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Assuming that within a… (Class 12 Biology SQP 2024-25, Q29)

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BiologyCBSE Class 12 · Sample paper, session 2024-25 · 4 marks · Case-based

CBSE Class 12 Biology sample paper 2024-25: question 29

From the official CBSE sample question paper for 2024-25, section D, question 29, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q29 · 4 marks · Case-basedCompetency-basedEvolution

Assuming that within a population of beetles where Hardy Weinberg conditions are met, the colour black (B) is dominant over the colour red (b). 40% of all beetles are red (bb).
Given this information, answer the questions below:

  1. A.

    What is the frequency of red beetles?

    1 mark
  2. B.

    Calculate is the percentage of beetles in the population that are heterozygous.

    2 marks
  3. C.

    What is the frequency of homozygous dominant individuals?

    1 mark

  4. OR

    D.

    Assuming that Hardy Wienberg conditions are met in the beetle population consisting of 1500 beetles. How many beetles would you expect to be black and red in colour respectively?

    1 mark

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

Red is recessive, so 40% red beetles gives you q².

Hint 2

Use p + q = 1, then the Hardy–Weinberg terms p², 2pq and q².

Hint 3

For D, use the given percentage of red beetles on the total of 1500.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

  1. A. Red beetles are bb, so the frequency of red beetles is q² = 0.40. The frequency of the recessive allele b is therefore q = √0.40 ≈ 0.63.

    1 mark

  2. B. Since p + q = 1, the frequency of allele B is p = 1 − 0.63 = 0.37.

    1 mark

  3. B. Frequency of heterozygotes = 2pq = 2 × 0.37 × 0.63 ≈ 0.47. So about 47% of the beetles are heterozygous (Bb).

    1 mark

  4. C. Frequency of homozygous dominant (BB) = p² = (0.37)² ≈ 0.14.

    1 mark

  5. D. Red beetles = 40% of 1500 = 0.4 × 1500 = 600.

    ½ mark

  6. D. Black beetles (BB + Bb) = 1500 − 600 = 900.

    ½ mark

Answer: A. q² = 0.4, so q ≈ 0.63; B. p = 0.37, 2pq ≈ 0.47, i.e. 47% heterozygous; C. p² ≈ 0.14; D. 900 black and 600 red.

Where the marks are

  • Finds q = √0.4 ≈ 0.63 from q² = 0.4 (1 mark)
  • Finds p = 1 − q = 0.37 (1 mark)
  • Calculates 2pq ≈ 0.47 (47%) (1 mark)
  • C: p² = 0.14 (1 mark) OR D: 600 red (0.5 mark) and 900 black (0.5 mark)

Revise the chapter

Practise this chapterBoard-style questions from Evolution, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

The question is transcribed from the official paper; figures are described in words. The solution, hints and chapter tag are ours. Spotted an error? Write tosupport@questionbuddy.in.