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Biology
Molecular Basis of Inheritance
200 questions from 38 papers, each with its answer.
- 200 questions
- 38 papers
- 5.67% of Biology
RNA Polymerases, Termination Factors and RNA Splicing
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Match List-I with List-II regarding transcription machinery and processing.
Pick the matching item from Column II for each row.
Column I
- A. RNA polymerase III
- B. Termination of transcription
- C. Splicing of Exons
- D. Tata box
Column II
- I. snRNPs
- II. Promotor
- III. Rho factor
- IV. SnRNAs, tRNA
Correct answer: A → IV, B → III, C → I, D → II
RNA polymerase III – transcribes tRNA, 5S rRNA and snRNAs Termination of transcription – Rho factor Splicing of exons – snRNPs TATA box – promoter region Correct match: A-IV, B-III, C-I, D-II.
Regulation of Gene Expression – Lac Operon
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In an E.coli strain $i$ gene gets mutated and its product can not bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome?
DNA Replication – Meselson and Stahl Experiment
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Ten E.coli with $^{15}N$-dsDNA are incubated in medium containing $^{14}N$ nucleotide. After 60 minutes, how many E.coli cells will have DNA totally free from $^{15}N$?
Translation
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During translation initiation in prokaryotes, a GTP molecule is needed in : -Initiation of translation in prokaryotes takes place in steps, with the help of initiation factors IF1, IF2 and IF3.
First the 30S subunit (with IF3) binds to the mRNA at the Shine–Dalgarno sequence. This step does not need GTP.
Next, the initiator tRNA (fMet-tRNA$^{fMet}$) is brought to the start codon AUG on the 30S–mRNA complex by IF2, which is bound to GTP (IF2–GTP–fMet-tRNA). So GTP is needed for the association of fMet-tRNA with the 30S–mRNA complex, forming the 30S initiation complex.
Finally the 50S subunit joins; the GTP bound to IF2 is hydrolysed to GDP + $P_i$ at this time, and the initiation factors are released.
Formation of formyl-met-tRNA (charging and formylation) uses ATP, not GTP.
So, as per the answer key, GTP is needed in the association of the 30S–mRNA complex with formyl-met-tRNA.
Genetic Code – Frameshift Mutations
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Under which of the following conditions will there be no change in the reading frame of following mRNA? $5'$AACAGCGGUGCUAUU$3'$
Structure of a transcription unit
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Which of the following statements are correct with reference to a transcription unit? I. A transcription unit in DNA is defined primarily by three regions: promoter, structural gene and terminator. II. The promoter is said to be located towards the 5'-end of the structural gene. III. The promoter is a DNA sequence that provides binding site for RNA polymerase. IV. The promoter defines the template and coding strands. V. The terminator is located towards the 3'-end of the coding strand and it defines the end of the process of transcription. Choose the CORRECT answer from the options given below:
DNA packaging - nucleosome and histones
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Which of the following statements are correct with reference to packaging of DNA helix? I. Histones are organized to form a unit of eight molecules called histone octamer. II. Histones are negatively charged basic proteins. III. Histones are rich in the basic amino acid residues - lysine and arginine. IV. The positively charged DNA is wrapped around the histone octamer to form nucleosome. V. The packaging of chromatin at higher levels requires an additional set of proteins called non-histone chromosomal proteins. Choose the CORRECT answer from the options given below:
Translation and its Inhibitors
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Which antibiotic inhibits interaction between tRNA and mRNA during bacterial protein synthesis ?Many antibiotics act by blocking a particular step of protein synthesis on the 70S ribosome of bacteria.
Neomycin is an aminoglycoside antibiotic. It binds to the 30S subunit and, according to the answer key of this paper, it inhibits the interaction between tRNA and mRNA (the codon-anticodon interaction).
Streptomycin, another aminoglycoside, inhibits the initiation of translation and causes misreading of the mRNA.
Erythromycin binds to the 50S subunit and inhibits the translocation of the ribosome along the mRNA.
Tetracycline inhibits the binding of aminoacyl-tRNA to the A site of the ribosome.
So the answer given is neomycin.
Structure of Chromosome
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Which of the following occurs more than one and less than five in a chromosome : -Each linear chromosome has two ends, and each end carries a telomere. So a chromosome has two telomeres (or four, counting both chromatids of a duplicated chromosome), i.e. more than one and less than five.
A chromosome normally has only one centromere.
Chromomeres are bead-like condensed regions along a chromosome, and their number is very large.
Chromatids are two per duplicated chromosome, but the answer key considers telomeres here.
So, as per the answer key, telomeres occur more than one and less than five in a chromosome.
Semiconservative Replication
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E. coli about to replicate was placed in a medium containing radioactive thymidine for five minutes. Then it was made to replicate in a normal medium. Which of the following observation shall be correct : -DNA replication is semiconservative: each daughter DNA molecule has one old (parental) strand and one newly synthesised strand.
During the five minutes in radioactive thymidine, the new strands being made take up the label, so the DNA formed has one radioactive (new) strand and one non-radioactive (old) strand.
When this cell then replicates in a normal medium, each labelled strand acts as a template for a new unlabelled strand. So DNA molecules carrying the label have only one radioactive strand.
So, as per the answer key, one strand of the DNA will be radioactive.
Gene Regulation in Eukaryotes
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At the time of organogenesis genes regulate the process at different levels and at different times due to :During development (organogenesis), different genes are switched on and off in different cells and at different times.
In eukaryotes, the coding sequences (exons) of a gene can be joined in different combinations by alternative splicing, so one gene can give different proteins in different tissues and at different stages.
As per the answer key, this regulation at different levels and times is attributed to exons.
(This answer is debatable: control of when and where genes are expressed is usually attributed to promoters and regulatory sequences.)
So, as per the answer key, the answer is exon.
Complementation Test
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A mutant strain of T4 bacteriophage, r-II, fails to lyse E. coli but when two strains r-IIx and r-IIy are mixed then they lyse E. coli. What may be the possible reason : -Seymour Benzer studied rII mutants of T4 phage, which cannot grow on E. coli strain K.
When two different rII mutants infect the same cell together, they can complement each other only if the mutations are in different functional units (cistrons A and B). Each phage supplies the product that the other lacks, so the bacterium is lysed.
If both mutations were in the same cistron, neither could supply that function and there would be no lysis.
So the two strains lyse E. coli together because they have mutations in different cistrons.
That is every question in Molecular Basis of Inheritance for this filter.
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