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Test: Biomolecules Level - 3 - Chemistry MCQ


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30 Questions MCQ Test Organic Chemistry - Test: Biomolecules Level - 3

Test: Biomolecules Level - 3 for Chemistry 2025 is part of Organic Chemistry preparation. The Test: Biomolecules Level - 3 questions and answers have been prepared according to the Chemistry exam syllabus.The Test: Biomolecules Level - 3 MCQs are made for Chemistry 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Biomolecules Level - 3 below.
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Test: Biomolecules Level - 3 - Question 1

The number of amino acids and number of peptides bonds in a linear tetrapeptide (made of different amino acids) are respectively:

Detailed Solution for Test: Biomolecules Level - 3 - Question 1

A linear tetrapeptide consists of four amino acids. Each amino acid is linked by a peptide bond, which forms during the process of protein synthesis.

Here's a breakdown of the components:

  • A linear tetrapeptide contains 4 amino acids.
  • To connect these amino acids, there are 3 peptide bonds formed.
  • Each peptide bond is created between two adjacent amino acids.

Thus, the total counts for a linear tetrapeptide made of different amino acids are:

  • 4 amino acids
  • 3 peptide bonds
Test: Biomolecules Level - 3 - Question 2

The aqueous solut ion of a carbohydrate gave dark blue colour with iodine solution. The carbohydrate is:

Detailed Solution for Test: Biomolecules Level - 3 - Question 2

Starch is a carbohydrate that reacts with iodine to produce a dark blue colour. This reaction is a characteristic test for starch, making it easy to identify.

  • Starch is a polysaccharide, meaning it is made up of many sugar units.
  • When iodine is added to a solution containing starch, it forms a complex that gives a distinct dark blue hue.
  • This reaction does not occur with other common carbohydrates like sucrose, fructose, or glucose.
  • Understanding this reaction helps in laboratory tests for starch presence in various samples.
Test: Biomolecules Level - 3 - Question 3

In the reactions, 

The major product X and Y respectively, are:

Test: Biomolecules Level - 3 - Question 4

Ident ify the correct set of stereochemical relat ionships amongst the following monosaccharides I-IV:


Test: Biomolecules Level - 3 - Question 5

In a double stranded DNA, if the sequence 5’AGATCC3’ appears on one strand of DNA, what sequence appears on the complementary strand?

Detailed Solution for Test: Biomolecules Level - 3 - Question 5

To find the complementary DNA strand, it is essential to understand base pairing.

  • In DNA, adenine (A) pairs with thymine (T).
  • Guanine (G) pairs with cytosine (C).

Given the sequence 5' AGATCC 3', the complementary bases are as follows:

  • A pairs with T
  • G pairs with C
  • A pairs with T
  • T pairs with A
  • C pairs with G
  • C pairs with G

Therefore, the complementary sequence is:

5' CCTAGA 3'

Test: Biomolecules Level - 3 - Question 6

Which one of the following compounds will form an osazone derivative:

Detailed Solution for Test: Biomolecules Level - 3 - Question 6

When sugars are reacted with excess of phenylhydrazine osazone formation takes place.Osazone will form in that sugar only where keto hydroxy group is present.

Thus CH3CH2COCH2OH will form an osazone derivative; as it has keto hydroxy group.

Test: Biomolecules Level - 3 - Question 7

An acidic amino acid among the following is:

Detailed Solution for Test: Biomolecules Level - 3 - Question 7

Acidic amino acids are those that contain a carboxyl group (-COOH) in addition to the amino group (-NH2). Among the amino acids, not all are classified as acidic.

The amino acids mentioned include:

  • Valine - A non-polar, branched-chain amino acid.
  • Proline - A unique amino acid that forms a ring structure.
  • Alanine - A non-polar amino acid, commonly found in proteins.

None of the above options are classified as acidic amino acids. The well-known acidic amino acids include:

  • Aspartic acid
  • Glutamic acid

Thus, the correct conclusion is that none of the listed amino acids are acidic.

