Mendelian Genetics — Part A (NEET Biology Class 12): Laws of Inheritance, Blood Groups & Ratios
Mendel & the Garden Pea
🎯 NEET priority: High-yield. Genetics is the single biggest scorer in NEET Biology — this chapter alone brings 8–10 questions a year. Every ratio (3:1, 1:2:1, 9:3:3:1) and every ABO genotype is fair game. Master it cold.
Why does a tall pea crossed with a dwarf one give only tall offspring — and then dwarfs reappear a generation later? Gregor Mendel answered this with seven years of patient pea-breeding, and his rules still anchor every genetics question NEET asks.
The man and the method
Mendel worked on the garden pea (Pisum sativum) for seven years (1856–1863) and was the first to apply statistics and mathematics to biology — his large sample sizes gave his data real credibility.
He chose 14 true-breeding lines — i.e. seven pairs of contrasting traits. A true-breeding line is one that, after continuous self-pollination, shows stable trait expression over many generations.
The seven characters (memorise the table)
Character | Dominant trait | Recessive trait |
Stem height | Tall | Dwarf |
Flower colour | Violet | White |
Flower position | Axial | Terminal |
Pod shape | Inflated | Constricted |
Pod colour | Green | Yellow |
Seed shape | Round | Wrinkled |
Seed colour | Yellow | Green |
Exam trap: NEET asks the exact pairs — note pod colour is green (dominant)/yellow (recessive), the opposite of seed colour yellow (dominant)/green (recessive). Mix these up and you lose the mark.
The Monohybrid Cross: Dominance, Segregation & the Test Cross
A monohybrid cross follows a single gene. Mendel crossed true-breeding tall (TT) × dwarf (tt) peas:
F1 = all Tall (Tt) — the dwarf trait vanishes.
Self-pollinate F1 → F2 = 3 Tall : 1 Dwarf (phenotypic 3:1), but genotypically 1 TT : 2 Tt : 1 tt = 1:2:1. The dwarf trait reappears, proving alleles don't blend.
The Punnett square (developed by Reginald C. Punnett) maps these outcomes by probability.
Mendel's first two laws
Law of Dominance: characters are controlled by discrete factors (genes) that occur in pairs; in a dissimilar pair, one factor (dominant) masks the other (recessive). This explains why F1 shows one trait and F2 gives the 3:1 ratio.
Law of Segregation: the two alleles of a pair separate (segregate) during gamete formation, so each gamete carries only one allele — there is no blending. This is also called the law of purity of gametes.
The test cross
A dominant phenotype (e.g. a tall plant) can be TT or Tt — you can't tell by looking. A test cross crosses it with the homozygous recessive (tt): if any recessive offspring appear, the unknown was heterozygous. It is the standard way to determine an unknown genotype.
When Dominance Isn't Complete: Incomplete Dominance, Codominance & ABO
Mendel's clean dominant/recessive picture has important exceptions — all heavily tested.
Incomplete dominance — a blended phenotype
Sometimes the F1 is intermediate between the parents. In the snapdragon / dog flower (Antirrhinum), true-breeding red (RR) × white (rr) gives a pink (Rr) F1. On selfing, F2 is 1 Red : 2 Pink : 1 White — the genotype ratio (1:2:1) is unchanged, but the phenotype ratio is no longer 3:1 because R is not fully dominant over r.
Codominance & the ABO blood groups
In codominance the F1 resembles both parents — both alleles express fully. The classic example is human ABO blood grouping, controlled by gene I with three alleles: IA, IB and i. IA and IB each make a different surface sugar; i makes none. When IA and IB are together (group AB), both sugars appear — that is codominance. There are 6 genotypes and 4 phenotypes.
Genotype | Blood group | Note |
IAIA or IAi | A | IA dominant over i |
IBIB or IBi | B | IB dominant over i |
IAIB | AB | codominance — both expressed |
ii | O | no sugar produced |
Multiple alleles: ABO is also the textbook case — three alleles govern one gene, yet any individual carries only two; multiple alleles are seen only at the population level.
The Dihybrid Cross & Independent Assortment
A dihybrid cross follows two genes at once. Mendel crossed round-yellow (RRYY) × wrinkled-green (rryy) seeds.
F1 = all round, yellow (RrYy).
F1 produces four gamete types — RY, Ry, rY, ry — each 25%.
Selfing F1 → F2 in the famous 9:3:3:1 ratio (9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green) across 16 squares.
Law of Independent Assortment: when two pairs of traits are combined, the segregation of one pair is independent of the other. That independence is exactly why 9:3:3:1 = (3 round : 1 wrinkled) × (3 yellow : 1 green).
Beyond Mendel: Polygenic Inheritance & Pleiotropy
Two more patterns extend Mendel — both NEET favourites.
