Chromosomes and Meiosis: The Physical Basis of Mendel's Laws

A fruit fly cell holds eight chromosomes. Its eggs and sperm hold four, one from each of four pairs, and which member of each pair goes in is, crossing over aside, settled at a single step of a single division.

In the previous lesson you met Mendel’s two laws: the two alleles of a gene separate into different gametes, and alleles of different genes enter gametes independently. Mendel inferred both from counts of peas. He never saw what did the separating. This lesson finds it. The answer is a cell division called meiosis, and once you can count the chromosomes in it, both laws read off the cell like a diagram.

Count before you divide: homologs, sisters and ploidy

Chromosomes live in the nucleus, the membrane-bound compartment that holds a eukaryotic cell’s DNA (cell compartments has more). Four terms do most of the work:

• Homologous chromosomes (homologs) are the two members of a pair, one inherited from each parent. They carry the same genes at the same loci, but can carry different alleles.
• Sister chromatids are the two identical copies of one chromosome made in S phase, still joined at the centromere.
• Ploidy is the number of chromosome sets in a cell. A diploid (2n) cell has two sets; a haploid (n) cell has one.
• The C value is the DNA content of one unreplicated haploid set. A diploid cell is 2C before S phase and 4C after it.

Think of a pair of shoes. The left and right shoes are homologs: same job, built separately, one from each parent. Now photocopy each shoe and staple the copy to it. Each stapled bundle is one chromosome made of two sister chromatids. The rule that saves most students on exams: count centromeres, not arms. A replicated chromosome is still one chromosome.

Meiosis I separates homologs; meiosis II separates sisters

Meiosis is one round of DNA replication followed by two divisions. In prophase I, homologs pair along their length, a process called synapsis; a pair of synapsed homologs is a bivalent, with four chromatids. At metaphase I each bivalent sits on the spindle with its two homologs facing opposite poles. Anaphase I pulls the homologs apart. There is no S phase before meiosis II, which then pulls the sisters apart, as a mitosis would.

Mitosis is different at the first step. Homologs do not synapse. Each chromosome lines up on its own, its sisters separate, and the two daughter cells keep both sets and match the parent. Meiosis halves the sets and, as you will see, shuffles them.

Worked problem. Drosophila melanogaster has 2n = 8, so n = 4. Assume normal divisions and track chromosomes, chromatids and C value.

Stage	Chromosomes	Chromatids	Bivalents	C value
G1, before S	8	8	0	2C
Prophase I, after S	8	16	4	4C
Each cell after meiosis I	4	8	0	2C
Each cell after meiosis II	4	4	0	1C

Step 1. S phase doubles chromatids and DNA but not chromosomes: 8 chromosomes, 8 × 2 = 16 chromatids, 4C. Step 2. Eight chromosomes in pairs make 8/2 = 4 bivalents. Step 3. Anaphase I sends one homolog of each pair to each pole: 4 chromosomes per cell, each still two chromatids, so 8 chromatids and 2C. The cell is now haploid, one set, even though its DNA content equals that of a G1 diploid cell. Step 4. Anaphase II splits the sisters: 4 chromosomes, 4 chromatids, 1C.

The common error is to say meiosis I separates sisters. It does not. If you find yourself calling a cell after meiosis I diploid, recount the centromeres.

What holds sisters together, and lets them part in two stages? A ring-shaped protein complex, cohesin, which links the sisters from S phase. In yeast and mammals as studied, cohesin on the chromosome arms is cut at anaphase I, while cohesin at the centromeres is protected by a protein called shugoshin until anaphase II.

One meiosis gives four haploid nuclei, but not always four gametes. In sperm production the four become four sperm. In egg production in humans and other animals, the divisions are unequal: one large egg and small polar bodies that do not become gametes.

Where Mendel’s laws live in the cell

Take a fly heterozygous Aa at one locus. A sits on one homolog, a on the other. After S phase one homolog carries A and A on its sisters, the other a and a. Anaphase I separates the homologs, so A and a go to different cells; anaphase II separates identical sisters. Two of the four products carry A and two carry a. That is the law of segregation, seen as chromosome movement.

There is a refinement. If a crossover falls between the locus and its centromere, each homolog leaves meiosis I carrying one A chromatid and one a chromatid, and A and a part at anaphase II instead. Either way, each final product gets exactly one allele.

Independent assortment comes from metaphase I. Each bivalent can face the spindle two ways, and how one pair faces does not affect how another does.

Worked problem, continued. Ignoring crossing over, the fly’s 4 bivalents give

2 × 2 × 2 × 2 = 2^4 = 16 possible chromosome combinations in a gamete

For humans, 2n = 46 and n = 23:

2^23 = 8,388,608, about 8.4 million

That is a count of combinations across many meioses, and it ignores crossing over, so it understates real variety. Now put alleles on it. A fly is Aa Bb, with the two loci on different chromosomes, and no crossing over. In any one meiosis the bivalents take one orientation, so the four products are either 2 AB + 2 ab or 2 Ab + 2 aB: only two kinds of gamete. The two orientations are equally likely, so across many meioses each of AB, Ab, aB and ab makes up 1/4 of the gametes. Mendel’s four equal classes are a property of the population of gametes, not of any single cell.

