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Case Based Questions Test: Biomolecules - 1 - NEET MCQ


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Case Based Questions Test: Biomolecules - 1 - Question 1

Read the passage given below and answer the following questions:

EVIDENCE FOR THE FIBROUS NATURE OF DNA
The basic chemical formula of DNA is now well established. As shown in Figure 1 it consists of a very long chain, the backbone of which is made up of alternate sugar and phosphate groups, joined together in regular 3’ 5’ phosphate di-ester linkages. To each sugar is attached a nitrogenous base, only four different kinds of which are commonly found in DNA. Two of these---adenine and guanine--- are purines, and the other two thymine and cytosine-are pyrimidines. A fifth base, 5-methyl cytosine, occurs in smaller amounts in certain organisms, and a sixth, 5-hydroxy-methyl-cytosine, is found instead of cytosine in the T even phages. It should be noted that the chain is unbranched, a consequence of the regular internucleotide linkage. On the other hand the sequence of the different nucleotides is, as far as can be ascertained, completely irregular. Thus, DNA has some features which are regular, and some which are irregular. A similar conception of the DNA molecule as a long thin fiber is obtained from physicochemical analysis involving sedimentation, diffusion, light scattering, and viscosity measurements. These techniques indicate that DNA is a very asymmetrical structure approximately 20 A wide and many thousands of angstroms long. Estimates of its molecular weight currently center between 5 X106 and X107 (approximately 3 X104 nucleotides). Surprisingly each of these measurements tend to suggest that the DNA is relatively rigid, a puzzling finding in view of the large number of single bonds (5 per nucleotide) in the phosphate-sugar back bone. Recently these indirect inferences have been confirmed by electron microscopy. 

Q. DNA molecule has ___________ internucleotide linkage and __________ sequence of the different nucleotides
 

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 1

DNA molecule has regular internucleotide linkage and irregular sequence of the different nucleotides.

Case Based Questions Test: Biomolecules - 1 - Question 2

Read the passage given below and answer the following questions:

EVIDENCE FOR THE FIBROUS NATURE OF DNA
The basic chemical formula of DNA is now well established. As shown in Figure 1 it consists of a very long chain, the backbone of which is made up of alternate sugar and phosphate groups, joined together in regular 3’ 5’ phosphate di-ester linkages. To each sugar is attached a nitrogenous base, only four different kinds of which are commonly found in DNA. Two of these---adenine and guanine--- are purines, and the other two thymine and cytosine-are pyrimidines. A fifth base, 5-methyl cytosine, occurs in smaller amounts in certain organisms, and a sixth, 5-hydroxy-methyl-cytosine, is found instead of cytosine in the T even phages. It should be noted that the chain is unbranched, a consequence of the regular internucleotide linkage. On the other hand the sequence of the different nucleotides is, as far as can be ascertained, completely irregular. Thus, DNA has some features which are regular, and some which are irregular. A similar conception of the DNA molecule as a long thin fiber is obtained from physicochemical analysis involving sedimentation, diffusion, light scattering, and viscosity measurements. These techniques indicate that DNA is a very asymmetrical structure approximately 20 A wide and many thousands of angstroms long. Estimates of its molecular weight currently center between 5 X106 and X107 (approximately 3 X104 nucleotides). Surprisingly each of these measurements tend to suggest that the DNA is relatively rigid, a puzzling finding in view of the large number of single bonds (5 per nucleotide) in the phosphate-sugar back bone. Recently these indirect inferences have been confirmed by electron microscopy. 

Q. Out of the four different kinds of nitrogenous bases which are commonly found in DNA, ___________ has been replaced in some organisms.
 

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 2

A chemical compound that is used to make one of the building blocks of DNA and RNA. It is a type of pyrimidine. i.e., cytosine.

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Case Based Questions Test: Biomolecules - 1 - Question 3

Read the passage given below and answer the following questions:

When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures remains intact.
Examples of denaturation of protein are coagulation of egg white on boiling, curding of milk, formation of cheese when an acid is added to milk.

