2nd PUC Biology Question Bank Chapter 5 Principles of Inheritance and Variation

You can Download Chapter 5 Principles of Inheritance and Variation Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 5 Principles of Inheritance and Variation

2nd PUC Biology Principles of Inheritance and Variation NCERT Text Book Questions and Answers

Question 1.
Mention the advantages of selecting pea plant for the experiment by Mendel.
Answer:
Mendel selected garden pea for his experimental work because it had the following advantages:

• The pea plants showed a number of well-defined contrasting characters.
• It has perfect bisexual flowers containing both male and female parts. The flowers are predominantly self-pollinating.
• Because of self fertilisation, plants are homozygous. It is, therefore, easy to get pure lines for several generations.
• It is an annual plant. Its short life cycle made it possible to study several generations within a short period.
• It is easy to cultivate.
• It is easy to cross because pollen from one plant can be introduced to the stigma of another plant.

Question 2.
Differentiate between:-
(a) Dominance and Recessive.
Answer:

Dominance	Recessive
1. Able to express even in presence of contrasting alleles. 2. Does not require another similar allele to produce its phenotype. 3. Produces complete polypeptide, protein, or enzyme. 4.  It is usually wild type allele.	1. Cannot express in presence of contrasting alleles. 2. Can produce its phenotype only along with a similar allele. 3. Forms incomplete products. 4.  Usually a mutant allele.

(b) Homozygous and Heterozygous.
Answer:

Homozygous	Heterozygous
1. Possesses similar alleles. 2.  2 types – Homozygous dominant and recessive. 3. Individual is pure for the trait. 4. On self breeding, a similar types of offspring are formed. 5. Only one type of gamete is formed	1. Possesses different alleles. 2.  Is of one type 3. Individual is seldom pure for the trait. 4. On breeding, 3 types of offspring are formed. 5. 2 types of gametes produced.

(c) Monohybrid and Dihybrid
Answer:

Monohybrid	Dihybrid
1. Cross made between individuals having contrasting traits in order to study the inheritance of a pair of alleles. 2.  Phenotypic monohybrid ratio in F2 generation is 3:1. 3. Genotypic monohybrid ratio in F2 generation is 2:1.	1. Cross made between individuals having contrasting traits in order to study the inheritance of 2 pairs of alleles. 2. Phenotypic dihybrid ratio is 9:3:3:1 3. Genotypic dihybrid ratio is 1:2:1:2:4:2:1:2:1

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
A diploid organism heterozygous for 4 loci will have the supported genetic constitution YyRr for two characters. The alleles Y-y and R-r will be present on different 4 loci. Each parent will produce four types of gametes – YR, Yr, yR, yr.

Question 4.
Explain the law of Dominance using a monohybrid cross.
Answer:
Law of Dominance:

• Characters are controlled by discrete units called factors.
• Factors occur in pairs.
• In a dissimilar pair of factors, one member of the pair dominates (dominant) the other (recessive) suppresses.

In a monohybrid cross:

Question 5.
Define and design a test cross.
Answer:
Test cross: An organism showing dominant phenotype (genotype is to be determined.) from F₂ is crossed with a recessive parent instead of self-crossing. Progenies of such a cross can be analysed to predict the genotype of the test organism. Eg: A tall and short plant taken into the experiment. From F₂ generation, TT or Tt could be tall  I plant. The recessive parent will be tt. So, if the test plant was TT, the resulting plant will only I tall. If the test plant was Tt, the resulting plant will be tall and short.

Question 6.
Using a Punnett square, workout distribution of phenotypic features in the F₁ generation after crossing between homozygous female and heterozygous male for a single locus.
Answer:

Question 7.
When the cross is made between a hybrid tall plant with yellow seeds (Tt Yy) and tall plant with green seeds (Tt yy), What proportion phenotype in offspring could be expected to be
(a) Tall and Green
(b) Dwarf and Green
Answer:

Question 8.
Two heterozygous plants are crossed. If 2 loci are linked what would the distribution of phenotypic features in F₁ generation in a dihybrid cross.
Answer:

Question 9.
Mention the contributions of TH Morgan in Genetics.
Answer:
Thomas Hunt Morgan (1866-1945) is called father of experimental genetics. He was an American scientists, famous for his experimental research with the fruit fly (Drosophila) by which he established the chromosome theory of heredity. He discovered presence of gene over chromosomes, chromosome theory of linkage, chromosome mapping, crossing over, criss-cross inheritance & mutability of genes. Morgan’s work played a key role in establishing the field of genetics. He received the nobel prize for physiology or Medicine in 1933.

