Molecular and Cellular Biology tutors

An illustrative GCSE practice result: a potato sample has an initial mass of 2.00 g and a final mass of 2.20 g after immersion in a solution. Comparable samples are treated consistently and blotted in the same way before weighing.

The gain is 0.20 g. Percentage change compares that gain with the initial mass:

(2.20 − 2.00) ÷ 2.00 × 100 = +10%.

“The potato gained 0.20%” confuses a mass with a percentage. “Sugar entered the cells” also goes beyond what this result shows. Under these controlled conditions, the gain is consistent with net water entry through partially permeable membranes by osmosis. A better response connects the calculation to that movement: “The mass increased by 10%, consistent with water entering the tissue by osmosis.”

For a follow-up, use a fresh illustrative result of 2.00 g to 1.80 g. The answer is −10%, consistent with net water loss under the same assumptions. Try the calculation and explanation before checking them.

At A-level, extend the explanation using water potential: net water movement is from higher to lower water potential across a partially permeable membrane. The mass result alone does not give a numerical water potential. This adds the depth required by AQA’s A-level cells content.

For an illustrative A-level task, compare two messenger RNA (mRNA) codons, UUU and UUC. A codon is a three-base instruction read during translation, the process of making a protein. In the standard genetic code, both of these codons encode the amino acid phenylalanine.

A learner might write: “One base changed, so a different amino acid must be added.” The code-table check corrects that reasoning. A more precise answer is: “Although the codon changed from UUU to UUC, the amino acid at this position stays phenylalanine.” This is an example of the genetic code being degenerate: more than one codon can specify the same amino acid.

Practise reading a supplied code table and tracing the sequence through transcription and translation. AQA’s genetic-information content requires this interpretation; its written papers do not require recall of specific codons. NCBI’s standard code table verifies this example, using T in place of RNA’s U in its display.

For a university-level extension, test the next claim: “The protein’s activity is unchanged.” The codon comparison does not measure activity. If the protein is an enzyme, a proposed test could compare reaction rates using equal amounts of each version under the same conditions, with repeated measurements. A better conclusion from the original evidence is: “The encoded amino acid is unchanged at this position; enzyme activity has not been tested.”

That distinction between a supported conclusion and a further hypothesis gives you a useful way to discuss research figures with a prospective tutor.