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Molecular Basis of Inheritance Test - 88

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Molecular Basis of Inheritance Test - 88
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  • Question 1
    1 / -0
    In a transcription, unit promoter is said to be located towards
    Solution
    • Promoter sequences are DNA sequences in the region in the gene where the RNA polymerase will bind and initiate the process of transcription. 
    • They are located upstream at the 5' end of the transcription initiation site in the structural gene. 
    • Promoter sequences define the direction of transcription and indicate the DNA strand that will be transcribed. 
    • The strand which is from 5'-3' is the non-template strand and the strand which runs from 3'-5' is known as the template strand. 

    Thus, the correct answer is option B.

  • Question 2
    1 / -0
    Assume that there are 6 types of nitrogen bases available and 40 types of amino acid are available for protein synthesis, then in genetic code each codon made up by minimum how many nitrogen codon made up by minimum how many nitrogen bases?
    Solution
    The genetic code consists of 64 triplets of nucleotides. These triplets are called as codons. With three exceptions, each codon encodes for one of the 20 amino acids used in the synthesis of proteins. That produces some redundancy in the code. Most of the amino acids being encoded by more than one codon. The genetic code can be expressed as either RNA codons or DNA codons. RNA codons occur in messenger RNA (mRNA) and are the codons that are actually "read" during the synthesis of polypeptides (the process called translation). But each mRNA molecule acquires its sequence of nucleotides by transcription from the corresponding gene. Because DNA sequencing has become so rapid and because most genes are now being discovered at the level of DNA before they are discovered as mRNA or as a protein product, it is extremely useful to have a table of codons expressed as DNA. So here are both. So, in genetic code each codon is made of 3 bases. Thus, option A is correct and other options are wrong.
  • Question 3
    1 / -0
    Which statement is correct?
    (A) Degeneracy of code is related to third member of codon.
    (B) Single codon codes for more than one amino acid.
    (C) In codons, first two bases are more specific.
    (D) In codons, third base is wobble.
    (E) Code is universal.
    Solution
    All the statements are correct except for (B) because, the genetic code expresses in a way that 64 Codons constitute it, as it occurs in triplets. According to the genetic code, three bases must be employed to encode the 20 standard amino acids used by living cells to build proteins. With four different nucleotides, a code of 2 nucleotides would allow for only a maximum of 42 = 16 amino acids. A code of 3 nucleotides could code for a maximum of 43 = 64 amino acids
  • Question 4
    1 / -0
    Out of 64 codons only 61 codes for the 20 different amino acids. This character of genetic code is called as
    Solution

    A) Correct option A
    B) Explanation of the correct option

    • The genetic code is degenerate which means that most amino acids are specified by more than one codon.
    • As we know that there are only four nucleotides namely deoxyadenosine, deoxyguanosine, deoxythymidine and deoxycytidine.
    • Further, there are 20 essential amino acids that are to be encoded by these four nucleotides.
    • A triplet code could make a genetic code for 64 different combinations (4 X 4 X 4) genetic code.
    • Thus, there are ample codons to specify 20 amino acids. 

  • Question 5
    1 / -0
    Arabidopsis thaliana and Caenorhabditis elegans has one thing common. Which is it?
    Solution
    Arabidopsis thaliana is a small flowering edible plant native to Eurasia. It was the first plant to have its genome sequenced for understanding the molecular biology of many plant traits.
    Caenorhabditis elegans is a free-living (not parasitic), transparent nematode (roundworm) that lives in temperate soil environments. It was the first multicellular organism to have its whole genome sequenced, and as of 2012, the only organism to have its connectome completed.
  • Question 6
    1 / -0
    DNA in the nucleus of rabbit egg is 1.8 picograms, its liver cell will have mass of DNA ______ picograms.
    Solution
    Egg and sperm are the germ cells that are produced by meiosis, a type of cell division with two consecutive nuclear division after one round of chromosome replication to produce four daughter cell with half the chromosome number as compared to the parent cell. Thus, egg and sperms are haploid cells in otherwise diploid organism. Rabbit is a diploid organism and its egg represent the haploid DNA content (1.8 picograms). Liver cells are diploid cells and hence, will have double the DNA content as compared to the egg cell, i.e., 1.8 X 2= 3.6 picograms. Thus, the correct answer is D.
  • Question 7
    1 / -0
    Synthesis of DNA can be stopped by incorporating artificial
    Solution
    • Primer is a stranded segment (complementary to the template) and has a free 3’ hydroxyl group to facilitate the addition of nucleotides by polymerase enzymes. 
    • DNA polymerase catalyzes the formation of the phosphodiester linkage between 3’ hydroxyl group of primer and $$\alpha$$-phosphate of the incoming nucleoside triphosphate. 
    • Dideoxyribonucleotide triphosphates (ddNTP) are modified nucleotide that has the same structure as a normal dNTP but with the exception of the 3' OH group being replaced by an H; thus not providing the free 3’OH group.
    • DNA polymerase senses the presence of modified monomers and immediately stops DNA synthesis. 
    • As termination of DNA synthesis is due to the absence of free 3’ OH group, any of four modified nucleotides produce the same effect. 

    Thus, the correct answer is option D.
  • Question 8
    1 / -0
    Nucleoside is 
    Solution

    Correct Option: A
    Explanation:

    • Nucleotides comprise a nitrogenous base, a pentose sugar and a phosphate. 
    • The molecule without the phosphate group is called a nucleoside. 
    • Deoxyadenosine, deoxyguanosine, deoxycytidine are few examples of nucleosides while adenylate, guanylate, uridylate, cytidylate are few examples of nucleotides. 
    • Hence, a nucleoside is a nitrogenous base + sugar.
  • Question 9
    1 / -0
    Broad spectrum antibotics like streptomycin bind to smaller subunits of ribosomes of prokaryotes thereby
    Solution
    Any antibiotic, irrespective of its interaction with 30S or 50S ribosomal subunits, inhibit proteins synthesis; option A does not specify effect of binding to 30S subunit. Broad spectrum antibiotics, streptomycin and neomycin, are aminoglycosides that bind to sites on the 30S subunit and causes the misreading of genetic code that leads to abnormal proteins; thus they inhibit peptide initiation and elongation, option C is correct. Antibiotics that attach to sites on the 50S subunit prevents the formation of peptide bonds (chloramphenicol) or inhibits translocation of the subunit during translation (erythromycin); option B is incorrect. Point mutation is caused by changes in single nucleotide of gene; substitution, deletion or insertion of bases are responsible for point mutation. Binding of antibiotics to ribosomal subunits does not affect the gene structure; option D is incorrect. Option C is correct. 
  • Question 10
    1 / -0
    If base sequence is GAG, AGG, GGA, CCA. It is the segment of 
    Solution
    Both DNA and RNA have four bases, that includes two purines and two pyrimidines, in the nucleotide chains. There are two purines namely adenine and guanine and three pyrimidines namely cytosine, thymine and uracil. The purines adenine and guanine and the pyrimidine cytosine are common to both DNA and RNA. DNA has the pyrimidine thymine; RNA has the pyrimidine uracil. Transfer-RNA has 7-15% post-transcriptionally modified bases like dihydrouridine (D), ribothymidine (T) and pseudouridine. In the given sequence, "GAG,AGG,GGA,CCA" thymine or uracil or the abnormal bases are not shown. It shows only the bases which are common to DNA/mRNA/tRNA; thus, it can be a sequence of DNA/mRNA/tRNA. 
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