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A student performed some… (Class 12 Biology 2026, Q30)

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BiologyCBSE Class 12 · Board exam 2026, set 57/1/1 · 4 marks · Case-based

CBSE Class 12 Biology 2026: question 30

Set in the CBSE Class 12 board exam of 2026, paper 57/1/1, section D, question 30, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q30 · 4 marks · Case-basedCompetency-basedPrinciples of Inheritance and Variation

A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny. Study the graphs and answer the questions :

  1. (a)

    What can you infer about the genotype of parents in crosses I and II ?

    1 mark

  2. OR

    (a)

    Which genetic cross is represented by these crosses ?

    1 mark
  3. (b)

    Looking at bar graph of cross III, identify the type of cross performed and its importance in genetics.

    2 marks
  4. (c)

    What conclusion can you draw from the results of bar graphs of crosses I and II ? Name the genetic principle illustrated.

    1 mark

Figure in the paper: Bar graph, y-axis Number of Plants (0-600), x-axis Crosses I, II, III; legend Tall (black), Dwarf (white). Cross I: Tall 500, Dwarf 0. Cross II: Tall 375, Dwarf 125. Cross III: Tall 250, Dwarf 250. See it in the official paper.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2026 board paper, set 57/1/1.

Hints

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

Hint 1

Work out the tall : dwarf ratio in each cross.

Hint 2

A 3 : 1 ratio and a 1 : 1 ratio each point to a particular parental combination.

Hint 3

Ask which phenotype appears in every cross and what that says about the alleles.

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) Cross I gives all tall progeny (500 : 0), so at least one parent is homozygous tall: TT × tt (also possible: TT × TT or TT × Tt).

    ½ mark

  2. Cross II gives tall : dwarf = 375 : 125 = 3 : 1, so both parents are heterozygous: Tt × Tt.

    ½ mark

  3. (a) OR: these crosses involve a single character (height), so they are monohybrid crosses.

    1 mark

  4. (b) Cross III gives 250 tall : 250 dwarf = 1 : 1, the result of a test cross (Tt × tt).

    1 mark

  5. Importance: a test cross crosses an organism showing the dominant phenotype with the homozygous recessive parent. It is used to find out whether the dominant organism is homozygous (TT) or heterozygous (Tt).

    1 mark

  6. (c) In Cross I only tall plants appear, and in Cross II tall plants outnumber dwarf plants 3 : 1. So the tall trait is dominant over the dwarf trait, and dwarfness is expressed only in the homozygous (tt) condition.

    ½ mark

  7. Principle illustrated: Law of Dominance.

    ½ mark

Answer: (a) Cross I: TT × tt (or TT × TT/Tt); Cross II: Tt × Tt, or the crosses are monohybrid crosses; (b) test cross, used to find the genotype of an organism with the dominant phenotype; (c) tall is dominant over dwarf, Law of Dominance

Where the marks are

  • Cross I genotype TT × tt/TT/Tt (½ mark)
  • Cross II genotype Tt × Tt (½ mark)
  • OR: monohybrid cross (1 mark)
  • Identifies a test cross (1 mark)
  • Importance: determining the genotype of a dominant phenotype (1 mark)
  • Tall is dominant over dwarf (½ mark)
  • Law of Dominance (½ mark)

Revise the chapter

Practise this chapterBoard-style questions from Principles of Inheritance and Variation, one at a time, marked against the answer key and marking points.Back to the full paper2026 board paper, set 57/1/1: 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.