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The lowest part of the ear… (Class 10 Science 2025, Q30)

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ScienceCBSE Class 10 · Board exam 2025, set 31/2/1 · 3 marks · Short answer

CBSE Class 10 Science 2025: question 30

Set in the CBSE Class 10 board exam of 2025, paper 31/2/1, section C, question 30, for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 3 marks · Short answerHeredity

The lowest part of the ear called earlobe, is closely attached to the side of the head in some of us (Figure 'X'), and not in others, called free earlobe (Figure 'Y'). Attached and free earlobes are two variants found in human populations. The gene for free earlobe is dominant over attached earlobes.

  1. (a)

    A man with attached earlobes marries a woman having free earlobes. 50% of their children have free earlobes and 50% have attached earlobes. Explain the inheritance of this trait and write the trait combinations of the progeny.

    0 marks
  2. (b)

    Write the gene combinations of the father and the mother in the above case.

    0 marks

Figure in the paper: Figure 'X' shows a close-up photograph of a human ear with the earlobe attached directly to the side of the head. Figure 'Y' shows a close-up photograph of a human ear with a free earlobe hanging separately below the attachment point. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2025 board paper, set 31/2/1.

Hints

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

Hint 1

A person showing a recessive trait must carry two recessive genes.

Hint 2

If some children are ff, each parent must have passed on one f. What must the mother's genotype be?

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) Let F = gene for free earlobe (dominant) and f = gene for attached earlobe (recessive). The attached-earlobe man must be ff. Because half the children have attached earlobes (ff), the free-earlobe woman must carry f, so she is Ff.
    Parents: woman Ff (free) × man ff (attached).

    ½ mark

  2. Gametes: woman gives F and f; man gives f and f.

    ½ mark

  3. Cross (F₁ progeny): F × f → Ff, F × f → Ff, f × f → ff, f × f → ff. Two children are Ff (free earlobe) and two are ff (attached earlobe).

    ½ mark

  4. Progeny: Ff : ff = 1 : 1, i.e. 50% free earlobe and 50% attached earlobe. The trait combinations are free earlobe (Ff) and attached earlobe (ff).

    ½ mark

  5. (b) Gene combination of father: ff

    ½ mark

  6. Gene combination of mother: Ff

    ½ mark

Answer: (a) Ff × ff gives progeny Ff (free) : ff (attached) = 50% : 50%. (b) Father ff, mother Ff.

Where the marks are

  • Parental genotypes Ff × ff (½ mark)
  • Correct gametes (½ mark)
  • Correct F₁ combinations (½ mark)
  • Ratio Ff : ff = 1 : 1 with phenotypes (½ mark)
  • Father ff (½ mark)
  • Mother Ff (½ mark)

Revise the chapter

  • Chapter: Heredity (notes, formulas, where marks are lost and every past question)
Practise this chapterBoard-style questions from Heredity, one at a time, marked against the answer key and marking points.Back to the full paper2025 board paper, set 31/2/1: 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.