Assignment Of Molecular Biology

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Assignment of Molecular Biology Topic (Conceptual, s Question) By; Nasir Hussain (PGB, M.Phil #05)

Submitted To; Dr. Gulam Muhammad Ali Dr. Omar Rashid Miss Farhat Nazeer

Plant Genomic & Biotechnology National University of Agricultural Sciences National Agricultural Research Centre, Islamabad.

2 Molecular Biology (Questions) 1. In eukaryotes, many genes may have to interact with each other, requiring more interacting elements than can fit around a single promoter. How this physical limitation is overcomed? ANS:

 Many structural genes respond to different cell signal and are regulated in integrated manner. The level of expression of these genes (integral genes) is controlled by sensor genes.

 Certain genes are regulated by cooperative bonding of regulatory protein.  The activity of regulators may also regulate.  Inn certain cases especially in this situation, the distant sites in chromosomes is used by regulators. Activators, repressor silence to regulate the gene expression

2.

E.coli is able to use foods other than glucose in the absence of available glucose. The falling levels of glucose cause an increase in which important component?

ANS: Glucose lowers the intracellular concentration of small molecules, cAMP. This molecule is the allosteric effecter for CAP. When glucose level is low than the cAMP levels increase and activate the lac genes.

3.

Consider a protein molecule. What kind of bonding force is primarily responsible for each of the following;

A.

The general shape of an alpha helix region of the protein?

ANS: Hydrogen bonding is mostly responsible for general shape of alpha helix region of protein. However certain other forces such as disulphide bonds between the cis residues also take part. Phophodiester interaction responsible for formation of S-S bonds.

B. The fact that the side chains of amino acids such as alanine (R= -CH3) and leucine (R= -C4H9) tend to be buried deep inside a folded protein? ANS: There is the hydrophobic interaction which causes of theses amino acids to bind to buried deep inside the folder protein. When tertiary structure of the protein completes it arrange itself in such a manner that hydrophobic amino acids are kept inside of the hydrophilic amino acid surrounding them

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C. The attachment of one amino acid to the next along a single peptide chain? ANS: The bond which attaches new coming amino acids to the growing polypeptide is peptide bond, a strong covalent bond.

4. Some viruses have ss RNA as their genetic material (Instead of ds DNA). There are two general strategies used by such RNA viruses for replicating; A. Ss RNA (That is, the viral RNA) → ds RNA → more ds RNA→ ss (viral) RNA.

B. Ss RNA → RNA – DNA hybrid (ds, 1 strand of each) → ds DNA → ss (viral) RNA. IS there any logical difference between these two pathways, in term of information flow? ANS: In term of information flow to next generation there is no difference in above two pathways. Only mode of action is different in both pathways as first directly flow into ds RNA and to ss RNA in next replication while second are involved in hybridization with ds DNA and then ss RNA in next replication 4.

When tryptophan is present in the medium, the transcription of tryptophan producing genes in E.coli is stopped by a helix-turn-helix regulator. At what site this motif binds.

ANS: The trp gene is auto regulated, when cellular concentration of tryptophan become high, it binds to trp repressor with high affinity and form repressor-trp complex causing a conformational change in repressor. This change repressor than easily and tightly bind to operator region resulting in the modulation of its function, here tryptophan act as co-repressor.

6. How it is that transcription can induce positive super coils (over winding) in part of the template? And which part of it? (Ignore other factors, such as topoisomerases). ANS: The binding of enzyme (RNA polymerase) to the DNA template results in the local unwinding of DNA helix as its pushes its way between the strands of double helix, RNA polymerase creates positive supercoils ahead of transcription site an negative supercoils behind it.

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7. Consider a single tRNA that can read both glutamine (Gln) codons. What is the anticodon? Make sure the sequence direction is clear. Assume that the E.coli wobble rules (Above) are used.

Base in anticodon, 1st position G C A U I

Base (s) in codon, 3rd position CU G U AG AUC

ANS: The mechanism by which tRNA can recognize more than one codon for a specific amino acid is described by the Wobble hypothesis in which the base at the5˝-end of the anticodon (the first base of the anticodon) is not as spatially defined as the other two bases. Movement of theta first base allows nontraditional base pairing with the 3˝ base of codon (last base of codon). This movement allows a single tRNA to recognize more than one codon. The anticodon will be 5˝GUU 3˝, according to the Wobble theory, when the first position or 3˝ base in anticodon is U. it can base pair with with both A and G at third position 3˝ - end of codon. The two codon are 5˝ CAA 3˝and 5˝ CAG 3˝. Therefore anticodon is 5˝ GUU 3˝.

7. Transcription factors appear to be unable to bind to a nucleosome, give reason? ANS: In order to transcribe a gene the RNA polymerase should have access to the promoter. Various levels of organization of chromatin can regulate the gene expression. Although RNA polymerase and enhancers can displace the histone and access to promoter, However high degree of organization can make the DNA unaccessible. The transcription factors on the other hand cannot accessible on a promoter if it is present within the nucleosome may be to repulsion between histone and transcription factors. ***************************************************************************