The NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance explains how genetic information is stored, copied, and passed on in living organisms. It describes the structure and function of DNA and RNA, showing that DNA is the primary genetic material in almost all living organisms and RNA plays important roles, mainly in some viruses. The Molecular Basis of Inheritance chapter also covers DNA replication, transcription, translation, and how genes are regulated. All the concepts are well-explained in the NCERT solutions.
This Story also Contains
The Molecular Basis of Inheritance Class 12 question answers are written by subject experts, making tough concepts like DNA replication, transcription, and translation easier to understand. Easy explanations for all the questions present in textbooks help students build a solid foundation in molecular genetics. The NCERT Solutions Class 12 Biology help students understand important topics and prepare for board exams effectively.
Students can get free PDFs of detailed answers to Chapter 5 from here. The Class 12 Biology Chapter 5 Molecular Basis of Inheritance question answer explain important topics like DNA replication, the genetic code, and how proteins are formed. These are clearly explained with diagrams and examples in the NCERT Solutions for Class 12.
Also Read,
Molecular basis of inheritance solutions explain complex topics like transcription and translation in a simple and step-by-step manner. The Class 12 Biology Chapter 5 Molecular Basis of Inheritance question answer includes detailed explanations that follow the NCERT guidelines, making it easier for students to revise and practice all important topics before exams.
Q1. Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil, and Cytosine.
Answer:
Nitrogenous bases- Adenine, thymine, uracil, and cytosine
Nucleosides- Cytidine and guanosine
Answer:
According to Chargaff's rule, A=T and G=C. As there are 20% cytosines, the guanine percentage will also be 20%. The G+C count is 20+20= 40%. Hence, the remaining 60% will be shared equally between adenine and thymine. So, the percentage of adenine and thymine is 30% each.
Q3. If the sequence of one strand of DNA is written as follows:
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of the complementary strands in 5′→3′ direction.
Answer:
About base pairs, the DNA strands are complementary to each other. So, if the sequence of DNA is
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
The sequence of the complementary strands will be
3'- TACGTACGTACGTACGTACGTACGTACG - 5'.
In the 5'→3' strand, it can be written as
5'- GCATGCATGCATGCATGCATGCATGCAT-3'
Q4. If the sequence of the coding strand in a transcription unit is written as follows:
5'-ATGCATGCATGCATGCATGCATGCATGC-3'
Write down the sequence of mRNA.
Answer:
The sequence of mRNA is the same as the coding strand of DNA, except that thymine is replaced by uracil. Thus, the sequence of mRNA will be
5' - AUGCAUGCAUGCAUGCAUGCAUGCAUGC - 3'.
Answer:
The characteristic of the DNA double helix that prompted Watson and Crick to speculate a semi-conservative mode of DNA replication is that the two DNA strands are antiparallel and complementary to one another in the sense of their base sequences. It follows from this organisation that DNA replication is semiconservative. When replication occurs, the two strands separate, and each strand serves as a template for the production of a new strand. By the completion of replication, parental types and recombinant types of DNA are produced.
Answer:
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), two types of nucleic acid polymerases are found
1. DNA-dependent DNA polymerase - It uses DNA as a template to synthesise new strands of DNA.
2. DNA-dependent RNA polymerase - It uses DNA as a template to synthesise RNA.
Answer:
Alfred Hershey and Martha Chase (1952) used a virus that infects bacteria, bacteriophages. They labelled DNA and protein with various radioactive isotopes. In their experiment, in one preparation, the protein component was rendered radioactive, and in the other, nucleic acid (DNA) was rendered radioactive. These two phage preparations were permitted to infect the culture of E. coli.
As soon as after infection, before cell lysis, the E. coli cells were mildly shaken in a blender, loosening the attached phage particles, and the culture was centrifuged. The denser infected bacterial cells were pelleted to the bottom, and lighter viral particles were in the supernatant. When a bacteriophage with radioactive DNA was used to infect E. coli, radioactivity was found in the pellet. If a bacteriophage with a radioactive protein coat was employed to infect E. coli, the radioactivity was present in most of the supernatant. They demonstrated in their experiment that protein does not enter the bacterial cell, whereas DNA gets transformed. Thus, they established that DNA is the genetic material.
