NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

Biology for Class 12 solution of class 12 Ncert, we have included all the chapter questions. Biology for Class 12 solution of class 12 Ncert, we have included all the chapter questions. NCERT Biology 12 solution will prove to be beneficial for the board exams as well as for the students who are preparing for NEET, UPCPMT and other Medical Entrance exam.   Biology 12 solution will prove to be beneficial for the board exams as well as for the students who are preparing for NEET, UPCPMT and other Medical Entrance exam.

  
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All Solutions ncert solutions for class 12 biology chapter 5 Provided by www.studyit.in at free of cost without any registrations. This biology class 12 Solutions has been completely prepared by our biology experts who provide complete solutions based entirely on the CBSE syllabus, there are a total of 14 chapters in this biology class 12 textbook whose chapter wise solution is given below.    
           

NCERT Biology Class 12 Solutions Chapter 5 will also save you a hefty coaching fee over time and will provide an authentic and accurate solution sitting at home which will be very useful in making you notes.These Solutions are part of NCERT Solutions for Class 12 Biology. Here we have given NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation.

NCERT Solutions for Class 12 Biology Chapter 5

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Solution:
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.

2. Differentiate between the following –
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid.
Ans: (a)The difference between dominance and recessive me:

(b) Differences between homozygous and heterozygous individuals :

(c) In breeding experiments when a cross is made between the individuals considering their same single character, it is called mono¬hybrid cross, while a cross is done consid¬ering two characters at fee same time is called dihybrid cross (Yellow Round * Green Wrinkled).

Question 2.
Differentiate between the following:

1. Dominant and Recessive
2. Homozygous and Heterozygous
3. Monohybrid and Dihybrid

Solution:

1. Differences between dominant and recessive genes are as follows :
   
   
2. Differences between homozygous and heterozygous are as follows :
   
   
   
3. Differences between monohybrid and dihybrid cross are as follows :
   Question 3.

A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Solution:
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.
Solution:
Law of dominance states that when a pair of alleles or allelomorphs are brought together in F1 hybrid, then only one of them expresses itself, masking the expression of other completely. Monohybrid cross was made to study simultaneous inheritance of a single pair of Mendelian factors. The cross in which only alternate forms of a single character are taken into consideration is called monohybrid cross. The trait which appeared in the F1 generation was called dominant and the other which did not appear in the F1 population was called recessive.

Thus, when a pair of alleles are brought together in F1 hybrid, then
only one of them expresses itself masking the expression of other completely. In the above example, in Tt – F1 hybrid (tall) only ‘T’ expresses itself so dominant and ‘t’ is masked so recessive. Thus, this’ proves and explains the law of dominance.

Question 5.

Define and design a tesr-cross.
Solution:
Crossing of F1 individual having dominant phenotype with its homozygous recessive parent is called test cross. The test cross is used to determine whether the individuals exhibiting dominant character are homozygous or heterozygous.
Example : When a tall plant (TT) is crossed with dwarf plant (tt) in the F1 , generation only tall plant (Tt) appear which is then crossed with homozygous recessive (tt) in a test cross.

In the given test cross between tall hetero-zygous F1 hybrid with dwarf homozygous recessive parent produces tall and dwarf progeny in equal proportion indicating that F : hybrids are heterozygous.

Question 6.

Using a Punnett square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Solution:
When a heterozygous male tall plant (Tt) is crossed with homozygous dominant female tall plant (TT), we get two types of gametes in males : half with T and half with t and in females, we get only one type of gametes i.e., T.

From the Punnett square it is seen that all the progeny in F generation are tall (Tt), 50% homozygous tall (TT) and 50% heterozygous tall (Tt).

Question 7.
When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

1. tall and green
2. dwarf and green

Solution:

Phenotypes of the offsprings –
Tall Yellow : 3
Tall Green : 3
Dwarf Green : 1
Dwarf Yellow : 1
(a) Proportion of tall and green is 3/8.
(b) Proportion of dwarf and green is 1/8.

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?
Solution:
Two heterozygous parents (i.e. GgLl and GgLl) are crossed and the two loci are linked then the cross will be

This means, if ‘G’ represent grey body (dorhinant), ‘g’ black body (recessive), ‘L’-long (dominant) and ‘I’-dwarf (recessive) then the distribution of phenotypic features in F1 generation will be 3 : 1 i.e. 3/4 will show the dominant feature, grey and long, either in homozygous (GGLL) or in heterozygous (GgLl) condition and 1/4 will show the recessive feature, black and dwarf (ggll).

Question 9.

Briefly mention the contribution of T.H. Morgan in genetics.
Solution:
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 an analysis, can be useful.
Solution:
A record of inheritance of certain genetic traits for two or more generations presented in the form of a diagram of family tree is called pedigree. Pedigree analysis is study of pedigree for the transmission of particular trait and finding the possibility of absence or presence of that trait in homozygous or heterozygous state in a particular individual. Pedigree analysis is useful for the following :

• It is useful for the genetic counsellors to advice intending couples about the possibility of having children with genetic defects like haemophilia, colour blindness, alkaptonuria, phenylketonuria, thalassemia, sickle cell anaemia (recessive traits), brachydactyly and syndactyly (dominant traits).
• Pedigree analysis indicates that Mendel’s principles are also applicable to human genetics with some modifications found out later like quantitative inheritance, sex linked characters and other linkages.
• It can indicate the origin of a trait in the ancestors, e.g., haemophilia appeared in Queen Victoria and spread in royal families of Europe through marriages.
• It helps to know the possibility of a recessive allele to create a disorder in the progeny like thalassemia, muscular dystrophy, haemophilia.
• It can indicate about the harm that a marriage between close relatives, may cause.
• It helps to identify whether a particular genetic disease is due to a recessive gene or a dominant gene.
• In certain cases it may help to identify the genotypes of offspring yet to be born.

