NEET Principles of Inheritance and Variation PYQs & Practice Questions

10 free NEET Biology questions on Principles of Inheritance and Variation, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Mendelian Genetics · Chromosomal Disorders · Sex Determination · Mutation · Incomplete Dominance · Linkage Groups · Terminology for Chromosome Number Variation · Gene Distance and Recombination Frequency · Identifying the Mode of Inheritance · Codon for Valine in Sickle Cell Anaemia · Co-dominance in the ABO Blood Group System · Interpreting a 9:3:3:1 Dihybrid Ratio

1. Q1NEET 2026ABO Blood Group Genetics

   For a person with blood group 'O', which of the following is not a possible combination of parents' blood group genotypes?

   • A. Father: IAi and Mother: IAi
   • B. Father: IBi and Mother: IBi
   • C. Father: IAIB and Mother: IAi
   • D. Father: IAi and Mother: IBi
   Show answer

   Answer: B — Father: IBi and Mother: IBi

   Blood group O requires genotype ii, meaning both parents must be able to contribute an 'i' allele. Father IAIB has only IA and IB alleles to contribute (no 'i' allele), so this parent can never contribute an 'i' allele to produce an ii offspring -- making option (3) impossible. In options (1), (2), and (4), both parents carry an 'i' allele that they can pass on, making an ii (blood group O) offspring possible.

2. Q2NEET 2026Chromosomal Disorders - Down's and Turner's Syndrome

   Given below are two statements: Statement I: Down's syndrome is caused by the absence of one of the X-chromosomes. Statement II: Turner's syndrome is caused by the presence of an additional copy of the chromosomes. In the light of the above statements, choose the correct answer from the options given below:

   • A. Both Statement I and Statement II are incorrect
   • B. Statement I is correct but Statement II is incorrect
   • C. Statement I is incorrect but Statement II is correct
   • D. Both Statement I and Statement II are correct
   Show answer

   Answer: —

   Down's syndrome is actually caused by the presence of an extra copy (trisomy) of chromosome 21, not the absence of an X-chromosome -- Statement I is incorrect (this description actually fits Turner's syndrome, which is monosomy of X). Turner's syndrome is actually caused by the ABSENCE of one X-chromosome (45, X0), not the presence of an additional chromosome (which would describe conditions like Klinefelter's or Down's syndrome) -- Statement II is incorrect. Both statements describe the conditions in reverse of reality.

3. Q3NEET 2025Polygenic Inheritance

   What is the pattern of inheritance for polygenic trait?

   • A. Mendelian inheritance pattern
   • B. Non-mendelian inheritance pattern
   • C. Autosomal dominant pattern
   • D. X-linked recessive inheritance pattern
   Show answer

   Answer: A — Mendelian inheritance pattern

   Polygenic traits are controlled by multiple genes contributing additively to the phenotype, showing continuous variation (e.g., height, skin colour) — this is a non-Mendelian inheritance pattern, unlike the clear-cut dominant/recessive patterns of Mendelian inheritance.

4. Q4NEET 2025Dihybrid Cross - Phenotypic Ratio in F2

   Genes R and Y follow independent assortment. If RRYY produce round yellow seeds and rryy produce wrinkled green seeds, what will be the phenotypic ratio of the F2 generation?

   • A. Phenotypic ratio - 1 : 2 : 1
   • B. Phenotypic ratio - 3 : 1
   • C. Phenotypic ratio - 9 : 3 : 3 : 1
   • D. Phenotypic ratio - 9 : 7
   Show answer

   Answer: B — Phenotypic ratio - 3 : 1

   In a dihybrid cross following independent assortment, the F2 generation shows round-yellow : round-green : wrinkled-yellow : wrinkled-green phenotypes in the ratio 9:3:3:1. Class 12th NCERT Page No.64.

Practise all 68 Principles of Inheritance and Variation questions with a timer

• Timed chapter test
• Score + weak topics
• Full NEET mocks & rank

Start free for 30 days

No payment, no card. Register in 2 minutes.

5. Q5NEET 2025Pedigree Analysis - X-linked Recessive Inheritance Probability

   With the help of given pedigree, find out the probability for the birth of a child having no disease and being a carrier (has the disease mutation in one allele of the gene) in F3 generation. [Pedigree: F0 - unaffected male x affected female; F1 - unaffected male x carrier female, carrier female x unaffected male; F2 - carrier female, carrier female, affected male x unaffected female, unaffected female; F3 - four children marked '?' from the carrier female x affected male cross]

   • A. 1/4
   • B. 1/2
   • C. 1/8
   • D. Zero
   Show answer

   Answer: —

   The pedigree pattern (disease in both sexes, carrier females, skips generations) indicates X-linked recessive inheritance. In F2, a carrier female (XᶜX) is crossed with an affected male (XᶜY). Gametes: Female gives Xᶜ or X; Male gives Xᶜ or Y. Possible F3 offspring: XᶜX (carrier female) = 1/4, XᶜXᶜ (affected female) = 1/4, XY (normal male) = 1/4, XᶜY (affected male) = 1/4. The probability of a child being unaffected AND a carrier (XᶜX) is 1/4. Class 12th NCERT Page No.73.

