Scientists place equal numbers of liver cells into two solutions, X and Y. They measure the initial urea concentrations and the mean net rate at which urea moves out of the cells.
| Solution | Urea concentration inside the cells (mmol dm⁻³) | Urea concentration in the surrounding solution (mmol dm⁻³) | Mean net rate of urea movement out of the cells (nmol min⁻¹) |
|---|---|---|---|
| X | 8.0 | 2.0 | 3.0 |
| Y | 8.0 | 6.0 | 1.0 |
Explain why the mean net rate of urea movement is greater in solution X than in solution Y.
Another sample of the liver cells is placed in a solution with a urea concentration of 8.0 mmol dm⁻³. The urea concentration inside these cells is also 8.0 mmol dm⁻³.
Describe the movement of urea molecules across the cell membranes.
194 exam-style questions on OCR GCSE Biology 2.1 Supplying the cell, covering 2.1.1 Transport across cell membranes, 2.1.2 Mitosis and the cell cycle, 2.1.3 Importance of cell differentiation, and 2.1.4 Stem cells. Each one has a worked solution and a mark scheme showing where the marks go.