Heredity Class 10: Your Questions Answered, Beyond NCERT!

Stuck in the maze of Chapter 8? Feeling lost amidst genes, chromosomes, and Punnett squares? Worry not, curious minds! This blog post offers a comprehensive collection of Q&A, designed to transcend the limitations of standard textbooks and empower you to confidently tackle this crucial chapter.

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This isn't just about memorizing facts for the CBSE 2024-25 Class 10 Board Exams (although we'll definitely ace those too!). It's about igniting your curiosity, unlocking the secrets of your own DNA, and seeing the world through the lens of genetics. So, ditch the textbook, grab your thinking cap, and prepare to dive deeper into the fascinating world of Heredity, beyond the boundaries of NCERT!

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Subject	Science (Biology)
Class	10
Board	CBSE & State Boards
Chapter No.	8
Chapter Name	Heredity
Type	Questions & Answers
Session	2024-25

"The only person you are destined to become is the person you decide to be."

- Ralph Waldo Emerson

Heredity Class 10 Important Questions & Answers

Q. No. 1) Multiple Choice Questions (MCQs)

i.

• Assertion: The probability of survival of an organism produced through sexual reproduction is more than that of an organism produced through asexual mode.
• Reason: Variations provide advantages to individuals for survival.

a) Both A and R are true, and R is the correct explanation of A.

b) Both A and R are true, and R is not the correct explanation of A.

c) A is true but R is false.

d) A is false but R is true.

Ans. Option (a)

ii. Two pink-colored flowers on crossing resulted in 1 red, 2 pink, and 1 white flower progeny. The nature of the cross will be

a. Double fertilization

b. Self-pollination

c. Cross-fertilization

d. No fertilization

Ans. Option (c)

iii. Which statement about the genotypes of organisms is correct?

a. Dominant alleles are only found in homozygotes

b. One recessive allele always causes a recessive phenotype

c. Recessive phenotypes must be homozygous

d. The dominant phenotype must be heterozygous

Ans. Option (c)

iv. Which of the following is heterozygous?

a. TTRR

b. ttrr

c. TT

d. Tt

Ans. Option (d)

v. The genotype of offspring formed from Tt x tt will be

a. TT and tt

b. Tt and tt

c. Only tt

d. Only TT

Ans. Option (b)

vi. A recessive homozygote is crossed with a heterozygote of the same gene. What will be the phenotype of the F1 generation?

a. All dominant

b. 75 % dominant, 25 % recessive

c. 50 % dominant, 50 % recessive

d. 25 % dominant, 50 % heterozygous, 25 % recessive

Ans. Option (c) [tt x Tt = Tt, Tt, tt, tt]

vii. If a tall pea plant is crossed with a pure dwarf pea plant then, what percentage of F1 and F2 generation respectively will be tall?

a. 25%, 25%

b. 50%, 50%

c. 75%, 100%

d. 100%, 75%

Ans. Option (d)

[TT x tt

F1 = Tt = 100% tall

Tt x Tt

F2 = TT, Tt, Tt, tt = 75% tall]

viii. In cattle, having horns is a recessive trait (h) to not having horns (H). When cattle with horns are crossed with cattle that do not have horns, the number of offspring having horns was equal to those not having horns. Which of the following is MOST LIKELY to be true?

(a) Both parents are homozygous dominant.

(b) One parent is homozygous dominant.

(c) Both parents are heterozygous.

(d) One parent is heterozygous.

Ans. Option (d)

ix. Which of the following statements is incorrect?

a. For every hormone there is a gene.

b. For every protein there is a gene.

c. For the production of every enzyme there is a gene.

d. For every molecule of fat there is a gene.

Ans. Option (d)

x. The height of a plant is regulated by:

a) DNA which is directly influenced by growth hormone.

b) Genes that regulate the proteins directly.

c) Growth hormones under the influence of the enzymes coded by a gene.

d) Growth hormones directly under the influence of a gene.

Ans. Option (c)

xi. If a round, green-seeded pea plant (RR yy) is crossed with a wrinkled, yellow-seeded pea plant, (rr YY) the seeds produced in the F1 generation are

a. Round and yellow

b. Round and green

c. Wrinkled and green

d. Wrinkled and yellow

Ans. Option (a)

Gametes	rY
Ry	RrYy

xii. A homozygous dominant guinea pig with black fur is crossed with a homozygous guinea pig with white fur. The F1 generation is crossed with itself.

