Nondisjunction in Meiosis I vs II: Worked Practice
A gamete with one extra chromosome can result from an error in either meiosis I or II. Its count alone doesn't identify the division. Following all four products helps: a meiosis I error affects both branches; an error in just one meiosis II cell leaves the other branch untouched.
Track one chromosome pair in a 2n = 6 species, then try eight nondisjunction practice problems. For normal counts, start with chromosome and chromatid counting.
Set up the pair you're following
Normally, homologs separate in meiosis I; sisters separate in meiosis II. Nondisjunction is a failure of that separation: both go to the same destination. OpenStax distinguishes the two errors.
Our invented species has three chromosome kinds: A, B, and C, with two homologs of each. Its haploid number is n = 3: a normal product receives one of each kind.
Here, n+1 means one extra chromosome, and n−1 means one missing chromosome. Neither describes a change by a whole chromosome set.
Label the A homologs A1 and A2, not alleles. Each replicated homolog remains one chromosome with two sister chromatids. Each chromatid contains one complete double-stranded nuclear DNA molecule; after separation, each counts as a daughter chromosome. The NHGRI chromatid definition describes the attached sister pair.
Every example assumes:
- One nondisjunction event involving A, either in meiosis I or in just one of the two meiosis II cells.
- Normal segregation of B and C, and normal behavior in every other division.
- One DNA replication before meiosis I; none between divisions.
- No chromosome loss, breakage, or extra replication. Counts refer to completed divisions, including cytokinesis.
Count nuclear DNA only, excluding mitochondrial and chloroplast DNA.
Our four-product schematic doesn't represent four functional human eggs: human oogenesis produces one egg and polar bodies through unequal divisions.
Meiosis I: both homologs enter one branch
After replication, the starting cell contains six chromosomes and twelve DNA molecules.
A1 and A2 both enter cell L during meiosis I; cell R receives neither. B and C separate normally: each cell gets one B and one C.
| After meiosis I | Replicated chromosomes present | Chromosomes | DNA molecules |
|---|---|---|---|
| Cell L | A1, A2, one B, one C | 4 | 8 |
| Cell R | One B, one C; no A | 2 | 4 |
Meiosis II proceeds normally. Each product from L receives a daughter chromosome from A1 and another from A2, plus B and C. Each product from R receives only B and C.
The four final counts are 4, 4, 2, 2: two n+1 products and two n−1 products. None has the normal chromosome complement. The products missing A aren't empty; the unaffected chromosomes are still there.
Meiosis II divides what's already in each branch. It can't send A back into cell R.
Meiosis II: keep the other branch in view
This time meiosis I is normal. Cell L receives replicated A1, B, and C; cell R receives replicated A2, B, and C. Each has three chromosomes containing six DNA molecules.
In cell L's meiosis II, both A1-derived sister chromatids go into product L1. Product L2 gets neither. B and C separate normally. Cell R completes meiosis II normally too.
| Final product | A-derived chromosomes | Other chromosomes | Total chromosomes | DNA molecules |
|---|---|---|---|---|
| L1 | 2 | One B, one C | 4 | 4 |
| L2 | 0 | One B, one C | 2 | 2 |
| R1 | 1 | One B, one C | 3 | 3 |
| R2 | 1 | One B, one C | 3 | 3 |
The four final counts are 4, 2, 3, 3: one n+1, one n−1, and two normal products. An error in R would move the abnormal products to that branch. “Normal” means the normal chromosome complement; it doesn't mean genetically identical.
MIT's introductory biology recitation confirms this distinction: homolog nondisjunction affects four products; chromatid nondisjunction in one daughter cell affects two.
A ledger that catches missing chromosomes
Both examples begin with twelve DNA molecules after replication. At the end, the totals remain twelve:
- Meiosis I error: 4 + 4 + 2 + 2 = 12.
- Meiosis II error: 4 + 2 + 3 + 3 = 12.
An extra chromosome in one product has a corresponding absence elsewhere; existing material is redistributed.
In these completed-stage counts, each chromosome after meiosis I contains two DNA molecules; each daughter chromosome after meiosis II contains one. OpenStax's meiosis account follows one replication through both divisions.
DNA-molecule count differs from DNA amount: chromosome lengths can differ, so four molecules needn't contain twice the DNA of two.
What the products let you infer
Given all four products from one meiosis, the known n, and our single-error assumptions, compare the complete set:
| Error | What goes to the same destination | Four final products | Products with a normal complement |
|---|---|---|---|
| Meiosis I | Both homologs of the affected pair | n+1, n+1, n−1, n−1 | 0 |
| Meiosis II in one cell | Both sisters of the affected chromosome | n+1, n−1, n, n | 2 |
A single n+1 or n−1 product is compatible with either division. Even one normal product only excludes the meiosis I error under this model; it doesn't establish that an error happened elsewhere. Mixing products from different meioses also removes the basis for this four-product comparison.
Don't diagnose the division from an arbitrary allele match: crossing over can change which alleles a chromatid carries. Here, follow chromosome counts and branches.