Question 10.
What is pedigree analysis? Suggest how such analysis can be useful.
Answer:
Pedigree analysis: Study of transmission of particular traits graphically over the present and the last few generations for finding out the possibility of their occurrence in future generations. So, analysis of traits in several generations of a family is called Pedigree Analysis.
Importance:

• To know the possibility of a recessive allele which may create disorder.
• Can indicate the origin of a trait in the ancestors.
• Analysis is used for genetic counseling.
• Extensively used in medical research.

Question 11.
A child has blood group O. If the father has blood group A and mother B, work out genotypes of parent and the possible genotypes of other offsprings.
Answer:
Blood group O can appear with 1° 1° only (2 recessive alleles).

Question 12.
How is sex determined in human beings?
Answer:
Chromosomal determination of sex in human beings is of XX-XY type. Human beings have 22 pairs of autosomes and one pair of sex chromosomes. The female possess two homomorphic (= isomorphic) sex chromosomes, named XX. The males contain two heteromorphic sex chromosomes, i.e., XY. All the ova formed by female are similar in their chromosome type (22 + X). Therefore, females are homogametic. The male gametes or sperms produced by human males are of two types, gymnosperms (22 + X) and androsperms (22 + Y). Human males are, therefore, heterogametic.

Sex of the offspring is determined at the time of fertilisation. Fertilisation of the egg (22 + X) with a gynosperm (22 + X) will produce a female child (44 + XX) while fertilisation with an androsperm (22 + Y) gives rise to male child (44 + XY). As the two types of sperms are produced in equal proportions, there are equal chances of getting a male or female child in a particular mating. As Y-chromosomes determine the male sex of the individual, it is also called androsome.

Question 13.
Explain the following terms with an example
(a) Co-dominance
(b) Incomplete dominance
Answer:
a. Codominance is a condition in which the F₁ generation resembles both parents, (e.g. Blood groups).
b. The type of inheritance in which the F₁ does not show the dominant character but shows an intermediate character between the dominant and recessive is called incomplete dominance.

Question 14.
What is point mutation? Give one example.
Answer:
When heritable alterations occur in a very small segment of DNA molecule i.e., a single nucleotide or nucleotide pair, then these types of mutations are called point mutations (also called gene mutations). The point mutations may occur due to inversion, substitution (transition and transversion), and frameshift (insertion and deletion) type of nucleotide change in the DNA or RNA. Phenylketonuria (PKU; Foiling 1934) is an inborn, autosomal, recessive metabolic disorder in which the homozygous recessive individual lacks the enzyme phenylalanine hydroxylase needed to change phenylalanine (amino acid) to tyrosine (amino acid) in the liver. It results in hyperphenylalaninemia which is characterized by the accumulation and excretion of phenylalanine, phenyl pyruvic acid, and related compounds.

The lack of the enzyme is due to the abnormal autosomal recessive gene on chromosome 12. This defective gene is due to substitution. Affected babies are normal at birth but within a few weeks, there is a rise (30 – 50 times) in plasma phenylalanine level which impairs brain development. Other symptoms are mental retardation, decreased pigmentation of hair and skin, and eczema.

Question 15.
Who had proposed the chromosomal theory of inheritance?
Answer:
Sutton and Boveri proposed the chromosomal theory of inheritance. The theory believes that chromosomes are vehicles of hereditary information that possess mendelian factors or genes and it is the chromosomes which segregate and assort independently during transmission from one generation to the next.

