Principles of Inheritance and Variation Exam Questions Class 12 Biology

Please see Chapter 5 Principles of Inheritance and Variation Exam Questions Class 12 Biology below. These important questions with solutions have been prepared based on the latest examination guidelines and syllabus issued by CBSE, NCERT, and KVS. We have provided Class 12 Biology Questions and answers for all chapters in your NCERT Book for Class 12 Biology. These solved problems for Principles of Inheritance and Variation in Class 12 Biology will help you to score more marks in upcoming examinations.

Exam Questions Chapter 5 Principles of Inheritance and Variation Class 12 Biology

Objective Type Questions

Question. Genetic is the branch of Biology that deals with
(a) Inheritance of characters from one generation to offspring
(b) variations of characters from one generation to offspring
(c) Both (a) and (b)
(d) None of these

Answer

Question. Passing of characters from one generation to other is known as
(a) Variation
(b) Selection
(c) Recombination
(d) Inheritance

Answer

Question. Sahiwal Cows are found in
(a) Punjab
(b) U.P
(c) Assam
(d) Rajasthan

Answer

Question. For how many years G. Mendel performed hybridization experiments
(a) 5 years
(b) 3 years
(c) 7 years
(d) 10 years

Answer

Question. True breeding line is one that have undergone continuous
(a) Cross pollination and shows stable trait over successive generations
(b) Self–pollination and shows stable trait over successive generations
(c) Artificial pollination and shows stable trait in alternating generations
(d) None of these.

Answer

Question. How many Pod characters were taken in Mendel’s experiment?
(a) One
(b) Two
(c) Zero
(d) Three

Answer

Question. Which of the following was not the contrasting recessive trait in the Mendel’s experiment?
(a) Dwarf plant
(b) Green seed colour
(c) Yellow pod colour
(d) Axial flower

Answer

Question. F₁generation stands for
(a) Fili 1 progeny
(b) Filial 1 progeny
(c) Functional 1 progeny
(d) None of these

Answer

Question. In monohybrid cross performed by Mendel emasculation was performed on
(a) Female plants
(b) Male plants
(c) Both plants
(d) None of the plant of parent generation

Answer

Question. In the monohybrid cross (TT × tt) the blending of contrasting trait was seen at
(a) F₁ generation
(b) F₂ generation
(c) In both F₁and F₂ generation
(d) Neither F₁ nor F₂ generation

Answer

Very Short Answer Questions

Question. A garden pea plant (A) produced inflated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits.
Ans. Inflated green pod is the dominant trait.

Question. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Ans. Male honeybee develops from unfertilised female gamete (Parthenogenesis) and thus has 16 chromosomes whereas female develops by fertilisation and thus has 32 chromosomes.

Question. A garden pea plant produced axial white flowers. Another of the same species produced terminal violet flowers. Identify the dominant traits.
Ans. Axial, violet flower.

Question. State what does aneuploidy lead to.
Ans. Aneuploidy leads to individuals with abnormal number of chromosomes. Some disorder due to aneuploidy are Down’s Syndrome, Turner’s Syndrome, Klinefelter’s Syndrome.

Question. A garden pea plant produced round green seeds. Another of the same species produced wrinkled yellow seeds. Identify the dominant traits.
Ans. Round, yellow seed are the dominant traits.

Question. How many kinds of phenotypes would you expect in F2 generation in a monohybrid cross?
Ans. Two (e.g., Tall and dwarf).

Question. How many chromosomes do drones of honeybee possess? Name the type of cell division involved in the production of sperms by them.
Ans. Drones possess 16 chromosomes. Mitosis is involved in the production of sperms.

Question. When a tall pea plant was self-pollinated, one-fourth of the progeny were dwarf. Give the genotype of the parent and dwarf progenies.
Ans. Genotype of parent is Tt and the genotype of dwarf progenies is tt.

Question. Give an example of a chromosomal disorder caused due to non-disjunction of autosomes.
Ans. Down’s Syndrome.

Question. Discuss is the genetic basis of wrinkled phenotype of pea seeds.
Ans. Wrinkled seed shape is a recessive trait. It expresses only under homozygous condition of alleles.

Short Answer Questions

Question. What are the characteristic features of a true-breeding line?
Ans. A true-breeding line for a trait is one that has undergone continuous self-pollination, showing a stability in the inheritance of the trait for several generations.