Add the normal gamete at fertilization
A normal gamete contributes n chromosomes, including one A. Fertilization combines the contributions; see the NHGRI explanation of haploid cells.
| Product used as a gamete | Normal partner | Zygote count | A chromosomes in zygote |
|---|---|---|---|
| n+1 = 4 | n = 3 | 2n+1 = 7 | 3 |
| n−1 = 2 | n = 3 | 2n−1 = 5 | 1 |
| n = 3 | n = 3 | 2n = 6 | 2 |
Three copies of A give trisomy for A; one gives monosomy for A. Both are aneuploidy, a gain or loss of individual chromosomes, as described by NHGRI.
Seven chromosomes means three A, two B, and two C. Triploidy requires three of every kind: 3n = 9. These counts don't predict development or survival.
Eight problems to solve before scrolling
Draw one starting cell → two meiosis I cells → four final products, keeping siblings grouped by branch. Use our stated assumptions.
1. Follow the missing pair
For 2n = 6, A1 and A2 both enter cell R during meiosis I. Give the chromosome and DNA-molecule counts in L and R after that division, then the chromosome counts in their four final products.
2. Change the affected branch
For 2n = 6, meiosis I is normal. A2's sisters undergo nondisjunction in cell R during meiosis II. List the number of A chromosomes and the total chromosome count in each final product.
3. Read a complete set
A species has 2n = 10. All four products of one meiosis have chromosome counts 6, 4, 5, and 5. Which division contains the error? Which products are siblings from the affected branch?
4. You only have one product
For 2n = 14, one final product contains eight chromosomes. Can its count identify meiosis I versus II nondisjunction? Give the two possible complete four-product patterns.
5. Find the broken ledger
A student predicts 4, 4, 3, and 2 final chromosomes after one meiosis I error in the 2n = 6 model. Explain what's wrong and repair the counts.
6. Calculate fertilization
For 2n = 12, a product missing the affected chromosome functions as a gamete and combines with a normal gamete. Give each gamete's chromosome count, the zygote count, and the number of copies of the affected chromosome in the zygote.
7. Extra chromosome or extra set?
For 2n = 8, an n+1 gamete combines with a normal gamete. Is the resulting zygote triploid? Give its chromosome count and the count a triploid zygote would have.
8. Test the assumptions
For 2n = 6, all four final products have counts 4, 2, 4, and 2, listed without sibling labels. A student labels this “meiosis II nondisjunction in one cell.” Does that fit? If meiosis I was normal, what additional error would produce this pattern?
Answers and explanations
1. The deficit follows cell L
L has 2 chromosomes / 4 DNA molecules; R has 4 chromosomes / 8 DNA molecules after meiosis I. L's final products have 2, 2 chromosomes; R's have 4, 4. B and C account for the two chromosomes in each deficient product.
2. Two normal products remain in L
L's products each have one A, three chromosomes total. R's have two A, four total, and zero A, two total. Their order within R can reverse: 3, 3, 4, 2.
3. An error in meiosis II
Here n = 5: n+1, n−1, n, n. The six- and four-chromosome products are siblings from the affected branch; both five-chromosome products come from the normal branch.
4. Both divisions remain possible
Here n = 7. A meiosis I error gives 8, 8, 6, 6; a meiosis II error gives 8, 6, 7, 7. Both include an eight-chromosome product.
5. One branch can't gain a normal sibling
The proposed total is 13 DNA molecules, exceeding the twelve after replication. Correct counts: 4, 4, 2, 2. Both products from the deficient cell lack A; normal meiosis II can't supply it to one sibling.
6. A monosomic zygote
Here n = 6. The deficient gamete has 5 chromosomes, and its normal partner has 6. The zygote has 11 = 2n−1, with one copy of the affected chromosome, supplied by the normal gamete. The other five kinds each have two copies.
7. Nine isn't three sets of four
Here n = 4. Fertilization gives 5 + 4 = 9 = 2n+1 chromosomes: three copies of one kind, two of each remaining kind. A triploid complement would have 12 = 3n, three copies of every kind.
8. Both meiosis II branches would need errors
One meiosis II error leaves two normal three-chromosome products, so the claim doesn't fit. After normal meiosis I, nondisjunction of the affected chromosome in both meiosis II cells could give 4, 2 from each branch. That requires two errors and falls outside our single-error model. Under the original assumptions, these unordered counts instead point to meiosis I: siblings would be grouped as 4, 4 and 2, 2.
Keep a card for the step you missed
Save the missed decision as a card. The guide to making better flashcards explains how to keep questions answerable on their own.
| Front | Back |
|---|---|
| Can one n+1 gamete identify meiosis I versus II nondisjunction? | No. Either division can produce n+1. All four products distinguish them if one chromosome pair is affected by one error and every other division is normal. |
| Meiosis I is normal, then one meiosis II cell has a nondisjunction event. Why are two products normal? | The other meiosis I cell completes meiosis II normally. |
| An n−1 gamete combines with a normal n gamete. What's the zygote count? | 2n−1. The normal partner adds one copy of the chromosome the first gamete lacks. |
Next time, change 2n and draw before calculating. For alleles in normal gametes, try monohybrid cross practice.