Principles of Inheritance and Variation VBQs Class 12 Biology

VBQs Principles of Inheritance and Variation Class 12 Biology with solutions has been provided below for standard students. We have provided chapter wise VBQ for Class 12 Biology with solutions. The following Principles of Inheritance and Variation Class 12 Biology value based questions with answers will come in your exams. Students should understand the concepts and learn the solved cased based VBQs provided below. This will help you to get better marks in class 12 examinations.

Principles of Inheritance and Variation VBQs Class 12 Biology

Very Short Answer Type Questions

Question. How does a test cross help to determine the genotype of an individual ?
Answer : Individual of unknown genotype crossed with recessive parent.
All dominant in progeny—Homozygosity, dominant to recessive ratio 1 : 1 in progeny—
Heterozygosity.
Detailed Answer :
Test-cross helps to determine the unknown genotype by crossing it with the recessive parent.
If in the progeny all are dominant type then the individual is homozygous and if in the progeny dominant to recessive ratio is 1 : 1, the individual is heterozygous.

Question. Give an example of a plant where the F₂ progeny of a monohybrid cross has same genotypic and phenotypic ratios.
Answer : Snapdragon/Antirrhinum majus/ Four O’ clock plants/Mirabilis jalapa. 

Question. In a dihybrid cross carried out by T. H. Morgan in Drosophila the F2 ratio deviated from that of Mendel’s dihybrid F₂ ratio. Give a reason.
Answer : Genes were linked/genes were on the same chromosome and closely associated 
Detailed Answer :
The F₂ ratio deviated from that of Mendel’s dihybrid F₂ ratio (9 : 3 : 3 : 1) in an experiment performed by Morgan on Drosophila because of Linkage.
The genes were linked as they were located on the same chromosome and closely associated.
Therefore, they failed to segregate at the time of gamete formation resulting in greater number of parental combinations and lesser number of new recombinations in F₂ generation, thereby deviating from the normal dihybrid Mendelian ratio.

Question. List any two characters of Pea plants used by Mendel in his experiments other than height of the plant and the colour of the seed.
Answer : Flower colour / Flower position / Pod shape / pod colour / Seed shape (Any two) 

Question. State a difference between a gene and an allele.
Answer : Gene : Contains information that is required to express a particular trait // unit of inheritance // segment of DNA called cistron // sequence of DNA coding for tRNA / rRNA / polypeptide / enzyme.
Allele : Genes which code for a pair of contrasting traits / different forms of the same gene / individual gene in a particular gene pair (for same character).
Detailed Answer :
Gene is a unit of inheritance, specific segment of DNA or a specific sequence of DNA coding for t-RNA/r-RNA, polypeptide or enzyme, that is transferred from the parent to the offspring. It controls the expression of a character.
Allele is an alternative form or forms of a single gene that lie on the same locus of homologous chromosomes. e.g. if a gene controls height then allele is which gives either tallness or dwarfness. 

Question. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel.
Answer : Round/Wrinkled, Yellow/Green. 
Detailed Answer :
Colour of seeds : Dominant – Yellow (YY), Recessive – Green (yy)
Shape of seeds : Dominant – Round (RR), Recessive – Wrinkled. (rr) 

Question. Name the stage of cell division where segregation of an independent pair of chromosomes occurs.
Answer : Anaphase-1 of Meiosis – 1. 1

Question. A garden pea plant produced axial white flowers.
Another of the same species produced terminal violet flowers. Identify the dominant traits.
Answer : Axial, violet flower. 
Detailed Answer :
In garden pea plant bearing axial & white flowers, axial position of the flower is a dominant trait while white flower is a recessive trait, whereas in plants having terminal violet flowers, the terminal position of the flower is a recessive trait while violet colour of the flower is a dominant trait.

Question. Write the percentage of the pea plants that would be homozygous recessive in F₂ generation when tall F₁ heterozygous pea plants are selfed.
Answer : 25% homozygous recessive pea plants will be obtained in F₂ generation when tall F₁ heterozygous pea plants are selfed as follows :

Question. In a dihybrid cross, when would the proportion of parental gene combinations be much higher than non-parental types, as experimentally shown by Morgan and his group ?
Answer : When the genes are linked i.e., when genes of dihybrid cross are closely situated on the same chromosome, the proportion of parental gene combinations will be much higher than nonparental types.

