A Level H2 Biology Cells Biomolecules Quiz

A-Level Biology H2 Quiz - Cells Biomolecules – ANSWERS

Total marks: 50
Marking points are shown in brackets; accept equivalent scientific wording.

Section A: Multiple Choice

1. C – Mitochondrion (1)
2. B – freely moving within a phospholipid bilayer (1)
3. C – Nucleotide – DNA (1)
4. A – increasing substrate concentration (1)
5. C – one original strand and one new strand (1)

Section B: Structured Questions

6.
(a)
P – Nucleus/nuclear envelope (1)
Q – Mitochondrion (1)
R – Rough endoplasmic reticulum (1)

(b)
Proteins from the RER are transported to the Golgi in vesicles (1). Within the Golgi, the proteins are modified (e.g. by addition of carbohydrate groups – glycosylation) (1). The modified proteins are then sorted and packaged into secretory vesicles for transport to the cell membrane or to other organelles (1).

7.
(a) The inhibitor does not change the activation energy; the peak height (activation energy) remains the same in the presence of inhibitor (1).
(Accept: activation energy is unchanged.)

(b) A non‑competitive inhibitor binds to an allosteric site (a site other than the active site) (1). This changes the shape (conformation) of the enzyme, including the active site, so the substrate can no longer bind (1). Since the inhibitor does not compete with the substrate, increasing substrate concentration cannot overcome the inhibition, and Vmax decreases (1).

8.
(a)
W – Phospholipid bilayer / phospholipid molecule (1)
X – (Integral/intrinsic) protein / channel protein / carrier protein (1)
Y – Glycoprotein / glycocalyx / carbohydrate chain (1)

(b) Component X (protein) can act as a channel or carrier for facilitated diffusion of ions/polar molecules across the membrane (1), or as a pump for active transport (1).

9.
(a) Solution C contains the highest concentration of reducing sugar (1). A brick‑red precipitate indicates a large amount of reducing sugar has reduced the Cu²⁺ ions in Benedict’s reagent to Cu⁺ (1).

(b) Solution A remained blue because it contained no (or a very low concentration of) reducing sugar, so no reduction of Cu²⁺ occurred (1).

10.
(a) Amylopectin (1). The molecule shows a branched structure / 1→6 glycosidic bonds at branch points (1).

(b) Amylopectin has many branches, which provides many free ends for rapid hydrolysis by enzymes (1). This allows glucose to be released quickly when energy is needed. The molecule is compact and insoluble, so it does not affect the water potential of the cell (1).

11.
(a) The rate of uptake increases rapidly at low glucose concentrations and then levels off (plateaus) at high concentrations, giving a hyperbolic curve (1). The curve does not increase linearly; it approaches a maximum rate (Vmax) (1).

(b) The pattern is characteristic of facilitated diffusion (1). The rate increases as more carrier/channel proteins are occupied, but once all carriers are saturated (occupied), the rate cannot increase further, hence the plateau (1). Since no metabolic energy is required and glucose moves down its concentration gradient, the mechanism is facilitated diffusion (1).

12.
(a) Hydrogen bond (1).

(b) Complementary base pairing (A with T, G with C) ensures that the sequence of the new strand is exactly determined by the template strand (1). When the double helix unwinds, each parental strand acts as a template; the specificity of hydrogen bonding ensures that only the correct nucleotide is added by DNA polymerase (1). This results in two identical daughter DNA molecules, preserving the genetic information accurately (1).

13.
Water is a polar molecule because the oxygen atom has a slight negative charge (δ⁻) and the hydrogen atoms have a slight positive charge (δ⁺) (1). This polarity allows water molecules to surround and separate charged ions (e.g. Na⁺, Cl⁻) and polar molecules (e.g. glucose) (1). The formation of hydration shells keeps the solutes in solution, enabling them to dissolve, react, and be transported in biological systems (1).

14.
(a) Triglyceride (or triacylglycerol) (1).

(b) Triglycerides are composed of a glycerol backbone esterified to three fatty acid chains (1). The long hydrocarbon tails contain a high proportion of carbon–hydrogen bonds and are highly reduced; they yield a large amount of energy per gram when oxidised in respiration. Being hydrophobic, triglycerides are stored as anhydrous fat droplets, which do not increase water content and therefore do not affect osmosis (1).

