Molecular Biology MCQ 2

21 Questions MCQ Test Mock Test Series of IIT JAM Biotechnology | Molecular Biology MCQ 2

This mock test of Molecular Biology MCQ 2 for IIT JAM helps you for every IIT JAM entrance exam. This contains 21 Multiple Choice Questions for IIT JAM Molecular Biology MCQ 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this Molecular Biology MCQ 2 quiz give you a good mix of easy questions and tough questions. IIT JAM students definitely take this Molecular Biology MCQ 2 exercise for a better result in the exam. You can find other Molecular Biology MCQ 2 extra questions, long questions & short questions for IIT JAM on EduRev as well by searching above.

Most ribozymes are not true enzymes because :-


One of the characteristics of the enzymes is that they are not consumed in the reaction they catalyse, but this is not true for ribozymes which get consumed in the reaction.


Which of the following proteins moves along a dsDNA, unwinds the strands ahead and degrade them, during homologous recombination is


Rec BCD is an enzyme which initiates recombinational repair, wherever there is double stranded break in DNA. RecBCD acts both as a helicase, which separates the two strands of DNA and nuclease which makes single stranded nicks in DNA. RecA is then recruited on to these single stranded nicks which proceed with the repair.


Which of the following codons represent the principle of degeneracy?


According to the principle of degeneracy of genetic codons, more than one codon code for the same amino acid. Out of the given options, only codon GGU and GGC code for the same amino acid i.e. Glycine. On the other hand AUG codes for methionine, AUC codes for isoleucine, UAA and UAG are stop codons, and UAC codes for tyrosine.


The ‘cap’ on a mRNA has a guanine ribonucleotide that is connected in a


One of the post- transcriptional modification of eukeryotic mRNA is the addition of cap at its 5' end. This cap consists of a guanine nucleotide connected  to mRNA via an unusual 5' to 5' triphosphate linkage. Also this guanine is methylated at position 7, thereby forming 7-methylguanosine cap.


Match the following


Griffith first showed the process of transformation through different strains of Streptococcus pneumoniae. Avery- MacLeod showed that DNA is the transforming factor in Griffith’s experiment. Hershey and Chase experiment using a bacteriophage further proved that DNA and not protein is the genetic material. Watson and Crick described the double helical model of the DNA.


Why does enzyme DNA polymerase needs a primer for synthesis of a new strand?


DNA polymerase cannot synthesize a new strand de novo. The enzyme functions by adding the nucleotides to the free 3’-OH groups of the preexisting nucleotide. Thus a primer, which is a short strand of RNA or DNA consisting of 18-20  bases. 


Gene regulation through RNA interference can be described by which of the following statement?


Small interfering RNAs are a class of double stranded RNA molecules, which play important roles in gene silencing through RNA interference pathway. Structurally, they are formed after long dsRNA is cleaved by enzyme Dicer. Once siRNA enters inside the cell through transfection, it is incorporated into other proteins to form RISC (RNA induced silencing complex). Inside the RISC, siRNA is unwound to form ssRNA. The ssRNA, which is a part of RISC scans and finds a complementary mRNA and induces its cleavage and thus silences the corresponding gene.


The Ames test is


The Ames test is a widely employed biological assay which is used to check whether the chemical under study can induce the mutations in the test organism or not. This test is neither employed to check the potency of various antibiotics, nor to determine the rate of DNA replication.


The methyl directed repair system in bacteria


Methyl directed repair is a type of mismatch repair system, whereby inserted, deleted and mismatched bases are detected and repaired. During this repair, the repair machinery recognises hemimethylated DNA (methylated placental strand and non-methylated daughter strand), removes the mismatched nucleotide from daughter strand and the gap so created is repaired by DNA polymerase III.


What is the role of eukaryotic RNA polymerase I?


In eukaryotes, different RNA polymerases are specialised for the synthesis of different types of RNA for Eg. RNA polymerase I synthesizes 28S, 18S and 5.85 rRNAs. RNA polymerase II synthesizes mRNAs and RNA polymerase III synthesizes tRNAs and 5Sr RNAs.


