CBSE Class 12 Biology Principles of Inheritance And Variation Question Bank

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.
Answer. Inflated green pod is the dominant trait.

Question. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason.
Answer. 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.
Answer. Axial, violet flower.

Question. State what does aneuploidy lead to.
Answer. 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.
Answer. Round, yellow seed are the dominant traits.

Question. How many kinds of phenotypes would you expect in F2 generation in a monohybrid cross?
Answer. 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.
Answer. 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.
Answer. 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.
Answer. Down’s Syndrome.

Question. Discuss is the genetic basis of wrinkled phenotype of pea seeds. 
Answer. 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?
Answer. 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.
Answer. 

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.
Answer.
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. 
Answer. 

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.
Answer. 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?
Answer. 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.
Answer. 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?
Answer. (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.
Answer. 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.
Answer. 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.
Answer. 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.
Answer. 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.
Answer. • 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
Answer. (a) Table 5.3: Differences between dominance and recessiveness

Question. Who proposed chromosomal theory of inheritance? Point out any two similarities in the behaviour of chromosomes and genes.
Answer. 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.
Answer. 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.
Answer. • 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. 
Answer. This kind of cross is observed in Mirabilis jalapa/Four o’clock plant/Antirrhinum majus.
For cross, refer to Fig. 5.4.
In heterozygous condition a single dominant gene is not sufficient to produce red colour therefore
it is a case of incomplete dominance.

Question. (a) Explain sex determination in humans.
(b) How do human males with ‘XXY’ abnormality suffer?
Answer. (a) Refer to Basic Concepts Point 13.
(b) 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.
Answer. 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? 
Answer. (a) Axial position (img 173)
(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) IAi      (ii) ii 
Answer. (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.
Answer. (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. Identify and write the correct statement :
(a) In grasshopper males two sex chromo- somes are X and Y types.
(b) In grasshopper males there exist XO type of sex - determinants. 
Answer. In grasshopper males, there exist XO type of sex-determinants.

Question. Identify the correct statement.
(a) Female  f many bird shasa pair of dissimilar ZW chromosomes, whilethe males possess a pair of similar ZZ chromosomes.
(b) Female of many birds has a pair of similar ZZ chromosomes, while the males possess a pair of dissimilar ZW chromosomes.
Answer. Statement (a) is correct. In birds, the male has two homomorphic sex chromosomes (ZZ) and is homogametic, and the female has two heteromorphic sex chromosomes (ZW) and is heterogametic. 

Question. Differentiate between ‘ZZ’ and ‘XY’ type of sex-determination mechanisms.
Answer. 

Question. Wr i t e t h e t yp es o f s ex-det er min a t io n mechanisms the following crosses show. Give an example of each type.
(a) Female XX with male XO
(b) Female ZW with male ZZ. 
Answer. (a) The type of sex determination mechanism shown in female XX with male XO is called male heterogamety. In this case male are heterogametic with half the male gametes carrying X-chromosome while the other half being devoid of it.
Example - Grasshopper
(b) The type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety because female produces two different types of gametes.
Example - Birds

Question. Explain why it is scientifically incorrect to blame the mother for bearing female child.
Answer. In humans, sex of the child is determined at the time of fertilisation. The female parent produces only one type of egg with X-chromosome. The male gametes are of two types with X-chromosome and Y-chromosome. Fertilisation of the egg with sperm carrying X-chromosome produces a female child while fertilisation with sperm carrying Y-chromosome give rise to male child. Thus sex of the child is determined by father and not by the mother.
Hence, it is scientifically incorrect to blame the mother for bearing female child.

Question. Do you agree to the perception in our society that the woman is responsibile for the gender. Substantiate your answer scientifically.
Answer. No, the perception in our society that the woman is responsible for the gender of the child is totally wrong. It is the male who is responsible for this. 

