Principles Of Inheritance And Variation Class 12 Biology Important Questions

Please refer to Principles Of Inheritance And Variation Class 12 Biology Important Questions with solutions provided below. These questions and answers have been provided for Class 12 Biology based on the latest syllabus and examination guidelines issued by CBSE, NCERT, and KVS. Students should learn these problem solutions as it will help them to gain more marks in examinations. We have provided Important Questions for Class 12 Biology for all chapters in your book. These Board exam questions have been designed by expert teachers of Standard 12.

Class 12 Biology Important Questions Principles Of Inheritance And Variation

Objective Type Questions

Question. Physical bases of inheritance is
(a) Gene
(b) DNA
(c) Protein
(d) None of these

Question. Extent by which a progeny differs from the parents is called
(a) Inheritance
(b) Selection
(c) Variations
(d) Heredity

Question. Sahiwal is a species of
(a) Pea plant
(b) Chick pea
(c) Cow
(d) Horse

Question. Gregor Mendel conducted hybridization experiments on which of the following
(a) Pisum sativum
(b) Chick pea
(c) Insects
(d) Dog plant

Question. Which of the following is incorrect?
(a) Mendel’s experiments had a small sampling size, which gave greater credibility to his data.
(b) Conformation of his inferences from experiments on successive generations of his test plant proved that his results point to a general rule of inheritance.
(c) He studied characters that were manifested as two opposite traits.
(d) He took seven characters with contrasting traits.

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

Question. Emasculation is
(a) Deletion of Y–chromosome in male
(b) Absence of male characters
(c) Removal of anthers
(d) Removal of stigma

Question. In Mendel’s experiment the study of inheritance of one gene (cress between tall and dwarf plant) showed
(a) 50% tall and 50% dwarf plants in F₁ progeny
(b) 75% tall and 25% dwarf plants in F₁ progeny
(c) All dwarf plants
(d) All Tall plants

Question. Purpose of doing emasculation was
(a) to perform only cross pollination between the parent generation
(b) to perform only self pollination between the parent generation
(c) to store the pollen of female plants
(d) None of these

Very Short Answer Questions

Question. Write the types of sex determination mechanisms the following crosses show. Give an example of each type.
(i) Female XX with Male XO
(ii) Female ZW with Male ZZ 
Ans. (i) Male heterogamety, Grasshopper
(ii) Female heterogamety, Birds

Question. Mention the type of allele that expresses itself only in homozygous state in an organism.
Ans. Recessive allele.

Question. Write the percentage of F2 homozygous and heterozygous populations in a typical monohybrid cross.
Ans. The ratio of a typical monohybrid cross is 1 : 2 : 1 where 50% are homozygous and 50% are heterozygous populations. (25% homozygous dominant, 25% homozygous recessive)

Question. Why do normal red blood cells become elongated sickle shaped structures in a person suffering from sickle cell anaemia? 
Ans. Due to point mutation, glutamic acid (Glu) is replaced by valine (Val) at the sixth position of b-globin chain of haemoglobin molecule. Under oxygen stress erythrocytes lose their circular shape and become sickle-shaped.

Question. Name the type of cross that would help to find the genotype of a pea plant bearing violet flowers. 
Ans. Test cross.

Question. A cross was carried out between two pea plants showing the contrasting traits of height of the plant. The result of the cross showed 50% of parental characters. Name the type of cross.
Ans. Test cross

Question. A human being suffering from Down’s syndrome shows trisomy of 21st chromosome. Mention the cause of this chromosomal abnormality. 
Ans. Due to non-disjunction i.e., 21st pair of chromosomes fail to separate during gametogenesis.
Therefore, the gamete possesses 24 chromosomes instead of 23. When such a gamete fuses with another gamete, the zygote will have three copies of chromosome 21 causing trisomy.

Question. Why, in a test cross, did Mendel cross a tall pea plant with a dwarf pea plant only?
Ans. To determine the genotype of the tall plant, whether it is homozygous dominant or heterozygous,as dwarfness is a recessive trait which is expressed only in homozygous condition and he was sure of genotype of dwarf plant.

Question. Name the stage of cell division where segregation of an independent pair of chromosomes occurs. 
Ans. Anaphase-I of Meiosis-I.