15.
(a) The human and mouse sequences differ at two positions: positions 5 and 6 (Pro/Asp and Glu/Ala) (1). The remaining eight amino acids are identical (1).

(b) The identical sequences between human and gorilla suggest a very close evolutionary relationship (recent common ancestor) (1). The greater number of differences between human and mouse indicates that humans and mice diverged earlier in evolutionary history (1).

Section C: Data-Based and Extended Response Questions

16.
(a) pH 7.0 (1)
(b) At pH 10.0, the pH is far from the enzyme’s optimum, causing disruption of hydrogen and ionic bonds that maintain the active site’s specific shape (denaturation) (1). The substrate can no longer bind effectively, so the rate of product formation falls sharply (1).
(c) Repeat the experiment at each pH at least twice more and calculate a mean value (1). (Accept: use a more precise buffer, control temperature more carefully, etc.)

17.
Model answer:

• Primary structure: Linear sequence of amino acids in a polypeptide chain, held together by peptide bonds (1).
• Secondary structure: Regular local folding patterns – α‑helices and β‑pleated sheets – stabilised by hydrogen bonds between the –NH and –C=O groups of the backbone (1).
• Tertiary structure: Overall three‑dimensional shape of a single polypeptide, maintained by interactions between R‑groups: disulfide bonds, ionic bonds, hydrophobic interactions, hydrogen bonds (1).
• Quaternary structure: Assembly of two or more polypeptide subunits (e.g. haemoglobin has four subunits) (1).
  A single amino acid substitution in the primary structure changes the sequence of R‑groups. In sickle‑cell haemoglobin, the substitution of glutamic acid (hydrophilic, charged) by valine (hydrophobic) at position 6 of the β‑globin chain (1) leads to a change in tertiary structure: the hydrophobic valine creates a “sticky” patch that causes haemoglobin molecules to aggregate into long rods when deoxygenated (1). This alters the quaternary structure and distorts the red blood cell into a sickle shape, impairing oxygen transport (1).

18.
(a) The carrier is heterozygous: they possess one allele for normal HbA and one allele for sickle‑cell HbS (1). Each allele produces a slightly different protein (HbA and HbS). Gel electrophoresis separates proteins mainly by charge and/or size; HbS has a different net charge (or molecular size) and therefore migrates to a different position from HbA (1). The presence of both proteins results in two distinct bands (1).
(b) In HbS, the amino acid glutamic acid (negatively charged) is replaced by valine (neutral/hydrophobic), changing the overall charge of the protein (1). (Accept: substitution of glutamic acid by valine.)

19.
(a) When lactose is absent, the regulatory gene produces the active repressor protein, which binds to the operator region (1). This blocks RNA polymerase from binding to the promoter, so transcription of the structural genes is prevented (1).
(b) When lactose is present, it is converted to allolactose, which acts as an inducer and binds to the repressor protein (1). The repressor changes shape and can no longer bind to the operator, allowing RNA polymerase to bind to the promoter and transcribe the structural genes (1).

20.
(a) At [S] = 5.0 mmol dm⁻³:
Rate without inhibitor = 35 μmol min⁻¹; rate with inhibitor = 28 μmol min⁻¹.
Percentage inhibition = [(35 – 28) / 35] × 100 = (7/35) × 100 = 20% (2 marks; 1 for working, 1 for correct answer).

(b) The data suggest competitive inhibition (1). In competitive inhibition, the inhibitor competes with the substrate for the active site. At low substrate concentration, the inhibitor has a large effect (rates reduced significantly), but at high substrate concentration, the substrate out‑competes the inhibitor, and the rates approach the uninhibited rate (Vmax almost reached). This is reflected in the data: at 10 mmol dm⁻³, the rates are nearly equal (40 vs 38 μmol min⁻¹) (1).
(If the student explains why non‑competitive inhibition is unlikely—e.g. Vmax would remain reduced even at high [S]—award credit.)

End of Answer Key