Important Questions for Class 12 Chapter 6: Molecular Basis of Inheritance

Genes are the basic unit of heredity. Most genes are made of strands of genetic material called DNA. DNA contains all the hereditary information of an individual. This information is passed on from one generation to the other in the form of homologous chromosomes. The DNA is transcribed into mRNA and translated into proteins. This is known as central dogma.

Explore important questions on the Molecular Basis of Inheritance for a better understanding of the concept.

Very Short Answer Type Questions

Q.1. State the function of histones in DNA packaging.

A.1.

  • They play a role in gene regulation.
  • They help the DNA to wind around it.
  • The histones are positively charged proteins, which can easily bind to the negatively charged DNA.

Also Read DNA packaging.

Q.2. What is the difference between heterochromatin and euchromatin?

A.2.

Heterochromatin is a tightly packed DNA, which can be identified when stained in an extreme nuclear stain.

Euchromatin is a lightly packed DNA, which can be identified when stained in a less nuclear stain.

More Details: Difference between heterochromatin and euchromatin

Q.3. Name any three viruses with RNA as the genetic material.

A.3. The viruses in which the genetic material is RNA is called the RNA virus. The three examples of the RNA virus.

  1. Influenza Virus.
  2. Hepatitis C Virus.
  3. Human Immunodeficiency Virus.

Explore More: Virus

Q.4. What is the reason for the discontinuous synthesis of DNA on one of the parental strands?

A.4. The biological process of DNA synthesis naturally occurs in 5′ to 3′ direction. In the double-stranded DNA, the strands are parallel and antiparallel to each other. During the synthesis of DNA, both the strands act as templates and only one (3′ to 5′ direction) can synthesise the parallel strand in 5’→3′ direction. The other strand 5′ to 3′ is synthesized in the opposite direction producing small stretches of DNA known as Okazaki fragments. This is the reason for the discontinuous synthesis of DNA on one of the parental strands.

Q.5. The sequence of the coding strand of DNA in a transcription unit is mentioned below.

3′ AATGCAGCTATTAGG 5′

Write the sequence for:

  1. Its complementary strand
  2. Its mRNA

A.5.

  1. The complementary strand is  5′ TTACGTCGATAATCC 3′
  2. The mRNA is  5′ AAU GCAGCUAUUAGG 3′

Q.6. What is DNA polymorphism?

A.6. DNA polymorphism is the variation in the DNA sequence arising due to mutation at non-coding sequences.

Q.7. Comment on the statement “Retroviruses do not follow central dogma.”

A.7. Retroviruses do not follow central dogma, because, they possess RNA as genetic material instead of the DNA, which is later converted into DNA by the enzyme reverse transcriptase.

Explore more: Retroviruses

Q.8. Sometimes, the young ones born have an extremely different set of eyes or limbs. Give a relevant explanation for the abnormality.

A.8. This abnormality is caused by many factors, including alcohol abuse by the mother during her pregnancy, medicine side effects or reactions caused to the womb,  environmental factors, such as maternal exposure to the chemicals, radiations, virus, and it can also be due to the genes and non-coordination in the regulation of expression in the set of genes associated with the development of organs.

Q.9. Explain about the dual polymerase present in E.coli.

A.9. The DNA polymerase present in E.coli is a DNA dependent polymerase. This DNA polymerase helps in the:

  1. Replication process.
  2. Performs the 5′ to 3′ polymerase activity as well as 3′ to 5′ exonuclease activity.
  3. The DNA polymerase III also has the ability to proofread the wrong nucleotides and substitutes it with the correct one.

Q.10. What are the functions of the :

  1. Methylated guanosine cap
  2. Poly-A tail

A.10.

  1. Methylated guanosine cap plays a primary role in the attachment of the mRNA to the smaller sub-units of the ribosome during translation initiation.

  2. The Poly-A tail functions by increasing the length of the mRNA and also provides longevity to the mRNA.

Q.11. Mention any two functions of AUG codon.

A.11. The AUG codon is also called the start codon. The two important functions of AUG codon include:

  1. It codes for methionine.
  2. It acts as an initiation codon for protein synthesis.

Q.12. What is the function of aminoacyl t-RNA synthetase?

A.12. Amino acyl t-RNA synthetase functions by attaching an appropriate amino acid on to the tRNA molecule.

Short Answer Type Questions

Q.1. Define the following term:

  1. Promoter
  2. tRNA
  3. Exons

A.1.

  1. Promoter– The promoter is the region and the transcription unit in DNA, which is located towards 5′ end of the structural gene. This promoter serves as the binding site for RNA polymerase for transcription. It also helps in transcription initiation.

  2. tRNA– tRNA or transfer RNA is one among the three major types of  RNA- Ribo Nucleic Acid. The tRNA is involved in the synthesis and translation of proteins in a cell. The tRNA also reads the genetic code on the RNA and transfers specific amino acids to mRNA on the ribosomes.

  3. Exons– Exons are defined as the coding sequences or expressed sequences or the coding regions present in a DNA sequence in eukaryotes. These sequences appear in mature or processed RNA, which are interrupted by the introns.

Q.2. Explain the role of the ribosomes in translation.

A.2. Ribosomes are the site of translation or the protein synthesis, which is made up of rRNA. It plays a major role during translation, which includes:

  1. It provides a binding site for transfer RNA (tRNA).
  2. The two ribosomal subunits assemble together and form a sandwich around the mRNA.

  3. When the ribosome encounters a stop codon, it dissociates and no more polypeptides are synthesized.
  4. With each triplet codon of mRNA moving through the ribosome, a specific tRNA with its own anticodon is recruited and a polypeptide is synthesized.

