Molecular Basis of Inheritance

1. In a DNA strand the nucleotides are linked together by:

a. Hlycosidic bonds

b. Phosphodiester bonds

c. Peptide bonds

d. Hydrogen bonds

Answer: b. phosphodiester bonds

Explanation: Nucleotides in a single DNA strand are covalently joined by 3'-5' phosphodiester bonds between the 3' hydroxyl of one sugar and the 5' phosphate of the next. Glycosidic bonds link the sugar to the base, and hydrogen bonds link the two separate strands together.

2. A nucleoside differs from a nucleotide. It lacks the:

a. Base

b. Sugar

c. Phosphate group

d. Hydroxyl group

Answer: c. phosphate group

Explanation: A nucleoside consists only of a nitrogenous base attached to a pentose sugar. It is only when a phosphate group is added to the 5' carbon of this sugar that the molecule becomes a nucleotide.

3. Both deoxyribose and ribose belong to a class of sugars called:

a. Trioses

b. Hexoses

c. Pentoses

d. Polysaccharides

Answer: c. pentoses

Explanation: Both ribose (found in RNA) and deoxyribose (found in DNA) are five-carbon monosaccharides. Sugars containing exactly five carbon atoms are classified structurally as pentoses.

4. The fact that a purine base always pairs through hydrogen bonds with a pyrimidine base in the DNA double helix leads to:

a. The antiparallel nature

b. The semiconservative nature

c. Uniform width throughout DNA

d. Uniform length in all DNA

Answer: c. uniform width throughout DNA

Explanation: A purine is a larger, double-ring structure, while a pyrimidine is a smaller, single-ring structure. The pairing of a large purine with a small pyrimidine ensures that the distance between the two sugar-phosphate backbones remains constant, creating a uniform width.

5. The net electric charge on DNA and histones is:

a. Both positive

b. Both negative

c. Negative and positive, respectively

d. Zero

Answer: c. negative and positive, respectively

Explanation: DNA carries a net negative charge due to the negatively charged phosphate groups in its backbone. Histones carry a net positive charge because they are rich in basic amino acids like lysine and arginine, allowing them to bind tightly to DNA.

6. The promoter site and the terminator site for transcription are located at:

a. 3' (downstream) end and 5' (upstream) end, respectively of the transcription unit

b. 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit

c. The 5' (upstream) end

d. The 3' (downstream) end

Answer: b. 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit

Explanation: By convention, positions in a transcription unit are defined with respect to the 5' to 3' coding strand. The promoter is located towards the 5' end (upstream) to initiate transcription, and the terminator is at the 3' end (downstream) to stop it.

7. Which of the following statements is the most appropriate for sickle cell anaemia?

a. It cannot be treated with iron supplements

b. It is a molecular disease

c. It confers resistance to acquiring malaria

d. All of the above

Answer: d. All of the above

Explanation: Sickle cell anaemia is a molecular disease caused by a point mutation affecting hemoglobin structure, meaning iron supplements cannot cure it. Interestingly, carriers of the sickle cell trait have an evolutionary advantage as their red blood cells resist malarial parasite infection.

8. Which of the following is true with respect to AUG?

a. It codes for methionine only

b. It is an initiation codon

c. It codes for methionine in both prokaryotes and eukaryotes

d. All of the above

Answer: d. All of the above

Explanation: The AUG codon serves a dual purpose in the genetic code. It acts universally as the start (initiation) codon for translation and specifically codes for the amino acid methionine in both prokaryotes and eukaryotes.  

9. The first genetic material could be:

a. Protein

b. Carbohydrates

c. DNA

d. RNA  

Answer: d. RNA

Explanation: According to the "RNA World" hypothesis, RNA was the first genetic material to evolve. This is because RNA has the unique ability to both store genetic information and act as a biological catalyst (ribozyme) for essential chemical reactions.  

10. With regard to mature mRNA in eukaryotes:

a. Exons and introns do not appear in the mature RNA

b. Exons appear but introns do not appear in the mature RNA

c. Introns appear but exons do not appear in the mature RNA

d. Both exons and introns appear in the mature RNA

Answer: b. exons appear but introns do not appear in the mature RNA

Explanation: In eukaryotes, the primary transcript (hnRNA) contains both coding sequences (exons) and non-coding sequences (introns). During post-transcriptional processing, introns are spliced out, and only exons are joined to form the mature mRNA.

