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AP® Biology · Unit 5 · Topic 5.2

Meiosis and Genetic Diversity: Study Guide (Biology 5.2)

How meiosis and fertilization create genetic variation: crossing over, independent assortment of homologous pairs, random fertilization, and what nondisjunction does to chromosome number.

Why the four cells are not the same

In Topic 5.1 you followed chromosomes through the two divisions of meiosis. If you look only at the numbers, the four haploid cells seem alike: each has n chromosomes. But they almost never carry the same genetic information. Brothers and sisters with the same two parents (unless they are identical twins) are genetically different for the same reason.

Three processes in sexual reproduction create this variation:

  1. Crossing over in prophase I.
  2. Independent (random) assortment of homologous pairs at metaphase I.
  3. Random fertilization of one gamete by another.

Variation matters because it gives a population many different combinations of alleles. When the environment changes, some combinations may survive and reproduce better than others. That is the raw material natural selection acts on, which you will meet in Unit 7.

Correct separation: a full set, mixed from both parents

Meiosis is built to deliver exactly one chromosome from each homologous pair to each gamete.

  • In meiosis I, the homologues of each pair separate. Each cell gets one member of every pair.
  • In meiosis II, the sister chromatids of each chromosome separate. Each final cell gets one chromatid of every chromosome.

The result is a haploid (1n) set: one copy of every chromosome, no more and no less. But which member of each pair a gamete receives (the one you inherited from your mother or the one from your father) varies. So every gamete is a mixture of maternal and paternal chromosomes. The rest of this guide explains how that mixing happens, and what goes wrong when separation fails.

Crossing over: new combinations on one chromosome

During prophase I, the homologues are held tightly together in synapsis. Two non-sister chromatids (one from each homologue) can break at matching points and rejoin with each other’s ends. Matching segments are swapped. This is crossing over, also called recombination. The point where the chromatids remain linked is visible as a chiasma (plural chiasmata).

Crossing over between non-sister chromatids Three panels. Panel 1: a pair of homologous chromosomes in prophase I. The two solid chromatids on the left carry alleles A and B; the two open chromatids on the right carry alleles a and b. Gene A is above gene B on each chromatid. Panel 2: the two inner, non-sister chromatids cross between the A and B positions, at a point labelled chiasma. Panel 3: the four chromatids after meiosis II, each in a different cell. From left: all solid, carrying A and B, labelled parental; solid top and open bottom, carrying A and b, labelled recombinant; open top and solid bottom, carrying a and B, labelled recombinant; all open, carrying a and b, labelled parental. A B a b A B a b chiasma A B A b a B a b parental recombinant recombinant parental 1. Synapsis in prophase I homologues paired 2. Crossing over non-sister chromatids swap 3. The four chromatids after meiosis II, one per cell
Figure 1. Crossing over between gene A and gene B. Solid chromatids came from one parent and open chromatids from the other. The swap makes two recombinant chromatids (Ab and aB) that neither parent's chromosome carried. Only one crossover is shown; each homologous pair usually has at least one, and often more.

Reading Figure 1. Before crossing over, one homologue carries A and B and the other carries a and b. After one crossover between the two genes, the four chromatids carry AB, Ab, aB and ab. The outer two are parental (unchanged). The inner two are recombinant: they combine alleles from both parents on a single chromatid.

Three consequences follow:

  • Crossing over does not change the number of chromosomes or genes. Each chromatid still has one allele of every gene.
  • It makes new combinations of existing alleles. It does not make new alleles; only mutation does that.
  • After crossing over, the two sister chromatids of a chromosome are no longer identical. So when they separate in meiosis II, the two cells get different genetic information. This is why all four products of meiosis can differ.

Independent assortment: shuffling whole chromosomes

At metaphase I, each homologous pair lines up at the plate with one homologue facing each pole. Which homologue faces which pole is random, and the orientation of one pair does not affect the orientation of any other pair.

