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Principles of Inheritance and Variation

Principles of Inheritance and VariationNEET Botany · Class 12 · NCERT Chapter 2

59 NEET previous-year questions on Principles of Inheritance and Variation, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.

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Principles of Inheritance and Variation (59)

A

Recessive epistasis

B

Dominant epistasis

C

Complementary genes

D

Inhibitory genes

Solution

Fruit colour in squash is an example of dominant epistasis, where one gene masks the expression of another gene. NCERT XII chapter Principles of Inheritance and Variation explains that in squash, the white fruit colour gene is epistatic over the yellow or green fruit colour genes, so option (b) is correct.

A

25 %

B

0 %

C

50 %

D

75 %

Solution

The man has a genotype of (since his father was colour blind, ) and the woman has a genotype of (since her mother was colour blind, ). The cross results in 50% of male children being (colour blind), so option (c) is correct.

A

0.4

B

0.5

C

0.6

D

0.7

Solution

The frequency of allele is calculated as . Substituting the values, , so option (c) is correct.

A

has 45 chromosomes with XO

B

has one additional X chromosome

C

exhibits male characters

D

is able to produce children with normal husband

Solution

A human female with Turner's syndrome has 45 chromosomes with an XO genotype. NCERT XII chapter Principles of Inheritance and Variation states that Turner's syndrome results from the absence of one X chromosome, leading to the 45,X karyotype, so option (a) is correct.

A

Gaseous nutrient cycle Carbon and Nitrogen Sedimentary nutrient cycle Sulphur and Phosphorus

B

Gaseous nutrient cycle Carbon and Sulphur Sedimentary nutrient cycle Nitrogen and Phosphorus

C

Gaseous nutrient cycle Nitrogen and Sulphur Sedimentary nutrient cycle Carbon and phosphorus

D

Gaseous nutrient cycle Sulphur and Phosphorus Sedimentary nutrient cycle Carbon and Nitrogen

Solution

Sulphur and phosphorus are found on earth crust in the form of rocks

A

0.5

B

1

C

Nil

D

0.25

Solution

Colour blindness is an X-linked recessive trait. The man is and the woman is . Their daughter will be . If the daughter marries a normal man (), their son has a 50% chance of being colour blind (). Thus, the probability of their grandson being colour blind is 0.5, so option (a) is correct.

A

T.H. Morgan

B

T. Boveri

C

G. Mendel

D

W. Sutton

Solution

The term "linkage" was coined by T.H. Morgan.

A

(b), (a), (d), (c)

B

(b), (a), (c), (d)

C

(a), (b), (c), (d)

D

(b), (c), (d), (a)

Solution

The sequence of event during meiosis are

A

Is expressed only primitive plants

B

Is a gene evolved during Pliocene

C

Controls a trait only in combination with another gene

D

Controls multiple traits in an individual

Solution

The gene which controls multiple traits in an individual.

A

Seed colour

B

Pod length

C

Seed shape

D

Flower position

Solution

Mendel did not selected Pod length as a character for study

A

Autosomal dominant

B

X-linked recessive

C

Autosomal recessive www.vedantu.com 9

D

X-linked dominant

Solution

The given pedigree represents inheritance of Autosomal recessive trait. `

A

Cell wall is absent in Animalia

B

Protista have photosynthetic and heterotrophic modes of nutrition

C

Some fungi are edible

D

Nuclear membrane is present in Monera

Solution

The members of kingdom-Monera are prokaryotes they lack nuclear membrane.

A

Recessive gene disorder

B

X – linked recessive gene disorder

C

Chromosomal disorder

D

Dominant gene disorder

Solution

Genes related with hemophilia are always present on X chromosome and it is a recessive gene disorder as it expresses itself in females when it comes a homozygous condition. It causes a defect in the clotting factor formation, thus a simple cut can bleed continuously leading to even death. Thus, it is also known as ‘Bleeders’ disease or ‘Royal Disease’ as Queen Victoria is a carrier for this disease.