Stage

Orientation of the two bivalents at metaphase I A and B face the same pole A and b face the same pole Each meiosis picks one at random Meioses tallied 10 50 200 1,000 Resample

cells

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ploidy of each cell

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chromosomes per cell

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chromatids per cell

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DNA per cell

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Rule to keep: anaphase I separates homologs and anaphase II separates sisters, so one meiosis gives two kinds of product, and only across many meioses with random orientation does each of AB, Ab, aB and ab make up 1/4. This is a sketch of the logic with two chromosome pairs and no crossing over.

The law holds only for loci that assort independently: those on different chromosomes, or far apart on one. Loci close together on the same chromosome tend to travel together, the subject of linkage and mapping.

Crossing over and the synaptonemal complex

Synapsis is held in place by the synaptonemal complex, a protein structure that zips homologs together along their length. Prophase I is long and divided into substages by what the chromosomes look like. Homologs pair in zygotene and are fully synapsed in pachytene. Crossing over, the exchange of segments between non-sister chromatids, is completed in pachytene, when there are four chromatids per bivalent. In diplotene the complex comes apart, homologs pull slightly away from each other, and they stay joined at chiasmata, the visible sites of earlier crossovers. Crossing over is a prophase I event. By metaphase I it is over, and it is not part of a normal mitosis.

Chiasmata matter for segregation as well as variety. In most organisms studied, a crossover plus the cohesion beyond it is what holds two homologs together until anaphase I, so the spindle can pull them to opposite poles. The molecular steps, which begin with deliberate breaks in DNA, are taught in DNA repair and recombination.

Crossover numbers vary even within a species. In single-sperm sequencing of 31,228 gametes from 20 men, the average number of crossovers per sperm ranged from 22.2 to 28.1 between donors. And the complex itself is less rigid than the zipper image suggests: in living C. elegans, its components behave like a liquid crystal, ordered yet able to flow. In maize, Harriet Creighton and Barbara McClintock tied the exchange of visibly marked chromosome segments to genetic crossing over in 1931; you will use that link to build maps.

Sutton, Boveri and the proof

By 1902 cytologists knew that chromosomes pair and are halved in number before gametes form. Walter Sutton, a graduate student with E. B. Wilson at Columbia, worked on the grasshopper Brachystola magna, whose chromosomes differ in size. He saw that they came in matched pairs, and in 1903 argued that maternal and paternal members pair and then separate, each pair orienting independently. That predicts both of Mendel’s laws. He also wrote that in his material he could not tell maternal from paternal members, so the idea remained a hypothesis.

Theodor Boveri came at it from sea urchins. In eggs fertilized by two sperm, chromosomes were distributed unevenly among three or four poles, and as a rule the resulting cells developed differently and abnormally. He concluded that development needs a particular combination of chromosomes, not merely a number, so different chromosomes carry different things. Neither man proved the theory alone. Sutton and Boveri are generally credited together, and the joined name came later.

The test of Sutton’s weak point came from Elinor Carothers in 1913. In grasshopper males with one heteromorphic pair, meaning one member visibly larger, and an unpaired X, she could see which homolog the X travelled with. It went with each about equally often. The strongest proof came in 1916, when Calvin Bridges showed in Drosophila that rare flies with exceptional X-linked inheritance had the exceptional chromosome sets his explanation predicted, a story for sex linkage and chromosome changes.

Even the human count needed work. The accepted diploid number was 48 until Joe Hin Tjio and Albert Levan reported 46 in 1956.

Design problem

You repeat Carothers’ kind of study. In a grasshopper, one homolog pair is heteromorphic, and the X is unpaired in males. You score 300 secondary spermatocytes: the X went to the pole with the larger homolog in 154 and with the smaller in 146. These counts are hypothetical. Does this support independent assortment, and what would dependence look like?

One way to do it

Expectation. If the X and the heteromorphic pair orient independently, the X goes with either homolog with probability 1/2: expected 150 and 150.

Test. Use the chi-square statistic, sum of (observed − expected)^2 / expected, taught fully in probability and crosses:

χ² = (154 − 150)^2/150 + (146 − 150)^2/150 = 16/150 + 16/150 = 32/150 = 0.21

With two classes there is 1 degree of freedom, and the critical value at P = 0.05 is 3.84. Since 0.21 is far below it, the counts are consistent with random, independent orientation.

What dependence would look like. A large excess in one direction, for example 220 to 80 (χ² = 65.3), that recurs in other males and other preparations. A single small deviation is sampling noise.

What this tests. It tests chromosomes directly, by sight, not alleles. It shows that two non-homologous chromosomes assort independently, which is what Mendel’s second law needs if genes sit on chromosomes. It does not by itself show that any gene is on these chromosomes.