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

CG Board (CGBSE / SCERT Chhattisgarh) Class 10 Science Solutions | 2026-27 New Edition (English & Hindi Medium)

📚 Chapter Overview:

  • Board: Chhattisgarh Board of Secondary Education (CGBSE) / CG SCERT Raipur
  • Class & Subject: Class 10 Science (विज्ञान) – 2026-27 Edition
  • Chapter 15: Heredity: From Parents to Offsprings (आनुवंशिकी: जनकों से संतानों तक)
  • CGBSE Board Weightage: 4 Marks (8 Academic Periods)
  • Medium: Bilingual Solutions (English Medium + हिंदी माध्यम)

🌐 Select Language / माध्यम चुनें:

Part 1

In-Text & Activity Questions

Q1

Activity1 Family Character Observation Questions Page 257 Observe…


📝 Question:

Activity-1 Family Character Observation Questions (Page 257): Observe Table-1 (Eye color, dimple chin, tongue rolling, toe length, ear lobe, thumb bending) and answer: (a) What are the characters present both in you and your parents? (b) Are there any characters that you share with your parents and grandparents? (c) What are the characters that are present in you and your grandparents but not in your parents? **(d) How do you think such characters may have passed on to you?

💡 Solution:

(a)** Heritable dominant characters like eye color, free earlobes, or tongue-rolling ability are often present in both children and parents.

  • (b) Traits such as dimple chin or thumb bending capability may be shared across three generations (grandparents, parents, and self).
  • (c) Recessive characters (such as attached earlobes or inability to roll tongue) can remain hidden (recessive) in parents but re-appear in the offspring if inherited from grandparents.
  • (d) These characters are passed on through hereditary units (genes/factors) carried in gametes (sperm and egg cells) across generations according to Mendel’s law of segregation.

Q2

Activity2 Seed Variation Questions Page 257 Observe the…


📝 Question:

Activity-2 Seed Variation Questions (Page 257): Observe the seeds in a pea or bean pod: (a) Are there any seeds that are same? **(b) What are the variations that you see in them?

💡 Solution:

(a)** No two seeds inside a pod are completely identical in every respect.

  • (b) Variations can be observed in their size, shape (round vs slightly flattened), surface texture, and slight shade of color (green to yellowish green).

Q3

Section 1521 Selection of Pea Plant Questions Page…


📝 Question:

Section 15.2.1 Selection of Pea Plant Questions (Page 259): (a) Why did Mendel carry out cross-pollination to obtain hybrid seeds? **(b) Why was Mendel studying mostly contrasting characters?

💡 Solution:

(a)** Mendel carried out artificial cross-pollination between pure parents with contrasting traits to combine two different hereditary factors into a single individual and observe how traits express in hybrid offspring (F1 generation).

  • (b) Contrasting characters (e.g., tall vs dwarf, purple vs white flower) were chosen because they have clear, unambiguous physical expressions that do not blend confusingly, making it easy to track, count, and analyze inheritance patterns mathematically.

Q4

Section 1522 F1 Generation Think Question Page 260…


📝 Question:

Section 15.2.2 F1 Generation Think Question (Page 260): Think why intermediate varieties (those between purple and white) may not have been obtained in F1 generation?

💡 Solution:

Intermediate varieties were not obtained because hereditary factors (genes) are discrete, particulate units that do not blend or mix together. In the presence of a dominant factor (V for purple flower), the recessive factor (v for white flower) is completely masked, resulting in only the dominant purple trait expressing fully in F1 (Vv).

Q5

Activity3 Coin Game Probability Question Page 262 If…


📝 Question:

Activity-3 Coin Game Probability Question (Page 262): If you would have played with the same coin, what would be the probability of getting a head or a tail?

💡 Solution:

A single coin has two mutually exclusive sides (Head and Tail). The theoretical probability of getting either a Head or a Tail in a single fair toss is

Probability = 1/2 = 0.5 = 50\%

Q6

Section 1531 Sickle Cell Anaemia Question Page 264…


📝 Question:

Section 15.3.1 Sickle Cell Anaemia Question (Page 264): What is the function of the haemoglobin protein?

💡 Solution:

Haemoglobin is an iron-containing respiratory pigment present inside Red Blood Cells (RBCs). Its primary function is to bind with oxygen in the lungs to form oxyhaemoglobin and transport oxygen to all living tissues of the body for respiration, as well as carry back a part of carbon dioxide.

Q7

Section 1532 Human Sex Determination Questions Page 266…


📝 Question:

Section 15.3.2 Human Sex Determination Questions (Page 266): (a) See Figure 3: What do you think are the probabilities of each of the four conditions XY, XY, XX and XX? (b) What do you think is the probability of having a female child? **(c) “The human male carries the Y chromosome so to have a male child a human male might have some role.” Do you agree with this statement? Why/why not?

💡 Solution:

(a) Out of 4 possible fertilization combinations (XX, XX, XY, XY), two are female (XX) and two are male (XY). The probability of each of the 4 conditions is 1 in 4 (25%)**.

