two monomers - releases water

Hydrolysis: breaking down of a polymer - requires water

Lipids - H on glycerol and OH on fatty acid react x3 to form 3 ester bonds and 3 H2O

Carbs - OH groups on C1 and C4 react to form a 1,4 glycosidic bond and H2O

Protein - OH group (in carboxyl group) and H (on amine group) react to form peptide bond and H2O

4
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What are the chemical elements that make up each biological molecule?

Protein - C,H,O,N,S

Lipids - C,H,O

Carbs - C,H,O

Nucleic acid - C,H,O,N,P

5
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Draw an alpha and beta glucose molecule. Describe the differences and similarities between them. 

Similarities:

Hexose ring structure

Same molecular formula - C6H12O6

Same structure on C2,3,4,5 + 6

Difference:

Alpha - C1 has same position of OH and H as C4 (OH below)

Beta - C1 has opposite position of OH and H as C4 (OH above)

<p><strong>Similarities</strong>:</p><p>Hexose ring structure </p><p>Same molecular formula - C<sub>6</sub>H<sub>12</sub>O<sub>6</sub></p><p>Same structure on C<sub>2,3,4,5 + 6</sub></p><p><strong>Difference</strong>:</p><p>Alpha - C<sub>1 </sub>has same position of OH and H as C<sub>4</sub> (OH below)</p><p>Beta - C<sub>1</sub> has opposite position of OH and H as C<sub>4</sub> (OH above)</p>
6
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Draw a ribose molecule. Describe the difference between glucose and ribose. 

Ribose only has 5 carbons - pentose molecule

<p>Ribose only has 5 carbons - pentose molecule </p>
7
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Describe how the structure of starch (amylose/amylopectin) relates to its function. 

Amylose: long chain of a-glucose molecules joined together by 1-4 glycosidic bonds to form an alpha helix (strengthened by H bonds

Function: Storage molecule in plants - compact, water insoluble

Amylopectin: Branched chain of a-glucose molecules with 1-4 bonds in chain and 1-6 bonds at branching points

Function: Storage molecule in plants - Compact, insoluble, branched so many free ends where glucose can be broken off for reactions (hydrolysis)

<p>Amylose: long chain of a-glucose molecules joined together by 1-4 glycosidic bonds to form an alpha helix (strengthened by H bonds </p><p>Function: Storage molecule in plants - compact, water insoluble </p><p>Amylopectin: Branched chain of a-glucose molecules with 1-4 bonds in chain and 1-6 bonds at branching points </p><p>Function: Storage molecule in plants - Compact, insoluble, branched so many free ends where glucose can be broken off for reactions (hydrolysis)</p>
8
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Describe how the structure of glycogen relates to its function. 

Structure: Branched chain of a-glucose molecules with 1-4 bonds in chain and 1-6 bonds at branching points. Branches more frequently than amylopectin

Function: Storage molecule in animals and fungi - compact, many free ends to release glucose (hydrolysis)

<p>Structure: Branched chain of a-glucose molecules with 1-4 bonds in chain and 1-6 bonds at branching points. Branches more frequently than amylopectin</p><p>Function: Storage molecule in animals and fungi - compact, many free ends to release glucose (hydrolysis)</p>
9
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Describe how the structure of cellulose relates to its function. 

Structure: chains of b-glucose where every other molecule is rotated 180º. Many chains joined together by H bonds to form large sheet of cellulose

Function: Cell wall of plants - strong (many H bonds) and water insoluble.

<p>Structure: chains of b-glucose where every other molecule is rotated 180º. Many chains joined together by H bonds to form large sheet of cellulose</p><p>Function: Cell wall of plants - strong (many H bonds) and water insoluble.</p>
10
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Compare and contrast the structure of triglycerides to phospholipids. 

Triglycerides: glycerol and 3 fatty acids joined by ester bonds, can be saturated or unsaturated (contains C=C), non-polar, stores energy.

Phospholipids: glycerol, 2 fatty acid and a phosphate group, hydrophilic head and hydrophobic tail, found in plasma membrane.