Question. What is the phenotype of the following ?
(i) I^Ai & (ii) ii
Answer : (i) Blood group A (Heterozygous) because the allele IA is dominant over allele i.
(ii) Blood group O (Homozygous condition)— because both alleles ii are recessive.

Question. Write the percentage of F₂ homozygous & heterozygous populations in a typical monohybrid cross.
Answer : 50% homozygous and 50% heterozygous. Out of 50% homozygous population 25% is homozygous dominant and 25% homozygous recessive. 

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

Short Answer Type Questions

Question. How does the gene I control ABO blood groups in humans ? Write the effect the gene has on the structure of red blood cells.
Answer : (i) Gene I has three different alleles I^A, I^B, i. 
(ii) I^A produces A type of sugar / Antigen → A group I^B produces B type of sugar / Antigen →  B group 
(iii) i – No sugar/Antigen – O group. 
(iv) Structure – sugar polymers protrude from the surface of plasma membrane of RBCs. 
Detailed Answer :
In humans, the ABO blood groups are controlled by a gene called I. It has three alleles, namely I^A, I^B and i. A person possesses any two of the three alleles. I^A and I^B are dominant over allele i. But I^A and I^B are codominant as they express themselves equally and independently, when present together. These three alleles yield six different combinations which give four type of blood groups. The allelic pair I^A I^A or I^A i yield blood group A, I^B I^B or I^Bi the blood group B, I^A I^B is blood group AB, and ii is blood group O.
The plasma membrane of red blood cells has sugar polymers that protrude out from its surface and the kind of sugar is regulated by the gene ‘I’ of ABO blood group. The alleles IA and IB produce A and B types of sugar, while i does not produce any sugar.

Question. Linkage or crossing-over of genes are alternatives of each other. Justify with the help of an example.
Answer : In Drosophila, a yellow bodied white eyed female was crossed with brown bodied red eyed male, F₁ progeny produced and intercrossed. 
The F₂ phenotypic ratio of Drosophila deviated significantly from Mendel’s 9 : 3 : 3 : 1, the genes for eye colour & body colour are closely located on the ‘X’ chromosome showing linkage & therefore inherited together, recombinants were formed due to crossing over but at low percentage.
Detailed Answer :
Linkage is the tendency of certain genes staying together during inheritance through generations without any change or separation. 
This is due to their location on the same chromosomes. The F₂ generation of Drosophila deviated from the Mendel’s 9 : 3 : 3 : 1 ratio when eye colour was considered. Morgan found that this was due to the fact that eye colour in Drosophila is a sex-linked character and its gene is located on X-chromosome. He also observed that eye colour and body colour are closely located on X-chromosome and hence show linkage. The linked characters are generally inherited together.
Linkage and crossing over are alternatives of each other as genes tend to remain together when they are located close to each other on the same chromosome. Crossing over between genes takes place only if they are located away from each other.

Question. Explain pleiotropy with the help of an example.
Answer : Effect of single gene on multiple phenotypic expressions. 
e.g. size of the starch grains produced and shape of the seeds in pea plant are controlled by a single gene // Phenylketonuria characterised by mental retardation and reduction in hair and skin pigmentation. 
Detailed Answer :
The ability of a gene to have multiple phenotypic effects because it influences more than one trait or a number of characters simultaneously is called pleiotropism and such genes are known as pleiotropic genes e.g. in man a gene producing the disease phenylketoneuria also produces a number of abnormal phenotypic traits such as short stature, mental retardation, widely spaced incisors, pigmented patches on skin etc. Another such example is the size of starch grains produced and shape of the seeds in pea plant are controlled by a single pleiotropic gene. In Drosophila the gene for vestigial wings also affects structure of reproductive organs and the bristles on the wings.

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.
Answer : (i) Tt × tt.

(ii) Test cross. 

