Exercises
Group the following as nitrogenous bases and nucleosides: Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer
Nitrogenous bases: adenine, thymine, uracil, cytosine. Nucleosides (base + pentose sugar): cytidine, guanosine.
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer
By Chargaff's rule, C = G and A = T. Given C = 20%
⇒ G = 20%
⇒ C + G = 40%
⇒ A + T = 100 − 40 = 60%
⇒ A = 60/2 = 30%.
If the sequence of one strand of DNA is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Write down the sequence of complementary strand in 5'→3' direction.
Answer
The complementary strand pairs A with T and G with C and runs antiparallel: 3'-TACGTACGTACGTACGTACGTACGTACG-5'. Written in the 5'
⇒ 3' direction it is:
5'-GCATGCATGCATGCATGCATGCATGCAT-3'
If the sequence of the coding strand in a transcription unit is written as follows: 5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Write down the sequence of mRNA.
Answer
The mRNA has the same sequence and polarity as the coding strand, with uracil in place of thymine:
5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer
The complementary base pairing between the two strands (A with T, G with C) of the double helix. Because the sequence of one strand fixes the sequence of the other, Watson and Crick proposed that the two strands separate and each acts as a template for synthesis of a new complementary strand. Each daughter DNA molecule therefore has one parental (old) strand and one newly synthesised strand – semi-conservative replication (later proved by Meselson and Stahl using ¹⁵N in E. coli).
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer
(i) DNA-dependent DNA polymerase: DNA template
⇒ DNA (replication). (ii) DNA-dependent RNA polymerase: DNA template
⇒ RNA (transcription); in eukaryotes RNA polymerase I (rRNAs 28S, 18S, 5.8S), II (hnRNA, precursor of mRNA) and III (tRNA, 5S rRNA, snRNAs). (iii) RNA-dependent DNA polymerase (reverse transcriptase): RNA template
⇒ DNA, found in retroviruses. (iv) RNA-dependent RNA polymerase (RNA replicase): RNA template
⇒ RNA, found in some RNA viruses.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer
They used radioactive labels that mark only one of the two molecules. DNA contains phosphorus but not sulphur; protein contains sulphur but not phosphorus. Bacteriophages grown on medium with radioactive phosphorus (³²P) had labelled DNA, while those grown with radioactive sulphur (³⁵S) had labelled protein. Each type was allowed to infect E. coli, then blending removed the viral coats and centrifugation separated the bacteria. Bacteria infected by ³²P-phages became radioactive, while with ³⁵S-phages the radioactivity stayed in the supernatant (viral coats), not in the bacteria. Thus DNA, not protein, enters the bacterium and is the genetic material.
Differentiate between the followings: (a) Repetitive DNA and Satellite DNA (b) mRNA and tRNA (c) Template strand and Coding strand
Answer
(a) Repetitive DNA consists of regions where a small stretch of DNA is repeated many times. When genomic DNA is subjected to density gradient centrifugation, the bulk DNA forms a major peak and the repetitive DNA separates as small peaks; this separated fraction is called satellite DNA. Satellite DNA is classified (micro-satellites, mini-satellites, etc.) by base composition, length and number of repeats; it does not code for proteins and shows high polymorphism, forming the basis of DNA fingerprinting.
(b) mRNA (messenger RNA) is a linear molecule that carries the genetic code (codons) from DNA to the ribosome and acts as the template for translation. tRNA (transfer RNA) is a small adapter molecule (clover-leaf in 2-D, inverted L in 3-D) that reads the code through its anticodon and carries a specific amino acid to the ribosome.
(c) Template strand has 3'
⇒ 5' polarity and is the strand that is actually transcribed into RNA. Coding strand has 5'
⇒ 3' polarity, is not transcribed, and has the same sequence as the RNA (with T in place of U); it is used as the reference when writing gene sequences.
List two essential roles of ribosome during translation.
Answer
(i) The ribosome binds the mRNA and provides the site (platform) where charged tRNAs line up by codon–anticodon pairing; its large subunit has two sites for successive aminoacyl-tRNAs to come close together. (ii) It catalyses peptide bond formation between adjacent amino acids – the 23S rRNA of bacteria acts as a ribozyme (peptidyl transferase). It also moves along the mRNA codon by codon.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer
Lactose (actually its product allolactose) is the inducer: it binds the repressor protein and inactivates it, so RNA polymerase can transcribe the lac genes (z, y, a) to make β-galactosidase, permease and transacetylase. β-galactosidase breaks lactose into glucose and galactose. As lactose is used up, no inducer is left; the free repressor again binds the operator and blocks transcription. Hence the operon shuts down – it is regulated by the availability of its substrate.
