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NEET Practice: Principles of Inheritance & Variation

A set of previous-year-pattern MCQs covering Mendel's laws, dihybrid crosses, incomplete dominance, linkage, sex determination and human genetic disorders — each with a worked solution and the NCERT reference.

ℹ️ These are original questions built on the way NEET has repeatedly tested this chapter — not a copy of any one official paper. Study them, then take the timed quiz to score yourself.

Solved Questions

Q1 · Monohybrid ratio
In a monohybrid cross between two heterozygous tall pea plants (Tt × Tt), what is the expected phenotypic ratio in the F₂ generation?
Answer: B. A Tt × Tt cross gives genotypes 1 TT : 2 Tt : 1 tt. Since T (tall) is completely dominant over t (dwarf), three genotypes (TT, Tt, Tt) show the tall phenotype and only tt is dwarf — a 3 : 1 phenotypic ratio. The 1 : 2 : 1 in option C is the genotypic ratio.
Q2 · Law of Independent Assortment
Mendel's Law of Independent Assortment is best demonstrated by a:
Answer: B. Independent assortment describes how alleles of two different genes segregate independently of one another during gamete formation. This can only be seen when two traits are followed together — i.e. a dihybrid cross (RrYy × RrYy), which gives the classic 9 : 3 : 3 : 1 F₂ ratio.
Q3 · Test cross
A test cross is performed between an organism showing the dominant phenotype and a:
Answer: B. A test cross crosses the individual in question with a homozygous recessive parent to reveal its unknown genotype. If any recessive offspring appear, the tested parent was heterozygous; if all offspring show the dominant trait, it was homozygous dominant.
Q4 · Incomplete dominance
In Mirabilis jalapa (four o'clock plant), a cross between red-flowered (RR) and white-flowered (rr) plants gives pink (Rr) F₁. This is an example of:
Answer: A. The heterozygote (Rr) shows an intermediate phenotype (pink), so neither allele is completely dominant — this is incomplete dominance. In co-dominance (e.g. AB blood group) both alleles express fully and separately, not as a blend.
Q5 · ABO blood groups
The ABO blood group system in humans is an example of multiple alleles and co-dominance. How many alleles control it and which are co-dominant?
Answer: B. The gene I has three alleles: Iᴬ, Iᴮ and i. Iᴬ and Iᴮ are both dominant over i but are co-dominant with each other, so an IᴬIᴮ person is blood group AB. This illustrates both multiple alleles and co-dominance in one gene.
Q6 · Dihybrid gametes
How many types of gametes can be produced by an individual with the genotype RrYy?
Answer: B. The number of gamete types = 2ⁿ, where n = number of heterozygous gene pairs. Here n = 2, so 2² = 4 gamete types: RY, Ry, rY, ry.
Q7 · Linkage
Genes located close together on the same chromosome tend to be inherited together and show reduced recombination. This phenomenon is called:
Answer: B. Linkage is the tendency of genes on the same chromosome to be inherited together. The closer two genes are, the tighter the linkage and the lower the frequency of recombination (crossing over) between them. Morgan demonstrated this in Drosophila.
Q8 · Sex determination
In humans, the sex of the offspring is determined by:
Answer: B. Human females are XX (produce only X-bearing ova) and males are XY. The sperm therefore decides sex: an X-bearing sperm gives a girl (XX), a Y-bearing sperm gives a boy (XY). This is the XY type of sex determination.
Q9 · Haemophilia
Haemophilia is a sex-linked recessive disorder. A carrier woman marries a normal man. What fraction of their sons is expected to be haemophilic?
Answer: B. The carrier mother is XᴴXʰ and the father is XᴴY. Sons receive Y from father and either Xᴴ or Xʰ from mother — so half the sons (XʰY) are haemophilic. Daughters get Xᴴ from father, so none are affected, though half are carriers.
Q10 · Down syndrome
Down syndrome in humans is caused by:
Answer: A. Down syndrome results from trisomy 21 — an extra copy of chromosome 21 (47 chromosomes total), usually due to non-disjunction during meiosis. Klinefelter's (XXY) involves an extra X; Turner's is monosomy X (45, X0).

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