NCERT Class 12 Biology Chapter 5 Principles of Inheritance and Variation

NCERT Class 12 Biology Chapter 5 Principles of Inheritance and Variation Solutions to each chapter is provided in the list so that you can easily browse through different chapters NCERT Class 12 Biology Chapter 5 Principles of Inheritance and Variation and select need one. NCERT Class 12 Biology Chapter 5 Principles of Inheritance and Variation Question Answers Download PDF. NCERT Biology Class 12 Solutions.

NCERT Class 12 Biology Chapter 5 Principles of Inheritance and Variation

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. CBSE Class 12 Biology Solutions are part of All Subject Solutions. Here we have given NCERT Class 12 Biology Chapter 5 Principles of Inheritance and Variation Notes, NCERT Class 12 Biology Textbook Solutions for All Chapters, You can practice these here.

Chapter: 5




Q. 1. Mention the advantages of selecting pea plant for experiment by Mendel. 

Ans: Mendel selected pea plants for his experiment because.

(i) Peas have seven visible contrasting characters like tall/dwarf plants, round/ wrinkled seeds, purple/white flowers etc.

(ii) Peas have bisexual flowers and therefore undergo self pollination easily.

(iii) Cross pollination can be easily achieved by emasculation in Peas.

(iv) Pea plants have a short life span and produce large number of progeny after a single cross.

Q. 2. Differentiate between the following:

(a) Dominance and Recessive.

Ans: Dominance and Recessive:

Dominant factor or allele expresses itself in the presence or absence of a recessive trait while recessive trait is able to express itself only in the absence of dominant trait.

Tall plants, round seed, violet flower etc are dominant characters in pea plant while dwarf plants, wrinkled seed, white flower etc are recessive traits in a pea plant.

(b) Homozygous and Heterozygous.

Ans: Homozygous and Heterozygous:

Homozygous – It has two similar alleles for a particular trait. It has either dominant or recessive but never both alleles e.g. RR or rr.It produces only one type of gamete. 

Heterozygous – It has two different alleles for a particular trait. It has both dominant and recessive alleles e.g. Rr. It produces two different types of gametes.

(c) Monohybrid and Dihybrid.

Ans: Monohybrid and Dihybrid:

Monohybrid is a cross that involves only one pair of contrasting traits e.g. cross between tall and dwarf pea plant.

Dihybrid involves cross between two pairs of contrasting traits e.g. cross between pea plants having yellow wrinkled seed with those having green round seeds.

Q. 3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Ans. No of gametes = 2n = 24 = 16.

Q. 4. Explain the Law of Dominance using a monohybrid cross.

Ans: Law of Dominance.

(i) Characters are controlled by discrete units called factors.

(ii) Factors occur in pairs.

(iii) In a dissimilar pair of factors one factor of pair dominates (dominant) the other factor (recessive).

F2 Phenotypic ratio = 3 (tall): 1(dwarf)

Genotypic ratio = 1 (TT): 2(Tt): 1 (tt)

Q. 5. Define and design a test-cross? 

Ans: Test cross is a cross between an organism with unknown genotype and recessive parent. It is used to find whether the individual is homozygous or heterozygous for a trait.

If the progenies produced by a test cross show 50% dominant trait and 50% recessive trait, then the unknown individual is heterozygous for a trait. If the progeny produced shows dominant trait, then the unknown individual is homozygous for a trait.

Q. 6. Using a Punnett square, work out the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Ans: Homozygous female/TTx Heterozygous male/Tt

Gametes T T Tt

TTT [tall]Tt [tall]
TTT [tall]Tt [tall]

Phenotype of first filial generation = all tall plants.

Q. 7. When a cross in made between tall plants with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be 

(a) Tall and green 

(b) Dwarf and green.


TtYy × Ttyy

Gametes TY Ty ty ty  Ty Ty ty ty


Progeny results are 6 TTYy/ tall and yellow; 6 Ttyy/ tall and green:2 TTYy/ dwarf and yellow; 2 ttyy/ dwarf and green plant.

Q. 8. Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?

Ans: If genes are tightly linked or associated [linkage] then they show little recombination.

If genes are loosely linked, theyshow higher recombinations or non parental gene combination.

Q.9. Briefly mention the contribution ofT.H. Morgan in genetics.

Ans: Morgan’s work is based on fruit flies Drosophila melanogaster). He performed hybrid crosses in Drosophila to show that inked genes are inherited together and are located on X-chromosome. His experiments proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

Q. 10. What is pedigree analysis? Suggest how such an analysis, can be useful. 

Ans: Analysis of inheritance of traits in several generations of family in the form of a family tree is called pedigree analysis.

Pedigree study is used to trace the inheritance of a specific trait, abnormality or disease in human genetics.

Q. 11. How is sex determined in human beings?

Ans: Human beings exhibit male heterogamy. Males produce two types of gametes, X and Y. Females produce only one type of gametes, X. The sex of the baby is determined by the type of male gamete that fuses with the female gamete. If the fertilizing sperm contains X chromosome, then the baby will be female and if the fertilizing sperm contains Y chromosome, then the baby will be male. Thus, it is the genetic makeup of the sperm that determines the sex of baby.

Q. 12. A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible. genotypes of the other offsprings.

Ans: If the father has blood group A and mother has blood group B child has blood group O then the possible genotype of the parents will be

Father: IA IOMother: LB IOChild: IO IO
Father: IA IO×Mother: IB IO

Q. 13. Explain the following terms with example

(a) Codominance.

Ans: Codominance:

(i) Both alleles of a gene express themselves simultaneously in a heterozygous state.

(ii) F1 generation resembles both parents.

(iii) E.g. ABO blood grouping in human beings.

(b) Incomplete dominance. 

Ans: Incomplete dominance:

(i) None of two alleles of a gene is dominant over each other.

(ii) Both alleles when present together produce a new phenotype which is intermediate between the independent expressions of two alleles.

(iii) It is seen in flower color of Mirabilis jalapa [4 o’ clock] and dog flower / snapdragon (Antirrhinum majus).

Q. 14. What is point mutation? Give one example.

Ans: Point mutation is a change in single base pair of DNA e.g. sickle cell anaemia.

Q. 15. Who had proposed the chromosomal theory of inheritance?

Ans: Sutton and Boveri proposed the chromosomal theory of inheritance.

Q. 16. Mention any two autosomal genetic disorders with their symptoms. 

Ans: Two autosomal genetic disorders are: 

(i) Sickle cell Anemia: It is an autosomal recessive disorder due to point mutation in the beta globin chain of hemoglobin. The disease is characterized by sickle shaped red blood cells, which are formed due to the mutant hemoglobin molecule.
(ii) Thalassemia: This is an autosomal recessive trait due to deletion of genes on chromosome 16 in á thalassemia and on chromosome 11 in â thalassemia. It shows reduced synthesis of globin chains of hemoglobin and anemia.

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