Previous year questions: Principles of Inheritance and Variation

The questions below come from actual CBSE board biology papers. Where the original printed options, all four are reproduced so the item looks like the paper. Answers use the key points of the CBSE marking schemes: genotype symbols, the ratio, the law and one line of reasoning. Each item carries the marks the paper printed, and report.md lists the source PDF for every question.

Q1 (CBSE Board 2023, 1 mark)

Which one of the following processes results in the production of recombinants in future generations? (i) Mutation (ii) Independent assortment during meiosis I (iii) Independent assortment during meiosis II (iv) Crossing over of bivalents (a) (iv) only (b) (ii) and (iv) (c) (i), (ii) and (iii) (d) (i), (ii), (iii) and (iv)

Answer: Option (b) (ii) and (iv). Recombinant chromosomes are produced when homologous chromosomes exchange segments during crossing over of bivalents and when the two chromosome pairs orient independently at meiosis I. Mutation creates new alleles, not new combinations of existing alleles.

Q2 (CBSE Board 2023, 1 mark)

Choose the set of sex determining chromosomes that indicates the correct sex of the respective organism. (a) Homozygous sex chromosomes (ZZ) determine female sex in birds (b) XO type of sex chromosomes determine male sex in grasshoppers (c) XXY condition in humans, as found in Turner syndrome, determines male sex (d) Homozygous sex chromosomes (XX) produce male sex in Drosophila

Answer: Option (b). In grasshoppers females are XX and males are XO, so the XO condition makes a male. In birds ZZ is male and ZW is female, so option (a) is wrong. XXY is Klinefelter syndrome, not Turner syndrome, so option (c) is wrong, and XX makes a female in Drosophila, so option (d) is wrong.

Q3 (CBSE Board 2023, 5 marks)

A true breeding tall pea plant with round seeds is crossed with a recessive dwarf pea plant having wrinkled seeds. Work out the cross up to the F2 generation giving the phenotypic ratios of the F1 and the F2 generation. State the Mendelian principle that can be derived with the help of such a cross.

Answer: Let tall be T, dwarf t, round R and wrinkled r. The parents are TTRR and ttrr, so the gametes are TR and tr and the F1 is TtRr, tall with round seeds, that is 100 percent of the F1. The F1 plants make four kinds of gametes, TR, Tr, tR and tr. The Punnett square of these gametes gives the F2 phenotypic ratio 9 tall round : 3 tall wrinkled : 3 dwarf round : 1 dwarf wrinkled. The cross proves the law of independent assortment, because new combinations of stem height and seed shape appear.

Q4 (CBSE Board 2023, 5 marks)

A homozygous tall pea plant with green seeds is crossed with a homozygous dwarf pea plant with yellow seeds. (i) Write the possible phenotype and genotype of the F1 generation. (ii) State the laws of Mendel that are proved true by the F1 generation. (iii) Mention the F2 phenotypic ratio along with the possible phenotypes. (iv) Write the genotypes of the male and female gametes produced by the F1 progeny.

Answer: Let T be tall and t dwarf, G green and g yellow. The parents are TTGG and ttgg, so F1 is TtGg, that is all tall with green seeds. The F1 generation proves the law of dominance, because only the dominant traits appear, and the law of segregation, because the F1 is heterozygous for both pairs. The F2 phenotypic ratio is 9 tall green : 3 tall yellow : 3 dwarf green : 1 dwarf yellow. The F1 plant produces four kinds of gametes, TG, Tg, tG and tg, in equal numbers.

Q5 (CBSE Board 2024, 1 mark)

A single gene that controls the expression of more than one trait is said to show (A) Multiple allelism (B) Polygenic inheritance (C) Incomplete dominance (D) Pleiotropism

Answer: Option (D) Pleiotropism. When one gene affects several traits, the phenomenon is pleiotropy. Multiple allelism means more than two alleles of a gene, polygenic inheritance means many genes control one trait, and incomplete dominance is an intermediate heterozygote.

