Molecular Basis of InheritanceNEET Botany · Class 12 · NCERT Chapter 3

Solution

The Tobacco mosaic virus (TMV) has a coiled RNA strand and capsomeres. NCERT XII chapter Molecular Basis of Inheritance describes TMV as a rod-shaped virus with a helical arrangement of RNA and protein subunits, so option (b) is correct.

Solution

Transcription involves writing information from DNA to mRNA, not tRNA. Translation, repressor proteins, and operons are correctly described in the other options, so option (a) is wrongly matched. NCERT XII chapter Molecular Basis of Inheritance provides these definitions.

Solution

Transformation was discovered by Griffith in his experiments with Streptococcus pneumoniae, demonstrating that a heritable factor could be transferred from one bacterium to another. NCERT XII chapter Molecular Basis of Inheritance credits Griffith with this discovery, so option (c) is correct.

Solution

Viruses typically contain either DNA or RNA enclosed in a protein coat called a capsid. They do not have a prokaryotic nucleus or a single chromosome, and they do not contain both DNA and RNA simultaneously, so option (a) is correct. NCERT XII chapter Molecular Basis of Inheritance describes the structure of viruses.

Solution

Recombinase enzymes are active during the pachytene stage of prophase I, facilitating crossing over between homologous chromosomes. NCERT XII chapter Molecular Basis of Inheritance describes this process, so option (a) is correct.

Solution

RNA synthesis occurs in the 5' to 3' direction, while the template DNA strand is read in the 3' to 5' direction. This is consistent with the mechanism of transcription described in NCERT XII Molecular Basis of Inheritance, so option (a) is correct.

Solution

Mucor has non-motile spore i.e. sporangiospores

Solution

Phosphodiester bond is formed between two nucleotides of nucleic acid. www.vedantu.com 13

Solution

Order of organisation of genetic material

Solution

Chargaff's rule is applicable only for DNA. www.vedantu.com 4

Solution

In viroids, RNA is of low molecular weight. www.vedantu.com 48

Solution

Lac operon becomes active only after inducing lactose as it is a substrate for the enzyme beta- galactosidase and it also regulates the switching on and off of the operon which cannot be done by glucose and galactose.

Solution

Hemophilia is a sex – linked recessive disease in which there is a problem in clotting of blood. Down’s syndrome (trisomy 21) is caused due to aneuploidy. Phenylketonuria is an autosomal recessive gene disorder. Sickle cell anaemia is autosomal recessive gene disorder.

Solution

AUG is start codon that codes for methionine whereas UGA, UAA and UAG are stop codons.

Solution

Hershey and Chase gave unequivocal proof which ended the debate between protein and DNA as genetic material.

Solution

Two DNA polymerase molecules work simultaneous at the DNA fork, one on the leading strand and the other on the lagging strand. Each Okazaki fragment is synthesized by DNA polymerase at lagging strand in 5′? → 3 ′? direction. New Okazaki fragments appear as the replication fork opens further. As the first Okazaki fragment appears away from the replication fork, the direction of elongation would be away from replication fork.

Solution

Spliceosomes are used in removal of introns during post-transcriptional processing of hnRNA in eukaryotes only as split genes are absent as prokaryotes.

Solution

Viroids are sub-viral agents as infectious RNA particles, without protein coat.

Solution

The association of H1 protein indicates the complete formation of nucleosome. Therefore the DNA is in condensed form. www.vedantu.com 24

Solution

DNA fragments are negatively charged because of phosphate group.

Solution

If deletion occurs at 901st position the remaining 98 bases specifying for 33 codons of amino acids will be altered.

Solution

rRNA is most abundant in animal cell. It constitutes 80% of total RNA of the cell.

Solution

Punnett square was developed by British geneticist Reginald Punnett to predict the outcomes of genetic crosses. NCERT XII chapter Molecular Basis of Inheritance covers the history and development of genetic tools, confirming option (b) as correct.

Solution

The experimental proof for semiconservative replication of DNA was first shown in bacteria, specifically $\text{E. coli}$, by Meselson and Stahl. NCERT XII chapter Molecular Basis of Inheritance describes this experiment, so option (b) is correct.

Solution

An operon consists of structural genes, an operator, and a promoter. An enhancer is not part of the operon structure but influences gene expression from a distance. NCERT XII chapter Molecular Basis of Inheritance defines the components of an operon, so option (a) is correct.

Solution

The transcribed mRNA sequence is complementary to the coding strand, with T replaced by U. Thus, AGGTATCGCAT transcribes to UCCAUAGCGUA, but since the mRNA is synthesized from the template strand, the correct sequence is the complement of the coding strand: AGGUAUCGCAU. Option (c) is correct.

Solution

Factual NCERT

Solution

Factual NCERT

Solution

Ref. XII NCERT Chapter- 16, Page No. 283

Solution

Out of NCERT Pinus seed can not germinate until unless mycorrhizal association is not developed.