  • (b) The overall probability of having a female child (XX) is 2 out of 4 = 50% (1/2).
  • (c) Yes, biologically agree because human females produce only one type of egg carrying an X chromosome, whereas males produce two types of sperms in 50:50 ratio: 50% carrying X and 50% carrying Y. The sex of the child depends entirely on whether an X-bearing sperm or a Y-bearing sperm fertilizes the egg. However, sperm fusion is a purely random natural chance event (50:50), so no parent consciously controls it.

Part 2

End-of-Chapter Exercise Solutions

Q1

Choose the right option


📝 Question:

Choose the right option

💡 Solution:

(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.

(ii) Mendel had studied certain contrasting characters of pea plants. Which character do you think was not chosen by Mendel?

  • (a) Tall and dwarf plants
  • (b) Yellow and green coloured seeds
  • (c) Terminal and axial flower
  • (d) Round and wrinkled stem

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).

(iii) Some people can roll their tongue (roller represented by RR/Rr) and this is an autosomal dominant character. Those who can’t roll have the recessive factors (Non roller represented by rr). A child is capable of rolling her tongue. Her brother fails to roll the tongue while two of her sisters can roll their tongues. If both of the parents can roll their tongue, what do you think would be the combination of factors in them?

  • (a) RR and RR
  • (b) Rr and Rr
  • (c) RR and rr
  • (d) rr and rr

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)

Q2

Human Blood Groups Questions Observe the table IsupAsup…


📝 Question:

Human Blood Groups Questions: Observe the table (IA IA, IA IO for A; IB IB, IB IO for B; IA IB for AB; IO IO for O) and answer: (a) Those having blood group A would have how many IA factors? (b) Those of blood group O should have how many IO? (c) Which factor do you think is recessive: IO, IA or IB? (d) ‘A recessive character represents pure parental generation.’ Justify this statement. **(e) A child of parents having blood group A has O group. What do you think are the factors that the parents have? What do you think could be the probable blood group type of a second child that they have?

💡 Solution:

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

  • (b) A person with blood group O must have two IO factors (homozygous IO IO).
  • (c) Factor IO is recessive, while IA and IB are dominant factors.
  • (d) A recessive character can express physically (phenotypically) only when both alleles in the pair are identical (homozygous condition, e.g., IO IO 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 IA IO.
  • Second child’s probable blood groups: Cross between IA IO × IA IO yields IA IA (Group A – 25%), IA IO (Group A – 50%), and II 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.

Q3

When a blue flowered plant Bb is crossed…


📝 Question:

When a blue flowered plant (Bb) is crossed with a white flowered plant (bb), what are the combinations possible in offspring generation?

💡 Solution:

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).

Q4

A variety of cow named Zeba is adapted…


📝 Question:

A variety of cow named ‘Zeba’ is adapted to survive in hot conditions. Another variety ‘Sahiwal’ gives 20,000 liters of milk every year. The variety ‘Angush’ has a well built body. If we want to acquire maximum milk from a variety of Cow here in Chhattisgarh, which varieties of cows do you think must be crossed? Give reasons for your answer

💡 Solution:

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.

Q5

How will our knowledge of heredity help in…


📝 Question:

How will our knowledge of heredity help in improving our crop plants to acquire maximum production?

💡 Solution:

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.

Q6

How can we choose suitable characters to obtain…


📝 Question:

How can we choose suitable characters to obtain new varieties of plants? (Hint: We had studied about plants of mustard family in Chapter 1, you may cite examples of other plants)

💡 Solution:

  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.

Q7

A farmer sowed seeds of purple flowered pea…


📝 Question:

A farmer sowed seeds of purple flowered pea plants and claimed that all plants in the following generation would have purple flowers. Was he right? Why, why not?

💡 Solution:

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.

Q8

We have been able to successfully obtain pink…


📝 Question:

We have been able to successfully obtain pink flowered Rose plants by crossing white and red coloured Rose plants respectively. Do you think that some characters do intermix? Why? Site some other examples of this type that are related to agriculture

💡 Solution:

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.

Q9

What would happen if you played the game…


📝 Question:

What would happen if you played the game of coins with just one coin? What is the probability of getting a head then?

💡 Solution:

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%.

Q10

Can we develop high yielding variety seeds by…


📝 Question:

Can we develop high yielding variety seeds by hybridization? Can such seeds be developed only by crossing pure varieties?

💡 Solution:

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.

Q11

Milk Digestion Lactose Intolerance Genetics Questions Page 269…


📝 Question:

Milk Digestion / Lactose Intolerance Genetics Questions (Page 269): The factor responsible for lactose-digesting enzyme is dominant (L). People failing to digest milk lack this factor (ll). (a) What would be the combination of factors of a child who fails to digest milk and milk products? What are characters represented by such factors called? (b) What is the combination of factors in the parents of such a child who are capable of digesting milk? **(c) What percent of children of such parents would be able to digest milk?

💡 Solution:

(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\%.

Q12

Green Gram Axial Flower Inheritance Questions Page 269…


📝 Question:

Green Gram Axial Flower Inheritance Questions (Page 269): In green gram, single flower at axial position is dominant (S), two flowers at axial position is recessive (s). (a) What will be the combination of factors in a plant bearing two flowers at axial position? (b) What will be the combination of factors in a pure plant bearing a single flower at axial position? **(c) When these two plants are crossed, what will be the characters and factor combinations in F1 and F2 generations?

💡 Solution:

(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).


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