Researchers identify a mutation in a gene that codes for an enzyme. The relevant sections of the two versions of the gene are shown.
| Version of gene | Base sequence |
|---|---|
| Original | CAA GGT ATG TTT ACG |
| Mutated | CAA GGT ATG TCT ACG |
The table shows the amino acids coded for by the triplets affected by the mutation.
| Triplet of bases | Amino acid coded for |
|---|---|
| TTT | Phenylalanine (Phe) |
| TCT | Serine (Ser) |
The two versions of the enzyme were tested using equal enzyme concentrations and equal volumes. The substrate concentrations and volumes were identical. Temperature, pH and all other reaction conditions were also kept the same.
| Version of enzyme | Mean rate of gaseous product formation (cm³ s⁻¹) |
|---|---|
| Original | 0.30 |
| Mutated | 0.12 |
Calculate the percentage decrease in the mean rate of gaseous product formation for the mutated enzyme compared with the original enzyme.
Show your working.
Explain how the mutation could have caused the decrease in the mean rate of gaseous product formation.
184 exam-style questions on OCR GCSE Biology 1.2 What happens in cells (and what do cells need)?, covering 1.2.1 Structure of DNA, 1.2.2 Nucleotides and complementary bases, 1.2.3 Protein synthesis, 1.2.4 DNA structure and the proteins made, 1.2.5 Investigating enzymatic reactions, and 1.2.6 Mechanism of enzyme action. Each one has a worked solution and a mark scheme showing where the marks go.