Question. During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the offspring populations were tall and dwarf in equal ratio. Work out a cross to show how it is possible.
Ans.

Principles of Inheritance and Variation Exam Questions Class 12 Biology

Question. With the help of a Punnett square, find the percentage of homozygous talls in a F2 population involving a true breeding tall and a true breeding dwarf pea plant.
Ans.
Percentage of homozygous tall = 1/4×100 = 25%

Question. With the help of a Punnett square, find the percentage of heterozygous individuals in a F2 population in a cross involving a true breeding pea plant with green pods and a true breeding pea plant with yellow pods respectively.
Ans.

Question. In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated by a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, Red = 138 Tall, White = 132
Dwarf, Red = 136 Dwarf, White = 128
Mention the genotypes of the two parents and of the types of four offsprings.
Ans. The result shows that the four types of offspring are in a ratio of 1 : 1 : 1 : 1. Such a result is observed in a test cross progeny of a dihybrid cross.
The cross can be represented as:
Parents: Tall and red (TtRr) × Dwarf and white (ttrr)
Offsprings:

Question. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Ans. Here, we apply the formula 2n where n = number of loci.
The organism is heterozygous for 4 loci,
n = 4
So, 2n = 24 = 2 × 2 ×2 × 2 = 16
∴ The organism will produce 16 types of gametes.

Question. In a typical monohybrid cross the F2 population ratio is written as 3:1 for phenotype but expressed as 1:2:1 for genotype. Explain with the help of an example.
Ans. This is a case of Mendel’s monohybrid cross.

Question. In snapdragon, a cross between true-breeding red flowered (RR) plants and true-breeding white flowered (rr) plants showed a progeny of plants with all pink flowers.
(a) The appearance of pink flowers is not known as blending. Why?
(b) What is this phenomenon known as?
Ans. (a) R (dominant allele red colour) is not completely dominant over r (recessive allele white colour). r maintains its originality and reappears in F2 generation. Therefore, it is not blending.
(b) Incomplete dominance.

QuestionQ. 10. The phenotypic and genotypic ratio in F2 generation are same in a certain kind of inheritance.
Name an organism in which it occurs and mention the kind of inheritance involved.
Ans. This kind of inheritance occurs in Mirabilis jalapa (4 O’clock plant) and the type of inheritance is called incomplete dominance.

Question. In a particular plant species, majority of the plants bear purple flowers. Very few plants bear white flowers. No intermediate colours are observed. If you are given a plant bearing purple flowers, how would you ascertain that it is a pure breed for that trait? Explain.
Ans. By test cross. Cross, purple flower plant with a (homozygous) recessive plant with white flowers,if all the flowers of the progeny are purple, the plant is homozygous dominant, i.e. pure breed.

Question. In snapdragon (Antirrhinum majus), a cross between varieties with red and white flowers produces all pink progeny. Explain how it is a case of incomplete dominance and not of blending inheritance.
Ans. In incomplete dominance, the genes of an allelomorphic pair are not expressed as dominant and recessive, but express themselves partially when present together in a hybrid and is an intermediate between the two genes. As a result an intermediate character is obtained. e.g., Two types of flowers occur in Mirabilis jalapa (4 o’ clock plant) and Antirrhinum majus (snapdragon/ dog flower). The red flower colour is due to gene RR, white flower colour is due to gene rr but pink flower colour appears in case of genotype Rr.)
It is not a case of blending inheritance because the parental characters reappear in the F2 generation without any modification.

Question. Explain the mechanism of sex determination in insects like Drosophila and grasshopper.
Ans. In grasshopper, the mechanism of sex determination is of the XO type. In females, the eggs bear a pair of X chromosomes along with the autosomes. Males contain only 1 X chromosome with autosomes. On the other hand, there are two types of sperms formed in males–one having a X chromosome and other without X chromosome. Hence, grasshopper shows male heterogamety.

