Principles of 05 Inheritance and Variation

055.1 Mendel’s Laws of Inheritance

5.2 Inheritance of One Gene

5.3 Inheritance of Two Genes

5.4 Sex Determination

5.5 Mutation

5.6 Genetic Disorders

8Maximum weightage is of Inheritance of One Gene.

8 Maximum number of VSA, SA I and LA type questions were asked from Inheritance of One Gene.

8 Maximum number of SA II type questions were asked from Genetic Disorders.

8 Genetics is a branch of biology that deals with the study of heredity and variations. Heredity is the study of transmission of characters from parents to o�spring or from one generation to the next. �e characters that are passed from one generation to the next are called hereditary characters. Variations may be

QUICK RECAP

Principles of Inheritance and Variation

Topicwise Analysis of Last 10 Years’ CBSE Board Questions

90 CBSE Chapterwise-Topicwise Biology

de�ned as the di�erences in characteristics shown by the individuals of a species and also by the o�spring or siblings of the same parents.

8 �e term genetics was �rst used by W. Bateson (1906).

8 Gregor Johann Mendel (1822-84) is called the Father of Genetics because he was the �rst to workout the patterns of heredity by performing experiments on garden pea plant (Pisum sativum).

GENETIC TERMINOLOGIES AND SYMBOLS

8 Character is a well de�ned morphological or physiological feature of an organism, for example plant height.

8 Trait is the distinguishing feature of a character, for example tallness or dwarfness.

8 Gene is the unit of genetic information that determines the biological character of an organism.

8 Allelomorphs or alleles refer to the two Mendelian factors which occur on the same locus on the two homologous chromosomes of an individual and control the expression of a character e.g., T and t, Y and y, R and r are pairs of alleles.

8 Term allele was given by W. Bateson.

8 Dominant allele is one of the factor of an allelic pair which can express itself whether present in homozygous or heterozygous state, e.g., T (tallness in pea), R (round seed in pea).

8 Recessive allele is the factor of an allelic pair which is unable to express its e�ect in the presence of its contrasting alternative form in a heterozygote, e.g., t in Tt.

8 �e diploid condition in which the alleles at a given locus are identical is called homozygous or pure, e.g., TT or tt.

8 Organism containing two di�erent alleles or individual containing both dominant and recessive genes of an allelic pair, e.g., Tt, is known as heterozygous or hybrid.

8 When only one allelic pair is considered in cross breeding, it is called monohybrid cross.

8 When two allelic pairs are considered for crossing, it is called dihybrid cross.

8 When there is involvement of more than two allelic pairs in a cross it is called polyhybrid cross.

8 F1 generation is the generation of hybrids produced from a cross between the genetically di�erent individuals called parents. For example, Tt individuals are produced in F1 generation from a cross between TT and tt parents.

8 F2 generation is the generation of individuals which arises as a result of inbreeding or inter-breeding amonst individuals of F1 generation.

8 Hybrid vigour or heterosis is the superiority of hybrid over either of its parents in one or more traits.

8 Genotype is the gene complement or genetic constitution of an individual with regard to one or more characters irrespective of whether the genes are expressed or not.

8 Phenotype is the observable characteristic of an organism which is determined by its genes.

8 �e portion or region on chromosome representing a single gene is called gene locus.

8 �e aggregate of all genes and their alleles present in an interbreeding population is known as the gene pool.

8 Pure line or pure breeding line is a strain of individuals homozygous for all genes considered. �e term was coined by Johannsen.

8 Punnett square is a checker board which was devised by R.C. Punnett and used to show the result of a cross between two organisms.

8 Genome is a constitution of all the genes contained in a single set of chromosomes, i.e., in a haploid nucleus. Each parent, through its reproductive cells, contributes its genome to its o�spring. A single genome is present in haploid cells, two in diploid cells and many in polyploid cells.

91Principles of Inheritance and Variation

MendelismMendel worked on garden pea, Pisum sativum. Mendel got success because he selected only pure breeding varieties of pea, selected distinct pairs of traits, considered one or two characters at a time, kept a well maintained record of every cross and used laws of probability for analysing results.

He deduced the following three laws:

Table : Traits studied by Mendel

Trait studied Dominant Recessive

1. Plant height Tall (T) Dwarf (t)

2. Flower position Axial (A) Terminal (a)

3. Pod colour Green (G) Yellow (g)

4. Pod shape Full or In�ated (I) Constricted (i)

5. Flower colour Violet (V) White (v)

6. Seed shape Round (R) Wrinkled (r)

7. Seed colour Yellow (Y) Green (y)

Back Cross and Test Cross

8 A cross of F1 hybrid with either of the two homozygous parents is known as back cross. When F1 o�spring are crossed with the dominant parents, all the F2 o�spring develop dominant character. On the other hand when F1 hybrids are crossed with recessive parent,

individuals with both the phenotypes appear in equal proportions.

8 While both the crosses are known as back cross, the second one is speci�ed as test cross. It is called so because it can be used to test genotype of a dominant phenotype. If dominant individual is pure (homozygous) it will produce only dominant trait in progeny.

Law of dominance

Every character is controlled by a gene (called factor by Mendel) which occurs in two alternative forms called alleles. Law of dominance states that only one factor expresses itself in F1 generation. �e factor/allele called dominant is able to express its e�ect while the factor called recessive remains suppressed in F1 generation. F2 generation expresses both the dominant and the hidden recessive factors in the ratio of 3 : 1.

Law of independent assortment

It states that the genes of di�erent characters located in di�erent pairs of chromosomes are independent of one another in their segregation during gamete formation. Mendelian recombinations were mainly due to independent assortment. �is priniciple was proved using dihybrid cross with phenotypic ratio 9 : 3 : 3 : 1.

Law of segregation

It states that when a pair of contrasting factors are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and separate again at the time of gamete formation. �is law is also known as “law of purity of gametes” because each gamete is pure in itself i.e., having either T (i.e., gene for tallness) or t (i.e., gene for dwarfness).

Mendel’s work has been formulated into 3 laws:


Modi�cation of e�ect of a gene under the in�uence of a non-allelic gene i.e., a gene at di�erent locus is termed as intergenic interaction.

Intergenic interaction

Gene interaction is the modi�cation of normal phenotypic expression of a gene due to either of its alleles or other non-allelic genes.

Gene interaction

Two types

In intragenic interaction, two alleles of a gene which are present on the same gene locus on the two homologous chromosomes, interact to produce modi�ed phenotype.

Intragenic interaction

It is the phenomenon where dominant allele does not completely express itself. �is was �rst studied in �ower colour of Mirabilis jalapa or four O’ clock plant. �e phenotypic as well as genotypic monohybrid ratio in F2 generation in incomplete dominance is 1 : 2 : 1 i.e., pure dominant: hybrid : pure recessive.

Incomplete dominance If two genes present on di�erent loci produce the same e�ect when present alone but interact to form a new trait when present together, they are called complementary genes. Complementary gene interaction is seen in �owers of sweet pea with F2 ratio 9 : 7.

Complementary genes

It is the phenomenon of two alleles lacking a dominant-recessive relationship where both of them express themselves together and equally in the organisms. �e codominant alleles are able to express themselves independently when present alone. �e phenotypic ratio is 1 : 2 : 1.

Codominance

�ese alleles are multiple forms (more than two alternatives) of a Mendelian factor or gene which occur on the same gene locus, distributed in di�erent organisms in the gene pool with an organism carrying only two alleles and gamete only one allele. ABO blood group system in human beings is an example of both codominance and multiple alleles. In human population, 3 di�erent alleles for ABO blood group systems are found IA, IB and IO or i. IA and IB are dominant over IO or i. IA and IB are responsible for A and B antigens (glycoproteins) while IO or i does not produce any of these A or B antigens. A person is having only two of these three alleles and blood type can be determined by their antigen types. Six genotype combinations are possible with these three alleles.

Multiple alleles

Supplementary genes are two non-allelic genes in which one type of gene produces its e�ect whether the other is present or not and the second (supplementary) gene produces its e�ect only in the presence of the �rst, usually forming a new trait. Interaction of supplementary genes in mice for coat colour gives F2 ratio 9 : 3 : 4.

Supplementary genes

Epistasis can be de�ned as the phenomenon of gene interaction wherein one gene interferes with the phenotypic expression of another gene or genes. �e gene or locus which suppresses or masks the action of a gene at another locus is called epistatic gene. �e gene or locus whose expression is suppressed by an epistatic gene is called hypostatic gene.�e dihybrid ratio for dominant epistasis is 12 : 3 : 1. Recessive epistasis dihybrid ratio is 9 : 3 : 4. Dominant recessive epistasis ratio is 13 : 3.

Epistasis

�ey are genes which control some vital functions of the organism and cause death of the organism in homozygous recessive condition. With the death of homozygous lethal, the monohybrid ratio becomes 2 : 1.

Lethal genes

Duplicate genes

Duplicate genes are two or more independent genes present on di�erent chromosomes which determine same phenotype whether present in homozygous or heterozygous state. �ey do not have cumulative e�ect. F2 ratio is 15 : 1.

Pleiotropic genesWhen a gene a�ects many aspects of phenotype or controls several phenotypes, it is said to be pleiotropic gene and this phenomenon is called pleiotropy. Pleiotropy is expressed in sickle cell anaemia, haemophilia, etc.


QUANTITATIVE AND QUALITATIVE INHERITANCE

VARIATIONS

8 Variations are di�erences found in morphological, physiological, cytological and behavioural traits of individuals belonging to same species, race and family. Hereditary variations are transmitted from generation to generation whereas environmental variations are temporary and do not relate with last or next generation.

CHROMOSOME THEORY OF INHERITANCE

8 Walter Sutton and �eodore Boveri (1902) postulated the “chromosome theory of inheritance”.

8 �e chromosome theory of inheritance states that the Mendelian factors or genes are located at speci�c loci on the chromosomes, and it is the chromosomes which segregate and assort independently during meiosis and recombine at the time of fertilisation in the zygote.

8 Salient features of this theory are : Bridge between one generation and the next fis through sperm and ovum. �e two must carry all the hereditary characters. Both the sperm and egg contribute equally in the heredity of the o�spring.Nucleus contains chromosomes. �erefore, fchromosomes must carry the hereditary traits. Every chromosome or chromosome pair has a de�nite role in the development of an individual.�e chromosomes retain their number, fstructure and individuality throughout the life of an organism and from generation to generation. �e two neither get lost nor mixed up. �ey behave as units.Both chromosomes as well as genes occur fin pairs in the somatic or diploid cells. A gamete contains only one chromosome of a type and only one of the two alleles of a trait.

LINKAGE

8 Linkage is the phenomenon of certain genes staying together and their inheritance from generation to generation without any change or separation due to their presence on the same chromosome. T. H. Morgan (1910) proved and de�ned linkage on the basis of his breeding experiments on fruit-�y, Drosophila melanogaster.

8 Genes are arranged in a linear fashion on the chromosome. Some genes situated in close proximity are always inherited together i.e., become linked. Strength of the linkage between two genes is inversely proportional to the distance between the two. All those genes which are located in the single chromosome, constitute a linkage group. �e number of linkage groups in a species corresponds to its haploid number of chromosomes, e.g., 23 linkage groups are present in man.


CROSSING OVER

8 Crossing over is recombination of genes due to exchange of genetic material between two synapsed homologous chromosomes. It is the mutual exchange of segments of genetic material between non-sister chromatids of two homologous chromosomes, so as to produce recombinations of genes. �e non-sister chromatids in which exchange of segments has occurred are called recombinants or cross-overs.

8 It results in new combinations of genes which

are di�erent from parents thus introduces variations. �e variations are helpful in struggle for existence and adaptability to changes in environment. Useful recombinations are picked up by breeders for development of improved varieties. �e frequency of crossing over is used for building linkage maps or chromosome maps.

SEX DETERMINATION

8 Sex determination is the method by which the distinction between males and females is established in a species.

SEX LINKED INHERITANCE

8 Sex-linked inheritance is the transmission of characters and their determining genes alongwith sex determining genes present on the sex chromosomes, X and Y which are inherited together from one generation to the next. X-linked traits are more apparent in males

than in females. Females generally function as carriers of X-linked disorders because recessive genes can express themselves in females only in the homozygous state. E.g., colour blindness, haemophilia etc.

8 X-linked traits follow the pattern of criss-cross inheritance where a parent passes

Types of linkageComplete linkage

In complete linkage the grouping of genes on a chromosome is not altered and is inherited as such from generation to generation without any cross-over. In such cases parental percentage is always 100% and recombinant percentage is zero percent (0%). It is rare but has been reported in certain cases like male Drosophila.

Incomplete linkageIn incomplete linkage, genes present in the same chromosome have a tendency to separate due to crossing over and hence produce recombinant progeny besides the parental type. It produces both parental and recombinant individuals. �e percentage of each parental type is more than 25% while that of each recombinant type is less than 25%.


the traits to the grandchild of the same sex through o�spring of the opposite sex, that is, father passes the traits to grandson through his daughter while the mother transfers traits to her grand daughter through her son.

MUTATION

8 Mutation is the sudden inheritable discontinuous variation which appears in an organism due to permanent changes in their genotypes. �e term mutation was coined by Hugo de Vries (1901).

Types of mutation

Chromosomal mutations�e change that occurs in the morphology of chromosomes resulting in change in number or sequence of gene without any change of ploidy.

Genomatic mutations

It involves the change in number of chromosomes. It is of 2 types: Euploidy and Aneuploidy.

Gene mutations

It is the sudden change in structure of a gene or cistron due to change in its nucleotide type or sequence.

Deletion

Loss of a segment (or part) from one or other end of chromosome (terminal) or within the chromosome (intercalary) followed by reunion of terminal segments.

Euploidy

An organism either loses a complete set or acquires one or more additional set of chromosomes over and above the two sets of diploid complement. It is of 2 types - monoploidy and polyploidy.