Question. Name the respective pattern of inheritance where F1 phenotype
(a) does not resemble either of the two parents and is in between the two.
(b) resembles only one of the two parents.
Ans. (a) Incomplete dominance (b) Dominance

Question. Name a human genetic disorder due to the following:
(i) An additional X-chromosome in a male
(ii) Deletion of one X-chromosome in a female
Ans. (i) Klinefelter’s Syndrom e (ii) Turner’s Syndrome

Short Answer Questions

Question. How would you find genotype of a tall pea plant bearing white flowers? Explain with the help of a cross. Name the type of cross you would use.
Ans. It can be done by a test cross. This is done by crossing the plant with homozygous recessive parent. If the ratio of progeny is 1 : 1, then the genotype of the plant is heterozygous.

Question. When a tall pea plant was selfed, it produced one-fourth of its progeny as dwarf. Explain with the help of a cross.
Ans. Production of one-fourth dwarf progeny on selfing of a tall pea plant indicates that the plant is heterozygous. This can be explained with the cross as follows: 

Question. A teacher wants his/her students to find the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible.
Ans. Test cross will be done.
• 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:

Question. When does a geneticist need to carry a test cross? How is it carried? 
Ans. Geneticists carry out a test cross to find out the genotype of the unknown parent. This is carried out by crossing the progeny with the homozygous recessive parent.

Question. Mendel crossed plants that bred true for yellow seeds with plants that bred true for green seeds.
All seeds in the F₁ generation were yellow. Work out the inheritance involved in this cross by using symbols for the trait. Which trait was dominant? 
Ans. 

Question. A cross was carried out between two pea plants showing the contrasting traits of height of the plant. The result of the cross showed 50% of parental characters.
(i) Work out the cross with the help of a Punnett square.
(ii) Name the type of the cross carried out.

Ans. (i)

(ii) Test cross

Question. What is a test cross? How can it decipher the heterozygosity of a plant?
Ans. A cross to analyse whether genotype of dominant individual is homozygous or heterozygous is called test cross.
On crossing with a recessive parent, if 50% of progeny have dominant trait and 50% have recessive trait then the plant is said to he heterozygous.

In the first cross the tall parent plant is heterozygous for the trait, in second cross tall parent plant is homozygous for the trait, hence the respective observation.

Question. How does a test cross help to determine the genotype of an individual? 
Ans. In a test cross the individual of unknown genotype is crossed with the recessive parent. If all progenies are dominant, then the genotype exhibits homozygosity and if the progenies have a dominant to recessive ratio 1 : 1, then the genotype exhibits heterozygosity.

Question. Human blood group is a good example of multiple allelism and co-dominance. Justify.
Ans. Multiple allelism: Generally in an individual or population, only two alleles of a trait govern the character, but in case of ABO blood group, three alleles IA, IB and i are found to govern blood group in human population. This is multiple allelism.
Co-dominance: Allele IA and IB when present in an individual, both being dominant express their own types of sugars or traits. Thus, exhibiting co-dominance.

Question. With the help of one example, explain the phenomena of co-dominance and multiple allelism in human population.
Ans. ABO blood group in human being is an example of multiple allelism and co-dominance. There are three alleles for the gene I, i.e., IA, IB, and i, thus, exhibiting multiple allelism.
When IA and IB are present together the blood group is AB. Both A and B blood groups are expressed. This is called co-dominance.

Question. Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross.
Ans. Sickle cell anaemia is an autosomal recessive trait in humans. The disease is controlled by a single pair of alleles HbA and HbS. Only the homozygous individuals HbS HbS show the diseased phenotype. The heterozygous individuals (HbAHbS) are carriers. 

Question. Recently a baby girl has been reported to suffer from haemophilia. How is it possible? Explain with the help of a cross.
Ans. It is possible to have a haemophilic girl if a carrier woman married a haemophilic man as shown here: (Image 168)

Question. Haemophilia is a sex-linked inheritance condition in humans where a simple cut causes nonstop bleeding. Study the pedigree chart showing the inheritance of haemophilia in a family. Answer the questions that follow: 

Give reasons which explain that haemophilia is
(i) sex-linked, and
(ii) caused by ‘X’-linked gene.
Ans. (i) Haemophilia is sex-linked because it shows transmission from unaffected carrier female to some of the male progeny.
(ii) Haemophilia is caused by ‘X’-linked gene because the heterozygous female for haemophilia may transmit the disease to sons. It appears more in males because of only one X chromosome.

