Five solved numerical problems on Cell Structure And Functions Class8 with step-by-step working for NEET and boards.

Cell Structure And Functions Class8: Numerical Problems Solved Step-by-Step

Numerical-style questions in biology usually look like data interpretation, percentage calculations, ratio problems from genetics, or quantitative facts from the cell as the basic unit of life — discovery, types, and organelles at Class 8 level. We’ll work through five solved problems together so you see exactly how to attack each one.

Question 1 — Quantitative Setup

Problem. In a typical human adult, blood constitutes about 8% of body weight. If a student weighs 60 kg and plasma forms 55% of blood by volume, estimate the plasma volume (assume blood density ≈ 1.06 g/mL).

Blood mass = $0.08 \times 60 = 4.8$ kg = $4800$ g.

Blood volume = $\dfrac{4800}{1.06} \approx 4528$ mL $\approx 4.53$ L.

Plasma volume = $0.55 \times 4528 \approx 2490$ mL, roughly $2.5$ L.

Plasma volume ≈ 2.49 L. Remember: plasma is ~55% of blood, formed elements (RBCs, WBCs, platelets) make up the remaining ~45%.

Question 2 — Ratio / Genetic Arithmetic

Problem. A monohybrid cross between two heterozygotes in Cell Structure And Functions Class8 context produces 800 offspring. How many are expected to show the recessive phenotype?

Heterozygous $\times$ heterozygous gives a 3:1 phenotypic ratio — $\tfrac{1}{4}$ recessive.

Recessive offspring = $\dfrac{1}{4} \times 800 = 200$ .

200 offspring show the recessive phenotype.

Question 3 — Rate / Flux Problem

Problem. A healthy kidney filters about 125 mL of plasma per minute (GFR). How much filtrate is formed in 24 hours, and how much of that is actually excreted as urine if the reabsorption efficiency is 99%?

$125 \text{ mL/min} \times 60 \times 24 = 180{,}000$ mL $= 180$ L per day.

Urine = $180 \times (1 - 0.99) = 1.8$ L per day.

Filtrate ≈ 180 L/day, urine ≈ 1.8 L/day. This is why the nephron’s counter-current mechanism matters so much — we can’t afford to lose that filtrate.

Question 4 — Percentage / Efficiency

Problem. If aerobic respiration of one glucose molecule theoretically yields 38 ATP and the free energy of glucose oxidation is 2870 kJ/mol while one ATP stores about 30.5 kJ/mol, what is the efficiency of energy capture?

$38 \times 30.5 = 1159$ kJ/mol.

Efficiency = $\dfrac{1159}{2870} \times 100 \approx 40.4\%$ .

About 40% — the rest is lost as heat, which keeps our body warm.

Question 5 — Data Interpretation

Problem. In a population of 10,000 people, the frequency of a recessive disease allele is 0.02. Using the Hardy–Weinberg principle, how many carriers (heterozygotes) are expected?

$q = 0.02$ , so $p = 0.98$ .

$2pq = 2 \times 0.98 \times 0.02 = 0.0392$ .

Carriers = $0.0392 \times 10{,}000 \approx 392$ .

Roughly 392 carriers. Notice how the carrier pool is much larger than the affected population — this is why recessive diseases persist.

For any biology numerical, write down the given data, convert units first, and then apply the formula. Most errors in Cell Structure And Functions Class8 numericals come from skipping the unit conversion step.

Quick Recap

Blood ≈ 8% of body weight; plasma ≈ 55% of blood.
Monohybrid heterozygous cross → 3:1 phenotypic ratio.
GFR ≈ 125 mL/min, roughly 180 L filtrate/day.
Aerobic respiration efficiency ≈ 40%.
Hardy–Weinberg: $p^2 + 2pq + q^2 = 1$ .

Keep these numbers on your fingertips — they show up in NEET and board papers every single year, and the cell as the basic unit of life — discovery, types, and organelles at Class 8 level is exactly where the examiner loves to plant numerical traps.