NCERT Solutions for Class 12 Science Biology Chapter 6 - Molecular Basis Of Inheritance

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In a DNA strand, the nucleotides are linked together by a 3'-5' phosphodiester bond to form a dinucleotide. Many nucleotides join together to form a polynucleotide chain.

Hence, the correct answer is option b.

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Nucleoside = Nucleotide – Phosphate group 

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Both deoxyribose and ribose sugars have 5-carbon atoms. Hence, they belong to the class of pentose sugar. 

Hence, the correct answer is option c.

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The width or diameter of the DNA strand is uniform or constant due to the base pairing of purine with pyrimidine through hydrogen bonds. 

Hence, the correct answer is option c.

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DNA is made up of a polynucleotide chain, that means it has a phosphate group attached to it. This gives DNA a negative charge. On the other hand, histones are rich in the basic amino acids such as lysine and arginine which impart positive charge to histone proteins.

Hence, the correct answer is option c.

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Sickle cell anaemia is a molecular disease, i.e. autosome linked recessive trait. It shows resistance to malaria infection due to the sickle-like structure of RBCs because of low oxygen tension and cannot be treated with iron supplements.

Hence, the correct answer is option d.

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AUG is the initiation codon or the start codon that initiates the translation process. It codes for amino acid methionine in both prokaryotes and eukaryotes.

Hence, the correct answer is option d.

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RNA was the first genetic material. According to RNA world hypothesis, it is evident that essential life processes such as metabolism, translation, splicing etc. evolved around RNA. DNA has evolved from RNA that make it more stable.

Hence, the correct answer is option d.

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Introns are the non-coding sequences whereas exons are the functional coding sequences. Splicing process removes the introns from the pre-mRNA and join exons to create functional RNAs. Thus, in mature RNA, exons appear but introns do not appear.

Hence, the correct answer is option b.

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In humans, chromosome 1 possess 2968 genes(highest) and chromosome Y possess 231 genes(lowest).

Hence, the correct answer is option c.

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Nucleotides are joined together by a 3'-5' phosphodiester bond to form a nucleotide. To prevent the polymerization of nucleotides, 3'OH group in deoxyribose sugar must be deleted or replaced, so that phosphodiester bond is not formed.

Hence, the correct answer is option b.

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DNA polymerase catalyzes the synthesis of DNA strands only in one direction at its 3'$\to$5' end. This is because nucleotides are added to the 3'-OH group of deoxyribose sugar. Thus, during replication, 3'$\to$5' strand of DNA is synthesized continuously and discontinuous synthesis of DNA occurs in the other strand.

Hence, the correct answer is option b.

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The elongation step in transcription is catalysed by the DNA dependent RNA polymerase.

Hence, the correct answer is option b.

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In prokaryotes, regulation of gene expression occurs at the level of transcription, while in eukaryotes it is regulated at various levels such as transcriptional level, processing level, transport or translational level.

Hence, the correct answer is option b.

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Regulatory proteins control the functions of structural gene and are known as regulatory genes. These can be promoters, terminators, operators or repressor. These protein interact with RNA polymerase and affect its function. They can act both as activators or repressors. 

Hence, the correct answer is option d.

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Human Genome Project (HGP) was a mega project that was started in 1990. Its goal was to sequence all $3\times {10}^9$ base pairs of human genome. But chromosome 1 was sequenced after the HGP had ended as it has most of the genes present i.e. approx 2968 genes.

Hence, the correct answer is option a.

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There are three major classes of RNA that are involved in gene expression.
These are rRNA, mRNA and tRNA.
rRNAs bind to protein molecules and form ribosomes.
mRNAs carry the genetic information for translation into protein.
tRNA carry amino acids to mRNA during protein synthesis.

Hence, the correct answer is option d.

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According to Chargaff's rule, the amount of adenine is equal to thymine and the amount of guanine is equal to cytosine. The ratio of adenine to thymine and that of guanine to cytosine is always equal to one. $\frac{\mathrm{A}}{\mathrm{T}}=\frac{\mathrm{G}}{\mathrm{C}}=1$
The given example shows that the percentage of adenine and guanine are not equal to thymine and cytosine respectively. Thus, it is single stranded DNA.

