Check out CBSE Class 10 Science HOTs Heredity Set 05 right here. Get complete High Order Thinking Skills (HOTS) questions and answers for Class 10 Science Chapter 08 Heredity. Created for the 2026-27 exam session, these analytical practice problems help learners master core ideas while following guidelines from CBSE, NCERT, and KVS.
Class 10 Science Chapter 08 Heredity HOTS Questions & Answers
Practicing Class 10 Science HOTS Questions is important for scoring high in Science. Use the detailed answers provided below to improve your problem-solving speed and Class 10 exam readiness.
HOTS Questions and Answers for Class 10 Science Chapter 08 Heredity
This page has 20 higher order thinking (HOTs) questions with answers for CBSE Class 10 Science Chapter 8, Heredity, collected from question banks and board-style practice. The questions ask you to reason about dominant and recessive traits, test crosses, blood group and eye colour inheritance, the role of the father in deciding the sex of a child, and the 9 : 3 : 3 : 1 ratio, and each answer is explained step by step.
Very Short Answer HOTS Questions
Question 1. Do genetic combination of mothers play a significant role in determining the sex of a new born?
Answer:
No. Every egg of a woman (XX) carries one X chromosome, so the mother always passes an X.
The sex of the baby is decided by the sperm: an X sperm gives a girl (XX) and a Y sperm gives a boy (XY). Since the father produces both kinds of sperms, the father's chromosome decides the sex.
In simple words: All eggs are X, so only the father's X or Y decides the sex.
Teacher's Note:
1. State that the mother contributes only X.
2. Name the father's chromosomes X and Y.
Question 2. A man with blood group 'A' marries a woman with blood group 'O' and their daughter has blood group O. Is this information enough to tell you which of the traits-blood group A or O is dominant? Why or why not?
Answer:
No, the information is not enough. Let us test both possibilities.
If A is dominant: the father can be heterozygous (A with O), the mother is OO, and the daughter can receive an O allele from each, so she has blood group O.
If O is dominant: the mother (O) can pass an O allele and the father (A, recessive) passes his allele, and the daughter can still show O.
Because the daughter can have blood group O in both cases, the cross alone cannot tell which trait is dominant. We need data on the blood groups of more children or more generations.
In simple words: The daughter's blood group O is possible whichever trait is dominant, so we cannot decide.
Teacher's Note:
1. Check both possibilities one by one.
2. Say that data of more generations is required.
Question 3. Why is recessive trait found in fewer individuals?
Answer:
A recessive trait is expressed only when an individual has two copies of the recessive allele (for example tt).
Individuals with even one dominant allele (TT or Tt) show the dominant trait. So only a smaller fraction of individuals, for example one-fourth in an F2 generation, show the recessive trait.
In simple words: Both parents must pass the recessive allele, so it shows less often.
Teacher's Note:
1. Link the answer to homozygous recessive (tt).
2. Give the one-fourth example from the F2 generation.
Question 4. A plant has two varieties, one with red petals and the other with white petals. When these two varieties are cross-pollinated, all the offsprings have red petals?
(a) Which gene is dominant?
(b) Choose suitable letters to represent the two genes.
Answer:
(a) The gene for red petals is dominant, because only red petals appear in the offspring (the first generation).
(b) Use R for the dominant red gene and r for the recessive white gene. Pure red plant RR × pure white plant rr gives all Rr (red).
In simple words: Red shows in all the offspring, so red is dominant (R) and white is recessive (r).
Teacher's Note:
1. Use a capital letter for the dominant gene.
2. Show the genotype Rr of the offspring.
(b) Both father and mother have the genes Ee in their cells. What colour are their eyes?
(c) Which combination of genes in the zygote will produce children with blue eyes?
(d) Which combinations of genes in the zygote will produce children with brown eyes?
Answer:
(a) A capital letter stands for the dominant gene: (i) recessive gene = e (blue); (ii) dominant gene = E (brown).
(b) Parents with Ee have brown eyes, because E is dominant over e.