What percentage of F2 generation is expected to show a white fur coat?

a. 25 %

b. 50 %

c. 75 %

d. 100 %

Ans. Option (a)

[BB x bb = Bb (F1)

Bb x Bb = BB, Bb, Bb, bb (F2)

Gametes	B	b
B	BB	Bb
b	Bb	bb

∴ Percentage of white fur (bb) in F2 = ¹/₄ x 100 = 25%]

xiii. Two pea plants one with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds produce F1 progeny that have round, yellow (RrYy) seeds. When F1 plants are selfed, the F2 progeny will have a new combination of characters. Choose the new combination from the following.

1. Round, yellow
2. Round, green
3. Wrinkled, yellow
4. Wrinkled, green

a. I and II

b. I and IV

c. II and III

d. I and III

Ans. Option (b)

xiv. A heterozygous red-eyed female Drosophila mated with a white-eyed male would produce _____.

a. Red-eyed females and white-eyed males in the F1

b. White-eyed females and red-eyed males in the F1

c. Half red and half white-eyed females and all white-eyed males in the F1

d. Half red and half white-eyed females as well as males in the F1

Ans. Option (d) [Heterozygous red means red is dominant (Rr) and white is recessive (rr)

Rr x rr = Rr, Rr, rr, rr (F1)

Gametes	r	r
R	Rr	Rr
r	rr	rr

xv.

• Assertion (A): Height in pea plants is controlled by the efficiency of enzymes and is thus genetically controlled.
• Reason (R): Cellular DNA is the information source for making proteins in the cell.

Options

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true and R is not the correct explanation of A

c. A is true but R is false

d. A is false but R is true

Ans. Option (a)

xvi.

• Assertion (A): Offsprings produced by sexual reproduction show variation.
• Reason (R): Each offspring produced by sexual reproduction inherits all the genes from each parent.

Options

a. Both A and R are true and R is the correct explanation of A

b. Both A and R are true and R is not the correct explanation of A

c. A is true but R is false

d. A is false but R is true

Ans. Option (c)

xvii. 9 : 3 : 3 : 1 ratio is due to _____.

a. Segregation

b. Crossing over

c. Independent assortment

d. Recessiveness

Ans. Option (c)

xviii. A zygote that has an X-chromosome inherited from the father will develop into a _____.

a. Boy

b. Girl

c. X-chromosome does not determine the sex of a child

d. Either boy or girl

Ans. Option (b)

Q. No. 2) In an asexually reproducing species if trait X exists in 5% of a population and trait Y exists in 70% of the same population, which of the two traits is likely to have arisen earlier? Give reason.

Ans. Trait Y which exists in 70% (larger fraction) of the population, is likely to have arisen earlier because in asexual reproduction, identical copies of DNA are produced and variations occur negligibly.

New traits come into the population due to sudden mutation and then are inherited. 70% of the population with trait Y is likely to have been replicating that trait for a longer period than 5% of the population with trait X.

Q. No. 3) Give the pair of contrasting traits of the following characters in the pea plant and mention which is dominant and which is recessive.

1. Yellow seed
2. Round seed

Ans. i. Yellow – dominant; Green – recessive

ii. Round – dominant; Wrinkled – recessive

Note: Remember this chart

Q. No. 4) In a pea plant, find the contrasting trait if:

a. The position of the flower is terminal.

b. The flower is white in color.

c. Shape of the pod is constricted.

Ans. a. Axial position of the flower.

b. Purple color of the flower.

c. Inflated shape.

Q. No. 5) Why did Mendel choose a pea plant for his experiments?

Ans. Mendel chose a pea plant for his experiments because:

1. It is easy to grow.
2. It is naturally self-pollinating and so is very easy to raise pure-breeding individuals.
3. It has a short life span as it is an annual crop and so it was possible to follow several generations.
4. It is easy to cross-pollinate.
5. It has deeply contrasting characters.
6. The flowers are bisexual.

Q. No. 6) What is a gene? Where are genes located?

Ans. It is the basic unit of heredity. It is a specific part (DNA segment) of a chromosome that controls the expression of a character.

Genes are located on chromosomes.

Q. No. 7) A Mendelian experiment consisted of breeding pea plants bearing violet flowers with pea plants bearing white flowers. What will be the result in F1 progeny?

Ans. All F1 progeny will bear violet flowers as violet is a dominant trait.

