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

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

Edited By Priyanka kumari | Updated on Aug 26, 2022 11:21 AM IST | #CBSE Class 12th

NCERT exemplar Class 12 Biology solutions chapter 5 Principles of Inheritance and Variation is an important chapter from an examination point of view. The experts in this field who have a deep understanding of the subject and chapter have prepared NCERT exemplar Class 12 Biology chapter 5 solutions to provide correct information. In order to answer the questions, the experts have prepared NCERT Class 12 Solutions using simple language. The students refer to this study material that is best and gives you an amazing learning experience. NCERT exemplar Class 12 Biology solutions chapter 5 pdf download is useful to access it offline.

Introduction of NCERT exemplar solutions for Class 12 Biology chapter 5 Principles of Inheritance and Variation

Multiple Choice Questions:

Question:1

All genes located on the same chromosome:
a. Form different groups depending upon their relative distance
b. Form one linkage group
c. Will not from any linkage groups
d. Form interactive groups that affect the phenotype

Answer:

The answer is the option (b) Form one linkage group
Explanation: Closer genes are more likely to form linkages, as the linkage is decided by the vicinity of two genes. Morgan showed that genes form one linkage group when they are located on the same chromosome. Thus, (b) is the answer.

Question:2

Conditions of a karyotype 2n +1, 2n –1 and 2n + 2, 2n – 2 are called:
a. Aneuploidy
b. Polyploidy
c. Allopolyploidy
d. Monosomy

Answer:

The answer is the option (a) Aneuploidy
Explanation: Aneuploidy is the condition in which loss or gain of chromosome is observed due to the failure of separation of chromatids during the cycle of cell division. Hence, (a) is the correct option

Question:3

Distance between the genes and the percentage of recombination shows:
a. a direct relationship
b. an inverse relationship
c. a parallel relationship
d. no relationship

Answer:

The answer is the option (b) An inverse relationship
Explanation: Reduced distance between genes increases the probability of recombination which shows an inverse relationship. Hence, (b) is the correct option.

Question:4

If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny, the disease is:
a. Autosomal dominant
b. Autosomal recessive
c. Sex-linked dominant
d. Sex-linked recessive

Answer:

The answer is the option (d) Sex-linked recessive
Explanation: Here, the X-chromosome has the flawed gene, but the carriers of the disease are females and it is manifested in males. Thus, it is referred to as sex-linked recessive disease. One example of such a disease is Haemophilia which is known as X-linked recessive disease.

Question:5

In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following triplets codes for valine?
a. G G G
b. A A G
c. G A A
d. G U G

Answer:

The answer is the option (d) G U G
Explanation: At the sixth codon of the beta globin gene, single base substitution takes place from GAG to GUG. Due to this, in the globin protein, replacement of glutamic acid with valine is seen. Hence, GUG is the code for valine.

Question:6

The person having genotype \left | A \right |B would show the blood group as AB. This is because of:
a. Pleiotropy
b. Co-dominance
c. Segregation
d. Incomplete dominance

Answer:

The answer is the option (b) Co-dominance
Explanation: Most commonly found in ABO blood grouping, co-dominance is when the F1 generation resembles both parent. Here, the result is AB group because of both A and B sugars being dominant.

Question:7

\frac{zz}{zw}type of sex determination is seen in:
a. Platypus
b. Snails
c. Cockroach
d. Peacock

Answer:

The answer is the option (d) Peacock
Explanation: Sex determination of the type \frac{ZZ}{ZW} is observed in birds. Males have ZZ combination and females have ZW combination.

Question:8

A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?
a. TT and Tt
b. Tt and Tt
c. TT and TT
d. Tt and tt

Answer:

The answer is the option (b) Tt and Tt
Explanation: In the case of Tt and Tt (option (b)), a few of the offspring will be dwarf (TT, Tt, tt) and most of them are going to be tall (TT, Tt). In case of option (a) and (c), the offspring will be all tall and in option (d), one of the parent is dwarf. Hence, (b) is the correct option.

Question:9

In a dihybrid cross, if you get 9:3:3:1 ratio it denotes that:
a. The alleles of two genes are interacting with each other
b. It is a multigenic inheritance
c. It is a case of multiple allelism
d. The alleles of two genes are segregating independently

Answer:

The answer is the option (d) The alleles of two genes are segregating independently.
Explanation: According to Mendel, this is the ideal dihybrid ratio. So, (d) is the appropriate option.

