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Molecular Basis of Inheritance Test - 1
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Molecular Basis of Inheritance Test - 1
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  • Question 1/10
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    Meselson and Stahl used which radio isotopes for their experiment?
    Solutions

    Concept-

    • Matthew Meselson and Franklin Stahl experiments that DNA replicates semi-conservatively.
    • Semi-conservative DNA means half of the DNA is conserved.

    Explanation-

    • They used heavy isotope (nitrogen-15) and Cairns used radioactive thymidine.
    • Due to the replication of DNA, two DNA molecules were found in which 50% radioactivity was discovered.
    • When these two DNA molecules were made to replicate, the next time four DNA molecules were produced. Out of this 4 DNA, 2 DNA molecules were radioactive and the other 2 are not.
    • And after that also, the  DNA molecules which is obtained were further replicated then also the number of radioactive DNA was 2.

     

    15N is the heavy isotope of nitrogen which is used by Meselson and Stahl.

    Additional Information

    • Hershey and Chase proved that DNA is the genetic material.
    • Hugo de vries ,Correns and Erich Von Tschermak rediscovered the mendel law.
    • Miller and Urey did the experiment that life could have come from pre exciting life.
  • Question 2/10
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    In the absence of Lactose, what is expected to happen according to lac operon model ?
    Solutions

    Concept-

    • Gene expression is the mechanism at the molecular level by which a gene can express itself in the phenotype of an organism.
    • The mechanism that stimulates the expression of certain genes and inhibits that of others is called the regulation of gene expression.
    • Francis Jacob and Monod proposed a mechanism called the operon model for the regulation of gene action in E.coli.
    • An operon is a part of genetic material that acts as a single regulated unit having one or more structural genes, a promoter gene, a regulator gene.
    • Operons are of two types- inducible and repressible.
    • The best-known operon is lac operon.

    Explanation-

    Explain the role of regulartory gene in a lac operon. Why is regulation of lac  operon called as negative regulation?

    • In the absence of lactose, the repressor protein binds to the operator region of the operon.
    • It prevents RNA polymerase from transcribing the operon.

    Thereby option B is the correct answer.

    Additional Information

    • In the presence of an inducer, such as lactose or allolactose, the repressor is inactivated by interaction with the inducer.
    • This allows RNA polymerase access to the promoter and transcription proceeds.
    • Essentially, regulation of lac operon can also be visualized as regulation of enzyme synthesis by its substrate. 
  • Question 3/10
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    Which of the following steps in transcription is catalyzed by RNA polymerase?
    Solutions

    Concept:

    • Transcription is the process of copying the genetic information from one strand of DNA onto the RNA.
    • It takes place with the help of the enzyme RNA polymerase (RNA pol).
    • It takes place at a defined region of DNA that is known as the transcription unit.
    • A transcriptional unit can be recognized by the following:
      • Template strand - It acts as the template for RNA pol to act on and has 3'→5' polarity.
      • Coding strand - It has 5'→3' polarity and is used as a reference strand for denoting upstream or downstream of a gene.
      • Structural gene - It contains the nucleotide sequence that actually codes for the required RNA.
      • Promoter - It is the nucleotide sequence that acts as a binding site for RNA pol enzyme.
      • Terminator - It is the nucleotide sequence that defines the end of transcription.
    • The promoter is located upstream of the gene, while the terminator is located downstream of the gene.
    • Upstream and downstream indicate towards the 5'-end and the 3'-end of coding strand respectively.

    Important PointsSteps of Transcription in Bacteria:

    • Initiation -
      • RNA pol enzyme binds at the promoter site of DNA with the help of sigma (σ) factor and initiates transcription.
    • Elongation -
      • The sigma factor is released and RNA pol moves along the DNA to untwist the DNA double helix.
      • RNA pol uses nucleoside triphosphates for polymerization of RNA that remains bound to the DNA in the transcription bubble.
      • Some of the newly synthesized RNA is released from the enzyme as a single strand.
      • Only a part of the RNA remains bound as DNA-RNA hybrid.
    • Termination -
      • When the RNA pol enzyme reaches the terminator regionRNA is unwound from the DNA.
      • It is carried out by the Rho (ρ) factor that binds to the terminator site.
      • The nascent RNA is released along with the enzyme.
      • The DNA double helix reforms and ρ factor dissociates, terminating the transcription process.

    Therefore, RNA polymerase catalyzes all the steps of transcription - initiation, elongation and termination.

