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Molecular Biology MCQ 1


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21 Questions MCQ Test Mock Test Series of IIT JAM Biotechnology | Molecular Biology MCQ 1

Molecular Biology MCQ 1 for IIT JAM 2022 is part of Mock Test Series of IIT JAM Biotechnology preparation. The Molecular Biology MCQ 1 questions and answers have been prepared according to the IIT JAM exam syllabus.The Molecular Biology MCQ 1 MCQs are made for IIT JAM 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Molecular Biology MCQ 1 below.
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Molecular Biology MCQ 1 - Question 1

Retrotransposons are different from DNA transposons in that they

Detailed Solution for Molecular Biology MCQ 1 - Question 1

Transposable element (TE) is a DNA sequence which can change its position in a genome, and often results in the duplication of the same genetic material. There are two types of TEs: (i) Retrotransposons and (ii) DNA transposons. Retrotransposons employ reverse transcription, whereby DNA sequences are first transcribed into RNA and then back into identical DNA sequences which are inserted in the genome. On the other hand, DNA transposons does not involve RNA intermediate, rather they encode transposase enzyme, which helps in their insertion or excision.

Molecular Biology MCQ 1 - Question 2

Alu element is a type of

Detailed Solution for Molecular Biology MCQ 1 - Question 2

An Alu element, a short stretch of DNA is the most abundant transposable element distributed throughout the human genome. Since their size is approximately 300 base pairs, so they are classified as short interspersed nuclear elements (SINEs).

Molecular Biology MCQ 1 - Question 3

In E.coli, which component of translation machinery carries out the function for the given statements.
(i) Recognition of termination codons UAG and UAA
(ii) Recruitment of f-met - tRNA on small ribosomal subunit

Detailed Solution for Molecular Biology MCQ 1 - Question 3

Prokaryotes have two class I release factors :-
(a)  RF1– recognises stop codons UAG and UAA
(b)  RF2– recognises stop codons UGA and UAA    
Out of the three translation initiation factors, initiation factor 2 (IF2) is a GTP ase, which in a GTP  bound state helps in the recriutment of f-met- tRNAif-met to the small ribosomal subnuit.

Molecular Biology MCQ 1 - Question 4

Deletion mutations would cause frameshift mutations. In which of the following would their effect be the most?

Detailed Solution for Molecular Biology MCQ 1 - Question 4

The codons are read in frames and each frame specifies an amino acid. Deletions would change these reading frames and the tRNAs would recognize wrong codons and an entirely different polypeptide chain would be formed. Thus, the effect would be seen during translation.

Molecular Biology MCQ 1 - Question 5

Match the transcription factors (Column A) with their functions (Column B)

Detailed Solution for Molecular Biology MCQ 1 - Question 5

TFIIB is a general transcription factor which helps in the formation of pre-initiation complex and helps in transcription initiation. 'IFIIA binds with TBP (TATA binding protein) subunit of TFIIH and thus stabilizes TBP-DNA binding. TFIIE helps in the recruitment of TFIIH to initiation complex. TFIIH helps in ATP dependent transition of pre-initiation complex to open complex i.e. promoter melting. 

Molecular Biology MCQ 1 - Question 6

Which of the following indicates that primitive life forms lacked both DNA and enzymes?

Detailed Solution for Molecular Biology MCQ 1 - Question 6

According to RNA world hypothesis, in the primitive life, RNA catalysts called ribozymes could have played key roles, eg. they could catalyse chemical reactions to copy themselves and such replicating RNA could pass genetic material from generation to generation, thereby fulfilling the most basic criteria for life. Thus, these two roles i.e., their catalytic nature and ability to act as a hereditary material justify the evidence for lack of DNA and enzymes in primitive cells.

Molecular Biology MCQ 1 - Question 7

Why do genetic mutations often lead to a disease?

Detailed Solution for Molecular Biology MCQ 1 - Question 7

Most of the protein functions depend upon the shape protein adopts, because that shape allows the catalytic groups to be oriented in an accurate way for the reaction to occur. Mutations often lead to a change in amino acid sequence of protein, as a result of which protein does not fold properly and its shape changes and with this change in shape, the function of protein is compromised.

Molecular Biology MCQ 1 - Question 8

If an intron fails to get removed from an mRNA, and it is 25 nucleotides long, then how many more amino acids would the final polypeptide contain?

Detailed Solution for Molecular Biology MCQ 1 - Question 8

25 nucleotides would mean 8 codons and 1 nucleotide. That means 8 more amino acids would be added to the polypeptide. So, the answer is 8

Molecular Biology MCQ 1 - Question 9

Match the translation inhibitor antibiotics(column A) with their mode of action (column B)

Detailed Solution for Molecular Biology MCQ 1 - Question 9

Chloramphenicol binds to 50S subunit of ribosomes and does not let the peptide bonds to form. Tetracycline binds to both 305 and 50S subunit of ribosomes and prevent the attachement of amino- acyl tRNAto A-site Diptheria toxin inhibits elongation by inactivating elongation factor. EF-Tu. Cyclohexemide is a eukaryotic translation inhibitor which inbits translation step by affecting peptdyl transferase reaction.

Molecular Biology MCQ 1 - Question 10

Which of the following is incorrect for tryptophan operon?

Detailed Solution for Molecular Biology MCQ 1 - Question 10

All the statements regarding tryptophan operon are correct, except statement (d) i.e. repressor binds to the operator in absence of tryptophan. Actually, when tryptophan is present in the medium, it binds to the repressor, changes its conformation and recruits it onto the operator, thereby stopping the expression of tryptophan operon.