Question:10

Which of the following will not result in variations among siblings?
a. Independent assortment of genes
b. Crossing over
c. Linkage
d. Mutation

Answer:

The answer is the option (c) Linkage
Explanation: Whenever two genes are in close proximity and located on same chromosome, linkage takes place irrespective of the variation.

Question:11

Mendel's Law of independent assortment holds good for genes situated on the:
a. non-homologous chromosomes
b. homologous chromosomes
c. extranuclear genetic element
d. same chromosome

Answer:

The answer is the option (a) Non-homologous chromosomes
Explanation: All the other three options (b), (c) and (d) do not exhibit independent assortment. Hence, (a) is the right answer.

Question:12

Occasionally, a single gene may express more than one effect. The the phenomenon is called:
a. multiple allelism
b. mosaicism
c. pleiotropy
d. polygeny

Answer:

The answer is the option (c) Pleiotropy
Explanation: Pleitropy is the phenomenon in which a single gene expresses more than one effect occasionally.

Question:13

In a certain taxon of insects, some have 17 chromosomes and the others have 18 chromosomes. The 17 and 18 chromosome-bearing organisms are:
a. males and females, respectively
b. females and males, respectively
c. all males
d. all females

Answer:

The answer is the option (a) Males and females, respectively
Explanation: Males have single X-chromosomes while females have pairs of X-chromosomes. This is XO sex determination which is seen in such insects. Hence, (a) is the answer.

Question:14

The inheritance pattern of a gene over generations among humans is studied by the pedigree analysis. Character studied in the pedigree analysis is equivalent to:
b. Mendelian trait
c. polygenic trait
d. maternal trait

Answer:

The answer is the option (b) Mendelian trait
Explanation: Pedigree analysis can be used to trace the Mendelian disorders and the pattern of their inheritance.

Question:15

It is said that Mendel proposed that the factor controlling any character is discrete and independent. His proposition was based on the
a. results of the F3 generation of a cross.
b. observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.
c. self-pollination of F1 offsprings
d. cross-pollination of F1 generation with recessive parent

Answer:

The answer is the option (b) observations that the offspring of a cross, made between the plants having two contrasting characters shows only one character without any blending.
Explanation: It was observed by Mendel that when two pairs of varying characters are analyzed during dihybrid cross, the behavior of a particular character is independent of another character. Hence, the correct answer is (b).

Question:16

Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the F1 heterozygote is crossed with the homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?
a. 1: 1 : 1: 1
b. 9 : 3 : 3: 1
c. 3: 1
d. 1: 1

Answer:

The answer is the option (a) 1:1:1:1
Explanation: The correct ratio for the given case is 1:1:1:1.

Question:17

In the F2 generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are:
a. phenotypes – 4; genotypes – 16
b. phenotypes – 9; genotypes – 4
c. phenotypes – 4; genotypes – 8
d. phenotypes – 4; genotypes – 9

Answer:

The answer is the option (d) Phenotypes – 4; genotypes - 9
Explanation: For example, consider a dihybrid cross between wrinkled green (rryy) and round yellow (RRYY). The phenotypes were as follows: round green, round yellow, wrinkled green and wrinkled yellow and the genotypes were: rryy, rrYy, rrYY, Rryy, RrYy, RrYY, RRyy, RRYy, and RRYY. Hence, correct option is (d).

Question:18

Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group, respectively. What would be the genotype of both mother and father?
a. Mother is homozygous for ‘A’ blood group and father is heterozygous for ‘B’
b. Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’
c. Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively
d. Both mother and father are homozygous for ‘A’ and ‘B’ blood group, respectively

Answer:

The answer is the option (c) Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively
Explanation: Offspring ii (O blood group) is the feasible phenotype of parents and IAi (A blood group) and IBi (B blood group) are the feasible genotype of the parents.

Very Short Answer Type Questions:

Question:1

What is the cross between the progeny of F1 and the homozygous recessive parent called? How is it useful?

Answer:

A test cross is a name for the cross between the homozygous recessive parent and the F1 progeny. The phenotype in F1 generation is determined by analysis of progenies of a test cross.

Question:2

Do you think Mendel's laws of inheritance would have been different if the characters that he chose were located on the same chromosome?