(c) Blue eyes need two recessive genes: ee.
(d) Brown eyes: EE and Ee.
| × | E | e |
|---|---|---|
| E | EE Brown | Ee Brown |
| e | Ee Brown | ee Blue |
In simple words: E is dominant, so EE and Ee give brown eyes and only ee gives blue eyes.
Teacher's Note:
1. Use the Punnett square to see all combinations.
2. Only ee gives blue eyes.
Question 6. A bacteria is haploid. Can we use the term homozygous and heterozygous for it? Explain why or why not.
Answer:
No. The terms homozygous and heterozygous describe the two alleles of a gene that a diploid organism has, one from each parent.
A haploid bacterium has only one copy of each gene, so it cannot have two similar or two different alleles for the same trait.
In simple words: One copy of each gene means no pair of alleles, so these terms do not apply.
Teacher's Note:
1. Link the terms to a pair of alleles.
2. Mention haploid means a single set.
Question 7. Why are asexually reproducing individuals capable of showing hereditary features?
Answer:
In asexual reproduction only one parent is involved, no gametes are formed and there is no fertilisation.
The cell copies its DNA and the copy passes directly to the offspring. Therefore the offspring inherit the parent's traits and show the hereditary features, with only small changes due to errors in DNA copying.
In simple words: The offspring get a copy of the parent's DNA, so they inherit its features.
Teacher's Note:
1. Mention DNA copying.
2. Note that variations are very small.
Short Answer HOTS Questions
Question 8. A child questioned his teacher that why do organisms resemble their parents more as compared to grandparents. In which way, will the teacher explain to the child?
Answer:
The teacher will explain that a child receives half of its genes (one set of chromosomes) from each parent.
A grandparent gives only about one-fourth of the genes, as the genes are passed through a parent, and in every generation the mixing of genes (and new changes in DNA) adds more differences. So the child resembles the parents more closely than the grandparents.
In simple words: The child gets 50% of its genes from each parent but only about 25% from each grandparent.
Teacher's Note:
1. Use the fractions one-half and one-fourth.
2. Mention mixing of genes in each generation.
(a) Why are no brown offsprings produced?
(b) If two of the black offsprings mate with each other, what kind of offspring would you expect and in what proportions? Give reason for your answer.
Answer:
Black is the dominant trait (B) and brown is the recessive trait (b). A pure black mouse BB × brown mouse bb gives all Bb.
| × | B | B |
|---|---|---|
| b | Bb Black | Bb Black |
| b | Bb Black | Bb Black |
(a) All the offspring get one B allele, which hides the b allele, so they are black. The brown trait is recessive and is not expressed.
(b) Bb × Bb gives BB, Bb, Bb and bb, as in the second square below.
| × | B | b |
|---|---|---|
| B | BB Black | Bb Black |
| b | Bb Black | bb Brown |
So we expect 3 black : 1 brown (genotypes 1 BB : 2 Bb : 1 bb), because the b alleles of the two parents can come together in one-fourth of the offspring.
In simple words: Black hides brown in the first cross; two black hybrids give 3 black and 1 brown.
Teacher's Note:
1. Name the dominant and the recessive trait first.
2. Show both crosses with Punnett squares.
Question 10. Pure-breed pea plants A are crossed with pure-breed pea plants B. It is found that the plants which look like A do not appear in F1 generation but re-emerge in F2 generation. Which of the plants A and B are: (i) tall and (ii) dwarf? Give reason for your answer.
Answer:
(i) Plants B are tall.
(ii) Plants A are dwarf.
Reason: The trait of A is absent in F1 and reappears in F2, so it is the recessive trait (dwarfness). The trait of B appears in F1, so it is the dominant trait (tallness).
In simple words: The trait that disappears in F1 and returns in F2 is recessive, so A is dwarf.
Teacher's Note:
1. Recessive traits hide in F1.
2. Give the reason with the word recessive.
Question 11. A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or Why not?
