Class 10 Science · Chapter 8

Heredity

120 Class 10 Science questions on Heredity from the NCERT textbook, NCERT Exemplar and CBSE board papers 2024–2026, each with its answer and a step-by-step solution.

CBSE Class 10 Science previous year papers, with answers →

Questions · page 4 of 6

1. Q61 CBSE 2026
   Identify ​the ​traits that cannot ​be transmitted from parents to the ​off-springs :

   1. (i)Swimming ​technique
   2. (ii)Blood ​group
   3. (iii)​Eye colour
   4. (iv)Dancing skills Options :

   1. (a)(i) and (ii)
   2. (b)(i) and (iv)
   3. (c)(iii) and (iv)
   4. (d)(ii) and (iii)
   Show answer

   (b) (i) and (iv)

   Step-by-step solution →
2. Q62 CBSE 2026
   When a human ​egg ​is fertilized by a ​sperm having ‘Y’ chromosome, ​the zygote ​has ​the following combination ​of chromosomes ​:

   1. (a)44 + XX
   2. (b)22 + XX
   3. (c)44 + XY
   4. (d)22 + XY
   Show answer

   (c) 44 + XY

   Step-by-step solution →
3. Q63 CBSE 2026
   Two pea plants, one ​with round green seeds (RRyy) and another with wrinkled yellow (rrYY) seeds ​were crossed with each other which produced $F_1$ progeny that have only ​round yellow (RrYy) ​seeds. When $F_1$ plants are self-pollinated, the $F_2$ progeny will have which new combination of characters, ​as compared to the parents ​:

   1. (i)Round, yellow
   2. (ii)Round, green
   3. (iii)​Wrinkled, yellow
   4. (iv)Wrinkled, green Correct options are ​:

   1. (a)(i) and (ii)
   2. (b)(i) and (iv)
   3. (c)(ii) and (iv)
   4. (d)(i) and (iii)
   Show answer

   (b) (i) and (iv)

   Step-by-step solution →
4. Q64 CBSE 2026
   Sex is determined by different factors in ​various species. However, in human ​beings, it is determined genetically. Which amongst the following option(s) is/are correct ​for ​human beings ​?

   1. (i)Gamete ​carrying X chromosome ​from female parent.
   2. (ii)Gamete carrying X chromosome from male parent.
   3. (iii)Gamete carrying ​Y chromosome from male parent.

   1. (a)(ii) and (iii)
   2. (b)(i) only
   3. (c)(i) and (iii)
   4. (d)(iii) only
   Show answer

   (a) (ii) and (iii)

   Step-by-step solution →
5. Q65 CBSE 2026
   Mendel crossed pea plants ​having yellow ​seeds (YY) with pea plants having ​green seeds (yy). Yellow seeds are obtained in $F_1$ ​progeny ​which undergoes self-pollination to produce $F_2$ progeny. ​Select the option showing ​the result of $F_2$ progeny.

   1. (a)3 green : 1 yellow
   2. (b)1 yellow : 1 green
   3. (c)3 yellow : 1 green
   4. (d)2 green : 2 yellow
   Show answer

   (c) 3 yellow : 1 green

   Step-by-step solution →
6. Q66 CBSE 2026
   Round, green seeded pea ​plant (RRyy) ​was crossed ​with ​wrinkled, yellow ​seeded pea ​plant (rrYY), and $F_1$ progeny is produced. All plants of $F_1$ progeny ​will have :

   1. (a)round and green seeds
   2. (b)round and yellow seeds
   3. (c)wrinkled and green seeds
   4. (d)wrinkled and yellow seeds
   Show answer

   (b) round and yellow seeds

   Step-by-step solution →
7. Q67 CBSE 2026
   Two pea plants one with round green seeds (RRyy) and ​another with wrinkled yellow seeds (rrYY) were crossed. ​All the seeds of ${\text{F}}_1$ progeny have round, yellow seeds (RrYy). When ${\text{F}}_1$ plants are ​self-pollinated, the ${\text{F}}_2$ progeny will have ​new ​combination of characters.

   1. (i)Round, yellow
   2. (ii)Round, green
   3. (iii)Wrinkled, yellow
   4. (iv)Wrinkled, green Choose the ​option that ​shows new combination of characters in ${\text{F}}_2$ ​progeny.

   1. (a)(i) and (ii)
   2. (b)(i) and (iv)
   3. (c)(iii) and (iv)
   4. (d)(i) and (iii)
   Show answer

   (b) (i) and (iv)

   Step-by-step solution →
8. Q68 CBSE 2026
   Question No. 8 & 9 consists of two ​statements – Assertion (A) ​and Reason (R). ​Answer these questions selecting the appropriate ​option from (a), ​(b), (c) and (d) given ​below :

   Assertion (A): Bacteria that can withstand heat have better chances of survival in a heat wave.

