Unit 1 - Biological molecules Flashcards
Leucyl-alanine is a dipeptide. Describe how a dipeptide is formed. (3)
A peptide bond (1) forms between the carboxyl group of one amino acid and the amino group of another amino acid (1). A condensation reaction takes place (1).
Myoglobin is a protein formed from a single polypeptide chain. Describe the tertiary structure of a protein like myoglobin. (2)
The secondary structure is coiled and folded further to form the protein’s final 3D structure (1). More bonds, including hydrogen bonds, ionic bonds and disulphide bridges, form between different parts of the polypeptide chain (1).
Describe the ‘induced fit’ model of enzyme action. (4)
The complementary substrate binds to the active site of the enzyme (1) to form an enzyme-substrate complex (1). As the substrate binds, the active site changes shape slightly, which provides a better fit (1). The substrate is broken down to form the product(s) (1).
Explain how a change in the amino acid sequence of an enzyme may prevent it from functioning properly. (2)
A change in the amino acid sequence of an enzyme may alter its tertiary structure (1). This changes the shape of the active site so that the substrate can’t bind to it (1).
Explain how a competitive inhibitor works. (3)
Competitive inhibitor molecules have a similar shape to the substrate molecules (1). They compete with the substrate molecules to bind to the active site of an enzyme (1). When an inhibitor molecule is bound to the active site it stops the substrate molecule from binding (1).
Explain how a non-competitive inhibitor works. (2)
Non-competitive inhibitor molecules bind to enzymes away from their active site (1). This causes the active site to change shape so the substrate molecule can no longer fit (1).
Triglycerides have a hydrophobic tail. Explain how this feature of a lipid is important for its function. (2)
The hydrophobic tails force them to clump together in the cytoplasm as insoluble droplets (1). This means they can be stored in cells, as a source of energy, without affecting the cell’s water potential (1).
Describe the structure of a phospholipid. (3)
Two fatty acid molecules (1) and a phosphate group (1) attached to one glycerol molecule (1).
Explain the difference between a saturated fatty acid and an unsaturated fatty acid. (2)
Saturated fatty acids don’t have any double bonds between their carbon atoms (1). Unsaturated fatty acids have one or more double bonds between their carbon atoms (1).
Maltose is a sugar. Describe how a molecule of maltose is formed. (3)
Two molecules of alpha-glucose (1), are joined by a glycosidic bond (1). A condensation reaction takes place (1).
Sugars can be classified as reducing or non-reducing. Describe the test used to identify a non-reducing sugar. Include different results you would expect to see if the test was positive or negative (5)
Take a new sample of the test solution, add dilute hydrochloric acid and heat it in a water bath that has been brought to a boil (1). Neutralise it with sodium hydrogen carbonate (1). Then add blue Benedict’s solution and heat it in a water bath that has been brought to a boil (1). If the test is positive for a non-reducing sugar, a brick-red precipitate will form (1). If the test is negative, the solution will stay blue (1).
Explain why chitin can be described as a polysaccharide. (1)
Because it is made up of chains of a monosaccharide (1).
Explain why chitin can be described as a polysaccharide. (1)
Because it is made up of chains of a monosaccharide (1).
Chitin is similar to cellulose in plants. Describe the ways in which cellulose and chitin are similar. (3)
Chitin and cellulose are both polysaccharides (1), made up of long unbranched chains (1). The chains are linked together by weak hydrogen bonds (1).
Chitin can be broken down by enzymes called chitinases, which catalyse hydrolysis reactions. Explain how these hydrolysis reactions break down chitin. (2)
A molecule of water (1) is used to break the glycosidic bond between monosaccharides in the chain (1).