Solution

Average phenotypes are selected in stabilizing selection

Solution

Glucocorticoids, such as cortisol, stimulate gluconeogenesis, the process of glucose synthesis from non-carbohydrate sources. This is a key function of glucocorticoids as described in NCERT XII Molecular Basis of Inheritance, making option (a) correct.

Solution

DNA helicase is the enzyme that unwinds and separates the DNA strands during transcription. NCERT XII chapter Molecular Basis of Inheritance describes DNA helicase as essential for breaking hydrogen bonds between base pairs, so option (b) is correct.

Solution

The first phase of translation involves the binding of mRNA to the ribosome, initiating the process of protein synthesis. NCERT XII chapter Molecular Basis of Inheritance describes this as the initial step, so option (a) is correct.

Solution

Adenine pairs with thymine through two hydrogen bonds, as described in the base-pairing rules of DNA structure in NCERT XI chapter Molecular Basis of Inheritance. Therefore, option (a) is correct.

Solution

Viroids are infectious agents composed of short strands of naked RNA without a protein coat. NCERT XII chapter Molecular Basis of Inheritance defines viroids as small, circular, single-stranded RNA molecules, so option (b) is correct.

Solution

The length of the DNA can be calculated using the formula $L=n\times d$, where $n$ is the number of base pairs and $d$ is the distance between two consecutive base pairs. Substituting the values, $L=6.6\times 10^9\times 0.34\times 10^{-9}\text{ m}=2.244\text{ m}\approx 2.2\text{ m}$. Therefore, option (c) is correct.

Solution

DNA stained with ethidium bromide appears as bright orange bands when viewed under UV radiation. Ethidium bromide intercalates into the DNA double helix and fluoresces under UV light, so option (b) is correct.

Solution

- Replication is the process of DNA making a copy of itself.
- Transcription is the synthesis of RNA from a DNA template.
- Translation is the process of synthesizing proteins from the mRNA.
- Protein is the final product of the central dogma. Option (c) correctly matches these processes.

Solution

RNA polymerase III transcribes tRNA, 5S rRNA, and small nuclear RNAs (snRNAs) in eukaryotes. This is detailed in NCERT XII chapter Molecular Basis of Inheritance, making option (b) correct.

Solution

The mutated gene will hybridize with the radioactive probe and appear clearly on a photographic film through autoradiography, as the probe has complementarity with the mutated DNA sequence. This is based on the principle of nucleic acid hybridization described in NCERT XII chapter Molecular Basis of Inheritance, so option (b) is correct.

Solution

DNA fingerprinting relies on identifying variations in polymorphic DNA, which are regions that show differences among individuals. NCERT XII chapter Molecular Basis of Inheritance explains that these polymorphisms are used for genetic identification, so option (d) is correct.

Solution

According to Chargaff's rules, the amount of adenine (A) equals thymine (T), and guanine (G) equals cytosine (C). If A is 30%, then T is also 30%. The total percentage of G and C is 40%, so G and C are each 20%. Therefore, option (c) is correct.

Solution

Endonucleases recognize and cut DNA at specific palindromic nucleotide sequences, which are sequences that read the same forwards and backwards. NCERT XII chapter Molecular Basis of Inheritance explains that restriction endonucleases identify these palindromic sequences to make precise cuts, so option (b) is correct.

Solution

siRNA (small interfering RNA) is not directly involved in protein synthesis. mRNA, tRNA, and rRNA are essential components of the translation process, as described in NCERT XII chapter Molecular Basis of Inheritance, so option (d) is correct.

Solution

DNA dependent RNA polymerase is the only enzyme that catalyzes the initiation, elongation, and termination of transcription in prokaryotes. NCERT XII chapter Molecular Basis of Inheritance describes this enzyme as essential for the entire transcription process, so option (b) is correct.

Solution

Statement I is incorrect because the codon ‘AUG’ codes only for methionine, not phenylalanine. Statement II is correct as both ‘AAA’ and ‘AAG’ codons code for the amino acid lysine, according to the genetic code table in NCERT XII Molecular Basis of Inheritance, so option (d) is correct.

Solution

When the small subunit of ribosom e encounters an mRNA, the process of translation of the mRNA to protein begins. This process is followed by the binding of bigger/larger subunit. t-RNA is activated by the addition of amino acid prior to the attachment of ribosome, in the first phase.

Solution

Polymorphism in DNA sequence is the basis of genetic mapping of human genome as well as of DNA fingerprinting.

Solution

Number of base pairs × distance between 2 consecutive base pairs = Length of DNA molecule x · 0.34 × 10–9 m = 1.1 m 9 1.1x 0.3 10 −= × = 3.6 × 109 ; 3.3 × 109 bp

Solution

From 10 parent E.coli cells 1st generation Therefore, after 60 minutes, 60 E.coli cells will have DNA totally free from 15N.

Solution

Upon exposure to UV radiation, DNA stained with ethidium bromide fluoresces a bright orange color. This is due to the intercalation of ethidium bromide into the DNA double helix, enhancing its fluorescence, as described in NCERT XII Molecular Basis of Inheritance, so option (a) is correct.