Q8. Differentiate between the following:
(a) Repetitive DNA and Satellite DNA
Answer:
Repetitive DNA | Satellite DNA |
Repetitive DNA refers to DNA sequences containing small segments that are repeated many times. | DNA sequences containing highly repetitive DNA. |
(b) mRNA and tRNA
Answer:
mRNA or messenger RNA | tRNA or transfer RNA |
It acts as the template for the translation of proteins. | tRNA acts as the adapter molecule that carries specific amino acids to mRNA for the synthesis of a polypeptide. |
(c) Template strand and Coding strand
Answer:
Template strand | Coding strand |
It is the template for the synthesis of mRNA during transcription | Strand of DNA having the same sequence as mRNA (thymine in DNA is replaced by uracil in RNA |
Its polarity is 3' to 5' | Its polarity is 5' to 3' |
Q9. List two essential roles of the ribosome during translation.
Answer:
The ribosome is a complex structure made up of ribonucleoproteins. It consists of two subunits, i.e., a larger subunit and a smaller subunit. The essential roles of ribosomes during translation are as follows:
1. The ribosome acts as the site of protein synthesis. The larger subunit of the ribosome acts as an amino acid binding site, while the small subunit attaches to the mRNA.
2. The ribosome acts as a catalyst for forming peptide bonds.
Answer:
The lac operon is a set of genes in bacteria that help them use lactose for energy.
It has three main genes that make enzymes to bring lactose into the cell and break it down.
A regulatory gene makes a repressor protein that blocks these genes when lactose is absent.
When lactose is present, it binds to the repressor, stopping it from blocking the genes.
This allows the enzymes to be made and lactose to be used.
When lactose is gone, the repressor blocks the genes again to save energy.
The operon is only active when lactose is available and glucose (a preferred sugar) is not.
Q11. Explain (in one or two lines) the function of the following:
(a) Promoter
Answer:
Function of promoter
The promoter gene refers to the site where the RNA polymerase enzyme binds and transcription of mRNA starts.
(b) tRNA
Answer:
Function of tRNA
tRNA plays a major role in the process of protein synthesis. It reads the genetic code present on mRNA.
Q12. Why is the Human Genome Project called a mega project?
Answer:
The Human Genome Project is called a mega project because its specific goal is to sequence every base pair in the human genome. HGP took approximately 13 years to accomplish in 2006. The main aim of HGP was to develop new technology and generate new information in genomic studies. Because of this project, many new areas have opened in genetics. Thus, HGP is a mega project.
Q13. What is DNA fingerprinting? Mention its application.
Answer:
DNA fingerprinting is a very easy and quick way to compare the DNA sequence of any two individuals. It includes identifying differences in some specific regions in DNA sequences called repetitive DNA sequences. In these regions, a small stretch of DNA is repeated many times, and they are specific to every individual. The technique of fingerprinting was initially developed by Alec Jeffrey.
Applications of DNA Fingerprinting
1. It is used in forensic science to identify individuals.
2. It can be used to establish paternity or maternity-related disputes.
3. DNA fingerprinting is used to establish evolutionary relationships between organisms.
Q14. Briefly describe the following:
(a) Transcription
Answer:
It is the process of replication of genetic information from a strand of DNA into mRNA. Transcription involves only one strand that is replicated into RNA. During replication, the adenine position is occupied by uracil. DNA transcription contains a transcription unit. The transcription unit has a promoter, the structural gene, and a terminator. The 3'→5' polarity strands serve as a template and are referred to as template strands, whereas the other strand is referred to as the coding strand. A schematic organisation of a transcription unit is represented below.
The promoter is located at the 5’ end, and it binds the enzyme RNA polymerase to start transcription. The sigma factor also helps in the initiation of the process of transcription. The terminator is located at 3’ end of the coding strand and usually defines the end of transcription where the rho factor will bind to terminate transcription.
14.(b) Polymorphism
Answer:
Polymorphism- It refers to a special kind of genetic variation in which nucleotide sequence can exist at a particular site in a DNA molecule. This heritable mutation is caused by a mutation in either the somatic cell or the germ cell. It ultimately results in the accumulation of various mutations at one site. Polymorphism brings revolution in the process of finding a location on the chromosome for disease-associated sequences and tracing human history.
14.(c) Translation
Answer:
Translation- The process of amino acid polymerisation to form a polypeptide chain is called translation. The sequence and order of the amino acids within a polypeptide chain rely on the sequence of mRNA. Translation consists of three steps, i.e., initiation, elongation, and termination. In the process of initiation, the ribosome attaches at the beginning codon, which is AUG. Ribosomes then continue to travel codon by codon along the mRNA to add onto the polypeptide chain. Ultimately, the release factors recognise the stop codon, which results in the end of translation and release of the polypeptide from the ribosome.
