MCAT content library / Biology & biochemistry / Meiosis and recombination: track chromosomes, DNA and alleles
Keep chromosome sets, DNA amount and allele combinations separate as you follow each division.
About 16 minutes, including the worked examples and checks.
Useful foundations
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Track three quantities separately
Meiosis problems often combine chromosome number, DNA amount and allele combinations. Those quantities change for different reasons. Copying DNA increases the amount of DNA; separating homologs reduces the number of chromosome sets; crossing over rearranges which alleles share a chromatid. A sound explanation keeps those operations distinct. These mechanics and their contribution to variation are explicit AAMC topics.1
Homologous chromosomes carry corresponding loci, although their alleles can differ. Sister chromatids arise by replication of one chromosome. Before sister separation, the two chromatids are treated as one duplicated chromosome; after separation, each is an individual chromosome. Thus, chromosome counting is not simply counting double-helical DNA molecules whenever replication has occurred.2
Use n for one complete chromosome set and C for the DNA amount in one unreplicated haploid genome. A diploid cell before DNA synthesis is 2n, 2C. After synthesis it is still 2n, now 4C. It has copied each chromosome, not acquired an additional homologous set. A haploid cell can likewise contain duplicated chromosomes without becoming diploid.
One replication, then two different separations
During prophase I, replicated homologs pair by synapsis. The synaptonemal complex helps organize close alignment, and exchanges can occur between nonsister chromatids. A paired unit contains two homologous chromosomes and four chromatids, so it is called a tetrad. Chiasmata reflect crossover connections that help maintain the association of homologs after the synaptonemal complex disassembles.3
Meiosis I separates homologous chromosomes while sister centromere regions remain associated. Each resulting nucleus receives one homolog from each pair: the reduction in chromosome sets has already occurred. Meiosis II separates sister chromatids. There is no new S phase between the two divisions. Crossing over may have made sisters genetically different along part of their length, but they remain sisters because they arose from replication of the same chromosome.
Contrast a conventional mitosis: duplicated chromosomes align without the same homolog-pairing program, and sisters separate in one division. Two diploid products can result from a diploid starting cell. In a standard completed meiosis, four haploid nuclear products result. This is a chromosome-accounting statement; it does not assume that all four products become functional gametes in every reproductive system.35
Worked chromosome and DNA accounting
Consider an original model organism with 2n = 4: two homologous pairs. For the following table, count each completed daughter nucleus or cell after division, not all descendants pooled together and not a cell temporarily in anaphase. Assume complete replication, normal segregation and no DNA loss.
| Stage, per completed cell/nucleus | Chromosomes | DNA molecules | DNA amount |
|---|---|---|---|
| Before S phase | 4 | 4 | 2C |
| After S phase, before meiosis I | 4 duplicated | 8 | 4C |
| After meiosis I | 2 duplicated | 4 | 2C |
| After meiosis II | 2 unduplicated | 2 | 1C |
The after-S row contains eight DNA molecules because each of four chromosomes has two chromatids. After meiosis I, a daughter receives two chromosomes, each still duplicated. Its DNA content is 2C, but its chromosome-set count is n. Saying “2C means diploid” would therefore misclassify it.
Suppose a stain gives fluorescence proportional to nuclear DNA mass, with 50 units for a pre-S diploid nucleus. The predicted values are 100 after S, 50 after meiosis I and 25 after meiosis II. The pre-S diploid nucleus and post-meiosis-I haploid nucleus have the same DNA mass in this model, despite different chromosome arrangements. Fluorescence alone cannot distinguish them.
There is also a useful conservation check. Immediately before meiosis I, the original cell has 4C DNA. At the end, four products each have 1C, giving the same total 4C. Treating each final product as 2C would silently invent another replication round. Fertilization of two normal 1C haploid gametes restores a 2n, 2C zygote before its next S phase.
Worked crossover: follow actual chromatids
On the first chromosome pair, the starting homologs carry AB and ab. The order is centromere–A–B. Replication gives two AB sisters and two ab sisters. One crossover occurs between A and B, involving the second AB chromatid and the first ab chromatid. Reciprocal exchange changes only those two chromatids:
| Chromatid identity | Before exchange | After exchange |
|---|---|---|
| AB homolog: sister 1 | AB | AB |
| AB homolog: sister 2 | AB | Ab |
| ab homolog: sister 1 | ab | aB |
| ab homolog: sister 2 | ab | ab |
Two chromatids retain parental marker combinations and two are recombinant. Crossing over has rearranged existing alleles; no new A, a, B or b allele was created. A crossover elsewhere on the chromosome need not recombine these particular markers. To predict marker combinations, locate the exchange relative to the loci and identify the participating chromatids.
The second chromosome pair carries C and c and has no crossover in this example. Choose one specific metaphase-I orientation: the C homolog moves to the same daughter as the homolog whose centromere is attached to the A-bearing chromatids. After the first division, that daughter contains AB and Ab sisters plus the two C sisters. The other contains aB and ab sisters plus the two c sisters. After meiosis II, the four products are ABC, AbC, aBc and abc.