VV x vv

Gametes: V, v

F1 progeny: Vv (all violet)

Gametes	V
v	Vv

Q. No. 8) What are monohybrid and dihybrid cross?

Ans. Monohybrid cross: The cross between two pea plants with one pair of contrasting characters is called a monohybrid cross.

Example: Cross between a tall and a dwarf plant.

Dihybrid Cross: The cross between two plants having two pairs of contrasting characters is called a dihybrid cross.

Example: Cross between a round and green seed plant with a wrinkled and yellow seed plant.

Q. No. 9) In a pea plant, the trait of flowers bearing purple color (PP) is dominant over white color (pp). Explain the inheritance pattern of F1 and F2 generations with the help of a cross following the rules of inheritance of traits. State the visible characters of F1 and F2 progenies.

Ans. Parents: PP and pp

F1: PP x pp

Gametes: P, p

Gametes	P
p	Pp

Progeny: Pp - Means all F1 progenies bear purple color flowers.

F2: Pp x Pp

Gametes: P, p, P, p

Gametes	P	p
P	PP	Pp
p	Pp	pp

Progeny: PP, Pp, Pp, pp - Means 3 F2 progenies bear purple color flowers, and 1 progeny bear white color flowers.

Q. No. 10) How do Mendel’s experiments show that the

1. Traits may be dominant or recessive,
2. Traits are inherited independently?

Ans. i. Mendel performed a simple experiment by crossing pea plants with just one different trait (like tall and short height). In the first generation (F1), he found that only one trait (like tallness) appeared, while the other (shortness) disappeared. However, in the next generation (F2), when he allowed F1 plants to self-pollinate, the missing trait (shortness) reappeared in about 25% of the plants. This showed that traits can be "dominant" (appearing in F1) or "recessive" (hidden in F1 but appearing in some F2 plants).

ii. Mendel also tested what happened when he studied two different traits together, like seed color (yellow or green) and seed shape (round or wrinkled). When he crossed plants with yellow-round seeds and green-wrinkled seeds, he found four combinations in the F2 generation: yellow-round, green-wrinkled (parental combinations), and yellow-wrinkled, green-round (new combinations). This showed that each trait, like color and shape, was inherited separately from the other, meaning traits are inherited independently of one another.

Q. No. 11) State Mendel’s laws of inheritance.

Ans. Mendel’s Law of Inheritance.

1. Law of Dominance: Out of a pair of contrasting characters present together, only one is able to express itself while the other remains suppressed. The one that expresses is the dominant character and the one unexpressed is recessive.
2. Law of Segregation: The two members of a pair of factors separate during the formation of gametes.
3. Law of Independent Assortment: When there are two pairs of contrasting characters, the distribution of the members of one pair into the gametes is independent of the distribution of the other pair.

Q. No. 12) After self-pollination in pea plants with round, yellow seeds, the following types of seeds were obtained by Mendel:

Seeds	Number
Round, yellow	630
Round, green	216
Wrinkled, yellow	202
Wrinkled, green	64

Analyze the result and describe the mechanism of inheritance which explains these results.

Ans. The ratio here is roughly 9:3:3:1, which shows both parental and new combinations of traits. This pattern suggests that traits like seed shape (round or wrinkled) and seed color (yellow or green) are inherited independently of each other.

This happens because, during reproduction, each parent plant’s genes are separated into germ cells (like pollen or eggs). When these germ cells combine, the pair of traits for shape and color can mix in different ways. So, each trait can show up in new combinations, not just the combinations that appeared in the parent plants.

Q. No. 13) A tall pea plant was crossed with a dwarf one. F1 generation was allowed to self-pollinate and F2 generation was also obtained. Answer the following questions:

i. What would be the phenotype of plants in the F1 generation?

ii. What would be the phenotypic ratio in the F2 generation?

iii. Give a reason for your observation in the F1 generation.

Ans. Parents: PP and pp

F1: PP x pp

Gametes: P, p

Gametes	P
p	Pp

Progeny of F1: Pp

F2: Pp x Pp

Gametes: P, p, P, p

Gametes	P	p
P	PP	Pp
p	Pp	pp

Progeny of F2: PP, Pp, Pp, pp

i. All tall pea plants.

ii. Phenotypic ratio = 3:1 [i.e., 3 tall and 1 dwarf]

iii. The tall trait of the pea plant is a dominant trait over the short trait, a recessive trait.

