Please refer to Principles of Inheritance and Variation Class 12 Biology Important Questions given below. These solved questions for Principles of Inheritance and Variation have been prepared based on the latest CBSE, NCERT and KVS syllabus and books issued for the current academic year. We have provided important examination questions for Class 12 Biology all chapters.
Class 12 Biology Principles of Inheritance and Variation Important Questions
Very Short Answer Questions
Question. The egg of an animal contains 10 chromosomes, of which one is X-chromosome. How many autosomes would there be in the karyotype of this animal?
Ans. There will be 9 pairs of autosomes in the karyotype of this animal.
Question. What is point mutation? Give one example.
Ans. Point mutation is a gene mutation that arises due to change in a single base pair of DNA.
Example: Sickle-cell anaemia.
A substitution of a single nitrogen base (GAG → GUG) at the sixth codon of the b-globin chain of haemoglobin molecule causes substitution of Glutamic acid by Valine at 6th position & thus the change in the shape of the RBC from biconcave disc to elongated spindle shaped, structure which results in sickle-cell anaemia.
Question. State a difference between a gene and an allele.
Ans. Gene contains information that is required to express a particular trait whereas alleles are alternating forms of a gene and are the code for a pair of contrasting traits for e.g., for plant height has two alleles – for tallness and dwarfness.
Question. Observe the pedigree chart and answer the following questions: (image 159)
(a) Identify whether the trait is sex-linked or autosomal.
(b) Give an example of a disease in human beings which shows such a pattern of inheritance.
Ans. (a) The trait is sex-linked.
(b) Haemophilia, Colour blindness (Any one)
Question. A haemophilic man marries a normal homozygous woman. What is the probability that their daughter will be haemophilic?
Ans. 0% because only one X chromosome will carry the haemophilia gene. So, she will be a carrier.
Question. A haemophilic son was born to normal parents. Give the genotypes of the parents and son.
Ans. Father : 44 + XY
Mother : 44 + XXh
Son : 44 + XhY.
(Xh= X chromosome with gene for haemophilia)
Question. State the chromosomal defect in individuals with Turner’s syndrome.
Ans. Monosomy of sex chromosome in females (XO condition).
Question. Name the event, during cell division cycle that results in the gain or loss of chromosome.
Ans. Failure of segregation of chromosomes.
Question. Name one autosomal dominant and one autosomal recessive Mendelian disorder in humans.
Ans. Huntington’s disease is an autosomal dominant disorder and sickle-cell anaemia is an autosomal recessive disorder.
Question. Why is it that the father never passes on the gene for haemophilia to his sons? Explain.
Ans. Haemophilia is a sex-linked recessive disease and the defective gene is present on X chromosome only and not on Y chromosome. Father never passes X chromosome to the son as father only contributes Y chromosome to the son.
Question. Who had proposed the chromosomal theory of inheritance?
Ans. In 1902, Walter Sutton and Theodore Boveri proposed the chromosomal theory of inheritance.
Short Answer Questions
Question. Mention any two autosomal genetic disorders with their symptoms.
Ans. Two autosomal genetic disorders are:
(a) Down’s Syndrome: It was first described by Langdon Down (1866). It is caused due to the presence of an additional copy of the chromosome number 21, i.e., trisomy (2n + 1).
(i) Short stature with small round head
(ii) Partially open mouth
(iii) Protruding furrowed tongue
(iv) Short neck
(v) Retarded mental development
(b) Phenylketonuria: It is an inborn error of metabolism. The affected individual lack an enzyme called phenylalanine hydroxylase that converts the amino acid phenylalanine into tyrosine.
As a result, phenylalanine gets accumulated and converted into phenylpyruvic acid and other derivatives in brain, causing mental retardation. These are also excreted through urine
due to their poor absorption by kidney.
Question. In order to obtain the F1 generation, Mendel pollinated a true-breeding, say, tall plant with a true-breeding dwarf plant. But for getting the F2 generation, he simply self-pollinated the tall F1 plants. Why?
Ans. All the F1 offsprings of the cross are heterozygous so allowing self-pollination is sufficient to raise F2 offspring. Also he intended to understand the inheritance of the selected trait over generations.
Question. (a) Why is human ABO blood group gene considered a good example of multiple alleles?
(b) Work out a cross up to F1 generation only, between a mother with blood group A (Homozygous) and the father with blood group B (Homozygous). Explain the pattern of inheritance exhibited.
Ans. (a) This is because more than two alleles govern the human ABO blood group gene. (Image 166)
The cross exhibits co-dominance. When the two alleles IA and IB are present together, both the alleles express each other equally forming the blood group AB.
