3.4.3 Genetic diversity via mutation and meiosis

Genetic variation enters a population two ways. DNA changes through mutation — typically one base at a time, occasionally a whole chromosome. And meiosis, the cell division that makes gametes, mixes parental chromosomes into combinations that have never existed before, before random fertilisation pairs the result.

Gene mutations change the DNA base sequence in one of two AQA-named ways.

A mutation is any change in the DNA base sequence. Two gene-level types are named at AQA: substitution and deletion. Both arise spontaneously during DNA replication. Mutagens can raise the mutation rate, but their mechanisms are outside the AQA 7402 spec at this code.

Write substitution or deletion. Point mutation alone scores zero — the named type is what earns the mark.

The asymmetry sits in the reading frame. A substitution at the third base of a codon often gives a synonymous codon because the genetic code is degenerate; the amino acid is unchanged and the protein's tertiary structure is unaffected. A deletion shifts every codon downstream of the cut, so most of the C-terminal sequence reads wrong. Where two sequences are stated to be the same length, deletion is ruled out — the answer must be substitution (or, at the chromosome level, inversion or translocation).

A mutation in a non-coding region does not alter the amino acid sequence because non-coding DNA does not encode amino acids. If it affects transcription (for instance, a promoter region), the credited chain is: less transcription, less mRNA, less protein translated. No active-site or tertiary-structure reasoning applies.

For a non-coding mutation, write the two-step chain: less transcription, less mRNA, less protein. Don't write tertiary structure, active site, or amino acid sequence — these belong to coding-region answers.

Chromosome mutations affect whole chromosomes or sets of chromosomes.

Chromosome mutations affect more than a single base. AQA names two: polyploidy and non-disjunction. Inversion and translocation appear as named alternatives where a sequence-length clue rules out a frameshift.

Spell non-disjunction correctly and write it, not addition, for an extra chromosome. Non-disfunction is rejected. Aneuploidy and trisomy are accepted alternatives where the context fits.

For diagram questions, non-disjunction of one chromosome affects only that chromosome. The other chromosomes still divide normally and must still appear in every daughter cell. A diagram that loses unrelated chromosomes alongside the non-disjoined pair is wrong.

Meiosis produces four haploid, genetically unique daughter cells in two divisions.

Meiosis is the cell division that produces gametes for sexual reproduction. One diploid parent cell yields four haploid daughter cells through two sequential nuclear divisions. Each daughter cell carries half the DNA of the parent. The genetic diversity that meiosis generates is established in meiosis I, not meiosis II.

Write half the DNA, not half the genetic material. AQA credits the DNA phrasing.

In a meiosis I diagram, the homologous chromosomes separate. Don't draw all four chromatids separating in meiosis I — that pattern belongs to meiosis II.

Meiosis II resembles a mitotic division. The sister chromatids of each chromosome separate to opposite poles, producing the final four cells. Meiosis II itself adds no further diversity; what variety the gametes carry was established by crossing over and independent assortment in meiosis I.

Three named mechanisms generate genetic diversity, and the list is fixed.

AQA examines three mechanisms of genetic diversity in sexually reproducing populations: crossing over, independent assortment, and random fertilisation. The list is fixed, and each mechanism has its own required vocabulary.

1. Early in meiosis I, homologous chromosomes pair up along their length to form bivalents.
2. Non-sister chromatids of the paired chromosomes make contact at points called chiasmata.
3. The chromatids exchange corresponding segments at each chiasma. Each chromatid emerges carrying a new combination of alleles on a single chromosome.

1. At metaphase of meiosis I, homologous chromosome pairs align along the spindle.
2. The orientation of each pair — which chromosome of the pair faces which pole — is random, and the orientations of different pairs are independent of each other.
3. The combination of maternal and paternal chromosomes in each daughter cell is therefore random. In humans, with 23 homologous pairs, this generates 2²³ possible combinations.

Write random fertilisation, not random mating. Random mating is a population-level concept and is not credited here.

Pitfall — The sources-of-variation answer has a fixed three-name list

The three named mechanisms are crossing over, independent assortment, and random fertilisation.

Each needs its own vocabulary anchor. Crossing over needs both homologous chromosomes and non-sister chromatids. Independent assortment needs random orientation of the pairs at metaphase I. Random fertilisation needs gametes.

The explicit rejections are: random mating, migration, epigenetics, non-disjunction, and mutation. None of these counts as a source-of-variation answer at AQA 7402.

The decision rule for which mechanism applies: if the new allele combination sits on one chromosome, it is crossing over. If it is a different combination of whole chromosomes across daughter cells, it is independent assortment.

Chi-squared tests whether observed counts differ from expected by chance.

The chi-squared test compares observed counts with the counts a hypothesis predicts. AQA examines it in inheritance contexts and in mutation/meiosis stems where genotype counts are tabulated. Chi-squared is formally introduced at 3.7, but the vocabulary discipline is identical wherever it appears, so the rules belong here at the point of use.

1. State the null hypothesis: any difference between observed and expected values is due to chance.
2. Calculate expected values by applying the null-hypothesis ratio to the relevant total, then calculate the chi-squared value from the observed and expected columns.
3. Compare the calculated value to the critical value at P = 0.05, using the correct degrees of freedom from the chi-squared distribution table.
4. Reject or retain the null hypothesis based on the comparison, then state the biological conclusion in plain language.

Write difference between observed and expected, not results due to chance. The word difference is a dedicated mark point on the probability statement.

Two arithmetic traps end the chi-squared question for many students. Copying observed values into the expected column gives a chi-squared of zero and earns no marks — the expected column applies the null-hypothesis ratio. Combining two data columns into a single inflated total then computing expected from that combined total inflates the expecteds wrongly.

Calculate expected values from the NEW total the question gives, not the original total. Apply the null-hypothesis ratio to the new total only.

Key terms

• mutation
• meiosis
• non-disjunction
• crossing over
• difference
• random fertilisation
• substitution
• deletion
• chromosomes
• homologous chromosomes