Substitution mutations

A nitrogenous base of triplet codon of DNA is replaced by another nitrogenous base or its derivative thus, changing the codon.Replacement of a purine/pyrimidine by another purine/pyrimidine in a polynucleotide chain is called transition.Replacement of purine by a pyrimidine or vice versa, is called transversion.

DuplicationPresence of same block of gene more than once in a haploid complement.

Translocation

Kind of interchromosomal rearrangement in which there is mutual exchange of chromosomal segments between non-homologous chromosomes.

Chromosome breaks at two points, the broken piece rotates through 180° and reunites in reverse order. When centromere is present in the inverted segment, it is called pericentric otherwise called paracentric inversion.

Aneuploidy

Addition or loss of one or more chromosomes to/from complete diploid chromosome complement. It is of 4 types: trisomy (2n+1), tetrasomy (2n + 2), monosomy (2n – 1) and nullisomy (2n – 2).

Frame-shi� mutation

�e mutations caused by the addition or deletion of nitrogenous bases in the DNA are known as frame-shi� mutations because these shi� the reading frame of codons from the site of change onwards. �ese are of 2 types : deletion mutation caused due to the loss or deletion of one or more nucleotides and insertion mutation caused by the addition of one or more extra nucleotides in a DNA molecule at one or more places.

Inversion

PEDIGREE ANALYSIS

8 Pedigree is a chart showing a record of inheritance of certain traits for two or more ancestral generations of human beings or

domesticated animals in the form of a diagram of family tree. Pedigree analysis is a system of analysis by following the movement and distribution of certain genetic traits. �is system has following conventions:


8 Pedigree analysis is useful in many ways like it helps to �ll up the possible genotypes by knowing the phenotypes only. It helps to study the pattern of inheritance of a dominant or a recessive trait. �e possible genetic makeup of a person for a trait can also be known with the help of pedigree.

HUMAN GENETIC DISORDERS

Chromosomal disorders

�ese are caused due to absence/excess or abnormal arrangement of one or more chromosomes. �ese are non-inheritable. Some chromosomal disorders have been discussed ahead in the table 1.

Gene related disorders

�ese disorders are determined by alteration or mutation in a single gene. �ey are transmitted to the o�spring as per Mendelian principles. �ese are also known as Mendelian disorders. Some common Mendelian or genetic disorders are discussed ahead in the table 2.

Human Genetic Disorders

Table 1 : Some known chromosomal abnormalities

Karyotype Common name of abnormality

Clinical symptoms

1. Trisomy 21 Down’s Syndrome Also known as Mongolian idiocy; oriental features, epicanthus skin fold above the eye, mental retardation (I.Q. usually below 50); short stature, protruding furrowed tongue, transverse palmar crease; susceptible to respiratory infections, etc.

2. XO Turner’s Syndrome Short stature, webbed neck, female with poorly developed breasts and degenerated ovaries and rudimentary sexual characterstics, with slight mental retardation.

3. XXY Klinefelter’s Syndrome

Male with degenerating testes, enlarged breasts.

Table 2 : Mendelian disorders

Disorder Mode of inheritance

Cause of disorder Clinical description

1. Cystic �brosis Autosomal recessive

Failure of chloride ion transport mechanism in cell surface membrane of epithelial cells.

High level of sweat electrolytes, pulmonary disease, cirrhosis of liver, pancreatic malfunction. Maldigestion of food with high fat content in stool.


2. Haemophilia X-linked recessive

Lack of blood coagulant. Chronic bleeding

3. Phenylketo-nuria (PKU)

Autosomal recessive

De�ciency of enzymes phenylalanine hydroxylase.

Leads to decreased pigmentation of hair, mental retardation.

4. Colour blindness

X-linked recessive

Absence of green cone or red cone pigment.

Inability to distinguish green colour and red colour.

5. Sickle cell-anaemia

Autosomal recessive

Formation of abnormal haemoglobin S in RBCs.

Abnormality of red blood cells caused by the presence of an inappropriate amino acid in the b-chain of haemoglobin molecule. Causes extreme distortion of shape (sickling) of RBC which leads to the premature destruction of cell.


5.1 Mendel’s Laws of Inheritance

VSA (1 mark)

1. A geneticist interested in studying variations and patterns of inheritance in living beings prefers to choose organisms for experiments with shorter life cycle. Provide a reason?

(Delhi 2015)

2. Mention any two contrasting traits with respect to seeds in pea plant that were studied by Mendel? (AI 2014)

3. A garden pea plant (A) produced in�ated yellow pod, and another plant (B) of the same species produced constricted green pods. Identify the dominant traits? (Delhi 2012)

4. Name the contrasting pod-related traits studied by Mendel in pea plant experiment?

(AI 2011C)

5. Mention two contrasting �ower related traits studied by Mendel in his pea plant experiments. (AI 2011C)

5.2 Inheritance of One Gene

VSA (1 mark)

6. State a di�erence between a gene and an allele. (Delhi 2016)

7. On what basis is the skin colour in humans considered polygenic? (AI 2015)

8. How many kinds of phenotypes would you expect in F2 generation in a monohybrid cross exhibiting co-dominance?

(Delhi 2014C)

9. 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.(AI 2012)

PREVIOUS YEARS MCQSPrevious Years’ CBSE Board Questions

10. How would you find the genotype of an organism exhibiting a dominant phenotype trait? (Delhi 2012C)

11. Write the percentage of the pea plants that would be homozygous recessive in F2 generation when tall F1 heterozygous pea plants are selfed?

(Delhi 2012C)

12. Write the percentage of the pea plants that would be heterozygous tall in F2 generation when tall heterozygous F1 pea plants are selfed. (Delhi 2012C)

13. In a test cross progeny of pea plants, all were bearing violet �owers. Give the genotypes of the parent pea plant. (AI 2012C)

14. Differentiate between dominance and co-dominance. (AI 2011)

15. Mention the type of allele that expresses itself only in homozygous state in organism.

(Foreign 2011)

16. When a tall plant was self-pollinated, one-fourth of the progeny were dwarf. Give the genotype of the parent and dwarf progenies. (AI 2008C)

SA I (2 marks)

17. When does a geneticist need to carry a test cross? How is it carried?

(Foreign 2015)18. State and explain the law of segregation as

proposed by Mendel in a monohybrid cross?(2/5, Foreign 2015)

19. Give an example of a gene responsible for multiple phenotypic expressions. What are such genes called ? State the cause that is responsible for such an e�ect?

(Foreign 2015)20. A cross was carried out between two pea plants

showing the contrasting traits of height of the plants. �e result of the cross showed 50% parental characters.


(a) Work out the cross with the help of a Punnett square.

(b) Name the type of the cross carried out. (Delhi 2014)

21. How does the gene ‘I’ control ABO blood groups in humans? Write the e�ect the gene has on the structure of red blood cells.

(Delhi 2014)

22. In snapdragon a cross between true-breeding red �owered (RR) plants and true-breeding white �owered (rr) plants showed a progeny of plants with all pink �owers.

(a) �e appearance of pink �owers is not known as blending. Why?

(b) What is this phenomenon known as?(AI 2014)

23. With the help of one example, explain the phenomena of co-dominance and multiple allelism in human population? (AI 2014)

24. Explain pleiotropy with the help of an example. (Foreign 2014)

25. In a cross between two tall pea plants some of the o�springs produced were dwarf. Show with the help of Punnett square how this is possible. (Delhi 2013)

26. A cross between a red �ower bearing plant and a white �ower bearing plant of Antirrhinum produced all plants having pink �owers. Work out a cross to explain how this is possible.

(AI 2013)27. 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. (AI 2013)

28. Work out a cross to �nd the genotype of a tall pea plant. Name the type of cross. (AI 2013)

29. Tallness of pea plant 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 plant in F2 shall be heterozygous? Give reasons. (AI 2011C)

30. How does a test-cross help in identifying the genotype of the organism? Explain.

(Delhi 2010)

31. When a tall pea plant was selfed, it produced one-fourth of its progeny as dwarf. Explain with the help of a cross. (Delhi 2010)

32. Inheritance pattern of ABO blood groups in humans shows dominance, codominance and multiple allelism. Explain each concept with the help of blood group genotypes.

(Delhi 2009)

33. Identify a, b and c in the table given below:

Pattern of inheritance

Monohybrid F1 phenotypic expression

1. Codominance a2. b �e progeny resembled

only one of the parents.3. Incomplete dominance

c

(AI 2009)

34. Provide genetic explanation for the observation in which the �ower colour in F1 generation of snapdragon did not resemble either of the two parents. However, the parental characters reappeared when F1 progenies were selfed.

(2/5, AI 2009C)35.

Look at the above diagram and answer the

following questions. (a) Write the genotypes of A, B, C, D. (b) Write the phenotypes of A, B, C, D. (c) Write phenotypic ratio of progeny. (d) Write genotypic ratio of progeny.

(AI 2009C)

36. A man with blood group A married a woman with B group. �ey have a son with AB blood group and a daughter with blood group O.


Workout the cross and show the possibility of such inheritance. (Delhi 2008)

37. A plant of Antirrhinum majus with red �owers was crossed with another plant of the same species with white flowers. The plants of the F1 generation bore pink �owers. Explain the pattern of inheritance with the help of a cross. (AI 2008)

38. A woman with blood group O married a man with AB group. Show the possible blood groups of the progeny. List the alleles involved in this inheritance. (AI 2008)

SA II (3 marks)

39. What is a test cross? How can it decipher the heterozygosity of a plant? (AI 2016)

40. A teacher wants his/her students to �nd the genotype of pea plants bearing purple coloured flowers in their school garden. Name and explain the cross that will make it possible. (Delhi 2015)

41. During a monohybrid cross involving a tall pea plant with a dwarf pea plant, the o�spring populations were tall and dwarf in equal ratio. Work out a cross to show how it is possible.

(AI 2015)

42. �e 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.

(AI 2015)

43. (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.

(Foreign 2014)44. Explain polygenic inheritance with the help

of a suitable example. (Delhi 2014C)

45. In pea plants, the colour of the �ower is either violet or white whereas human skin colour shows many gradations. Explain giving reasons how it is possible. (Delhi 2013C)

46. Human population shown variations in blood groups. Explain the genetic basis for this variation seen in the population. (AI 2013C)

47. Explain the mechanism of inheritance of the progeny produced when two Antirrhinum plants bearing pink �owers were crossed.

(AI 2012)

48. (a) Explain the phenomena of multiple allelism and co-dominance taking ABO blood group as an example.

(b) What is the phenotype of the following:(i) IA i(ii) ii (AI 2012)

49. (a) Explain a monohybrid cross taking seed coat colour as a trait in Pisum sativum. Work out the cross upto F2 generation.

(b) State the laws of inheritance that can be derived from such a cross. (3/5, AI 2012)

50. “Multiple alleles can be found only when population studies are made.” Explain with the help of an example in humans. (Delhi 2012C)

51. How are dominance, codominance and incomplete dominance patterns of inheritance di�erent from each other? (Delhi 2011)

52. A pea plant with purple �owers was crossed with white �owers producing 50 plants with only purple �owers. On sel�ng, these plants produced 482 plants with purple �owers and 162 with white �owers. What genetic mechanism accounts for these results? Explain.

(Delhi 2011)53. When tall pea plants were selfed, some of the

o�spring were dwarf. Explain with the help of a Punnett square.

(Delhi 2011C)

54. (a) Tallness of pea plant is a dominant trait; dwarfness is the alternate recessive trait. A pure tall pea plant is crossed to a dwarf one. Work out the cross to show what fraction of the tall plants in F2 generation is heterozygous?

(b) State any one law of Mendel which can be derived from this cross.

(Delhi 2008C)


LA (5 marks)

55. (a) What is polygenic inheritance? Explain with the help of a suitable example.

(b) How are pleiotropic inheritance di�erent form polygenic pattern of inheritance?

(AI 2015)

56. (a) How are Mendelian inheritance, polygenic inheritance and pleiotropy di�erent from each other?

(b) Explain polygenic inheritance pattern with the help of a suitable example. (AI 2015)

57. (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. (Foreign 2015)

58. (a) During a cross involving true breeding red �owered and true breeding white �owered 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. (Foreign 2015)

59. How do “pleiotropy”, “incomplete dominance”, “co-dominance” and “polygenic inheritance” deviate from the observation made by Mendel? Explain with the help of one example for each. (Delhi 2015C)

60. (a) A couple with blood groups ‘A’ and ‘B’ respectively have a child with blood group ‘O’. Work out a cross to show how it is possible and the probable blood groups that can be expected in their other o�-springs.

(b) Explain the genetic basis of blood groups in human population. (AI 2015C)

61. Work out a monohybrid cross upto F2 generation between two pea plants and two Antirrhinum plants both having contrasting traits with respect to colour of �ower. Comment on the pattern of inheritance in the crosses carried above.

(AI 2014C)

62. (a) Differentiate between dominance and co-dominance.

(b) Explain co-dominance taking an example of human blood groups in the population.

(AI 2013)

63. Di�erentiate between the following: (a) Polygenic inheritance and pleiotropy. (b) Dominance, co-dominance and incomplete

dominance. (AI 2013C)

64. 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? (Delhi 2012)

65. (a) List the three di�erent allelic forms of gene ‘I’ in humans. Explain the di�erent 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) �ey produce children with blood group

“A” only.(ii) �ey produce children some with “O”

blood group and some with “A” blood group.

(Delhi 2012)

66. (a) Work out a cross upto F2 generation between two pure breed pea plants, one bearing violet �owers and the other white �owers.

(b) (i) Name this type of cross.(ii) State the di�erent laws of Mendel that can

be derived from such a cross. (AI 2012C)67. State and explain with the help of a cross, the

law of segregation as proposed by Mendel. (AI 2012C)

68. (a) Explain the genetic basis of ABO - blood groups in human population.