Question. A cross between a red flower-bearing plant and a white flower-bearing plant of Antirrhinum produced all plants having pink flowers. Work out a cross to explain how this is possible.
Answer : A cross between a red flower bearing plant and a white flower bearing plant of Antirrhinum produced all plants having pink flowers. This type of cross shows that red flower colour is not completely dominant over white colour flowers i.e., it is a case of incomplete dominance. Cross to explain how this is possible :

Question. In a cross between two tall pea plants, some of the offsprings produced were dwarf. Show with the help of the Punnett Square how this is possible.
Answer : In a cross between two tall pea plants, some of the offsprings produced were dwarf. It indicates that parent pea plants were heterozygous for tallness (Tt) i.e., they contain a recessive gene (t) for dwarfness from each of the parent plant.

25% plants were dwarf. Phenotypic ratio – Tall :
Dwarf :: 3 : 1 Genotypic ratio – 1 : 2 : 1
Punnett square : Punnett square is a graphic representation of the probabilities of all the possible genotypes and phenotypes of offsprings in a cross. 

Question. Tallness of pea plants is a dominant trait, while dwarfness is the alternate recessive trait. When a pure-line tall is crossed with pure-line dwarf, what fraction of tall plants in F2 shall be heterozygous ?
Give reasons.
Answer : Pure line tall is crossed with pure line dwarf.

Phenotypic ratio = Tall : dwarf
                               3 : 1
Genotypic ratio = TT : Tt : tt
                           1 : 2 : 1
Two third of tall progenies are heterozygous because gene for tallness (T) is dominant and expresses itself in heterozygous condition.
A Punnet square is used to understand a typical monohybrid cross between tall and dwarf plants.

Question. Explain co-dominance with the help of one example.
Answer : When the dominant alleles of the same gene which are contributed by both parents are expressed (called co-dominance) // F₁ generation resembles both the parents :
In human blood group :

OR
In human red blood cells, alleles I^A and I^B of gene I are both dominant, when I^A & I^B are present together in an individual both are expressed as I^A I^B, (AB blood group).

Long Answer Type Questions

Question. Aneuploidy of chromosomes in human beings results in certain disorders. Draw out the possibilities of the karyotype in common disorders of this kind in human beings and its consequences in individuals. 
Answer : Down’s syndrome, Turner’s syndrome, Klinefelter’s syndrome are the common examples of Aneuploidy of chromosomes in human beings.
Down’s syndrome results in the gain of extra copy of chromosome 21- trisomy. 
Turner’s syndrome results due to loss of an X chromosome in human females- XO monosomy.
Klinefelter’s Syndrome is caused due to the presence of an additional copy of X-chromosome resulting into XXY condition.
Down’s Syndrome: The affected individual is:
(a) Short statured with small round head.
(b) Furrowed tongue and partially open mouth.
(c) Palm is broad with characteristic palm crease.
(d) Physical, psychomotor and mental development is retarded. 
Klinefelter’s Syndrome: The affected individual is:
(a) A male with development of breast i.e. Gynecomastia.
(b) Such individuals are sterile. 
Turner’s Syndrome: The affected individual shows following characters:
(a) Females are sterile as ovaries are rudimentary.
(b) Lack of other secondary sexual characters.

Question. Thalassemia and Haemophilia are both Mendelian disorders related to blood. Write the symptoms of the diseases. Explain with the help of crosses the difference in the inheritance pattern of the two diseases.
Answer : Thalassemia- Anaemia
Haemophilia-Non stop bleeding
Haemophilia- Sex linked recessive disorder, is generally passed on from (carrier) mother to some of her sons/ from affected father to daughter (carrier).

(Note : Any one cross, one mark to be given if the entire diagram (cross) is correct)
Thalassemia – Autosome linked recessive blood disease, inheritance is like Mendelian inheritance pattern.

Detailed Answer :
Difference in inheritance pattern :
Haemophilia : It is a sex-linked (X-linked) recessive disorder, inherited from haemophilic father (X^hY) or carrier mother (X^hX). Females are haemophilic only in homozygous double recessive state (X^hX^h) but such females die before birth.
Thalassemia : It is caused by haemolytic anaemia.
It shows autosomal recessive pattern of inheritance and is controlled by two pairs of alleles (HBA1 & HBA2). The effect is more pronounced when the defective gene occur in homozygous state, causing thalassemia major. In heterozygous state the adverse effect of thalassemia is minor. The trait (Thalassemia) is inherited as autosomal recessive.
This is found equally in both males and females.
However the defective alleles for thalassemia in both males & females unlike haemophilia expresses itself only when it is in homozygous condition. The heterozygotes for recessive trait remain unnoticed but act as heterozygous carriers.