Hence, the correct answer is option b.

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Viruses such as retroviruses have RNA as their genome. In these viruses, reverse transcriptase enzyme generates a complementary DNA(cDNA) from an RNA template. This process is called reverse transcription. 

Hence, the correct answer is option c.

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Messelson and Stahl observed that the DNA of the first generation was hybrid $\left({\mathrm{N}}^{15} \mathrm{and} {\mathrm{N}}^{14}\right).$ DNA extracted from the culture in the second generation after 40 minutes was 50% ${\mathrm{N}}^{15}{\mathrm{N}}^{15}$ and 50%${\mathrm{N}}^{14}{\mathrm{N}}^{14}.$ After the DNA extracted from the culture in third generation after 60 minutes had 25% intermediate and 75% light in 1 : 3 ratio. In the fourth generation after 50 minutes it had 12.5% intermediate $\left({\mathrm{N}}^{15}{\mathrm{N}}^{14}\right)$ and 87.5% light $\left({\mathrm{N}}^{14}{\mathrm{N}}^{14}\right)$ that is in 0 : 1 : 7 ratio.

Hence, the correct answer is option d.

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Hence, the correct answer is option a.

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Transcription of the structural gene is assisted by RNA Polymerase. However, RNA polymerase helps in recognition of promoter sequence and Rho factor helps in recognition of terminator sequence. RNA polymerase is only capable of catalysing the elongation process thus transcribes structural gene only.

Hence, the correct answer is option d.
Disclaimer : Solution given in NCERT exemplar is option c which is incorrect.

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In tRNA, the amino acid binding site (AA-binding site) lies at the 3'-end opposite the anticodon and has a CCA-OH group. This site is unique to each amino acid.

Hence, the correct answer is option b.

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The ribosome is responsible for protein synthesis and is made up of proteins and structural RNAs. Ribosome has two subunits : a large subunit and a small subunit. The mRNA to protein translation begins when a small subunit comes in contact with mRNA.

Hence, the correct answer is option a.

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In E.coli, the lac operon gets switched on when lactose is present and it binds to the repressor. Lactose acts as an inducer and binds to the repressor and inactivates it. The repressor thus, fails to bind to the operator region. RNA polymerase then binds to the operator and initiates transcription. 

Hence, the correct answer is option a.

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Histones are the basic proteins that are rich in arginine and lysine and have positive charge. Histones form an octamer to which negatively charged DNA is wrapped around and forms  a nucleosome. Thus, it helps in packaging of DNA in eukaryotes.

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Heterochromatin are densely packed, darkly stained and transcriptionally inactive chromatin, whereas euchromatin is loosely packed, lightly stained and transcriptionally active chromatin.

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DNA polymerase uses a DNA template to catalyze the synthesis of DNA molecules by speeding the polymerization of deoxynucleotides in
5'$\to$3' direction. It also proofreads the DNA strand if any error/mutation occurs during replication.

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DNA dependent DNA polymerase synthesizes the DNA strand in 5'$\to$3' direction only as nucleotides are added at 3'OH group of the sugar. However, the replication is continuous on the leading strand (3'$\to$5') but is discontinuous on the lagging strand (5'$\to$3'). The fragments called okazaki fragments produced during dicontinuous synthesis are joined later by DNA ligase enzyme.

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(a) Its complementary strand
5'–TTACGTCGATAATACC–3'
(b) The mRNA 
5'–AAUGCAGCUAUUAGG–3'

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Polymorphism is variation at genetic level that arises due to mutation. Thus, polymorphism is referred to as the inheritable mutation observed in a population at high frequency.
This identification tool has its application in forensic science. It is the basis of paternity testing or identification of criminals etc.

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This occurs due to disturbance in co-ordinated regulation of expression of sets of genes that are associated with organ development.

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During replication, only deoxyribonucleotides added because DNA polymerase is a highly specific enzyme and recognize only deoxyribonucleotides.