Answer:

Yes, the results would have been completely different if the characters chosen were located on the same chromosome. In his experiments on Drosophila, Morgan found that the ratio of phenotype in the F1 generation was different than that in Mendel’s observation because the characters were on the same chromosome.

Question:3

Enlist the steps of controlled cross-pollination. Would emasculation be needed in a cucurbit plant? Give reasons for your answer.

Answer:

Controlled cross-pollination involves the following steps:

  1. Emasculation

  2. Transfer of pollen from a different flower

  3. Pollination

  4. Fertilization

Emasculation is necessary in the case of a monoecious plant, but it is not needed in the plants of Cucurbitaceae as they have unisexual flowers.

Question:4

A person has to perform crosses to study the inheritance of a few traits/characters. What should be the criteria for selecting the organisms?

Answer:

The following criteria should be followed for selecting the organisms:

  • Easy hybridization should be possible.

  • The sets of contrasting characters are chosen should be easily recognizable.

  • The organism should have a short life cycle for convenient and early completion of the study.

Question:5

The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion based on the pedigree.

Answer:

The breeding between parents is shown in the first row, and the number of progenies is depicted by the second row. The figure shows two female and three male children with one girl and one boy affected by some kind of genetic disease. From the figure, it can be concluded that the nature of the trait is recessive, and autosome linked in nature.

Question:6

To obtain the F1 generation Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But forgetting the F2 generation, he simply self-pollinated the tall F1 plants. Why?

Answer:

To prevent the masking of the effect of inheritance to F2 generation by character set, Mendel self-pollinated the tall F1 plants. This is because all plants of the F1 generation turned out to be tall and he could not understand the fate of the recessive character.

Question:7

Genes contain the information that is required to express a particular trait.” Explain.

Answer:

The fact that genes are responsible for inheritance was discovered after the discovery of chromosomes. Before that, genes were not in knowledge of the scientists. Only after Mendel’s period, it was known that the information of the expression of a particular trait is contained in the genes.

Question:8

How are the alleles of a particular gene differ from each other? Explain its significance.

Answer:

Alleles are genes which code for a pair of contrasting characters, and these are different from each other within a pair. However, the slight difference is not noticeable as characters or phenotype, but as the absence or presence of a molecule of a substance like sugar polymer. The substance is present on the gene I which regulates the ABO blood grouping.
Alleles are vital because they help in identifying whether a particular trait is recessive or dominant or even co-dominant (in some cases).

Question:9

In a monohybrid cross of plants with red and white-flowered plants, Mendel got only red-flowered plants. On self-pollinating these F1 plants got both red and white-flowered plants in 3:1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of the parental generation.

Answer:

The usage of particular alphabets is only meant for convenience. Generally, the character is shown by the first letter of a particular trait, and a different case of the same letter is used to show the contrasting character. Conventionally, a capital letter depicts the dominant trait, and the lower case letter depicts the recessive trait.

Question:10

For the expression of traits, genes provide only the potentiality, and the environment provides the opportunity. Comment on the veracity of the statement.

Answer:

Genes carry the potentiality for the expression of a trait, i.e. phenotype and are responsible for inheritance. Phenotype depends on many factors. If a trait gets combined with a dominant one, it may get recessive, and the survival depends on the natural selection passage. Therefore, the opportunity for expression of a particular trait is provided by and depends on the environment.

Question:12

In our society, a woman is often blamed for not bearing a male child. Do you think it is right? Justify.

Answer:

The Y chromosome, which is present in males, determines the sex of the human child. Hence, it is evident that the father is responsible for having a male or a female child. Therefore, a mother should not at all be blamed for the same. However, the probability of X or Y chromosome, making its way to the zygote is equal, i.e. 50-50. Hence, no one is to be blamed or rewarded for that. A female child is as important as a male child.

Question:13

Discuss the genetic basis of wrinkled phenotype of pea seed.

Answer:

Generally, seeds dry before dispersal and germination. The amount of starch in the seed decides whether the seed will be round or wrinkled, and the relative amount of starch is looked after by a gene. If there is less than required starch in the seed, it will be wrinkled. In pea plants, a mutant gene’s enzymes are responsible for reducing the amount of starch in seeds and hence, the seeds become wrinkled.

Question:14

Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?

Answer:

Alleles are present only in pairs because maximum organisms are diploid. Thus, an individual will have only two alleles for the character even after showing multiple allelism. For example, the ABO blood grouping is governed by alleles I, IA and IB.