Answer:
No, we cannot say. The finding fits both possibilities.
If light eye colour is recessive (ee), two light-eyed parents can only have light-eyed children. If light eye colour is dominant, light-eyed parents (EE or Ee) can also have light-eyed children.
To decide, we need more information, such as the eye colour of children when both parents have dark eyes, and the data of at least three generations.
In simple words: Light-eyed children from light-eyed parents is possible either way, so the data is not enough.
Teacher's Note:
1. Test both cases, dominant and recessive.
2. Say what extra data is needed.
Question 12. A woman has only daughters. Analyse the situation genetically and provide a suitable explanation.
Answer:
A woman has two X chromosomes, so all her eggs carry X. Her husband (XY) makes two kinds of sperms, half with X and half with Y.
A daughter is born when an X sperm fertilises the egg. For each child the chance of an X sperm is 50%, so it is quite possible by chance that all the children are girls.
The sex of the children depends on the father's sperm, so the woman is not responsible for having only daughters.
In simple words: It is a matter of chance which sperm fertilises the egg; the mother cannot decide.
Teacher's Note:
1. State that eggs always carry X.
2. Remove blame from the mother; the father's gamete decides.
Answer:
Yes, by a test cross: cross the individual with a plant having the recessive trait (tt).
If it is homozygous (TT): TT × tt gives all Tt, so all the offspring are tall.
| × | T | T |
|---|---|---|
| t | Tt | Tt |
| t | Tt | Tt |
If it is heterozygous (Tt): Tt × tt gives Tt and tt in equal numbers, so half the offspring are tall and half are dwarf.
| × | T | t |
|---|---|---|
| t | Tt Tall | tt Dwarf |
| t | Tt Tall | tt Dwarf |
A dwarf offspring shows that the tall parent was heterozygous.
In simple words: Cross it with a recessive plant: all tall offspring means TT; some dwarf offspring means Tt.
Teacher's Note:
1. The recessive parent is called the test parent.
2. Show both possible crosses.
Question 14. Sahil has black hair, his mother also has black hair but his father has brown hair. How can you explain on the basis of transmission of genes for hair colour from the mother and father?
Note: It is given that black hair colour (B) is dominant over brown hair colour (b).
Answer:
The father has brown hair, so his genotype is bb and he can pass only b.
Sahil has black hair, so he must have the dominant allele B. He got B from his mother, so her genotype is BB or Bb.
Sahil's genotype is therefore Bb: B from the mother and b from the father. The b allele is hidden by the dominant B, so he has black hair.
In simple words: Sahil gets B from his mother and b from his father, and B hides b.
Teacher's Note:
1. Start from the recessive parent, the father (bb).
2. Write the genotype of Sahil as Bb.
(A) How many short pea plants will be found in the progeny in the F1 generation?
(B) Plants that reproduce through asexual reproduction give rise to similar and not identical offsprings. What could be the most likely reason for the above statement to be true?
Answer:
(A) TT × Tt: the gametes are T, T and T, t.
| × | T | T |
|---|---|---|
| T | TT | TT |
| t | Tt | Tt |
The offspring are TT or Tt, so all of them are tall. The number of short plants in F1 is 0 (50% TT and 50% Tt).
(B) DNA copying during asexual reproduction is not 100% accurate. Small errors (mutations) in the new copies of DNA cause minor differences, so the offspring are similar but not identical to the parent.
In simple words: No short plant can appear, because every offspring gets at least one T.
Teacher's Note:
1. A short plant needs the genotype tt.
2. Link the asexual variation to errors in DNA copying.
Long Answer HOTS Questions
(A) Assuming that the gene for plant height obeys the Mendel's laws of inheritance, indicate the genotypes and phenotypes of all the possible parent pairs that could have dwarf offspring.
(B) Using any of the parent pairs mentioned by you in (A), perform a cross to show the genotypes of the offspring that might arise in the next generation.