Polygenic inheritance — traits on a gradient
Many traits aren't "either/or" but vary across a continuous range (human height, skin colour). These are controlled by three or more genes and are called polygenic traits.
The effect of each allele is additive, and the environment also contributes. For skin colour (genes A, B, C), the most dominant alleles (AABBCC) give the darkest skin and all-recessive (aabbcc) the lightest; intermediate counts give intermediate shades.
Pleiotropy — one gene, many effects
A pleiotropic gene produces multiple phenotypic effects, usually by acting on a metabolic pathway. The example is phenylketonuria (PKU): a single mutation in the gene for the enzyme phenylalanine hydroxylase causes both mental retardation and reduced hair/skin pigmentation.
Why this matters for NEET
Ratio recall is automatic marks: 3:1 (monohybrid), 1:2:1 (genotype / incomplete dominance), 9:3:3:1 (dihybrid).
ABO is asked every which way — genotype↔blood group, codominance vs multiple alleles, "which genotype gives O" (only ii).
Classic traps: incomplete dominance changes phenotype ratio but not genotype ratio; pleiotropy = one gene→many traits, NOT many genes→one trait (that's polygenic).
Test Yourself: MCQs, PYQs & Active Recall
Answer these, then close the article and do an Active Recall. Reveal each answer only after you commit to one.
Practice Questions
Q1. In a monohybrid cross, the F2 genotypic ratio is:
(a) 1:1
(b) 3:1
(c) 9:3:3:1
(d) 1:2:1
Show answer
Answer: (d) — F2 genotypes are 1 TT : 2 Tt : 1 tt (1:2:1); the phenotypic ratio is 3:1.
Q2. Mendel studied how many pairs of contrasting traits in pea?
(a) 5
(b) 14
(c) 10
(d) 7
Show answer
Answer: (d) — Seven pairs (from 14 true-breeding lines): stem height, flower colour, flower position, pod shape, pod colour, seed shape, seed colour.
Q3. The Law of Segregation is also known as the law of:
(a) Linkage
(b) Purity of gametes
(c) Dominance
(d) Independent assortment
Show answer
Answer: (b) — A gamete carries only one allele of a pair (no blending), so it is 'pure' — the law of purity of gametes.
Q4. A test cross is performed by crossing the individual with a:
(a) F1 hybrid
(b) Homozygous recessive
(c) Homozygous dominant
(d) Heterozygote
Show answer
Answer: (b) — Crossing the unknown with the homozygous recessive reveals its genotype from the offspring.
Q5. Incomplete dominance in Antirrhinum gives an F2 phenotypic ratio of:
(a) 9:3:3:1
(b) 2:1
(c) 3:1
(d) 1:2:1
Show answer
Answer: (d) — RR red, Rr pink, rr white → F2 is 1 red : 2 pink : 1 white (1:2:1); phenotype ratio equals the genotype ratio here.
Q6. Blood group AB is an example of:
(a) Pleiotropy
(b) Incomplete dominance
(c) Epistasis
(d) Codominance
Show answer
Answer: (d) — In I^A I^B, both I^A and I^B express their own surface sugar — codominance (the F1 resembles both parents).
Q7. How many genotypes and phenotypes exist for human ABO blood groups?
(a) 6 genotypes, 4 phenotypes
(b) 6 genotypes, 6 phenotypes
(c) 4 genotypes, 6 phenotypes
(d) 3 genotypes, 3 phenotypes
Show answer
Answer: (a) — Three alleles (I^A, I^B, i) give 6 genotypes and 4 phenotypes (A, B, AB, O).
Q8. The genotype of a person with blood group O is:
(a) I^B i
(b) ii
(c) I^A I^B
(d) I^A i
Show answer
Answer: (b) — Only ii gives group O — allele i produces no surface sugar.
Q9. The dihybrid F2 phenotypic ratio in Mendel's cross was:
(a) 1:1:1:1
(b) 3:1
(c) 1:2:1
(d) 9:3:3:1
Show answer
Answer: (d) — 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green = 9:3:3:1, derived from (3:1)(3:1).
Q10. Human skin colour and height are examples of:
(a) Polygenic inheritance
(b) Pleiotropy
(c) Codominance
(d) Incomplete dominance
Show answer
Answer: (a) — Both are continuous traits controlled by three or more genes with additive allele effects — polygenic inheritance.
Q11. Phenylketonuria, where one gene affects pigmentation and mental development, illustrates:
(a) Pleiotropy
(b) Polygenic inheritance
(c) Multiple alleles
(d) Codominance
Show answer
Answer: (a) — A single phenylalanine-hydroxylase gene producing multiple effects is pleiotropy (one gene → many traits).
Q12. The Punnett square was developed by:
(a) Walter Sutton
(b) Gregor Mendel
(c) T.H. Morgan
(d) Reginald Punnett
Show answer
Answer: (d) — Reginald C. Punnett, a British geneticist, devised the square to predict offspring genotype probabilities.