Test: Biomolecules Level - 3 - Question 8

Ribose and 2-deoxyribose can be different iated by:

Detailed Solution for Test: Biomolecules Level - 3 - Question 8

Ribose and 2-deoxyribose can be distinguished based on their chemical reactions with specific reagents. Here are the key points regarding their differentiation:

  • Fehling’s reagent is used to test for reducing sugars.
  • Tollen’s reagent also identifies reducing sugars by producing a silver mirror in the presence of aldehydes.
  • Osazone formation occurs when sugars react with phenylhydrazine, forming characteristic crystalline structures.
  • Barfoed’s reagent differentiates between monosaccharides and disaccharides through a specific reaction.

Among these methods, osazone formation is particularly effective for distinguishing ribose from 2-deoxyribose due to the unique crystalline structures formed by each sugar.

Test: Biomolecules Level - 3 - Question 9

The number of disulphide linkage present in insulin are:

Detailed Solution for Test: Biomolecules Level - 3 - Question 9

Insulin contains a total of three disulphide linkages. These linkages are essential for maintaining the structure and stability of the insulin molecule. Here’s a brief overview:

  • Disulphide linkages are formed between the amino acid cysteine residues.
  • They play a crucial role in the folding and function of insulin.
  • Insulin is composed of two chains, A and B, linked by these disulphide bonds.
  • The presence of three disulphide linkages ensures the proper configuration necessary for insulin's biological activity.
Test: Biomolecules Level - 3 - Question 10

Which one of the following is an essential amino acid:

Detailed Solution for Test: Biomolecules Level - 3 - Question 10

Amino acids are the building blocks of proteins, and they play a crucial role in many bodily functions. Among them, some are classified as essential amino acids, meaning that the body cannot produce them and they must be obtained through diet.

The essential amino acids include:

  • Histidine
  • Isoleucine
  • Leucine
  • Lysine
  • Methionine
  • Phenylalanine
  • Threonine
  • Tryptophan
  • Valine

In contrast, some amino acids are classified as non-essential, meaning the body can synthesise them. Examples include:

  • Tyrosine
  • Proline
  • Glycine

Among the options provided, Methionine is an essential amino acid, while Tyrosine, Proline, and Glycine are non-essential. This distinction is vital for maintaining proper nutritional balance and supporting various physiological functions.

Test: Biomolecules Level - 3 - Question 11

The pKa values for the three ionizable groups X, Y and Z or Glutamic acid are 4.3, 9.7 and 2.2 respectively:

The isoelectric point for the amino acid is:  

Test: Biomolecules Level - 3 - Question 12

The reagent ‘oxine’ commonly used in analytical chemistry is:

Detailed Solution for Test: Biomolecules Level - 3 - Question 12

Oxine is a reagent frequently employed in analytical chemistry. It is particularly known for its role in complexation reactions with metal ions. Here are some key points regarding oxine:

  • Structure: Oxine, also known as 8-hydroxyquinoline, has a unique structure that allows it to effectively bind with various metals.
  • Application: It is commonly used to detect and quantify metal ions such as aluminium, copper, and iron.
  • Method: The process typically involves forming a coloured complex, which can be measured using spectrophotometry.
  • Benefits: Its sensitivity and specificity make it a valuable tool in environmental and industrial analysis.
Test: Biomolecules Level - 3 - Question 13

The structure of D-(+)-glucose is:

The structure of L-(–)-glucose is:

Test: Biomolecules Level - 3 - Question 14

Which of the following compounds will behave as a reducing sugar in an aqueous KOH solution:

Detailed Solution for Test: Biomolecules Level - 3 - Question 14

Reducing sugars are carbohydrates that can donate electrons to other molecules, which is a key feature in various chemical reactions. In an aqueous KOH solution, certain compounds can exhibit this behaviour.

To identify which compounds behave as reducing sugars, consider the following characteristics:

  • Free aldehyde or ketone groups: Compounds with these functional groups can readily reduce other substances.
  • Structural changes: The presence of hydroxyl groups can enhance reactivity in alkaline conditions.
  • Ring structures: Many sugars exist in cyclic forms that can open to reveal reactive groups.

Among the provided compounds, the one that qualifies as a reducing sugar will have the necessary functional groups and structural features that allow it to participate in redox reactions.