Answer:
(A) A dwarf offspring (tt) must get one t from each parent. The possible parent pairs are:
| Parent pair (genotypes) | Phenotypes |
|---|---|
| Tt × Tt | Tall × Tall |
| Tt × tt | Tall × Dwarf |
| tt × tt | Dwarf × Dwarf |
(B) Cross Tt × Tt:
| × | T | t |
|---|---|---|
| T | TT Tall | Tt Tall |
| t | Tt Tall | tt Dwarf |
Genotypes of the offspring: TT, Tt, Tt, tt in the ratio 1 : 2 : 1, that is 75% tall and 25% dwarf. (Tt × tt would give 50% tall Tt and 50% dwarf tt; tt × tt gives all dwarf.)
In simple words: Both parents must carry at least one t allele to have a dwarf (tt) child.
Teacher's Note:
1. List all three parent pairs in part (A).
2. Draw the Punnett square for one pair.
(A) Describe the steps Mendel took to ensure that the seeds he planted would only produce dwarf pea plants.
(B) The allele for tallness in pea plants is T. The allele for dwarfism in pea plants is t. Draw a genetic diagram to show the genotypes of the offspring of a heterozygous tall pea plant and a homozygous dwarf pea plant.
Answer:
(A) Dwarfness is recessive, so every dwarf plant is tt (pure). Mendel took these steps:
1. He selected only the dwarf plants.
2. He bred them together (or let them self-pollinate) and kept the flowers covered so that insects could not bring pollen from tall plants.
3. From the offspring he again selected only dwarf plants.
4. He repeated the process for many generations until only dwarf plants were obtained. These seeds always give dwarf plants.
(B) Heterozygous tall (Tt) × homozygous dwarf (tt):
| × | T | t |
|---|---|---|
| t | Tt Tall | tt Dwarf |
| t | Tt Tall | tt Dwarf |
Offspring: Tt, Tt, tt, tt, so 50% tall (Tt) and 50% dwarf (tt), a 1 : 1 ratio.
In simple words: Keep breeding only dwarf plants (and keep insects away) to get a pure dwarf line.
Teacher's Note:
1. Mention selection of dwarf plants and repeating the cycle.
2. Check the gametes: Tt gives T and t; tt gives only t.
(A) What should be the genotype of the parent plants to give violet and white flowers in equal proportions? Justify your answer.
(B) Simi self-pollinates two heterozygous plants having violet flowers and obtains 100 plants in the F1 generation.
(i) What would be the number of violet and white flowers respectively in F1 generation?
(ii) What would be the number of violet and white flowers respectively in F1 generation if she crosses one homozygous plant with violet flowers and the other heterozygous plant with violet flowers?
Answer:
Violet is dominant (V) and white is recessive (v), because the cross gave all violet flowers.
(A) The parents should be Vv (violet) × vv (white). This test cross gives Vv and vv in equal numbers, that is, 50% violet and 50% white.
| × | V | v |
|---|---|---|
| v | Vv Violet | vv White |
| v | Vv Violet | vv White |
(B)(i) Vv × Vv gives VV, Vv, Vv, vv, which is 3 violet : 1 white. Out of 100 plants: 75 violet and 25 white.
| × | V | v |
|---|---|---|
| V | VV Violet | Vv Violet |
| v | Vv Violet | vv White |
(ii) VV × Vv gives VV and Vv only, so all the plants are violet: 100 violet and 0 white.
In simple words: A heterozygous plant crossed with a white plant gives equal violet and white; two hybrids give 3 : 1.
Teacher's Note:
1. Use the dominant allele V and the recessive allele v.
2. Convert the ratio to numbers out of 100.
Question 19. Mendel used a number of contrasting visible characters of garden peas. He also explained that each trait is determined by a pair of factors.
(A) (i) Which term has replaced factors for the representation in a trait?
(ii) Where are these factors found in gametes?
(B) What is the number of chromosomes present in an ovum and fertilised egg?
(C) What do the following stand for?
(i) XX
(ii) XY
Which one of them produces heterogametes and which contributes towards the sex determination in the human progeny?
