Random Fertilisation & Chromosome Combinations Revision Notes | A Level AQA Biology

1. Why Genetic Variation Matters — and Where Meiosis Fits In

Every individual in a sexually reproducing species is genetically unique. That uniqueness does not appear by accident — it is the direct result of two powerful mechanisms: the shuffling of chromosomes during meiosis, and the random collision of gametes at fertilisation. This lesson focuses on the numbers behind that variation and on exactly where in a life cycle meiosis must be occurring.

Before calculating anything, you need a firm anchor: meiosis is the only cell division process that halves the chromosome number, converting a diploid cell (2n) into haploid gametes (n). Wherever a life cycle transitions from diploid to haploid, meiosis is happening at that point.

Recognising Meiosis in an Unfamiliar Life Cycle

AQA questions sometimes present an organism you have never studied — a fern, a fungus, or a moss — and ask you to identify where meiosis occurs. The logic is always the same: find the step where the chromosome number drops from 2n to n. That transition is meiosis, regardless of what the organism is called.

• If the diagram shows a diploid (2n) structure producing haploid (n) spores or gametes, meiosis occurs at that production step.
• If the diagram shows a haploid (n) structure producing more haploid cells by mitosis, those divisions are mitosis — not meiosis.
• Fertilisation always restores the diploid number: n + n → 2n. Meiosis must occur somewhere after fertilisation to bring the number back down before the next generation.

With that identification skill secured, you can now explore why meiosis generates so many different gametes — and why the numbers involved are staggeringly large.

2. Independent Assortment and the $2^n$ Expression

Picture a human cell entering meiosis. It contains 23 pairs of homologous chromosomes — one chromosome in each pair inherited from the mother, one from the father. At metaphase I, each pair lines up at the cell's equator, and the orientation of each pair is entirely random. The maternal chromosome of pair 1 might end up on the left or the right — and that choice is made independently of every other pair.

This is independent assortment. Because each of the n pairs has two possible orientations, and the choices are independent, the total number of genetically different gametes that can result is $2^n$, where n is the haploid number of the organism.

For humans, n = 23, giving $2^{23}$ = 8,388,608 possible chromosome combinations from meiosis alone — before crossing over is even considered. That figure applies equally to the gametes produced by each parent.

3. Deriving the Formula for Random Fertilisation

Meiosis produces an enormous range of gametes, but the variation does not stop there. At fertilisation, any one gamete from one parent can fuse with any one gamete from the other parent. Because both parents independently produce $2^n$ different gametes, the number of genetically distinct zygotes that can result is found by multiplying the two together.

This is the key derivation AQA expects you to be able to perform. Start from what you already know: each parent produces $2^n$ gamete types. The number of zygote combinations equals the number of gamete types from parent 1 multiplied by the number from parent 2.

Worked example: Deriving the Random Fertilisation Formula

Derive a formula for the total number of different chromosome combinations possible after random fertilisation of two gametes, given that each parent has a haploid number of n.

1. Each parent produces $2^n$ genetically different gametes (from independent assortment, no crossing over). — This is the $2^n$ expression you already know. Both parents have the same haploid number n.
2. Total zygote combinations $= 2^n \times 2^n$ — Any gamete from parent 1 can fuse with any gamete from parent 2 — these are independent events, so you multiply.
3. $2^n \times 2^n = 2^{n+n} = 2^{2n}$ — When multiplying powers with the same base, add the exponents. This gives the compact derived formula.
4. $$\text{Total combinations after random fertilisation} = 2^{2n}$$ — This formula assumes no crossing over and that both parents have the same haploid number n.

Answer: $2^{2n}$, where n is the haploid chromosome number of each parent.

For humans, $2n = 46$, so $n = 23$ and the total number of possible zygote combinations is $2^{46}$. That is approximately $7 \times 10^{13}$ — vastly more than the number of humans who have ever lived. This figure illustrates why no two sexually reproduced individuals (other than identical twins) are genetically identical.

