Grade 9 : Science - 3. Reproduction (Set 5) | Concept of Maths

Question 1

1. An initial population of $N_0 = 100$ Amoeba cells undergoes binary fission once every $2$ hours. However, due to an environmental toxin introduced at $t = 0$, $20\%$ of the existing population is eliminated at the end of every $4$ hours. Determine the net population of Amoeba at the end of $8$ hours.

A) $800$ cells

B) $1024$ cells

C) $1600$ cells

D) $512$ cells

Answer

Correct Answer: B

Solution: Let's track the population step-by-step: - At $t = 0$: $100$ cells. - At $t = 2$ hours: After one cycle of binary fission, population $= 100 \times 2 = 200$ cells. - At $t = 4$ hours: After second binary fission, population $= 200 \times 2 = 400$ cells. Immediately, $20\%$ are killed: $400 \times (1 - 0.20) = 320$ cells. - At $t = 6$ hours: After third binary fission, population $= 320 \times 2 = 640$ cells. - At $t = 8$ hours: After fourth binary fission, population $= 640 \times 2 = 1280$ cells. Immediately, $20\%$ are killed: $1280 \times 0.80 = 1024$ cells. Therefore, the net population is $1024$ cells.

Question 2

2. A biology student performs an experiment on Hydra regeneration. They slice a mature Hydra horizontally into three distinct zones: Segment A (top, containing tentacles and hypostome), Segment B (middle, containing the body column with a developing bud), and Segment C (basal disc). Assuming optimal environmental conditions, which of the following outcomes is most biologically accurate regarding regeneration?

A) Only Segment B will survive because it contains the budding zone; the other segments will undergo autolysis.

B) All three segments (A, B, and C) will regenerate into fully functional individuals, but Segment B will complete regeneration faster due to active cell division in the budding region.

C) Segment A will regenerate tentacles downwards, Segment C will regenerate a hypostome upwards, but Segment B cannot regenerate because its cells are already committed to budding.

D) Regeneration occurs only in Segment C since stem cells are localized exclusively in the basal disc.

Question 3

3. In a genetics experiment on flowering plants, a mutated strain of Arabidopsis thaliana exhibits self-pollination but has a genetic defect where pollen tube elongation is slowed down by $50\%$. Under standard conditions, wild-type pollen tubes take $5$ hours to reach the ovary. If the female gametophyte (ovule) remains viable for a maximum of $9$ hours after pollination, will fertilization occur in this mutant strain, and what is the margin of time?

A) Yes, fertilization occurs with a surplus margin of $4$ hours.

B) Yes, fertilization occurs exactly at $9$ hours with $0$ hours margin.

C) No, fertilization fails because the mutant pollen tube takes $10$ hours, which exceeds the $9$-hour viability window by $1$ hour.

D) No, fertilization fails because self-pollination is completely blocked by self-incompatibility genes in all mutants.

Answer

Correct Answer: C

Solution: Let $v$ be the rate of elongation of wild-type pollen. The time taken is $t_{wt} = \frac{d}{v} = 5\text{ hours}$. For the mutant, the rate of elongation is $50\%$ slower, meaning $v_{mut} = 0.5v$. The time taken by the mutant pollen tube is: $$t_{mut} = \frac{d}{0.5v} = 2 \times 5 = 10\text{ hours}$$ Since the ovule remains viable for only $9$ hours, the mutant pollen tube arrives $10 - 9 = 1$ hour too late. Fertilization fails.

Question 4

4. In the human male reproductive system, the path of sperm transportation involves several ducts. If a surgical blockage or ligation is specifically performed at the level of the vasa efferentia (ductules leading out of the testis), which of the following physiological consequences will occur?

A) Sperm will fail to travel from the testes to the epididymis, rendering semen completely azoospermic (lacking sperm) in the long run.

B) Seminal fluid production from the seminal vesicles and prostate gland will completely cease.

C) Sperm will remain trapped in the vas deferens and fail to mix with prostatic secretions.

D) Testosterone production by the Leydig cells will drop to zero, halting the development of secondary sexual characteristics.

Answer

Correct Answer: A

Solution: The vasa efferentia connect the rete testis to the epididymis. If these tubules are blocked, sperm cannot leave the testes to reach the epididymis (where they undergo maturation and are stored) or the vas deferens. This results in azoospermia (absence of sperm in the ejaculate). Leydig cells secrete testosterone directly into the blood, so hormonal levels remain unaffected.

Question 5

5. During cellular reproduction, DNA copying is highly precise but not error-free. Suppose a cell has a replication error rate of $2.5 \times 10^{-7}$ mutations per base pair per replication cycle. If the total genome size of the reproducing organism is $4 \times 10^7$ base pairs, what is the expected average number of mutations introduced into the offspring's genome per replication cycle?

A) $0.1$ mutations

B) $1.0$ mutation

C) $10$ mutations

D) $100$ mutations

Answer

Correct Answer: C

Solution: The expected average number of mutations can be calculated by multiplying the replication error rate per base pair by the total number of base pairs in the genome: $$\text{Expected Mutations} = \text{Error Rate} \times \text{Genome Size}$$ $$\text{Expected Mutations} = (2.5 \times 10^{-7}) \times (4 \times 10^7) = 2.5 \times 4 = 10$$ Thus, $10$ mutations are expected on average per replication cycle.

Question 6

6. The human female menstrual cycle is governed by intricate hormonal feedback loops. Consider a standard 28-day cycle where ovulation occurs at day 14. If a patient exhibits an abnormally low surge of Luteinizing Hormone (LH) around day 13, what is the most likely physiological consequence?