Question. The ma le f r uit f ly and fema le fow l are heterogametic while the female fruit fly and the male fowl are homogametic. Why are they called so?
Answer. Male fruit fly and female fowl produce two types of gametes whereas female fruit fly and male fowl produce only one type of gamete.
(i) The type of sex determination mechanism shown in female XX with male XY is called male heterogamety because male produces two different types of gametes. Example - Drosophila.
(ii) The type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety because female produces two different types of gametes.
Example - Birds 

Question. (a) Why are grasshopper and Drosophila said to show male heterogamety? Explain.
(b) Explain female heterogamety with the help of an example. 
Answer. (a) In male heterogamety, males produce two different types of gametes. In human and Drosophila the males have one X and one Y chromosome, whereas in grasshopper the male have only one X-chromosome (XO type). Thus, the males of these organisms show male heterogamety as they produce (i) gametes either with or without X-chromosome or (ii) some gametes with X-chromosome and some with Y-chromosome. (b) In some organisms female produce two different types of gametes. This is termed as female heterogamety. In birds and some reptiles female has two different sex chromosomes (one Z and one W chromosome) whereas male has a pair of same chromosome (a pair of Z-chromosomes). 

Question. Explain the sex determination mechanism in humans. How is it different in birds?
Answer. Chromosomal determination of sex in human beings is of XX-XY type. Human beings have 22 pairs of autosomes and one pair of sex chromosomes. The female possess two homomorphic (= isomorphic) sex chromosomes, named XX. The males contain two heteromorphic sex chromosomes, i.e., XY. All the ova formed by female are similar in their chromosome type (22 + X). Therefore, females are homogametic. The male gametes or sperms produced by human males are of two types, gynosperms (22 + X) and androsperms (22 + Y). Human males are therefore, heterogametic. Sex of the offspring is determined at the time of fertilisation. Fertilisation of the egg (22 + X) with a gynosperm (22 + X) will produce a female child (44 + XX) while fertilisation with an androsperm (22 + Y) gives rise to male child (44 + XY). As the two types of sperms are produced in equal proportions, there are equal chances of getting a male or female child in a particular mating. As Y-chromosomes determines the male sex of the In case of birds, the type of sex determination is ZW-ZZ type. Female has two different sex chromosomes (Z and W) whereas as male has a pair of same chromosomes (ZZ), therefore, in birds, sex is determined by female.

Question. Explain the mechanism of sex determination in insects like Drosophila and grasshopper.
(b) How does it differ from sex determination in birds and honey bees ? 
Answer. In Drosophila, XX–YY type of sex determina- tion has been found, in which the female is homogametic and produces only one type of eggs (22 + X). The male gametes are of two types, androsperms (22 + Y) and gynosperms (22 + X). They are produced in equal proportion. Fertilisation of the egg (22 + X) with a gynosperm (22 + X) will produce a female child (44 + XX) while fertilisation with an androsperm (22 + Y) gives rise to male child (44 + XY). In roundworms and some insects (true bugs, grasshoppers, cockroaches), the type of sex determination is XX – X0 type. The females have two sex chromosomes, XX, while the males have only one sex chromosome. There is no second sex chromosome. Therefore, the males are designated as X0. 

Question. Name the event during cell divison cycle that results in the gain or loss of chromosome.
Answer. Mutation

 Question. The cell division involved in gamete formation is not of the same type in different organisms. Justify.
Answer. The parent organism may be haploid or diploid but the gametes produced by them are always haploid. This is because of the different cell division that take place during gamete formation. The diploid parents undergo meiosis to produce haploid gametes whereas the haploid parents undergo mitosis to produce haploid gametes. 

Question. Give an example of a human disorder that is caused due to a single gene mutation.
Answer. Sickle cell anaemia is due to inheritance of a defective allele coding for -globin. It results in the transformation of HbA into HbS in which glutamic acid is replaced by valine at sixth position in each of two -chains of haemoglobin. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG

Question. State the chromosomal defect in individuals with Turner’s syndrome. 
Answer. Turner’s syndrome is due to monosomy. It occurs due to union of an allosome free egg (22 + 0) and a normal X sperm or a normal egg and an allosome free sperm (22 + 0). The individual has 2n = 45 chromosomes (44 + X0) instead of 46.

