Molecular Biology MCQ 1 - Biotechnology Engineering (BT) MCQ

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.

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).

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- tRNA_i^f-met to the small ribosomal subnuit.

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.

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. 

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.

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.

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

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.

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.

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.

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'

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.

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.

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 

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.

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.

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

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%.

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%

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