Genetics – Chromosomal Basis of Inheritance: NEET MCQs with Answers

15 free questions on Genetics – Chromosomal Basis of Inheritance for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Chromosomal Basis of Inheritance: Chromosome theory of inheritance; Chromosomes and genes; Mutations

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Question 1

Previous-year question
Select the correct match. [NEET 2018]
  1. Ribozyme – Nucleic acid
  2. F2 × recessive parent – Dihybrid cross
  3. G. Mendel – Transformation
  4. T. H. Morgan – Transduction
Show answer
Answer: (A) Ribozyme – Nucleic acid
Explanation: A ribozyme is an RNA molecule, i.e. a nucleic acid, that acts as a biological catalyst. A cross with the recessive parent is a test cross, transformation was discovered by Griffith and transduction by Zinder and Lederberg.

Question 2

Previous-year question
Experimental verification of the chromosomal theory of inheritance was done by [NEET 2020]
  1. Sutton
  2. Boveri
  3. Morgan
  4. Mendel
Show answer
Answer: (C) Morgan
Explanation: Sutton and Boveri proposed the chromosome theory of inheritance by noting the parallel behaviour of genes and chromosomes. T. H. Morgan provided the experimental proof through his breeding work on Drosophila melanogaster.

Question 3

Previous-year question
What map unit (centimorgan) is adopted in the construction of genetic maps? [NEET 2019]
  1. A unit of distance between two expressed genes, representing 10% crossing over
  2. A unit of distance between two expressed genes, representing 100% crossing over
  3. A unit of distance between genes on chromosomes, representing 1% crossover
  4. A unit of distance between genes on chromosomes, representing 50% crossover
Show answer
Answer: (C) A unit of distance between genes on chromosomes, representing 1% crossover
Explanation: One centimorgan or map unit is defined as that distance between two genes for which the recombination frequency is 1%. Map distances are additive and are used to build linkage maps.

Question 4

Previous-year question
The frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes was explained by [NEET 2019]
  1. T. H. Morgan
  2. Gregor J. Mendel
  3. Alfred Sturtevant
  4. Sutton Boveri
Show answer
Answer: (C) Alfred Sturtevant
Explanation: Alfred Sturtevant, a student of Morgan, used recombination frequencies to determine the linear order of genes and their relative distances, and thus prepared the first genetic map of a chromosome.

Question 5

Previous-year question
A cell at the telophase stage is observed by a student in a plant brought from the field. He tells his teacher that this cell is not like other cells at telophase, as there is no formation of a cell plate and the cell therefore contains a higher number of chromosomes compared with other dividing cells. This would result in [NEET 2016]
  1. Aneuploidy
  2. Polyploidy
  3. Somaclonal variation
  4. Polyteny
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Answer: (B) Polyploidy
Explanation: Failure of cytokinesis after the chromosomes have already duplicated and separated leaves both sets in a single cell, doubling the entire chromosome complement. Addition of whole sets of chromosomes is polyploidy.

Question 6

Previous-year question
The term 'linkage' was coined by [AIPMT 2015]
  1. T. Boveri
  2. G. Mendel
  3. W. Sutton
  4. T. H. Morgan
Show answer
Answer: (D) T. H. Morgan
Explanation: T. H. Morgan, working on Drosophila melanogaster, coined the term linkage for the physical association of genes on a chromosome, and recombination for the generation of non-parental combinations.

Question 7

Previous-year question
Which one of the following statements about a testcross involving F1 dihybrid flies is correct: more parental-type offspring were produced than recombinant-type offspring. This indicates [NEET 2016]
  1. The two genes are located on two different chromosomes
  2. Chromosomes failed to separate during meiosis
  3. The two genes are linked and present on the same chromosome
  4. Both characters are controlled by more than one gene
Show answer
Answer: (C) The two genes are linked and present on the same chromosome
Explanation: An excess of parental combinations shows that the two genes tend to be transmitted together, which happens only when they lie on the same chromosome. Independently assorting genes would give parental and recombinant types in equal numbers.
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Question 8

Previous-year question
Haploids are able to express both recessive and dominant alleles/mutations because there are [AIPMT 1988]
  1. many alleles for each gene
  2. two alleles for each gene
  3. only one allele for each gene in the individual
  4. only one allele in a gene
Show answer
Answer: (C) only one allele for each gene in the individual
Explanation: In haploids there is only one allele for each gene, so there is no second allele to mask it. That is why all mutations, whether dominant or recessive, are expressed and haploids are excellent material for mutation studies.