Question:15

How does a mutagen induce mutation? Explain with example.

Answer:

The mutation is induced by physical and chemical factors called mutagens. During replication, the base-pair sequence in the DNA is altered for inducing mutation.

Short Answer Type Questions:

Question:1

In a Mendelian monohybrid cross, the F2 generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.

Answer:

The alleles segregate and go to each of the parents when plants of F1 generation reproduce without the cross. The segregation is random with equal probability of a particular allele going to either of the parental gametes. Thus, genotype and phenotype in the F2 generation are identical. So, there will be 50% of plants with pure genotype and 50% of plants with mixed genotype.

Question:2

Can a child have blood group O if his parents have blood group ‘A’ and ‘B’? Explain.

Answer:

Yes, it is possible for a child to have blood group O if the parents have blood group A and B. If genotypes of the parents are IAi and IBi, then gametes can have one of the genotype from i, IA or IB. The child’s genotype will be ii if fertilization of gametes occurs with i; hence the child will have blood group O.

Question:3

What is Down’s syndrome? Give its symptoms and cause. Why is it that the chance of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?

Answer:

First discovered and described by Langdon Down in 1866, Down’s syndrome refers to the presence of an additional copy of 21st chromosome. It is a chromosomal disorder. A person with this condition has:

  • distinct facial appearance with partially opened mouth and furrowed tongue

  • small round head

  • delayed physical and mental development

  • short stature

  • broad palm

According to studies, with advancing maternal age, the chances of Down’s syndrome increases in the child. This is because cells get older with the age of the mother and the ova are present since birth in females. Various physio-chemical exposures also lead to increased chances.

Question:4

How was it concluded that genes are located on chromosomes?

Answer:

The behaviour of chromosome and genes was studied during meiosis by Walter Sutton and Theodor Boveri, and they observed similar movement in chromosome and gene. Hence, it was proposed by them that genes are located on chromosomes.

Question:5

A plant with red flowers was crossed with another plant with yellow flowers. If F1 showed all flowers orange in colour, explain the inheritance.

Answer:

Incomplete dominance of a particular trait over another is seen sometimes which results in the manifestation of both characters in some way. For example, consider genotype of red flowers to be RR and that of yellow flowers to be rr. In F1 generation, all progenies have genotype as Rr. Hence, orange flowers are produced by all plants in F1 generation because red colour does not completely dominate the yellow colour.

Question:6

What are the characteristic features of a true-breeding line?

Answer:

The characteristic features of a true-breeding line are:

  • Throughout several generations, stable trait inheritance and expression of characters is seen.

  • Self-pollination

Question:8

Why is the frequency of red-green colour blindness is many times higher in males than that in the females?

Answer:

The genes with photo-pigments are located on X-chromosome, and if the gene is destroyed or missing, it results in colour blindness. Chances of this disease are much higher in males than in females because males have only one X-chromosome; hence there is no second X-chromosome to compensate for the defect, like in females. Females have colour blindness if the defect is in both her X-chromosomes.

Question:9

If a father and son are both defective in red-green colour vision, is it likely that the son inherited the trait from his father? Comment.

Answer:

X-chromosome contains the genes for colour blindness. However, in the male child, the X-chromosome is contributed by the mother. Therefore, the son inherits the trait from the mother even if his father is suffering from colour blindness.

Question:10

Discuss why Drosophila has been used extensively for genetical studies.

Answer:

Drosophila is ideal for genetical studies because:

  • Its life cycle is completed in two weeks.

  • Many offspring are produced through a single mating.

  • It can be in a laboratory.

  • It exhibits sexual dimorphism and many hereditary variations.

Question:11

How do genes and chromosomes share similarity from genetical studies?

Answer:

From the point of view of genetical studies, the similarities in genes and chromosomes are:

  • Both genes and chromosomes are found in pairs.

  • Independent segregation is observed in independent pairs.

  • Segregation takes place when gamete is formed, and only one pair is transmitted to the gamete.

Question:12

What is recombination? Discuss the applications of recombination from genetic engineering.

Answer:

Recombination is a combination of genes which is different from parental gene combination. It can occur either naturally during meiosis or artificially through genetic engineering, which is used for recombining different species for producing beneficial products for humans. For example, production of BT cotton, BT brinjal, and vaccines like hepatitis B was done with genetic engineering.