Question 16.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Phenylketonuria: This inborn error of metabolism is also inherited as the autosomal recessive trait. The affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine. As a result, this phenylalanine is accumulated and converted into phenyl pyruvic acid and other derivatives. Accumulation of these in the brain results in mental retardation.

Sickle cell anemia: This is an autosome liked recessive trait that can be transmitted from parents to the offspring when both the partners are carriers for the gene of heterozygous.

2nd PUC Biology Principles of Inheritance and Variation Additional Questions and Answers

2nd PUC Biology Principles of Inheritance and Variation One Mark Questions

Question 1.
Give the meaning of the term allele.
Answer:
It represents a pair of genes of 2 alter-natives of the same character and occupies the same position (locus) on the homologous chromosome. OR The various forms of genes are called alleles.

Question 2.
What is known as the alternative forms of a gene that can occupy the same locus on a particular chromosome and that controls the same character?
Answer:
Allelomorph.

Question 3.
Differentiate between mutants and mutons.
Answer:
Mutants: The organism showing mutation.
Mutons: The smallest segment of DNA which can undergo mutation.

Question 4.
What do you mean by heredity material which is found in the outer part of the chromosome?
Answer:
Plasmogene.

Question 5.
What term did Mendel use for what we now call the genes? [Hots]
Answer:
Factors or determination.

Question 6.
What do you say when a gene shows a dominant effect?
Answer:
Dominant gene.

Question 7.
Define Phenotype.
Answer:
The observable or external characteristics of an organism, constitute its phenotype.

Question 8.
What is meant by true breeding?
Answer:
True breeding means that the parents will also pass-down a specific phenotypic trait to their offspring.

Question 9.
Mendel studied seven traits in garden pea. Which one or more of the following were recessive?
Wrinkled seed, axial flower, yellow colour of pod, tall, purple flower.
Answer:
Wrinkled seed

Question 10.
What do you mean by Fj in mendels law?
Answer:
First fertile generation.

Question 11.
Name the plant in which Mendel performed his experiment.
Answer:
Pea plant – Pisum sativum.

Question 12.
Who gave the name ’X’ body? Which genes are select sex determination in human beings?
Answer:
Herman Henking suggested the name and occur.es the sex determination in humans by XX and XY chromosome.

Question 13.
Mendel observed 2 kinds of ratios, i.e., 3:1 and 1:2:1 in F₂ generation in his experiments on the garden pea. Name these two kinds of ratios respectively.
Answer:

• 3:1 – Phenotypic ratio
• 1:2:1 – Genotypic ratio.

Question 14.
Give one example of Co-dominancy.
Answer:
‘A’, ‘B’ and ‘O’ genes are co-dominant which found in blood group.

Question 15.
What is epistasis?
Answer:
It is the interaction between two different genes (non-allelic genes) where one gene masks the effect of another gene.

Question 16.
What do you mean by the diagrammatic representation of hereditary characters which shows specific characters from generation-to-generation?
Answer:
Pedigree analysis.

Question 17.
In man, four types of blood groups A, B, AB, and O are controlled by three alleles of a gene. Name the type of inheritance involved in the blood group.
Answer:
Multiple allelic inheritances.

Question 18.
Give the name of any allopolyploid species which is originated in a natural way.
Answer:
Wheat (Triticum astivum).

Question 19.
What is the phenotypic and genotypic ratio of incomplete dominance?
Answer:
The phenotypic and genotypic ratios are same i.e. 1AA : 2Aa : laa

Question 20.
When are individuals exhibiting morphological characteristics of both sexes in drosophila called this method?
Answer:
Gynandromorphy.

Question 21.
Name the pigment that gives colour to the skin of man.
Answer:
Melanin

Question 22.
Names a specific DNA segment which is functioned as a unit of heredity.
Answer:
Gene.

Question 23.
What are complementary genes?
Answer:
These are the two independent pairs of genes that interact to produce a trait together but each dominant gene alone does not show its effect.

Question 24.
What is the name of the arms of a chromosome?
Answer:
Chromatid.

Question 25.
Name the pigments that control skin colouration in man. [Hots]
Answer:
Melanin.