14. (d) Bioinformatic
Bioinformatics- It is a recent and very effective area in the field of biology. In this discipline, the knowledge from the DNA sequences is used for solving various doubts regarding organisms that can not be studied in real time. Therefore, we can say that we utilise the biological information stored in the DNA of an organism.
Also, check NCERT Books and NCERT Syllabus here:
To solve the questions effectively, one needs to understand the concepts first. The chapter has certain key terms, such as replication, transcription, translation, and operon. Students can make notes from the Class 12 Biology Chapter 5 Molecular Basis of Inheritance question answer, which helps in highlighting all important points and definitions. These solutions have all the solved exercise questions, with to-the-point answers and necessary information. Students should practice the questions given in the Class 12 Biology Chapter 5 Molecular Basis of Inheritance Solutions to have an overall idea of the chapter and build confidence before the exams.
NCERT Solutions for Class 12- Subject-wise
Given below is a solved practice question from this chapter. To understand the complex processes such as transcription, translation, lac operon, etc, students need to refer to the NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance.
Question 1. In a DNA strand, the nucleotides are linked together by:
Options:
1. Glycosidic bonds
2. Phosphodiester bonds
3. Peptide bonds
4. Hydrogen bonds
Answer:
In a DNA molecule, nucleotides are joined through phosphodiester bonds, which are crucial covalent connections between the 3'-hydroxyl group of sugar in one nucleotide and the 5'-phosphate group of the sugar in the next.
This bonding pattern generates the DNA's sugar-phosphate backbone, ensuring a clear directionality from 5' to 3' ends. The significance of these bonds lies in their contribution to DNA's structural integrity and facilitation of nucleotide chain elongation.
Hence, the correct answer is option 2) Phosphodiester bonds.
NCERT Exemplar Class 12 Solutions
Students gain a deeper understanding of how the genetic material carries information and directs the functioning of living organisms.
Learners explore the detailed structure and features of DNA, including the double-helix model.
They understand the mechanism of replication through the Molecular Basis of Inheritance NCERT Solutions.
Study transcription, translation, and regulation of gene expression in both prokaryotes and eukaryotes.
Gain knowledge about important experiments like Griffith’s, Hershey-Chase, and Meselson-Stahl through Molecular Basis of Inheritance Class 12 question answer for conceptual clarity.
Explore the concept of the Human Genome Project and its significance in understanding genetic material.
Studying concepts beyond the NCERT will help in gaining conceptual clarity. It also allows students to answer application-based questions easily, especially for competitive exams or school exams. Some extra concepts that can be studied for NEET, along with the NCERT Solutions for Class 12 Biology Chapter 5 Molecular Basis of Inheritance, are given below.
Topics | NEET | NCERT |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ | |
✅ | ✅ |
Below mentioned are the chapter-wise solutions:
Frequently Asked Questions (FAQs)
DNA replication is semi-conservative, where each strand serves as a template. Enzymes like DNA helicase, DNA polymerase, and ligase assist in unwinding, copying, and sealing the DNA strands.
Transcription: Formation of mRNA from DNA inside the nucleus.
Translation: Conversion of mRNA sequence into a protein at ribosomes.
Students can find the NCERT Solutions for Class 12 Biology Chapter 5, Molecular basis of inheritance, from the Careers360 page, which is
https://school.careers360.com/ncert/ncert-solutions-class-12-biology-chapter-5
The important topics which are covered in the NCERT Solutions for Class 12 Biology Chapter 5 Molecular basis of inheritance are given below:
DNA structure and replication
RNA and its types
Transcription, translation, and the genetic code
Regulation of gene expression
Human Genome Project and DNA fingerprinting
The genetic code is a set of three-letter codons in mRNA that specify which amino acids will be assembled into a protein during translation.
On Question asked by student community
Hello,
The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.
Hope this information is useful to you.
Hello,
Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.
Hope this information is useful to you.
Hello Pruthvi,
Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.
The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.
For more details about the KCET Exam preparation,
CLICK HERE.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.
Hello
For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.
This ebook serves as a valuable study guide for NEET 2025 exam.
This e-book offers NEET PYQ and serves as an indispensable NEET study material.
As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
As per latest syllabus. Chemistry formulas, equations, & laws of class 11 & 12th chapters
As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE
As per latest syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters