25 questions in 25 minutes on the NTA computer-based test interface, all from Principles of Inheritance and Variation. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Principles of Inheritance and Variation, of which 19 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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NEET CBT mock test: common questions
How many questions are in this Principles of Inheritance and Variation mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 19 of them are NEET previous-year questions.
Is this the same interface as the real NEET CBT?
Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Principles of Inheritance and Variation?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
All 25 questions with answers and solutions
The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
Show all 25 questions with answers and solutions
Botany
- Q1Down’s syndrome is caused by an extra copy of chromosome number 21. What percentage of offspring produced by an affected (trisomy-21) mother and a normal father would be affected by this disorder?
- 100%
- 50%
- 25%
- 75%
Answer: (B) 50%
A trisomy-21 mother produces 50% normal eggs and 50% eggs carrying an extra chromosome 21. Fertilised by normal sperm, about 50% of offspring receive the extra chromosome and are affected.
- Q2What is the genetic disorder in which an individual has an overall masculine development, gynaecomastia and is sterile?
- Down’s syndrome
- Edward syndrome
- Turner’s syndrome
- Klinefelter’s syndrome
Answer: (D) Klinefelter’s syndrome
Klinefelter’s syndrome (47, XXY) gives an overall masculine but sterile individual with gynaecomastia (breast development) due to the extra X. Turner’s (XO) is female; Down’s is trisomy 21.
- Q3In Mendel’s dihybrid cross with garden pea, round seed shape (RR) was dominant over wrinkled (rr) and yellow cotyledon (YY) over green (yy). What are the expected phenotypes in the F2 generation of the cross RRYY x rryy?
- Only wrinkled seeds with yellow cotyledons
- Round seeds with yellow cotyledons, round with green, wrinkled with yellow and wrinkled with green
- Only round seeds with green cotyledons
- Only round seeds with green cotyledons and wrinkled with yellow
Answer: (B) Round seeds with yellow cotyledons, round with green, wrinkled with yellow and wrinkled with green
RRYY x rryy gives F1 RrYy (round, yellow). Selfing the F1 gives an F2 with all four phenotype combinations in 9:3:3:1 (round-yellow, round-green, wrinkled-yellow, wrinkled-green). So all four phenotypes appear.
- Q4A human male produces sperms with the genotypes AB, Ab, aB and ab pertaining to two diallelic characters in equal proportions. What is the corresponding genotype of this person?
- AaBb
- AABB
- AaBB
- AABb
Answer: (A) AaBb
Four equally frequent gamete types (AB, Ab, aB, ab) require both gene pairs to be heterozygous, i.e. genotype AaBb (2² = 4 gametes). Any homozygous pair would reduce the number of gamete types.
- Q5Which one of the following characters was NOT considered by Mendel in his experiments on pea?
- Seed colour (Green or Yellow)
- Pod shape (Inflated or Constricted)
- Trichomes (Glandular or Non-glandular)
- Stem height (Tall or Dwarf)
Answer: (C) Trichomes (Glandular or Non-glandular)
Mendel’s seven characters were stem height, flower colour, flower position, pod shape, pod colour, seed shape and seed colour. Trichome type was NOT among them, so B is the trait he did not study.
- Q6In a monohybrid cross, the F2 genotypic ratio of Tt × Tt is:
- 1:2:1
- 3:1
- 1:1
- 9:3:3:1
Answer: (A) 1:2:1
Tt × Tt gives 1 TT : 2 Tt : 1 tt (genotypic ratio 1:2:1). The 3:1 is the phenotypic ratio (tall:dwarf).
- Q7Fruit colour in squash is an example of:
- dominant epistasis
- recessive epistasis
- complementary genes
- inhibitory genes
Answer: (A) dominant epistasis
In squash, a dominant allele (W) for white masks the colour gene, giving a 12:3:1 ratio – dominant epistasis, where the dominant epistatic gene suppresses the hypostatic colour gene.
- Q8The chromosomal theory of inheritance was proposed independently around 1902 by which pair of scientists?