   Reason (R): Accumulation of variations in a species increases the chances of its survival in changing environment.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
   3. (c)Assertion (A) is true, but Reason (R) is false.
   4. (d)Assertion (A) is false, but Reason (R) is true.
   Show answer

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

   Step-by-step solution →
9. Q69 CBSE 2026
   For Questions number 8 and 9, ​two statements ​are given ​— one labelled ​as Assertion (A) and the ​other labelled as Reason (R). Select the ​correct answer to these questions from the codes ​(a), (b), ​(c) and (d) as given below.

   Assertion (A): In some reptiles, the temperature at which the fertilized eggs are kept determines whether the egg will develop into a male or a female.

   Reason (R): In human beings, the sex is determined genetically.

   1. (a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
   2. (b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
   3. (c)Assertion (A) is true, but Reason (R) is false.
   4. (d)Assertion (A) is false, but Reason (R) is true.
   Show answer

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

   Step-by-step solution →
10. Q70 CBSE 2026
    “The sex ​of ​the children in ​the ​human beings will be determined by what they inherit ​from ​their father.” Explain this statement.
    Show answer

    Every egg carries X, but a sperm carries X or Y, so the sex chromosome inherited from the father decides the child's sex.

    One of the twenty three pairs of human chromosomes is the sex-chromosome pair: a woman has $XX$ and a man has $XY$. Because a gamete carries one chromosome of each pair, every egg the mother makes carries an X, while the father makes two kinds of sperm in equal numbers — half carrying X and half carrying Y. If an X carrying sperm fertilises the egg the child is $XX$, a girl; if a Y carrying sperm fertilises it the child is $XY$, a boy. The mother’s contribution is an X in either case, so the only thing that differs between a son and a daughter is the sex chromosome that came from the father — which is why the sex of the child is determined by what it inherits from its father.

    Step-by-step solution →
11. Q71 CBSE 2026
    Do all variations in a species ​have ​equal chances ​of surviving ​in the ​environment in ​which ​they find themselves ? Explain.
    Show answer

    No; individuals with variations beneficial in that environment survive, e.g. heat-resistant bacteria when the water warms.

    No. Depending on the nature of their variations, different individuals have different kinds of advantages in a given environment, so variations do not have equal chances of survival: individuals with variations that are beneficial for survival in that environment survive and pass those variations on to their progeny. For example, if a population of bacteria lived in temperate waters and the water temperature rose because of global warming, most of these bacteria would die, but the few variants resistant to heat would survive and grow further. Similarly, in a population of red beetles living on green bushes, a green variant cannot be seen by crows, so it survives and leaves more offspring. This selection of variants by environmental factors forms the basis of evolution.

    Step-by-step solution →
12. Q72 CBSE 2026
    Describe ​how ​the ​sex of a ​new born ​individual is determined in ​humans.
    Show answer

    Every egg carries X; an X sperm gives a girl and a Y sperm a boy, so the father's sperm decides the sex.

    The sex of a new born child in human beings is decided by the sex chromosome it receives from its father. A human body cell has 23 pairs of chromosomes; 22 pairs are alike in both sexes and the remaining pair is the pair of sex chromosomes. A woman has a matched pair, XX, and a man has a mismatched pair, XY. When germ cells are formed each gamete gets only one chromosome of a pair, so every egg produced by the mother carries an X chromosome, while the father produces two kinds of sperm in equal numbers, half carrying X and half carrying Y. If a sperm carrying X fertilises the egg the zygote is XX and the child is a girl; if a sperm carrying Y fertilises it the zygote is XY and the child is a boy. Since the mother's contribution is the same either way, the sex of the child depends on the father, and a girl and a boy are equally likely.

    Step-by-step solution →
13. Q73 CBSE 2026
    Mendel ​used a number of ​easily observable ​contrasting characters ​of garden ​pea plant. List any two pairs of ​such characters. ​Also mention the dominant ​and recessive character in each pair.
    Show answer

    Tall (dominant) and dwarf (recessive) stem height; round (dominant) and wrinkled (recessive) seed shape.

    Any two of Mendel's seven pairs, for example: (1) Height of the stem: tall is dominant and dwarf (short) is recessive. (2) Shape of the seed: round is dominant and wrinkled is recessive. Other correct pairs: colour of the seed (yellow dominant, green recessive); colour of the flower (violet dominant, white recessive); position of the flower (axial dominant, terminal recessive); shape of the pod (inflated dominant, constricted recessive); colour of the pod (green dominant, yellow recessive).