Test: Biomolecules Level - 3 - Question 15

The complementary DNA sequence of the given DNA 5' - G - A - A - T - T - C - 3' is:

Detailed Solution for Test: Biomolecules Level - 3 - Question 15

Complementary DNA sequence:

  • The given DNA sequence is 5' - G - A - A - T - T - C - 3'.
  • To find the complementary sequence, replace each base with its pair:
    • G pairs with C
    • A pairs with T
    • T pairs with A
    • C pairs with G
  • The complementary strand is written in the opposite direction, resulting in 3' - C - T - T - A - A - G - 5'.

The correct complementary sequence is 3' - C - T - T - A - A - G - 5'.

Test: Biomolecules Level - 3 - Question 16

α-Farnesene shown below is a 

Test: Biomolecules Level - 3 - Question 17

The correct relationship within each pair of the natural product is

Detailed Solution for Test: Biomolecules Level - 3 - Question 17

Camphor is classified as a terpene, while insulin is a type of protein. Nicotine belongs to the alkaloid category, and streptomycin is known as a carbohydrate.

  • Camphor: A natural compound primarily found in camphor trees, widely used for its aromatic properties.
  • Insulin: A crucial hormone that regulates blood sugar levels, essential for metabolism.
  • Nicotine: A stimulant found in tobacco, recognised for its psychoactive effects.
  • Streptomycin: An important antibiotic used to treat bacterial infections.

This classification helps in understanding the different characteristics and applications of these substances in various fields, such as medicine and biochemistry.

Test: Biomolecules Level - 3 - Question 18

Which one among the following is a Sesquiterpene:

Detailed Solution for Test: Biomolecules Level - 3 - Question 18

Sesquiterpenes are a class of terpenes composed of three isoprene units, resulting in a molecular formula of C15H24. They are commonly found in various plants and are notable for their strong aromas and potential health benefits.

  • Sesquiterpenes are often responsible for the fragrance of essential oils.
  • They play a crucial role in plant communication and defence mechanisms.
  • These compounds are studied for their medicinal properties, including anti-inflammatory and antimicrobial effects.
  • Examples of sesquiterpenes include caryophyllene and humulene.

Identification of sesquiterpenes can be done through various methods including:

  • Gas chromatography for separation and analysis.
  • Mass spectrometry for determining molecular structure.

In summary, sesquiterpenes are vital compounds in nature with significant applications in both industry and health.

Test: Biomolecules Level - 3 - Question 19

The compound given below is a: 

Test: Biomolecules Level - 3 - Question 20

One sequence of amino acids respects for long distances in silk protein. Complete hydrolysis of one mole of a fragment with this sequence gives 2 mol alanine, 3 mol glycine, and 1 mol serine. Partial hydrolysis yields Ala-Gly-Ala, Gly-Ala-Gly, Gly-Ser-Gly, and Ser-Gly-Ala peptides. What is the amino acid repeat:

Detailed Solution for Test: Biomolecules Level - 3 - Question 20

In silk proteins, a specific sequence of amino acids is notable for its long-distance properties. The complete hydrolysis of one mole of a fragment with this sequence yields:

  • 2 moles of alanine
  • 3 moles of glycine
  • 1 mole of serine

During partial hydrolysis, the following peptides are produced:

  • Ala-Gly-Ala
  • Gly-Ala-Gly
  • Gly-Ser-Gly
  • Ser-Gly-Ala

To determine the amino acid repeat sequence, we analyse the quantities of each amino acid:

  • Alanine (Ala): 2
  • Glycine (Gly): 3
  • Serine (Ser): 1

This suggests that the sequence likely includes a pattern that accommodates these ratios. The possible sequences are compared, and the one that best fits the required quantities is:

Gly-Gly-Gly-Ala-Ala-Ser

This sequence aligns with the hydrolysis results, confirming it as the amino acid repeat.

Test: Biomolecules Level - 3 - Question 21

Tyrosine, at pH 12, has the following structure:

Detailed Solution for Test: Biomolecules Level - 3 - Question 21

Tyrosine, when at a pH of 12, adopts a specific structure that can be visualised in the following images:

  • A:

    Structure A

  • B:

    Structure B

  • C:

    Structure C

  • D:

    Structure D

Among these structures, the correct representation of tyrosine at a high pH is illustrated in option B.