(b) How do ABO blood groups explain the phenomenon of dominance and co-dominance? (AI 2012C)


69. Pea seeds with BB alleles have round seeds and large starch grains, while seeds with bb alleles have wrinkled seeds with small starch grains. Work out the cross between these two parents. explain the phenotypic ratio of the progeny with respect to seed shape and the starch grain size of the progeny produced.

(AI 2012C)

70. (a) Explain multiple allelism in humans with the help of a suitable example.

(b) Multiple alleles can be found only when population studies are made. Justify.

(Delhi 2011C)

71. Describe the mechanism of pattern of inheritance of ABO blood groups in humans. (AI 2011C)

72. When a garden pea plant with violet �owers was crossed with another plant with white �owers, 50% of the progeny bore violet �owers.

(a) Work out the cross. (b) Name the type of cross and mention its

signi�cance. (c) How does the inheritance pattern of �ower

colour in snapdragon di�er from the above? (Delhi 2010C)

73. When a garden pea plant with green pods was cross pollinated with another plant with yellow pods, 50% of the progeny bore green pods.

(a) Work out the cross to illustrate this. (b) How do you refer to this type of cross?

Why is such a cross done? (AI 2010C)

74. Inheritance pattern of �ower colour in garden pea plant and snapdragon di�ers. Why is this difference observed? Explain showing the crosses upto F2 generation. (Delhi 2009)

75. You are given a red �ower-bearing pea plant and a red �ower bearing snapdragon plant. How would you �nd the genotypes of these two plants with respect to the colour of the �ower? Explain with the help of crosses. Comment upon the pattern of inheritance seen in these two plants. (Delhi 2009)

76. A particular garden pea plant produces only violet �owers.

(a) It is homozygous dominant for the trait or heterozygous?

(b) How would you ensure its genotype? Explain with the help of crosses.

(AI 2009)

77. With the help of one example each provide genetic explanation for the following observa-tions :

(a) F1 - generation resembles one of two parents.

(b) F1 - generation resembles both the parents. (Delhi 2009C)

78. A snapdragon plant homozygous for red �ower when crossed with a white �owered plant of the same species produced pink �owers in F1 generation.

(a) What is this phenotypic expression called?

(b) Work out the cross to show the F2 generation when F1 was self - pollinated. Give the phenotypic and genotypic ratios of F2 generation.

(c) How do you compare the F2 phenotypic and genotypic ratios with those of Mendelian monohybrid F2 ratios? (AI 2008)

79. Given below is a table showing the genotypes and the phenotypes of blood groups in the human population :

S.No Genotype Phenotype

1 (W) A

2 IBIO (Y)

3 IA IB (Z)

4 (X) O

(a) Identify the genotypes (W) and (X) and the phenotypes (Y) and (Z).

(b) How is co-dominance different from incomplete dominance and dominance?

(c) Name the pattern of inheritance exhibited by the phenotypes (Y) and (Z) in the table.

(AI 2008C)


5.3 Inheritance of Two GenesVSA (1 mark)

80. Name the stage of cell division where segregation of an independent pair of chromosome occurs. (AI 2014)

81. A garden pea plant produced axial white �owers. Another of the same species produced terminal violet �owers. Identify the dominant traits. (Delhi, AI 2012)

82. 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?

(AI 2012)

SA I (2 marks)

83. How would you �nd genotype of a tall pea plant bearing white �owers ? Explain with the help of a cross. Name the type of cross you would use. (Delhi 2016)

84. Why did T.H.Morgan select Drosophila melanogaster to study sex linked genes for his lab experiments ? (Foreign 2015)

85. Write the scienti�c name of the fruit �y. Why did Morgan prefer to work with fruit-�ies for his experiments? State any three reasons.

OR Linkage and crossing over of genes are

alternative of each other. Justify with the help of an example. (AI 2014)

86. Study the �gures given below and answer the question.

Identify in which of the given crosses, the strength of linkage between the genes is higher. Give reasons in support of your answer.

(Foreign 2014)

87. How is the phenotypic ratio of F2 generation in a dihybrid cross is di�erent from monohybrid cross? (2/5, AI 2012)

88. In a dihybrid cross white eyed, yellow bodied female Drosophila crossed with red eyed, brown bodied male Drosophila produced in F2 generation 1.3 percent recombinants and 98.7 percent progeny with parental type combinations. �is observation of Morgan deviated from Mendelian F2 phenotypic dihybrid ratio. Explain, giving reasons, Morgan’s observations. (Foreign 2011)

SA II (3 marks)

89. 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 di�erent from the law of segregation?

(3/5, Foreign 2015)

90. Mendel published his work on inheritance of characters in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in accepting his work. (Delhi 2014)

91. Explain with the help of a suitable example the inheritance of a trait where two di�erent dominant alleles of a trait express themselves simultaneously in the progeny. Name this kind of inheritance pattern. (AI 2014)

92. Morgan carried out several dihybrid crosses in Drosophila and found F2 - ratios deviated very signi�cantly from the expected Mendelian ratio. Explain his �ndings with the help of one example. (Delhi 2014C)

93. In pea plant let, symbol Y represent dominant yellow; symbol y, the recessive green; symbol R, the round seed shape and symbol r, the wrinkle seed shape. A typical Mendelian dihybrid cross was carried out in pea plants. Write the genotypes of

(a) Homozygous dominant and recessive parents

(b) Gametes produced by both the parents


(c) F1 o�spring (d) Gametees produced by F1 o�spring (AI 2011C)

94. How did Morgan explain linkage of genes? (3/5, AI 2011C)

95. During the studies on genes in Drosophila that were sex-linked, T.H. Morgan found F2 population phenotypic ratios deviated from expected 9 : 3 : 3 : 1. Explain the conclusion he arrived at. (Delhi 2010)

96. You are given tall pea plants with yellow seeds whose genotypes are unknown. How would you �nd the genotype of these plants? Explain with the help of cross. (3/5, AI 2009)

97. State the three principles of Mendel’s law of inheritance. (AI 2009C)

LA (5 marks)

98. Give a genetic explanation for the following cross. When a tall pea plant with rounds seeds was crossed with a dwarf pea plant with wrinkled seeds then all the individual of F1 populations were tall with round seeds. However, sel�ng among F1 population led to a 9:3:3:1 phenotypic ratio. (AI 2016)

99. (a) Dihybrid cross between two garden pea plant one homozygous tall with round seeds and the other dwarf with wrinkled seeds was carried.(i) Write the genotype and phenotype of the

F1 progeny obtained from this cross.(ii) Give the di�erent types of gametes of the

F1 progeny.(iii) Write the phenotypes and its ratios of the

F2 generation obtained in this cross along with the explanation provided by Mendel.

(b) How were the observations of F2 progeny of dihybrid crosses in Drosophila by Morgan different from that of Mendel carried in pea plants? Explain giving reasons.

(Delhi 2015C)

100. A tall pea plant bearing violet flowers is given with its unknown genotypes. Explain by working out the crosses how would you �nd

the correct genotypes with respect to the two traits mentioned only by “sel�ng” the given plants. (AI 2015C)

101. A pea plant producing yellow coloured and round seeds is given with unknown genotypes. Explain how you would find the correct genotypes of the plants with respect to the two traits mentioned. Work out the cross and name it. (AI 2015C)

102. Workout a typical Mendelian dihybrid cross and state the law that he derived from it.

(AI 2014)

103. 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. (Foreign 2014)

104. (a) Work out cross between a tall pea plant bearing violet �owers (heterozygous for both) with a dwarf pea plant having white �owers. Write the genotypes and phenotypes of the progeny along with their ratios.

(b) Name such a cross and state its importance. (Delhi 2014C)

105. (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? (AI 2013)

106. Using Punnett square show the F2 results of a dihybrid cross where the pure bred parents have contrasting traits with reference to seed shape and seed colour in Pisum sativum. Give the phenotypic ratio. (Delhi 2013C)

107. Using Punnett square show the results of F2 generation of dihybrid cross where the parents have contrasting traits with reference to pod colour and seed shape in Pisum sativum. Give the phenotypic ratio. (AI 2013C)


108. (a) A garden pea plant bearing terminal, violet �owers, when crossed with another pea plant bearing axial, violet �owers, produced axial, violet �owers and axial, white �owers 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. (Delhi 2012C)

109. (a) A true breeding homozygous pea plant with green pods and axial �owers as dominant characters, is crossed with a recessive homozygous pea plant with yellow pods and terminal �owers. Work out the cross upto F2 generation respectively.

(b) State the Mendelian principle which can be derived from such a cross and not from monohybrid cross. (AI 2011)

110. (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.

(AI 2010)

111. (a) A true breeding pea plant, homozygous for in�ated green pods is crossed with another pea plant with constricted yellow pods (�gg). What would be the phenotype and genotype of F1 and F2 generations? Give the phenotype ratio of F2 generation.

(b) State the generalisation proposed by Mendel on the basis of the above mentioned cross. (Delhi 2008)

112. A true breeding pea plant homozygous for axial violet �owers is crossed with another pea plant with terminal white �owers (aavv).

(a) What would be the phenotype and genotype of F1 and F2 generations?

(b) Give the phenotype ratio of F2 generations. (c) List the Mendel’s generalisations that can

be derived from the above cross. (Delhi 2008)

113. A tall pea plant with yellow seeds (heterozyous for both the traits) is crossed with a dwarf pea plant with green seeds. Using a Punnett square

work out the cross to show the phenotypes and the genotypes of F1 generation.

(AI 2008)114. A homozygous tall pea plant with green seeds

is crossed with a dwarf pea plant with yellow seeds.

(a) What would be the phenotype and genotype of F1?

(b) Work out the phenotypic ratio of F2 generation with the help of a Punnett square.

(AI 2008)

115. Let ‘Y’ be the genotypic symbol for dominant yellow seed colour, symbol ‘y’ for recessive green seed colour, symbol ‘R’ for dominant round shape of seed and symbol ‘r’ for recessive wrinkled seed shape in garden pea. Using these symbols explain the Mendel’s law of independent assortment.

(AI 2008C)

5.4 Sex Determination VSA (1 mark)

116. A male honeybee has 16 chromosomes whereas its female has 32 chromosomes. Give one reason. (AI 2016)

117. How many chromosomes do drones of honeybee possess? Name the type of cell division involved in the production of sperms by them. (AI 2015)

118. Identify and write the correct statement : (a) Drosophila male has one X and one Y

chromosome. (b) Drosophila male has two X chromosomes.

(AI 2014)

119. 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. (AI 2014)

120. Identify the correct statement. (a) Female of many birds has a pair

of dissimilar ZW chromosomes, while the 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.

(AI 2014C)

SA I (2 marks)

121. Differentiate between male and female heterogamety. (Delhi 2015)

122. Explain mechanism of sex-determination in birds. (Delhi 2015)

123. Di�erentiate between ‘ZZ’ and ‘XY’ type of sex-determination mechanisms.

(Delhi 2015)

124. Write the types of sex-determination mechanisms the following crosses show. Give an example of each type.

(a) Female XX with male XO (b) Female ZW with male ZZ. (Delhi 2014)

125. Explain why it is scienti�cally incorrect to blame the mother for bearing female child.

(Delhi 2013C)

126. Do you agree to the perception in our society that the woman is responsibile for the gender. Substantiate your answer scienti�cally.

(2/5, AI 2011C)

127. The male fruit fly and female fowl are heterogametic while the female fruit �y and the male fowl are homogametic. Why are they called so? (Delhi 2008)

SA II (3 marks)

128. (a) Why are grasshopper and Drosophila said to show male heterogamety? Explain.

(b) Explain female heterogamety with the help of an example. (AI 2010)

129. Explain the sex determination mechanism in humans. How is it di�erent in birds?

(AI 2010)

130. Explain the mechanism of sex determination in insects like Drosophila and grasshopper.

(AI 2010)

LA (5 marks)

131. (a) How is sex determined in humans ?

(b) How does it di�er from sex determination in birds and honey bees ? (Delhi 2014C)

132. (a) E xpl a in t he me chanism of s ex-determination in humans.

(b) Di�erentiate between male heterogamety and female heterogamety with the help of one example of each. (AI 2013)

5.5 Mutation VSA (1 mark)

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

(Delhi 2011)

SA I (2 marks)134. �e cell division involved in gamete formation

is not of the same type in di�erent organisms. Justify. (AI 2011)

5.6 Genetic Disorders VSA (1 mark)

135. Give an example of a human disorder that is caused due to a single gene mutation.

(AI 2016) 136. State the chromosomal defect in individuals

with Turner’s syndrome. (Delhi 2015C)

137. Write the chromosomal defect in individuals a�ected with Klinefelter’s syndrome.

(AI 2015C)138. Why do normal red blood cells become

elongated sickle shaped structures in a person su�ering from sickle cell anaemia?

(Foreign 2014)

139. Name one autosomal dominant and one autosomal recessive Mendelian disorder in human. (Delhi 2010)

140. Write the genotype of (a) an individual who is carrier of sickle cell anaemia gene but apparently una�ected, and (b) an individual a�ected with the disease. (AI 2010)

141. A human being su�ering from Down’s syndrome shows trisomy of 21st chromosome. Mention the cause of this chromosomal abnormality.

(AI 2010)


142. �e son of a haemophilic man may not get this genetic disorder. Mention the reason.

(Delhi 2010C)

143. Write the cause of Down’s syndrome in humans. (AI 2010C)

144. A haemophilic son was born to normal parents. Give the genotypes of the parents.

(Delhi 2009C)

SA I (2 marks)

145. Why is the possibility of a human female su�ering from haemophilia rare ? Explain.

(Foreign 2014)

146.