Question 26.
Who first observed the ‘X’ chromosome?
Answer:
Henking. [Hots]

Question 27.
Why ‘X’ chromosome is called a sex chromosome?
Answer:
It involves the sex determination of an individual.

Question 28.
Name the disease caused by inadequate or defective nutrition.
Answer:
Dystrophy.

Question 29.
Define heterogamety. Give an example.
Answer:
It is a phenomenon in which organisms produce 2 types of (more than one type) gametes.
Eg: Human male, male drosophila / female fowl.

Question 30.
What is the exception of Mendel’s Independent assortment theory?
Answer:
Linkage.

Question 31.
Give the term for the factors which cause mutation.
Answer:
Mutagens.

Question 32.
Mention 2 sex-linked Mendelian disorders.
Answer:
Haemophilia and colour blindness

Question 33.
Name some diseases which can be avoided in the progeny through pedigree analysis of parents.
Answer:
Colour blindness, Tuberculosis, Turner’s syndrome.

Question 34.
Give the reason for Down’s syndrome.
Answer:
Trisomy of 21st chromosome.

Question 35.
Define trisomic condition.
Answer:
When a particular chromosome is present in 3 copies in a diploid cell, the condition is called trisomic condition.

36.
Define monosomic condition.
Answer:
When a particular chromosome is present in a single copy in a diploid cell, the condition is called a monosomic condition.

2nd PUC Biology Principles of Inheritance and Variation Two Marks Questions

Question 1.
Define test cross. What is its significance?
Answer:
It is the cross between an individual with a dominant phenotype with an individual homozygous recessive for the trait. It is used to determine the genotype of an individual for any character trait.

Question 2.
What are multiple alleles? Give an example.
Answer:
When a gene exists in more than 2 allelic forms, the alleles are called multiple alleles.
Eg: The gene ‘I’ controls the human blood group. It exist in 3 different alleleic form i.e., I^A I^B and i.

Question 3.
Write T.H.Morgan’s contribution to genetics.
Answer:

• Morgan conducted experiments in Drosophila melanogaster and discovered sex linkage.
• He also found that linked genes may show the phenomenon called linkage where the recombinants in a test than 50%. cross progeny are less.

Question 4.
Differentiate between monohybrid and dihybrid.
Answer:

Monohybrid	Dihybrid
It is a cross, where 2 forms of the single trait are hybridised.	It is a cross where 2 forms of 2 different traits are hybridised.

Question 5.
Differentiate between genotypes and phenotype.
Answer:

Genotype	Phenotype
(a) It is the total genetic constitution of an individual.	(a) It is the external appearance of an individual.
(b) It is the expression of the genome or more specifically the alleles present at one locus.	(b) It is the expression of genotype produced under the influence of an environment.

Question 6.
Differentiate between test cross and reciprocal cross.
Answer:

Test cross	Reciprocal cross
It is a cross between, F1 hybrid and recessive parent. It confirms the purity of the F1 hybrid whether it is homozygous or heterozygous.	It is the second cross involving the same strains carried by sexes opposite to those in the first cross. It is able to distinguish between nuclear chromosomal and sex-linked inheritance.

Question 7.
Is not possible to study the inheritance of traits in humans in the same way as in peas? Give 2 main reasons for it and the alternative method used for such study.[Hots]
Answer:
Mendel’s laws are not applicable for human beings because:

• In human the generation time is too long and produces a small progeny. It creates difficulty in the statistical computation of any generation of individuals.
• Human cannot be crossed at will. Pedigree analysis is another method used to study the family instances and the transmission of particular trait generation after generation.

Question 8.
What is back cross?
Answer:
When an intercross done between 2 genetically different parents, a hybrid is produced which may be homozygous or heterozygous. To determine the purity of parents and to test genotypes of F₁ hybrid, a cross is made between F₁, hybrid and of the parent. Such cross is known as backcross.