A

1 : 2 : 1 :: Tall homozygous : Tall heterozygous : Dwarf

B

1 : 2 : 1 :: Tall heterozygous : Tall homozygous : Dwarf www.vedantu.com 55

C

3 : 1 :: Tall : Dwarf

D

3 : 1 :: Dwarf : Tall

Solution

Phenotypic ratio : tall : dwarf Genotypic ratio : TT : Tt : tt 1 : 2 : 1

A

(a) (b) (c) (d) (ii) (i) (iv) (iii)

B

(a) (b) (c) (d) (ii) (iii) (iv) (i)

C

(a) (b) (c) (d) (iv) (i) (ii) (iii)

D

(a) (b) (c) (d) (iv) (iii) (i) (ii)

Solution

www.vedantu.com 58 Dominance – In a heterozygous organism only one allele expresses itself. Codominance – In a heterozygous organism both alleles express themselves equally. Pleiotropy – A single gene influences many characters. Polygenic Inheritance – Many genes govern a single character.

A

The two genes are located on two different chromosomes

B

Chromosomes failed to separate during meiosis

C

The two genes are linked and present on the same chromosome

D

Both of the characters are controlled by more than one gene

Solution

If a plant genotype Aa Bb is crossed with aabb then Independent Assortment would result in production of 4 type of offsprings in equal proportion. Aa Bb – gametes AB Ab aB ab Aa bb – Gametes ab ab ab ab Offspring according to independent assortment AaBb Aabb aaBb aabb 1 : 1 : 1 : 1 (Parental) (Recombinants) (Parental) Since parental percentage is more than recombinants it is due to linkage between genes A and B.

A

Down's syndrome

B

Klinefelter's syndrome

C

Turner's syndrome

D

Sickle cell anemia

Solution

Down’s syndrome is caused by non-disjunction of 21 st chromosome.

A

3 genotypes ; 3 phenotypes

B

3 genotypes ; 4 phenotypes

C

4 genotypes ; 3 phenotypes

D

4 genotypes ; 4 phenotypes

Solution

AB AII Ii Husband Wife× I A I B I A I A I A I A I B iI i A Ii B + Number of genotypes = 4 Number of phenotypes = 3 I AIA and IAi = A IAIB = AB IBi = B

A

Stem – Tall or Dwarf

B

Trichomes – Glandular or non-glandular

C

Seed – Green or Yellow

D

Pod – Inflated or Constricted

Solution

During his experiments Mendel studied seven characters. Nature of trichomes i.e., glandular or non-glandular was not considered by Mendel. www.vedantu.com 27

A

1856 - 1863

B

1840 - 1850

C

1857 - 1869

D

1870 - 1877

Solution

Mendel conducted hybridization experiments on Pea plant for 7 years between 1856 to 1863 and his data was published in 1865 (according to NCERT).

A

Only grandchildren

B

Only sons

C

Only daughters

D

Both sons and daughters

Solution

An X-linked condition on one of the woman's X chromosomes can be inherited by both her sons and daughters. Sons receive their X chromosome from their mother, while daughters receive one X chromosome from each parent, so option (d) is correct.

A

b, d and e

B

a, b and c

C

b, c and e

D

a, c and e

Solution

The inheritance of blood groups in humans is characterized by dominance, co-dominance, and multiple alleles. These concepts are explained in NCERT XII chapter Principles of Inheritance and Variation, making option (b) correct.

A

One fuses with egg and other fuses with central cell nuclei.

B

One fuses with egg, other(s) degenerate(s) in the synergid.

C

All fuse with the egg.

D

one fuses with egg. other(s) fuse(s) with synergid nucleus.

Solution

Ref. XII NCERT Chapter- 02, Page No. 34

A

Lippes Loop, Multiload 375

B

Vaults, LNG-20

C

Multiload 375, Progestasert

D

Progestasert, LNG-20

Solution

Factual NCERT

A

0.16(AA); 0.36 (Aa); 0.48 (aa)

B

0.36(AA); 0.48 (Aa); 0.16 (aa)

C

0.16(AA); 0.24 (Aa); 0.36 (aa)

D

0.16(AA); 0.48 (Aa); 0.36 (aa)

Solution

Given (𝑝)=0.4 (𝑝+𝑞=1) Hence (𝑞)=0.6 To find Homozygous dominant = 𝑝2 =(0.4)2=0.16 Heterozygous = 2pq = 2 × 0.4 × 0.6 = 0.48 Homozygous recessive = 𝑞2 = (0.6)2 = 0.36

A

(ii) (iii) (iv) (iv)

B

(i) (ii) (iii) (iv)

C

(iii) (ii) (i) (iv)

D

(ii) (i) (iii) (iv)

Solution

Ref. XI NCERT Chapter- 02, Page No. 22

A

Ligases - Join the two DNA molecules

B

Polymerases - Break the DNA into fragments

C

Nucleases - Separate the two strands of DNA

D

Exonucleases - Make cuts at specific positions within DNA

Solution

Ligases join two DNA molecules by forming a phosphodiester bond between the 3' hydroxyl and 5' phosphate ends. NCERT XII chapter Principles of Inheritance and Variation describes ligases as essential for DNA replication and repair, so option (a) is correct.