Question:13

What is artificial selection? Do you think it affects the process of natural selection? How?

Answer:

The selective breeding of animals and plants is called artificial selection. It is done to include advantageous traits in plants and animals, and it can be natural or artificial. However, ethically, artificial selection is wrong. But according to the law of survival of the fittest, only the varieties which are fit to survive will survive in artificial selection. Therefore, it will not affect the process of natural selection.

Question:14

With the help of an example, differentiate between incomplete dominance and co-dominance.

Answer:

Incomplete dominance

Co-dominance

In the F1 generation, there is a partial manifestation of phenotypes from both the parents.

In the F1 generation, there is a complete manifestation of phenotypes from both the parents.

Example: F1 generation results in pink flowers when snapdragon plants with red flowers and white flowers were crossed.

Example: Co-dominance is shown in ABO blood grouping in humans.


Question:15

It is said that the harmful alleles get eliminated from the population over some time, yet sickle cell anaemia is persisting in the human population. Why?

Answer:

A single pair of allele, i.e. HbA and HbS, controls the sickle cell anaemia. The offspring may suffer from this disease if both parents are heterozygous (HbAHbS). But the offspring should be homozygous (HbSHbS) because heterozygous carry this disease. However, it still persists in humans because, in adaptation, heterozygous individuals are advantageous.

Long Answer Type Questions:

Question:1

In a plant, tallness is dominant over dwarfness, and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is the standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?

Answer:

The cross between a dwarf plant with white flowers (ttrr) and tall plant with red flowers (TTRR) is depicted by the Punnett Square shown below:

Plants of the F1 generation will produce red flowers and will be tall. When they are permitted to self-pollinate, F2 generation’s phenotype can be depicted by following Punnett Square.
Here, the ratio is 9:3:3:1, which is the standard dihybrid ratio, also shown as:

  • Tall plant red flower = 9

  • Tall plant white flower = 3

  • Dwarf plant red flower = 3

  • Dwarf plant white flower = 1

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This ratio stands true only when the contrasting characters have their genes on distinct chromosomes. This is because of the fact that there will be interaction if the characters are on the same chromosome.

Question:2

a. In humans, males are heterogametic, and females are homogametic. Explain. Are there any examples where males are homogametic and females heterogametic?

b. Also describe as to, who determines the sex of an unborn child? Mention whether the temperature has a role in sex determination.

Answer:

1. In humans, males are called heterogametic because they have X and Y chromosomes in their 23rd pair of the chromosome. On the other hand, females are called homogametic as they have two same chromosomes, i.e. XX chromosomes in their 23rd pair of the chromosome. However, in birds, the scenario is quite the opposite. Male birds have ZZ chromosomes, and female birds have ZW chromosomes. Hence, in cases like these, males are homogametic, and females are heterogametic.
2. The sex of an unborn child is determined by the 23rd pair of chromosome in humans. The males have X and Y chromosomes and females have X and X chromosomes in that pair. So, when the ovum is fertilized with the sperm containing X chromosome, the result is a female child, and when it is fertilized with the Y chromosome, the result is a male child. In some animals like crocodiles, the sex is determined by temperature. Male crocodiles are born if eggs are brooded at high temperatures.

Question:3

A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be the probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.

Answer:

The pedigree analysis shown below depicts the colour blindness in offspring:

  • The F1 generation depicts two female children and one male child – One of the female children might be a carrier, and the male child will suffer from colour blindness.

  • The P generation depicts a normal husband and a carrier wife.

About 8% of the males are colour blind, while only 0.5% of the females are colour blind. This is because only one X-chromosome is present in males. If that one X-chromosome carries the defect, the boy will suffer from colour blindness. But in females, there are two X chromosomes which help in the compensation of deficiency in one of the chromosomes. Thus, females are generally carriers of the disease and rarely the sufferer.

Question:4

Discuss in detail the contributions of Morgan and Sturvant in the area of genetics.

Answer:

Sturtevant was one of the students of Morgan. Morgan has carried out several experiments related to genetics. Some of their contributions are:

  • In the various dihybrid crosses of Drosophila carried out by Morgan, it was observed by him that the phenotypic ratio did not resemble the standard ratio, which was observed by Mendel. Morgan and his group noticed that the genes did not show independent segregation when they were located on the same chromosome as they were aware that genes were located on the X chromosome. Morgan termed the interaction of genes on the same chromosome as linkage and generation of non-parental combination as recombination. On the same chromosome, the probability of parental combination is much higher than that of non-parental combination.