Question. A colour-blind man marries a woman with normal vision whose father was colour-blind.
Work out a cross to show the genotype of the couple and their respective sons.
Ans. The father of normal woman is colour-blind, so the woman will be carrier, i.e., XXC. 

Question. Marriage between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man, to their surprise, the grandson was also haemophilic.
(a) Represent this cross in the form of a pedigree chart. Give the genotypes of the daughter and her husband.
(b) Write the conclusion you draw from the inheritance pattern of this disease.
Ans. (a) 

(b) Sex-linked recessive inheritance pattern.

Question. 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 the two loci are completely linked, then there would be no segregation and the F1 generation will exhibit parental characteristics only. But if the two loci are incompletely linked, then segregation would occur partly and the F1 generation will exhibit both parental and recombinant characteristics but the recombinants will be in a very small proportion depending on distance between gases.

Long Answer Questions

Question. For flower colour in pea, the allele for purple flower (P) is dominant to the allele for white flower (p). A purple flowered plant therefore could be of genotype PP or Pp. What genetic cross would you make to determine the genotype of a purple flowered plant? Explain how your cross gives you the correct genotype of the purple flowered plant? 
Ans. The genotype of a purple flowered plant can be determined by conducting a test cross i.e., crossing the purple flowered plant with homozygous recessive individual i.e., pp.

If the F1 generation produces purple and white flowers in 1:1 ratio, the parent would be heterozygous, i.e., Pp.

Question. How are dominance, co-dominance and incomplete dominance patterns of inheritance different from each other?
Ans. Dominance: It is a phenomenon in which when two contrasting alleles are present together, only one expresses itself and is called dominant whereas the other which does not express itself is called recessive e.g., Tt – ‘T’ is dominant over t (dwarfness).
Co-dominance: It is a phenomenon in which when two contrasting alleles are present together, both the alleles express themselves e.g., IA IB genotype gives blood group AB.++ Incomplete dominance: It is a phenomenon in which when two contrasting alleles are present together neither of the alleles is dominant over other and the phenotype formed is intermediate of the two alleles. e.g.,

Red flower × White flower → Pink flower colour
   RR    ×    rr                           Rr

Question. Both Down’s syndrome and Turner’s syndrome are examples of chromosomal disorders. Cite the differences between the two. 
Ans. Differences between Down’s syndrome and Turner’s syndrome 

S.No.	Down’s syndrome	Turner’s syndrome
(i)	It is a trisomy of chromosome number 21.	It is a monosomy of the X-chromosome
(ii)	It can occur in either males or females.	It can occur only in females.
(iii)	The total number of chromosomes in the genome is 47.	The total number of chromosomes in the genome is 45.

Question. During his studies on genes in Drosophila that were sex-linked. T.H. Morgan found population phenotypic ratios deviated from expected 9 : 3 : 3 : 1. Explain the conclusion he arrived at.
Ans. (i) He observed that when the two genes in a dihybrid cross are located on the same chromosome,the proportion of parental gene combinations in the progeny was much higher than the nonparental or recombination of genes.
(ii) Morgan and his group found that when genes were grouped on the same chromosome, some genes are tightly linked and show less recombination.
(iii) When the genes are loosely linked they show higher recombination.

Question. A red-eyed heterozygous female fruit fly is crossed with a red-eyed male. Work out all possible genotypes and phenotypes of the progeny. Comment on the pattern of inheritance of eye colour in fruit flies. 

(ii) The gene for eye colour is sex-linked and is present on X chromosome.
The character passes into the male from female and the male passes it to the female in the next generation. Male has only one X-chromosome and one Y-chromosome with no corresponding allele.

In humans, male heterogamety is observed. They exhibit XY type of sex determination.

Question. (i) How does a chromosomal disorder differ from a Mendelian disorder?
(ii) Name any two chromosomal aberration associated disorders.
(iii) List the characteristics of the disorders mentioned above that help in their diagnosis.