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Other enzymes that are used in DNA replication are:-
(1) Helicase - unwinds the DNA double helix.
(2) Topoisomerase - removes the supercoiling of DNA.
(3) Primase - synthesizes RNA primer.
(4) Telomerase - synthesizes telomeres at the end of chromosomes.
 

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Three viruses which have RNA as genetic material are-TMV (tobacco mosaic virus), HIV (Human immunodeficiency virus) and influenza virus.

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In 1928, Frederick Griffith performed a series of experiments using pneumonia bacteria Streptococcus pneumoniae and mice. Streptococcus were grown on culture media in which strain R II produced rough colonies and strain S III produced smooth colonies. Mice infected with S strain die from pneumonia infection due to the presence of mucous coat while R strains do not die. Griffith was able to kill bacteria by heating them. Heat killed S strain bacteria injected into the mice did not kill the bacteria. But when the mixture of heat killed S and live R bacteria was injected, the mice died. 
     S strain $\to$ Inject into $\to$ Mice 
(Heat killed)       Mice            Live

   S strain   +    R strain   $\to$  Inject into   $\to$ Mice 
(heat killed)      (live)               mice                die

He thus concluded that Heat killed S-strain transferred some transforming principle to the live R strain which enabled the R strain to produce a polysaccharide coat.
He assumed this transforming principle as genetic material. Thus, this principle helped in identification of DNA as genetic material.   
 

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Oswald T Avery, Colin Macleod and Maclyn McCarty performed the experiment in vitro to identify the biochemical nature of transforming principle in Griffith's experiment. They purified DNA, RNA and protein from the heat killed S-Strain cells, and noticed that digesting enzymes (RNase, protease) did not affect the transformation of non-virulent R Strain. They found DNase inhibited transformation suggesting that DNA is the cause of transformation.

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Messelson and Stahl performed the experiment in 1958 and grew E.coli in a medium containing heavy isotope of nitrogen $\left({\mathrm{N}}^{15}\right)$ as the only source of nitrogen for many generations. This is because heavy isotope of nitrogen gets easily separated from regular DNA in the cesium chloride (CsCl) density gradient centrifugation. Thus, use of lighter and heavier nitrogen in DNA simplified the process of identifying DNA transfer in successive generations.

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A cistron is a segment of DNA that codes for a polypeptide. It is a type of gene. Monocistronic is referred to as a stretch of replicating DNA that contains only one gen/cistron while polycistronic refers to as a stretch of replicating DNA that contains more than one cistron. Monocistronic is found in eukaryotes and polycistronic is found in prokaryotes.

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Six features of human genome are:-
(i)  The genome has around 3164.7 million nucleotide bases.
(ii) 99.9% of the nucleotide bases are same in all humans.
(iii) The largest gene is Dystrophin having 2.4 million bases.
(iv) Chromosome 1 has most genes (2968) and y has the least genes (231).
(v) Less than 2% of the genome has the coding sequence for proteins.
(vi) Only 50% of the total discovered genes have known functions.

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DNA replication is a tremendously energy-intensive process, so DNA segments do not copy in one go. Unwinding creates tension in the molecule as uncoiled parts start forming super coils due to the interaction of exposed nucleotides. During replication, helicase enzyme unwinds the DNA at ori site (origin of replication) and a small stretch is unzipped. This gives a y shaped structure called replication fork.
Two functions of the monomers (dNTPs) play are:
(i) They provide energy for polymerisation reaction and forms phosphodiester linkages with exposed nucleotides
(ii) They act as substrates and provide deoxyribonucleotides for DNA replication.

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Central Dogma of biology is DNA → RNA → Protein. Since, retroviruses have RNA as their genetic material instead of DNA, they do not follow central dogma. Their RNA is converted to DNA by the enzyme reverse transcriptase.

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Distance between two consecutive base pairs: 0.44 nm or 0.44 × 10^–9
Length of the DNA = 2×10⁹ × 0.44 × 10^–9

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If histones were mutated and made rich in acidic amino acids, they will not be able to coil around DNA on them as they will become negatively charged. Basicity gives them a positive charge through which they are easily wound to negatively charged DNA. So if histones become negatively charged they will repel DNA and packaging of DNA in eukaryotes would not happen, hence chromatin fibre will not be formed.