�is is the pedigree of a family tracing the movement of the gene for haemophilia. Explain the pattern of inheritance of the disease in the family. (Delhi 2013C)

147. A relevant portion of b - chain of haemoglobin of a normal human is given below :

�e codon for sixth amino acid is GAG. �e

sixth codon GAG mutates to GAA as a result of mutation ‘A’ and into GUG as a result of mutation ‘B’. Haemoglobin structure did not change as a result of mutation ‘A’ whereas haemoglobin structure changed because of mutation ‘B’ leading to sickle shaped RBCs. Explain giving reasons how could mutation ‘B’ change the haemoglobin structure and not mutation ‘A’. (AI 2011)

148. Why is haemophilia generally observed in human males?

Explain the conditions under which a human female can be haemophilic. (2/5, AI 2011C)

SA II (3 marks)

149. Give an example of an autosomal recessive trait in humans. Explain its pattern of inheritance with the help of a cross. (Delhi 2016)

150. A couple with normal vision bear a colourblind child. Work out a cross to show how it is possible and mention the sex of the a�ected child. (Delhi 2016)

151. (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.

(Foreign 2015)

152. Which chromosomes carry the mutant genes causing thalassemia in humans? What are the problems caused by these mutant genes?

(Delhi 2015C)

153. If there is a history of haemophilia in the family, the chances of male members becoming haemophilic are more than that of the female.

(a) Why is it so? (b) Write the symptoms of the disease.

(AI 2015C)

154. A colourblind child is born to a normal couple. Work out a cross to show how it is possible. Mention the sex of this child.

(Delhi 2014)

155. A cross 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 of the inheritance pattern of this disease.

(Delhi 2014)

156. Why is pedigree analysis done in the study of human genetics? State the conclusions that can be drawn from it. (AI 2014)

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


No. Syndrome Cause Charact-eristics of a�ected

individuals

sexmale/

female/both

1. Down’s Trisomyof 21

‘a’ (i) (ii) ‘b’

2. ‘c’ XXY Overall masculine development

‘d’

3. Turner’s 45 with XO

‘e’ (i), (ii) ‘f ’

(AI 2014)

158. Why is haemophilia rare in human females? Mention a clinical symptom for the disease.

(AI 2014)

159. Name a blood related autosomal Mendelian disorder. Why is it called Mendelian disorder? How is the disorder tansmitted from parents to o�springs? (AI 2014C)

160. Why are human females rarely haemophilic? Explain. How do haemophilic patients su�er? (AI 2013)

161. Given below is the representation of amino acid composition of the relevant translated portion of b-chain of haemoglobin, related to the shape of human red blood cells.

....CTC...Gene ... GAG...

mRNA...GAG...

Val His LeuLeu �r Pro GluGlu GluGlu

1 23 4 5 6 7

HbA peptide (a) Is this representation indicating a normal

human or a sufferer from certain related genetic disease? Give reason in support of your answer.

(b) What di�erence would be noticed in the phenotype of the normal and the su�erer related to this gene?

(c) Who are likely to su�er more from the defect related to the gene represented the

males, the females or both males and females equally? And why? (Delhi 2012)

162. Why is it that the father never passes the gene for haemophilia to his sons? Explain.

(AI 2012)

163. (a) Sickle celled anaemia in humans is a result of point mutation. Explain.

(b) Write the genotypes of both the parents who have produced a sickle celled anaemic o�spring. (Delhi 2011)

164. Name a disorder, give the karyotype and write the symptoms a human su�ers from as a result of monosomy of the sex chromosome.

(Foreign 2011)

165. Name a disorder, give the karyotype and write the symptoms a human su�er from as a result of an additional X-chromosome.

(Foreign 2011)

166. Name the genetic disorder caused by trisomy of 21st chromosome in humans. Write the diagnostic features of the disorder.

(Delhi 2011C)

167. Name the genetic disorder caused by an extra ‘X’ chromosome in a human male. State the diagnostic features of the individual su�ering from it. (Delhi 2011C)

168. (a) Name the genetic disorder in a human female having 44 + XO karyotype. Mention the diagnostic features of the disorder.

(b) Explain the cause of such chromosomal disorder. (Delhi 2011C)

169. Study the following pedigree chart of a family starting with mother with AB blood group and father with O blood group.

(a) Mention the blood group as well as its genotype of the offspring numbered 1 in generation II.


(b) Write the possible blood groups as well as their genotypes of the o�springs numbered 2 and 3 in generation III.

(Delhi 2010C)

170. Study the given pedigree chart showing the pattern of blood group inheritance in a family.

XAB

AY

A

BA

O

(a) Give the genotype of the following :(i) parents(ii) the individual ‘X’ in second generation

(b) State the possible blood groups of the individual ‘Y’ in third generation.

(c) How does the inheritance of this blood group explain co-dominance?

(AI 2010C)

171. Haemophilia is sex linked recessive disorder of humans. �e pedigree chart given below shows the inheritance of haemophilia in one family. Study the pattern of inheritance and answer the questions given.

12

21

313 14 1510 11

3 4 5 6 7 8 9

(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. �e member numbered 15 has recently married the member numbered 14. What is the probability that their �rst child will be a haemophilic male?

(Delhi 2009)

172. Recently a girl baby has been reported to su�er from haemophilia. How is it possible? Explain with the help of a cross.

(AI 2009)

173.

Study the pedigree chart given above, showing

the inheritance pattern of blood groups in a family and answer the following questions.

(a) Give the possible genotypes of the individuals 1 and 2.

(b) Which antigen or antigens will be present on the plasma membranes of the RBCs of individuals 5 and 9?

(c) Give the genotypes of the individuals 3 and 4. (AI 2009C)

174. Explain the pattern of inheritance of haemo-philia in humans. Why is the possibility of a human female becoming haemophilic extremely rare? Explain. (Delhi 2008)

175. 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 possible children who could have born to them. (Delhi 2008)

176. (a) A normal couple gave birth to one haemophilic son and two normal daughters. Work out the cross to show the genotypes of the parents and their progeny.

(b) Give the possible genotypes of the parents who can give birth to haemophilic daughters. (Delhi 2008C)

177. How is it that human females are rarely haemophilic? Explain. (AI 2008C)

LA (5 marks)

178. (a) Why are colourblindness and thalassemia categorised as Mendelian disorders? Write the symptoms of these diseases seen in people su�ering from them.

(b) About 8% of human male population suffers from colourblindness whereas only about 0.4% of human female population su�ers from this disease. Write an explanation to show how it is possible. (AI 2015)


179. Write the type and location of the gene causing thalassemia in humans. State the cause and symptoms of the disease. How is sickle cell anaemia di�erent from this disease?

(Foreign 2014)180. A child su�ering from thalassemia is born to a

normal couple. But the mother is being blamed by the family for delivering a sick baby.

(a) What is thalassemia ? (b) How would you counsel the family not

to blame the mother for delivering a child su�ering from this disease? Explain.

(c) List the values your counselling can propagate in the families. (Delhi 2013)

181. Why is thalassemia categorised as a Mendelian disorder ? Write the symptoms and explain the cause of the disease. How does it di�er from sickle cell anaemia? (AI 2013C)

182. (a) State the cause and symptoms of Down’s syndrome. Name and explain the event responsible for causing this syndrome.

(b) Haemophilia and thalassemia are both examples of Mendelian disorder, but show di�erence in their inheritance pattern. Explain how. (AI 2013C)

183. (a) Why is haemophilia generally observed in human males? Explain the conditions under which a human female can be haemophiliac.

(b) A pregnant human female was advised to undergo M.T.P. It was diagnosed by her doctor that the foetus she is carrying has developed from a zygote formed by an XX-egg fertilised by Y-carrying sperm. Why was she advised to undergo M.T.P?

(AI 2011)

184. (a) How does a chromosomal disorder di�er from a Mendelian disorder?

(b) Name any two chromosomal aberration associated disorders.

(c) List the characteristics of the disorders mentioned above that help in their diagnosis. (Delhi 2010)

185. Explain the causes, inheritance pattern and symptoms of any two Mendelian genetic disorders. (AI 2010)

186. Study the given pedigree chart and answer the question that follow.

I

II

III

(a) Is the trait recessive or dominant ? (b) Is the trait sex-linked or autosomal ? (c) Give the genotypes of the parents in

generation I and of their third and fourth child in generation II. (AI 2008)

187. Study the pedigree chart given below showing the inheritance pattern of a human trait and answer the question that follow:

(a) Give the genotype of the parents shown

in generation I and of the son and daughter shown in generation II.

(b) Give the genotype of the daughters shown in generation III.

(c) Is the trait sex-linked or autosomal? Justify your answer. (Foreign 2008)


1. Living beings with shorter life cycles are preferred by geneticists for studying variations and patterns of inheritance because such organisms complete their life cycle in short duration and produce large number of progenies in less time span e.g., pea plant used in Mendel’s experiments.2. Seed traits studied by Mendel in pea plant were: (i) Seed shape – Smooth/Round (R) Wrinkled (r) (ii) Seed (cotyledon) – Yellow (Y) Colour Green (y)3. In�ated and green colour pod are dominant traits over constricted and yellow colour pod which are recessive traits.4. Traits related to pod studied by Mendel were pod shape and pod colour. In�ated (I) shaped and green coloured (G) pod were dominant traits while constricted (i) pod and yellow coloured (g) pod were recessive traits.5. Traits related to �ower studied by Mendel were �ower colour [violet/red (V or R) dominant over white (v or r)] and �ower position [Axial (A) dominant over terminal (a)].6. Di�erences between a gene and an allele are as follows :

Gene Allele(i) A gene is the unit

of DNA responsible for the appearance and inheritance of a character.

An allele is one of the forms in which a gene can exist. Normally there are two alleles for a given gene that are located at the same locus in the homologous chromosomes.

(ii) It refers to section of DNA that controls certain traits e.g., eye colour, blood group type, skin colour, etc.

It refers to speci�c variation of a gene e.g., blue eyes, green eyes, blood group A, blood group B etc.

7. �e skin colour of human is controlled by three genes where the dominant alleles have cumulative e�ect. Each dominant allele expresses a part or unit

Detailed Solutions

of the trait (skin colour). Such type of genes are called polygenes and their inheritance is called as polygenic inheritance. So, the skin colour of human is a polygenic trait.

8. �ree types of phenotypes are obtained in a monohybrid cross exhibiting co-dominance. E.g., coat colour in cattle. In F2 generation three types of phenotypes were obtained – red, roan and white coat colour.9. (a) Incomplete dominance is the condition in which the F1 phenotype does not resemble both the parents and is in between the two.(b) Complete dominance is the condition in which the F1 phenotype resembles the dominant parent i.e., one of the two parents.10. By performing a test cross, one can �nd the genotype of an organism exhibiting a dominant phenotypic trait.11. 25% of homozygous recessive is obtained in F2 generation. It can be illustrated as

12. 50% of heterozygous tall pea plant is obtained in F2 generation. For more refer to answer 11.13. Genotypes of the parent pea plant : Homozygous dominant (violet – VV) and homozygous recessive (white – vv).14.

Dominance Co-dominance(i) �e e�ect of

dominant allele is conspicuous.

�e e�ect of both the alleles is equally conspicuous.

(ii) Dominant allele produce its e�ect even in the presence of recessive allele.

Both the alleles produce their e�ect independently.


15. Recessive allele, e.g., tt represents dwarf plant.16. Genotype of parent = TtGenotype of dwarf progeny = tt

17. To determine the genotype of a plant i.e., whether the individual is exhibiting dominant character is homozygous or heterozygous, a test cross is carried out by a geneticist. �e individual having dominant phenotype is crossed with its homozygous recessive parent. If heterozygous tall is crossed with homozygous recessive parent, tall and dwarf will be produced, in equal proportion while if homozygous tall is crossed with homozygous recessive, the upcoming progenies will contain all tall plant.18. Principle of segregation states that, “when a pair of contrasting factor or gene are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and separate at the time of gamete formation”, i.e, allele pairs segregate during gamete formation and the paired condition is restored by random fusion of gametes during fertilisation. �e above law is also known as “law of purity of gametes” because each gamete is pure in itself.

19. Multiple alleles is the presence of more than two alleles of a gene. �ey are produced due to repeated mutation of the same gene but in di�erent directions and show meristic type of germinal variations, e.g., eye colour in Drosophila. Multiple alleles occur on the same gene locus of the same chromosome or its homologue and are responsible for multiple phenotypic expression.For example, the wild type of allele for red eye colour (w+ or W) in Drosophila melanogaster mutated to form allele for white eye (w). Further mutations in both have produced as much as 15 alleles which are recessive to wild type and dominant over white eye (w) but have incomplete intermediate dominance over one another.20. Two contrasting characters of height are tall and dwarf. In the given cross, if 50% of the progeny shows parental characters, then it must be a cross between a heterozygous tall and a homozygous recessive dwarf parent.

(a) Parent : Tall Dwarf

Genotype : Tt × t t

Gametes : T t t t

F1 generation : TTt

Tall

TtTall

ttDwarf

ttDwarf

tt

t

Phenotypic ratio ⇒ Tall : Dwarf = 1 : 1

(b) �is type of cross is known as test cross.

21. ABO blood groups are controlled by the gene I. �e plasma membrane of the red blood cells has sugar polymers that protrude from its surface. �e kind of sugar is controlled by the gene I. �e gene I has three alleles IA, IB and i. �e alleles IA and IB

produce a slightly di�erent form of the sugar while allele i does not produce any sugar. Each person possesses any two of the three I gene alleles. IA and IB are completely dominant over i. When IA and IB are present together they both express their own types of sugar because of codominance. Hence red blood cells have both A and B types of sugars. Since there are three di�erent alleles, there are six di�erent combinations of these three alleles that are possible, and therefore, a total of six di�erent genotypes are there in human ABO blood group.

22. (a) When a cross is made between a red �owered plant with a white �owered plant of snapdragon, the F1 hybrid has pink �owers. When the F1 individual was self pollinated F2 individuals were obtained bearing red, pink and white �owers in the ratio 1 : 2 : 1. It is not a case of blending inheritance because the parental characters appear in the F2 generation without any change. It is due to law of seggregation which states that the members of the allelic pair that remained together in the parent, segregate during gamete formation and only one factor enters a gamete.(b) In this neither of the two alleles of a gene is completely dominant over the other, hence the phenomenon is known as incomplete dominance.