Explore more: Ribosomes

Q.3. What is cistron? Differentiate between monocistronic and polycistronic transcription units.

A.3. Cistron can be defined as a segment of DNA coding for a specific polypeptide during translation.

The differences between monocistronic and polycistronic transcription units are:

Monocistronic Units Polycistronic Units
It codes for only one protein. It codes for several proteins.
It is the structural gene in a transcription unit in eukaryotes. It is the structural gene in a transcription unit mostly in bacteria or prokaryotes.
Eukaryotic mRNAs are monocistronic. Prokaryotic mRNAs are polycistronic.

Q.4. List out the enzymes involved in DNA replication along with their functions.

A.4. The five enzymes, which are involved in DNA replication are listed below along with their functions:

  1. Ligase- This enzyme functions by joining two strands of DNA.
  2. Topoisomerase- This enzyme functions by resolving the supercoiling of DNA.
  3. Helicase- This enzyme functions by unwinding the DNA strands during replication.
  4. Primase- This enzyme functions by the synthesis of RNA primer during DNA replication.
  5. DNA Polymerase- This enzyme functions by the synthesis of DNA molecules during replication.

Q.5. What is the Human Genome Project? List out the important features of this project.

A.5. The Human Genome Project is an international scientific project started with the aim of identifying all the genes in the human DNA from both a physical and functional perspective. The features of the Human Genome Project are:

  1. On average, a gene is made up of 3000 nucleotides.
  2. The human genome has 3164.7 million nucleotide bases.
  3. Proteins are coded by less than 2 per cent of the genome.
  4.  The largest human being dystrophia has 2.4 million bases.
  5. The functions of 50 per cent of the genes are not discovered yet.
  6. There are a large number of repeated sequences in the human genome.
  7. The maximum genes are present in the chromosome I whereas least are present in chromosome Y.

Explore more: Human Genome Project

Q.6. How are the abnormal haemoglobin molecules formed? What are the consequences of such abnormality?

A.6. The abnormal haemoglobin molecules are the molecules, which are inherited from the parents to their offsprings. These molecules are formed due to point mutation. The p-globin chain of the haemoglobin molecule is mutated, which replaces glutamic acid with valine at the sixth position. As a result, the erythrocytes became sickle-shaped and the oxygen-carrying capacity of the cells is inhibited.

The consequences of abnormal haemoglobin molecules lead to a deficiency of oxygen in the body cells, which is also responsible for the various blood-related disorders.

Q.7. Is it possible to use DNA probes such as VNTRs in the DNA fingerprinting of a bacteriophage?

A.7. No, VNTRs cannot be used in the DNA fingerprinting of a bacteriophage. Because the genome of a bacteriophage is very small and do not contain the repetitive and coding sequences similar to the VNTRs.

Also Refer: VNTRs

Q.8. Why does the lac operon show a low level of expression all the time?

A.8.

Lactose can be transferred from the medium into the bacterial cell only in the presence of permease. In the absence of permease, lactose is not transported into the cell and it fails to act as an inducer, thereby, showing a low level of expression.

Q.9. Can alternate splicing of exons enable a structural gene to code for several iso-proteins from one and the same gene? Give reasons.

A.9. The alternate splicing of exons is sex-specific, tissue-specific, and even developmental stage-specific. The alternate splicing helps in encoding a single gene for several iso-proteins. If this splicing is absent there would have been new genes for every protein or isoprotein.

Q.10. Mutation in a single base in a gene may not always result in a gain or loss of a function. Comment.

A.10. Due to the degeneracy of codons mutation at the third base of codons do not result in any phenotypic change.

Long Answer Type Questions

Q.1. Enumerate the post-transcriptional modifications in a eukaryotic mRNA.

A.1. Transcription is the process of conversion of DNA to mRNA. The post-transcriptional modifications include:

  • Capping at 5’-end
  • Poly-A tail at 3’-end
  • mRNA splicing

The 5’ cap protects the RNA from ribonucleases. The poly-A tail protects the mRNA from enzymatic degradation. The introns are spliced during mRNA splicing and the exons are joined together to form a continuous sequence that codes for a functional protein.

Q.2. Explain the process of translation.

A.2. The translation is the process of protein synthesis in which the mRNA is used to synthesize proteins. The mRNA sequence is decoded to specify the amino acid of a polypeptide. The process of translation is carried out in the following steps:

  • Initiation
  • Elongation
  • Termination

Q.3. Explain the process of DNA fingerprinting.

A.3. DNA fingerprinting is a technique that is used to analyze the genetic makeup of living beings. It is widely used in paternal disputes to identify the biological parents of the child, and also to identify the criminal during forensic investigations.

Extended Reading: DNA fingerprinting

Q.4. What is an operon? Explain an inducible operon.

A.4. An operon is the functional unit of DNA that contains a cluster of genes controlled by a single promoter. It consists of the following components:

  • The DNA fragment that transcribes the mRNA.

  • A promoter where the RNA polymerase binds and initiates the transcription.

  • An operator that is a DNA sequence adjacent to the promoter where the repressor protein binds.

  • Regulator gene that codes for a repressor protein

  • Inducer that prevents the repressor from binding to the operator.

The lac operon of E.coli is an inducible operon.

Q.5. Explain the process of DNA replication.

A.5. DNA replication is a biological process of producing two identical strands of DNA from the original strand. The original strand is known as the parent strand and the new strands are known as the daughter strands. This is achieved by a number of enzymes such as DNA polymerase, helicase, primase, topoisomerase, and ligase. For more information on Molecular Basis Of Inheritance and Biology related topics keep visiting BYJU’S website. You can also refer to the BYJU’S app for further details.

Read More: DNA replication

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1 Comment

  1. Great questions thanks for helping me in preparing my test

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