The single most reliable trap in meiosis questions: swapping events between meiosis I and meiosis II (or worse, importing mitosis events into meiosis I). If the question says "synapsis" or "crossing over," the answer is prophase I of meiosis — never mitosis, never meiosis II.
What meiosis does. Meiosis is a reductional division that converts a diploid (2n) cell into four haploid (n) daughter cells. It occurs in germ cells and consists of two sequential divisions: meiosis I (reductional — homologues separate) and meiosis II (equational — sister chromatids separate). NCERT Class 11 Biology Chapter 10, page 125, identifies meiosis as the basis of gamete formation and genetic variation.
Meiosis I phases:
Meiosis II resembles mitosis: sister chromatids separate at anaphase II. No DNA replication between meiosis I and II.
Significance: genetic variation (crossing over + independent assortment), maintenance of chromosome number across generations, provides raw material for evolution.
Watch-out: Sister chromatids vs homologues — confusing which structure separates at which anaphase is the direct path to losing marks.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
During which sub-stage of prophase I does synapsis occur?
Answer: D. Synapsis (pairing of homologous chromosomes to form bivalents) occurs during zygotene of prophase I (NCERT Class 11 Biology Chapter 10, page 126).
Why A is wrong: A is wrong because leptotene involves chromosome condensation only — homologues have not yet paired. (trap: sequence confusion within prophase I sub-stages)
Why B is wrong: B is wrong because diplotene is when homologues begin to separate and chiasmata become visible — synapsis is already complete. (trap: mixing later prophase I events with earlier ones)
Why C is wrong: C is wrong because pachytene follows synapsis — crossing over occurs at this stage, not the initial pairing. (trap: conflating pairing with crossing over)
How many functional daughter cells are produced at the end of meiosis in spermatogenesis?
Answer: A. Meiosis produces four haploid daughter cells from one diploid parent cell. In spermatogenesis, all four are functional spermatids (NCERT Class 11 Biology Chapter 10).
Why B is wrong: B is wrong because while meiosis produces haploid cells, the total number is four (two divisions), not two. (trap: forgetting meiosis II)
Why C is wrong: C is wrong because meiosis halves the chromosome number — daughter cells are haploid, not diploid. (trap: confusing meiosis with two rounds of mitosis)
Why D is wrong: D is wrong because meiosis is a reductional division — daughter cells are haploid, not diploid, and there are four, not two. (trap: confusing meiosis outcome with mitosis outcome)
Chiasmata are first clearly visible during which sub-stage of prophase I?
Answer: B. Chiasmata become clearly visible during diplotene, when homologous chromosomes begin to separate but remain connected at crossover points (NCERT Class 11 Biology Chapter 10).
Why A is wrong: A is wrong because zygotene is the stage of synapsis (pairing) — crossing over has not yet occurred. (trap: jumping ahead in the prophase I sequence)
Why C is wrong: C is wrong because although crossing over physically occurs during pachytene, chiasmata become visible only when homologues start to repel in diplotene. (trap: confusing when crossing over happens vs when chiasmata are seen)
Why D is wrong: D is wrong because diakinesis involves terminalization of chiasmata (they move toward chromosome ends), not their first appearance. (trap: confusing terminalization with initial visibility)
A cell with 2n = 14 chromosomes enters meiosis. At the end of anaphase I, how many chromosomes does each pole receive?
Answer: D. At anaphase I, homologous chromosomes separate (not sister chromatids). Each pole receives n = 7 chromosomes, and each chromosome still consists of two sister chromatids joined at the centromere.
Why A is wrong: A is wrong on both counts — the number should halve to 7, and chromatids remain paired until anaphase II. (trap: double error combining no-reduction with premature chromatid separation)
Why B is wrong: B is wrong because sister chromatids have NOT separated at anaphase I — that happens at anaphase II. Each chromosome at this point still has two chromatids. (trap: confusing anaphase I with anaphase II — the core meiosis trap)
Why C is wrong: C is wrong because anaphase I is reductional — chromosome number halves from 2n to n at each pole. 14 chromosomes at a pole means no reduction occurred. (trap: treating meiosis I as equational)
Which of the following events is exclusive to meiosis and does NOT occur in mitosis?
Answer: B. Crossing over between non-sister chromatids of homologous chromosomes occurs only during prophase I of meiosis. It requires bivalent formation (synapsis), which does not occur in mitosis. (trap: mitosis vs meiosis stages)
Why A is wrong: A is wrong because chromosome condensation occurs in prophase of both mitosis and meiosis — it is a general feature of cell division. (trap: assuming all prophase events are meiosis-specific)
Why C is wrong: C is wrong because spindle fiber attachment to kinetochores occurs in both mitosis (metaphase) and meiosis (metaphase I and II). (trap: confusing a universal mechanism with a meiosis-specific one)
Why D is wrong: D is wrong because sister chromatid separation occurs in anaphase of mitosis AND anaphase II of meiosis — it is common to both. (trap: forgetting that meiosis II is essentially equational like mitosis)
A student observes a cell where bivalents are aligned at the metaphase plate. This cell is in:
Answer: A. Bivalents (paired homologous chromosomes) at the metaphase plate is diagnostic of metaphase I of meiosis. In mitosis and meiosis II, individual chromosomes (not bivalents) align at the plate.
