A Level H1 Biology Practice Paper 1

TuitionGoWhere Practice Paper - Biology H1 A-Level (Answer Key)

Total Marks: 80

Section A: Multiple Choice Questions [20 marks]

1. B - Phospholipids and proteins can move laterally within the membrane [2]

2. C - Peroxisomes [2]

3. B - Protein synthesis and modification [2]

4. C - Competitive inhibition can be overcome by increasing substrate concentration [2]

5. B - The active site is complementary to the substrate [2]

6. C - Starch [2]

7. C - Peptide bond [2]

8. B - Regulates membrane fluidity [2]

9. B - The sequence of amino acids [2]

10. D - Active transport [2]

Section B: Structured Response Questions [60 marks]

Question 1 [12 marks]

(a) Identify the organelles labeled A, C, and D. [3]

Answer: A: Nucleus [1] C: Rough endoplasmic reticulum / Rough ER [1] D: Golgi apparatus / Golgi body [1]

(b) Explain why liver cells contain large numbers of organelle B (mitochondria). [3]

Answer: Liver cells are metabolically very active [1] and require large amounts of ATP [1] for processes such as protein synthesis, detoxification, and gluconeogenesis [1].

Marking notes: Accept any reference to high metabolic activity and ATP requirement. Examples of liver functions are not essential but add clarity.

(c) Describe the role of organelle C in protein synthesis and explain how its structure is adapted for this function. [4]

Answer: Role: Rough ER synthesizes proteins that are destined for secretion or membrane incorporation [1]

Structural adaptations:

• Ribosomes attached to the surface provide sites for protein synthesis [1]
• Large surface area due to flattened sacs (cisternae) allows for extensive protein synthesis [1]
• Connected to nuclear envelope allowing direct transfer of mRNA from nucleus [1]

(d) State two functions of organelle E (peroxisomes) in liver cells. [2]

Answer: (i) Detoxification of harmful substances / breakdown of hydrogen peroxide [1] (ii) β-oxidation of fatty acids / breakdown of fats [1]

Alternative acceptable answers: Synthesis of bile acids, breakdown of purines

Question 2 [15 marks]

(a) Plot a graph of the results on the grid provided. [4]

Marking criteria:

• Correct axes labels with units [1]
• Appropriate scale [1]
• Accurate plotting of points [1]
• Smooth curve drawn [1]

(b) From your graph, determine the optimum pH for pepsin activity. [1]

Answer: pH 2.0 [1]

(c) Explain why pepsin activity decreases at pH values above and below the optimum. [4]

Answer: Above optimum pH: The pH is too high/alkaline [1], causing changes in the enzyme's tertiary structure/denaturation [1]

Below optimum pH: The pH is too low/acidic [1], causing changes in the enzyme's active site shape so substrate cannot bind effectively [1]

Marking notes: Must mention structural changes to enzyme for full marks.

(d) Explain why the pH optimum of pepsin is suitable for its location in the digestive system. [2]

Answer: Pepsin is found in the stomach [1] where the pH is approximately 1.5-2.0 due to hydrochloric acid secretion [1].

(e) Predict what would happen to pepsin activity if the enzyme was moved to the small intestine (pH 8.5). Justify your answer. [4]

Answer: Pepsin activity would be very low or zero [1]. The pH 8.5 is much higher than pepsin's optimum pH of 2.0 [1]. This alkaline pH would denature the enzyme [1] by disrupting the bonds maintaining its tertiary structure, changing the active site shape [1].

Question 3 [18 marks]

(a) Label the hydrophilic head and hydrophobic tails on the phospholipid molecule in Fig. 2. [2]

Answer: Correct labeling of hydrophilic head (phosphate group end) [1] Correct labeling of hydrophobic tails (fatty acid chains) [1]

(b) Explain why phospholipids spontaneously form bilayers in aqueous solutions. [3]

Answer: Hydrophilic heads are attracted to water [1], while hydrophobic tails are repelled by water [1]. Bilayer formation minimizes contact between hydrophobic tails and water while maximizing contact between hydrophilic heads and water [1].

(c) Describe three ways in which small molecules can cross phospholipid bilayers, giving an example of each. [6]

Answer: Method 1: Simple diffusion [1] Example: Oxygen, carbon dioxide, or small lipid-soluble molecules [1]

Method 2: Facilitated diffusion [1] Example: Glucose (through glucose transporters) or ions (through channel proteins) [1]

Method 3: Active transport [1] Example: Sodium ions (through sodium-potassium pump) or calcium ions [1]

(d) Explain how the fluid mosaic model accounts for the selective permeability of cell membranes. [4]

Answer: The phospholipid bilayer is selectively permeable [1] - small, non-polar molecules can pass through easily while large or polar molecules cannot [1]. Membrane proteins provide specific pathways [1] for substances that cannot cross the lipid bilayer, such as channel proteins for ions and carrier proteins for glucose [1].

(e) State two factors that affect membrane fluidity and explain how each factor influences fluidity. [3]

Answer: Factor 1: Temperature [1] Effect: Higher temperature increases fluidity by increasing molecular motion [0.5]

Factor 2: Cholesterol content [1] Effect: Cholesterol decreases fluidity by restricting phospholipid movement [0.5]

Alternative acceptable factors: Fatty acid saturation, fatty acid chain length

Question 4 [15 marks]

(a) Compare the structure and function of starch and cellulose. [8]

Answer:

Structure: Starch: Made of α-glucose monomers [1] joined by α-1,4 and α-1,6 glycosidic bonds [1]. Forms helical/coiled structure (amylose) and branched structure (amylopectin) [1]

Cellulose: Made of β-glucose monomers [1] joined by β-1,4 glycosidic bonds [1]. Forms straight, unbranched chains that can form hydrogen bonds between chains [1]

Function: Starch: Energy storage in plants [1]

Cellulose: Structural support in plant cell walls [1]

(b) Explain why humans can digest starch but not cellulose. [4]

Answer: Humans have amylase enzymes [1] that can break the α-1,4 glycosidic bonds in starch [1]. Humans do not have cellulase enzymes [1] that can break the β-1,4 glycosidic bonds in cellulose [1].

Marking notes: Must mention specific enzymes and bond types for full marks.

(c) Describe the role of cellulose in plant cell walls and explain how its structure makes it suitable for this role. [3]

Answer: Role: Provides structural support and strength to plant cells [1]

Structural suitability: Straight chains can pack closely together [1] and form hydrogen bonds between chains, creating strong microfibrils that resist tension [1].

Marking notes: Accept references to preventing cell bursting, maintaining cell shape, or providing rigidity.

Overall Marking Guidelines

Grade Boundaries (Suggested):

• A: 70-80 marks (87.5-100%)
• B: 60-69 marks (75-86.25%)
• C: 50-59 marks (62.5-74.75%)
• D: 40-49 marks (50-62.25%)
• E: 30-39 marks (37.5-49.75%)

Common Marking Points:

• Award marks for correct scientific terminology
• Accept alternative correct explanations where appropriate
• Deduct marks for incorrect spelling of key scientific terms
• Look for clear, logical explanations that demonstrate understanding
• Partial marks may be awarded for incomplete but correct responses