NEET Previous Year Questions (PYQs)
Real NEET previous-year questions on this chapter, with explanations in our own words.
Q13. ABO blood grouping in humans is an example of: (NEET PYQ)
(a) Both codominance and multiple alleles
(b) Incomplete dominance
(c) Only codominance
(d) Only multiple alleles
Show answer
Answer: (a) — Three alleles (I^A,I^B,i) = multiple alleles; I^A and I^B both expressing in AB = codominance. ABO shows both.
Q14. In Antirrhinum, a cross between red (RR) and white (rr) flowers gives a pink F1. This is: (NEET PYQ)
(a) Codominance
(b) Complete dominance
(c) Pleiotropy
(d) Incomplete dominance
Show answer
Answer: (d) — Neither allele is fully dominant, so the heterozygote (Rr) is an intermediate pink — incomplete dominance.
Q15. A man of blood group AB marries a woman of blood group O. The possible blood groups of children are: (NEET PYQ)
(a) A, B, AB and O
(b) Only AB
(c) AB and O
(d) A and B
Show answer
Answer: (d) — AB (I^A I^B) × O (ii) → children are I^A i (d) or I^B i (c). Only A and B are possible.
Q16. The number of different types of gametes produced by a plant of genotype RrYy is: (NEET PYQ)
(a) Eight
(b) Four
(c) Two
(d) Sixteen
Show answer
Answer: (b) — RrYy produces four gamete types — RY, Ry, rY, ry — by independent assortment.
Q17. Which of Mendel's laws is based on the separation of alleles during gamete formation? (NEET PYQ)
(a) Law of linkage
(b) Law of dominance
(c) Law of segregation
(d) Law of independent assortment
Show answer
Answer: (c) — The Law of Segregation: paired alleles separate so each gamete gets only one (purity of gametes).
Q18. A single gene affecting multiple phenotypic traits is termed: (NEET PYQ)
(a) Pleiotropy
(b) Epistasis
(c) Polygeny
(d) Codominance
Show answer
Answer: (a) — Pleiotropy — one gene, many effects (e.g. phenylketonuria, sickle-cell). Polygeny is the reverse (many genes → one trait).
Q19. The phenotypic ratio of a Mendelian dihybrid cross is: (NEET PYQ)
(a) 9:3:3:1
(b) 1:1:1:1
(c) 3:1
(d) 1:2:1
Show answer
Answer: (a) — 9:3:3:1 in F2 — the product of two independent 3:1 monohybrid ratios.
Q20. Test cross is a cross between F1 hybrid and: (NEET PYQ)
(a) Another F1
(b) Homozygous dominant parent
(c) Homozygous recessive parent
(d) F2 individual
Show answer
Answer: (c) — Crossing with the homozygous recessive parent reveals whether the dominant individual is homozygous or heterozygous.
Q21. Mendelian genetics fails to explain continuous variation such as human height because such traits are: (NEET PYQ)
(a) Pleiotropic
(b) Polygenic
(c) Sex-linked
(d) Codominant
Show answer
Answer: (b) — Continuous traits like height and skin colour are polygenic — many genes with additive effects plus environment.
Q22. If both parents are heterozygous for blood group (I^A i and I^B i), the chance of an O-group child is: (NEET PYQ)
(a) 25%
(b) 0%
(c) 50%
(d) 75%
Show answer
Answer: (a) — I^A i × I^B i → I^A I^B, I^A i, I^B i, ii — one of four (25%) is ii = group O.
Q23. Walter Sutton and Theodore Boveri are associated with the: (NEET PYQ)
(a) Lac operon
(b) Theory of natural selection
(c) Law of dominance
(d) Chromosomal theory of inheritance
Show answer
Answer: (d) — They noted chromosome behaviour parallels gene behaviour and proposed the chromosomal theory of inheritance (1902).
Q24. In incomplete dominance, the F2 ratio differs from a normal monohybrid cross in its: (NEET PYQ)
(a) Genotypic ratio only
(b) Neither ratio
(c) Both ratios
(d) Phenotypic ratio only
Show answer
Answer: (d) — Genotype stays 1:2:1, but the phenotype becomes 1:2:1 (not 3:1) because the heterozygote looks different.
Active Recall Prompt
Write everything you can recall, naming each topic first. Cover: Mendel (pea, seven traits, true-breeding); the monohybrid cross (3:1, 1:2:1, Punnett, test cross); the Law of Dominance and Law of Segregation; incomplete dominance (Antirrhinum) and codominance (ABO blood groups, genotypes); the dihybrid cross (9:3:3:1, independent assortment); and polygenic inheritance and pleiotropy (PKU). Begin each fact with its topic and end it with a full stop.