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| Chapter | 6. Molecular Basis Of Inheritance |
| Subject | Biology |
| Class | Twelve |
| Category | NCERT Solutions for Class 12 |
Our NCERT Solutions for Class 12 Biology is a great resource for students preparing for boards or competitive exams, such as the NEET. These Biology Class 12 Solutions are made by expert faculties, keeping the latest curriculum in mind. Besides helping students with understanding the concepts of Biology, these solutions are also helpful in writing accurate answers that are vital to score full marks in examinations. On this page, we have given the Class 12 Biology Chapter 6: Molecular Basis Of Inheritance Solutions.
NCERT Solutions for Class 12 Biology Chapter 6
Molecular Basis Of Inheritance Solutions
Q1) Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer)
| Nitrogenous base | Nucleosides |
|---|---|
| Adenine | Cytidine |
| Thymine | Guanosine |
| Uracil | |
| Cytosine |
Q2) If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer) The percent of cytosine = 20 (given) % of C = % of G
Therefore, the percent of guanine = 20
The percent of thymine + adenine will be 100 – (20 + 20) = 60
Therefore, the percent of adenine will be 60/2 = 30
Q3) If the sequence of one strand of DNA is written as follows:
5′ -ATGCATGCATGCATGCATGCATGCATGC-3′
Write down the sequence of complementary strand in 5’→3′ direction.
Answer) With regards to base sequence, DNA strands are complementary to each other. Therefore, if the sequence of one strand of DNA is written as:
5′ -ATGCATGCATGCATGCATGCATGCATGC-3′
The sequence of the complementary strand in 3′ -> 5′ is as follows:
3′ – TACGTACGTACGTACGTACGTACGTACG – 5′
Subsequently, sequence of complementary strand in 5’→3′ direction is 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)
If the coding strand in a transcription unit is
5’− ATGCATGCATGCATGCATGCATGCATGC-3’
Then, the template strand in 3’ to 5’ direction would be
3’ − TACGTACGTACGTACGTACGTACGTACG-5’
It is known that the sequence of mRNA is same as the coding strand of DNA.
However, in RNA, thymine is replaced by uracil.
Hence, the sequence of mRNA will be
5’ − AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3’
Q5) Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer) The property of DNA double helix led Watson and Crick are:
- Two strands running opposite to each other, wherein bases will always pair with their counterpart-A with T and G with C (specific pairing).
- If H bonds break and bases of one strand lie exposed, unpaired, they will easily pair up with free nucleotides as well.
This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. where the two strands separate and act as a template for the synthesis of a new complementary strand.
Q6) Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer)
- DNA-dependent DNA polymerase uses a DNA template to catalyse the polymerisation deoxynucleotides.
- DNA-dependent RNA polymerase catalyses the transcription of all types of RNA (in bacteria).
- DNA-dependent RNA polymerase-I transcribes rRNA.
- DNA-dependent RNA polymerase-II transcribes the precursor of mRNA (hnRNA).
- DNA-dependent RNA polymerase-III transcribes tRNA.
Q7) How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer) To prove DNA is the genetic material, Hershey and Chase studied and worked on bacteriophage and E.Coli. To label protein coat and DNA of the bacteriophage, they made use of different radioactive isotopes. In a medium containing radioactive phosphorous (32P), they cultivated some bacteriophages to detect DNA and few more on a medium comprising radioactive sulphur (35S) to detect protein. The labelled radioactive phages were then made to infect the bacteria – E.coli. Once infected, the protein coat of the bacteriophage was segregated from the bacterial cell by mixing and then subjecting to the centrifugation process.
It was observed that in the supernatant, as the protein coat was lighter, the infected bacteria got settled at the bottom of the centrifuge tube. In case I – supernatant was observed to be radioactive indicating that protein did not enter in the bacterial cell when infected. But in case II – the bacterial cells were radioactive as they possess radioactive DNA. Thus, it was proved that DNA is a genetic material as it was transferred from virus to bacteria.
Q8) Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand
Answer a)
| Repetitive DNA | Satellite DNA |
|---|---|
| Repetitive DNA are DNA sequences that contain small segments, which are repeated many times. | Satellite DNA are DNA sequences that contain highly repetitive DNA. |
Answer b)
| mRNA | tRNA |
|---|---|
| mRNA or messenger RNA acts as a template for the process of transcription. | tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide. |
| It is a linear molecule. | It has clover leaf shape. |
Answer c)
| Template strand | Coding strand |
|---|---|
| Template strand of DNA acts as a template for the synthesis of mRNA during transcription. | Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA). |
| It runs from 3’ to 5’. | It runs from 5’to 3’. |
Q9) List two essential roles of ribosome during translation.
Answer) The important functions of ribosome during translation are as follows:
- Ribosomes are sites where synthesis of proteins occurs from individual amino acids. It consists of two subunits – larger subunit serves as an amino acid binding site whereas smaller subunit attaches to the mRNA forming a protein synthesizing complex
- Since large subunit of ribosome has two different sites to attach to tRNA, it facilitates amino acids to come closer for peptide bond formation. Also, ribosome behaves as a catalyst for the formation of peptide bond. Example – 23s r-RNA acts as a ribozyme in bacteria
Q10) In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer) Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolize lactose into glucose and galactose.
In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolized into glucose and galactose.
After sometime, when the level of inducer decreases as it is completely metabolized by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.
Q11) Explain (in one or two lines) the function of the followings:
(a) Promoter
(b) tRNA
(c) Exons
Answer)
- a. Promoter: Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.
- b. tRNA: tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.
- c. Exons: Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.
Q12) Why is the Human Genome project called a mega project?
Answer) The Human genome project was ought to be a mega project as the scale and the goals of the project were humongous. It had goals to sequence every base pair located in the human genome which took around 13 years for completion and accomplished in 2003. This large scale project intended to develop new technologies and produce new information in the stream of genomic studies. Subsequently, it unlocked scope for several new areas and possibilities such as in the stream of biotechnology, genetics, medical sciences etc which hints at comprehending different aspects of human biology.
Q13) What is DNA fingerprinting? Mention its application.
Answer) DNA fingerprinting is a technique used to identify and analyze the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.
Applications:
- It is used in forensic science to identify potential crime suspects.
- It is used to establish paternity and family relationships.
- It is used to identify and protect the commercial varieties of crops and livestock.
- It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.
Q14) Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Answer)
a) Transcription: The process of copying genetic information from one strand of the DNA into RNA is known as transcription. RNA is assembled simply based on complementarity of the DNA strand, only uracil is substituted in place of thymine. Only a small segment of DNA that codes for a polypeptide is copied.
b) Polymorphism: The variation in DNA arising through mutation at non-coding sequences is known as Polymorphism. Such variations are unique to specific sites of DNA and can occur due to deletion, insertion or substitution of bases. It can be observed by making fragments of DNA sample and separating them through electrophoresis. The polymorphism in a DNA sequence is the basis of genetic mapping of the human genome as well as DNA fingerprinting.
c) Translation: It refers to the process of polymerisation of amino acids to form a polypeptide. The order and sequence of amino acids are defined by the sequence of bases in the mRNA. It occurs in cytoplasm in both prokaryotes and eukaryotes.
d) Bioinformatics: It is the application of computer science and information technology which deals with handling, storing of huge information of genomics, processing information, analyzing data and creating new knowledge.
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