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RNA is more liable and prone to degradation because of the presence of 2'OH group in its ribose sugar. Hence, heat killed S-strain may not have retained its ability to transform the R-strain into virulent form, if RNA was its genetic material.

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Use of ¹⁵N will be inappropriate because method of detection of ³²P and ¹⁵N different. (³²P being a radioactive isotope while ¹⁵N is non-radioactive but is the heavier isotope of Nitrogen). Even if ¹⁵N was radioactive then its presence would have been detected, both inside the cell (¹⁵N incorporated as nitrogenous base in DNA) as well as in the supernatant, because ¹⁵N would also get incorporated in amino group of amino acids in proteins. Hence, the use of ¹⁵N would not give any conclusive results.

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Some amino acids are coded by more than one codon (known as degeneracy of codons), hence, on deducing a nucleotide sequence from an amino acid sequence, multiple nucleotide sequence will be obtained, e.g. Ile (Isoleucine) has three codons AUU, AUC, AUA. Hence, a dipeptide Met-Ile can have the following nucleotide sequence.
i) AUG - AUU
ii) AUG - AUC
iii) AUG - AUA
If we deduce amino acid sequence from the above nucleotide sequences, all the three will code for Met-Ile.

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The statement is correct, because of degeneracy of codons, mutations at third base of codon, usually does not result into any change in phenotype. This is called silent mutations. On the other hand, if codon is changed in away that now it specifies another amino acid, it may other the protein function as it happens in case of B-globulin of haemoglobin protein. Where a substitution of valine instead of glutamic acid causes change in its structure & function, and resulting into sickle-cell trait.

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When the expression of lac operon is completely absent, enzyme permease will not be synthesized. Lactose present in the external medium can only enter the bacterium when it has permease within it. If lactose cannot enter the cell, then it cannot act as inducers. Hence, cannot relieve the lac operon from its repressed state.

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The sequencing of the human genome opened new windows for the treatment of various genetic disorders because it led to a better knowledge of genetic disorders. A better understanding on diagnosis, treatment and prevention of genetic disorders is possible.

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It is because the repeated sequences make up very large portion of the human genome. Repetitive sequences are stretches of DNA sequences that are repeated many times, sometimes hundred to thousand times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution. 

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Human genome helps to find out the complete genome sequence of the human.
Its advantages are:
1) It would lead to better diagnosis, treatment and prevention of genetic disorders.
2) A better understanding of the DNA gene sequence would lead to a better understanding of biological systems.
Its disadvantages are:
1) Patenting of the genetic test results can be done. In a turn of which gene patenting can also be done.
2) Genetic disorders which can't be treated might be discovered.

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Bacteriophage does not have repetitive sequences such as VNTRs in its genome. Its genome is very small and have all the coding sequence. DNA fingerprinting is not done on bacteriophages.

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No polymerisation will occur as the 3'OH group on sugar is not present to add a new nucleotide for forming ester bonds.

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Watson and Crick had developed the model of DNA by using the following information:-
i) Chargaff's law which states that Adenine forms bond with thymine and guanine forms bond with cytosine; i.e. A = T and C ≡ G. 
ii) Wilkins and Rosalind Franklin's work of DNA crystal's X-ray diffraction studies about DNA's physical structure.
Watson & Crick finally proposed a double helix structure of DNA.
Their contributions are
i) discovered pattern of complementary base pairing
ii) semi-conservative replication

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(i) Methylated guanine cap helps in binding of mRNA to smaller ribosomal sub-unit during initiation of translation.
(ii) Poly-A-tail provides longevity to mRNAs life. Tail length and longevity of mRNA are positively corelated.

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Functional mRNA of structural genes do not include all of its exons. This alternate splicing of exons is sex-specific, tissue-specific and developmental stage specific. By such alternate splicing of exons, a single gene may encode for several isoproteins/proteins of similar class. In absence of such a kind of splicing, there should have been new genes for every protein. Such an extravagancy has been avoided in natural phenomena by way of alternate splicing.

