Ask & Answer: India's Largest Education Community

After the Christ University BSc selection rounds have been concluded, selected candidates must report to the allocated campus with a parent or parents. They must bring along the documents that are specified by the university. Candidates selected for BSc admission at Christ Bannerghatta Campus must carry the required original documents and a set of black and white photocopies of the same. The list of the required documents is given below:

• Class 10 and Class 11 final mark sheet or Class 12 result
• Transfer Certificate (TC)
• Migration Certificate (MIG)
• Two Passport Size Photographs (in formal dress with white background)
• Two Copies of Payment Acknowledgment Receipt
• Valid ID Proof: Aadhaar, PAN, etc.

To select prospective candidates for BSc admission at Christ Bannerghatta Campus, the university conducts a holistic selection process. It includes four rounds. To participate in these rounds, candidates must apply online through the university's official website. They must also fulfil the course-specific eligibility criteria prescribed by the university. The Christ Bannerghatta BSc selection rounds are as follows:

1. Entrance Test (ET): A 90-minute entrance test, unless specified otherwise
2. Skill Assessment (SA): Candidates are required to write an essay/ article on a topic
3. Micro Presentation (MP): Candidates are required to talk for 90 seconds on a topic given by the panel
4. Personal Interview (PI): A 10-minute interview to assess the overall personality of the candidates

This is a Long answer type Questions as classified in NCERT Exemplar

When DNA is fully degraded, a pentose sugar, phosphoric acid, and nitrogen-containing heterocyclic compounds known as bases are produced. Below is a diagram illustrating the structure of sugar and phosphoric acid.

DNA contains four bases adenine (A), guanine (G), cytosine (C) and thymine (T). The structures of all bases are shown below.

The attachment of a base to the 1'-position of sugar produces a nucleoside. When a nucleoside attaches to phosphoric acid at the sugar moiety's 5? position, a nucleotide is formed. Nucleotides are linked through the phosphodiester linkage between the 5' and 3' carbon atoms of the pentose sugar.

In DNA, two nucleic acid chains wrap around each other and are held together by hydrogen bonds between their bases.

This is a Long answer type Questions as classified in NCERT Exemplar

Proteins can have a single polypeptide chain or many polypeptide chains as their basic structure. A considerable number of α - amino acids are joined in a particular sequence in each polypeptide chain. The specific order in which the various α− amino acids found in a protein are linked to one another is the protein's primary structure. Any change to the main structure, i.e. the amino acid sequence, results in a different protein

Protein secondary structure: Secondary structure defines how polypeptide chains are folded or structured. As a result, the shape or conformation of the protein molecule is determined. This is explained by the planar geometry of the peptide bond and the hydrogen link between the and groups of separate peptide bonds.

An α - helix is a polypeptide chain that twists into a right-handed screw (helix), with the −NH group of each amino acid residue hydrogen bonded to the −C=O group of the next turn of the helix. The a-helix structure is also known as the 3.613 helix structure.

β- pleated sheet structure of proteins: The polypeptide chains are zigzagged side by side in this shape, with alternative R groups on the same side spaced at fixed intervals. The two nearby polypeptide chains are held together by intermolecular H-bonds. This type of interbounded chain can be used to make a sheet. These sheets are then stacked one on top of the other to create a three-dimensional structure, similar to the pages of this book.

.

This is a Long answer type Questions as classified in NCERT Exemplar

Ans: The following evidences are used to assign the given structure:

Molecular formula; of the molecule glucose is found to be C₆H₁₂O₆. 

Straight chain structure: When an aqueous solution of glucose reacts with sodium amalgam, sorbitol (a hexahydric alcohol) is produced by the reaction.

Glucose after a prolonged time of heating with hydroiodic acid and red phosphorus at 100^oC gives a mixture of n-hexane and 2-iodohexane as two products. The reaction is depicted below:

Five hydroxyl groups: When glucose is acetylated in the presence of acetic anhydride, it produces pentaacetate after the reaction. This confirms the presence of five -OH hydroxyl groups in the glucose molecule. The presence of two or more -OH groups on the same carbon atom is known to make the molecules unstable.

Presence of a single primary alcoholic group: When oxidized with concentrated nitric acid, glucose and gluconic acid, these all molecules yield the same product, dicarboxylic acid, saccharic acid, or glucaric acid. The main alcoholic group (CH₂OH) is always present at the end of the carbon chain in the molecule.

Presence of an aldehyde group: Glucose molecule combines with the hydroxylamine (NH₂OH) to produce glucose CHO oxime as the product. This implies that glucose has carbonyl (CHOH)₄ (>C=O) groups as it is needed to react in order to produce an oxime product.

This is a Long answer type Questions as classified in NCERT Exemplar

Ans: The open chain structure of glucose explained the majority of its properties. However, it fails to explain the following facts:

I Although glucose has an aldehydic group, it does not undergo all aldehydic reactions.

(a) Glucose does not react with ammonia, for example.

(b) Glucose does not create an addition product when it combines with sodium bisulphite.

(c) Glucose does not pass the Schiff's test or the 2, 4-DNP test, unlike other aldehydes.

(ii) Glucose forms an oxime when it reacts with hydroxylamine, but glucose pentaacetate does not. This proves that glucose pentaacetate contains no aldehyde groups.

(iii) There are two stereoisomeric forms of glucose: glucose and glucose. The melting temperatures and optical rotations of these two crystalline forms are different.

(iv) Mutarotation occurs in an aqueous Ans of glucose, which means that its specific rotation decreases when compared to glucose and increases when compared to glucose.

(v) Glucose produces isomeric methyl glucosides. Two isomeric monomethyl compounds are generated when glucose is burnt with methanol in the presence of dry hydrogen chloride gas.