Question. Two independent monohybrid crosses were carried out involving a tall pea plant with a dwarf pea plant. In the first cross, the offspring population had equal number of tall and dwarf plants, whereas in the second cross it was different. Work out the crosses, and explain giving reasons for the difference in the offspring populations.
Work out a cross to find the genotype of a tall pea plant. Name the type of cross.
Ans. This type of cross called a test cross.
Question. A, B and D are three independently assorting genes with their recessive alleles a, b and d, respectively. A cross was made between individuals of AabbDD genotype and aabbdd. Explain the type of genotypes of the offspring produced.
Question. Why is pedigree analysis done in the study of human genetics? State the conclusions that can be drawn from it.
Ans. Pedigree analysis is done because control crosses are not possible in case of humans beings.
This can be useful for analysis of traits, in several generations of a family, to trace pattern of inheritance to check whether the trait is dominant or recessive or sex-linked or not.
Question. A man with blood group A married a woman with B group. They have a son with AB blood group and a daughter with blood group O. Work out the cross and show the possibility of such inheritance. (Image 166)
Thus, the F1 progeny can have all the four possible blood groups, i.e., A, B, AB and O.
Question. The pedigree chart given below, present a particular generation which shows a trait irrespective of sexes (i.e., present in both male and female). Neither of the parents of the particular generation shows the trait. Draw your conclusion on the basis of the pedigree.
Ans. The trait is autosome linked and recessive in nature. Both the parents are carriers (i.e., heterozygous). Hence, among the offsprings only few show the trait irrespective of sex. The other offsprings are either normal or carrier.
Question. Differentiate between male and female heterogamety.
Ans. Differences between male heterogamety and female heterogamety
|S.No.||Male heterogamety||Female heterogamety|
|(i)||Male produces two types of gametes (while female produces only one type of gamete)||Female produces two types of gametes (while male produces only one type of gamete)|
|(ii)||XY and XO type are two types of male heterogamety||ZW type is a type of heterogamety|
|(iii)||Example, male grasshopper produce gametes of two types––X and O.||Example, female birds produce gametes of two types––Z and W.|
Question. Explain mechanism of sex determination in birds.
Ans. In birds, females are heterogemetic and males are homogametic. Females have one Z sex chromosome and one W sex chromosome. Males have a pair of Z sex chromosome. If Z sperm fertilises Z ovum, a male offspring is produced, and if Z sperm fertilises W ovum a female offspring is produced.
Question. Both haemophilia and thalassemia are blood related disorders in humans. Write their causes and the difference between the two. Name the category of genetic disorder they both come under.
Ans. Differences between haemophilia and thalassemia
|(i)||Cause||Single protein is involved in the clotting of blood is affected.||Defects in the synthesis of globin leading to formation of abnormal haemoglobin.|
|(ii)||Genetic disorder||Sex-linked recessive disorder.||Autosomal recessive disorder.|
|(iii)||Difference||Blood does not clot due to lack of clotting factors.||Results in anaemia (abnormal or lack of haemoglobin).|
Question. Name the phenomenon that leads to situations like ‘XO’ abnormality in humans. How do humans with ‘XO’ abnormality suffer? Explain.
Ans. Absence of one X chromosome due to non segregation of chromatids during cell division leads to XO abnormality. These are sterile female with rudimentary ovaries. They have shield-shaped thorax, webbed neck, poor development of breasts, short stature, small uterus and puffy fingers.
Question. Differentiate between “ZZ” and “XY” type of sex-determination mechanisms.
Ans. ZZ type is seen in birds. The males are homogametic (ZZ) and females are heterogametic (ZY).
Sex is determined by the type of egg getting fertilised.
XY type is seen in human beings The males are heterogametic (XY) and females homogametic (XX). Sex is determined by the type of sperm fertilising the ovum.
Question. Which chromosome carries the mutated gene causing b-thalassemia? What are the problems caused by the mutation?
Ans. Chromosome number 11 carries the mutant gene causing b-thalassemia. It causes formation of abnormal haemoglobin molecules, resulting into anaemia.
Question. Haemophilia is a sex-linked recessive disorder of humans. The pedigree chart given below shows the inheritance of haemophilia in one family. Study the pattern of inheritance and answer the questions given.
(a) Give all the possible genotypes of the members 4, 5 and 6 in the pedigree chart.
(b) A blood test shows that the individual 14 is a carrier of haemophilia. The member numbered 15 has recently married the member numbered 14. What is the probability that
their first child will be a haemophilic male?