23. In the ABO system, there are four blood groups A, B, AB and O. ABO blood groups are controlled by gene I. �e gene I has three alleles IA, IB and i. �is phenomenon is known as multiple allelism. IA and IB are completely dominant over i. When IA and IB are present together they both express themselves and produce blood group AB. �is phenomenon is known as codominance.24. �e ability of a gene to have multiple phenotypic e�ect because it in�uences a number of characters simultaneously is known as pleiotropy. �e gene having a multiple phenotypic e�ect because of its ability to control expression of two or more characters is called pleiotropic gene. For example, in cotton a gene for the lint also in�uences the height of plant, size of the boll, number of ovules and viability of seeds.25. Tall plants may either have genotype TT or Tt. Two tall pea plants that produce some dwarf plants among their progenies must be heterozygous with the genotype Tt, because TT plants cannot produce dwarf o�springs as they lack the allele for dwarfness (t) and hence cannot transfer it to the progeny. Besides, both of them should have a ‘t’ allele as dwarfness is expressed in homozygous (tt) condition only. It can be expressed using Punnett square as follows:

26. Refer to answer 22.27. Monohybrid cross is the cross that considers only one trait, e.g., cross of a yellow seeded pea plant

with green seeded pea plant. If pea plant with yellow seed coat is crossed with pea plant having green seed coat then in the F1 generation all the plants will produce yellow seeds. On sel�ng F2 generation will be as shown below:

×

28. Test cross is a cross used to identify whether an individual is homozygous or heterozygous for dominant character. �e individual is crossed with homozygous recessive parent for the trait being investigated. Tall plant could have two possible genotypes : TT and Tt Case I : Tall (homozygous) pea plant crossed with dwarf pea plant : Parent : Tt × tt

Gametes : T T t t Progeny :

If plant produces tall plants as o�spring, then the genotype of plant is TT i.e., homozygous tall plant.Case II : Tall (heterozygous) pea plant is crossed with dwarf pea plant.


If plant produces both tall and dwarf plants in the ratio of 1 : 1 then genotype of plant is TT, i.e., heterozygous tall pea plant.29.

Fraction of heterozygous tall plants in F2 = 12

.

�is can be explained by law of segregation which states that the members of the allelic pair that remained together in the parent, segregate during gamete formation and only one factor enters a gamete.30. Refer to answer 28.31. Refer to answer 25.32. Dominance : �e alleles IA and IB both are dominant over allele i as IA and IB form antigen A and antigen B respectively but i does not form any antigen. Codominance : Both the alleles IA and IB are codominant as both of them are able to express themselves in the presence of each other in blood group AB (IAIB) by forming antigens A and B.Multiple allelism : It is the phenomenon of occurrence of a gene in more than two allelic forms on the same locus. �e ABO blood groups in humans are determined by three di�erent allelic forms IA, IB and i.33. (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.34. Refer to answer 22.35. (a) A – Tt B – TT C – Tt D – tt (b) A – Tall B – Tall C – Tall D – Dwarf

(c) Phenotypic ratio is 3 : 1; Tall : Dwarf (d) Genotypic ratio of progeny is TT Tt tt 1 : 2 : 1 Homozygous : Heterozygous : Dwarf tall tall36. Parents

GenotypeGametes

Man WomanI iA

IA , iGenotype

IB , iI IA B I iA I iB iiAB A B OBlood group

I iB×

Parents must be heterozygous since blood group O appears in progeny. �e progeny can have all the four blood groups A, B, AB and O. �ere are three alleles of the gene controlling blood group character, i.e., IA, IB and i. IA and IB are dominant over i and together they are codominant to each other.37. Refer to answer 22.38.

Possible blood groups of the progeny are A and B.�e gene for blood group exists in three allelic forms IA, IB and i. Blood grouping is an example of multiple allelism. 39. Refer to answer 28.40. �e genotype of pea plant that whether an individual for purple coloured �owers is homozygous or heterozygous, can be determined by test cross. Purple coloured �owers in pea plant is a dominant trait. If the individual is homozygous dominant, then all o�springs will be 100% dominant. In case of heterozygous individual, o�springs will be 50% dominant and 50% recessive. �is can be explained as follows: When plant is homozygous dominant.


When plant is heterozygous dominant.

41. Refer to answer 20.

42. �e phenotypic and genotypic ratio of 1 : 2 : 1 in F2 progeny of a monohybrid cross is seen in incomplete dominance. Incomplete dominance is the phenomenon where dominant allele does not completely express itself. �is phenomenon has been observed in �ower colour of Mirabilis jalapa or four O’ clock plant. �e phenotypic as well as genotypic monohybrid ratio in F2 generation in incomplete dominance is 1 : 2 : 1 i.e., pure dominant : hybrid : pure recessive. F1 generation expresses a phenotype which is intermediate between those of the parent.

Parents :

Gametes :

F generation :1

F generation :2

Phenotypic ratio :

When a cross is made between a red �owered plant and a white �owered plant of Mirabilis jalapa, the F1 hybrid has pink �owers. When the F1 individual was self pollinated, F2 individuals were obtained bearing red, pink and white �owers in the ratio 1 : 2 : 1.

43. (a) Whenever Mendel carried out a cross between plants for a contrasting trait he found that only one trait out of the two appears in the F1

generation. He concluded that the trait which is expressed in F1 is dominant while the one which remains hidden is recessive. He also said that characters are controlled by discrete unit called factors which occur in pair.(b) In a diploid organism, there are two copies of each gene, i.e., pair of alleles. �ese two alleles are not always identical, as in a heterozygote. One of them may be modi�ed due to mutation. �e unmodi�ed functional allele that represents the original phenotype behaves as dominant allele and codes for functional protein. �e mutated non-functional allele behaves as recessive allele and codes for mutant or non-functional protein. �e phenotype of the organism will only be dependent on the functioning of the unmodi�ed allele. Hence, in a heterozygote, the dominant allele will express itself whereas recessive allele will remain hidden.44. Refer to answer 7.45. Flower colour in pea plant is a case of Mendelian inheritance where only one of the parental trait appears at F1. �e contrasting trait did not show any blending. Human skin colour is an example of polygenic inheritance. �e inheritance is controlled by three genes in which the dominant alleles have cumulative e�ect with each dominant allele expressing a part or unit of the trait.46. ABO blood groups are controlled by gene I. �e gene I has three alleles IA, IB and i. �is phenomenon is known as multiple allelism.�e blood groups and their possible genotypes are given below in the table :

47. Refer to answer 22.48. (a) Refer to answer 23.(b) (i) IA i – blood group A, (ii) ii – blood group O

49. (a) Refer to answer 27.(b) Law of dominance and law of segregation can be derived from such a cross.

50. Human population is characterised by the presence of di�erent blood groups although this trait is controlled by a single gene which indicates that


more than two alleles of the gene are involved in the inheritance of this trait i.e., multiple allelism. Since, an individual contains only two alleles of a gene, therefore to study multiple allelism, population survey is required.�e gene involved in the inheritance of human blood group possesses four alleles which are present in di�erent combinations that results in six genotypes and four phenotypes.51. In dominance, F1 is similar to the dominant parent, phenotypic ratio is di�erent from genotypic ratio. In incomplete dominance, F1 is di�erent from either of the two parents. Phenotypic and genotypic ratios are the same. In codominance, the e�ect of both the alleles is equally conspicuous. Both the alleles produce their e�ect independently.

52. In pea plant purple colour is dominant over the white colour. �e cross between the two can be shown as below : Parents

Gametes

Purple flowerPP

White flowerpp

Purple flowerSelfing

Genotype×

P p

Pp

P p

P PPp

Pp

Pp pp

F1

F2

Purple �ower : White �ower 3 : 1Mendel’s law of dominance and law of seggregation can be derived from this cross. Law of dominance states that when individuals di�ering in a pair of contrasting characters are crossed, the character that appears in the F1 hybrid is dominant over the alternate form that remain hidden. Principle of segregation states that, “when a pair of contrasting factor or gene are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and separate at the time of gamete formation”, i.e, allele pairs segregate during gamete formation and the paired condition

is restored by random fusion of gametes during fertilisation.53. Refer to answer 25.54. Refer to answer 29. 55. (a) Refer to answer 7.(b) Polygenic inheritance is a type of inheritance controlled by one or more genes in which the dominant alleles have cumulative e�ect with each dominant allele expressing a part or unit of the trait, the full trait being shown only when all the dominant alleles are present. Here a cross between two pure breeding parents does not produce dominant trait of one parent but instead an intermediate trait is exhibited. Similarly in F2 generation apart from the two parental types there are several intermediate types which link the two parental traits. E.g., kernel colour in wheat, cob length in maize, skin colour in human beings, etc.Pleiotropy is the ability of a gene to have multiple phenotypic e�ect because it in�uences a number of characters simultaneously. Pleiotropy is due to e�ect of the gene on two or more inter-related metabolic pathways that contribute to formation of di�erent phenotypes. It is not essential that all the traits are equally in�uenced. Sometimes the e�ect of a pleiotropic gene is more evident in case of one trait (major e�ect) and less evident in case of others (secondary e�ect). E.g., in cotton a gene for the lint also in�uence the height of the plant, size of the boll, number of ovules and viability of seeds.56. (a) Mendelian Inheritance : Mendelian inheritance is a type of inheritance controlled by one or more genes in which only dominant trait was expressed in the F1 generation while at the F2 stage both the traits were expressed. �e contrasting traits did not show any blending at either F1 or F2 stage on the basis of this Mendel proposed three laws: (a) Law of Dominance, (b) Law of Segregation and (c) Law of Independent Assortment.Polygenic Inheritance : Polygenic inheritance is a type of inheritance controlled by one or more genes in which the dominant alleles have cumulative e�ect with each dominant allele expressing a part or unit of the trait, the full trait being shown only when all the dominant alleles are present. Pleiotropy: �e ability of a gene to have multiple phenotypic e�ect because it in�uences a number of


characters simultaneously is known as pleiotropy. Pleiotropy is due to e�ect of the gene on two or more inter-related metabolic pathways that contribute to formation of di�erent phenotypes. (b) Human skin colour is an example of polygenic inheritance. Human skin colour is caused by pigment melanin. �e quantity of melanin is due to three pairs of polygenes (A, B and C). If black or very dark (AABBCC) and white or very light (aabbcc) individuals marry, the o�springs or individuals of F1 generation show intermediate colour o�en called mulatto (AaBbCc). When two such individuals of intermediate colour marry, the skin colour of the children will vary from very dark or black to very light or white. A total of eight allele combinations is possible in the gametes forming 27 distinct genotypes distributed into 7 phenotypes-1 very dark, 6 dark, 15 fairly dark, 20 intermediate, 15 fairly light, 6 light and 1 very light.

57. (a) Law of dominance : According to this law, characters are controlled by discrete units called factors, which occur in pairs with one member of the pair dominating over the other dissimilar pair. �is law explains expression of only one of the parental character in F1 generation.�is can be explained by the following cross :

Parents Yellow seed coat(GG)

Green seed coat(gg)

G

Gg

g

Yellow seed coatF generation1

Gametes

×

In the given cross, the trait producing yellow seeds is dominant over the trait producing green seeds. In F1 generation all o�springs showed yellow colour of seed (dominant character) and no green seeds plants were obtained.(b)

Incomplete dominance

Co-dominance Mendelian inheritance

F1 Dominant trait is incompletely expressed.

Expressed phenotype is combination of two phenotypes and their alleles.

Dominant trait is completely expressed.

F2 �e F2 generation of snapdragon �ower consists of three types of �owers - red, pink and white in the ratio of 1:2:1.

�e F2 generation in human blood group may have upto 4 phenotypes – ‘A’, ‘B’, ‘AB’ and ‘O’. With blood group AB showing codominance.

F2 progeny has 3:1 ratio i.e., dominant trait : recessive trait.

58. (a) Refer to answer 22.(b) Refer to answer 56 (b).59. Mendel discovered laws of inheritance. According to law of dominance when two individuals of a species, di�ering in a pair of contrasting forms of a trait are crossed, the form of the trait that appears in the F1 hybrid is dominant and the alternate form that remains hidden, is called recessive.Incomplete dominance and co-dominance are exception to this law. Incomplete dominance is the phenomenon where none of the two contrasting alleles or factors is dominant. �e expression of the character in a hybrid or F1 individual is intermediate or a �ne mixture of the expression of the two factors. As seen in Mirabilis jalapa where when two types of plants having �ower colour in pure state red and white are crossed, the hybrid or F1 generation have pink �owers. Co-dominance is the phenomenon of expression of both the alleles in a heterozygote, i.e., both alleles are able to express themselves independently when present together. E.g., hair colour in cattle. When red cattle are crossed with white cattle, the hybrid of F1 generation are of roan colour i.e., having a dark coat interspered with white hair. According to Mendel one gene control the expression of one character only. Pleiotropy is exception to this. �e ability of a gene to have multiple phenotypic e�ect because it in�uences a number of characters simultaneously is known as pleiotropy. �e gene having a multiple phenotypic e�ect because of its ability to control expression of two or more characters is called pleiotropic gene. For example, in cotton a gene for the lint also in�uences the height of plant, size of the boll, number of ovules and viability of seeds.