Solution

Alfred Hershey and Martha Chase provided unequivocal proof that DNA is the genetic material through their bacteriophage infection experiments. NCERT XII chapter Molecular Basis of Inheritance highlights their work as a key piece of evidence, so option (c) is correct.

Solution

Expressed Sequence Tags (ESTs) are short subsequences of transcribed cDNA, representing all genes that are expressed as RNA. NCERT XII chapter Molecular Basis of Inheritance defines ESTs as partial sequences of cDNA, so option (b) is correct.

Solution

RNA polymerase III is responsible for the transcription of tRNA, 5S rRNA, and snRNA in eukaryotes. This is detailed in NCERT XII chapter Molecular Basis of Inheritance, making option (c) correct.

Solution

A transcription unit of DNA is defined primarily by the three regions in the DNA: (i) A promoter (ii) The structural gene (iii) A terminator The promoter is said to be located towards 5 ′-end (upstream) of the structural gene (the reference is made with respect to the polarity of coding strand) The terminator is located towards 3′-end (downstream) of the coding strand. - 48 - NEET (UG)-2024 (Code-Q1)

Solution

The correct answer is option (2). The first restriction endonuclease – Hind II, whose functioning depends on a specific DNA nucleotide sequence was isolated. It was found that Hind II always cut DNA molecules at a particular point by recognising sequence of six base pairs. Option (1), (3) and (4) are incorrect because they have either more than 6 or less than 6 bp.

Solution

Frederick Griffith series of experiment witness miraculous transformation in the bacteria. The elucidation of Lac operon was a result of a close association between geneticist, Francois Jacob and a biochemist, Jacques Monod. Meselson and Stahl gave semi-conservative mode of DNA replication. Har Gobind Khorana developed chemical method to define combination of bases in genetic code.

Solution

In Prokaryotes, like E.coli during replication, the DNA dependent DNA polymerase catalyse polymerization only in one direction, that is 5’ → 3’ ZOOLOGY SECTION-A

Solution

Template DNA is : 3’TACATGGCAAATATCCATTCA5’ 5’AUGUACCGUUUAUAGGUAAGU3’ m-RNA

Solution

In RNA world, RNA was the first genetic material as there are enough evidences to suggest that essential life processes (such as metabolism, translation, splicing, etc) evolved around RNA. RNA used to act as a genetic material as well as catalyst (there are some important biochemical reaction in living systems that are catalysed by RNA catalysts not by protein enzymes) so, statement I is correct statement II is also correct as DNA being double stranded and having complementary strands further resists changes by evolving a process of repair.

Solution

Chromosome 1 has the highest number of genes in the human genome, with over 2000 genes. NCERT XII chapter Molecular Basis of Inheritance provides this information, making option (c) correct.

Solution

Both transfer RNAs and ribosomal RNA interact with mRNA.

RNA interference (RNAi) takes place in all eukaryotic organisms as a method of cellular defence.

Solution

Post-transcriptional events in eukaryotic cells include the removal of introns and joining of exons (splicing), addition of a methyl group at the 5' end (capping), and addition of adenine residues at the 3' end (polyadenylation). Base pairing of two complementary RNAs is not a post-transcriptional event, so option (b) is correct.

Solution

The cutting of DNA by restriction endonucleases results in the fragments of DNA. These fragments can be separated by a technique known as gel electrophoresis.

The separated bands of DNA are cut out from the agarose gel and extracted from the gel piece. This step is known as elution. The DNA fragments purified in this way are used in constructing rDNA by joining them with cloning vectors.

• In gel electrophoresis, the DNA fragments separate (resolve) according to their size through sieving effect provided by the agarose gel. Hence, the smaller the fragment size, the farther it moves from cathode towards anode.

Solution

RNAi (RNA interference) takes place in all eukaryotic organisms as a method of cellular defense. This method involves silencing of a specific mRNA due to a complementary dsRNA molecule that binds to and prevents translation of the mRNA.

Solution

- Alfred Hershey and Martha Chase confirmed DNA as the genetic material (IV).
- Euchromatin is loosely packed and light-stained (III).
- Frederick Griffith worked with $\text{Streptococcus pneumoniae}$ (I).
- Heterochromatin is densely packed and dark-stained (II).

Solution

George Gamow, a physicist proposed that genetic code for amino acids should be made up of three nucleotides.

Solution

The process of copying genetic information from one strand of the DNA into RNA is known as transcription. It occurs in the cytoplasm with the help of transcribing enzyme.

Transport of pre-mRNA to cytoplasm prior to splicing is a part of transcription.

The primary transcript is converted into functional mRNA after post transcriptional processing involves 3 steps as follows-

• Modification of 5′ end by capping,

• Tailing,

• Splicing.

Base pairing of two complementary RNA is not on event of post-transcription. Hence, statements B, C, D are post-transcriptional modification events in eukaryotic cell.

Solution

The $\rho$ (rho) factor is a protein that plays a crucial role in the termination of transcription by facilitating the dissociation of the RNA polymerase from the DNA template. NCERT XII chapter Molecular Basis of Inheritance describes the function of the $\rho$ factor in transcription termination, so option (c) is correct.