Question 9

Previous-year question
Out of A═T, G≡C pairing, bases of DNA may exist in alternate valency state owing to arrangement called [AIPMT 1994]
  1. analogue substitution
  2. tautomerisational mutation
  3. frameshift mutation
  4. point mutation
Show answer
Answer: (B) tautomerisational mutation
Explanation: Tautomerisation occurs through rearrangement of electrons and protons within the molecule. Because of it the purines and pyrimidines of DNA and RNA can exist in alternate forms called tautomers, which pair abnormally and cause mutation.

Question 10

Previous-year question
The most striking example of point mutation is found in a disease called [AIPMT 1995]
  1. thalasemia
  2. night blindness
  3. Down's syndrome
  4. sickle-cell anaemia
Show answer
Answer: (D) sickle-cell anaemia
Explanation: Sickle-cell anaemia is the classic example of a point mutation. A single triplet CTC is changed to CAC in the Hbᴬ gene on chromosome 11, so that valine replaces glutamic acid at the sixth position of the β-chain of haemoglobin.

Question 11

Previous-year question
HJ Muller was awarded Nobel Prize for his [AIPMT 1996]
  1. discovery that chemicals can induce gene mutations
  2. discovery that ionizing radiations can induce gene mutations
  3. work on gene mapping in Drosophila
  4. efforts to prevent the use of nuclear weapons
Show answer
Answer: (B) discovery that ionizing radiations can induce gene mutations
Explanation: H. J. Muller received the Nobel Prize for demonstrating that ionizing radiation (X-rays) can induce gene mutations. His work established radiation as a powerful artificial mutagen.

Question 12

Previous-year question
Different mutations referrable to the same locus of chromosome give rise to [AIPMT 1997]
  1. pseudoalleles
  2. polygenes
  3. oncogenes
  4. multiple alleles
Show answer
Answer: (D) multiple alleles
Explanation: Multiple allelism is the simultaneous occurrence of more than two alleles at a given gene locus. Any mutation occurring within a gene at the same locus gives rise to a new form, i.e. a new allele of that gene.

Question 13

Previous-year question
The formation of multivalents at meiosis in diploid organism is due to [AIPMT 1998]
  1. monosomy
  2. inversion
  3. deletion
  4. reciprocal translocation
Show answer
Answer: (D) reciprocal translocation
Explanation: Reciprocal translocation involves the mutual exchange of segments between two non-homologous chromosomes. During meiosis the four chromosomes involved pair together to form a multivalent (usually a cross-shaped quadrivalent).

Question 14

Previous-year question
Loss of an X-chromosome in a particular cell, during its development, results into [AIPMT 1998]
  1. diploid individual
  2. triploid individual
  3. gynandromorphs
  4. Both (a) and (b)
Show answer
Answer: (C) gynandromorphs
Explanation: Gynandromorphs are individuals in which one part of the body is female while another part is male. They arise from an irregularity in mitosis, such as loss of an X-chromosome at the first cleavage of the zygote.

Question 15

Previous-year question
Which base is responsible for hotspots for spontaneous point mutations? [AIPMT 1998]
  1. Guanine
  2. Adenine
  3. 5-bromouracil
  4. 5-methylcytosine
Show answer
Answer: (D) 5-methylcytosine
Explanation: 5-methylcytosine is the classical hotspot for spontaneous point mutation because its deamination produces thymine directly, creating a mismatch that is often not repaired. 5-bromouracil is an induced mutagen, not a natural DNA base.

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