Question 9.
How would you find the genotype of an organism exhibiting a dominant phenotypic trait?[Delhi 2008]
Answer:
First a test cross will be done between the dominant phenotypic individual (F₁ hybrid) with the recessive phenotypic individual (parent). If the individuals are homozygous dominant, all the individuals in the progeny

If the individuals are heterozygous, the progeny will show dominant phenotype and recessive phenotype in the ratio 1:1.

Question 10.
Give the structure of E. coli chromosome.
Answer:
Isolated E. coli chromosomes display many individually supercoiled loops emanating from a central region. These supercoiled loops were hypothesized to be topological domains. These domains are sufficient to generate the observed precision of E. coli chromosome structure. Its chromosomal DNA has been completely sequenced by lab researchers. E. coli has a single chromosome with about 4,600 kb, about 4,300 potential coding sequences, and only about 1,800 known E. coli proteins.

Question 11.
What is meant by chromosomal abberation? In which type of cells it commonly occurs?
Answer:
Alternation in chromosomes is due to the deletion/duplication/addition of a segment of DNA is called chromosomal abberation. Commonly occurs in cancerous cells.

Question 12.
What is a modifier gene?
Answer:
Modifier genes, instead of making the effects of another gene. A gene can modify the expression of a second gene. In mice, coat colour is controlled by the B gene. THeB allele conditions black coat colour and is dominant to the b allele that produces a brown coat.

Question 13.
Classify the following into chromosomal and Mendelian disorders.
(a) Cystic fibrosis
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Hemophilia
Answer:
(a) Cystic fibrosis – Mendelian disorder
(b) Turner’s syndrome – Chromosomal disorder
(c) Klinefelter’s syndrome – Chromosomal disorder
(d) Haemophilia – Mendelian disorder

Question 14.
Describe the causes of hereditary variation.
Or
Why do hereditary variations originate?
Answer:
Any change in the structure of the gene of an organism will produce heritable variations. The chief causes of these variations are as follows:

(i) Genetic recombination: The exchange of chromosomal segments of genes takes place during reduction division which causes recombination of genes. These recombinant organisms fuse to form a zygote. The genetic structure of these zygotes is different from their parents. Adults formed from this zygote will show variations.

(ii) Changes in the number of chromosomes: Changes in the number of chromosomes will also cause genetic variations, e.g., modern varieties of wheat contain 16,21 and 42 chromosomes. These varieties are produced from their parents having only 7 chromosomes.

(iii) Changes in the structure of genes: Changes in the structure of genes will also cause genetic variations, e.g., 30 varieties of wheat is the best example of genetic variations. Like this, if any changes in the genes of human beings which are related to pigment production, will inhibit the production of these pigments and thus man become albino.

Question 15.
Explain how an XXY individual can arise in humans.
Answer:
An XXY individual can arise if an abnormal egg with XX chromosomes fertilizes with a normal sperm carrying Y’ chromosome.

Question 16.
Why are women normally carriers of sex-linked diseases? Write with an example.
Or
Men are generally suffering from color-blindness but women are only carriers of this disease. Explain the reason.
Answer:
The gene for color-blindness is recessive and carried by X-chromosomes. A color-blind man has a single recessive gene (X^CY). The gene for normal vision is dominant. When sex chromosomes (X) of a man contain genes for color-blindness (X^C), they express themselves and produce color-blindness in them. A carrier woman contains one recessive gene which remains suppressed due to the presence of the dominant gene of normal vision on the other sex chromosomes (X^CX). These women play important role in the conduction of genes of color-blindness. Hence, they are carrier (X^CX). Women become color-blind only when both chromosomes contain the genes of color-blindness (X^CX^C).

Question 17.
Differentiate between haemophilia and sickle cell anemia.
Answer:

Haemophilia	Sickle cell anaemia
1. It is due to recessive defective allele present on X – chromosome. 2.  It is a sex-linked ‘disorder. 3.  A protein necessary for clotting of blood is deficient.	1. It is due to point mutation i.e., a single base pair change leading to a change in amino acid. 2. It is an autosomal disorder. 3. The defective haemoglobin leads to a change in the shape of RBCs, which become a sickle cell.