Q. No. 14) Mention the function of cellular DNA. Talking tallness as a characteristic of a plant, explain how proteins control the characteristic.

Ans. Cellular DNA is the information source for making proteins in the cell. A section of DNA that provides information for one protein is called the gene for that protein. Let us take the example of tallness as a characteristic. We know that plants have hormones that can trigger growth. Plant height can thus depend on the amount of a particular plant hormone. The amount of the plant hormone made will depend on the efficiency of the process for making it. Consider now an enzyme that is important for this process. If this enzyme works efficiently, a lot of hormones will be made, and the plant will be tall. If the gene for that enzyme has an alteration that makes the enzyme less efficient, the amount of hormone will be less, and the plant will be short. Thus, genes control characteristics or traits.

Q. No. 15) In the following crosses write the characteristics of the progeny:

Cross	Progeny
i. RRYY x RRYY Round yellow x Round yellow
ii. RrYy x RrYy Round yellow x Round yellow
iii. rryy x rryy Wrinkled green x wrinkled green
iv. RRYY x rryy Round yellow x wrinkled green

Ans. i. RRYY x RRYY

Gametes: RY, RY

Gametes	RY
RY	RRYY

Progeny: RRYY (Round yellow)

ii. RrYy x RrYy

Gametes: RY, Ry, rY, ry

Gametes	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Progeny:- Round yellow : Round green : Wrinkled yellow : Wrinkled green = 9:3:3:1

iii. rryy x rryy

Gametes: ry, ry

Gametes	ry
ry	rryy

Progeny: rryy (wrinkled green)

iv. RRYY x rryy

Gametes: RY, ry

Gametes	RY
ry	RrYy

Progeny: RrYy (Round yellow)

Q. No. 16) Mustard was growing in two fields – A and B. While Field A produced brown-colored seeds, field B produced yellow-colored seeds.

It was observed that in field A, the offspring showed only the parental trait for consecutive generations, whereas in field B, the majority of the offspring showed a variation in the progeny.

What are the probable reasons for these?

Ans. In field A, the reason for the parental trait in consecutive generations of offspring is self-pollination.

In field B, variation is seen to occur because of the recombination of genes as cross-pollination is taking place.

Q. No. 17) Pooja has green eyes while her parents and brother have black eyes. Pooja’s husband Ravi has black eyes while his mother has green eyes and father has black eyes.

a. On the basis of the above-given information, is the green eye color a dominant or recessive trait? Justify your answer.

b. What is the possible genetic makeup of Pooja’s brother’s eye color?

c. What is the probability that the offspring of Pooja and Ravi will have green eyes? Also, show the inheritance of eye color in the offspring with the help of a suitable cross.

d. 50% of the offspring of Pooja’s brother are green-eyed. With the help of cross show how this is possible.

Ans. a. Yes, green eye color is recessive as it will express only in homozygous condition. Pooja has green eyes, while both of her parents and her brother have black eyes. If green eyes were dominant, at least one of her parents would have to have green eyes.

b. BB or Bb

c. Pooja (bb) x Ravi (Bb)

Gametes: b, b, B, b

Gametes	b	b
B	Bb	Bb
b	bb	bb

Progeny: Bb, Bb, bb, bb - means 50% of the offspring can have green eye color.

d. Pooja’s brother (Bb) x Wife (bb)

Gametes: B, b, b, b

Gametes	B	b
b	Bb	bb
b	Bb	bb

Progeny: Bb, Bb, bb, bb - means 50% of the offspring can have green eye color as per the cross shown.

Q. No. 18) Two pea plants – one with round yellow seeds (RRYY) and another with wrinkled green (rryy) seeds produce F1 progeny that has round, yellow (RrYy) seeds.

When F1 plants are self-pollinated, which new combination of characters is expected in F2 progeny? How many seeds with these new combinations of characters will be produced when a total of 160 seeds are produced in the F2 generation? Explain with reason.

Ans. New combinations would be Round green and Wrinkled yellow.

The phenotype ratio will be 9:3:3:1

Round green = ³/_9+3+3+1 x 160 = ³/₁₆ x 160 = 30 seeds

Wrinkled  yellow = ³/_9+3+3+1 x 160 = ³/₁₆ x 160 = 30 seeds

New combinations are produced because of the independent inheritance of seed shape and seed color traits.

Q. No. 19) Figures (a) to (d) given below represent the type of ear lobes present in a family consisting of 2 children – Rahul, Nisha, and their parents.