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Tandemness in repeats provides many copies of the sequence for finger-printing and variability in nitrogen base sequences present in them. Being individual-specific, this proves to be useful in the process of DNA fingerprinting.

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Hershey and Chase Experiment to Confirm DNA as the Genetic Material

Hershey and Chase worked on bacteriophages (viruses that infect bacteria). When a bacteriophage infects a bacterium, the viral genetic material gets attached to the bacterial genetic material and bacteria then treat the viral genetic material as their own to synthesize more viral particles.
Hershey and Chase worked to discover whether it was a protein or DNA that entered the bacteria from the virus. They labelled some phages with radioactive sulphur and others with radioactive phosphorus. These radioactive phages were used to infect E. coli.
E.coli was then blended and centrifuged to remove viral particles. It was observed that bacteria with radioactive DNA were radioactive while those with radioactive proteins lost their radioactivity.
This showed that it is the DNA that enters the bacteria from viruses and not proteins. Hence, it was concluded that DNA is the genetic material.
If both DNA and proteins contained phosphorus and sulphur then the bacterial cell will show not show any radioactivity and it would be difficult to identify the genetic material among DNA and protein. 

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Desired criteria in a molecule that can act as genetic material - 

• It should be able to replicate (duplicate to produce its identical copy).
• It should be chemically and structurally stable.
• It should have scope for changes that are essential for evolution.
• It should follow the Mendelian principles of inheritance.

• DNA is less reactive and mutates slowly while RNA is more reactive and mutates faster than DNA. 
• Presence of uracil decreases stabilty of RNA while presence of thymine in place of uracil confers additional stability to DNA.
• DNA is double-stranded having complementary strand, which resists the changes by repair mechanism. RNA is a single stranded molecule. 
• In RNA, a 2′ OH group is present at every nucleotide. This makes RNA unstable and degradable. 

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Post-transcriptional modifications of a eukaryotic mRNA- 

The precursor of mRNA, i.e. hnRNA, contains both introns and exons. Introns are removed and exons are joined by a process called splicing. The remaining mRNA is processed in two ways:

Capping − In this, methyl guanosine triphosphate is added to the 5′ end of hnRNA.
Tailing − In this, adenylate residues are added to the 3′ end of hnRNA.

When hnRNA is fully processed, it is known as mRNA, which is transported out of the nucleus to get translated.

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TRANSLATION

The mRNA contains the genetic information, which is translated into the amino acid sequence with help of tRNA. Amino acids are polymerised to form a polypeptide. Amino acids are joined by a peptide bond.

First of all, charging of tRNA (amino-acylation of tRNA) takes place. In this, amino acids are activated in the presence of ATP and are linked to their corresponding tRNA.

Ribosomes are the workbenches for translation. Ribosomes have 2 subunits: a large subunit and a small subunit.

Smaller subunit comes in contact with mRNA to initiate the process of translation.

Translational unit in an mRNA is the region flanked by start codon and stop codon.

Untranslated regions (UTR) are the regions on mRNA that are not themselves translated, but are required for efficient translation process. They may be present before start codon (5′ UTR) or after stop codon (3′ UTR).

Initiation- Initiator tRNA recognises the start codon.
Then t-RNA-amino acid complexes bind to their corresponding codon on the mRNA and base pairing occurs between codon on mRNA and tRNA anticodon.
Elongation- tRNA moves from codon to codon on the mRNA and amino acids are added one by one. 
Termination- Release factor binds to stop codon to terminate the translation. 

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An arrangement where a polycistronic gene is regulated by a common promoter and regulatory genes is known as an operon. Lac operon, trp operon, his operon, val operon are examples of such systems. 

The lac operon consists of three structural genes  (z, y and a), and one regulatory gene- i gene (here, i stands for inhibitor).
The z-gene codes for beta-galactosidase, the y-gene codes for permease, and the a-gene encodes transacetylase.
All genes involved in the lac operon are required for the metabolism of lactose.
Lactose acts as an inducer for the lac operon since it regulates the switching on and off of the operon.
If lactose is provided to the growth media of bacteria in absence of any other carbon source, then it is transported inside the cells by permease. For permease to be present and lactose to enter the cells, a low level of expression of the lac operon must be present all the time.