6. Q6NEET 2024Mendel's Law of Dominance - Explanatory Statements

   Which one of the following can be explained on the basis of Mendel's Law of Dominance? (A) Out of one pair of factors one is dominant and the other is recessive. (B) Alleles do not show any expression and both the characters appear as such in F2 generation. (C) Factors occur in pairs in normal diploid plants. (D) The discrete unit controlling a particular character is called factor. (E) The expression of only one of the parental characters is found in a monohybrid cross. Choose the correct answer from the options given below:

   • A. B, C and D only
   • B. A, B, C, D and E
   • C. A, B and C only
   • D. A, C, D and E only
   Show answer

   Answer: C — A, B and C only

   The Law of Dominance explains: one factor of a pair is dominant and other recessive (A); factors occur in pairs (C); the discrete unit controlling a character is called a factor (D); and only one parental character is expressed in a monohybrid cross (E). Statement B is incorrect/not part of this law (alleles don't both appear unexpressed in F2 - it describes something else entirely, like codominance absence). So A, C, D and E only. 12th NCERT Page No. 59.

7. Q7NEET 2024Incomplete Dominance - Snapdragon Flower Color Cross

   A pink flowered Snapdragon plant was crossed with a red flowered Snapdragon plant. What type of phenotype/s is/are expected in the progeny?

   • A. Only pink flowered plants
   • B. Red, Pink as well as white flowered plants
   • C. Only red flowered plants
   • D. Red flowered as well as pink flowered plants
   Show answer

   Answer: C — Only red flowered plants

   In Snapdragon, flower colour shows incomplete dominance where red (RR) and white (rr) produce pink (Rr) in F1. Crossing pink (Rr, heterozygous) with red (RR, homozygous dominant) gives offspring in ratio 1 RR (red) : 1 Rr (pink), so both red and pink flowered plants are expected, no white. 12th NCERT Page No. 60.

8. Q8NEET 2024Test Cross

   In a plant, black seed color (BB/Bb) is dominant over white seed color (bb). In order to find out the genotype of the black seed plant, with which of the following genotype will you cross it?

   • A. Bb
   • B. BB/Bb
   • C. BB
   • D. bb
   Show answer

   Answer: C — BB

   To determine whether an organism showing the dominant phenotype is homozygous or heterozygous, it is crossed with the homozygous recessive parent (bb). This is called a test cross — if all offspring show the dominant trait, the parent is BB; if a 1:1 ratio results, the parent is Bb.

Practise all 68 Principles of Inheritance and Variation questions with a timer

• Timed chapter test
• Score + weak topics
• Full NEET mocks & rank

Start free for 30 days

No payment, no card. Register in 2 minutes.

9. Q9NEET 2024Genetics Terminology

   Match List I with List II. List I: (A) Two or more alternative forms of a gene (B) Cross of F1 progeny with homozygous recessive parent (C) Cross of F1 progeny with any of the parents (D) Number of chromosome sets in plant List II: (I) Back cross (II) Ploidy (III) Allele (IV) Test cross Choose the correct answer from the options given below:

   • A. A-III, B-IV, C-I, D-II
   • B. A-IV, B-III, C-II, D-I
   • C. A-I, B-II, C-III, D-IV
   • D. A-II, B-I, C-III, D-IV
   Show answer

   Answer: —

   Alleles are alternative forms of a gene (A-III). A cross of F1 with homozygous recessive parent is a test cross (B-IV). A cross of F1 with any parent is a back cross (C-I). The number of chromosome sets is ploidy (D-II). So A-III, B-IV, C-I, D-II is correct.

10. Q10NEET 2024Meiosis - Prophase I Sub-stages

    Match List I with List II. List I (Sub phases of prophase I): (A) Diakinesis (B) Pachytene (C) Zygotene (D) Leptotene List II (Specific characters): (I) Synaptonemal complex formation (II) Completion of terminalisation of chiasmata (III) Chromosomes look like thin threads (IV) Appearance of recombination nodules Choose the correct answer from the options given below:

    • A. A-II, B-IV, C-I, D-III
    • B. A-IV, B-III, C-II, D-I
    • C. A-IV, B-II, C-III, D-I
    • D. A-I, B-II, C-IV, D-III
    Show answer

    Answer: —

    Diakinesis involves completion of terminalisation of chiasmata (A-II). Pachytene shows appearance of recombination nodules (B-IV). Zygotene involves synaptonemal complex formation (C-I). Leptotene shows chromosomes looking like thin threads (D-III). So A-II, B-IV, C-I, D-III is correct.