Polygenic Inheritance is a type of inheritance controlled by one or more genes in which the dominant alleles have cumulative e�ect with each dominant allele expressing a part or unit of the trait, the full trait being shown only when all the dominant alleles are present. �e genes involved in quantitative inheritance are called polygenes and inheritance called as polygenic inheritance. E.g., human skin colour. Human skin colour is caused by pigment called melanin. �e quantity of melanin is due to three pairs of polygenes (A, B and C). If black or very dark (AABBCC) and white or very light (aabbcc) individuals marry, the o�spring show intermediate colour called mulatto (AaBbCc). When two such individuals of intermediate colour marry, the skin colour of the children will vary from very dark or black to very light or white. A total of eight allele combinations is possible in the gametes forming 27 distinct genotypes distributed into 7 phenotypes.60. (a) Refer to answer 36.(b) ABO blood groups are controlled by the gene I (also called L) located on 9th chromosome that has 3 multiple alleles, out of which any two are found in a person. �ese groups show Mendelian inheritance (Bernstein, 1924). �e IA and IB alleles produce enzyme called glycosyltransferase for the synthesis of sugars. �e sugars are attached to lipids and produce glycolipids. �ese glycolipids then associate with membrane of RBC to form blood group antigens. Allelle i does not produce any enzyme/antigen.Antigenic precursor H is present in RBC mem-brane. Allele IA produces a-N-acetylgalactosamyl transferase which adds a-N-acetylgalactosamine to sugar part of H to form A antigen. �e allele IB produces a-D-galactosyltransferase which adds galactose into H to form B antigen. In case of IAIB heterozygote, both the enzymes are produced. �erefore, both A and B antigens are formed.61. In case of pea plantParents

Gametes

F1

Red flowerRR

White flowerrr

R rRr

(Red flower)

RRRed flower

RrRed flower

RrRed flower

rrWhite flowerr

R

R rSelfing

Genotype×

F2

Genotype : RR : Rr : rr 1 : 2 : 1 Phenotype : Red : white 3 : 1 In case of Antirrhinum plant

Genotype : RR : Rr : rr 1 : 2 : 1 Phenotype : Red : Pink : White 1 : 2 : 1�e inheritance pattern of �ower colour in garden pea plant is an example of complete dominance whereas inheritance pattern of �ower colour in Antirrhinum is an example of incomplete dominance.62. (a)

Dominance Co-dominance(i) F1 is similar to the

dominant parent.F1 is di�erent from either of the two parents.

(ii) In F1 hybrid, the dominant trait is completely expressed.

In F1 hybird, both the alleles express themselves independently.

(b) Codominance is a phenomenon in which alleles do not show dominance-recessive relationship and are able to express themselves independently when present together. In human blood groups both the alleles IA and IB are codominant as both of them are able to express themselves in the presence of each other in blood group AB (IAIB) by forming antigens A and B.63. (a) Refer to answer 55 (b).(b) Refer to answer 59.64. �e starch synthesis in pea plants is controlled by a single gene. It has two alleles B and b. Homozygous for BB produced large starch grains as compared to that produced by plants which are homozygous for


bb. A�er maturation it was observed that BB seeds were round and bb were wrinkled. When they were crossed the result and progeny were intermediate size Bb seeds showing round seeds. �e cross involved is:

Genotype : BB : Bb : bb 1 : 2 : 1 Phenotype : 1 Large : 2 Intermediate : 1 SmallBut in case of seed shape the phenotype is 3 : 1; Round : Wrinkled.Deviation from Mendel’s law of dominance : If starch grain size is considered as the phenotype, then the alleles show incomplete dominance. �us, dominance is not an autonomous feature of a gene, it depends on gene product and production of particular phenotype from this product.

65. (a) In humans, ABO system of blood group is a case of multiple allelism, codominance and dominance. In the ABO system, there are four blood groups A, B, AB and O. ABO blood groups are controlled by gene I. �e gene I has three alleles IA, IB and i. �is phenomenon is known as multiple allelism. IA and IB are completely dominant over i. �is is known as dominance. When IA and IB are present together they both express themselves and produce blood group AB. �is phenomenon is known as codominance. �e blood groups and their possible genotypes are given below in the table :

Blood group Genotypes (possible)

A I I I iorA A A

B I I I iorB B B

AB I I A B

O ii

(b) (i) Yes, there may be a possibility of progeny having blood group ‘A’ only. It can be explained as follows:

(ii) Yes, this may be the possibility when woman is heterozygote for blood group ‘A’. It can be explained as follows:

66. (a) Refer to answer 52.(b) (i) Monohybrid cross(ii) Two laws of inheritance can be derived from such a cross. �ese are given below: Law of dominance: According to this law, characters are controlled by discrete units called factors, which occur in pairs with one member of the pair dominating over the other dissimilar pair. �is law explains expression of only one of the parental character in F1 generation.Law of segregation : Principle of segregation states that, “when a pair of contrasting factor or gene are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and, separate at the time of gamete formation”. �e above law is also known as “law of purity of gametes” because each gamete is pure in itself.

67. Law of segregation states that, “when a pair of contrasting factor or gene are brought together in a hybrid; these factors do not blend or mix up but simply associate themselves and remain together and, separate at the time of gamete formation”. �e above law is also known as “law of purity of gametes” because each gamete is pure in itself i.e., having either G (i.e., gene for yellow seed coat) or g (i.e., gene for green seed coat) as illustrated below.


68. (a) Refer to answer 60 (b).(b) Refer to answer 65 (a).69. Refer to answer 64.

70. (a) Refer to answer 46.(b) Refer to answer 50.

71. Refer to answer 65(a).72. (a) 50% of violet �ower appear only if the parent is heterozygous for violet �ower. �e cross can be as follows :

�e plants in F1 will bear purple and white coloured �owers in the ratio 1 : 1.(b) �e cross is test cross. A test cross is performed to �nd out the genotype of the unknown plant.(c) Refer to answer 61.73. (a) In pea plant, green pod is dominant over yellow pod. 50% of dominant trait will only occur in progeny when parent is heterozygous for dominant trait. �e cross can be illustrated as follows :

(b) Refer to answer 72 (b).74. Refer to answer 61.75. A test cross is required to �nd out the genotype of both the plants.Case I In garden pea :

From the above crosses, it is clear that if the F1 generation contains all red coloured �owers, then the genotype of the given plant is RR (homozygous dominant); whereas if the F1 generation contains red and white �owers in the ratio of 1 : 1, then the genotype of the given plant is Rr (heterozygous dominant).Case II In snapdragon:(i)

r

×

R

(Pink flowers)

RRRed flower White flower

rrGenotypes

Gametes

RrF generation1

(ii)

From the above crosses, it is clear that if the F1 generation contains all pink coloured �owers, then the genotype of the given plant is RR (homozygous dominant); whereas if the F1 generation contains both pink and white �owers, then the genotype of the given plant is Rr.


�e inheritance pattern of �ower colour in garden pea plant is an example of complete dominance whereas inheritance pattern of �ower colour in snapdragon is an example of incomplete dominance.76. (a) Homozygous dominant. (b) By performing test cross, genotype of the given plant can be determined.

77. (a) In dominance, F1 resembles one of the two parents, i.e., it resembles dominant trait of the parent. Example : When violet �ower coloured pea plant is crossed with white �ower coloured pea plant, the F1 is violet coloured pea plant. �is can be illustrated as given below :

Genetic explanation : When two individuals of a species, di�ering in a pair of contrasting forms of a trait are crossed, the form of the trait that appears in the F1 hybrid is dominant i.e., it resemble one of the two parents. (b) In codominance both the alleles express themselves independently. Hence, in case of codominance, F1 resembles both the parent. E.g., roan coat colour in cattle. �e cross can be illustrated as given below :

Genetic explanation : Codominance is a pheno-menon in which alleles which do not show dominance-recessive relationship are able to express themselves independently when present together.78. (a) �is phenotypic expression is called incomplete dominance.(b) Refer to answer 22 (b).(c) Mendelian phenotypic F2 monohybrid ratio = 3 : 1Mendelian genotypic F2 monohybrid ratio = 1 : 2 : 179. (a) Genotypes : W – IA I° or IAIA and X – I°I°Phenotypes : Y – B and Z – AB(b) Refer to answer 51.(c) ‘Y’ exhibits dominance while ‘Z’ exhibits codominance pattern of inheritance.80. During anaphase of meiosis I segregation of an independent pairs of chromosomes occur.81. Axial �ower position is dominant over terminal �ower position. Violet colour is dominant over white colour.82. When the genes involved are linked, the proportion of parental gene combinations be much higher than non-parental types in a dihybrid cross.83. Test cross is a cross used to identify whether an individual is homozygous or heterozygous for dominant character. �e individual is crossed with homozygous recessive parent for the trait being investigated.Tall plant with white �owers could have two possible genotypes : TTpp and Ttpp (white colour of �ower in pea is a recessive trait, so it will be ‘pp’ in pea plant).Case I : Tall (homozygous) pea plant with white �owers crossed with dwarf pea plant with white �owers.

If plant produces all tall plants with white �owers as o�spring, then genotype of plant is TTpp i.e., homozygous tall plant with white �owers.Case II : Tall (heterozygous) pea plant with white �owers is crossed with dwarf pea plant with white �owers.


If plant produces both tall plant with white �owers and dwarf plant with white �owers, then genotype of plant is Ttpp i.e., heterozygous tall pea plant with white �owers.84. T.H. Morgan select Drosophila melanogaster to study sex-linked genes because of following reasons:(i) �ey could be grown on simple synthetic medium in the laboratory.(ii) �ey complete their life cycle in about two weeks.(iii) A single mating could produce a large number of progeny �ies.(iv) �ere was a clear di�erentiation of sexes - the male and female �ies are easily distinguishable.(v) �ey have many types of hereditary variations that can be seen with lower microscope.

85. Drosophila melanogaster is fruit �y. Also, refer to answer 84.

OR

Linkage is the tendency of two di�erent genes on the same chromosome to remain together during the separation of homologous chromosomes at meiosis. Linked genes do not exhibit the dihybrid ratio of 9:3:3:1. It produces o�spring with parental characters. Crossing over is the exchange of genes occurring during meiotic prophase I to break old linkage and establish new ones. It produces recombination resulting in new varieties. �us, they are alternative of one another i.e., if linkage is present in between genes, no crossing over occurs between them and if crossing over occurs between the two genes, they are not linked. Example : In Drosophila a yellow bodied white eyed female was crossed with brown bodied red eyed male, F1 progeny produced and

intercrossed. �e F2 phenotypic ratio of Drosophila deviate signi�cantly from Mendel’s 9:3:3:1. �is signify that the genes for eye colour and body colour are closely located on the ‘X’ chromosome and are linked. �erefore, inherited together. Recombinants were formed due to crossing over but at low percentage.

86. In cross A, the strength of linkage between the genes is higher. �e distance between the linked genes in the chromosome determines the strength of linkage. �e closely located genes show stronger linkage than the distant genes, because the latter are more likely to undergo crossing over than the former.

87. In a monohybrid cross, the phenotypic ratio of F2 generation is 3:1 whereas in dihybrid cross, the phenotypic ratio of F2 generation is 9:3:3:1.

88. By conducting the given cross, Morgan conclude that the genes for eye colour and body colour are linked. �e linked genes do not show independent assortment but remain together and are inherited, thereby producing only parental type of progeny.

89. Mendelian F2 phenotypic ratio in a dihybrid cross is 9:3:3:1. Law proposed by Mendel on the basis of this ratio is law of independent assortment. It states that in the inheritance of two pairs of contrasting characters, the factors of each pair of characters segregate independently of the factors of the other pair of characters. It is di�erent from law of segregation as law of segregation states that the members of the allelic pair that remained together in the parent, segregate during gamete formation and only one factor enters a gamete.

90. �e following are the three reasons that led to the delay in acceptance of Mendel’s work:(i) Lack of communication and publicity in those days.(ii) His concept of factors (genes) as stable and discrete units that controlled expression of traits and, of the pair of alleles that did not blend with each other was not accepted in the light of variations occurring continuously in nature.(iii) Mendel’s approach to explain biological phenomenon with the help of mathematics was also not accepted.


91.

�is kind of inheritance is known as two gene inheritance represented by a dihybrid cross. In two gene inheritance both the traits express themselves simultaneously in the progeny.92. Refer to answer 85.93. (a) Homozygous dominant = YYRRHomozygous recessive = yyrr(b) Gametes produced by both the parents = YR and yr(c) F1 = YyRr(d) Gametes produced by F1 o�spring =YR, Yr, yR and yr.94. Refer to answer 85.95. During his studies on genes in Drosophila, Morgan found that the phenotypic ratio of F2 population deviates from expected 9:3:3:1 ratio. On the basis of his study he conclude the chromosome theory of linkage which states that:– Linked genes are genes which stay together during

transmission from generation to generation and occur on the same chromosome.

– Genes are arranged in a linear fashion on the chromosome.

– Genes tend to maintain original parental combination of alleles with the exception of an occasional crossing over.

– Crossing over occurs due to weakening of linkage between two genes.

– Strength of the linkage between two genes is inversely proportional to the distance between

the two, i.e., two linked genes show higher frequency of crossing over if the distance between them is higher and lower frequency if the distance is small.

96. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character. �e individual is crossed with homozygous recessive parent for the trait being investigated. Tall plant with yellow seeds could have four possible genotypes : TTYY, TtYy, TTYy or TtYY.Case I : Tall and yellow seed (homozygous) pea plant crossed with dwarf and green seed pea plant.

If plant produce all tall plants with yellow seeds as o�spring, then genotype of plant is TTYY i.e,. homozygous for both the trait.Case II : Homozygous tall plant with yellow seed (heterozygous) crossed with recessive parent.

TyTYTtYyty

ty TtYy

Ttyy

TtyyTall Yellow

Tall Yellow

Tall green

Tall green

F generation:1

TTYyParents : × ttyy

Gametes : TY Ty ty ty

If plant produces tall plant with yellow seeds and tall plant with green seeds in the ratio of 1:1, then the genotype of plant is TTYy.Case III : Heterozygous tall plant with yellow seeds (homozygous) crossed with recessive parent.

Parents :

Gametes :

F generation :1

If plant produces tall plant with yellow seeds and dwarf plant with yellow seeds in the ratio of 1:1 then the genotype of the plant is TtYY.