  • Sturtevant discovered that on the same chromosome, recombination or lack of recombination was largely dependent on the relative distance between the two genes. There was no recombination when genes were tightly linked, and chances of recombination were higher when genes had a large distance between them.

Contributions made by Morgan and his group, including Sturtevant, contributed hugely to the development of the genetic mapping.

Question:5

Define aneuploidy. How is it different from polyploidy? Describe the individuals having followed chromosomal abnormalities.
a. Trisomy of 21st Chromosome
b. XXY
c. XO

Answer:

Aneuploidy is a condition in which gain or loss of chromosome is seen during the cell division due to the failure of chromatid segregation. On the other hand, an increase in the number of chromosome set caused due to the failure of cytokinesis after telophase is called polyploidy. This condition is mostly seen in plants and rarely seen in animals.

  1. Trisomy of 21st chromosome: First discovered and described by Langdon Down in 1866, it refers to the presence of an additional copy of 21st chromosome. It is known as Down’s syndrome. A person with this condition has a distinct facial appearance with partially opened mouth and small round head. Physical and mental development is delayed in such individuals.

  2. XXY: In this condition, a karyotype of 47 is caused due to an additional copy of the X chromosome. It is a genetic disorder and is called Klinefelter’s syndrome. In this case, three chromosomes (XXY) are present in the 23rd set of chromosomes. A person suffering from this disorder is sterile and has enlarged breasts (gynecomastia) with overall masculine development.

  3. XO: Lack of an X chromosome causes this genetic disorder of ploidy of 45 (XO). It is known as Turner’s syndrome. Such females have rudimentary ovaries and are sterile. Absence of secondary sexual characters is also seen.

The chapter includes Mendel’s Laws of Inheritance, Law of Dominance, Law of Segregation, Law of Independent Assortment, Chromosomal Theory of Inheritance, Mutation, Down’s syndrome, Turner’s syndrome, and Klinefelter’s syndrome.

As per class 12 Biology NCERT exemplar solutions chapter 5, the key principles of Mendelian inheritance are summed up by Mendel’s three laws. The solutions will also help you to excel in the concepts and acquire comprehensive knowledge about the various types of questions. The solutions are exercise wise and will help you to write a proper solution.

Major Subtopics in NCERT Exemplar Class 12 Biology Solutions Chapter 5

● 5.1 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 Disorder

NCERT Exemplar Class 12 Biology Solutions Chapter 5 Principles of Inheritance and Variation - Learning Outcome

In Class 12 Biology NCERT exemplar solutions chapter 5 will help the students to learn and study for the students as they need to know when it comes to learning about the different principles of inheritance and variation. Every topic in NCERT exemplar Class 12 Biology chapter 5 solutions is very well explained.

The subject deals with the inheritance, as well as the variation of characters. Inheritance is the process by which characters are passed. Genetics which deals with the inheritance, , heredity, and variation of characters from parents to offspring. One of the causes of variation is also sexual reproduction. Overall, NCERT exemplar Class 12 Biology solutions chapter 5 along with giving information is also fun and interesting at the same time.

NCERT Exemplar Class 12 Biology Chapter Wise Links:

Important Topics in NCERT Exemplar Class 12 Biology Solutions Chapter 5

- Hereditary is a process of transmission of heritable traits from parents to their offsprings along with different topics covered in the chapter. The topics are detailed in the NCERT exemplar solutions for Class 12 Biology chapter 5.

- NCERT exemplar Class 12 Biology chapter 5 solutions provides you all the essential information that one needs to know while preparing for the examination.

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Frequently Asked Question (FAQs)

1. Who can benefit from this NCERT exemplar Class 12 Biology solutions chapter 5?

The students who are preparing for board exams and also for entrance exams can benefit from this.


2. How to download NCERT Exemplar Class 12 Biology solutions chapter 5?

You can download it in pdf format from the solutions page itself or from a different link.

3. What can one learn from this NCERT exemplar solutions for Class 12 Biology chapter 5?

The students learn about the different principles of inheritance and variation.


4. How are these questions solved?

The teachers use the practical knowledge with the given information that is provided to them.

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Have a question related to CBSE Class 12th ?