Question. Mention the advantages of selecting pea plant for experiment by Mendel.
Ans. • Mendel selected garden pea as his experimental material because of the following reasons:
(i) It is an annual plant with a short life-cycle. So, several generations can be studied within a short period.
(ii) It has perfect bisexual flowers containing both male and female parts.
(iii) The flowers are predominantly self-pollinating. It is easy to get pure line for several generations.
(iv) It is easy to cross-pollinate them because pollens from one plant can be introduced to the stigma of another plant by removing the anthers.
(v) Pea plant produces a large number of seeds in one generation.
(vi) Pea plants could easily be raised, maintained and handled.
(vii) A number of easily detectable contrasting characters/

Question. Differentiate between the following:
(a) Dominance and recessiveness
(b) Homozygous and heterozygous
(c) Monohybrid and dihybrid
Ans. (a) Table 5.3: Differences between dominance and recessiveness

S.No.	Dominance	Recessive
(i)	The phenomenon where one allele expresses itself even in the presence of other allele.	The phenomenon where an allele expresses itself in the absence of its dominant allele but remains masked in its presence.
(ii)	Dominant allele forms a complete functional enzyme due to which complete polypeptide is formed to express completely.	Recessive allele forms incomplete or defective, or non-functional polypeptide enzyme, due to which non-functional polypeptide is formed and fails to express completely.

(b) Differences between homozygous and heterozygous

S.No.	Homozygous	Heterozygous
(i)	When both alleles of a gene are similar, then the individual is called homozygous.	When both alleles of a gene are dissimilar, then the individual is called heterozygous.
(ii)	The genotype is expressed as TT or tt, i.e., they contain either both dominant alleles or both recessive alleles.	The genotype is expressed as Tt, i.e., they contain one dominant allele and one recessive allele.
(iii)	They are true breeding, leading to pure lines.	They are not true breeding.
(iv)	The gametes produced by them are similar in genotype.	The gametes produced by them are of two types, one with dominant allele and other with recessive allele.

(c) Differences between monohybrid and dihybrid

S.No.	Monohybrid	Dihybrid
(i)	It is the cross between two individuals considering a single contrasting trait or character at a time.	It is the cross between two individuals taking two contrasting traits or characters at a time.
(ii)	It helps to study the inheritance of a pair of allele.	It helps to study the inheritance of two pairs of allele.
(iii)	The phenotypic ratio in F2 generation is 3 : 1.	The phenotypic ratio in F2 generation is 9 : 3 : 3 : 1.
(iv)	The genotypic ratio in F2 generation is 1 : 2 : 1.	The genotypic ratio in F2 generation is 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1.

Question. Who proposed chromosomal theory of inheritance? Point out any two similarities in the behaviour of chromosomes and genes.
Ans. It was proposed by Sutton and Boveri.
Similarities:
(i) Both genes and chromosomes occur in pairs in a diploid cell (2n).
(ii) Both of them separate out during gametogenesis to enter into different gametes.
(iii) Paired condition is again restored by fusion of gametes. (Any two)

Question. Explain the law of dominance using a monohybrid cross.
Ans. Law of dominance states that when two different allelomorphic forms (genes) are present in an organism, only one expresses itself in F1 generation which is called dominant gene while the other which does not show its effect and remains masked is called recessive gene.
When a cross is performed between two individuals taking a single contrasting character at a time, it is called a monohybrid cross.
The character ‘height’ in pea plant has two alleles ‘T’ and ‘t’. ‘T’ exhibits tallness whereas ‘t’ exhibits dwarfness. When a pure tall (TT) pea plant is crossed with a pure dwarf (tt) plant, in the
F1 generation hybrid (heterozygous) ‘Tt’ is obtained, which is tall due to the presence of allele ‘T’.
This shows that tallness is dominant over dwarfness which remain unexpressed in generation.
Thus, this cross explains law of dominance.
For cross

Question. Define and design a test cross.
Ans.

• It is a method devised by Mendel to determine the genotype of an organism.
• In this cross, the organism with dominant phenotype (but unknown genotype) is crossed with the recessive individual.
• In a monohybrid cross between violet colour flower (W) and white colour flower (w), the F1 hybrid was violet colour flower. The test crosses are:

Long Answer Questions

Question. The F2 progeny of a monohybrid cross showed phenotypic and genotypic ratio as 1 : 2 : 1,unlike that of Mendel’s monohybrid F2 ratio. With the help of a suitable example, work out a cross and explain how it is possible.
Ans. This kind of cross is observed in Mirabilis jalapa/Four o’clock plant/Antirrhinum majus.
For cross,

In heterozygous condition a single dominant gene is not sufficient to produce red colour therefore
it is a case of incomplete dominance.