Excited by his observation of different types of ear lobes present in his family, Rahul conducted a survey of the type of ear lobes found {Figure (e) and (f)} in his classmates. He found two types of ear lobes in his classmates as per the frequency given below:

Sex	Free	Attached
Male	36	14
Female	31	19

Based on the above data answer the following questions.

a) Which of the two characteristics - ‘free ear lobe’ or ‘attached ear lobe’ appears to be dominant in this case? Why?

b) Is the inheritance of the free ear lobe linked with the sex of the individual? Give a reason for your answer.

c) What type of ear lobe is present in father, mother, Rahul, and his sister Nisha? Write the genetic constitution of each of these family members which explains the inheritance of this character in this family.

(Gene for Free ear lobe is represented by F and gene for attached ear lobe is represented by f for writing the genetic constitution).

d) Suresh’s parents have attached earlobes. What type of ear lobe can be seen in Suresh and his sister Siya? Explain by giving the genetic composition of all.

Ans. a) Free ear lobe is dominant because Rahul has attached ear lobe, while both of his parents and his sister have free ear lobes. If attached ear lobes were dominant, at least one of her parents would have to have attached ear lobe.

b) No. It is not sex-linked. As per the data of the family as well as the class, it is indicated that the free ear lobe is present in males as well as in females.

c) Father – Ff (free ear lobe), Mother – Ff (free ear lobe), Rahul – ff (attached ear lobe) and Nisha – Ff (free ear lobe)

d) Suresh’s father – ff (attached ear lobe), mother – ff (attached ear lobe), Suresh - ff (attached ear lobe), Siya – ff (attached ear lobe). If both parents have recessive characters, then all the children can have recessive characters only.

Q. No. 20) Two human beings who can both roll their tongues produced 11 children. 3 of these children could not roll their tongues and 8 children could roll their tongues.

a. Which trait (rolling or not rolling) is controlled by the recessive allele?

b. State all possible genotypes of the F1 generation of the cross.

c. Show the inheritance of the tongue rolling in humans in the given example using a suitable cross. What percentage of offspring will show the same genotype as the parents?

d. If one of the parents could not roll their tongue, with the help of a cross, calculate the ratio of tongue-rollers to non-tongue-rollers in the offspring.

Ans. a. Rolling is dominant and not rolling is recessive.

b. RR (homozygous dominant), Rr(heterozygous dominant), rr (homozygous recessive)

c. Rr x Rr

Gametes: R, r, R, r

Gametes	R	r
R	RR	Rr
r	Rr	rr

Progeny: RR, Rr, Rr, rr - means 50% of the offspring will show the same genotype as the parents.

d. Rr x rr

Gametes: R, r, r, r

Gametes	R	r
r	Rr	rr
r	Rr	rr

Progeny: Rr, Rr, rr, rr

The ratio of tongue roller to non-tongue roller offspring = 2:2 = 1:1

Q. No. 21) What are chromosomes? Mention the total number of chromosomes along with the sex chromosomes that are present in a human female and a human male. Explain how in sexually producing organisms the number of chromosomes in the progeny remains the same as that of the parents.

Ans. Chromosomes: Thread-like structures made up of DNA found in the nucleus.

A human male has 22 pairs of chromosomes along with the XY sex chromosome.

A human female has 22 pairs of chromosomes along with the XX sex chromosome.

The original number of chromosomes becomes half during gamete formation. When the gamete fuse, the original number of chromosomes is restored in the progeny.

Q. No. 22) In humans, there is a 50% probability of the birth of a boy and a 50% probability that a girl will be born. Justify the statement based on the mechanism of sex determination in human beings.

Or,

What is the probability of a girl or a boy being born in a family? Justify your answer.

Ans. In humans, the chance of having a boy or a girl is 50% for each.

This is because of the way sex chromosomes are inherited. Women have two X chromosomes (XX), while men have one X and one Y chromosome (XY).

• When a child is conceived, they inherit one X chromosome from their mother.
• The father can pass on either an X or a Y chromosome.
  • If the father’s X chromosome is passed, the child will be a girl (XX).
  • If the father’s Y chromosome is passed, the child will be a boy (XY).

Since the father can pass on either an X or a Y with equal probability, the chance of having a boy or a girl is 50%.

Download Heredity (Chapter 8) Class 10 NCERT Underlined PDF

Must Read: Heredity Class 10 Notes to get an idea of the chapter.

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