CG Board Class 10 Science Chapter 15 Solutions: Heredity: From Parents to Offsprings (आनुवंशिकी: जनकों से संतानों तक) (2026-27)

(i) What do you think may have been the reason for Mendel’s success?**

• (a) Clear planning based on reasoning, experimenting, recording observations etc. and duly writing article regarding the same.
• (b) Focusing on certain clear contrasting characters.
• (c) Use of mathematical calculations.
• (d) All above options

(d) All above options

Explanation: Mendel’s success was due to systematic experimental planning, meticulous record-keeping, selection of sharp contrasting pea traits, and mathematical/statistical analysis of results.

Answer: (c) Terminal and axial flower / (Official textbook key option (c)) is specified).

Explanation: The textbook official key marks option (c). Physically, stem shape (round/wrinkled stem) was not a trait studied by Mendel (stem trait was length: tall/dwarf).

Answer: (b) Rr and Rr

Explanation: Since one brother is a non-roller (rr), he must have received one recessive factor (r) from each parent. Since both parents are rollers, both parents must be heterozygous rollers (Rr).

: (b) Rr Rr (Rr and Rr)

(a) A person with blood group A can have either two I^A factors (homozygous I^A I^A) or one I^A factor** (heterozygous I^A I^O).

• (b) A person with blood group O must have two I^O factors (homozygous I^O I^O).
• (c) Factor I^O is recessive, while I^A and I^B are dominant factors.
• (d) A recessive character can express physically (phenotypically) only when both alleles in the pair are identical (homozygous condition, e.g., I^O I^O or vv or rr). Since no dominant allele is present to mask it, it represents a pure homozygous state like the pure parental generation.
• (e)
• Parents’ factors: Both parents must be heterozygous for blood group A, having factor combination I^A I^O.
• Second child’s probable blood groups: Cross between I^A I^O × I^A I^O yields I^A I^A (Group A – 25%), I^A I^O (Group A – 50%), and I^I O (Group O – 25%).
• Thus, the second child has a 75% (3/4) chance of Blood Group A and a 25% (1/4) chance of Blood Group O.

Parents:** Blue flower (Bb) × White flower (bb)

• Gametes: From Bb → B and b; From bb → b
• Cross/Punnett Square:
• B × b → Bb (Blue flower – Heterozygous)
• B × b → Bb (Blue flower – Heterozygous)
• b × b → bb (White flower – Homozygous)
• b × b → bb (White flower – Homozygous)
• Ratio: 50% Blue flowered (Bb) and 50% White flowered (bb) (Ratio 1:1).

Cross Required: Cross ‘Sahiwal’ with ‘Zeba’**.

• Reason: Chhattisgarh has a hot tropical climate.

1. Sahiwal provides the desirable genetic trait for high milk yield (20,000 liters/year).
2. Zeba provides the desirable genetic trait for heat tolerance and adaptation to survive in hot conditions.

• Hybridization between Sahiwal and Zeba will yield a new hybrid variety that combines high milk production with heat resistance suitable for Chhattisgarh’s climate.

Knowledge of heredity helps in crop improvement through selective breeding and hybridization

1. Combining desirable traits: Crossing two plant varieties allows us to combine traits like high grain yield, disease resistance, drought tolerance, and early maturity into a single hybrid crop.
2. Developing pure lines: Understanding homozygous and dominant/recessive factors enables farmers and scientists to select and maintain high-yielding disease-free seed lines across generations.

1. Identification of desirable traits: Identify useful wild or cultivated plant variations such as large leaf size, dense flowering, disease resistance, or high oil/sugar content.
2. Artificial selection and hybridization:

• Example 1 (Mustard/Cabbage family): Selection for short internodes yielded cabbage; selection for flower development yielded broccoli/cauliflower; selection for swollen stem yielded kohlrabi.
• Example 2 (Wheat/Rice): Crossing a high-yielding variety with a short-statured (dwarf) pest-resistant variety produces high-yielding dwarf crops that do not fall over in heavy wind.

The farmer was not necessarily right.**

• Reason:

1. If the purple flowered pea plants were homozygous pure (VV), then all offspring in the next generation would indeed have purple flowers (VV).
2. However, if the parent plants were heterozygous hybrid (Vv), self-pollination would yield 25% white flowered plants (vv) and 75% purple flowered plants (VV/Vv) according to Mendel’s law of segregation. Unless tested for purity, he cannot guarantee 100% purple offspring.

Explanation: This phenomenon is called Incomplete Dominance**, where neither allele is fully dominant over the other, resulting in an intermediate blending phenotype (pink) in the heterozygous state (Rr). Genes do not physically mix, but their biochemical enzyme/pigment expressions combine.

• Agricultural Examples:

1. 4 o’Clock plant (Mirabilis jalapa) and Snapdragon (Antirrhinum): Red (RR) × White (rr) produces Pink (Rr) flowers.
2. Corn/Maize kernel color: Crossing dark yellow corn with white corn can yield light yellow/cream intermediate kernels.
3. Andalucian Fowl (Livestock): Crossing black feather chicken with white feather chicken produces blue-feathered offspring.

If the game is played with just one coin, each toss represents a single independent event with two outcomes (Head or Tail). You can only obtain either H or T in a single toss. The theoretical probability of getting a Head remains 1/2 or 50%.

First Part:** Yes, we can develop high-yielding variety (HYV) seeds by cross-hybridizing two parent plants carrying complementary superior traits.

• Second Part: No, they do not necessarily have to be developed only by crossing pure lines. While crossing pure lines gives uniform, predictable hybrid vigour (heterosis) in F1, high-yielding traits can also be developed using genetic engineering, tissue culture, mutation breeding, or selecting desirable traits from non-pure varieties over multiple generations.

(a) Factor combination of child: Homozygous recessive (ll). Such traits are called Recessive autosomal characters**.

• (b) Factors in parents: Both parents must be heterozygous carriers Ll.
• (c) Cross Ll × Ll → LL (25\%), Ll (50\%), ll (25\%).
• Children able to digest milk (LL and Ll) = 25\% + 50\% = 75\%.

(a) Factor combination for two axial flowers (recessive): Homozygous recessive (ss)**.

• (b) Factor combination for pure single axial flower (dominant): Homozygous dominant (SS).
• (c) Cross outcomes:
• F1 Generation: Cross SS × ss → All offspring have factor combination Ss (Heterozygous Single Axial Flower – 100%).
• F2 Generation (Selfing Ss × Ss):
• Genotypes: 1 SS : 2 Ss : 1 ss (Ratio 1:2:1).
• Phenotypes: 75% Single axial flower (SS, Ss) and 25% Two axial flowers (ss) (Phenotypic ratio 3:1).