- Sutton and Boveri
- Bateson and Punnett
- Watson and Crick
- T. H. Morgan
Answer: (A) Sutton and Boveri
Walter Sutton and Theodor Boveri independently (around 1902) proposed the chromosomal theory of inheritance, noting that chromosome behaviour during meiosis parallels Mendel’s factors. Morgan later verified it experimentally.
- Q9Klinefelter’s syndrome and Turner’s syndrome are characterised respectively by the karyotypes:
- Trisomy 21 and Trisomy 18
- XO (male) and XXY (female)
- XYY (male) and XXX (female)
- XXY (male) and XO (female)
Answer: (D) XXY (male) and XO (female)
Klinefelter’s = 47, XXY (sterile male with some feminine features); Turner’s = 45, XO (sterile female). Both arise from non-disjunction of sex chromosomes.
- Q10Which one of the following most appropriately describes haemophilia?
- Dominant gene disorder
- Chromosomal disorder
- X-linked recessive gene disorder
- Y-linked recessive gene disorder
Answer: (C) X-linked recessive gene disorder
Haemophilia is an X-linked recessive single-gene (Mendelian) disorder of blood clotting, showing criss-cross inheritance. It is not a chromosomal-number disorder.
- Q11Two non-allelic genes produce a new phenotype when present together but fail to do so independently. This is called:
- non-complementary gene
- epistasis
- complementary gene
- polygene
Answer: (C) complementary gene
Complementary genes are two different genes that must both be present (each in dominant form) to produce a phenotype; alone, neither is expressed. This gives the modified 9:7 dihybrid ratio.
- Q12Two genes show a recombination frequency of 8%. The map distance between them is:
- 80 cM
- 8 cM
- 16 cM
- 0.8 cM
Answer: (B) 8 cM
1% recombination = 1 centimorgan (map unit). A recombination frequency of 8% therefore equals a map distance of 8 cM.
- Q13The ABO blood group system illustrates multiple allelism because the gene I has three alleles. Yet a single person can carry at most how many of these alleles?
- One
- Three
- Six
- Two
Answer: (D) Two
Multiple alleles exist in the population (I^A, I^B, i), but any diploid individual has only two alleles of a gene (one on each homologous chromosome).
- Q14Two linked genes a and b show 20% recombination. The individuals of a dihybrid cross between ++/++ x ab/ab will show gametes in the ratio:
- ++ 80 : ab 20
- ++ 30 : ab 30 : +a 20 : +b 20
- ++ 50 : ab 50
- ++ 40 : ab 40 : +a 10 : +b 10
Answer: (D) ++ 40 : ab 40 : +a 10 : +b 10
The F1 (++/ab) shows 20% recombination, so 20% of gametes are recombinant (split equally into +a and +b = 10% each) and 80% are parental (split into ++ and ab = 40% each). Hence ++ 40 : ab 40 : +a 10 : +b 10.
- Q15Mental retardation in man, associated with a sex-chromosomal abnormality, is usually due to:
- large increase in Y-complement
- moderate increase in Y-complement
- increase in X-complement
- reduction in X-complement
Answer: (C) increase in X-complement
An increase in the X-complement (extra X chromosomes, as in Klinefelter’s XXY/XXXY) is associated with mental impairment and sterility. It is the surplus of X-chromosomes, not Y, that produces these effects.
- Q16A gene locus has two alleles A and a. If the frequency of the dominant allele A is 0.4 in a population at Hardy-Weinberg equilibrium, the frequencies of homozygous dominant (AA), heterozygous (Aa) and homozygous recessive (aa) individuals respectively are:
- 0.16 (AA); 0.36 (Aa); 0.48 (aa)
- 0.16 (AA); 0.48 (Aa); 0.36 (aa)
- 0.38 (AA); 0.48 (Aa); 0.16 (aa)
- 0.16 (AA); 0.24 (Aa); 0.36 (aa)
Answer: (B) 0.16 (AA); 0.48 (Aa); 0.36 (aa)
p(A) = 0.4 so q(a) = 0.6. AA = p² = 0.16, Aa = 2pq = 2(0.4)(0.6) = 0.48, aa = q² = 0.36. These sum to 1.0.