Ans. (i) Differences between Mendelian disorder and chromosomal disorder

S.No.	Mendelian disorder	Chromosomal disorder
(i)	This disorder is mainly due to alteration or mutation in the single gene.	This disorder is caused due to absence or excess or abnormal arrangement of one or more chromosomes.
(ii)	This follows Mendel’s principles of inheritance.	This does not follow Mendel’s principles of inheritance.
(iii)	This may be recessive or dominant in nature	This is always dominant in nature.
(iv)	For example, haemophilia, sickle-cell anaemia.	For example, Turner’s syndrome.

(ii) Two chromosomal aberration-associated disorders are Down’s syndrome and Klinefelter’s syndrome.
(iii) (a) Down’s syndrome: The individuals have overall masculine development but they express feminine development like development of breast, i.e., gynaecomastia. They are sterile.
(b) Klinefelter’s syndrome: The females are sterile as ovaries are rudimentary. Other secondary sexual characters are also lacking.

Question. Identify ‘a’, ‘b’, ‘c’, ‘d’, ‘e’ and ‘f’ in the table given below: 

S.No.	Haemophilia	Sickle-cell anaemia
(i)	It is a sex-linked recessive disorder.	It is an autosomal linked recessive trait.
(ii)	The gene for haemophilia is located on X-chromosome.	The disease is controlled by a single pair of allele HbA and HbS.
(iii)	More males suffer from haemophilia than females because in males single gene for the defect is able to express. Females suffer from this disease only in homozygous condition,i.e., XcXc.	Only the homozygous individuals for HbS, i.e., HbS HbS show the diseased phenotype.
(iv)	The defective alleles produce non-functional protein which later form a non-functional cascade of proteins involved in blood clotting.	Due to point mutation Glutamic acid (Glu) is replaced by Valine (Val) at sixth positions of beta globin chain of haemoglobin molecule.

Ans. (a) Short stature/small round head/furrowed tongue/partially open mouth/ mental development retarded.
(b) Both
(c) Klinefelter’s syndrome
(d) Male
(e) (i) Sterile ovaries;i i)( Lack of secondary sexual characters.
(f) Female

Question. (a) A pea plant bearing axial flowers is crossed with a pea plant bearing terminal flowers. The cross is carried out to find the genotype of the pea plant bearing axial flowers. Work out the cross to show the conclusions you arrive at.
(b) State the Mendel’s law of inheritance that is universally acceptable. 
Ans. (a) (i) If the plant is homozygous for the dominant trait.

Conclusion: If all progeny show axial flowers (dominant) the plant is homozygous (AA),If 50% of progeny show axial flower (Dominant) and 50% terminal flower (Recessive) the plant is heterozygous.

(b) Law of Segregation is universally accepted. It states that allelic pair segregate (separates) during gamete formation.

Question. In the case of snapdragon (Antirrhinum majus) a plant with red flowers was crossed with another plant with white flowers. Trace the inheritance of flower colour up to F2 generation indicating the genotype and phenotype at each level. What special feature do you notice in the genotype and phenotype ratio in F2 generation?
Ans. Comment: This is a case of Mendelian deviation and that shows incomplete dominance as red and white both are not expressed but produce pink trait in F1. Here, both the genotypic and phenotypic ratio are 1 : 2 : 1.

Question. (a) Provide genetic explanation for the observation in which the flower colour in F1 generation of snapdragon did not resemble either of the two parents. However, the parental characters reappeared when F1 progenies were selfed.
(b) State the three principles of Mendel’s law of inheritance.
Ans. (a) This is an exception to Mendel’s principle of dominance and can be explained by the phenomenon of ‘Incomplete dominance’. It is a phenomenon where none of the two contrasting alleles or factors are dominant. The expression of the character in a hybrid or F1 individual is intermediate or a fine mixture of expression of the two factors (pink flowers in this case from two parents with red and white flowers). This may be considered as an example of quantitative inheritance where only a single gene pair is involved. F2 phenotypic ratio is 1 : 2 : 1, similar to the genotypic ratio, in which the parental characters also reappear.