Question. Write the chromosomal defect in individuals affected with Klinefelter’s syndrome.
Answer. Klinefelter’s syndrome is caused by union of an abnormal XX egg and a normal Y sperm or normal X and abnormal XY sperm. The individual has 47 (44 + XXY) chromosomes.

Question. Why do normal red blood cells become elongated sickle shaped structures in a person suffering from sickle cell anaemia?
Answer. Sickle-cell anaemia is caused by the formation of an abnormal haemoglobin called haemoglobin-S. The mutant haemoglobin undergoes polymerisation under low oxygen pressure and causes change in the RBCs from biconcave disc to elongated sickle-shaped structure.

Question. Name one autosomal dominant and one autosomal recessive Mendelian disorder in human.
Answer. Huntington’s disease is an autosomal dominant and sickle-cell anaemia is an autosomal recessive Mendelian disorder.

Question. Write the genotype of (a) an individual who is carrier of sickle cell anaemia gene but apparently unaffected, and (b) an individual affected with the disease. 
Answer. (a) HbAHbS
(b) HbSHbS

Question. A human being suffering from Down’s syndrome shows trisomy of 21st chromosome. Mention the cause of this chromosomal abnormality.
Answer. Down’s syndrome is an autosomal aneuploidy, caused by presence of an extra chromosome number 21. Both the chromosomes of the 21st pair pass into a single egg due to non-disjunction during oogenesis. Thus the egg possess 24 chromosomes instead of 23 and offspring has 47 chromosomes (45 + XY in male, 45 + XX in female) instead of 46. Down’s syndrome is also called 21- trisomy.

Question. The son of a haemophilic man may not get this genetic disorder. Mention the reason.
Answer. The gene for haemophilia is located on the X-chromosome. A male receives the X-chromosome from his mother so haemophilic father does not pass X-chromosome or haemophilia to his son.

 Question. A haemophilic son was born to normal parents.
Give the genotypes of the parents.
Answer. Genotypes of parents : XXh and XY

Question. Why is the possibility of a human female suffering from haemophilia rare ? Explain.
Answer. Haemophilia is genetically due to the presence of a recessive sex linked gene ‘h’, carried by X chromosome. It is generally observed in males as a single gene for the defect is able to express itself as the Y chromosome is devoid of any corresponding allele (XhY). Women will suffer from this disorder when a carrier woman (XXh) marries with haemophilic man (XhY). 50% girl babies will be carriers (XXh) while the remaining 50% will be haemophilic (XhXh).

Question. Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross.
Answer. Autosomal recessive traits are expressed only when autosomal recessive genes are present in homozygous condition. For example, sickle cell anaemia in humans is an autosomal recessive trait/ disorder. In this disorder, the erythrocytes become sickle shaped under oxygen deficiency. This occurs due to formation of abnormal haemoglobin-S (Hbs). It is formed when the glutamic acid present at 6th amino acid in -chain of normal Hb-A, is replaced by valine.
Sickle cell anaemia can be transmitted from parents to the offspring when both the parents are carrier for the gene or are heterozygous. When two sickle celled heterozygotes marry they may give birth to three types of children—homozygous normal, heterozygous carrier and homozygous sickle celled in the ratio of 1 : 2 : 1. However, homozygous sickle- celled individuals (HbSHbS) die in childhood (before reproductive age) due to acute anaemia. This can be shown by the given cross:

Question. (a) Name the kind of diseases/disorders that are likely to occur in humans if
(i) mutation in the gene that codes for an enzyme phenylalanine hydrolase occurs
(ii) there is an extra copy of chromosome 21 (iii) the karyotype is XXY.
(b) Mention any one symptom of the diseases/disorders named above.
Answer. (a) (i) Phenylketonuria
(ii) Down’s syndrome
(iii) Klinefelter’s syndrome
(b) Symptoms:
(i) Phenylketonuria-mental retardation
(ii) Down’s syndrome-partially open mouth with furrowed tongue
(iii) Klinefelter’s syndrome-development of feminine characters like development of breasts in male.

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