4. Applying the Formulas — Worked Calculations

Knowing the formulas is one thing; applying them accurately under exam conditions is another. The steps below walk through two calculations — one using $2^n$ and one using $2^{2n}$ — with the kind of working that earns method marks.

Worked example: Calculating Gamete Combinations for a Given Organism

A species of flowering plant has a haploid chromosome number of n = 7. Calculate the number of different chromosome combinations possible in its gametes, assuming no crossing over. Show your working.

1. Number of gamete combinations $= 2^n$ — This formula applies to one parent producing gametes via meiosis with independent assortment and no crossing over. [M1]
2. $= 2^7$ — n = 7 is given directly as the haploid number. [M1]
3. $2^1 = 2$ $2^2 = 2 \times 2 = 4$ $2^3 = 4 \times 2 = 8$ $2^4 = 8 \times 2 = 16$ $2^5 = 16 \times 2 = 32$ $2^6 = 32 \times 2 = 64$ $2^7 = 64 \times 2 = 128$ — Expanding each power step by step avoids arithmetic errors. [A1]

Answer: 128 different chromosome combinations are possible in the gametes of this plant.

Worked example: Calculating Zygote Combinations After Random Fertilisation

Using the same flowering plant (n = 7), calculate the total number of different chromosome combinations possible in zygotes formed by random fertilisation. Show your working and state the formula used.

1. Total zygote combinations $= 2^{2n}$ — Derived from $2^n \times 2^n$, where each parent independently produces $2^n$ gamete types. [M1]
2. $2n = 2 \times 7 = 14$ — Multiply the haploid number by 2 to find the exponent. [M1]
3. $2^7 = 128$ (from previous calculation) $2^{14} = 2^7 \times 2^7 = 128 \times 128$ $128 \times 128 = 128 \times 100 + 128 \times 28$ $= 12800 + 3584 = 16384$ — Using the result from the previous calculation and expanding the multiplication avoids errors. [A1]
4. $2^{14} = 16384$; this is $128^2$, and $128 \times 128 = 16384$ ✓ — The zygote figure must be the square of the gamete figure when both parents have the same n. If your zygote answer is less than your gamete answer, recheck.

Answer: 16,384 different chromosome combinations are possible in the zygotes of this plant.

5. How Random Fertilisation Increases Genetic Variation

The calculation above shows the scale of variation, but the exam also asks you to explain the mechanism. Random fertilisation increases genetic variation because any haploid gamete produced by one parent can fuse with any haploid gamete produced by the other parent. Since each gamete already carries a unique combination of maternal and paternal chromosomes (due to independent assortment), the resulting zygote inherits a combination that is statistically unlikely to have ever existed before.

This is multiplicative, not additive. Meiosis alone generates $2^n$ gamete types per parent. Random fertilisation multiplies those possibilities together, producing $2^{2n}$ possible zygote genotypes. Each additional source of variation — independent assortment, crossing over, and random fertilisation — compounds the others rather than adding to them.

The biological consequence is profound: within a sexually reproducing population, virtually every individual carries a unique genotype. This genetic diversity is the raw material on which natural selection acts, underpinning the capacity of a species to adapt to changing environments over generations.

6. Exam Technique and Common Mistakes

The calculation questions on this topic are straightforward once you know the formulas, but there are several predictable errors that cost marks. The guidance below covers the most common ones alongside the language and structure that AQA mark schemes reward.

For longer 'explain' or 'evaluate' questions about sources of genetic variation, structure your answer by addressing each mechanism separately: first independent assortment (with the $2^n$ figure), then crossing over (which increases variation further beyond $2^n$), and finally random fertilisation (which multiplies the possibilities to $2^{2n}$). Examiners award marks for each distinct mechanism named and explained with a causal link.

Taken together, the $2^n$ and $2^{2n}$ expressions capture the quantitative scale of genetic variation arising from meiosis and random fertilisation — satisfying the core AQA requirement to calculate and derive these values, and to explain the biological mechanism that makes each zygote genetically distinct.