Mastering the variations in chromosome number (n) and DNA content (C) throughout the cell cycle is crucial. This document offers a comprehensive, upgraded set of numerical problems, complete with step-by-step solutions, visual aids, and a specialized section on common pitfalls that often trip up students in competitive exams (like NEET, AIIMS, and Olympiads).
Before diving into the problems, you must understand the distinction between chromosome number and DNA content:
- 'n' (Ploidy): Represents the number of sets of chromosomes. A diploid cell is
2n, and a haploid cell is n. The number of chromosomes is strictly counted by the number of centromeres.
- 'C' (DNA Content): Represents the amount of DNA in a haploid genome. A typical diploid cell in G1 has
2C DNA.
[!IMPORTANT]
The Golden Rule: The number of chromosomes only changes when centromeres split (Anaphase of mitosis / Anaphase II of meiosis) or when homologous chromosomes separate into different cells (Meiosis I). DNA content doubles during the S phase and halves during cytokinesis.
Mitosis (Somatic Cells) Flow:
- G1 Phase: Chr: 2n | DNA: 2C
- S Phase / G2 Phase / Prophase / Metaphase: Chr: 2n | DNA: 4C
- Anaphase: Chr: 4n (temporarily) | DNA: 4C
- Telophase & Cytokinesis: Yields two identical Daughter Cells (Chr: 2n | DNA: 2C each).
Meiosis (Germ Cells) Flow:
- G1 Phase: Chr: 2n | DNA: 2C
- S / G2 / Prophase I / Metaphase I: Chr: 2n | DNA: 4C
- ↓ (Homologous Chr. separate)
- Anaphase I: Chr: 2n | DNA: 4C
- End of Meiosis I (2 Cells): Chr: n | DNA: 2C (per cell)
- Prophase II / Metaphase II: Chr: n | DNA: 2C (per cell)
- Anaphase II: Chr: 2n (temporarily) | DNA: 2C (per cell)
- End of Meiosis II (4 Cells): Chr: n | DNA: C (per cell)
A diploid somatic cell of Drosophila melanogaster (fruit fly) has 8 chromosomes (2n = 8). The total DNA content in its G1 phase is 12 picograms (pg). Calculate the number of chromosomes and the amount of DNA per cell at the following stages:
- G2 phase
- Metaphase of Mitosis
- Anaphase of Mitosis
- After Cytokinesis (in each daughter cell)
Solution (Step-by-Step):
- Initial State (G1 phase): Chromosomes = 8, DNA = 12 pg.
- Step 1 (G2 phase): The cell has passed through S phase, where DNA replicates. The chromosome number remains the same because sister chromatids are held together at the same centromere.
- Chromosomes: 8
- DNA: 12 pg × 2 = 24 pg
- Step 2 (Metaphase of Mitosis): Chromosomes align at the equator. No division of centromeres has occurred yet.
- Chromosomes: 8
- DNA: 24 pg
- Step 3 (Anaphase of Mitosis): The centromeres split. Each sister chromatid becomes an independent chromosome. The cell temporarily has double the number of chromosomes within a single boundary before cytokinesis.
- Chromosomes: 8 × 2 = 16
- DNA: 24 pg (No DNA is created or destroyed, just partitioned)
- Step 4 (After Cytokinesis): The cell divides into two identical daughter cells, splitting the chromosomes and DNA equally.
- Chromosomes: 16 / 2 = 8
- DNA: 24 pg / 2 = 12 pg
A primary spermatocyte in a human male (2n = 46) is about to undergo meiosis. If the DNA content of a single human sperm cell is 3.3 pg, what are the chromosome number, DNA content, and the number of chromatids per cell in:
- Prophase I
- Metaphase II
- A mature spermatozoon
Solution (Step-by-Step):
- Context Check: A mature spermatozoon is a haploid gamete (n). Thus, n = 23, and C (DNA of haploid genome) = 3.3 pg.
- Initial State (G1 of Primary Spermatocyte): It is diploid (2n = 46), with 2C DNA = 6.6 pg. After S phase (entering Meiosis I), it has 2n chromosomes, 4C DNA, and each chromosome has 2 chromatids.
- Step 1 (Prophase I):
- Chromosomes: 46 (still a single diploid cell)
- DNA Content: 4C = 4 × 3.3 = 13.2 pg
- Chromatids: 46 chromosomes × 2 = 92 chromatids (forming 23 bivalents/tetrads)
- Step 2 (Metaphase II): Meiosis I is complete. Homologous chromosomes have separated, reducing the chromosome number by half. However, centromeres have not split, so each chromosome still has 2 chromatids.
- Chromosomes: 46 / 2 = 23 (haploid, n)
- DNA Content: 13.2 / 2 = 6.6 pg (2C)
- Chromatids: 23 chromosomes × 2 = 46 chromatids
- Step 3 (Mature Spermatozoon): After Meiosis II and differentiation. Sister chromatids have separated.