11. The human chromosome with the highest and least number of genes in them are respectively:

a. Chromosome 21 and Y

b. Chromosome 1 and X

c. Chromosome 1 and Y

d. Chromosome X and Y

Answer: c. Chromosome 1 and Y

Explanation: Data from the Human Genome Project revealed that Chromosome 1, the largest human chromosome, contains the maximum number of genes (2968). Conversely, the Y chromosome is the smallest and possesses the fewest genes (231).

12. Who amongst the following scientists had no contribution in the development of the double helix model for the structure of DNA?

a. Rosalind Franklin

b. Maurice Wilkins

c. Erwin Chargaff

d. Meselson and Stahl

Answer: d. Meselson and Stahl

Explanation: Watson and Crick developed the double helix model using X-ray diffraction data from Franklin and Wilkins, and base-pairing rules from Chargaff. Meselson and Stahl later performed experiments proving the semi-conservative replication of DNA, not its structural model.  

13. DNA is a polymer of nucleotides which are linked to each other by 3'-5' phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?

a. Replace purine with pyrimidines

b. Remove/Replace 3' OH group in deoxy ribose

c. Remove/Replace 2' OH group with some other group in deoxy ribose

d. Both 'b' and 'c'  

Answer: b. Remove/Replace 3' OH group in deoxy ribose

Explanation: DNA polymerase strictly requires a free 3'-OH group on the growing strand to attach the 5'-phosphate of the incoming nucleotide. If this 3'-OH is removed (as in dideoxynucleotides), phosphodiester bond formation becomes impossible, permanently halting polymerization.

14. Discontinuous synthesis of DNA occurs in one strand, because:

a. DNA molecule being synthesised is very long

b. DNA dependent DNA polymerase catalyses polymerisation only in one direction (5'->3')

c. It is a more efficient process

d. DNA ligase joins the short stretches of DNA

Answer: b. DNA dependent DNA polymerase catalyses polymerisation only in one direction (5'->3')

Explanation: Because the two strands of a DNA double helix are antiparallel, and DNA polymerase can only synthesize in the 5' to 3' direction, one strand (the lagging strand) must be synthesized backwards in small pieces called Okazaki fragments as the replication fork opens.

15. Which of the following steps in transcription is catalysed by RNA polymerase?

a. Initiation

b. Elongation

c. Termination

d. All of the above

Answer: b. Elongation

Explanation: While the RNA polymerase enzyme complex is involved in the whole process, the core enzyme itself is only capable of catalyzing the elongation of the RNA chain. Initiation and termination specifically require the transient binding of additional factors (like sigma and rho, respectively).

16. Control of gene expression in prokaryotes take place at the level of:

a. DNA-replication

b. Transcription

c. Translation

d. None of the above

Answer: b. Transcription

Explanation: In prokaryotes, because there is no nuclear membrane separating transcription and translation, the processes are coupled. Therefore, the primary and most efficient point to regulate gene expression is by controlling the initiation of transcription.

17. Which of the following statements is correct about the role of regulatory proteins in transcription in prokaryotes?

a. They only increase expression

b. They only decrease expression

c. They interact with RNA polymerase but do not affect the expression

d. They can act both as activators and as repressors

Answer: d. They can act both as activators and as repressors

Explanation: Regulatory proteins influence the ability of RNA polymerase to bind to the promoter. They can act as positive regulators (activators) that promote transcription, or as negative regulators (repressors) that bind to operators and block transcription.

18. Which was the last human chromosome to be completely sequenced:

a. Chromosome 1

b. Chromosome 11

c. Chromosome 21

d. Chromosome X

Answer: a. Chromosome 1

Explanation: Chromosome 1 is the largest chromosome in the human genome, containing massive amounts of highly repetitive DNA sequences. Due to this complexity, its sequencing was finalized last, completed in May 2006, marking a major milestone of the Human Genome Project.

19. Which of the following are the functions of RNA?

a. It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides.

b. It carries amino acids to ribosomes.

c. It is a constituent component of ribosomes.

d. All of the above.

Answer: d. All of the above.

Explanation: RNA plays diverse roles in cellular function: messenger RNA (mRNA) carries genetic codes, transfer RNA (tRNA) delivers specific amino acids to the ribosome, and ribosomal RNA (rRNA) provides the structural and catalytic framework of the ribosome itself.  

