A Level H2 Biology Practice Paper 1

TuitionGoWhere Practice Paper - Biology H2 A-Level - Mark Scheme

Total Marks: 75

Section A [25 marks]

Question 1

(a) (i) Identify the structures labelled A, B, C and D. [4]

• A: Nucleus [1]
• B: Mitochondrion [1]
• C: Rough endoplasmic reticulum / RER [1]
• D: Ribosomes [1]

(ii) State the magnification used if the actual diameter of structure A is 8 μm and its diameter in the micrograph is 32 mm. [2]

• Convert units: 32 mm = 32,000 μm [1]
• Magnification = 32,000 ÷ 8 = ×4000 [1]

(b) With reference to Fig. 1.1, explain how structure B is adapted for its role in aerobic respiration. [4]

• Inner membrane folded into cristae [1]
• Increases surface area for electron transport chain / ATP synthesis [1]
• Matrix contains enzymes for Krebs cycle [1]
• Double membrane allows compartmentalization / proton gradient formation [1]

Question 2

(a) Describe and explain how gel electrophoresis is used to diagnose sickle cell anaemia. [4]

• Electric field / potential difference applied across gel [1]
• Proteins move according to charge and molecular mass [1]
• Normal haemoglobin (HbA) and sickle haemoglobin (HbS) migrate differently [1]
• Different migration patterns allow identification of genotype [1]

(b) With reference to Fig. 2.1, explain the different numbers of fragments seen in different individuals. [3]

• Different alleles produce proteins of different sizes/charges [1]
• Heterozygotes carry two different alleles, so show two bands [1]
• Homozygotes carry identical alleles, so show one band [1]

(c) Identify which individual(s) would be expected to show symptoms of sickle cell anaemia. Explain your answer. [2]

• Individual(s): Individual 3 [1]
• Explanation: Shows only one band in position of HbS / is homozygous for sickle cell allele [1]

Question 3

(a) Describe how you would test for the presence of starch. [2]

• Add iodine solution [1]
• Blue-black colour indicates presence of starch [1]

(b) (i) Explain the shape of the curve between points X and Y. [3]

• Increasing substrate concentration increases rate of reaction [1]
• More substrate molecules available for enzyme active sites [1]
• More enzyme-substrate complexes formed per unit time [1]

(ii) Explain why the curve levels off after point Y. [2]

• All enzyme active sites are occupied / enzyme is saturated [1]
• Further increase in substrate concentration has no effect on rate [1]

Section B [25 marks]

Question 4

(a) The lac operon codes for inducible enzymes. Explain why it is advantageous for E. coli to have an inducible operon for lactose metabolism. [3]

• Enzymes only produced when lactose is present [1]
• Saves energy/resources when lactose is absent [1]
• Allows rapid response to lactose availability [1]

(b) (i) lactose is absent from the growth medium [4]

• Repressor protein binds to operator [1]
• RNA polymerase cannot bind to promoter / transcription blocked [1]
• No mRNA produced [1]
• No enzymes for lactose metabolism produced [1]

(ii) lactose is present in the growth medium [4]

• Lactose acts as inducer / binds to repressor protein [1]
• Repressor protein changes shape / cannot bind to operator [1]
• RNA polymerase can bind to promoter [1]
• Transcription occurs / enzymes for lactose metabolism produced [1]

Question 5

(a) Explain why oxygen concentration decreases after the addition of glucose at point A. [3]

• Glucose provides substrate for respiration [1]
• Oxygen is used as final electron acceptor in electron transport chain [1]
• Rate of oxygen consumption increases [1]

(b) Sodium azide is an inhibitor of cytochrome oxidase in the electron transport chain. With reference to Fig. 5.1, explain the effect of adding sodium azide at point B. [4]

• Electron transport chain is blocked [1]
• Oxygen cannot accept electrons / is not reduced [1]
• Oxygen consumption stops [1]
• Oxygen concentration increases as it is no longer being used [1]

(c) Calculate the rate of oxygen consumption between 2 and 4 minutes. [2]

• Change in oxygen concentration = (reading at 4 min) - (reading at 2 min) [1]
• Rate = change ÷ time = change ÷ 2 minutes [1] Note: Specific values depend on graph scale provided

Question 6

Describe the structure of proteins, explaining how the different levels of structure contribute to protein function. [10]

Mark scheme:

• Primary structure: sequence of amino acids joined by peptide bonds [1]
• Determines all higher levels of structure [1]
• Secondary structure: α-helices and β-pleated sheets [1]
• Held by hydrogen bonds between backbone atoms [1]
• Provides structural stability [1]
• Tertiary structure: 3D folding of polypeptide chain [1]
• Held by various bonds (hydrogen, ionic, disulfide, hydrophobic interactions) [1]
• Determines active site shape in enzymes / binding sites [1]
• Quaternary structure: association of multiple polypeptide chains [1]
• Examples: haemoglobin (4 subunits), enzymes with multiple subunits [1]

Section C [25 marks]

Question 7

(a) (i) Using appropriate symbols, write the genotypes of the parent generation. [2]

• Male: C^B C^B [1]
• Female: C^W C^W [1]

(ii) Explain the phenotype of the F1 generation. [2]

• All F1 are C^B C^W (heterozygous) [1]
• Codominance results in blue feathers (intermediate phenotype) [1]

(b) (i) State the null hypothesis for this test. [1]

• The observed results fit the expected 1:2:1 ratio [1]

(ii) Calculate the expected numbers for each phenotype class in the F2 generation. [3]

• Total F2 offspring = 89 + 183 + 94 = 366 [1]
• Expected black: 366 ÷ 4 = 91.5 [1]
• Expected blue: 366 ÷ 2 = 183 [1]
• Expected splashed white: 366 ÷ 4 = 91.5 [1]

(iii) State the conclusion of this test and explain what this means. [2]

• Conclusion: Accept null hypothesis / no significant difference [1]
• Explanation: Observed results fit expected 1:2:1 ratio / genes assort independently [1]

Question 8

(a) With reference to Fig. 8.1, explain the role of electrons as they move from photosystem II to photosystem I. [4]

• Electrons are excited by light energy at PSII [1]
• Move through electron transport chain [1]
• Energy released is used to pump H+ into thylakoid lumen [1]
• Creates proton gradient for ATP synthesis [1]

(b) Explain how the light-dependent reactions of photosynthesis produce ATP. [6]

• Light energy excites electrons in photosystems [1]
• Electrons move through electron transport chain [1]
• Energy released pumps H+ from stroma into thylakoid lumen [1]
• Creates proton gradient across thylakoid membrane [1]
• H+ flow back through ATP synthase [1]
• ADP + Pi → ATP (chemiosmosis) [1]

(c) Describe and explain the effects of an increase in oxygen concentration on the rate of photosynthesis in C3 plants. [5]

• Rate of photosynthesis decreases [1]
• RuBisCO catalyses both carboxylation (with CO2) and oxygenation (with O2) [1]
• High oxygen concentration favours oxygenation of RuBP [1]
• This leads to photorespiration [1]
• Photorespiration produces no ATP or NADPH, reducing net photosynthesis [1]