<p>Triglycerides: glycerol and 3 fatty acids joined by ester bonds, can be saturated or unsaturated (contains C=C), non-polar, stores energy. </p><p>Phospholipids: glycerol, 2 fatty acid and a phosphate group, hydrophilic head and hydrophobic tail, found in plasma membrane. </p>
11
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Explain the differences between saturated and unsaturated fatty acids 

Saturated - all C-C, no C=C, higher melting point as they are more compact, so bonds harder to break

Unsaturated - contains C=C, lower melting point as they are less compact, so bonds easier to break

12
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Describe the general structure of an amino acid. 

Amine group - NH2

Central Carbon - CH

R group - a range of chemical groups different in each amino acids. 20 different amino acids naturally occuring in the human body.

Carboxyl group - COOH

<p>Amine group - NH<sub>2</sub></p><p>Central Carbon - CH</p><p>R group - a range of chemical groups different in each amino acids. 20 different amino acids naturally occuring in the human body. </p><p>Carboxyl group - COOH</p>
13
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Explain how a peptide bond forms.

Condensation reaction: bond forms between the C in carboxyl group and N in amine group, releasing H2O and forming peptide bond

Broken via hydrolysis reaction.

<p>Condensation reaction: bond forms between the C in carboxyl group and N in amine group, releasing H<sub>2</sub>O and forming peptide bond </p><p>Broken via hydrolysis reaction. </p>
14
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Outline the four different levels of protein structure. 

Primary - the sequence of amino acids in a polypeptide chain. Determines all other levels of structure

Secondary - Polypeptide chain fold into a helix or b pleated sheet due to hydrogen bonding within the peptide chain.

Tertiary structure - 3D shape held in place by bond between the R groups in the polypeptide chains: covalent disulphide, ionic bonds, hydrophobic and hydrophilic interactions. Vital for function.

Quaternary - NOT ALL HAVE: Composed of 2+ polypeptide chains/subunits interacting and the interactions between peptide chains and prosthetic groups

15
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Compare conjugated, globular and fibrous protein structures providing examples of each.

Conjugated - contains a prosthetic group ie. non-protein component. Example: Haemoglobin - contain Fe2+ in haem group

Globular proteins - compact, water soluble, spherical shape. Hydrophilic R groups on outside, hydrophobic inside. Example: Insulin - transported in bloodstream

Fibrous proteins - Long and insoluble. contain many hydrophobic R groups. Repetitive and organise structure with simpler 3D structure. Examples: Keratin, Elastin and Collagen - all have strength and flexibilty function so must be strong

<p>Conjugated - contains a prosthetic group ie. non-protein component. Example: Haemoglobin - contain Fe<sup>2+</sup> in haem group</p><p>Globular proteins - compact, water soluble, spherical shape. Hydrophilic R groups on outside, hydrophobic inside. Example: Insulin - transported in bloodstream</p><p>Fibrous proteins - Long and insoluble. contain many hydrophobic R groups. Repetitive and organise structure with simpler 3D structure. Examples: Keratin, Elastin and Collagen - all have strength and flexibilty function so must be strong </p>
16
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Explain how to chemically test for reducing and non reducing sugars

Benedict’s test for reducing sugars:

Add equal parts Benedict’s solution to sample and warm in water bath (~80ºC)

Benedict’s test for non - reducing sugars:

Add 2cm of ~1.5M HCl and heat in water bath for 3 mins (to hydrolyse glycosidic bonds)

Cool and neutralise with NaHCO3, checking with pH paper

Carry out Benedict’s test as above

Results:

Red - Large, Orange - moderate, Green - traces, Blue - none

17
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Explain how to chemically test for proteins, lipids and starch

Proteins - Biuret test for proteins

Add Biuret A and B. Lilac colour indicates protein present

Lipids - Emulsion test

Add ethanol, shake and add distilled water. Milky white emulsion indicates presence of lipid.

Starch - Iodine test

Add iodine solution. Blue black colour indicates presence of starch.