Daily Practice Problems

📝 SolutionNucellus — 2n. It is the parental (sporophytic) tissue of the ovule that surrounds the embryo sac.

Megaspore mother cell (MMC) — 2n. It is a diploid cell before meiosis; meiosis then produces four haploid megaspores.

Endosperm — 3n. Formed by triple fusion: one male gamete (n) fuses with the two polar nuclei (n + n) of the central cell, giving the primary endosperm nucleus (3n).

Egg cell — n. A haploid product of the functional megaspore.

Trap: Option (b) marks the MMC as haploid — but 'mother cell' always means the diploid cell before meiosis, not after.

📝 SolutionAfter ovulation, the ruptured Graafian follicle becomes the corpus luteum, which secretes large amounts of progesterone — essential for maintaining the thickened endometrium in readiness for implantation.

If fertilisation does not occur:
1. The corpus luteum degenerates into the corpus albicans
2. Progesterone levels fall sharply
3. Without progesterone, the endometrium cannot be maintained and breaks down
4. This causes menstruation (menstrual flow)

If fertilisation does occur: the developing embryo secretes hCG, which rescues the corpus luteum. It keeps producing progesterone, the endometrium is maintained, and menstruation stops. (hCG is exactly what a pregnancy test detects.)

📝 SolutionLactational amenorrhea ('absence of menstruation during breastfeeding') is based on this fact: during the period of intense lactation following parturition, the mother does not usually undergo a normal ovarian cycle — so ovulation does not occur, and conception is very unlikely.

It is effective only up to a maximum of 6 months after childbirth.

Advantages: no side effects at all — it is entirely natural.
Limitation: the chance of failure rises significantly beyond 6 months, so it cannot be relied on as a long-term method.

Trap: Option (b) stretches it to 12 months, which is exactly the misconception the NCERT text warns against.

📝 SolutionColour blindness is an X-linked recessive trait.

The man: colour-blind $\Rightarrow$ genotype $\mathrm{X^c Y}$

The woman: she has normal vision, but her father was colour-blind ($\mathrm{X^c Y}$). A father passes his only X to every daughter — so she must have received $\mathrm{X^c}$ from him. Since she is not colour-blind herself, her other X must be normal.
$\Rightarrow$ She is a carrier: $\mathrm{X^C X^c}$

The cross: $\mathrm{X^c Y} \times \mathrm{X^C X^c}$

Offspring:
• $\mathrm{X^C X^c}$ — carrier daughter (normal vision)
• $\mathrm{X^c X^c}$ — colour-blind daughter
• $\mathrm{X^C Y}$ — normal son
• $\mathrm{X^c Y}$ — colour-blind son

Among the sons, half are colour-blind $\Rightarrow$ probability $= \mathbf{1/2}$

Trap: Option (b) — 1/4 — answers a different question: 'what fraction of all children are colour-blind sons?' Here we are told the child is a son, so the sample is restricted to sons only. Read the conditional carefully.

(Note: half the daughters here are colour-blind too — unusual, and only possible because the father is affected.)

📝 SolutionThe whole experiment depends on finding elements that are present in one molecule and absent from the other — that is what makes the labels distinguishable.

DNA: its sugar-phosphate backbone is rich in phosphorus. DNA contains no sulphur.
Protein: the amino acids cysteine and methionine contain sulphur. Protein contains no phosphorus.

So:
• $^{32}\mathrm{P}$ labels only DNA
• $^{35}\mathrm{S}$ labels only protein

The result: Hershey and Chase infected bacteria with labelled bacteriophages, then blended and centrifuged them. Radioactive $^{32}$P was found inside the bacterial cells (DNA entered), while $^{35}$S remained outside in the supernatant with the empty phage coats (protein did not enter).

Conclusion: DNA — not protein — is the genetic material that passes from virus to host.