Biochemistry-I MCQs | BIO202 MCQs | Set 1
Biochemistry-I MCQs | BIO202 MCQs | Set 1
MCQs (Multiple Choice Questions)
1) Glucose residues in amylose are linked by _________.
a) `alpha` (1,4) linkage `beta`
b) (1,4) linkage `alpha`
c) (1,6) linkage
d) None of these
Correct Answer:
The correct answer is 'b'.
Explanation:
Amylose, a type of starch, is composed of glucose molecules linked together by α(1,4) glycosidic bonds. This means that the glucose molecules are connected through a covalent bond between the first carbon atom (anomeric carbon) of one glucose molecule and the fourth carbon atom of the next glucose molecule, with the oxygen linkage at the first carbon being in the α configuration. This specific type of linkage gives amylose its characteristic structure, with a linear chain of glucose units.
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2) The number of carbon atoms in lysine is
a) 4
b) 6
c) 8
d) 10
Correct Answer:
The correct answer is 'b'.
Explanation:
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3) If the carbonyl group is present at the end of the monosaccharide then it is called _________.
b) Aldose
d) None of these
Correct Answer:
The correct answer is 'b'.
Explanation:
An aldehyde sugar, also known as an aldose, is a type of monosaccharide where the carbonyl group (a functional group containing a carbon atom double-bonded to oxygen and single-bonded to hydrogen) is present at the end of the carbon chain. In aldoses, the carbonyl group is typically an aldehyde group (-CHO), and this distinguishes them from ketoses, where the carbonyl group is located within the carbon chain.
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4) The number of stereoisomers for a molecule containing only one chiral carbon is _____.
a) 16
b) 2
c) 4
d) 8
Correct Answer:
The correct answer is 'b'.
Explanation:
A molecule containing only one chiral carbon, also known as a stereocenter or asymmetric carbon, can exist in two different stereoisomeric forms: the enantiomer and the original molecule (which is not optically active). These two stereoisomers are mirror images of each other and cannot be superimposed. Therefore, the correct answer is b) 2. This is a fundamental principle in stereochemistry where a single chiral center gives rise to a pair of enantiomers.
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5) Which of the following glycosidic linkage exists between two glucose units of maltose?
a) alpha 1,4
b) beta 1,4
c) alpha 1,6
d) alpha1,4 and alpha 1,6
Correct Answer:
The correct answer is 'a'.
Explanation:
Maltose is a disaccharide composed of two glucose molecules. In maltose, the two glucose units are linked together by an α(1→4) glycosidic bond. This means that the carbon atom at position 1 of one glucose molecule is bonded to the carbon atom at position 4 of the other glucose molecule with an α linkage. This specific glycosidic bond is characteristic of maltose, and it's why the answer is "alpha 1,4".
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6) Amino sugar is formed by the replacement of the hydroxyl group at _____ of the parent monosaccharide with amino group.
a) C-1
b) C-2
c) C-3
d) C-4
Correct Answer:
The correct answer is 'b'.
Explanation:
Amino sugars, such as glucosamine and galactosamine, are formed by the replacement of the hydroxyl group at the carbon atom labeled as "C-2" of the parent monosaccharide with an amino group. This substitution at the second carbon atom (C-2) is a characteristic feature of amino sugars, and it distinguishes them from regular monosaccharides. So, the correct answer is "C-2."
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7) Which one of the following is a structural homopolysaccharide?
a) Cellulose
b) Starch
c) Glycogen
d) All of these
Correct Answer:
The correct answer is 'a'.
Explanation:
Cellulose is a structural homopolysaccharide. It is composed of repeating units of glucose, and all the glucose molecules in cellulose are connected by β(1→4) glycosidic linkages. This straight-chain structure and the strength of cellulose make it a major component of plant cell walls, providing rigidity and structural support to plant cells. In contrast, starch and glycogen are energy storage polysaccharides made of glucose but have different branching and linkages, and they serve different functions. Starch is the primary energy storage polysaccharide in plants, while glycogen is the main storage polysaccharide in animals. So, the correct answer is "Cellulose."
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8) Which one of the following amino acids is not specified by three letter codon?
a) Serine
b) Proline
c) Selenocysteine
d) Methionine
Correct Answer:
The correct answer is 'c'.
Explanation:
Selenocysteine is the amino acid that is not typically specified by a standard three-letter codon. Instead, selenocysteine is specified by a unique recoding mechanism involving a UGA codon. In the standard genetic code, UGA is a stop codon, signaling the end of protein synthesis. However, in the presence of specific sequences and factors, the UGA codon can be redefined as selenocysteine. This process requires a specialized tRNA molecule and other factors to incorporate selenocysteine into the growing polypeptide chain. Selenocysteine is a rare amino acid in proteins and plays specific roles in certain enzymes and proteins, particularly those involved in redox reactions. This unique encoding mechanism distinguishes selenocysteine from the amino acids specified by typical three-letter codons.
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9) Acid dissociation constants are designated as _________.
a) pKa
b) K
c) Pa
d) None of these
Correct Answer:
The correct answer is 'a'.
Explanation:
Acid dissociation constants (Ka) are often expressed in terms of their negative logarithm, known as pKa, to make them more convenient to work with and compare. The pKa value represents the acidity or basicity of a compound and provides information about the strength of an acid or base. The lower the pKa value, the stronger the acid, and the higher the pKa value, the weaker the acid. So, pKa is the commonly used designation for acid dissociation constants.
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10) The pI of glutamate is ________.
a) 1.2
b) 5.6
c) 3.22
d) 6.7
Correct Answer:
The correct answer is 'c'.
Explanation:
The pI (isoelectric point) of an amino acid, such as glutamate, is the pH at which the amino acid carries no net electrical charge. At this pH, the number of positively charged amino groups (NH3+) and negatively charged carboxyl groups (COO-) is equal, resulting in a neutral molecule.
Glutamate, like other amino acids, has two ionizable groups: the amino group (-NH2) and the carboxyl group (-COOH). The pI is calculated as the average of the pKa values of these two groups.
For glutamate:
- The pKa value for the amino group (NH2) is around 2.19.
- The pKa value for the carboxyl group (COOH) is around 4.25.
Taking the average of these pKa values gives a pI of approximately 3.22 (option c), which is the pH at which glutamate is electrically neutral.
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