Test: Biomolecules Level - 3 - Question 22

A disaccharide does not give a posit ive test for Tollen’s reagent. Upon acidic hydrolysis, it gives an equimolar mixture of two different monosaccharides, both the which can be oxidized by bromine water. This disaccharide is:

Detailed Solution for Test: Biomolecules Level - 3 - Question 22

Bromine water oxidizes aldose sugars only not ketose sugar. “A” gives one aldose & one Ketose sugar. B & C produces two moles of same monosaccharides. Only “D” gives two different aldose sugars on hydrolysis.

Test: Biomolecules Level - 3 - Question 23

Match the following compounds with their respective classes

Test: Biomolecules Level - 3 - Question 24

Predict the products X and Y of the following peptide ligation react ion:

Test: Biomolecules Level - 3 - Question 25

Which of the following monomers do you expect to find in a RNA molecule:

Detailed Solution for Test: Biomolecules Level - 3 - Question 25

RNA molecules are made up of specific building blocks known as monomers. The primary type of monomer found in RNA is called a nucleotide. Each nucleotide consists of three components:

  • A sugar molecule, specifically ribose.
  • A phosphate group.
  • A nitrogenous base, which can be one of the following: adenine (A), guanine (G), cytosine (C), or uracil (U).

Unlike DNA, which contains thymine, RNA substitutes uracil as its base. This difference plays a crucial role in the structure and function of RNA. Understanding the components of RNA is essential for grasping how it carries genetic information and participates in protein synthesis.

Test: Biomolecules Level - 3 - Question 26

In the following react ion, the major products X and Y are:

Test: Biomolecules Level - 3 - Question 27

Among the following, the incorrect statement(s) is/are:

Detailed Solution for Test: Biomolecules Level - 3 - Question 27

Incorrect statements regarding the given options are as follows:

  • Guanine is indeed a purine nucleobase, so this statement is correct.
  • Glycine and proline are not achiral amino acids. Glycine is unique as it has two hydrogen atoms attached to its alpha carbon, making it achiral. However, proline is classified as chiral due to its cyclic structure.
  • DNA does contain glycosidic bonds and pentose sugars, confirming this statement as correct.
  • Sucrose is a non-reducing sugar because it does not have a free aldehyde or ketone group, making this statement accurate.

In conclusion, the incorrect statement is about glycine and proline being achiral.

Test: Biomolecules Level - 3 - Question 28

All natural amino acids, except one, react with cold nitrous acid and produce nitrogen gas. Which one of the following is that odd amino acid:

Detailed Solution for Test: Biomolecules Level - 3 - Question 28

All natural amino acids, except one, react with cold nitrous acid and produce nitrogen gas.

The amino acid that does not react is proline. Here are some key points to understand this:

  • Proline is unique among amino acids due to its cyclic structure.
  • This structure affects how it interacts with nitrous acid.
  • Most amino acids release nitrogen gas when treated with cold nitrous acid.
  • Proline, however, does not produce nitrogen gas in this reaction.

Understanding these differences is crucial in biochemistry, where the reactions of amino acids can influence various biological processes.

Test: Biomolecules Level - 3 - Question 29

Which one of the following statement is correct:

Detailed Solution for Test: Biomolecules Level - 3 - Question 29

Naturally occurring DNA is typically found in the beta configuration. This means that the structure of DNA in living organisms is stabilised in a way that is crucial for its functions.

Here are some key points regarding DNA and nucleic acids:

  • Nucleic acids are not derived from proteins; instead, they are the building blocks of genetic information.
  • Proteins do not store genetic information; their primary role is to perform functions within the cell.
  • Vitamins generally do not act as enzymes; they often serve as co-factors or co-enzymes that assist in various biochemical reactions.

In summary, the correct statement is that naturally occurring DNA has a beta configuration.

Test: Biomolecules Level - 3 - Question 30

 Most abundant RNA in the cell

Detailed Solution for Test: Biomolecules Level - 3 - Question 30

Ans: a

Explanation: Ribosomal ribonucleic acid (rRNA) is the most abundant type of RNA. It constitutes the material inside the ribosome and directly helps in the translation of mRNA into proteins.

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