Molecular Biology MCQ 1 - Question 11

During mRNA splicing:

Detailed Solution for Molecular Biology MCQ 1 - Question 11

mRNA splicing is a process whereby introns are  excised  and exons are joined to from mature mRNA. There are 2 types of introns ® Group I and Group II. Splicing process involves non- ATP dependent 2 transesterification reactions, whereby 3'OH of  free G nucleotide attacks 5’ splice site and intron junction in Group I introns and an 2'OH of an internal A nucleotide attacks the some junction in Group II introns.

Molecular Biology MCQ 1 - Question 12

What would be mRNA from the following DNA sequence?
5' G A T C A C C G A T C A T 3'

Detailed Solution for Molecular Biology MCQ 1 - Question 12

The DNA strand given in the question is non-template strand and the information to form mRNA lies in the template strand which in complementary to non- template strand. So the sequence of template strand would be 3’ C T A G T G G C T A G T A 5' and mRNA sequence being complementary to template strand would be
5’ G A U C A C C G A U C A U 3' or 3' U A C U A G C C A C U A G 5'

*Multiple options can be correct
Molecular Biology MCQ 1 - Question 13

Which of the following functions does DNA packaging serve?

Detailed Solution for Molecular Biology MCQ 1 - Question 13

DNA packaging serves all the purposes in an organism, except that it helps the DNA replicate easily. DNA packaging makes the DNA compact, thereby making it inaccessible for replication / transcription machinery. DNA needs to be unpackaged for efficient replication to occur within the cell.

*Multiple options can be correct
Molecular Biology MCQ 1 - Question 14

Which of the following events occur when E. coli is straved for tryptophan ?

Detailed Solution for Molecular Biology MCQ 1 - Question 14

In the absence of tryptophan, the ribosomes would stall at trp codons in the leader sequence present in region1. This allows region 2 and 3 to pair and from an anti-termination loop, which allows transcription to continue into trp EDCBA genes, which code the enzymes for tryptophan synthesis. Since tryptophan is not present, so tryptophanyl- tRNA would not be formed, hence options b and c are  correct.

*Multiple options can be correct
Molecular Biology MCQ 1 - Question 15

Binding of RNA polymerase to its specific promoter in E.coli:-

Detailed Solution for Molecular Biology MCQ 1 - Question 15

In E.coli, RNA polymerase is recruited onto the promoters of genes with the help of sigma factor, which plays a role in the promoter identification. Near  the transcription start site, six  nucleotide rich A+T base pair sequence (TATA sequence) is present, which because of its lower melting temperature, initiates the unwinding of ds DNA, so that transcription machinery can act on it 

*Multiple options can be correct
Molecular Biology MCQ 1 - Question 16

Consider the following statements
P) The length of the DNA is directly proportional to the gene density
Q) DNA proofreading by RNA polymerase is done by backtracking
R)  Splicing of nuclear mRNA occurs in the nucleus

Detailed Solution for Molecular Biology MCQ 1 - Question 16

Out of the given statements, statement Q and R are true and statement P is false. The length of DNA does not directly relate with  gene density. Rather, the longer is DNA, greater are the chances of it containing non-coding regions or introns. DNA proofreading by RNA polymerase involves backtracking,  whereby RNA polymerase removes abnormal or mismatched nucleotides. Splicing is a process, whereby introns are removed from the primary transcript and it occurs in nucleus of the cell either co-transcriptionally or post-transcriptionally.

*Multiple options can be correct
Molecular Biology MCQ 1 - Question 17

Which of the following statements are true regarding lac operon ?

Detailed Solution for Molecular Biology MCQ 1 - Question 17

The lac operon of E.coli contains genes involved in lactose utilization. In the absence of lactose, it is present in a represeed state due to the binding of lac repressor on operator site. Also, the presence of glucose prevents this operon from turning on by a complex process called catabolite represion. The expression of lac operon genes is leaky, even in the absence of operon activation, some amounts of gene products are always formed within the cell.

*Answer can only contain numeric values
Molecular Biology MCQ 1 - Question 18

The number of base pairs in a double helical B-DNA molecule is 5430. The number of turns in the molecule will be ____________ [Answer in integer]


Detailed Solution for Molecular Biology MCQ 1 - Question 18

No of base pairs per turn = 10
Therefore number of turns = 5430/10 = 543

*Answer can only contain numeric values
Molecular Biology MCQ 1 - Question 19

If the base composition of A in a dsDNA molecule is 20%, the amount of G (in %) present in this DNA would be ______________ [Answer in integer]


Detailed Solution for Molecular Biology MCQ 1 - Question 19

According to Chargaff’s rule, a 1: 1 ratio of purines to pyrimidines must be present in any ds DNA.
% A = %T and %G = %C
If A = 20% and T will also be 20%. Further, 100%-40% = 60%.
Since G and C are present in equal amounts, 60/2 = 30%.
Therefore, amount of G present will be 30%.

*Answer can only contain numeric values
Molecular Biology MCQ 1 - Question 20

E.coli cells were grown in a medium containing heavy nitrogen and were later transferred to a medium containing light nitrogen. The percent of DNA expected to have been made up of light nitrogen after three generations ___________ [Answer upto one decimal place]


Detailed Solution for Molecular Biology MCQ 1 - Question 20

After 3 generations, a total of 8 double helices or 16 single strands of DNA will be obtained. Out of these, 2 of them are heterogenous and are made up of one heavy and one light nitrogen strand. Thus, 14 strands contain light nitrogen out of 16, which is equal to 87.5%

*Answer can only contain numeric values
Molecular Biology MCQ 1 - Question 21

If the number of amino acids in a protein is 226 , then the number of nucleotides coding for the protein will be ____________ [Answer in integer]


Detailed Solution for Molecular Biology MCQ 1 - Question 21

3 nucleotides code for 1 amino acid.
Therefore, to code 226 amino acids, 226*3 = 678 nucleotides will be required.

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