20. While analysing the DNA of an organism a total number of 5386 nucleotides were found out of which the proportion of different bases were: Adenine = 29%, Guanine = 17%, Cytosine = 32%, Thymine = 17%. Considering the Chargaff's rule it can be concluded that:

a. It is a double stranded circular DNA

b. It is single stranded DNA

c. It is a double stranded linear DNA

d. No conclusion can be drawn  

Answer: b. It is single stranded DNA

Explanation: Chargaff's rule states that in double-stranded DNA, the percentage of Adenine must equal Thymine (A=T), and Guanine must equal Cytosine (G=C). Because A (29%) ≠ T (17%) and G (17%) ≠ C (32%) in this sample, the DNA must be single-stranded.

21. In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called:

a. A-DNA

b. B-DNA

c. cDNA

d. rDNA

Answer: c. cDNA

Explanation: DNA generated from a single-stranded RNA template via the enzyme reverse transcriptase is called complementary DNA (cDNA). This mechanism is characteristically used by retroviruses, such as HIV, to integrate their genetic material into host cells.

22. If Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of N15/N15 : N15/N14 : N14/N14 containing DNA in the fourth generation would be:

a. 1:1:0

b. 1:4:0

c. 0:1:3

d. 0:1:7

Answer: d. 0:1:7

Explanation: Because DNA replication is semi-conservative, the original heavy N15 strands are never lost but are diluted. By generation 4 (16 DNA molecules total), there are 2 hybrid (N15/N14) molecules and 14 completely light (N14/N14) molecules, giving a ratio of 0:2:14, which simplifies to 0:1:7.  

23. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is: 5'-ATGAATG-3', the sequence of bases in its RNA transcript would be;

a. 5'-AUGAAUG-3'

b. 5'-UACUUAC-3'

c. 5'-CAUUCAU-3'

d. 5'-GUAAGUA-3'  

Answer: a. 5'-AUGAAUG-3'

Explanation: The newly synthesized RNA transcript shares the exact same sequence and 5' to 3' polarity as the DNA coding strand. The only biochemical difference is that RNA contains Uracil (U) in place of Thymine (T).

24. The RNA polymerase holoenzyme transcribes:

a. The promoter, structural gene and the terminator region

b. The promoter and the terminator region

c. The structural gene and the terminator region

d. The structural gene only.

Answer: d. the structural gene only.

Explanation: The promoter sequence acts strictly as a binding site for the RNA polymerase and is not transcribed into RNA. While the polymerase reaches the terminator region to stop the process, it is fundamentally transcribing the sequence of the structural gene.

25. If the base sequence of a codon in mRNA is 5'-AUG-3', the tRNA pairing with it must be:

a. 5'UAC3'

b. 5'CAU - 3'

c. 5'-AUG-3'

d. 5'-GUA-3'

Answer: b. 5'CAU - 3'

Explanation: The anticodon on the tRNA pairs with the mRNA codon in a complementary and antiparallel manner. The complement to 5'-AUG-3' is 3'-UAC-5', which when read in the standard 5' to 3' direction is written as 5'-CAU-3'.

26. The amino acid attaches to the tRNA at its:

a. 5' - end

b. 3' - end

c. Anti codon site

d. DHU loop

Answer: b. 3' - end

Explanation: All tRNA molecules possess a universally conserved CCA sequence at their 3' end, forming an acceptor stem. Specific enzymes (aminoacyl-tRNA synthetases) covalently attach the correct amino acid to the hydroxyl group of this 3' terminal adenosine.

27. To initiate translation, the mRNA first binds to:

a. The larger ribosomal sub-unit

b. The smaller ribosomal sub-unit.

c. The whole ribosome

d. No such specificity exists.

Answer: b. The smaller ribosomal sub-unit.

Explanation: Translation initiation is a step-wise process. The mRNA first binds to the smaller ribosomal subunit (along with the initiator tRNA) to establish the correct reading frame before the larger subunit joins to form the functional ribosome complex.

28. In E.coli, the lac operon gets switched on when:

a. Lactose is present and it binds to the repressor

b. Repressor binds to operator

c. RNA polymerase binds to the operator

d. lactose is present and it binds to RNA polymerase

Answer: a. lactose is present and it binds to the repressor

Explanation: Lactose acts as an inducer for the operon. When present, it binds to the active repressor protein, causing a conformational change that prevents the repressor from binding to the operator, thereby freeing the promoter for RNA polymerase to initiate transcription.