Question 18.
Differentiate between Turners and Klinefelter’s syndrome.
Answer:

Turner’s Syndrome	Klinefelter’s Syndrome
1. The individual is female.	1. The individual is male.
2. The individual has one X chromosome less i.e., she has 45 chromosomes.	2. The individual has an extra X chromosome i.e., 47 chromosomes.
3.The individual has a short stature with an underdeveloped feminine character.	3. The individual has a tall stature with a feminised character.

Question 19.
Describe Mendel’s observation on the hybridization experiments on the garden pea.
Answer:

• The F hybrids always showed one of the parental forms of the trait and there was no blending of traits.
• Both the parental form of traits appeared without any change / blending in the F₂ generation.
• The form of the trait that appeared in the F₁ generation is called the dominant character and it appeared in the F₂ generation about 3 times in frequency as that of its alternate form.

Question 20.
Why Mendel selected pea plants for his experiments?
Answer:
Mendel selected pea plants on the basis of the following characters:

• The Pea plant exhibited a number of contrasting characters.
• It normally undergoes self-pollination but can be cross-pollinated manually.
• It is an annual plant and yield’s results in a year’s time, so it can be observed for many generations.
• Many varieties were available with observable alternative forms for a trait.
• Pure varieties of pea were available which always bred true.
• A large number of seeds are produced per plant which can be easily handled and cultivated.

Question 21.
List any 7 traits in garden pea which Mendel studied in his breeding
Answer:

Question 22.
What is Klinefelter’s syndrome? Also, describe their symptoms.
Answer:
Klinefelter’s syndrome: It is a sex chromosomal syndrome. It is found only in men. They possess 47 chromosomes in their cells (22 pairs autosomes + XXY sex chromosomes). One extra X chromosome is found in them, thus, there is Trisomy in the sex-chromosome.

Question 23.
A pea plant with Purple flowers was crossed with a plant with white flowers producing 40 plants with only purple flowers. On selfing, these plants produced 470 plants with purple flowers 162 with white flowers. What genetic mechanism accounts for these results?
Answer:

F₂ generation 470 purple: 162 white. In the F₁ generation only purple flower-producing plants appeared. This means the purple colour is dominant. Which doesn’t allow the white colour to express itself.

In F₂ generation, purple and white coloured flowers were produced in the ratio 3:1. Here the parental character of white again reappeared in about one-fourth of the progeny. This occurs due to the segregation of genes in the gamete formation. This represents the law of segregation and the monohybrid ratio.

Question 24.
In human beings, blue eye colour is recessive to brown eye colour. A brown-eyed man has a blue-eyed mother
(a) What is the genotype of the man and his mother?
(b) What are the possible genotype of his father?
(c) If a man marries a blue-eyed woman, what are the possible genotypes of their offsprings?
Answer:
As per the given condition, the brown eye colour is dominant over the blue eye colour.

(a) The mother’s genotype must be ‘bb’ as she is recessive for blue coloured eye. The man is brown-eyed (dominant). Its possible genotype must be ‘Bb’ as he is procuring one of the recessive gene from his mother.

(b) As the genotype of the man is ‘Bb’ so the possible genotypes of his father may be BB or Bb

Question 25.
How do you relate dominance, codominance, and incomplete dominance in the inheritance of character?
Answer:

• Dominance: It is the phenomenon in which one of the alleles of a gene expresses itself in the hybrid / heterozygous condition. While the other is suppressed (recessive) in the presence of the other.
• Codominance: It is the phenomenon in which 2 alleles of a gene are equally dominant and express themselves in the presence of the other
• Incomplete dominance: It is the phenomenon in which neither of two alleles of a gene is completely dominant over the other and the hybrid is intermediate between the two parents.

Question 26.
Describe the XO- type of sex determination. [HOTS]
Answer:
XO – type of sex determination occurs in certain insects like grasshoppers. The males have only one X – chromosome and hence they have one chromosome less than females. All the ova contains autosomes and one X chromosome. 50% of the sperm contains one X chromosome besides the autosomes, while 50% of sperm don’t have X – chromosomes. Sex of the individual is determined by the type of sperm fertilizing ovum. Egg fertilized by sperm having an X- chromosome developed into female. Egg fertilized by sperm having no chromosome develop into male.