Question. (i) Why are thalassemia and haemophilia categorized as Mendelian disorders ? Write the symptoms of these diseases. Explain their pattern of inheritance in humans.
(ii) Write the genotypes of the normal parents producing a haemophilic son.
Answer : (i) Both are caused due to alteration/mutation, in a single gene and follow Mendelian pattern of inheritance. 
Symptoms
Thalassemia-anaemia (caused due to defective/abnormal Hb).
Haemophilia-non stop bleeding even in minor injury. 
Pattern of inheritance
Thalassemia-autosomal recessive inheritance pattern, inherited from heterozygous/parent carrier. 
Haemophilia-X linked recessive inheritance, inherited from a haemophilic father/carrier mother (females are rarely haemophilic).
(ii) X^hX-Mother
Detailed Answer :
(i) Thalassemia and haemophilia are categorised as Mendelian disorders because they occur by mutation in a single gene. Their mode of inheritance follows the principles of Mendelian genetics.
Mendelian disorders can be :
(a) Autosomal dominant (muscular dystrophy)
(b) Autosomal recessive (thalassemia)
(c) Sex linked (haemophilia)
Symptoms of thalassemia :
Thalassemia minor results only in mild anaemia, characterised by low haemoglobin level.
Thalassemia major is also known as Cooley’s anaemia. In this disease, affected infants are normal but as they reach 6 to 9 months of age, they develop severe anaemia, skeletal deformities, jaundice, fatigue etc.
Symptoms of Haemophilia : Persons suffering from this disease does not develop a proper blood clotting mechanism. A haemophilic patient suffers from non-stop bleeding even on a simple cut, which may lead to death.
Patterns of inheritance of Thalassemia :
Pairs of alleles Hb^A and HbT controls the expression of this disease.
Conditions for Thalasemia :
Hb^A and Hb^A: Normal
Hb^A and Hb^T: Carrier
Hb^T and Hb^T : Diseased
Let us assume that parents are carriers of betathalassemia.

Patterns of inheritance of Haemophilia :
It is X-linked genetic disorder. Compared to females, males have higher chances of getting affected because females have XX chromosomes while males have only one X with Y chromosomes. Thus, for a female to get affected by haemophilia, she has to have the mutant gene on both the X-chromosomes while males can be affected if they carry it on the single X-chromosome.
Conditions for haemophilia :
XX, XX – Normal
X^hY – Haemophilic
X^hX – Carrier
X^hX^h-Haemophilic
Let us assume that a carrier female (X^hX) is married to a normal male :

(ii) When normal male marries a carrier female (she is considered normal as she contains the mutant gene
on one of her X-chromosomes), they can produce a haemophilic son. So, the genotype of the parents
would be XY and X^hX.
Parents : XY (Father) × XhX (Mother)
Offspring :

Question. Why is thalassaemia categorized as a Mendelian disorder ? Write the symptoms and explain the causes of the disease. How does it differ from sicklecell anaemia ? 
Answer : Thalassaemia is categorized as a Mendelian disorder because it is determined by single allelic mutation.
It involves the genes HBA1 and HBA2, inherited in a Mendelian recessive fashion.
Symptoms : People with thalassaemia make less haemoglobin and less circulating red blood cells than normal human beings, which results in mild or severe anaemia.
Cause : Both α and β-thalassaemias are often inherited in an autosomal recessive fashion, although this is not always the case. Thalassaemias are a group of disorders caused by defects in the synthesis of globin polypeptide. Absence or reduced synthesis of one of the globin chains results in an excess of the other. In this situation, free globin chains, which are insoluble, accumulate inside the red blood cells and form precipitates which damage the cells, causing cell lysis, resulting in anaemia.
There are two main types of thalassaemias in which synthesis of α or β globin is defective.
In thalassaemia, patients have a deficiency of either α or β globin. But patients with sickle-cell anaemia have a specific mutant form of globin, causing production of abnormal red blood cells. Sickle- cell anaemia is caused by the mutant recessive allele on chromosome 11.