Question.
(a) How do human males with ‘XXY’ abnormality suffer?
Ans. (a) The XXY individual suffers from Klinefelter’s syndrome.

Question. A pea plant with purple flowers was crossed with white flowers producing 50 plants with only purple flowers. On selfing, these plants produced 482 plants with purple flowers and 162 with white flowers. What genetic mechanism accounts for these results? Explain.
Ans. The gene for purple flowers is dominant over that of white flowers. So, when two pure varieties are crossed, the F1 generation has only purple flowers and on selfing, the flowers are produced in a 3 : 1 ratio in F2 generation.

This result is obtained due to segregation of the alleles at the time of gametogenesis. The alleles remain together in a zygote but during gamete formation, they segregate such that the gametes carry only one allele.

Question. A cross is made between different homozygous pea plants for contrasting flower positions.
(a) Find out the position of flowers in F1 generation on the basis of genotypes.
(b) Work out the cross upto F2 generation.
(c) Compute the relative fraction of various genotypes in the F2 generation?
Ans. (a) Axial position
(b)

Question. (a) Explain the phenomena of dominance, multiple allelism and co-dominance taking ABO blood group as an example.
(b) What is the phenotype of the following?
(i) I^Ai (ii) ii
Ans. (a) Dominance: The alleles IA and IB both are dominant over allele i as IA and IB form antigens A and B, respectively, but i does not form any antigen.
Multiple allelism: It is the phenomenon of occurrence of a gene in more than two allelic forms on the same locus. In ABO blood group in humans, one gene I has three alleles IA, IB and IO/i.
Co-dominance: It is the phenomena in which both alleles express themselves when present together. We inherit any two alleles for the blood group. When the genotype is IAIB the individual has AB blood group since both IA and IB equally influence the formation of antigens A and B.
(b) (i) IAi — A blood group.
(ii) ii — O blood group.

Question. When snapdragon plant bearing pink colour flower was selfed, it was found that; 69 plants were having red coloured flowers. What would be the number of plants bearing pink flower and white flower? Show with the help of Punnett square. Identify the principle of inheritance involved in this experiment.
Ans. (a) There will be 138 pink flower bearing plants and 69 white flower bearing plants according to the ratio 1 : 2 : 1.
(b) P ink (Rr) selfing

Phenotypic ratio— red : pink : white
1 : 2 : 1
(c) The principle involved in the experiment is incomplete dominance.

Question. You are given a red flower-bearing pea plant and a red flower-bearing snapdragon plant. How would you find the genotypes of these two plants with respect to the colour of the flower?
Explain with the help of crosses. Comment upon the pattern of inheritance seen in these two plants.
Ans. A test cross is required to find out the genotype of both the plants.
(a) Garden pea

If the F1 generation plants have all red flowers, the genotype of the parent plant will be homozygous dominant and if the F1 generation plants have red and white flowers in the ratio
of 1 : 1, then the genotype of the parent plant is heterozygous dominant. This inheritance follows the Mendelian law of dominance.
(b) In snapdragon:

The parent plant will be homozygous for flower colour because a heterozygous plant will have pink flowers due to the phenomenon of incomplete dominance.

Question. (a) You are given tall pea plants with yellow seeds whose genotypes are unknown. How would you find the genotype of these plants? Explain with the help of cross.
(b) Identify a, b and c in the table given below: (image 183)

Ans. (a) Test cross will be performed to know the genotype of these plants. (image 184)

If all the plants of F1 generation are tall with yellow seeds, then the phenotype of the parent is homozygous dominant (case i). If the plants in F1 generation are in the ratio of 1 : 1 : 1 : 1, then the parent plant is heterozygous dominant.
(b) a–Both the forms of a trait are equally expressed in F1 generation.
b–Dominance.
c– Phenotypic expression of F1 generation is somewhat intermediate between the two parental forms of a trait.

