AP Biology · Unit 5 · Topic 5.2
Independent assortment, crossing over and random fertilisation. Enumerate the gametes yourself for small numbers, then see what the same rule gives for 23 pairs.
Learning objective 5.2.A Explain how meiosis generates genetic diversity.
At metaphase I each homologous pair straddles the plate, and which homologue faces which pole is settled separately for that pair. Flip any pair and only that pair changes — which is what independent means, and why the counts multiply.
Move the slider from one pair to two and watch the grid. Every gamete that already existed appears twice in the new list — once taking M from the new pair and once taking P. Adding a pair does not add two gametes; it doubles the ones you already had.
Look at the grid with four pairs. Half of all sixteen gametes take M from the first pair. Now look only at those eight: half of them take P from the second pair — the same proportion as in the whole list. Knowing what one pair did tells you nothing about any other, and that is exactly the condition under which counts multiply instead of adding.
Each arrangement sends one combination to the left pole and the exact opposite to the right. So a single meiosis does not sample the whole list — it produces one pair of complementary gametes out of it, and a different cell in the same organism will produce a different pair.
That is why the number in the readout is what an organism can make, not what one cell makes.
Independent assortment moves whole chromosomes, so it can never separate two genes that sit on the same one. Crossing over can. Follow two genes, A/a and B/b, on one homologous pair.
A crossover happens between one chromatid of one homologue and one chromatid of the other. The other two chromatids are not involved and come through unchanged. That is why a bivalent with a single crossover yields two parental and two recombinant products rather than four recombinant ones — a detail that turns up in exam questions far more often than its size suggests.
Count the letters in the readout before and after. There are two A, two a, two B and two b either way. Crossing over changes which chromosome each allele is sitting on; it does not change which alleles exist. Making a new allele is mutation, and that is a different process.
Switch the crossover off and read the four chromatids. Only AB and ab appear, however the pairs are arranged at the plate, because these two genes travel together on the same chromosome. No amount of independent assortment will produce Ab, since assortment can only choose which chromosome goes where, never rebuild one.
The doubling you watched in the grid does not stop. Drag along the number of pairs and read what the same rule gives — the vertical axis is logarithmic, because nothing else would fit on a page.
The 223 figure counts independent assortment only. The figure for random fertilisation multiplies two of those. Crossing over multiplies again by an amount that is not fixed, because the number and position of crossovers differ from one meiosis to the next — so any single total you are quoted is a floor with an unstated assumption attached, not a measurement.
The defensible statement in an exam answer is the mechanism plus the arithmetic you can justify: 2n from assortment, squared by fertilisation, multiplied further and indefinitely by crossing over.
Answer these in your own words before you open the answers.
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| Independent assortment | Crossing over |
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A question that asks how a particular new combination arose is asking you to choose a column. If the genes are on different chromosomes, assortment is enough; if they are on the same one, only crossing over will do it.