A

Mendel

B

Sutton

C

Boveri

D

Morgan

Solution

Thomas Hunt Morgan provided experimental verification of the chromosomal theory of inheritance through his work with Drosophila melanogaster. NCERT XII chapter Principles of Inheritance and Variation credits Morgan for this contribution, so option (d) is correct.

A

The gene (I) has three alleles.

B

A person will have only two of the three alleles.

C

When I A and I B are present together, they express same type of sugar.

D

Allele ‘i’ does not produce any sugar.

Solution

The gene ‘I’ has three alleles: , , and . A person will have only two of these alleles. When and are present together, they express different types of sugars, not the same type. The allele does not produce any sugar. Therefore, option (c) is incorrect.

A

Haemophilia - Y linked

B

Phenylketonuria - Autosomal dominant trait

C

Sickle cell anaemia - Autosomal recessive trait, chromosome-11

D

Thalassemia - X linked

Solution

- Haemophilia is X-linked recessive.
- Phenylketonuria is autosomal recessive.
- Sickle cell anaemia is an autosomal recessive trait linked to chromosome 11.
- Thalassemia is autosomal recessive, not X-linked.

Therefore, option (c) is correct.

A

4

B

2

C

14

D

8

Solution

Mendel selected 14 true breeding pea plant varieties as pairs, each differing in one character with contrasting traits. This is detailed in NCERT XII chapter Principles of Inheritance and Variation, so option (c) is correct.

A

Bullet square

B

Punch square

C

Punnett square

D

Net square

Solution

The production of gametes, formation of zygotes, and the F1 and F2 generations are illustrated using a Punnett square. This diagram helps predict the genotypic and phenotypic ratios of offspring, so option (c) is correct.

A

Mature sieve tube elements possess a conspicuous nucleus and usual cytoplasmic organelles.

B

Microbodies are present both in plant and animal cells.

C

The perinuclear space forms a barrier between the materials present inside the nucleus and that of the cytoplasm.

D

Nuclear pores act as passages for proteins and RNA molecules in both directions between nucleus and cytoplasm.

Solution

Mature sieve tube elements lack a nucleus and most cytoplasmic organelles, making option (a) incorrect. NCERT XI chapter Cell: The Unit of Life describes sieve tube elements as enucleated and having a minimal cytoplasm.

A

Kinetin

B

Infrared rays

C

Gamma rays

D

Zeatin

Solution

Mutations in plant cells can be induced by gamma rays, which are ionizing radiation known to cause genetic changes. NCERT XII chapter Principles of Inheritance and Variation discusses the role of physical mutagens like gamma rays in inducing mutations, so option (c) is correct.

A

Pyramid of biomass in sea is generally inverted.

B

Pyramid of biomass in sea is generally upright.

C

Pyramid of energy is always upright.

D

Pyramid of numbers in a grassland ecosystem is upright.

Solution

The pyramid of biomass in the sea is generally inverted because the biomass of consumers exceeds that of producers. NCERT XII chapter Ecosystem explains that in aquatic ecosystems, the biomass of phytoplankton is less than that of zooplankton, making option (b) incorrect.

A

During aerobic respiration, role of oxygen is limited to the terminal stage.

B

In ETC (Electron Transport Chain), one molecule of NADH + H + gives rise to 2 ATP molecules, and one FADH 2 gives rise to 3 ATP molecules.

C

ATP is synthesized through complex V.

D

Oxidation-reduction reactions produce proton gradient in respiration.

Solution

In the Electron Transport Chain (ETC), one molecule of NADH + H gives rise to 3 ATP molecules, and one FADH gives rise to 2 ATP molecules. This is a key concept in cellular respiration as described in NCERT XI chapter on Respiration in Plants, making option (b) incorrect.

A

In capping, methyl guanosine triphosphate is added to the 3 9 end of hnRNA.

B

RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.

C

The coding strand in a transcription unit is copied to an mRNA.

D

Split gene arrangement is characteristic of prokaryotes.

Solution

RNA polymerase binds with the Rho factor to terminate transcription in bacteria. This is a key mechanism of transcription termination in prokaryotes, as described in NCERT XII chapter Principles of Inheritance and Variation, making option (b) correct.