Which statement(s) of protein remain(s) intact during denaturation process?

Case Based Questions Test: Biomolecules - 1 - Question 4

Read the passage given below and answer the following questions:

When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures remains intact.
Examples of denaturation of protein are coagulation of egg white on boiling, curding of milk, formation of cheese when an acid is added to milk.

Q. Secondary structure of protein refers to v

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 4

Secondary structure, refers to regular l folded structures that form within a polypeptide due to interactions between atoms of the backbone. Both structures are held in shape by hydrogen bonds, which form between the carbonyl O of one amino acid and the amino H of another.

Case Based Questions Test: Biomolecules - 1 - Question 5

Read the passage given below and answer the following questions:

When a protein in its native form, is subjected to physical changes like change in temperature or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. This is called denaturation of protein.
The denaturation causes change in secondary and tertiary structures but primary structures remains intact.
Examples of denaturation of protein are coagulation of egg white on boiling, curding of milk, formation of cheese when an acid is added to milk.

Q. Mark the wrong statement about denaturation of proteins.

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 5

Due to denaturation, globules unfold and helix get uncoiled and protein loses its biological activity. Thus globular proteins are converted into fibrous proteins.
Hence, (C) is wrong.

Case Based Questions Test: Biomolecules - 1 - Question 6

Read the passage given below and answer the following questions:

The two monosaccharides are joined together by an oxide linkage formed by the loss of a water molecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage. In disaccharides, if the reducing groups of monosaccharides i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars, e.g., sucrose. On the other hand, sugars in which these functional groups are free, are called reducing sugars, for example, maltose and lactose.

A non reducing disaccharide ‘A’ on hydrolysis with dilute acid gives an equimolar mixture of D–(+)– glucose and D-(-)-Fructose.

 

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 6

It indicates that 5’-OH groups are present which react with acetic acid to give glucose pentaacetate.

Case Based Questions Test: Biomolecules - 1 - Question 7

Read the passage given below and answer the following questions:

The sequence of bases in m-RNA are read in a serial order in groups of three at a time. Each triplet of nucleotides (having a specific sequence of bases) is known as codon. Each codon specifies one amino acid. Many amino acids have more than one codons. The amino acids are brought to the mRNA by another type of RNA and called tRNA. Each amino acid has atleast one corresponding tRNA. At one end of the tRNA molecule is a trinucleotide base sequence that is complementary to some trinucleotide base sequence on mRNA.

Q. Each triplet of nucleotides is called:

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 7

Each triplet of nucleotides, which have a specific sequence of bases, is called codon.

Case Based Questions Test: Biomolecules - 1 - Question 8

Read the passage given below and answer the following questions:

The sequence of bases in m-RNA are read in a serial order in groups of three at a time. Each triplet of nucleotides (having a specific sequence of bases) is known as codon. Each codon specifies one amino acid. Many amino acids have more than one codons. The amino acids are brought to the mRNA by another type of RNA and called tRNA. Each amino acid has atleast one corresponding tRNA. At one end of the tRNA molecule is a trinucleotide base sequence that is complementary to some trinucleotide base sequence on mRNA.

Q. In mRNA, the complementary bases of AAT is:

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 8

In mRNA
A ::: U
T ::: A
So, the complementary bases of AAT in mRNA is UUA

Case Based Questions Test: Biomolecules - 1 - Question 9

Read the passage given below and answer the following questions:

Adenosine triphosphate (ATP) is the energy carrying molecule found in the cells of all living things. ATP captures chemical energy obtained from the breakdown of food molecules and releases it to fuel other cellular processes. ATP is a nucleotide that consists of three main structures: the nitrogenous base, adenine; the sugar, ribose; and a chain of three phosphate groups bound to ribose. The phosphate tail of ATP is the actual power source which the cell taps. Available energy is contained in the bonds between the phosphates and is released when they are broken, which occurs through the addition of a water molecule (a process called hydrolysis). Usually only the outer phosphate is removed from ATP to yield energy; when this occurs ATP is converted to adenosine diphosphate (ADP), the form of the nucleotide having only two phosphates.
The importance of ATP (adenosine triphosphate) as the main source of chemical energy in living matter and its involvement in cellular processes has long been recognized. The primary mechanism whereby higher organisms, including humans, generate ATP is through mitochondrial oxidative phosphorylation. For the majority of organs, the main metabolic fuel is glucose, which in the presence of oxygen undergoes complete combustion to CO2 and H2O: C6H12O6 + 6O2 → 6O2 + 6H2O + energy
The free energy (ΔG) liberated in this exergonic (ΔG is negative) reaction is partially trapped as ATP in two consecutive processes: glycolysis (cytosol) and oxidative phosphorylation (mitochondria). The first produces 2 mol of ATP per mol of glucose, and the second 36 mol of ATP per mol of glucose. Thus, oxidative phosphorylation yields 17-18 times as much useful energy in the form of ATP as can be obtained from the same amount of glucose by glycolysis alone.
The efficiency of glucose metabolism is the ratio of amount of energy produced when 1 mol of glucose oxidised in cell to the enthalpy of combustion of glucose. The energy lost in the process is in the form of heat. This heat is responsible for keeping us warm.

Q. Nearly 95% of the energy released during cellular respiration is due to:

Case Based Questions Test: Biomolecules - 1 - Question 10

Read the passage given below and answer the following questions:

Adenosine triphosphate (ATP) is the energy carrying molecule found in the cells of all living things. ATP captures chemical energy obtained from the breakdown of food molecules and releases it to fuel other cellular processes. ATP is a nucleotide that consists of three main structures: the nitrogenous base, adenine; the sugar, ribose; and a chain of three phosphate groups bound to ribose. The phosphate tail of ATP is the actual power source which the cell taps. Available energy is contained in the bonds between the phosphates and is released when they are broken, which occurs through the addition of a water molecule (a process called hydrolysis). Usually only the outer phosphate is removed from ATP to yield energy; when this occurs ATP is converted to adenosine diphosphate (ADP), the form of the nucleotide having only two phosphates.
The importance of ATP (adenosine triphosphate) as the main source of chemical energy in living matter and its involvement in cellular processes has long been recognized. The primary mechanism whereby higher organisms, including humans, generate ATP is through mitochondrial oxidative phosphorylation. For the majority of organs, the main metabolic fuel is glucose, which in the presence of oxygen undergoes complete combustion to CO2 and H2O: C6H12O6 + 6O2 → 6O2 + 6H2O + energy
The free energy (ΔG) liberated in this exergonic (ΔG is negative) reaction is partially trapped as ATP in two consecutive processes: glycolysis (cytosol) and oxidative phosphorylation (mitochondria). The first produces 2 mol of ATP per mol of glucose, and the second 36 mol of ATP per mol of glucose. Thus, oxidative phosphorylation yields 17-18 times as much useful energy in the form of ATP as can be obtained from the same amount of glucose by glycolysis alone.
The efficiency of glucose metabolism is the ratio of amount of energy produced when 1 mol of glucose oxidised in cell to the enthalpy of combustion of glucose. The energy lost in the process is in the form of heat. This heat is responsible for keeping us warm.

Q. What is the efficiency of glucose metabolism if 1 mole of glucose gives 38ATP energy? (Given: The enthalpy of combustion of glucose is 686 kcal, 1ATP = 7.3kcal)

Detailed Solution for Case Based Questions Test: Biomolecules - 1 - Question 10

Glucose catabolism yields a Total of 38 ATP. 38 ATP x 7.3 kcal/mol ATP = 262 kcal. Glucose has 686 kcal. Thus the efficiency of glucose metabolism is 262/686 x 100 = 38%.

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