Why B is wrong: B is wrong because mitosis has no bivalent formation — homologues do not pair. Individual chromosomes line up at the metaphase plate. (trap: mitosis vs meiosis stages — importing meiosis structures into mitosis)
Why C is wrong: C is wrong because by meiosis II, homologues have already separated — only individual chromosomes (each with two sister chromatids) align at the plate, not bivalents. (trap: confusing meiosis I with meiosis II alignment)
Why D is wrong: D is wrong because at anaphase I, homologues are actively separating — they are no longer aligned at the metaphase plate. (trap: confusing the stage sequence)
A student claims that sister chromatids separate during anaphase I of meiosis. What is the correct rebuttal?
Answer: C. Anaphase I is reductional — homologous chromosomes (not sister chromatids) move to opposite poles. Sister chromatid separation occurs at anaphase II. This is the central distinction between the two meiotic divisions. (mistake: mistake: sister chromatids vs homologues)
Why A is wrong: A is wrong because crossing over involves exchange of segments between non-sister chromatids of homologues — it does not separate sister chromatids from each other. (trap: confusing physical exchange with chromatid separation)
Why B is wrong: B is wrong because sister chromatids absolutely exist in meiosis — after S phase, each chromosome consists of two sister chromatids joined at the centromere. They separate at anaphase II. (trap: denying the existence of a structure to avoid the real confusion)
Why D is wrong: D is wrong because simultaneous separation of homologues and sisters does not occur — meiosis I separates homologues while centromeres remain intact, preserving sister chromatid linkage until meiosis II. (trap: mistake: sister chromatids vs homologues)
In a cell with 2n = 8, if crossing over occurs at one chiasma per bivalent during meiosis I, how many genetically distinct types of gametes can potentially be produced considering ONLY independent assortment (ignoring crossing over variability)?
Answer: C. With 2n = 8, n = 4 bivalents align at metaphase I. Independent assortment gives 2^n = 2^4 = 16 possible combinations of maternal and paternal chromosomes in the gametes. The crossing over adds further variety but the question specifies to consider only independent assortment.
Why A is wrong: A is wrong because 2^n with n = 4 gives 16, not 4. Calculating 2^2 = 4 suggests confusing the number of bivalents with something else. (trap: arithmetic error in applying 2^n — possibly using n/2 instead of n)
Why B is wrong: B is wrong because 2^3 = 8 suggests using n = 3 instead of n = 4. With 2n = 8, n = 4 bivalents form, so the correct calculation is 2^4. (trap: off-by-one error in determining n from 2n)
Why D is wrong: D is wrong because 2^5 = 32 would require 5 bivalents (2n = 10). With 2n = 8, only 4 bivalents form. (trap: using 2n instead of n in the exponent, or miscounting chromosome number)
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Given
• Diploid number: 2n = 24• Number of bivalents with one chiasma each: 3
Required
(a) Total bivalents at metaphase I
(b) Chromosome number per cell after telophase I
(c) Total chromatids participating in crossing over
Concept
During meiosis I, all homologous chromosome pairs form bivalents at metaphase I (regardless of whether crossing over occurred). Telophase I produces haploid cells. Each chiasma involves an exchange between two non-sister chromatids (one from each homologue).
Formula / Rule
• Number of bivalents = n = 2n/2• Chromosome number after meiosis I = n (haploid)• Each chiasma involves 2 non-sister chromatids (out of 4 chromatids in the bivalent)
Substitution
(a) n = 24/2 = 12 bivalents
(b) Chromosomes per cell after telophase I = n = 12
(c) 3 chiasmata × 2 chromatids per chiasma = 6 chromatids
Calculation
(a) 12 bivalents align at metaphase I
(b) 12 chromosomes per daughter cell (each chromosome = 2 sister chromatids at this point)
(c) 6 chromatids involved in crossing over
Note on exact values: The numbers 24, 3, 2, and 12 are exact counting integers and do not affect significant-figure considerations.
Final answer
(a) 12 bivalents
(b) 12 chromosomes per cell
(c) 6 chromatids
Common trap
Confusing "chromosomes" with "chromatids" after telophase I. Each cell has 12 chromosomes, but each chromosome still consists of 2 sister chromatids (24 chromatids total per cell) — the sister chromatids separate only in meiosis II. Also: students sometimes count all 4 chromatids of a bivalent as "involved" in crossing over when only 2 non-sister chromatids exchange segments per chiasma.
Similar NEET-style question
A plant cell with 2n = 16 undergoes meiosis. If independent assortment is the only source of variation, how many genetically different gamete types are theoretically possible? (Answer: 2^8 = 256)
Reductional (I) + equational (II): 2n → n. Prophase I sub-stages: leptotene, zygotene (synapsis), pachytene (crossing over), diplotene (chiasmata), diakinesis. Independent assortment + crossing over generate diversity.
-- NCERT Class 11 Biology, Ch. 10, p. 126These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
Synapsis, crossing over, chiasmata are meiosis-only (prophase I). Equational mitosis has none.
Question describes a phase event and asks which division it belongs to.
Synapsis/CO/chiasmata = meiosis I. Sister chromatid separation = mitosis or meiosis II. Homologue separation = meiosis I.
Root cause: term confusion
Sister chromatids = two copies after S phase, joined at centromere. Homologues = paired chromosomes (one maternal, one paternal). Anaphase I separates homologues; Anaphase II separates sisters.
More in Cell: Structure and Function: 12 exam traps and mistakes · 1 formula · 1 question pattern from its other lessons.
10 questions from NEET 2020, 2021, 2022, 2023, 2024, 2026. Answers verified against NTA official keys.
Which of the following stages of meiosis involves division of centromere?
The appearance of recombination nodules on homologous chromosomes during meiosis characterizes :
Regarding Meiosis, which of the statements is incorrect?
Which stage of meiotic prophase shows terminalisation of chiasmata as its distinctive feature?
Dissolution of the synaptonemal complex occurs during :
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