A

Large colorless empty - Subsidiary cells cells in the epidermis of grass leaves

B

In dicot leaves, vascular - Conjunctive bundles are surrounded tissue by large thick-walled cells

C

Cells of medullary rays - Interfascicular that form part of cambium cambial ring

D

Loose parenchyma cells - Spongy rupturing the epidermis parenchyma and forming a lens- shaped opening in bark

Solution

- Large colorless empty cells in the epidermis of grass leaves are subsidiary cells, which assist in the opening and closing of stomata.
- Option (a) is correct as it accurately describes the function and location of subsidiary cells.

A

Fusion of two cells is called Karyogamy.

B

Fusion of protoplasms between two motile on non-motile gametes is called plasmogamy.

C

Organisms that depend on living plants are called saprophytes.

D

Some of the organisms can fix atmospheric nitrogen in specialized cells called sheath cells.

Solution

Fusion of protoplasms between two motile or non-motile gametes is called plasmogamy, as stated in NCERT XII chapter Principles of Inheritance and Variation. Option (b) is correct.

A

Both ATP and NADPH + H + are synthesized during non-cyclic photophosphorylation.

B

Stroma lamellae have PS I only and lack NADP reductase.

C

Grana lamellae have both PS I and PS II.

D

Cyclic photophosphorylation involves both PS I and PS II.

Solution

Cyclic photophosphorylation involves only PS I, not PS II. NCERT XI chapter Photosynthesis in Higher Plants states that cyclic photophosphorylation generates ATP but not NADPH, and it occurs without the involvement of PS II, so option (d) is incorrect.

A

50%

B

75%

C

25%

D

100%

Solution

In a cross between two heterozygous individuals (Aa x Aa), the expected phenotypic ratio is 3:1, with 25% of the progeny being homozygous recessive (aa) and thus diseased. NCERT XII chapter Principles of Inheritance and Variation explains this using a Punnett square, so option (c) is correct.

A

Alkaloids - Codeine

B

Toxin - Abrin

C

Lectins - Concanavalin A

D

Drugs - Ricin

Solution

Ricin is a toxin, not a drug. Alkaloids include codeine, toxins include abrin, and lectins include concanavalin A, as per NCERT XII Principles of Inheritance and Variation. Therefore, option (d) is incorrect.

A

Drosophila

B

Birds

C

Grasshoppers

D

Monkeys

Solution

Grasshopper is an example of XO type of sex determination in which the males have only one X-chromosome besides the autosomes, whereas females have a pair of X-chromosomes.

A

Both Statement I and Statement II are correct

B

Both Statement I and Statement II are incorrect

C

Statement I is correct but Statement II is incorrect

D

Statement I is incorrect but Statement II is correct

Solution

Gregor J. Mendel, conducted hybridisation experiments on garden peas and selected 14 true breeding pea plant varieties (seven contrasting traits). Contrasting traits studied were smooth or wrinkled seeds, yellow or green seeds, inflated on constricted pods, green or yellow pods, tall or dwarf plants, violet or white flowers and axial or terminal flower positions. SECTION-B

A

Sickle cell anaemia

B

Myotonic dystrophy

C

Haemophilia

D

Thalessemia

Solution

Haemophilia is a X-linked recessive disorder. Thalassemia is an autosomal recessive disorder. Sickle cell anaemia is an autosomal recessive disorder. Myotonic dystrophy is an autosomal dominant disorder i.e. it occurs due to the presence of autosomal linked dominant trait.

A

Both (A) and (R) are correct and (R) is the correct explanation of (A)

B

Both (A) and (R) are correct but (R) is not the correct explanation of (A)

C

(A) is correct but (R) is not correct

D

(A) is not correct but (R) is correct

Solution

Closely located genes do not show independent assortment. Mendel’s law of independent assortment holds good for those genes which are located on different chromosomes.

A

25%

B

50%

C

75%

D

100%

Solution

If mother of man is colourblind, then man will also be colourblind as colour blindness is a X -linked recessive trait and shows criss-cross inheritance. - 63 - NEET (UG)-2022 (Code- Q1)

A

a, d, b, c

B

d, b, a, c

C

a, b, c, d

D

a, c, b, d

Solution

1% recombination frequency = 1 centi Morgan To place the genes on a linear chromosome, decreasing order of recombination frequency will be considered.

A

more than two genes affecting a single character.

