The Hardest A Level Biology Questions: Tricky Exam Problems Explained

Ready to test yourself against some of the hardest A Level Biology questions? This guide breaks down tricky exam problems step by step, helping you master difficult topics, avoid common mistakes, and build the problem-solving skills needed for top grades in A Level Biology.

Problem 1: Haemoglobin Amino Acid Count

Haemoglobin contains two alpha (α) chains and two beta (β) chains. The number of amino acids in each beta chain is 3.546% greater than in each alpha chain. Each alpha chain contains 141 amino acids.

Calculate the total number of amino acids in one haemoglobin molecule. Give your answer to the nearest whole number.

Answer

Step 1: Find the number of amino acids in one beta chain

141 × 1.03546 = 146.0 amino acids (to nearest whole number) = 146

Step 2: Find the total number of amino acids in one haemoglobin molecule

Haemoglobin has 2 alpha chains and 2 beta chains, so:

(2 × 141) + (2 × 146) = 282 + 292 = 574 amino acids

Problem 2: Cookie Respiration Data

The table shows the volume of carbon dioxide produced and oxygen consumed by organisms fed on two different cookies.

Cookie	CO₂ produced (cm³/min)	O₂ consumed (cm³/min)
1	13.3	13.6
2	3.0	13.9

Which cookie contains more protein? [3]

Answer

Step 1: Calculate the respiratory quotient (RQ) for each cookie

RQ = CO₂ produced / O₂ consumed

Cookie 1: RQ = 13.3 / 13.6 = 0.978 Cookie 2: RQ = 3.0 / 13.9 = 0.216

Step 2: Interpret the RQ values

• RQ of carbohydrates = 1.0

• RQ of fats = ~0.7

• RQ of protein = ~0.9

  
  

Step 3: Compare to known RQ values

Cookie 1 RQ of 0.978 is very close to 1.0, suggesting it is mostly carbohydrate.

Cookie 2 RQ of 0.216 is far below even the fat RQ of 0.7, which indicates a large amount of fat AND protein is being respired — protein respiration produces relatively little CO₂ compared to O₂ consumed, pulling the RQ down significantly.

Step 4: Conclusion

Cookie 2 contains more protein, because its RQ is much lower than expected for carbohydrate or fat alone, indicating that a significant proportion of the substrate being respired is protein, which has a low RQ due to producing much less CO₂ relative to O₂ consumed.

Problem 3: Glucose Absorption in the Small Intestine

Explain how the transport of sodium ions is involved in the absorption of glucose from the small intestine. (5)

Answer

Step 1: Sodium is actively pumped out of the epithelial cell

Sodium-potassium (Na⁺/K⁺) ATPase pumps on the basolateral membrane actively transport sodium ions out of the epithelial cell into the blood, using ATP. This creates a low sodium concentration inside the epithelial cell.

Step 2: A sodium concentration gradient is established

Because sodium is continuously pumped out, the sodium concentration inside the epithelial cell remains lower than in the lumen of the small intestine. This creates a concentration gradient for sodium from the lumen into the cell.

Step 3: Sodium moves into the epithelial cell via the co-transporter

Sodium moves down its concentration gradient from the lumen into the epithelial cell through a sodium-glucose co-transporter (SGLT1) protein on the apical membrane. This is facilitated diffusion — no ATP is directly used at this step.

Step 4: Glucose is carried in alongside sodium

The co-transporter only functions when both sodium and glucose bind to it simultaneously. So as sodium moves down its gradient into the cell, glucose is carried in with it — even if glucose concentration inside the cell is already high. This is secondary active transport.

Step 5: Glucose exits the epithelial cell into the blood

Glucose moves out of the epithelial cell across the basolateral membrane into the blood via facilitated diffusion through GLUT2 carrier proteins, down its concentration gradient.

Problem 4: DNA Base Ratio

Two strands of DNA in a gene contain 168 guanine bases.

G = 4(A + T) - C

Use this information to calculate the maximum number of amino acids coded for by this gene. (2 marks)

Answer

Step 1: Apply Chargaff's rules

Since G = C and A = T in double stranded DNA:

G = C = 168

Step 2: Use the formula to find (A + T)

G = 4(A + T) - C

168 = 4(A + T) - 168

336 = 4(A + T)

A + T = 84

Step 3: Find the total number of bases

Total bases = (A + T) + (G + C) = 84 + (168 + 168) = 84 + 336 = 420

Step 4: Find the number of codons

Each codon = 3 bases, and DNA is double stranded so only one strand is used as the template:

Number of codons = (420 / 2) / 3 = 210 / 3 = 70 codons

Step 5: Find the maximum number of amino acids

One codon codes for one amino acid, but one codon is a stop codon and does not code for an amino acid:

Maximum number of amino acids = 70 - 1 = 69 amino acids

Problem 5: Explaining Heart Rate Changes

Describe how chemoreceptors and baroreceptors bring about changes to the heart rate during exercise. (3)

Answer

Chemoreceptors

During exercise, CO₂ levels in the blood rise, lowering blood pH. Chemoreceptors detect this drop in pH and send nerve impulses to the cardiovascular centre in the medulla oblongata, which responds by sending impulses via the sympathetic nervous system to the sinoatrial node (SAN), increasing heart rate.

Baroreceptors

During exercise, increased heart rate and cardiac output cause blood pressure to rise. Baroreceptors in the aortic arch and carotid sinus detect this increase in blood pressure and send impulses to the cardiovascular centre, which responds by sending impulses via the parasympathetic nervous system to the SAN to moderate and regulate the rise in heart rate, preventing it from increasing excessively.