Question. (a) Differentiate between dominance and co-dominance.
(b) Explain co-dominance taking an example of human blood groups in the population.
Ans. (a) Dominance: It is a phenomenon in which when two contrasting alleles are present together, only one expresses itself and is called dominant whereas the other which does not express itself is called recessive.
Co-dominance: It is a phenomenon in which when two contrasting alleles are present together, both of the alleles express themselves.
(b) Co-dominance
• The alleles which are able to express themselves independently, even when present together are called co-dominant alleles and this biological phenomenon is called co-dominance
• ABO blood groups are controlled by gene I. Gene I has three alleles IA, IB and IO/i.
• IA and IB produce RBC surface antigens which are sugar polymers A and B, respectively, whereas i does not produce any antigen.
• IA and IB are dominant over i hence IA and IB are dominant alleles and i is recessive allele as in IAi and IBi.
• When IA and IB are present together, both express equally and produce both the surface antigens A and B, hence show co-dominance.
• Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.

Question. (a) Four children with four different blood groups are born to parents where the mother has blood group ‘A’ and the father has blood group ‘B’. Work out the cross to show the genotypes of the parents and all four children.
(b) Explain the contribution of Alfred Sturtevant in ‘Chromosome mapping’.
Ans. (a) Parents

(b) Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between genes and mapped their position on the chromosome.

Question. (a) Write the blood group of people with genotype IAIB. Give reasons in support of your answer.
(b) In one family, the four children each have a different blood group. Their mother has blood group A and their father has blood group B. Work out a cross to explain how it is possible.
Ans. (a) Blood group AB. Both the alleles IA and IB are co-dominant and express themselves completely.
(b) A cross is carried out between heterozygous father (for blood group B) and heterozygous mother (of blood group A) to get four children with different blood groups.

All the four blood groups are controlled by three allelic genes IA, IB, i and thus it shows phenomena of multiple allelism. Both IA and IB are dominant over i. However, when together,both are dominant and show the phenomena of co-dominance forming the blood group AB. Six genotypes are possible with combination of these three alleles.

Question. (a) List the three different allelic forms of gene ‘I’ in humans. Explain the different phenotypic expressions, controlled by these three forms.
(b) A woman with blood group ‘A’ marries a man with blood group ‘O’. Discuss the possibilities of the inheritance of the blood groups in the following starting with ‘yes’ or ‘no’ for each:
(i) They produce children with blood group ‘A’ only.
(ii) They produce children some with ‘O’ blood group and some with ‘A’ blood group.

Ans. (a) The three different allelic forms are: IA, IB, IO/i.
The alleles which are able to express themselves independently, even when present together are called co-dominant alleles and this biological phenomenon is called co-dominance For example, ABO blood grouping in humans.
• ABO blood groups are controlled by gene I. Gene I has three alleles IA, IB and IO/i.
• IA and IB produce RBC surface antigens which are sugar polymers A and B, respectively, whereas i does not produce any antigen.
• IA and IB are dominant over i hence IA and IB are dominant alleles and i is recessive allele as in IAi and IBi.
• When IA and IB are present together, both express equally and produce both the surface antigens A and B, hence show co-dominance.
• Since humans are diploid, each person possesses any two of the three ‘I’ gene alleles, resulting into six different genotypic combinations and four phenotypic expressions.
(b) (i) Yes; when both the parent are homozygous 

(ii) Yes; when the woman is heterozygous

Question. (a) How are Mendelian inheritance, polygenic inheritance and pleiotropy different from each other?
(b) Explain polygenic inheritance pattern with the help of a suitable example. 
Ans (a) Differences between Mendelian inheritance, polygenic inheritance and pleitropy

Mendelian Inheritance	Polygenic inheritance	Pleiotropy
One gene controls one trait/character/phenotype	Two or more genes influence the expression of one trait/ character/phenotype	One genes controls the expression of more than one traits/characters/phenotypes

(b) Human height or skin colour are examples of polygenic inheritance. Height trait is controlled by at least three gene pairs. Additive effect of alleles contributes to the phenotypic expression of the trait. The more is the number of dominant alleles, more pronounced is the phenotypic expression or more is the height. The recessive alleles are less pronounced in the phenotypic expression.