  • Question 4/10
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    If a four nucleotide sequence code for an amino acid instead of three theoretically how many unique amino acids could be coded by such a system:-
    Solutions
    Concept:
    • Central dogma of molecular biology states that flow of genetic information takes place from DNA to RNA to proteins.
    • Transcription of DNA to RNA is based on the complementarity of their strands.
    • However, no such complementarity exists between the RNA and the proteins for translation to take place.
    • Evidences show that changes in nucleic acids also caused changes in the amino acids of the proteins.
    • This suggested the existence of a Genetic Code that directs the synthesis of amino acids that form the proteins.
    • It was George Gamow who suggested that if the 4 nitrogen bases of nucleic acids have to code for the 20 non-essential amino acids produced in the body, the code should constitute a combination of bases.
    • The nitrogen bases include Adenine (A), Guanine (G), Cytosine (C) and Uracil (U) for an RNA.
    • A permutation combination of 2 bases would give: 42 = 16 amino acids only.
    • So, it has to be a combination of 3 bases giving: 43 = 64 codons.

    Important Points

    • From the above points, we can get the formula as:

    (No. of N-bases available)No. of bases in per sequence code = No. of codons

    • Therefore, if we take 4 nucleotide sequence code,

    No. of codons = 44 = 256

    • Similarly if we take a 5 sequence code, we will get 45 = 1024 codons.
    • Please note that these are only theoretical possibilities.
  • Question 5/10
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    Histone octamer does not include
    Solutions

    Concept-

    • There is a set of positively charged, basic proteins called histones.

    Explanation-

    • The nucleosome, a section of DNA wrapped around a core of proteins, is the fundamental subunit of chromatin.
    • The chain of nucleosomes is then compacted further and forms a highly organized complex of DNA and protein called a chromosome.
    • Each nucleosome is comprised of a little less than two turns of DNA wrapped around a set of eight proteins called histones, which are known as a histone octamer. 
    • Each histone octamer is composed of two copies each of the histone proteins H2A, H2B, H3, and H4

    The number of cytosine bases in a DNA molecule is equal to the number of  .......... bases.

    Thereby, H1 is not a part of this octamer. It is connected to the wrapped DNA. 

    Additional Information

    • In the nucleus of each cell, the DNA molecule is packaged into thread-like structures called chromosomes.
    • DNA is the basic building block of the genetic material found in all living organisms.
    • A very long chain of DNA forms a chromosome.
  • Question 6/10
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    A nitrogenous base linked to a pentose sugar is known as
    Solutions
    Concept:
    • Deoxyribonucleic acid or DNA is a polynucleotide of deoxyribonucleotides.
    • Ribonucleic acid or RNA is a polynucleotide of ribonucleotides.
    • Thus nucleotides are the basic units of these molecules.
    • A nucleotide has 3 components -
    1. Pentose sugar - It is ribose in RNA and deoxyribose in DNA.
    2. Nitrogenous base - It is a nitrogen-containing molecule with the properties of a base. It is of 2 types: 
      • Purine - It includes Adenine (A) and Guanine (G).
      • Pyrimidine - It includes Cytosine (C), Thymine (T) and Uracil (U).
    3. Phosphate group
    • Thymine is present only in DNA and gets replaced by Uracil in RNA.
    • Therefore, a DNA strand will contain A, T, G and C bases, while RNA will contain A, U, G and C.

    Important Points

    • Nucleoside -
      • It is formed when a nitrogenous base is linked to the pentose sugar by N-glycosidic linkage.
      • The nucleosides are named as deoxyadenosinedeoxyguanosinedeoxycytidine or deoxythymidine, depending on the nitrogenous base.
    • ​Nucleotide -
      • It is formed when a phosphate group is linked to the 5'-OH of a nucleoside.
      • The linkage present is phosphoester linkage.
    • Polynucleotide -
      • It is the polymer of nucleotides formed by joining multiple nucleotides in a chain-like manner.
      • The linkage between two nucleotides is phosphodiester linkage.
    • Ribose -
      • It is the type of pentose sugar present in RNA.

  • Question 7/10
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    If the sequence of bases in a DNA strand is ATGCGCTGA, what would be the sequence of bases on its complementary strand?
    Solutions
    Key Points

    According to Watson and Crick (1953), DNA is a double helical structure with the following features:

    • It is made up of 2 polynucleotide strands.
    • The two chains have anti-parallel polarity: one has 5'→3' and the other has 3'→5'.
    • The two strands are known as complementary strands.
    • The bases of two strands project inside and pair through H-bonds and form base pairs.
    • Adenine pairs with Thymine by 2 H-bonds and Guanine with Cytosine by 3 H-bonds.
    • Thus, purine is always paired with pyrimidine, creating uniform distance between both strands.
    • It is a right-handed helix (clockwise) with a pitch of 3.4nm and about 10bp in each turn.