Reversing the second pair’s orientation produces ABc, Abc, aBC and abC instead. The independent orientation of different homologous pairs generates combinations across chromosomes. Crossing over generates combinations within a homologous pair. These mechanisms are related sources of variation, but they are not the same event.4
Choose the correct population for a probability
Without crossing over, n independently orienting homologous pairs can generate 2n parental-chromosome combinations across many meioses. That count describes possible combinations, not the number of products from one meiosis. One standard meiosis still produces four haploid nuclei. Likewise, observing two recombinant chromatids in the specified tetrad does not establish the population recombination frequency for the two loci; that requires many meiotic events and their distribution of crossovers.
For any changed case, label homolog identity and sister identity first. Mark the crossover position, carry the centromeres through meiosis I, and separate sisters in meiosis II. Finally, state whether you are reporting DNA mass, chromosome count per product, allele content of one product, or possible outcomes across a population. Most counting contradictions disappear once the unit being counted is explicit.
ACTIVE PRACTICE · MEIOSIS AND RECOMBINATION
Track Chromosomes, DNA Molecules, and Segregation Independently
The original organism has 2n = 4. Values describe one completed cell or nucleus at each stage, not all descendants pooled. A separate single crossover involves one AB and one ab sister chromatid between A and B, giving AB, Ab, aB and ab.
Chromosomes: ; DNA molecules: ; DNA amount: .
Expected DNA-fluorescence units at this stage (pre-S diploid = 50):.
One original meiosis produces four chromatids/gametes carrying: ABC, AbC, aBc, abc.
Single-crossover chromatids are specifically assigned as in the lesson. Equal fluorescence between pre-S and post-I nuclei does not mean equal ploidy. The four gametes shown are one event, not a population distribution.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A model organism has four homologous chromosome pairs and genotype AaBbCcDD, with one scored locus on each pair. Crossing over is absent, segregation and independent orientation are normal, and all gametes are viable. Across many meioses, how many different gamete genotypes can occur at these four scored loci?
A. 4
Review this choice. Four is the number of nuclear products from one standard meiosis, not the number of possible marker combinations across many meioses.
B. 8
Correct. The three heterozygous loci each contribute two possibilities. The DD locus contributes only D, so 2 × 2 × 2 × 1 = 8.
C. 12
Review this choice. Multiplying the three variable loci by four products does not enumerate independent allele combinations.
D. 16
Review this choice. There are four chromosome pairs, but the two homologs at the scored D locus carry the same allele. Counting their parental origins separately does not create a new scored gamete genotype.
The principle to carry forward
Distinguish possible parental-chromosome combinations from distinguishable allele combinations at the markers being scored.
Check 2
A species has 2n = 6. One diploid cell completes DNA replication and meiosis I normally. For one resulting daughter cell, how many chromosomes and DNA molecules are present before meiosis II?
A. 3 chromosomes and 3 DNA molecules
Review this choice. There are three chromosomes, but each remains duplicated and therefore contains two DNA molecules.
B. 3 chromosomes and 6 DNA molecules
Correct. One homolog from each of three pairs gives three chromosomes. Each still has two chromatids, giving six DNA molecules.
C. 6 chromosomes and 6 DNA molecules
Review this choice. Six chromosomes would retain both homologous sets in this daughter rather than reflecting normal reductional division.
D. 6 chromosomes and 12 DNA molecules
Review this choice. This describes the starting diploid cell after S phase, before homologs are distributed.
The principle to carry forward
After meiosis I, a nucleus is haploid even though each chromosome still contains two chromatids.
Check 3
Two replicated homologs carry AB and ab, with loci ordered centromere–A–B–chromosome end. Exactly one reciprocal crossover occurs between B and the chromosome end in two nonsister chromatids. No mutation or other crossover occurs. Considering only markers A and B, which combinations are present among the four chromatids afterward?
A. AB and ab only
Correct. The exchange is distal to both markers, so it does not separate A from B or a from b. Distal DNA is exchanged while the scored marker combinations remain parental.
B. AB, Ab, aB and ab
Review this choice. All four marker combinations would result from a suitable crossover between A and B, not beyond both markers.
C. Ab and aB only
Review this choice. The given exchange does not recombine A and B, and a single crossover would not convert all four chromatids into recombinant marker types.
D. AA and bb only
Review this choice. Each chromatid carries one allele at A and one at B; AA and bb do not describe the two loci on one chromatid.
The principle to carry forward
A crossover is not automatically detectable as recombination between the particular markers being scored.
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Related lessons
- DNA replication and repair: follow strands, fragments and errors
- Mendelian inheritance: calculate from the cross and its assumptions
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Sources
1. AAMC: B/B Content Category 1C
2. OpenStax Biology 2e: Cell Division
3. OpenStax Biology 2e: The Process of Meiosis