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Principles of Inheritance and Variation Class 12 Biology MCQ With Solutions

Update on: 02 Feb 2024, 11:01 AM

The Principles of Inheritance and Variation Class 12 Biology MCQ with Answers has been organized according to the latest exam pattern in the CBSE Class 12 Biology syllabus for the upcoming CBSE Class 12 Biology Exam. Principles of Inheritance and Variation covers topics like mendelian genetics, chromosomal Theort of inheritance, Linkage and sex determination, Mutation and Genetic Disorders, These questions are designed to help students understand the concept thoroughly. They are helpful for revision and for grasping the concepts effectively. The MCQ questions for Chapter 4 Principles of Inheritance and Variation cover in-text book questions and important topics to help students assess their preparation level for CBSE board exams.

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CBSE Class 12 Chapter 4 Principles of Inheritance and Variation MCQs

Check below the  CBSE Class 12 Biology MCQ with answers and Clear your idea of Chapter 4 Principles of Inheritance and Variation.

Q.1. In a pedigree analysis, a mutation in recessive gene would be expressed only in

(a) recessive homozygous condition
(b) recessive heterozygous condition
(c) dominant homozygous condition
(d) both (a) and (b)

Answer
(a): Diploid organisms carry two copies of each gene, they may carry identical alleles, that is, being homozygous for a gene, or carry different alleles, that is, being heterozygous for a gene. In a recessive mutation, both alleles must be mutant in order for the mutant phenotype to be observed; therefore, the individual must be homozygous for the mutant allele to show the mutant phenotype.

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Q.2. Linked genes on homologous chromosome can be separated due to

(a) recombination (b) inversion (c) point mutation (d) none of these.

Answer
(a): Linked genes can be separated by recombination in which homologous chromosomes exchange genetic information during meiosis; this results in parental, or non recombinant genotypes, as well as a smaller proportion of recombinant genotypes.

 

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Q.3. The non-disjunction of 21st pair of autosomal chromosome leads to

(a) Sickle cell anaemia
(b) Klinefelter’s syndrome
(c) Turner’s syndrome
(d) Down’s syndrome.

Answer

(d): Down’s syndrome is usually due to non-disjunction of chromosome 21 that results in an embryo with three copies of chromosome 21 instead of the usual two. Prior to or at conception, a pair of 21st chromosomes in either the sperm or the egg fails to separate.

 

Q.4. In a Mendelian dihybrid cross, the phenotypic ratio obtained in F2 progeny is

(a) 9 : 7
(b) 9 : 3 : 3 : 1
(c) 9 : 3 : 4
(d) 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.

Answer
(b): According to the law of independent assortment, a phenotypic ratio among the progeny of a dihybrid cross is 9 : 3 : 3 : 1 .

 

Q.5. The human blood groups are determined by

(a) 4 alleles in which IA is dominant.
(b) 3 alleles in which IA and IB are codominant.
(c) 3 alleles in which none is dominant.
(d) 3 alleles in which IA and i are codominant.

Answer
(b): Human blood groups are determined by three alleles, viz., IA, IB and i. IA, IB are co-dominant and i is recessive

 

Q.6. In a cross between a male and female, both heterozygous for sickle cell anaemia gene, what percentage of the progeny will be diseased?

(a) 100% (b) 50% (c) 75% (d) 25%

Answer
(d): Sickle cell anaemia is an autosomal recessive disorder.
Answer 6

 

Q.7. On selfing a heterozygous round and yellow seeded plants, the production of heterozygous for both the traits would be

(a) 9/16 (b) 3/16
(c) 1/16 (d) 4/16

Answer
(d): The heterozygous for both seed shape and colour (RrYy) produced during dihybrid cross are four out of sixteen
selfing a heterozygous round and yellow seeded plants

 

Q.8. The production of gametes by the parents, formation of zygotes, the F1 and F2 plants, can be understood from a diagram called

(a) net square (b) bullet square (c) punch square (d) Punnett square.

Answer
(d): The prodcuction of gametes by the parents, formation of the zygotes, the F1 and F2 plants can be understood from a diagram called Punnett square developed by a British geneticist, Reginald C. Punnett.

 

Q.9. Which of the following does not follow Mendel’s law of dominance?

(a) Pisum sativum
(b) Mirabilis jalapa
(c) Lathyrus odoratus
(d) Oenothera lamarckiana

Answer
(b): Mirabilis jalapa shows incomplete dominance which is an exception to law of dominance. Incomplete dominance is the condition when a dominant allele, or form of a gene, does not completely mask the effects of a recessive allele, and the organism’s resulting physical appearance shows a blending of both alleles.

 

Q.10. The cross between white eyed female Drosophila and a red eyed male Drosophila would lead to the production of

(a) red eyed males and red eyed females
(b) white eyed males and white eyed females
(c) red eyed females and white eyed males
(d) white eyed males and red eyed females.

Answer
(c): The cross of white eyed female with red eyed male gives red eyed females and white eyed males. Rarely this cross may give all white eyed females and red eyed males. This results from non-separation of two homologous X chromosomes during anaphase I of meiosis. Both X chromosomes go together to same pole resulting in formation of one XXY i.e., white eyed female and red eyed males. This phenomenon is known as non-disjunction of two X chromosomes.

 

We hope the MCQs for the CBSE Class 12 Biology Chapter 4 on Principles of Inheritance and Variation are helpful for your board exam preparation.

Keep learning and stay updated with us for more CBSE exam updates.


 

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