Topoisomerases :


Topoisomerases are the enzymes which are involved in underwinding or overwinding of DNA. All these enzymes bind to double stranded DNA, make an incision in one or both strands of DNA, pass the DNA strands through one another and then seal the breaks. These enzymes always require energy, but the source can be NAD or ATP.


Formation of Z-DNA is favored by


Z-DNA is one of the biologically active double helical structures. It is a left handed helix and is formed under conditions of high salt concentration. Structurally, it consists of alternating purine and pyrimidine residues and is formed at a pH around 7.4 

*Multiple options can be correct

Which of the following are properties of enzyme telomerase ?


Telomerase is an enzyme which adds TTAGGG repeact sequences at the 3' end of telomeres in eukaryotes and prevents telomere shortening. This enzyme has its own RNA, which serves as a template for telomere synthesis. Telomerase works in a cyclic fashion, after every cycle, it realigns the 3' end of telomeres to the template and repeat the process. It has no role in the synthesis of RNA primers    

*Multiple options can be correct

Which of the following statements are incorrect?


Out of the given options, statements 1 and 3 are incorrect. All the parts of DNA are replicated equally, similarly, once the final mRNA is formed after post transcriptional modifications, the entire mRNA is translated equally. But, we can say that according to the needs of a cell, some parts of DNA are transcribed more, some transcribed less and some not transcribed at all.

*Multiple options can be correct

Which of the following are needed in the transcription process?


RNA polymerase itself unwinds the DNA and does not need another helicase enzyme for this purpose.

*Multiple options can be correct

What are the correct statements about tRNA?


The rRNA has complementary regions where it pairs, and the other regions that do not pair form loops. Thus, there are 4 complementary regions in the tRNA which pair up to form double stranded regions. One of the loops has an anticodon at one end which recognizes the codon of mRNA during the process of translation.

*Multiple options can be correct

Which of the following statements are correct?


The heteronuclear RNA or hnRNA is the RNA before splicing has occurred. They are pre-mRNAs that contain both introns and exons, i.e. the total DNA information in a contiguous (adjacent) manner. UTRs or untranslated regions are present either upstream(5’-UTR) or downstream (3’-UTR) to the coding segment of the gene. These do not form part of the proteins but are very much part of the exons and are retained after splicing.

*Answer can only contain numeric values

Protein synthesis rates in prokaryotes are limited by the rate of mRNA synthesis. If RNA synthesis occurs at the rate of 50 nucleotides/sec, then rate of protein synthesis occurs at ________ amino acids/sec.


If the nucleotide is synthesized at a rate of 50  nucleotides per second, then the number of codons would be 50/3. This corresponds to 16 codons and 2 nucleotides. Hence, a peptide of 16 amino acids would form in one sec. Thus, the rate at which protein is synthesized would be 16 amino acids/sec

*Answer can only contain numeric values

Approximate molecular weight (kDa) of the product after translation of a 390 bases mRNA will be __________KDa


There are 390 bases in mRNA and they lead to the synthesis of 390/3 = 130 amino acids after translation. Average weight of an amino acid is 110 Daltons. Hence the weight of translated product is 130*110 = 14300 Daltons or ~14KDa.

*Answer can only contain numeric values

Multiple RNA polymerase transcribes a DNA template, unwinding about 1.5 turns of DNA template per transcription bubble. From the structural information of classical B-DNA the number of transcription bubbles for a 180 base pair DNA molecule will be____________ 


Number of base pairs present per turn are 10. Hence in a stretch of 180 base pairs we get 180/10 = 18 turns. 1.5 turns makes one transcription bubble hence when there are 18 turns we have 18/1.5 = 12 transcription bubbles.

*Answer can only contain numeric values

The value of linking number for a closed circular DNA molecule of 2000 bp, when DNA is underwound by 20 enzymatic turnovers by DNA gysase (+ ATP) _____________ [Answer in Integer]
Assumption: There are 10 base pairs/ turn in relaxed DNA


When DNA is relaxed

DNA gyrase underwinds the DNA by 2 turns in each cycle
So. it would underwind the DNA by 2 x 20 = 40 nuns in 20 cycles
Hence, the new linking number would be Lk = Lk0 - 40 = 160

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