Case IV : Tall plant with yellow seed (heterozygous for both) crossed with dwarf plant with green seed.

If plant produces all the four types i.e., tall yellow, tall green, dwarf yellow and dwarf green as the o�spring then the genotype of plant is TtYy.97. �e three principles of Mendel’s law of inheritance are : Law of dominance : It explains that when two individuals of a species, di�ering in a pair of contrasting forms of a trait are crossed, the form of the trait that appears in the F1 hybrid is dominant and the alternate form that remains hidden, is called recessive.Law of segregation : It states that the members of the allelic pair that remained together in the parent, segregate during gamete formation and only one factor enters a gamete.Law of independent assortment : It states that in the inheritance of two pairs of contrasting characters, the factors of each pair of characters segregate independently of the factors of the other pair of characters.98.

When F1 seeds were grown into plants, F2 seeds were obtained which showed all the four possible combinations, i.e., (i) tall and round seeds (ii) tall and wrinkled seeds, (iii) dwarf and round seeds, and (iv) dwarf and wrinkled seeds in 9 : 3 : 3 : 1 ratio.Genetic explanation : �e genes of di�erent characters located in di�erent pairs of chromosomes are independent of one another in their segregation during gamete formation. �is law is called as law of independent assortment.99. (a) Refer to answer 98.(b) Morgan observed that the F2 ratio obtained in the cross deviates signi�cantly from 9 : 3 : 3 : 1 ratio i.e., Mendelian ratio. �is is because the genes are linked. �ey are carried on the same chromosome and are inherited together. Linkage was not observed by Mendel as the characters he chose were unlinked genes.100. Tall plant with violet coloured �ower could have four possible genotypes : TTVV, TtVV, TTVv and TtVv.Case I : Homozygous tall plant with violet colour (homozygous) �ower is selfed.

If plant produce all tall plants with violet �owers as o�spring, then genotype of plant is TTVV.Case II : Heterozygous tall plant with (homozygous) violet coloured �ower is selfed.

If plant produces tall plants with violet �owers and dwarf plants with violet �owers in the ratio of 3:1 as o�spring, then the genotype of parent is TtVV.Case III : Homozygous tall plant with violet (heterozygous) �ower is selfed.


If above mentioned ratio is obtained in the progeny, then the genotype is YyRR.Case III : Homozygous yellow coloured round (heterozygous) seed pea plant is crossed with green coloured wrinkled seed pea plant.

Yellow round seed : Yellow wrinkled seed 1 : 1

If above mentioned ratio is obtained in the progeny, then the genotype is YYRr.Case IV : Heterozygous pea plant for yellow and round seed is crossed with pea plant having green and wrinkled seed.

If above mentioned ratio is obtained in the progeny, then the genotype is YyRr.102. Refer to answer 98.103. (a), (b) Refer to answer 101 case IV.(c) Refer to answer 72 (b).

104. (a) white:

:

×

(b) Refer to answer 72(b).

105. (a) Refer to answer 98.(b) Refer to answer 99 (b)

If the plant produces tall plant with violet �ower and tall plant with white �ower in the ratio of 3 : 1 as o�spring then the genotype of parent is TTVv.Case IV: Tall plant with violet �ower (heterozygous for both the trait) is selfed.Parents : TtVv × TtVvGametes : TV, Tv, TV, Tv, tV, tV, tv tv

TvTVTTVVTV

Tv TTVv

TTVv

TTvvTall Violet Tall Violet

Tall Violet Tall WhiteTtVVtV

tv TtVv

TtVv

TtvvTall Violet Tall Violet

Tall Violet Tall White

TtVV

TtVvTall Violet

Tall VioletttVV

ttVvDwarf ioletV

Dwarf Violet

TtVv

TtvvTall Violet

Tall WhitettVv

ttvvDwarf ioletV

Dwarf white

tvtVF :1

Tall iolet :v Tall hite :w Dwarf iolet :v Dwarf white9 3 3 1: : :

If the above given ratio is obtained, then the genotype is TtVv.

101. Test cross is used to identify whether an individual is homozygous or heterozygous for dominant character. �e individual is crossed with homozygous recessive parent for the trait being investigated. Yellow coloured round seed pea plant could have four possible genotypes: YYRR, YyRR, YYRr and Yy Rr.Case I : Yellow coloured round seed (homozygous) pea plant is crossed with green coloured wrinkled seed pea plant.

Parents : YYRR × yyrrGametes :

Progeny : YyRrAll yellow and round seeds.

If plant produced all yellow coloured round seeds as o�spring, then genotype of parent is YYRR.Case II : Yellow (heterozygous) coloured round (homozygous) seed pea plants is crossed with green coloured wrinkled seed pea plant.

Yellow round seed : Green round seed 1 : 1

YR yr


106.

107.

GrGRGGRRGR

Gr GGRrGGRrGGrr

GgRRgRgr GgRr

GgRrGgrr

GgRRGgRrggRRggRr

GgRrGgrrggRrggrr

grgRF :2

Green podwith round

seed

Yellow podwith wrinkled

seedGGRR ggrrParents

Gg RrGreen pod with

round seedSelfing

Green pod

9

with roundseed

Green pod

3

with wrinkledYellow

3seed

Yellow pod with

1

wrinkledseed

Phenotypicratio

: : : :round

: : :seed

podwith

108. (a) Axial �ower position is dominant over terminal and violet colour �ower is dominant over white. Since in the progeny only axial violet and axial white �owers are produced, this means that plant is heterozygous for violet colour. �e cross can be explained as:

aaVvAAVvParents: Axial violetGametes :

Terminal violetAV Av aV av

(b) Refer to answer 89.

109. (a)

(b) Refer to answer 89.110. (a) According to law of independent assortment the two factors of each character assort or separate independent of the factors of other characters at the time of gamete formation and get randomly re-arranged in the o�spring producing both parental and new combination of traits.(b) �e principle or law of independent assortment can be studied by means of dihybrid cross between heterozygous parents having YyRr genotype. �is can be demonstrated through Punnett square as follows:

Phenotypic ratio Yellow round : 9 Yellow wrinkled : 3 Green round : 3 Green wrinkled : 1

�us, the phenotypic ratio of a dihybrid cross is 9 : 3 : 3 : 1. �e occurrence of four types of plants (parental types) in the F2 generation of dihybrid cross shows that the factors of each of the two characters assort independent of the other as if the other pair of factor are not present.


111. (a)

Phenotypic ratio – In�ated : In�ated : Constricted : Constricted green yellow green yellow pods pods pods pod 9 : 3 : 3 : 1

In F1 generation all the individuals had in�ated green pods.

(b) �e generalisation proposed by Mendel was law of independent assortment which states that when two pairs of traits are combined in hybrid, the factors of every character segregate independently of the factors of other pairs of characters.

112. (a)

�e phenotype of F1 generation is axial violet �owers and genotype is AaVv.

(b) �e phenotypic ratio of F2 generation is Axial : Axial : Terminal : Terminal violet white violet white �owers �owers �owers �ower 9 : 3 : 3 : 1

(c) Refer to answer 111 (b).

113. Refer to answer 96 case IV.

114.

(a) �e phenotype of F1 is tall plants with yellow seeds. F1 genotype is TtYy.

(b) Phenotypic ratio of F2 generation is as follows : Tall plants with yellow seeds = 9 Tall plants with green seeds = 3 Dwarf plants with yellow seeds = 3 Dwarf plants with green seeds = 1

�e ratio is 9 : 3 : 3 : 1

115. Law of independent assortment states that in the inheritance of two pairs of contrasting characters, the factors of each pair of characters segregate independently of the factors of the other pair of characters. It can be explained as when a dihybrid cross is made between a pure round yellow seeded pea plant (RRYY) with wrinkled green seeded pea plant (rryy). F1 generations of hybrid plants are all round and yellow seeded. F1 generation are allowed to self breed and produce F2 generation. F2 generation has four types of plants-round yellow, wrinkled yellow, round green and wrinkled green in the ratio of 9 : 3 : 3 : 1.


116. In honeybees an unfertilised egg develops into a male and a fertilised egg develops into a female. �erefore, the female is diploid (2n), and the male is haploid (n).

117. Drones of honeybees are haploid and possess 16 chromosomes. Mitosis is involved in the production of sperms.

118. Drosophila male has one X and one Y chromosome.

119. In grasshopper males, there exist XO type of sex-determinants.

120. 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.

121. �e type of sex determination mechanism shown in female XX with male XY is called male heterogamety because male produces two di�erent types of gametes. Example - Drosophila�e type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety because female produces two di�erent types of gametes. Example - Birds

122. Birds have ZW - ZZ type of sex determination mechanism. In this type the male has two homomorphic sex chromosomes (ZZ) and is

homogametic, and the female has two heteromorphic sex chromosomes (ZW) and is heterogametic. �ere are, thus, two types of eggs: with Z and with W, and only one type of sperms, i.e., each with Z. Fertilisation of an egg with Z chromosome by a sperm with Z chromosome gives a zygote with ZZ chromosomes (male). Fertilisation of an egg with W chromosome by a sperm with Z chromosome yields a zygote with ZW chromosomes (female).

bird.

:

:

:

123. ZZ-sex determination XY-sex determination

1. �is is chromosomal sex determination where females are heterogametic i.e, they produce two type of gametes, (ZW)while the male are homogametic i.e, they produce similar type of gamete (ZZ).

�is is chromosomal sex determination where male are heterogametic, i.e, they produce two types of gametes (XY), while the females are homogametic i.e,. they possess similar type of gamete (XX).

2. �e females produce two type of eggs (A + Z) and (A + W), while the males produce only one type of sperm (A + Z).

�e females produce one type of egg (X) while the males produce two types of sperms (X) and (Y).

3. Organisms that have the ZZ type sex-determination mechanism are birds, �sh, some reptiles, etc.

Organisms that have XY type of sex-determination mechanism are humans and Drosophila.

124. (a) �e 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) �e type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety because female produces two di�erent types of gametes.Example - Birds125. In humans, sex of the child is determined at the time of fertilisation. �e female parent produces only one type of egg with X-chromosome. �e 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. �us sex of the child is determined by father and not by the mother.Hence, it is scienti�cally incorrect to blame the mother for bearing female child.126. 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. For more refer to answer 125.127. Male fruit �y and female fowl produce two types of gametes whereas female fruit �y and male fowl produce only one type of gamete.(i) �e type of sex determination mechanism shown in female XX with male XY is called male heterogamety because male produces two di�erent types of gametes. Example - Drosophila.(ii) �e type of sex determination mechanism shown in female ZW with male ZZ is female heterogamety because female produces two di�erent types of gametes.Example - Birds

128. (a) In male heterogamety, males produce two di�erent 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). �us, 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 di�erent types of gametes. �is is termed as female heterogamety. In birds and some reptiles female has two di�erent sex chromosomes (one Z and one W chromosome) whereas male has a pair of same chromosome (a pair of Z-chromosomes).

129. 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. �e female possess two homomorphic (= isomorphic) sex chromosomes, named XX. �e males contain two heteromorphic sex chromosomes, i.e., XY. All the ova formed by female are similar in their chromosome type (22 + X). �erefore, females are homogametic. �e 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 o�spring 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 individual, it is also called androsome.

In case of birds, the type of sex determination is ZW-ZZ type. Female has two di�erent sex chromosomes (Z and W) whereas as male has a pair of same chromosomes (ZZ), therefore, in birds, sex is determined by female.130. 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). �e male gametes are of two types, androsperms (22 + Y) and gynosperms (22 + X). �ey 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. �e females have two sex chromosomes, XX, while the males have only one sex chromosome. �ere is no second sex chromosome. �erefore, the males are designated as X0.131. (a) Refer to answer 129.(b) Refer to answer 129.Sex determination in honeybeeIn honeybees haploid-diploid mechanism of sex determination has been found. It is a unique phenomenon in which an unfertilised egg develops into a male and a fertilised egg develops into a female. �erefore, the female is diploid (2n), and the male is haploid (n).132. (a) Refer to answer 129.(b) Refer to answer 121.

133. Mutation134. �e parent organism may be haploid or diploid but the gametes produced by them are always haploid. �is is because of the di�erent cell division that take place during gamete formation. �e diploid parents undergo meiosis to produce haploid gametes whereas the haploid parents undergo mitosis to produce haploid gametes.135. Sickle cell anaemia is due to inheritance of a defective allele coding for b-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 b-chains of haemoglobin. �e 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.136. 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). �e individual has 2n = 45 chromosomes (44 + X0) instead of 46.

137. Klinefelter’s syndrome is caused by union of an abnormal XX egg and a normal Y sperm or normal X and abnormal XY sperm. �e individual has 47 (44 + XXY) chromosomes.

138. Sickle-cell anaemia is caused by the formation of an abnormal haemoglobin called haemoglobin-S. �e mutant haemoglobin undergoes polymerisation under low oxygen pressure and causes change in the RBCs from biconcave disc to elongated sickle-shaped structure.

139. Huntington’s disease is an autosomal dominant and sickle-cell anaemia is an autosomal recessive Mendelian disorder.

140. (a) HbAHbS

(b) HbSHbS

141. 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. �us the egg possess 24 chromosomes instead of 23 and o�spring has 47 chromosomes (45 + XY in male, 45 + XX in female) instead of 46. Down’s syndrome is also called 21- trisomy.

142. �e 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.143. Refer to answer 141.144. Genotypes of parents : XXh and XY145. 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 su�er 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).146. Haemophilia is a sex-linked, recessively inherited disorder whose gene is present on the X chromosome. �e pedigree shows criss-cross inheritance. Here, a parent passes the traits to the grand child of the same sex through o�spring of the opposite sex, i.e., in the given case, father passes the traits to grandson through his daughter.