Question 27.
Describe the different types of mutation.
Answer:

• A mutation is a phenomenon arising from the alteration of DNA sequences or structure or number of chromosomes.
• Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in the alteration in chromosome structure, which is called a chromosomal aberration.
• When a mutation arises due to a change in single base pair of DNA, it is known as a point mutation. Eg: Sickle cell anaemia. Deletion or addition of base pairs of DNA causes frameshift mutation.

2nd PUC Biology Principles of Inheritance and Variation Five Marks Questions

Question 1.
What is incomplete dominance? Describe with one example.
Answer:
When 2 parents are intercrossed with each other, the hybrid that produced doesn’t resemble either of the parents, but mid-way between 2 parents. The phenomenon of incomplete dominance occurs in the Four O’clock plant (Mirabilis Jalopa) and Snapdragon (Antirrhinum Majus). A cross between red-flowered (RR) and a white-flowered (rr) plant yields the hybrid flowered (Rr) one which is of pink colour in F, generation.

The pink hybrids on crossing with each other give the usual Mendelian ratio 1 : 2 : 1 (one red : 2 pink : 1 white). In this, the gene “R is not completely dominant over the allele ‘r’. The heterozygous Rr synthesizes only half pigment, so they are pink in colour.

Question 2.
In Snapdragons, tall (DD) is dominant over dwarf (dd) and Red (RR) is incompletely dominant over white (rr), the hybrid being pink. A pure tall white is crossed to a pure dwarf red and F₂ is self-fertilized. Give the expected genotype and phenotype in F₁ and F₂ generations.
Answer:

Question 3.
Sex determination is based on particular chromosomes in both birds and humans. State two points of difference between their mechanisms of sex determination.
Answer:
i. In birds female has dissimilar chromosomes (ZW) and male has two similar ZZ chromosomes.
But in humans female has XX (homogametic) and male has XY (heterogametic)
ii. In humans the sperm is responsible for the sex of child but in birds egg is responsible for the sex of the chicks.

Question 4.
Study the Pedigree analysis chart and answer the question given below: [CBSE 2008]
Answer:

(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 3rd and 4th child in Generation II.
Answer:
(a) The trait is recessive
(b) The trait is autosomal
(c) Parent Aa and Aa
3rd child is II generation – aa 4th child is II generation – Aa

Question 5.
Study the given pedigree chart to show the inheritance pattern of a human trait and answer questions given below.
(a) Give the genotype of a parent in 1st generation and of son and daughter were shown in 2nd generation.
(b) give the genotype of daughters shown in 3rd generation.
(c) Is the trait sex-linked or autosomal. Justify your answer.
Answer:
(a) Genotypes of parents in Generation 1. Male – Aa Female – Aa Son (Generation II) – Aa Daughter (Generation II) – aa.
(b) Genotype of the daughter in Generation III – Aa
(c) It is an autosomal trait and not sex-linked because if it is sex-linked, the daughter in generation II cannot have it.

Question 6.
A girl baby has been reported to suffer from haemophilia. How is it possible? Explain with the help of a cross.
Answer:
(a) Haemophilia is due to the presence of a recessive defective allele on the X – chromosome.
(b) A female with XX – chromosome must be homozygous for the disease to appear.
(c) She must receive one of the alleles from her haemophilie father (X^hY) and the other from her mother who is also an haemophilie or at least a carrier i.e. the heterozygous for the disease XX^h.

Question 7.
How are Pedigree charts prepared?
Answer:

• Squares denote males.
• Circles denote females.
• Mating is shown by a horizontal line connecting a male symbol with a female symbol.
• Offsprings symbols are rearranged from left to right in the order of their birth and connected by a horizontal line below the parents and this line is connected to the parent’s marriage line by a vertical line.
• A solid or blackened symbol represents the individual with the trait under study.
• An open/clear symbol represents the absence of a trait under study.