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BiologyCBSE Class 12 · Board exam 2024, set 57/1/1 · 5 marks · Long answer

CBSE Class 12 Biology 2024: question 32(b)

Set in the CBSE Class 12 board exam of 2024, paper 57/1/1, section E, question 32(b), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q32(b) · 5 marks · Long answerPrinciples of Inheritance and Variation

A tall pea plant bearing violet flowers with unknown genotype is given. Find the genotype by working out different crosses only by selfing the plants. Write the genotypic and phenotypic ratios of each cross shown by you.

Source: official question paper from cbse.gov.in (PDF). See it in the full 2024 board paper, set 57/1/1.

Hints

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

Hint 1

A plant that shows both dominant traits may be homozygous or heterozygous for each gene. How many combinations does that give?

Hint 2

Self each possible genotype with a Punnett square.

Hint 3

Look at which recessive phenotype (dwarf, white, or both) appears in the progeny; that reveals the genotype.

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. Possible genotypes. Tall (T) is dominant over dwarf (t), and violet (V) is dominant over white (v). A tall, violet plant can therefore have any one of four genotypes: TTVV, TTVv, TtVV, TtVv. Selfing the plant and studying its progeny tells us which one it is.

    2 marks
  2. Case I: TTVV × TTVV (selfing)

    Gametes: TV only, so all progeny are TTVV.

    • Genotypic ratio: all TTVV (1)
    • Phenotypic ratio: all tall violet (1)

    If no segregation occurs, the plant is homozygous for both traits (TTVV).

  3. Case II: TtVV × TtVV

    Gametes: TV, tV

    ♀ \ ♂TVtV
    TVTTVV (tall violet)TtVV (tall violet)
    tVTtVV (tall violet)ttVV (dwarf violet)
    • Genotypic ratio: TTVV : TtVV : ttVV = 1 : 2 : 1
    • Phenotypic ratio: tall violet : dwarf violet = 3 : 1

    Dwarf plants appear but no white-flowered ones, so the plant is TtVV.

    1 mark
  4. Case III: TtVv × TtVv

    Gametes: TV, Tv, tV, tv

    ♀ \ ♂TVTvtVtv
    TVTTVVTTVvTtVVTtVv
    TvTTVvTTvvTtVvTtvv
    tVTtVVTtVvttVVttVv
    tvTtVvTtvvttVvttvv
    • Phenotypic ratio: tall violet : tall white : dwarf violet : dwarf white = 9 : 3 : 3 : 1
    • Genotypic ratio: TTVV : TTVv : TtVV : TtVv : TTvv : Ttvv : ttVV : ttVv : ttvv = 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1

    All four phenotypes appear, so the plant is heterozygous for both traits (TtVv).

    1 mark
  5. Case IV: TTVv × TTVv

    Gametes: TV, Tv

    ♀ \ ♂TVTv
    TVTTVV (tall violet)TTVv (tall violet)
    TvTTVv (tall violet)TTvv (tall white)
    • Genotypic ratio: TTVV : TTVv : TTvv = 1 : 2 : 1
    • Phenotypic ratio: tall violet : tall white = 3 : 1

    White-flowered plants appear but no dwarf ones, so the plant is TTVv.

    1 mark

Answer:

The plant can be one of four genotypes, each identified by the progeny of selfing:

  • TTVV: all tall violet (1 genotype).
  • TtVV: genotypes TTVV : TtVV : ttVV = 1 : 2 : 1; phenotypes tall violet : dwarf violet = 3 : 1.
  • TTVv: genotypes TTVV : TTVv : TTvv = 1 : 2 : 1; phenotypes tall violet : tall white = 3 : 1.
  • TtVv: genotypes 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1; phenotypes 9 : 3 : 3 : 1.

Where the marks are

  • Lists the four possible genotypes TTVV, TTVv, TtVV and TtVv (½ mark each, 2 marks)
  • For any three selfing crosses, shows gametes and progeny with the correct genotypic and phenotypic ratios (1 mark per cross, 3 marks)
  • Correct ratios: TtVV gives 1:2:1 and 3:1; TTVv gives 1:2:1 and 3:1; TtVv gives 9:3:3:1 and 1:2:2:4:1:2:1:2:1; TTVV gives all tall violet (included in the above)

Revise the chapter

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.