147. Mutation A leads to change in the codon GAG to GAA. �is does not lead to the change in haemoglobin structure because GAG and GAA both codes for amino acid glutamic acid, i.e, the mutation A does not change the amino acid. In mutation B, the codon GAG is changed to GUG, where GUG codes for valine, while the original codon GAG codes for glutamic acid; hence, there is a change in the haemoglobin structure and it leads to sickle-cell anaemia. 148. Refer to answer 145.


149. 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 de�ciency. �is occurs due to formation of abnormal haemoglobin-S (Hbs). It is formed when the glutamic acid present at 6th amino acid in b-chain of normal Hb-A, is replaced by valine.Sickle cell anaemia can be transmitted from parents to the o�spring 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. �is can be shown by the given cross:

150. Colourblindness is a X-linked recessive disorder which shows transmission from carrier female to male progeny and hence, usually males are a�ected and females remain carriers.If the given couple is normal with a colourblind child, their genotypes will be Male FemaleParents : XY × XcX

Gametes : X Y X Y

Hence, a colourblind male child is born to the given couple.

151. (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.152. �alassemia is an autosomal, recessively inherited disorder. �e defect can occur due to mutation or deletion of the genes controlling the formation of globin chains (commonly a and b) of haemoglobin.a thalassemia is caused by the defective formation of a-globin which is controlled by two genes HBA1 and HBA2 present on chromosome 16. �e mutant gene cause anaemia, jaundice, hepatoseplenomegaly and bone changes. All the defective alleles kill the foetus resulting in still birth or death soon a�er delivery.b thalassemia is caused due to decreased synthesis of b globin. �e defect is due to alleles of HBB gene present on chromosome 11. It results in severe haemolytic anaemia, hepatosplenomegaly, cardiac enlargement and skeletal deformities.

153. (a) Refer to answer 145.(b) Symptoms of haemophilia: Clotting of blood is a�ected as one of the factors needed for clotting is not synthesised, so in a haemophilic individual, even a simple cut leads to non-stop bleeding.

154. Refer to answer 150.

155. (a)

(b) It a sex-linked inheritance showing criss-cross pattern where a parent passes the traits to the grand child. Here , the female is the carrier of the disease.

156. Pedigree analysis is study of pedigree for the transmission of particular trait. It is done to study human genetics because control crosses are not


possible in human being as the generation time is more in humans. Pedigree analysis is useful in following ways: (i) It is useful in �nding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual and his family members.(ii) It is useful in detecting genetic defects like haemophilia, colourblindness, alkaptonuria, phenylketonuria, thalassemia, sickle cell anaemia (recessive traits), brachydactyly and syndactyly (dominant traits).(iii) It helps to detect sex-linked characters and other linkages.

157. (a) (i) Partially opened mouth with furrowed tongue.(ii) Palm is broad with palm crease(b) Both (c) Klinefelter’s (d) Male(e) (i) Sterile female with poorly developed ovaries and under developed breasts.(ii) Webbed neck and broad chest (f) Female

158. Refer to answer 145.�e patient having haemophilia will continue to bleed even from a minor cut since he/she does not possess the natural phenomenon of blood clotting due to the absence of antihaemophilic globulin or factor responsible for clotting.

159. Sickle-cell anaemia is a blood related autosomal Mendelian disorder. It is called Mendelian disorder because it is transmitted to the o�spring as per Mendelian principles. �e gene for sickle-celled erythrocytes is represented by Hbs while that of normal erythrocytes is written as HbA. �e homozygotes for the two types are Hbs Hbs and HbA HbA. �e heterozygotes are written as HbA HbS. When two sickle cell heterozygotes marry they produce three types of children–homozygous normal, heterozygous carrier and homozygous sickle celled in the ratio of 1 : 2 :1. However, homozygous sickle-celled individuals (Hbs Hbs) die in childhood (before reproductive age) due to acute anaemia. �erefore, a ratio of one normal to two carriers is obtained.

160. Refer to answer 145 and 158.161. (a) �e representation indicates a normal human. HbA is a normal peptide with glutamic acid at the sixth position of beta globin chain.(b) �e normal individual has biconcave, disc-like RBCs whereas the su�erer of the disease has elongated sickle-shaped RBCs.(c) Both males and females su�er equally because sickle-cell anaemia is an autosomal disease and not a sex-linked one. �e sickle-shaped RBC will cause equal oxygen de�ciency in both males and females.162. Refer to answer 142.163. (a) Sickle cell anaemia is caused by the formation of an abnormal haemoglobin called haemoglobin-S. Haemoglobin-S di�ers from normal haemoglobin-A in only one amino acid – 6thamino acid of b-chain. Here, glutamic acid is replaced by valine due to substitution (trans-version) of T by A in the second position of the triplet codon (CTC) which is changed to CAC. �e gene is situated on chromosome 11. �e codon CTC is transcribed into GAG (coding for glutamic acid) but due to substitutions of T by A the new codon CAC is transcribed into GUG that codes for valine. Hence it is a result of point mutation.(b) Refer to answer 159.

164. Turner’s syndrome is a disorder where the individual has 22 pairs of autosomes and XO sex chromosomes i.e., 45 chromosomes. So, the karyotype will be 44 +XO.Symptoms : (i) Sterile females(ii) Rudimentary ovaries(iii) Lack of secondary sexual characters(iv) Webbed neck and broad chest(v) Underdeveloped breasts


165. Klinefelter’s syndrome is a disorder caused due to an additional X - chromosome. �e individual has 22 pairs of autosomes and XXY sex chromosomes i.e., the karyotype is 44 + XXY.Symptoms: (i) �e individual is a male.(ii) �e male shows development of feminine characters like development of breasts.(iii) Body hair is sparse.(iv) �e individual is sterile.166. Refer to answer 157. 167. Refer to answer 165.168. (a) Refer to answer 164.(b) Refer to answer 136.169. (a) O�spring numbered 1 has blood group B with genotype IOIB.

(b) For o�spring numbered 2, the possible blood groups as well as their genotype would be O (IOIO) or A (IOIA) as shown:

�ere can be two cases for o�spring numbered 3 to know the possible blood groups as well as their genotypes.Case I : When parent is homozygous for A i.e, IAIA

Case II : When parent is heterozygous for A i.e, IOIA

170.

(a) (i) Parents : IAIO and IBIO

(ii) Since the o�spring of X is a individual with blood group A, so the genotype of individual X can be IOIO

or IOIA or IAIA. It can be explained as follows:Case I: When genotype of X is IOIO

Case II : When genotype of X is IOIA

Case III : When genotype of X is IAIA

(b) �e possible blood groups of individual ‘Y’ in third generation may be A or O. �is can be illustrated as given below:Case I : When an individual with blood group O (IOIO) marries with an individual with blood group A (IAIA) :�e o�spring will have blood group A with genotype IO IA.Case II :When an individual with blood group O (IO IO) marries a person with blood group A (IOIA)�e o�spring can either have blood group O with genotype IOIO or A with genotype IAIO.


(c) �e pedigree shown occurrence of blood group AB in the family. Blood group AB is an example of co-dominance. Both the alleles IA and IB are codominant as both of them are able to express themselves in the presence of each other in blood group AB by forming antigens A and B.171. (a) 4 - X Xh

5 - XhY 6 - X Y(b) �e probability of their �rst child being a haemophilic male is 25%.

(X X)h (XY)

c

172. Haemophilia is an X-linked recessive disorder. It is possible to have a haemophilic girl when a carrier woman marries a haemophilic man.

173. (a) Individual 1 – IOIB

Individual 2 – IOIA

(b) Individual 5 will have both antigens A and B while individual 9 will neither have the two on the plasma membranes of the RBCs.(c) Individual 3 – IBIO

Individual 4 – IAIO

174. Haemophilia is a sex-linked recessive disorder. Haemophilia (= hemophilia) is genetically due to the presence of a recessive sex linked gene h, carried by X chromosome. A female becomes haemophilic only when both its X chromosomes carry the gene (XhXh). However, such females generally die before birth because the combination of these two recessive

alleles is lethal. A female having only one allele for haemophilia (XXh) appears normal because the allele for normal blood clotting present on the other X chromosome is dominant. Such females are known as carriers. In case of males, a single gene for the defect is able to express itself as the Y chromosome is devoid of any corresponding allele (XhY).�e possibility of human female becoming haemophilic is extremely rare because she has to be homozygous recessive for the trait, i.e., her father must be a haemophilic and mother must be atleast a carrier.

175. �e doctor must have used pedigree analysis which refers to the analysis of distribution and movement of traits in a series of generations of a family.Since the non-haemophilic parents may give rise to a haemophilic child, the genotypes of them should be:Father : XY (normal)Mother : XXh (carrier/heterozygous, non-haemophilic)

�e progeny can consist of the following genotypes and phenotypes : XX XXh XY XhY Normal : Carrier : Normal: Haemophilic female female male male 176. (a) Refer to answer 175.(b) Refer to answer 172.177. Refer to answer 174.

178. (a) Colourblindness and thalassemia are categorised as Mendelian disorders because of the following reasons:(i) �ey are mainly due to alteration or mutation in a single gene.(ii) �ese disorders are transmitted to the o�spring in the same line as Mendelian principles of inheritance, i.e., by the parents who are carriers and are apparently normal.


(iii) �e pattern of inheritance of these disorders can be traced in a family by pedigree analysis.Symptoms of Colourblindness : �e person fails to discriminate between red and green colour due to the defect in either red or/and green cone cells of retina.Symptoms of �alassemia : �e person su�ers from anaemia as the synthesis of either alpha globin chain(s) or beta globin chain (s) of haemoglobin is impaired.(b) Colourblindness is a X-linked recessive disorder which shows transmission from carrier female to male progeny. In females colour blindness appears only when both the sex chromosomes carry the recessive gene (XcXc). �e females have normal vision but function as carrier if a single recessive gene for colourblindness is present (XXc). However , in human males the defect appears in the presence of a single recessive gene (XcY) because Y-chromosome of male does not carry any gene for colour vision. As a result colour blindness is more common in males (8%) as compared to females (0.4%). 179. Refer to answers 152, 178 and 138.180. (a) �alassemia is a disease characterised by reduced synthesis of either the a or b chains of haemoglobin and likewise designated as a or b thalassemia. It may lead to haemolytic anaemia.(b) �alassemia is an autosome linked recessive blood disease transmitted from parents to the o�spring when both the partners are una�ected carrier for the gene i.e. have heterozygous condition for thalassemia gene.A child born to a diseased mother and normal father will have only one gene for the disease and will be an una�ected carrier. A thalassemic baby can be born only to a carrier couple. �us, mother only cannot be blamed for the birth of a thalassemic baby, its father is equally responsible for the disease.(c) �e counselling can encourage families to understand that only mother is not responsible for any disease or defect found in the baby. A baby inherits its characters from both of its parents not just only from mother. It will help to upgrade condition of women in such orthodox families.181. Refer to answers 178, 152 and 138.182. (a) Refer to answer 141.Symptoms of Down’s syndrome : It is characterised by round face, broad forehead, permanently open mouth, protruding tongue, projecting lower lip, short neck, �at hands and stubby (small) �ngers, many ‘loops’ on �nger tips, coarse and straight hair, furrowed tongue, broad palm with characteristic

palmer crease, which runs all the way across the palm and monogolian type eye lid fold (epicanthus). �e victim has little intelligence.(b) Inheritance pattern of �alassemia : �alassemia is an autosomal, recessively inherited disorder, transmitted to the o�spring when both the parents are heterozygous/carriers for the disease.Alpha thalassemia is controlled by two closely linked genes, HBA1 and HBA2, located on chromosome 16. Beta thalassemia is controlled by a single gene, HBB, located on chromosome 11.Inheritance pattern of Haemophilia : Haemophilia is a sex-linked, recessively inherited disorder, whose gene is present on the X chromosome. It a�ects more males than females, because a male has only one X chromosome and the female has two X chromosomes and has to be homozygous recessive for the disease to develop.183. (a) Refer to answer 174.(b) M.T.P (Medical Termination of Pregnancy) is an intentional or voluntary termination of pregnancy before full term. �e pregnant human female was advised to undergo MTP because, the zygote is formed by an additional copy of X-chromosome resulting into a karyotype of 47 with XXY. Such an individual will develop Klinefelter’s syndrome. Such persons are sterile males with undeveloped testes, mental retardation, female like sparse body hair, and knock knees, long limbs and with some female characteristics such as feminine pitched voice and enlarged breasts (gynaecomastia).

184. (a)Mendelian disorder

Chromosomal disorder

(i) �is disorder is mainly due to alteration or mutation in the single gene.

�is disorder is caused due to absence or excess or abnormal arrangement of one or more chromosomes.

(ii) �is follows Mendel’s principles of inheritance.

�is does not follow Mendel’s principles of inheritance.

(iii) �is may be recessive or dominant in nature.

�is is always dominant in nature.

(iv) For example, haemo-philia, sickle-cell anaemia.

For example, Turner’s syndrome.


(b) Two chromosomal aberration-associated disorders are Down’s syndrome and Klinefelter’s syndrome.(c) (i) Down’s syndrome: �e individuals have overall masculine development but they express feminine development like development of breast, i.e., gynaecomastia. �ey are sterile.(ii) Klinefelter’s syndrome: �e females are sterile as ovaries are rudimentary. Other secondary sexual characters are also lacking.185. Refer to answer 178 and 152.186. (a) It is a recessive trait.(b) It is an autosomal trait.

(c) Generation I → Aa and Aa Generation II → �ird child - aa Fourth child - Aa187. (a) Genotype of the parents in generation I : Aa and Aa; Son (Generation II) – Aa; Daughter (Generation II) – aa(b) Genotypes of the daughters shown in generation III – Aa and aa.(c) It is an autosomal trait, because if the sex-linked trait has to appear in the daughter (generation III), the father must have it; but he does not show the trait and so it is not sex-linked.

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Principles of 05 Inheritance and Variation