Question 11.
How is sex determined in human beings?
Solution:
Chromosomal determination of sex in human beings is of XX-XY type. Human beings have 22 pairs of autosomes and one pair of sex chromosomes. The female possess two homomorphic (= isomorphic) sex chromosomes, named XX. The males contain two heteromorphic sex chromosomes, i.e., XY. All the ova formed by female are similar in their chromosome type (22 + X). Therefore, females are homogametic. The male gametes or sperms produced by human males are of two types, gynosperms (22 + X) and androsperms (22 + Y). Human males are therefore, heterogametic. Sex of the offspring is determined at the time of fertilisation. Fertilisation of the egg (22 + X) with a gynosperm (22 + X) will produce a female child (44 + XX) while fertilisation with an androsperm (22 + Y) gives rise to male child (44 + XY). As the two types of sperms are produced in equal proportions, there are equal chances of getting a male or female child in a particular mating. As Y-chromosomes determines the male sex of the individual, it is also called androsome.

Question 12.
A child has blood group O. If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Solution:
If the father has blood group A i.e., IAIA (homozygous) and mother has blood group B i.e., IBIB (homozygous) then all the offsprings will have blood group AB (IAIB) and not blood group O.

Thus the genotypes of the parents of child with blood group O will be IAi and IBi There is possibility of 3 other types of blood group of offsprings besides O blood group offspring. They are IAi (blood group A). IBi (blood group B) and IAIB (blood group AB).

Question 13.
Explain the following terms with example:

1. Codominance
2. Incomplete dominance

Solution:

1. Codominance (1 : 2 : 1) — It is the phenomenon of two alleles (different forms of a Mendelian factor present on the same gene locus on homologous chromosomes) lacking dominant- recessive relationship and are able to express themselves independently when present together.
   Example – AB blood group : Alleles for blood group A(IA) and blood group B(IB) are codominant so that when they come together in an individual, they produce blood group AB. It is characterised by the presence of both antigen A (from IA) and antigen B (from IB) over the surface of erythrocytes.
   
   
2. Incomplete dominance (1 : 2 : 1) – It is the phenomenon where none of the two contrasting alleles being dominant so that expression in the hybrid is intermediate between the expressions of the two alleles in homozygous state. F2 phenotypic ratio is 1 : 2 : 1, similar to genotypic ratio. Example-In Mirabilis jalapa (Four o’clock) and Antirrhinum majus (Snapdragon or dog flower), there are two types of flower colour generation are of three types- red, pink and white flowered in the ratio of 1 : 2 : 1. The pink colour apparently appears either due to mixing of red and white colours (incomplete dominance).
   

Question 14.
What is point mutation? Give one example.
Solution:
When heritable alterations occur in a very small segment of DNA molecule i.e., a single nucleotide or nucleotide pair, then these type of mutations are called point mutations (also called as 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 liver. It results in hyperphenylalaninemia which is characterised by accumulation and excretion of phenylalanine, phenylpyruvic acid and related compounds. 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 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 the inheritance?
Solution:
Sutton and Boveri proposed the chromosomal theory of the inheritance. The theory believes that chromosomes are vehicles of hereditary information 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
Solution:

• Cystic fibrosis is an autosomal recessive disorder of infants, children and young adults that is due to a recessive autosomal allele present on chromosome 7. It is common in Caucasian Northern Europeans and White North Americans. The disease gets its name from the fibrous cysts that appear in the pancreas. In 70% of cases, it is due to deletion of three bases. It produces a defective glycoprotein. The defective glycoprotein causes formation of thick mucus in skin, lungs, pancreas, liver and other secretory organs. Accumulation of thick mucus in lungs results in obstruction of airways. Because of it the disease was also called mucoviscoides, Mucus deposition in pancreas blocks secretion of pancreatic juice. There is maldigestion of food with high fat content in stool. Liver may undergo cirrhosis and there is impaired production of bile. Vasa deferentia of males undergo atrophy.
• Huntington’s disease or Huntington’s chorea is a dominantly autosomal inherited disorder in which muscle and mental deterioration occurs. There is gradual loss of motor control resulting in uncontrollable shaking and dance like movements (chorea). The brain shrinks between 20-30% in size followed by slurring of speech, loss of memory and hallucinations. Life expectancy averages 15 years from the onset of symptoms. This disorder does not occur till the age of 25 to 55. The defective gene is dominant autosomal, located on chromosome 4. This defective gene has 42 -100 repeats of CAG instead of 10-34 repeats in normal gene. The frequency of this disorder is 1 in 10000 to 1 in 20000.

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