Two possible arrangements of two homologous pairs at metaphase I Two rows. Row 1, Arrangement 1: a cell at metaphase I with a dashed equator; the solid long and solid short chromosomes are on the left of the equator and the open long and open short on the right. Arrows lead to two gametes: one with long solid plus short solid, one with long open plus short open. Row 2, Arrangement 2: the solid long and open short are on the left and the open long and solid short are on the right. Its gametes are long solid plus short open, and long open plus short solid. A note at the bottom says two ways to line up, two gamete types each, two squared equals four gamete types. Arrangement 1 long solid + short solid long open + short open Arrangement 2 long solid + short open long open + short solid Two ways to line up, two gamete types each: 2 × 2 = 2² = 4 gamete types
Figure 2. Independent assortment in a cell with n = 2 (two homologous pairs). Solid chromosomes came from one parent and open chromosomes from the other. The two possible arrangements at metaphase I give four kinds of gamete. Crossing over is not shown.

Reading Figure 2. With two pairs, there are two ways to line up. Each way gives two kinds of gamete, so four kinds are possible in total: two that are all from one parent and two that are mixed. Every extra pair doubles the number of combinations. The general rule is:

number of chromosome combinations in gametes = 2ⁿ, where n is the haploid number (the number of homologous pairs).

For humans, n = 23, so one person can make 2²³ = 8 388 608 (about 8.4 million) different combinations of whole chromosomes, before crossing over is even counted.

Random fertilization: multiplying the variation

Any one of a father’s possible sperm can fuse with any one of a mother’s possible eggs. For one human couple, ignoring crossing over:

2²³ × 2²³ = 2⁴⁶ ≈ 7.0 × 10¹³ (about 70 trillion) possible chromosome combinations in a child.

Crossing over adds even more, because it creates combinations of alleles within chromosomes. Taken together, crossing over, independent assortment and random fertilization make it almost certain that every child of the same parents (except identical twins) is genetically unique.

When separation fails: nondisjunction

Sometimes homologues or sister chromatids fail to separate. This is nondisjunction. The gametes that result are not haploid: some have an extra chromosome (n + 1) and some are missing one (n − 1).

Where it happens What fails to separate Gametes from one cell
Meiosis I one homologous pair goes to the same pole all four abnormal: two n + 1, two n − 1
Meiosis II sister chromatids in one of the two cells two normal (n), one n + 1, one n − 1

If an abnormal gamete fuses with a normal one, the zygote has an abnormal chromosome number (aneuploidy): 2n + 1 (a trisomy, three copies of one chromosome) or 2n − 1 (a monosomy, one copy). Such changes usually have large effects on development. In humans, an extra copy of chromosome 21 (trisomy 21) causes Down syndrome. You will meet the effects of chromosome changes on phenotype again later in the course.

Worked example 1: counting combinations

Question. A fictional beetle species has 2n = 8. Ignore crossing over. (a) How many different combinations of chromosomes can one beetle’s gametes contain? (b) What is the probability that a gamete contains only chromosomes the beetle inherited from its mother? (c) How many different chromosome combinations are possible in the offspring of one pair of beetles? (d) Explain why the true number of genetically different offspring is larger.

(a) 2n = 8, so n = 4 homologous pairs. Each pair can go either way, independently: 2 × 2 × 2 × 2 = 2⁴ = 16 combinations.

(b) For each pair, the chance that the maternal homologue goes into a given gamete is 1/2. The pairs are independent, so multiply: (1/2)⁴ = 1/16 = 0.0625 (6.25%). Check: this is 1 of the 16 combinations from (a).

(c) Each offspring gets one of 16 combinations from its mother and one of 16 from its father: 16 × 16 = 256 combinations.

(d) Crossing over in prophase I makes recombinant chromatids, so even two gametes with the “same” combination of whole chromosomes can carry different combinations of alleles along each chromosome.

Common slips. Using 2n (8) or n² (16 here only by coincidence: 4² = 16 = 2⁴) instead of 2ⁿ. Test your rule on a different number: for n = 3, 2ⁿ = 8 but n² = 9. Only 2ⁿ is correct.

