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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
Hershey-Chase experiment
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If DNA contained sulfur instead of phosphorus and proteins contained phosphorus instead of sulfur, what would have been the outcome of Hershey and Chase experiment?The genetic material (DNA) enters the bacterial cell. If DNA contained sulfur, radioactive sulfur would be found inside the bacterial cells.
Lac operon
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In lac operon, z gene codes for:The z gene codes for $\beta$-galactosidase, the y gene for permease and the a gene for transacetylase.
DNA replication
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Given below are two statements. Statement I: DNA polymerases catalyses polymerisation only in one direction, that is $5'\rightarrow3'$ Statement II: During replication of DNA, on one strand the replication is continuous while on the other strand it is discontinuous. In the light of the above statements, choose the correct answer from the options given belowDNA-dependent DNA polymerase catalyses polymerisation only in the $5'\rightarrow3'$ direction. Hence replication is continuous on the template strand with polarity $3'\rightarrow5'$ and discontinuous on the template strand with polarity $5'\rightarrow3'$. Both statements are correct.
Lac operon genes
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Match List - I with List - II. Choose the correct answer.
Pick the matching item from Column II for each row.
Column I
- A. In lac operon i gene codes for
- B. In lac operon z gene codes for
- C. In lac operon y gene codes for
- D. In lac operon a gene codes for
Column II
- I. transacetylase
- II. permease
- III. $\beta$-galactosidase
- IV. Repressor
Correct answer: A → IV, B → III, C → II, D → I
In the lac operon, the regulator i gene codes for the repressor protein; the structural genes z, y and a code for $\beta$-galactosidase, permease and transacetylase respectively.
Length of DNA in different organisms
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Match List - I with List - II. Choose the correct answer.
Pick the matching item from Column II for each row.
Column I
- A. Bacteriophage $\phi$X174
- B. Bacteriophage Lambda
- C. Escherichia coli
- D. Haploid content of human DNA
Column II
- I. 48502 base pairs
- II. 5386 nucleotides
- III. $3.3\times10^9$ base pairs
- IV. $4.6\times10^6$ base pairs
Correct answer: A → II, B → I, C → IV, D → III
Bacteriophage $\phi$X174 has 5386 nucleotides (ssDNA); bacteriophage lambda has 48502 base pairs; Escherichia coli has $4.6\times10^6$ bp; the haploid content of human DNA is $3.3\times10^9$ bp.
Codon-anticodon pairing
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Against the codon 5′ UAC 3′, what would be the sequence of anticodon on tRNA?The anticodon pairs antiparallel with the codon. For mRNA codon 5′ UAC 3′, the anticodon is 3′ AUG 5′, i.e. written 5′ GUA 3′.
Chargaff's rules
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If A and C make 30% and 20% of DNA, respectively, what will be the percentage composition of T and G?According to Chargaff's rules, A = T and G = C. So T = 30% and G = 20%.
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