Q6 (CBSE Board 2024, 1 mark)

A person with trisomy of the 21st chromosome shows (i) Furrowed tongue (ii) Characteristic palm crease (iii) Rudimentary ovaries (iv) Gynaecomastia Select the correct option: (A) (ii) and (iv) (B) (i), (ii) and (iv) (C) (ii) and (iii) (D) (i) and (ii)

Answer: Option (D) (i) and (ii). Down syndrome is trisomy 21, and its features include a furrowed tongue and a characteristic palm crease, along with retarded physical and mental development. Rudimentary ovaries belong to Turner syndrome, and gynaecomastia is seen in Klinefelter syndrome.

Q7 (CBSE Board 2024, 1 mark)

Assertion (A): Linked genes do not show a dihybrid F2 ratio of 9 : 3 : 3 : 1. Reason (R): Linked genes do not undergo independent assortment. (a) Both A and R are true and R is the correct explanation of A (b) Both A and R are true but R is not the correct explanation of A (c) A is true but R is false (d) A is false but R is true

Answer: Option (a). Linked genes are carried on the same chromosome, so they tend to stay together and do not assort independently. That is why the parental combinations dominate and the 9 : 3 : 3 : 1 ratio breaks down. The reason explains the assertion.

Q8 (CBSE Board 2024, 5 marks)

Work out a dihybrid cross up to the F2 generation between pea plants bearing violet coloured axial flowers and white coloured terminal flowers using Punnett square. Give their F2 phenotypic ratio. State the law of inheritance derived from such a cross.

Answer: The parents are homozygous violet axial VVAA and white terminal vvaa. The F1 is VvAa, all violet with axial flowers. The F1 makes four gametes, VA, Va, vA and va, and the Punnett square of 16 boxes gives 9 violet axial : 3 violet terminal : 3 white axial : 1 white terminal. The law derived is the law of independent assortment, because the two pairs of traits combine in new ways in the F2.

Q9 (CBSE Board 2025, 1 mark)

In a pedigree chart, a cross between a normal couple resulted in a son who was haemophilic and a normal daughter. When the daughter married a normal man, their son was also haemophilic. Choose the option that indicates the correct inheritance of the trait. (A) Autosome linked dominant trait (B) Sex linked dominant trait (C) Autosomal recessive trait (D) Sex linked recessive trait

Answer: Option (D) Sex linked recessive trait. Haemophilia is carried on the X chromosome, so a carrier mother can have an affected son while both parents look normal. The affected son of the normal daughter confirms that she is a carrier, which fits X linked recessive inheritance.

Q10 (CBSE Board 2025, 1 mark)

In a pedigree chart, two unaffected individuals have a child with the trait. What is the most likely mode of inheritance for this trait? (A) Autosomal dominant (B) Autosomal recessive (C) X linked dominant (D) X linked recessive (inference from the given data)

Answer: Option (B) Autosomal recessive. For unaffected parents to have an affected child, both parents must carry a hidden recessive allele. If the trait were dominant, at least one parent would have to show it. Autosomal recessive is the usual explanation when both sexes can be affected equally.

How the marking scheme awards marks here

For a cross question, the marking scheme gives marks separately for the parental genotypes, the gametes, the Punnett square, the F1 and the F2 ratios, and the name of the law. Missing the square costs more than missing the final ratio, because the square is the working. For an MCQ or assertion-reason item, the option letter alone is not enough when the question carries a reason, so add one line of justification. For a pedigree item, state the mode and the reason in separate sentences: recessive if unaffected parents have an affected child, X linked recessive if mainly males are affected through carrier mothers.

Numbers that commonly appear

Monohybrid F2: 3 : 1; genotypic 1 : 2 : 1. Dihybrid F2: 9 : 3 : 3 : 1. Incomplete dominance F2: 1 : 2 : 1. ABO groups: three alleles A, B, O with A and B codominant. Down syndrome: 47 chromosomes with trisomy 21. Klinefelter: 47, XXY. Turner: 45, X0. Haemophilia and colour blindness: X linked recessive. Sickle cell anaemia, thalassemia and phenylketonuria: autosomal recessive. These are the numbers and names that repeat across papers, so keep them on one page for revision.