Worked example 2: nondisjunction and a testable question

Question. A fictional animal has 2n = 6. (a) Give the chromosome number of each of the four gametes if one homologous pair fails to separate in meiosis I. (b) Repeat for a failure of sister chromatids to separate in one cell in meiosis II. (c) Each abnormal gamete fuses with a normal gamete. Give the possible chromosome numbers of the zygotes. (d) A fictional study measured the percentage of eggs with the wrong chromosome number in female mice of different ages:

Age of female / months 3 6 9 12
Eggs with wrong chromosome number / % 1.5 2.5 6.0 14.0

Pose a testable question based on these data, and outline how you would test it.

(a) Meiosis I. n = 3. The pair that fails goes to one cell, so one cell after meiosis I has 4 chromosomes and the other has 2. Meiosis II then proceeds normally in both, so the gametes have 4, 4, 2 and 2 chromosomes. All four are abnormal.

(b) Meiosis II. Meiosis I is normal: two cells with 3 chromosomes each. In one cell a pair of sister chromatids goes to the same pole, giving 4 and 2. The other cell divides normally. Gametes: 3, 3, 4 and 2. Half are abnormal.

(c) A normal gamete has 3 chromosomes. 4 + 3 = 7 (2n + 1, trisomy). 2 + 3 = 5 (2n − 1, monosomy). Normal gametes from (b) give 3 + 3 = 6, a normal zygote.

(d) Posing a testable question. The data show that the percentage of abnormal eggs rises with age, from 1.5% at 3 months to 14.0% at 12 months (about nine times higher). A good testable question names one variable to change and one to measure:

  • Question: “Does the age of a female mouse affect the percentage of her eggs that have an abnormal chromosome number?”
  • Independent variable: age of the female (for example 3, 6, 9 and 12 months).
  • Dependent variable: the percentage of eggs with n + 1 or n − 1 chromosomes, found by counting chromosomes in each egg.
  • Controlled variables: strain of mouse, diet and housing, and the method and person used to count chromosomes.
  • Replication: many eggs from several females at each age, so a pattern is not caused by one unusual animal.

Interpretation. The table alone shows a pattern but not a cause. The data were not collected to rule out other differences between the age groups. A controlled study like the one outlined, or a further question about which division fails more often in older females, would test the idea directly. Notice that a testable question can be answered with data you could actually collect; “Why is meiosis less accurate in old age?” is a good starting point for thinking, but it is not yet testable as written.

Common misconceptions

  • “Crossing over happens between sister chromatids.” It happens between non-sister chromatids of a homologous pair. Swapping between identical sisters would change nothing.
  • “Crossing over creates new alleles.” It creates new combinations of existing alleles. Mutation creates new alleles.
  • “Independent assortment happens because sister chromatids separate randomly.” The key random event is the orientation of each homologous pair at metaphase I.
  • “The number of gamete combinations is 2n or n².” It is 2ⁿ, where n is the number of homologous pairs.
  • “Fertilization just restores the diploid number.” It also adds variation, because any gamete can meet any other.
  • “Nondisjunction in meiosis I and meiosis II have the same effect.” In meiosis I all four gametes are abnormal; in meiosis II only two are.
  • “Meiosis is the only source of genetic variation.” Mutation is the original source of new alleles. Meiosis and fertilization reshuffle them.

Where this leads

The shuffling of chromosomes in meiosis is the physical basis of Mendel’s laws. Topic 5.3, Mendelian genetics, uses it to predict the outcomes of crosses. Test yourself with the practice questions, then use the revision notes and the checklist to consolidate.

Quick questions

Does crossing over create new alleles?

No. It creates new combinations of alleles that already exist on the two homologues. New alleles come from mutation.

Why does nondisjunction in meiosis I affect all four gametes, but in meiosis II only two?

In meiosis I, a whole homologous pair goes to one cell, so both cells from the first division are wrong, and so are all four products. In meiosis II, the error happens in only one of the two cells, so the other cell still makes two normal gametes.

Do I need to know the life cycles of ferns, mosses or other organisms?

No. The course states that the details of sexual life cycles in different plants and animals are beyond the scope of the exam. Focus on what happens to chromosomes in meiosis and fertilization.

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