Question. Let us assume in a given plant the genotype symbol “Y” stands for dominant yellow seed colour and “y” for recessive green seed colour; symbol “R” for round seed shape and “r” for wrinkled seeds. Two homozygous parents (plants) with genotypes “RRYY” and “rryy” are crossed and their F1-generation progeny is then selfed. What shall be the
(a) Phenotype of F1-progeny
(b) Genotype of F1-progeny
(c) Gamete genotypes of F1-progeny
(d) Phenotypic ratio of F2 population
(e) Phenotypic ratio of yellow seed to green seed and round seed to wrinkled seed in F2 population.
Ans.
(a) Phenotype of F1-progeny: Round seeds that are yellow in colour
(b) Genotype of F1-progeny: RrYy
(c) Gamete genotypes of F1-progeny: RY, Ry, rY and ry
(d) Phenotypic ratio of F2 population: 9 : 3 : 3 : 1.
Nine round-yellow seeds; three round-green seeds; three wrinkled-yellow seeds; one wrinkled-green seed.
(e) Phenotypic ratio of yellow seed to green seed and round seed to wrinkled seed in F2 population:
Yellow seed to green seed = 3 : 1
Round seed to wrinkled seed = 3 : 1

Question. A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be the probability of her (a) sons (b) daughters to be colour blind? Explain with the help of pedigree chart.
Ans. The genotypes of parents are:

All daughters are normal visioned and 50% of sons are likely to be colour blind.

Question. (a) State and explain the law of segregation as proposed by Mendel in a monohybrid cross.
(b) Write the Mendelian F2 phenotypic ratio in a dihybrid cross. State the law that he proposed on the basis of this ratio. How is this law different from the law of segregation?
Ans. (a) (ii) Law of segregation or law of purity of gametes
• This law states that the factors or alleles of a pair segregate from each other during gamete formation,such that a gamete receives only one of the two factors. They do not show any blending but simply remain together.
• Homozygous parent produces all gametes that are similar, heterozygous parent produces two types of gametes, each having one allele in equal proportion.
(iii) Law of independent assortment
• According to this law the two factors of each character assort or separate out independent of the factors of other characters at the time of gamete formation and get randomly rearranged in the offsprings producing both parental and new combinations of characters.
• When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

(b) The F2 phenotypic ratio is 9:3:3:1. On the basis of this ratio Mendel proposed Law of Independent Assortment.
(iii) Law of independent assortment.
• According to this law the two factors of each character assort or separate out independent of the factors of other characters at the time of gamete formation and get randomly rearranged in the offsprings producing both parental and new combinations of characters.
• When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters.

Question. (a) State and explain the law of dominance as proposed by Mendel.
(b) How would phenotypes of monohybrid F1 and F2 progeny showing incomplete dominance in snapdragon and co-dominance in human blood group be different from Mendelian monohybrid F1 and F2 progeny? Explain.
Ans. (a) This law states that when two alternative forms of a trait or character (genes or alleles) are present in an organism, only one factor expresses itself in F1 progeny and is called dominant while the other that remains masked is called recessive.
The characters are controlled by discrete units called factors. These factors occur in pairs.

(b)

Question. (a) Write the conclusions Mendel arrived, at on dominance of traits on the basis of monohybrid crosses that he carried out in pea plants.
(b) Explain why a recessive allele is unable to express itself in a heterozygous state.
Ans. (a) Mendel concluded that:
(i) Characters are controlled by discrete units called factors.
(ii) Factors occur in pair.
(iii) In a dissimilar pair of factors one member of the pair dominates/only one of the parental character is expressed in a monohybrid cross in the F1 and both are expressed in the F2.
(b) The alleles are present on homologous chromosomes. The recessive allele does not code for its product or codes for a defective product. The other allele remains normal and thus expresses itself.

Question. (a) A true breeding homozygous pea plant with green pods and axial flowers as dominant characters, is crossed with a recessive homozygous pea plant with yellow pods and terminal flowers. Work out the cross up to F2 generation giving the phenotypic ratios of F1 and F2 generation respectively.
(b) State the Mendelian principle which can be derived from such a cross and not from monohybrid cross.

Ans. (a)

(b) From the above cross law of independent assortment can be derived which states that when two pairs of traits are combined in a hybrid, segregation of one pair of character is
independent of the other pair of characters.

Question. A homozygous tall pea plant with green seeds is crossed with a dwarf pea plant with yellow seeds:
(i) What would be the phenotype and genotype of F1?
(ii) Work out the phenotypic ratio of F2 generation with the help of a Punnett square.

Ans.