    Step-by-step solution →
14. Q74 CBSE 2026
    Use the following information for Pea plants ​: Dominant allele ​for tall plants – T Recessive allele for short plants – t Dominant allele for purple ​flowers – ​P Recessive allele for white flowers – p

    1. (a)​Perform a cross between pure plant dominant for both the traits with another plant which is recessive for both the traits and work out the progeny upto $F_2$ generation.
    2. (b)What is the probability of ​obtaining ​tall plants with white flowers in $F_2$ progeny ?
    Show answer

    ${\text{F}}_2$ shows $9:3:3:1$, and the probability of a tall plant with white flowers is $\frac{3}{16}=0.1875$.

    (a) The pure dominant parent is $TTPP$ (tall, purple) and the parent recessive for both traits is $ttpp$ (short, white). Their only gametes are $TP$ and $tp$, so the whole $F_1$ is $TtPp$ - every plant tall with purple flowers. Self-pollinating the $F_1$ gives $TtPp\times TtPp$; each parent now forms four kinds of gamete, $TP$, $Tp$, $tP$ and $tp$, and the sixteen boxes of the Punnett square give an $F_2$ of $9$ tall purple $:3$ tall white $:3$ short purple $:1$ short white (genotypes $1TTPP:2TTPp:2TtPP:4TtPp:1TTpp:2Ttpp:1ttPP:2ttPp:1ttpp$). (b) Tall plants with white flowers occupy $3$ of the $16$ boxes, so the probability is $\frac{3}{16}$, that is $0.1875$ or $18.75\%$. Their genotypes are $1TTpp$ and $2Ttpp$.

    Step-by-step solution →
15. Q75 CBSE 2026
    1. (a)A couple are parents to 4 daughters in a ​sequence, ​and do not have any son. Does this ​indicate ​that the husband does not ​produced Y-chromosome bearing sperms ​? Explain.
    2. (b)What are the chances of this couple ​bearing yet another daughter ? Show with the help ​of a cross.
    Show answer

    No: the father makes X and Y sperms equally and sex is by chance; another daughter has a 1 in 2 (50%) chance.

    (a) No. A man (XY) produces two kinds of sperm in equal numbers: half carry X and half carry Y, while every egg carries X. Which sperm fertilises the egg is a matter of chance at each conception, so four daughters mean only that an X-bearing sperm happened to fertilise the egg four times. That is 1 of the 16 equally likely birth orders for four children, and it does not show that the husband lacks Y-bearing sperm. (b) The chance is 1 in 2 (50%). Cross: $\text{XY}\times \text{XX}$ gives gametes X and Y (sperm) and X (egg), and offspring XX (girl) and XY (boy) in the ratio $1:1$, whatever the earlier children were.

    Step-by-step solution →
16. Q76 CBSE 2026
    A blue colour flower (BB) ​bearing plant is crossed with ​a plant with white coloured flowers (bb).

    1. (a)State the colour of the flowers ​you would expect in their ${\text{F}}_1$ progeny.
    2. (b)What ​must be the percentage of white ​flower plants in the ${\text{F}}_2$ generation, if the ${\text{F}}_1$ plants are self-pollinated ?
    3. (c)State the expected ratio of genotypes ​BB and ​bb in ${\text{F}}_2$ progeny.
    Show answer

    ${\text{F}}_1$ is all blue ($Bb$); $25\%$ of ${\text{F}}_2$ are white, and $BB:bb=1:1$.

    (a) All the ${\text{F}}_1$ plants ($\text{Bb}$) bear blue flowers, because blue ($\text{B}$) is dominant over white ($\text{b}$). (b) $25\%$ of the ${\text{F}}_2$ plants bear white flowers ($\text{bb}$, $1$ in every $4$). (c) $\text{BB}:\text{bb}=1:1$ (each is $\frac{1}{4}$ of the ${\text{F}}_2$; the full genotype ratio is $\text{BB}:\text{Bb}:\text{bb}=1:2:1$).

    Step-by-step solution →
17. Q77 CBSE 2026
    Based on Mendel’s experiment ​show ​the cross ​between round, yellow seeds ​of pea plant (RRYY) with ​wrinkled and green seeds of ​pea plant (rryy) upto ${\text{F}}_2$ generation. Write the number of ​seeds/plants ​with following genetic (Genotype) makeup in ${\text{F}}_2$ generation :

    1. (i)RRYY
    2. (ii)rryy
    Show answer

    In ${\text{F}}_2$, $\text{RRYY}$ is 1 out of 16 and $\text{rryy}$ is 1 out of 16; the phenotypic ratio is $9:3:3:1$.