Explain (in one or two lines) the function of the followings: (a) Promoter (b) tRNA (c) Exons
Answer
(a) Promoter: a DNA sequence at the 5' end (upstream) of the structural gene that provides the binding site for RNA polymerase and decides which strand is the template, thus starting transcription. (b) tRNA: adapter molecule that reads the mRNA codon by its anticodon and brings the corresponding amino acid to the ribosome. (c) Exons: coding (expressed) sequences of a split gene that remain in mature mRNA after splicing and are translated into protein.
Why is the Human Genome project called a mega project?
Answer
(i) It aimed to sequence about 3 × 10⁹ base pairs of human DNA and identify all the 20,000–25,000 genes. (ii) At the initial estimate of US $3 per bp, the cost would be about 9 billion US dollars. (iii) The data would fill about 3300 books of 1000 pages each with 1000 letters per page. (iv) Storage, retrieval and analysis of this huge data needed high-speed computers, leading to the new field of bioinformatics. (v) It also had to address ethical, legal and social issues (ELSI), and involved many countries and scientists over 13 years (1990–2003). Its size, cost and scope make it a mega project.
What is DNA fingerprinting? Mention its application.
Answer
DNA fingerprinting is a quick technique to compare the DNA of individuals by finding differences in repetitive DNA (satellite DNA, such as VNTRs) that show high polymorphism. Developed by Alec Jeffreys, it involves isolating DNA, digesting it with restriction enzymes, separating fragments by electrophoresis, blotting to a membrane, hybridising with a labelled VNTR probe and detecting bands by autoradiography; the band pattern is unique to each person except identical twins.
Applications: (i) forensic science – identifying criminals from blood, hair, skin, saliva or sperm; (ii) paternity testing in disputes, as polymorphisms are inherited; (iii) determining population and genetic diversity; (iv) studies in evolutionary biology.
Briefly describe the following: (a) Transcription (b) Polymorphism (c) Translation (d) Bioinformatics
Answer
(a) Transcription: copying of genetic information from one strand of DNA (template) into RNA by DNA-dependent RNA polymerase. It has three steps – initiation (RNA polymerase binds the promoter; in bacteria with the σ factor), elongation (ribonucleoside triphosphates added complementary to template) and termination (at the terminator; with the ρ factor in bacteria). In eukaryotes the primary transcript (hnRNA) undergoes splicing (removal of introns), capping (methyl guanosine triphosphate at 5' end) and tailing (poly-A tail at 3' end) to form mRNA.
(b) Polymorphism: variation at the genetic level, arising from mutations. An inheritable mutation in a population at a frequency greater than 0.01 is called a DNA polymorphism. Such variation, especially in non-coding repetitive DNA, accumulates over generations and is the basis of genetic mapping and DNA fingerprinting.
(c) Translation: polymerisation of amino acids into a polypeptide following the sequence of codons on mRNA, on ribosomes. Amino acids are first activated with ATP and attached to their tRNAs (charging/aminoacylation). Initiation starts at AUG with the initiator tRNA; during elongation, codon–anticodon pairing brings successive amino acids and peptide bonds form as the ribosome moves along the mRNA; at a stop codon (UAA, UAG, UGA) a release factor ends translation and the polypeptide is released.
(d) Bioinformatics: the field that uses computers and software to store, retrieve, analyse and interpret large biological data such as DNA and protein sequences. It developed rapidly with the Human Genome Project.
Important Extra Questions
Why is DNA a better genetic material than RNA?
Answer
DNA is chemically less reactive and structurally more stable (no 2'-OH on sugar, thymine instead of uracil, double-stranded), and it mutates more slowly; RNA is reactive and easily degraded.
What are the salient features of the genetic code?
Answer
Triplet; 61 codons code for amino acids and 3 are stop codons; unambiguous and specific; degenerate (some amino acids have more than one codon); read continuously without punctuation; nearly universal; AUG is the start codon and also codes for methionine.
What is a nucleosome?
Answer
The repeating unit of chromatin: about 200 bp of negatively charged DNA wrapped around a positively charged histone octamer.
Why is DNA replication continuous on one strand and discontinuous on the other?
Answer
DNA polymerase works only in the 5'
⇒ 3' direction, so the strand with 3'
⇒ 5' template is copied continuously, while the other is made in short Okazaki fragments joined by DNA ligase.
What is the function of the σ and ρ factors in bacterial transcription?
Answer
The σ (sigma) factor helps RNA polymerase start (initiation) and the ρ (rho) factor helps it stop (termination).