Regulation in Absence of Inducer-
In absence of inducer, i gene transcribes to synthesise repressor mRNA, which translates to form repressor.
This repressor binds with the operator region of operon and prevents RNA polymerase to transcribe genes − z, y, and a (negative regulation). Therefore, in absence of the products of these genes, metabolism of lactose ceases.




Regulation in Presence of Inducer-
Inducer binds with the protein product of gene i (repressor) and inactivates it. This inactivated repressor is unable to inactivate RNA polymerase enzyme and z, y, and a genes synthesise their respective mRNA, which in turn gets translated to form β-galactosidase, permease, and transacetylase. In presence of all these enzymes, the metabolism of lactose proceeds in a normal manner.

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DNA fingerprinting can solve the problem of paternity disputes for a child. It is a method for comparing the DNA sequences of any two individuals. 
As it is a daunting task to sequence and compare all the bases of DNA in two individuals. So, instead of considering the entire genome, certain specific regions called repetitive DNA sequences are used for comparative study.
Repetitive DNA is separated from bulk genomic DNA since it appears as a distinct peak during density gradient centrifugation.
The major peaks are formed by bulk DNA while smaller peaks are formed by the satellite DNA. 
Satellites are of two types—micro-satellites and mini satellites, depending upon the base composition, length of segment and the number of repetitive units. Satellites do not code for proteins, but have a major role to play in DNA fingerprinting.
Polymorphism is actually a result of mutation. A germ cell mutation (which can pass on to the next generation through sexual reproduction) gives rise to polymorphism in populations. In other words, an inheritable mutation if observed in higher frequencies in a population is known as polymorphism.
Polymorphisms arise normally in non-coding sequences because mutations in non-coding sequences do not affect an individual’s reproductive ability.

Methodology of DNA fingerprinting-

• VNTR (variable number of tandem repeats) are satellite DNAs that show high degree of polymorphism.
• VNTRs are used as probes in DNA fingerprinting.
• First of all, DNA from an individual is isolated and cut with restriction endonucleases.
• Fragments are separated according to their size and molecular weight on gel electrophoresis.
• Fragments separated on electrophoresis gel are blotted (immobilised) on a synthetic membrane such as nylon or nitrocellulose.
• Immobilised fragments are hybridised with a VNTR probe.
• Hybridised DNA fragments can be detected by autoradiography.
• VNTRs vary in size from 0.1 to 20 kb.
• Hence, in the autoradiogram, band of different sizes will be obtained.
• These bands are characteristic for an individual. They are different in each individual, except identical twins.

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There are two methods used in sequencing the human genome- Identifying ESTs (Expressed sequence Tags) and sequence annotation. 

ESTs − As the name suggests, this refers to the part of DNA that is expressed, i.e. transcribed, as mRNA and translated into proteins thereafter. It basically focuses on sequencing the part denoting a gene.

Annotation − In this approach, entire genome (coding + non-coding) is sequenced and later on function is assigned to each region in the genome.

Genome Sequencing

• DNA from the cells is isolated and is randomly broken into fragments of smaller sizes.
• These fragments are cloned into suitable host using vectors.
• Cloned fragments amplify in the host. Amplification facilitates an easy sequencing.
• Common vectors used − BAC (Bacterial artificial chromosomes) and YAC (Yeast artificial chromosomes)
• Common hosts − Bacteria and yeasts
• Automated sequencers are used to sequence these smaller fragments (Sanger sequencing).
• The sequences so obtained are arranged based on overlapping regions within them (alignment).
• Alignment of the sequences is also done automatically by computer programs.
• Then these sequences are annotated and assigned to each chromosome. 

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Markers that are used in DNA fingerprinting- 

• Amplified Fragment Length Polymorphisms (AFLPs)
• Random Amplified Polymorphic DNAs (RAPDs)
• Restriction Fragment Length Polymorphisms (RFLPs)
• Simple Sequence Repeats (SSRs) (microsatellites)
• Single Nucleotide Polymorphisms (SNPs) 

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Option (d) is the correct result.