    $\text{P}$: $\text{RRYY}$ (round, yellow) $\times$ $\text{rryy}$ (wrinkled, green); gametes $\text{RY}$ and $\text{ry}$. ${\text{F}}_1$: all $\text{RrYy}$ — round and yellow seeds. ${\text{F}}_1$ selfed ($\text{RrYy}\times \text{RrYy}$): each plant forms four kinds of gametes, $\text{RY}$, $\text{Ry}$, $\text{rY}$, $\text{ry}$, and the ${\text{F}}_2$ checkerboard is $$\begin{array}{ccccc}& \text{RY}& \text{Ry}& \text{rY}& \text{ry}\\ \text{RY}& \text{RRYY}& \text{RRYy}& \text{RrYY}& \text{RrYy}\\ \text{Ry}& \text{RRYy}& \text{RRyy}& \text{RrYy}& \text{Rryy}\\ \text{rY}& \text{RrYY}& \text{RrYy}& \text{rrYY}& \text{rrYy}\\ \text{ry}& \text{RrYy}& \text{Rryy}& \text{rrYy}& \text{rryy}\end{array}$$ Round yellow : round green : wrinkled yellow : wrinkled green $=9:3:3:1$. (i) $\text{RRYY}$: $1$ out of the $16$ (one box of the checkerboard). (ii) $\text{rryy}$: $1$ out of the $16$ (one box of the checkerboard).

    Step-by-step solution →
18. Q78 CBSE 2026
    A tall pea plant (TT) is ​crossed with a dwarf ​pea plant (tt). ​Find out ${\text{F}}_1$ and ${\text{F}}_2$ progeny. Also mention the processes that lead to the ​formation of ${\text{F}}_1$ and ${\text{F}}_2$ progeny ​during ​the cross.
    Show answer

    ${\text{F}}_1$ is all tall ($\text{Tt}$); ${\text{F}}_2$ has tall and dwarf plants in the ratio $3:1$; ${\text{F}}_1$ forms by cross-pollination, ${\text{F}}_2$ by self-pollination.

    ${\text{F}}_1$ progeny: every plant is $\text{Tt}$ and all of them are **tall**, because $\text{T}$ (tallness) is dominant over $\text{t}$ (dwarfness). ${\text{F}}_2$ progeny: $\text{TT}$, $\text{Tt}$, $\text{Tt}$ and $\text{tt}$, that is a genotypic ratio of $1:2:1$, giving tall and dwarf plants in the phenotypic ratio $3:1$ (three tall to one dwarf). Processes: the ${\text{F}}_1$ is formed by **cross-pollination** (cross-fertilisation) between the tall $\text{TT}$ plant and the dwarf $\text{tt}$ plant; the ${\text{F}}_2$ is formed by **self-pollination** of the ${\text{F}}_1$ ($\text{Tt}$) plants.

    Step-by-step solution →
19. Q79 CBSE 2026
    Study the given table and answer the following questions :

    1. (a)Depict ​the information given in ​the table in the form of a ​cross.
    2. (b)Deduce ​the $F_2$ phenotypic ratio/$F_2$ ratio ​showing ​external look as well as ​genotypic /genetic makeup, ratio.

    Cross between a red-flowered and a white-flowered pea plant

    Parents	$F_1$	$F_2$
    A pea plant with red flowers was crossed with a pea plant with white flowers.	30 seeds were sown and all bore plants with red flowers.	44 seeds were sown, 33 plants bore red flowers and 11 plants bore white flowers.

    Show answer

    The cross $RR\times rr$ gives all $Rr$ red in ${\text{F}}_1$; ${\text{F}}_2$ shows red : white $=3:1$ and $RR:Rr:rr=1:2:1$.

    (a) The cross is $RR$ (the red-flowered parent) $\times$ $rr$ (the white-flowered parent). Their gametes are $R$ and $r$, so the whole $F_1$ is $Rr$ and red, which is the 30 red plants recorded. Self-pollinating the $F_1$ gives $Rr\times Rr$, whose offspring are $RR$, $Rr$, $Rr$ and $rr$. (b) $F_2$ phenotypic ratio (external look) $=3$ red $:1$ white, which is the printed $33:11$; $F_2$ genotypic ratio (genetic make-up) $=1RR:2Rr:1rr$, that is $11:22:11$ of the 44 plants.

    Step-by-step solution →
20. Q80 CBSE 2026
    1. (a)​Giving two examples, justify ​the statement that the sex ​of the offspring in all animals is not always ​genetically determined.
    2. (b)Differentiate between the sex chromosomes ​of male and female human ​beings.
    Show answer

    Sex can be set by egg temperature (turtles) or by changing sex (snails); females have $XX$ and males $XY$ sex chromosomes.

    (a) Sex is not always decided by genes. First, in some reptiles, such as many turtles, the temperature at which the fertilised eggs are incubated decides whether the young develop into males or females. Second, in snails an individual can change its sex during its life. In both cases sex is decided or changed after fertilisation, so it is not fixed by the genes inherited. (b) Human females have a perfect (matched) pair of sex chromosomes, both X, so they are XX. Human males have a mismatched pair, one normal-sized X and one short Y, so they are XY. Hence all eggs carry X, while half the sperms carry X and half carry Y.

    Step-by-step solution →