Answer:
(A)(i) The term genes (the different forms of a gene are called alleles) has replaced Mendel's factors.
(ii) The factors (genes) are located on the chromosomes. A gamete carries only one of the two factors of each pair, since the two separate when gametes are formed.
(B) An ovum has 23 chromosomes and a fertilised egg (zygote) has 46 chromosomes, as 23 + 23 = 46.
(C)(i) XX is the sex chromosome pair of a human female. (ii) XY is the sex chromosome pair of a human male.
XY (the male) produces two kinds of gametes (heterogametes), sperms with X and sperms with Y, and the male gamete contributes towards determining the sex of the child. A female (XX) produces only one kind of egg.
In simple words: Factors are genes on chromosomes; the father (XY) makes two kinds of sperms and decides the child's sex.
Teacher's Note:
1. Give 23 and 46 for the chromosome numbers.
2. Use the word heterogametic for the male.
Question 20. Mendel's dihybrid cross resulted in 4 types of individuals. What was the rationale behind Mendel deducing independent assortment from the observation?
Answer:
In the dihybrid cross of round yellow (RRYY) and wrinkled green (rryy) plants, the F1 plants were all round yellow (RrYy). On selfing, the F2 plants were of four types in the ratio 9 : 3 : 3 : 1: round yellow, round green, wrinkled yellow and wrinkled green.
If the two traits were inherited together, only the parental types (round yellow and wrinkled green) would appear, in the ratio 3 : 1. The two new combinations, round green and wrinkled yellow, appeared, which means that seed shape and seed colour separate and recombine freely.
Also, each trait taken separately gives the 3 : 1 ratio (12 round : 4 wrinkled and 12 yellow : 4 green out of 16), and 3 : 1 × 3 : 1 gives 9 : 3 : 3 : 1. Therefore the factors of the two traits are inherited independently of each other.
In simple words: The appearance of new combinations, with a 9 : 3 : 3 : 1 ratio, shows that the two traits are inherited independently.
Teacher's Note:
1. Mention the two new combinations.
2. Use the 3 : 1 ratio of each trait to explain 9 : 3 : 3 : 1.
Free study material for Science
Higher Order Thinking Skills (HOTS) for Class 10 Science Chapter 08 Heredity
About Chapter 08 Heredity HOTS for Class 10 Science
Strengthen your preparation for Class 10 Science examinations with specialized Chapter 08 Heredity HOTS worksheets. These structured problems are curated to match official CBSE guidelines and help tackle difficult board-level questions.
Step-by-Step Answers for Chapter 08 Heredity
All analytical exercises for Chapter 08 Heredity are structured around standard CBSE textbooks. Cross-reference your answers with our professional step-by-step guides to ensure complete conceptual accuracy.
Complete Your Chapter Revision
Wrap up your chapter revision by testing your speed with our interactive Science quizzes. Access all curriculum-aligned study resources online without any restrictions.
FAQs
You can download the teacher-verified PDF for CBSE Class 10 Science HOTs Heredity Set 05 from StudiesToday.com. These questions have been prepared for Class 10 Science to help students learn high-level application and analytical skills required for the 2026-27 exams.
In the 2026 pattern, 50% of the marks are for competency-based questions. Our CBSE Class 10 Science HOTs Heredity Set 05 are to apply basic theory to real-world to help Class 10 students to solve case studies and assertion-reasoning questions in Science.
Unlike direct questions that test memory, CBSE Class 10 Science HOTs Heredity Set 05 require out-of-the-box thinking as Class 10 Science HOTS questions focus on understanding data and identifying logical errors.
After reading all conceots in Science, practice CBSE Class 10 Science HOTs Heredity Set 05 by breaking down the problem into smaller logical steps.
Yes, we provide detailed, step-by-step solutions for CBSE Class 10 Science HOTs Heredity Set 05. These solutions highlight the analytical reasoning and logical steps to help students prepare as per CBSE marking scheme.