- Chromosomes: 23 (n)
- DNA Content: 3.3 pg (C)
- Chromatids: 0 (Once separated, they are referred to as distinct chromosomes, not chromatids).
In an angiosperm, the root tip cell has 24 chromosomes and 20 pg of DNA in the G1 phase. Calculate the number of chromosomes and DNA content in:
- The microspore mother cell (MMC) at G2 phase.
- The generative cell of the pollen grain.
- The endosperm cell of the mature seed at G1 phase.
Solution (Step-by-Step):
- Initial State: Root tip is somatic (diploid, 2n). 2n = 24. 2C = 20 pg. Therefore, n = 12, and C = 10 pg.
- Step 1 (MMC at G2 Phase): The MMC is a diploid cell (2n) preparing for meiosis. In G2, it has replicated its DNA.
- Chromosomes: 2n = 24
- DNA Content: 4C = 40 pg
- Step 2 (Generative Cell): Formed via meiosis (yielding haploid microspores) followed by mitosis. It is a haploid cell (n) in the G1 state.
- Chromosomes: n = 12
- DNA Content: C = 10 pg
- Step 3 (Endosperm Cell at G1): In angiosperms, the endosperm is triploid (3n), formed by triple fusion (one haploid male gamete + two haploid polar nuclei).
- Chromosomes: 3n = 3 × 12 = 36
- DNA Content: 3C = 3 × 10 = 30 pg (since it's asked at G1 phase, before any subsequent DNA replication).
An organism has a DNA content of 100 pg and 40 chromosomes in a meiocyte at the end of S-phase. Calculate:
- The number of bivalents in Zygotene.
- The number of DNA molecules in a daughter cell post-Meiosis I.
- The number of chromosomes in the G1 phase of the original meiocyte.
Solution (Step-by-Step):
- Analyze the Given: A meiocyte at the end of S-phase is in a 2n, 4C state.
- Therefore, 2n = 40. n = 20.
- 4C = 100 pg.
- Step 1 (Bivalents in Zygotene): A bivalent is a pair of homologous chromosomes. Since there are 40 total chromosomes, they form pairs.
- Step 2 (DNA molecules post-Meiosis I): After Meiosis I, the cell is 'n' but '2C'. Each chromosome has 2 chromatids, and each chromatid represents one double-stranded DNA molecule.
- Chromosomes: 20. Each has 2 chromatids = 40 chromatids.
- Number of DNA molecules: 40 (Total DNA content would be 50 pg).
- Step 3 (Chromosomes in G1 phase): G1 phase is before DNA replication. The chromosome number is the same as the diploid state.
- Chromosomes: 40 (DNA content would be 2C = 50 pg).
These questions specifically target rote memorization. Read them carefully!
[!WARNING]
Common Pitfall: Students often confuse "chromatids" with "chromosomes" in Anaphase. Remember: A centromere is the structural unit of a chromosome. If a structure has one centromere, it is ONE chromosome, regardless of whether it has one or two chromatids.
Question: A cell with 2n=10 undergoes mitosis. How many chromosomes and chromatids are present during Anaphase?
The Trap: Students often say 10 chromosomes and 20 chromatids (assuming they haven't split) OR 20 chromosomes and 20 chromatids.
The Solution: During Anaphase, the centromeres split. The 10 chromosomes (each with 2 chromatids) are pulled apart. The cell temporarily has 20 chromosomes. However, because the chromatids have separated, they are no longer called "sister chromatids" but individual chromosomes. Therefore, there are 0 chromatids in Anaphase.
Question: A hepatocyte (liver cell) exits the cell cycle and enters the G0 phase. If its G2 DNA content is 60 pg, what is its DNA content and chromosome ploidy in G0?
The Trap: Students might guess it's halved again or stays the same as G2.
The Solution: Cells enter G0 from the G1 phase, not from G2. Therefore, they have the DNA content and chromosome number of a G1 cell. If G2 (4C) = 60 pg, then G1 (2C) = 30 pg. The ploidy is diploid (2n), and the DNA content is 30 pg.
Question: An organism has 2n=20. Compare the number of chromosomes present in a single cell during Anaphase I and Anaphase II.
The Trap: Students assume Anaphase always doubles the chromosome number in the cell.
The Solution:
- Anaphase I: Homologous chromosomes separate, but centromeres do not split. The cell still contains 20 chromosomes (moving to opposite poles).
- Anaphase II: The cell starting Meiosis II has n=10 chromosomes. During Anaphase II, the centromeres do split. Therefore, the cell temporarily has 10 × 2 = 20 chromosomes.
- Result: Interestingly, both Anaphase I and Anaphase II of this organism will have 20 chromosomes within the cell membrane at that precise moment, though the DNA content (4C vs 2C) differs dramatically!
BioNotes