Question
Monosaccharides combine to form polymers.
(a) Identify one similarity and one difference between the structures of starch and cellulose.
Similarity:
Difference:
(b) Outline why humans cannot digest cellulose.
Answer/Explanation
Answer:
(a) Similarity:
polymers of glucose
OR
«1-4» glycosidic «links»
Difference:
starch contains α-glucose AND cellulose contains β-glucose
OR
starch contains α AND cellulose contains β «1-4» glycosidic links
OR
starch «may contain» 1-6 glycosidic links AND cellulose does not
OR
starch «may be» branched AND cellulose unbranched
(b) lack of cellulase/enzyme
Question
The amino acids in a protein can be separated using paper chromatography. The Rf values using a solvent of butanol and ethanoic acid are given.
(a) The following diagram shows a chromatogram.
(i) Determine the identity of the amino acid creating spot C by calculating the \(R_f\) value from the chromatogram.
Identify of spot C:
(ii) Predict, referring to the structure of the amino acids, whether spot X on the chromatogram in part (a)(i) is more likely to be serine or phenylalanine. Use the table of \(R_f\) values and section 33 of the data booklet.
(b) One role of proteins in the body is to catalyse reactions. Describe how enzymes catalyse reactions in the body.
(c) State one industrial use of enzymes.
(d) Explain how a non-competitive inhibitor affects the Michaelis constant, \(K_m\), and \(V_{max}\) of a reaction. Refer to the reaction between the inhibitor and the enzyme in your answer.
Effect on \(K_m\):
Effect on \(V_{max}\):
Explanation for \(K_m\):
Explanation for \(V_{max}\)
(e) Determine the concentration, in mol \(dm^{-3}\), for a protein sample with absorbance of 0.50 at 240nm. Use section 1 of the data booklet.
Molar extinction coefficient = 0.75\(dm^3\) \(cm^{-1}\) \(mol^{-1}\)
Path length = 1.0cm
Answer/Explanation
Answer:
(a) (i) Identity of spot C:
leucine
(ii) serine AND more polar «than phenylalanine»
OR
serine AND OH group «on side chain»
hydrogen bonding/greater affinity with stationary phase
OR
less soluble/poor affinity in solvent/mobile phase
(b) bind to substrate at active site
«provide» alternative pathway with lower «activation» energy
(c) additives to detergents/washing powders/liquids
OR
breakdown oil spills/industrial waste
(d) Effect on \(K_m\): remains the same/no change AND
Effect on \(V_{max}\): decreases/reduced
Explanation for \(K_m\):
no decrease in affinity of enzyme for substrate
Explanation for \(V_{max}\):
binds at allosteric site
OR
binds away from active site
OR
changes shape of active site
OR
renders active site ineffective
(e) « \(log_{10} \frac{I_0}{I} = \varepsilon lc\) and \(log_{10}\) Io/I = A »
\(\varepsilon \) Ic = 0.50
«c = 0.50 / (0.75 \(dm^3\) \(cm^{−1}\) \(mol^{−1}\) x1 cm)»
0.67 « mol \(dm^{-3}\) »
Question
Outline the meaning of the term xenobiotic.
Answer/Explanation
Answer:
chemicals found in an organism that are not normally present