B

presence of several alleles of a single gene controlling a single crossover.

C

presence of two alleles, each of the two genes controlling a single trait.

D

a single gene affecting multiple phenotypic expression.

Solution

Pleiotropism is the phenomenon where a single gene affects multiple phenotypic traits. NCERT XII chapter Principles of Inheritance and Variation defines pleiotropism as the ability of a single gene to have multiple effects, so option (d) is correct.

A

Henking

B

Thomas Hunt Morgan

C

Sutton and Boveri

D

Alfred Sturtevant

Solution

Alfred Sturtevant used the frequency of recombination between gene pairs to map their positions on chromosomes, pioneering the concept of genetic mapping. NCERT XII chapter Principles of Inheritance and Variation credits Sturtevant with this method, so option (d) is correct.

A

A-I, B-II, C-III, D-IV

B

A-II, B-I, C-III, D-IV

C

A-III, B-IV, C-I, D-II

D

A-IV, B-III, C-II, D-I

Solution

A. Two or more alternative forms of gene are called alleles. B. Cross of F1 progeny with homozygous recessive parent is a test cross. C. Cross of F1 progeny with any of the parents is a back cross. D. Number of chromosome sets in plant is called ploidy.

A

BB

B

bb

C

Bb

D

BB/Bb

Solution

To determine the genotype of a black seed colour at F2, the black seed from F2 is crossed with the white seed colour. This is called a test cross. ∴ To determine the genotype of (BB/Bb) black seed we need to cross them with white seed i.e. bb.

A

A, B and C only

B

A, C, D and E only

C

B, C and D only

D

A, B, C, D and E

Solution

According to Law of Dominance (1) Characters are controlled by discrete units called factors (2) Factors occur in pairs (3) In a dissimilar pair of factors one member of the pair dominates (dominant) the other recessive The law of dominance is used to explain the expression of only one of the parental characters in a monohybrid cross. Law of segregation is based on the fact that the allel es do not show any expression and both the characters are recovered as such in F2 generation

A

Only red flowered plants

B

Red flowered as well as pink flowered plants

C

Only pink flowered plants

D

Red, Pink as well as white flowered plants

Solution

Pink colour flower in snapdragon have genotype Rr Red flowered snapdragon have genotype RR when they both are crossed ♂ ♀ R R Phenotype R RR RR Red : Pink : White r Rr Rr 2 2 0 So the progeny that we get are red and pink flowered plants only

A

Genetic recombination

B

Genetic drift

C

Gene migration

D

Constant gene pool

Solution

The correct answer is option (4) as a constant gene pool will not disturb the Hardy-Weinberg equilibrium. Option (1), (2) & (3) will affect the equilibrium leading to evolution.

A

A only

B

B only

C

C & B only

D

D & E only

Solution

Genotype of father with blood group B+ = IBi/iIB Genotype of mother with blood group A+ = IAi/iIA Genotype of child with blood group O+ = ii Hence only ‘A’ is correct.

A

Mendelian inheritance pattern

B

Non-mendelian inheritance pattern

C

Autosomal dominant pattern

D

X-linked recessive inheritance pattern

Solution

Polygenic inheritance refers to the inheritance of a trait controlled by two or more genes. When human disorders are determined by mutation in the single gene then they are transmitted to the offspring as per Mendelian principle. Polygenic trait shows non-Mendelian inheritance pattern.

A

A-II, B-IV, C-I, D-III

B

A-IV, B-II, C-I, D-III

C

A-IV, B-III, C-I, D-II

D

A-III, B-II, C-IV, D-I

Solution

The unequivocal proof that DNA is the genetic material came from the experiment of Alfred Hershey and Martha Chase.

Euchromatin are lightly stained region with loosely packed chromatin fibre.

Frederick Griffith performed series of experiments by selecting the different strains of Streptococcus pneumoniae.

Heterochromatin are darkly stained region with tightly packed chromatin fibre.

A

Phenotypic ratio - 1 : 2 : 1

B

Phenotypic ratio - 3 : 1

C

Phenotypic ratio - 9 : 3 : 3 : 1

D

Phenotypic ratio - 9 : 7

Solution

Sol. A classical dihybrid cross performed by Mendel involves.

A cross which was made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded plant (rryy). Yellow colour is dominant over green and round seed shape over wrinkled seed shape.

Phenotypic ratio in F2 generation

9 : 3 : 3 : 1

Round yellow Round green Wrinkled yellow Wrinkled green

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