Ans. (a) Genotypes of member 4—XX or XXh
Genotype of membe
r 5—XhY and Genotype of member 6—XY
(b) The probability of first child to be a haemophilic male is 25%.
50% sons will be colour-blind and rest 50% will be normal.
50% daughters will be colour-blind and rest 50% will be carriers.
Question. Give the chromosomal constitution and the resulting sex in each of the following syndromes:
(i) Turner’s syndrome
(ii) Klinefelter’s syndrome
Ans. (i) XO, female
(ii) XXY, male with female characters
Question. How is sex determined in human beings?
Ans. • Humans show XY type of sex determining mechanism.
• Out of 23 pair of chromosomes, 22 are autosomes (same in both males and females).
• Females have a pair of X-chromosomes.
• Males have an X and a Y chromosome.
• During spermatogenesis males produce two types of gametes with e•al probability – sperm carrying either X or Y chromosome.
• During oogenesis, females produce only one types of gamete – having X chromosome.
• An ovum fertilised by the sperm carrying X-chromosome develops into a female (XX) and an ovum fertilised by the sperm carrying Y-chromosome develops into a male (XY).
Long Answer Questions
Question. What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?
Ans. Down’s syndrome is a human genetic disorder caused due to trisomy of chromosome 21. Such individuals are aneuploid and have 47 chromosomes (2n + 1). The symptoms include mental retardation, growth abnormalities, constantly open mouth, dwarfness, etc. The reason for the disorder is the non-disjunction (failure to separate) of homologous chromosome of pair 21 during meiotic division in the ovum.
The chance of having a child with Down’s syndrome increase with the age of the mother (40+) because ova are present in females since their birth and therefore older cells are more prone to chromosomal non-disjunction because of various physicochemical exposures during the mother’s life-time.
Question. Explain how does trisomy of 21st chromosome occur in humans. List any four characteristic features in an individual suffering from it.
Ans. Cause: Additional copy of chromosome number 21 or trisomy of chromosome 21.
(i) Short statured with small round head.
(ii) Partially open mouth with protruding furrowed tongue.
(iii) Palm is broad with characteristic palm crease.
(iv) Physical, psychomotor and mental development retarded.
Question. Explain the mechanism of ‘sex determination’ in birds. How does it differ from that of human beings?
Ans. In birds, female heterogamety is observed. They exhibit ZW type of sex determination.
Question. List any four symptoms shown by Klinefelter’s syndrome sufferer. Explain the cause of this disease.
Ans. (ii) Klinefelter’s syndrome
Cause: Presence of an additional copy of X chromosome resulting in the karyotype 44+XXY. i.e., 47 chromosomes.
(i) Sex of the individual is masculine but possess feminine characters.
(ii) Gynaecomastia, i.e., development of breasts.
(iii) Poor beard growth and often sterile.
(iv) Feminine pitched voice.
(v) They are sterile.
(vi) Tall stature.
Question. (a) How does mutation occur?
(b) Differentiate between point mutation and frameshift mutation.
Ans. (a) Mutation occurs due to loss by deletion or gain by insertion/duplication/addition or change in position of DNA segments or chromosomes.
(b) Mutation due to change in a single base pair of DNA is point mutation.
Insertion or deletion of one or two bases change the reading frame from the point of insertion or deletion. This is called as frameshift mutation.
Question. A non-haemophilic couple was informed by their doctor that there is possibility of a haemophilic child be born to them. Explain the basis on which the doctor conveyed this information. Give the genotypes and the phenotypes of all the possible children who could be born to them.
Ans. On the basis of pedigree analysis, the doctor conveyed this information. Pedigree analysis is a strong tool, which is utilised to trace the inheritance of a specific trait, abnormality or disease. Since, both the parents are non-haemophilic, their genotypes will be:
Question. (a) Explain Mendel’s law of independent assortment by taking a suitable example.
(b) How did Morgan show the deviation in inheritance pattern in Drosophila with respect to this law?
Ans. (a) 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.
The Punnett square can be effectively used to understand the independent segregation of the two pairs of genes during meiosis and the production of eggs and pollen in the F1 (RrYy) plant. Consider the segregation of one pair of genes R and r. Fifty per cent of the gametes have the gene R and the other 50 per cent have gene r. Now besides each gamete having either R or r, it should also have the allele Y or y. The important thing to remember here is that segregation of 50 per cent R and 50 per cent r is independent from the segregation of 50 per cent Y and 50 per cent y. Therefore, 50 per cent of the r bearing gamete has Y and the other 50 per cent has y. Similarly, 50 per cent of the R bearing gamete has Y and the other 50 per cent has y. Thus there are four genotypes of gametes (four types of pollen and four types of eggs). The four types are RY, Ry, rY and ry each with a frequency of 25 per cent or 1/4th of the total gametes produced.