Here are some options you can explore to get admission in a good school even though admissions might be closed for many:

  • Waitlist: Many schools maintain waitlists after their initial application rounds close.  If a student who secured a seat decides not to join, the school might reach out to students on the waitlist.  So, even if the application deadline has passed,  it might be worth inquiring with schools you're interested in if they have a waitlist and if they would consider adding you to it.

  • Schools with ongoing admissions: Some schools, due to transfers or other reasons, might still have seats available even after the main admission rush.  Reach out to the schools directly to see if they have any open seats in 10th grade.

  • Consider other good schools: There might be other schools in your area that have a good reputation but weren't on your initial list. Research these schools and see if they have any seats available.

  • Focus on excelling in your current school: If you can't find a new school this year, focus on doing well in your current school. Maintain good grades and get involved in extracurricular activities. This will strengthen your application for next year if you decide to try transferring again.


Best CBSE schools in Delhi: Click Here


In India, the design and coding fields offer exciting career options that can leverage your interest in both. Here's how you can navigate this path:

Choosing Your Stream:

  • Graphic Design Focus: Consider a Bachelor's degree in Graphic Design or a design diploma. Build a strong portfolio showcasing your creative skills. Learn the basics of HTML, CSS, and JavaScript to understand web development better. Many online resources and bootcamps offer these introductory courses.

  • Coding Focus: Pursue a Computer Science degree or a coding bootcamp in India. These programs are intensive but can equip you with strong coding skills quickly. While building your coding prowess, take online courses in graphic design principles and UI/UX design.

Engineering Subjects (for a Degree):

  • Information Technology (IT): This offers a good mix of web development, networking, and database management, all valuable for web design/development roles.

  • Human-Computer Interaction (HCI): This is a specialized field that bridges the gap between design and computer science, focusing on how users interact with technology. It's a perfect choice if you're interested in both aspects.

  • Passing NIOS in October 2024 will make you eligible for NIT admissions in 2025 . NIT admissions are based on your performance in entrance exams like JEE Main, which typically happen in January and April. These exams consider the previous year's Class 12th board results (or equivalent exams like NIOS).

Here's why 2025 is more likely:

  • JEE Main 2024 Admissions: The application process for NITs through JEE Main 2024 is likely complete by now (May 2024). They consider your 2023 Class 12th marks (CBSE in this case).
  • NIOS Results: Since NIOS results typically come out after the NIT admission process, your October 2024 NIOS marks wouldn't be available for JEE Main 2024.

Looking Ahead (2025 Admissions):

  • Focus on JEE Main: Since you have a computer science background, focus on preparing for JEE Main 2025. This exam tests your knowledge in Physics, Chemistry, and Mathematics, crucial for engineering programs at NITs.
  • NIOS Preparation: Utilize the time between now and October 2024 to prepare for your NIOS exams.
  • Eligibility Criteria: Remember, NITs typically require a minimum of 75% marks in Class 12th (or equivalent) for general category students (65% for SC/ST). Ensure you meet this criteria in your NIOS exams.

Yes, scoring above 99.9 percentile in CAT significantly increases your chances of getting a call from IIM Bangalore,  with your academic background. Here's why:

  • High CAT Score: A score exceeding  99.9 percentile is exceptional and puts you amongst the top candidates vying for admission. IIM Bangalore prioritizes  CAT scores heavily in the shortlisting process.

  • Strong Academics: Your 96% in CBSE 12th and a B.Tech degree demonstrate a solid academic foundation, which IIM Bangalore also considers during shortlisting.

However, the shortlisting process is multifaceted:

  • Other Factors: IIM Bangalore considers other factors beyond CAT scores, such as your work experience (if any), XAT score (if you appear for it), academic diversity, gender diversity, and performance in the interview and Written Ability Test (WAT) stages (if shortlisted).

Here's what you can do to strengthen your application:

  • Focus on WAT and PI: If you receive a shortlist, prepare extensively for the Written Ability Test (WAT) and Personal Interview (PI). These stages assess your communication, soft skills, leadership potential, and suitability for the program.

  • Work Experience (if applicable): If you have work experience, highlight your achievements and how they align with your chosen IIM Bangalore program.

Overall, with a stellar CAT score and a strong academic background, you have a very good chance of getting a call from IIM Bangalore. But remember to prepare comprehensively for the other stages of the selection process.

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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