- Q17If two persons with AB blood group marry and have a sufficiently large number of children, these children could be classified as A blood group : AB blood group : B blood group in a 1 : 2 : 1 ratio. This is an example of:
- codominance
- partial dominance
- incomplete dominance
- complete dominance
Answer: (A) codominance
I^A I^B x I^A I^B gives 1 I^A I^A (A) : 2 I^A I^B (AB) : 1 I^B I^B (B). Because both I^A and I^B express fully (the AB class shows both antigens), this is codominance, not incomplete dominance (which would give a blended phenotype).
- Q18A person with 47 chromosomes due to an additional sex chromosome (karyotype 44 + XXY) suffers from a condition called:
- Down’s syndrome
- Turner’s syndrome
- Super-female
- Klinefelter’s syndrome
Answer: (D) Klinefelter’s syndrome
An extra X giving 47 chromosomes (44 autosomes + XXY) is Klinefelter’s syndrome – a sterile male with gynaecomastia, arising from sex-chromosome non-disjunction. Turner’s (XO) has 45 chromosomes; Down’s involves an extra chromosome 21, not a sex chromosome.
- Q19Which one of the following pea characters studied by Mendel was dominant?
- Terminal flower position
- Wrinkled seed
- Green seed colour
- Green pod colour
Answer: (D) Green pod colour
Green pod colour is dominant over yellow pod in pea. The other options are recessive: green seed (yellow is dominant), terminal flower (axial is dominant) and wrinkled seed (round is dominant).
- Q20A polygenic (quantitative) inheritance in human beings is exemplified by:
- phenylketonuria
- sickle-cell anaemia
- colour blindness
- skin colour
Answer: (D) skin colour
Human skin colour is controlled by several genes acting additively (Davenport’s hypothesis), giving continuous variation – the defining feature of polygenic inheritance. The other three are single-gene (Mendelian) conditions.
- Q21In a cross between a male and female, both heterozygous for the sickle-cell anaemia gene, what percentage of the progeny will be diseased (suffer from sickle-cell anaemia)?
- 50%
- 75%
- 100%
- 25%
Answer: (D) 25%
Sickle-cell anaemia is autosomal recessive (HbS HbS). Carrier x carrier (HbA HbS x HbA HbS) gives 1 HbA HbA : 2 HbA HbS : 1 HbS HbS. Only the homozygous HbS HbS (25%) actually suffers the disease.
- Q22Hybridisation between Tt and tt gives rise to progeny in the phenotypic ratio of:
- 1 : 2 : 1
- 1 : 1
- 4 : 1
- 1 : 2
Answer: (B) 1 : 1
Tt x tt is a test cross. Tt gives T or t; tt gives only t. Offspring: Tt (tall) and tt (dwarf) in a 1:1 ratio.
- Q23A single gene controlling more than one phenotypic trait is an example of:
- Codominance
- Pleiotropy
- Polygeny
- Epistasis
Answer: (B) Pleiotropy
Pleiotropy = one gene affecting multiple traits (e.g. phenylketonuria affecting mental ability and pigmentation). Polygeny is the reverse (many genes → one trait).
- Q24The AB blood group in humans is a classic example of:
- Incomplete dominance
- Epistasis
- Codominance
- Pleiotropy
Answer: (C) Codominance
In group AB both I^A and I^B alleles are fully and simultaneously expressed (both A and B antigens appear), which is codominance, not blending.
- Q25Thalassemia and sickle-cell anaemia are both caused by a problem in globin molecule synthesis. The correct statement is:
- Sickle-cell anaemia is due to a quantitative problem of globin molecules
- Both are due to a quantitative defect in globin chain synthesis
- Both are due to a qualitative defect in globin chain synthesis
- Thalassemia is due to less synthesis of globin molecules
Answer: (D) Thalassemia is due to less synthesis of globin molecules
Thalassemia is a QUANTITATIVE defect – reduced synthesis of normal globin chains. Sickle-cell anaemia is a QUALITATIVE defect – a structurally abnormal globin (Glu to Val). So only C is correct.