    Explanation:

    • The complementary bases always pair as:
      • A with T
      • G with C

    • Therefore, if the nucleotide sequence is ATGCGCTGA, the complementary strand sequence will be TACGCGACT.
  • Question 8/10
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    Griffith's experiment on 'transforming principle' was conducted with which of the following bacteria?
    Solutions
    Key Points
    • DNA was discovered by Friedrich Meischer (1869) as an acidic substance present in the nucleus.
    • By 1926, the search for genetic material was already narrowed down to the chromosomes present in the nucleus.
    • But the molecule acting as the genetic material was yet to be determined.
    • Frederick Griffith (1928) conducted a series of experiments with Streptococcus pneumoniae in search for the genetic material.
    • He found that S. pneumoniae produced 2 types of strains when grown in culture medium:
      • S strain- produced smooth shiny colonies because of a mucous polysaccharide coating.
      • R strain- produced rough colonies that lacked the mucous coating.

    Experiment:

    Observations:

    • Mice injected with S-strain died of pneumonia infection, while mice injected with R-strain did not.
    • Therefore, S-strain was the virulent (infectious) form.
    • Heat-killed S-strain bacteria did not cause any infection, so the mice lived.
    • When heat-killed S-strain bacteria were injected along with R-strain, the mice died and live S-strain bacteria were recovered from the dead mice.

    Conclusions:

    • From the above observations, Griffith concluded that R-strain had somehow been "transformed" by the heat-killed S-strain to become virulent.
    • Some substance was transferred from the heat-killed S-strain to the R-strain that enabled the R-strain bacteria to produce the mucous coating and become virulent.
    • This substance was named as the 'transforming principle'.

    Drawback:

    • The biochemical nature of the genetic material could not be defined.
  • Question 9/10
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    During mRNA synthesis, the unwanted DNA regions are removed and amino acid coding regions are joined by the process of:
    Solutions

    Concept:

    • The process of copying genetic information from one strand of DNA into RNA is termed transcription or mRNA synthesis.
    • It is completed in three steps Initiation, Elongation & Termination
    • The mRNA formed after transcription is non-functional, containing both the coding region, exon, and the non-coding part, intron. This type of RNA is called heterogenous RNA or hnRNA.
    • This hnRNA is unstable and cannot be directly translated into proteins. It has to undergo some post-transcriptional modifications to become stable for translation.
    • The post-transcriptional modification includes CappingTailing & Splicing.

    Explanation:

    • Capping: The process of addition of an unusual nucleotide, 7 methyl guanosine to the 5' end of the hnRNA is called capping. It forms a cap-like structure on the 5' end of hnRNA.
    • Tailing: The process of addition of 200-300 nucleotides of adenylic acid to the 3' end of hnRNA is called tailing. This tail is called poly-A tail.
    • Splicing: After capping & tailing now the hnRNA undergoes a process where the non-coding regions, introns are removed and the coding region exons are joined together to form the functional mRNA by the process called splicing.
    • The removal of introns is done with the help of the ribonuclease enzyme and the exons are joined with the help of enzyme DNA ligase.

    Additional Information

    • Termination: The transcription is completed in three steps Initiation, Elongation & Termination. Termination is the final step. Once the RNA polymerase reaches the termination region of DNA, the RNA polymerase is separated from the DNA-RNA hybrid, as a result, nascent RNA separates. This process is called termination which is facilitated by a termination factor ρ (rho).
    • Methylation: DNA methylation is a biological process by which methyl groups are added to the DNA molecule. Methylation can change the activity of a DNA segment without changing the sequence. When located in a gene promoter, DNA methylation typically acts to repress gene transcription. 
  • Question 10/10
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    The technique for DNA fingerprinting was developed by
    Solutions
    Key Points
    • DNA fingerprinting was developed by Alec Jeffreys in 1984.
    • It is based on DNA polymorphism.
    • Humans are almost 99.9% identical and the rest 0.1% constitutes about 3 million differences in our genome.
    • DNA polymorphism - These are inheritable mutations that occur at a high frequency in a population.
    • Such variations occur frequently in non-coding DNA sequences and keep accumulating over generations.
    • These mutations are the basis of variability or polymorphism in a population.
    • Satellite DNA shows high degree of polymorphism and hence, was used by Jeffreys for DNA fingerprinting.
    • The DNA probes used were VNTRs.
    • VNTR- Variable Number of Tandem Repeats or VNTRs are composed of 8-80 bp sequences which are tandemly repeated to form 1-30kb length of DNA.

    Procedure:

    1. Isolation of DNA
    2. Digestion of DNA by restriction endonuclease
    3. Separation of DNA fragments by gel electrophoresis
    4. Transferring of separated DNA fragments to synthetic membranes like nylon and nitrocellulose - Southern Blotting technique
    5. Hybridization by using radio-labelled VNTR probes.
    6. Detection of hybridized DNA fragments by autoradiography.

    Additional Information

    • Frederick Sanger - gave the sequencing method for determination of amino acid sequences in proteins.
    • Frederick Griffith - performed the transforming principle experiment for determining the genetic material.
    • Jacob and Monod - proposed the lac operon model for transcriptionally-regulated gene expression.
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