Question. Work out a typical Mendelian dihybrid cross and state the law that he derived from it.
For the dihybrid cross Mendel derived the law of Independent Assortment: It 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) State the cause and symptoms of colour-blindness in humans.
(b) Statistical data has shown that 8% of the human males are colour-blind whereas only 0.4% of females are colour-blind. Explain giving reasons how is it so.
Ans. (a) Colour-blindness is a sex-linked recessive disorder.
Its symptoms are failure to discriminate between red and green colour.
(b) Since males have only one X chromosomes, hence one gene for colour blindness, so if present in any one parent will always be expressed, whereas in female it will be expressed only if it is present on both the X chromosome or when both parents are carrying gene for colour blindness.
Question. Inheritance pattern of flower colour in garden pea plant and snapdragon differs. Why is this difference observed? Explain showing the crosses up to F2 generation.
Ans. Inheritance pattern of flower colour in garden pea follows principle of dominance whereas inheritance in snapdragon shows incomplete dominance.
Inheritance of flower colour in garden pea plant:
Phenotypic ratio—3 : 1
Genotypic ratio—1 : 2 : 1.
Inheritance of flower colour in snapdragon:
Phenotypic ratio—1 : 2 : 1
Genotypic ratio—1 : 2 : 1.
Question. (a) State the law of independent assortment.
(b) Using Punnett square demonstrate the law of independent assortment in a dihybrid cross involving two heterozygous parents.
Ans. (a) 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.
Question. (a) Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross up to F2 generation.
(b) State the laws of inheritance that can be derived from such a cross.
(c) How is the phenotypic ratio of F2 generation different in a dihybrid cross?
(b) Law of Dominance: In a contrasting pair of factors, one member of the pair dominates (dominant) the other is recessive.
Law of Segregation: Factors or allele of pair separate from each other such that gamete receives only one of the two factors.
(c) Phenotypic ratio of F2 in monohybrid cross is 3 : 1 whereas in a dihybrid cross the phenotypic ratio is 9 : 3 : 3 : 1.
Question. A cross was carried out between a pea plant heterozygous for round and yellow seeds with a pea plant having wrinkled and green seeds.
(a) Show the cross in a Punnett square.
(b) Write the phenotype of the progeny of this cross.
(c) What is this cross known as? State the purpose of conducting such a cross.
(b) Both the phenotypic and genotypic ratio are same, i.e., 1: 1 : 1 : 1.
(c) This cross is known as test cross.
Question. (a) A garden pea plant bearing terminal, violet flowers, when crossed with another pea plant bearing axial, violet flowers, produced axial, violet flower and axial, white flowers in the ratio of 3 : 1. Work out the cross showing the genotypes of the parent pea plants and their progeny.
(b) Name and state the law that can be derived from this cross and not from a monohybrid cross.
(b) Law of Independent Assortment: When two pairs of traits are combined in a hybrid,segregation of one pair of characters is independent of the other pair of character.
Question. What is the inheritance pattern observed in the size of starch grains and seed shape of Pisum sativum? Workout the monohybrid cross showing the above traits. How does this pattern of inheritance deviate from that of Mendelian law of dominance?
Ans. A single gene controls the size of the starch grains and seed shape of Pisum sativum.
With respect to size of starch grains it shows 3 forms-big, Intermediate and small as in incomplete dominance but with respect to seed shape it follows Mendelian law of Dominance showing either round or wrinkled.
Question. (a) During a cross involving true breeding red-flowered and true breeding white-flowered snapdragon plants, the F1 progeny did not show any of the parental traits, while they reappeared in F2 progenies. Explain the mechanism using Punnett Square.
(b) Explain polygenic inheritance with the help of an example.
Ans. (a) • It is a phenomenon in which the F1 hybrid exhibits characters intermediate of the parental genes.
• Here, the phenotypic ratio deviates from the Mendel’s monohybrid ratio.
• It is seen in flower colours of Mirabilis jalapa (4 O’ clock plant) and Antirrhinum majus (snapdragon),where red colour is due to genetic constitution RR, white colour is due to genetic constitution rr and pink colour is due to genetic constitution Rr.
(b) • It is a type of inheritance, in which a trait is controlled by three or more genes. Such traits are called polygenic traits.
• The phenotype reflects contribution of each allele and is also influenced by the environment.
• For example, human skin colour. Suppose three genes A, B and C control skin colour with A, B, C being the dominant alleles and a, b, c being the recessive alleles.