Chemistry GS TIFR- Tata Institute of Fundamental Research Questions 6-10_ Solved

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6. If you have a polypeptide of size 11 kilo Daltons, what would be the mass of the open reading frame that codes for the polypeptide? The average mass of an amino acid and a nucleotide is considered to be 110 Daltons and 330 Daltons, respectively.

(i) 99990

(ii) 99000

(iii) 90900

(iv) 99900

Answer: To determine the mass of the open reading frame (ORF) that codes for the given polypeptide, we need to consider the fact that each amino acid in the polypeptide is coded for by three nucleotides (a codon). Therefore, we can calculate the mass of the ORF by dividing the mass of the polypeptide by the number of amino acids and then multiplying by three to get the number of nucleotides in the ORF.

Number of amino acids in polypeptide = 11,000 / 110 = 100

Mass of ORF = (100 x 3) x 330 = 99,000 Daltons

Therefore, the mass of the ORF that codes for a polypeptide of size 11 kilo Daltons is 99,000 Daltons, assuming that each amino acid is coded for by three nucleotides and the average mass of an amino acid and a nucleotide are 110 Daltons and 330 Daltons, respectively.

7. When the velocity of enzyme activity is plotted against substrate concentration, which of the following is obtained?

(i) Hyperbolic curve

(ii) Parabola

(iii) Straight line with positive slope

(iv) Straight line with negative slope

Answer: When the velocity of enzyme activity is plotted against substrate concentration, a typical hyperbolic curve is obtained, known as a Michaelis-Menten curve. At low substrate concentrations, the velocity of enzyme activity is directly proportional to the substrate concentration. As the substrate concentration increases, the velocity of enzyme activity approaches a maximum level, called Vmax. At this point, the enzyme is saturated with substrate, and adding more substrate will not increase the velocity of enzyme activity.

The shape of the curve is due to the fact that the rate of the enzymatic reaction is limited by the rate of the formation of the enzyme-substrate complex, which is dependent on the concentration of both the enzyme and the substrate. The Michaelis-Menten equation describes the relationship between the velocity of the reaction, the substrate concentration, and the kinetic constants of the enzyme, which can be derived from the curve.

8. Select one of the below sets in which all the molecules show a microwave rotational spectrum?

(i) H2, CH4, SF6

(ii) HCl, CH3Cl, CH2Cl2, H2O

(iii) H2, HCl, CH4, CH3Cl

(iv) CH2Cl2, H2O, SF6

Answer: The set of molecules that all show a microwave rotational spectrum is:

HCl, CH3Cl, CH2Cl2, H2O

Microwave rotational spectroscopy is a technique that is used to study the rotational motion of molecules. This technique is particularly useful for studying molecules that have a permanent electric dipole moment, such as those that contain polar covalent bonds.

Out of the given sets, the molecules HCl, CH3Cl, CH2Cl2, and H2O all have a permanent electric dipole moment and therefore exhibit a microwave rotational spectrum.

H2 and SF6 are non-polar molecules and do not exhibit a microwave rotational spectrum. While CH4 is a polar molecule, it does not have a permanent electric dipole moment due to its symmetrical tetrahedral shape, and therefore it also does not exhibit a microwave rotational spectrum.

9. A tyrosine-to-histidine mutation in the active site of an enzyme significantly affects its activity. The reaction is likely to be

(i) Acid-catalysed

(ii) Base-catalysed

(iii) Free radical-catalyzed

(iv) None of the above

Answer: The tyrosine-to-histidine mutation in the active site of an enzyme is likely to affect the enzyme's activity because histidine is a stronger base than tyrosine. This change can affect the enzyme's ability to act as either an acid or a base catalyst, depending on the specific reaction and the role of the active site residue in that reaction.

If the active site residue is involved in donating a proton to the substrate or accepting a proton from the substrate, the mutation could affect the acid-base catalysis of the enzyme. In this case, the mutation to histidine could make the enzyme a stronger base and affect the rate of the reaction.

On the other hand, if the active site residue is involved in stabilizing a radical intermediate during the reaction, the mutation to histidine could affect the free radical catalysis of the enzyme. In this case, the mutation could affect the enzyme's ability to stabilize the intermediate and, thus, the overall rate of the reaction.

Therefore, the answer to the question depends on the specific reaction and the role of the active site residue in that reaction.

10. If a branched polysaccharide (e.g. amylopectin) has 7 branch points, how many free -OH ends are available for reduction?

(i) 7

(ii) 1

(iii) 6

(iv) 0

Solution: Amylopectin is a branched polysaccharide composed of glucose units linked together by alpha-1,4-glycosidic bonds in the linear chain, and alpha-1,6-glycosidic bonds at the branch points.

The number of free -OH ends available for a reduction in amylopectin can be calculated by considering the number of non-reducing ends and reducing ends in the molecule.

Non-reducing ends are the ends of the molecule that terminate in a glycosidic bond while reducing ends are the ends that terminate in a free -OH group that can undergo reduction reactions.

In amylopectin, there is only one reducing end, located at the end of each branch. Therefore, the total number of reducing ends in amylopectin is equal to the number of branch points.

Since it is stated that amylopectin has 7 branch points, there are 7 reducing ends, and therefore 7 free -OH ends available for reduction.

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