Heredity CBQs — CBSE Class 10 (2027) with step-wise solutions

Sample Heredity CBQs

1. Question 1MCQ1 mark
   In pea plants, the allele for tallness (T) is dominant over the allele for dwarfness (t). A student at a Krishi Vigyan Kendra crossed a tall pea plant with a dwarf pea plant and raised 101 plants from the seeds: 52 were tall and 49 were dwarf. Which of the following correctly gives the genotypes of the two parent plants?
   • TT × tt
   • Tt × tt
   • Tt × Tt
   • TT × Tt
   Show step-wise solution

   1. [1 mark] Tt × tt. The dwarf parent must be tt; the progeny are tall and dwarf in about 1 : 1 (52 : 49), so the tall parent must carry the recessive allele, i.e. it is Tt (Tt × tt gives Tt and tt in equal numbers; TT × tt would give only tall plants).

   Answer: Tt × tt

2. Question 2Assertion-Reason1 mark
   Assertion (A): Bougainvillea plants raised from stem cuttings of one parent plant show more variation among themselves than bougainvillea plants raised from seeds of that same parent. Reason (R): Seeds are formed after fertilisation, in which a male gamete and a female gamete bring together a new combination of genes.
   • (A) Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
   • (B) Both Assertion and Reason are true but Reason is NOT the correct explanation of Assertion.
   • (C) Assertion is true but Reason is false.
   • (D) Assertion is false but Reason is true.
   Show step-wise solution

   1. [1 mark] (D) A is false: cuttings are vegetative (asexual) propagation, so plants from them carry the parent's DNA and are nearly identical, while seed-raised plants vary more. R is true: fertilisation combines genes from two gametes, which is exactly why seedlings vary.

   Answer: (D) Assertion is false but Reason is true.

3. Question 3Short answer2 marks
   A gardener in Ooty crossed a pure-breeding pea plant having green pods (GG) with a pure-breeding pea plant having yellow pods (gg). All the plants of the F1 generation had green pods. (a) Which pod colour is dominant? Write the genotype of the F1 plants. (b) The F1 plants were self-pollinated and 800 F2 plants were raised. How many of these are expected to have yellow pods, and how many are expected to be pure-breeding green-podded (GG) plants?
   Show step-wise solution

   1. [1 mark] (a) Green pod colour is dominant, because only green pods appear in F1 although the yellow factor is present; the F1 genotype is Gg.
   2. [0.5 mark] (b) Gg × Gg gives GG : Gg : gg = 1 : 2 : 1; yellow-podded plants (gg) = 1/4 × 800 = 200.
   3. [0.5 mark] (b) Pure-breeding green-podded plants (GG) = 1/4 × 800 = 200.

   Answer: (a) Green; Gg. (b) 200 yellow-podded plants; 200 pure-breeding green (GG) plants.

4. Question 4Case study4 marks
   The eco-club of a school in Pune repeated two of Mendel's experiments in the school garden. Pure-breeding pea plants with contrasting forms of a trait were crossed, the F1 plants were self-pollinated and the F2 plants were counted. The students recorded: Trait | Form 1 | Form 2 Flower position | Axial: 612 | Terminal: 198 Pod shape | Inflated: 735 | Constricted: 245 In each cross, all the F1 plants showed only Form 1 of the trait. (i) Which flower position is dominant? Give a reason based on the observations. (ii) What is the ratio of inflated to constricted pods in F2, and what does it tell you about the genotype of the F1 plants? (iii) The students wish to know whether a particular axial-flowered F2 plant is pure-breeding. Suggest how they can find this out by self-pollination, and state the result expected if the plant is pure-breeding and if it is a hybrid. (Use A for the axial allele and a for the terminal allele.)
   Show step-wise solution

   1. [1 mark] (i) Axial flower position is dominant: all F1 plants were axial, and in F2 axial plants are about three times the terminal ones (612 : 198 ≈ 3 : 1); terminal reappears only in F2, so it is recessive.
   2. [1 mark] (ii) 735 : 245 = 3 : 1; a 3 : 1 ratio in F2 shows that the F1 plants were heterozygous (hybrid, e.g. Ii) for pod shape.
   3. [1 mark] (iii) Self-pollinate the chosen axial-flowered plant and raise plants from its seeds; if all the progeny have axial flowers, the plant is pure-breeding (AA).
   4. [1 mark] (iii) If some progeny have terminal flowers (axial : terminal ≈ 3 : 1), the plant is a hybrid (Aa).

   Answer: (i) Axial. (ii) 3 : 1; F1 plants were heterozygous. (iii) Self-pollinate: all axial progeny means pure-breeding; axial and terminal in about 3 : 1 means hybrid.

5. Question 5Long answer5 marks
   In pea plants, round seed shape (R) is dominant over wrinkled seed shape (r), and yellow seed colour (Y) is dominant over green seed colour (y). Meera, a student at an agricultural college, crossed a pure-breeding plant having round green seeds with a pure-breeding plant having wrinkled yellow seeds. (a) Write the genotypes of the two parent plants and the gametes each of them produces. (b) Write the genotype and phenotype of the F1 plants. (c) The F1 plants were self-pollinated. Write the phenotypes that appear in the F2 generation and their ratio. (d) If 1600 F2 seeds were obtained, how many are expected to be round and yellow, and how many wrinkled and green? (e) What do the round yellow and wrinkled green seeds of the F2 generation show about the inheritance of seed shape and seed colour?
   Show step-wise solution

   1. [1 mark] (a) Round green parent = RRyy, producing gametes Ry; wrinkled yellow parent = rrYY, producing gametes rY.
   2. [1 mark] (b) F1 genotype = RrYy; all F1 plants have round yellow seeds, since round and yellow are dominant.
   3. [1 mark] (c) F2 phenotypes: round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1.
   4. [1 mark] (d) Round yellow = 9/16 × 1600 = 900 seeds; wrinkled green = 1/16 × 1600 = 100 seeds.
   5. [1 mark] (e) Round yellow and wrinkled green are combinations not present in either parent; their appearance shows that seed shape and seed colour are inherited independently of each other.

   Answer: Parents RRyy (Ry) × rrYY (rY); F1 RrYy, round yellow; F2 9 : 3 : 3 : 1; 900 round yellow and 100 wrinkled green; the two traits are inherited independently.

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