A) Rapid proliferation of the endometrium leading to early menstruation.

B) Failure of the Graafian follicle to rupture, preventing ovulation.

C) Overproduction of progesterone leading to immediate pseudo-pregnancy.

D) Hyper-activation of the corpus luteum, extending the cycle to 45 days.

Answer

Correct Answer: B

Solution: The LH surge (a sharp peak in Luteinizing Hormone around day 13-14) is the primary endocrine trigger that causes the mature Graafian follicle to rupture and release the secondary oocyte into the fallopian tube (ovulation). Without a sufficient LH surge, ovulation fails to occur.

Question 7

7. A sporocarp of the fungus Rhizopus contains multiple sporangia. In an experiment, a student counts $8$ mature sporangia, each containing approximately $5 \times 10^4$ asexual spores. If only $0.04\%$ of the released spores find a suitable moist bread substrate and successfully germinate into new mycelia, calculate the total number of new Rhizopus colonies formed.

A) $16$ colonies

B) $160$ colonies

C) $1,600$ colonies

D) $16,000$ colonies

Answer

Correct Answer: B

Solution: First, calculate the total number of spores produced: $$\text{Total Spores} = 8 \times (5 \times 10^4) = 40 \times 10^4 = 400,000\text{ spores}$$ Now, calculate the number of spores that successfully germinate ($0.04\%$): $$\text{Germinated Spores} = 400,000 \times \frac{0.04}{100} = 400,000 \times 0.0004 = 160$$ Therefore, $160$ new colonies are formed.

Question 8

8. In angiosperm sexual reproduction, triple fusion is a unique feature. If the somatic cells of a certain species of mustard plant have a diploid chromosome number of $2n = 20$, determine the chromosome numbers of: (i) the vegetative cell of the pollen grain, (ii) the synergid cell, and (iii) the primary endosperm nucleus (PEN).

A) (i) $10$, (ii) $10$, (iii) $30$

B) (i) $10$, (ii) $20$, (iii) $30$

C) (i) $20$, (ii) $10$, (iii) $40$

D) (i) $20$, (ii) $20$, (iii) $30$

Answer

Correct Answer: A

Solution: Given diploid number $2n = 20$, the haploid chromosome number is $n = 10$. - (i) The vegetative cell of the pollen grain is haploid ($n = 10$). - (ii) The synergid cell inside the embryo sac is also haploid ($n = 10$). - (iii) The primary endosperm nucleus (PEN) is formed by triple fusion (one haploid male gamete ($n$) + two haploid polar nuclei ($2n$)), making it triploid ($3n = 30$). Thus, the correct sequence is $10$, $10$, $30$.

Question 9

9. Consider the mechanisms of different contraceptive methods. Method X (e.g., Copper-T IUD) prevents implantation, while Method Y (e.g., Oral Contraceptive Pills) prevents ovulation. If a couple uses Method X, but fertilisation accidentally occurs due to atypical sperm migration, why does the pregnancy still fail to establish?

A) The copper ions released by the IUD suppress embryonic cleavage, turning the zygote into a haploid state.

B) The IUD creates a localized inflammatory response in the endometrium, making it hostile for blastocyst implantation.

C) The IUD blocks the fallopian tube, forcing the zygote to undergo binary fission in the ovary.

D) The IUD suppresses the pituitary gland from releasing progesterone.

Answer

Correct Answer: B

Solution: An Intrauterine Device (IUD) like Copper-T primarily acts by increasing the phagocytosis of sperms within the uterus and suppressing sperm motility. However, if fertilization does occur, the presence of the foreign body (IUD) along with copper ions induces a localized cellular and inflammatory response in the endometrium. This alters the lining of the uterus, making it completely hostile for the blastocyst to attach (implant), thereby preventing pregnancy.

Question 10

10. A student observes a healthy filament of the green alga Spirogyra consisting of exactly $15$ cells. Due to mechanical disturbance from a water current, the filament fragments into three parts containing $4$, $5$, and $6$ cells respectively. Over the next few days, each fragment grows back to a length of $15$ cells. What is the total number of cell divisions (mitosis) that occurred collectively across all three fragments to achieve this growth?

A) $15$ divisions

B) $20$ divisions

C) $30$ divisions

D) $45$ divisions

Answer

Correct Answer: C

Solution: Growth in Spirogyra filaments occurs through mitotic cell divisions. Let's calculate the number of new cells added to each fragment: - Fragment 1: starts with $4$ cells, grows to $15$ cells. New cells added $= 15 - 4 = 11$. - Fragment 2: starts with $5$ cells, grows to $15$ cells. New cells added $= 15 - 5 = 10$. - Fragment 3: starts with $6$ cells, grows to $15$ cells. New cells added $= 15 - 6 = 9$. Since each new cell added to a linear filament requires exactly one mitotic division, the total number of cell divisions is: $$\text{Total Divisions} = 11 + 10 + 9 = 30\text{ divisions}$$

Accepted Answer

Correct Answer: B

Solution: Hydra exhibits incredible regenerative capabilities because its body is rich in specialized interstitial stem cells. When cut, all three segments (A, B, and C) can successfully regenerate into whole, independent individuals. Segment B, which has high mitotic activity in its budding zone, undergoes rapid cellular proliferation to complete its structure faster than highly specialized polar ends.

Grade 9 : Science - 3. Reproduction (Set 5) | Concept of Maths