Zoology MCQs for NEET — Practice Questions with Answers

(a) The structure of DNA molecule is a double helical structure. In this structure double helix are made up of two right handed helical polynucleotide chains which are held together by H-bonds. In these helixes the adenine (A) base is linked with thymine (T) by two H-bonds and guanine (G) is linked with cytocine (C) by three H-bonds as A=T, and G$\equiv$C.

(c) In the process of digestion the proteins present in food material are hydrolysed to amino acid. In this process two enzymes pepsin and trypsin are involve as follows :

$\mathrm{Proteins} \underset{(\mathrm{Enzyme} \mathrm{A})}{\overset{\mathrm{Pepsin}}{\to }} \mathrm{Polypeptide} \underset{(\mathrm{Enzyme} \mathrm{B})}{\overset{\mathrm{Trypsin}}{\to }} \mathrm{Amino} \mathrm{acid}$

(b) The helical structure of protein is stabilised by hydrogen bonds between amide group of the same peptide chain. These bonds are formed by -NH- group of one unit and oxygen of carbonyl group of the third unit. It takes 3.6 amino acid to complete one turn of the helix to enable. Such H-bonding and a 13 membering is formed by H-bonding. This H-bonding responsible for holding helix in a position.

(b) The a-d-glucose and b-D-glucose are isomers called anomers. The isomer having the hydroxyl group(-OH) on the right is called a-d-glucose and one having the hydroxyl group(-OH) on the left is called b-D-glucose. Such pairs of optical isomers which differ in the configuration only around C₁ atom are called anomers.

(a) Glycolysis is the first stage in the oxidation of glucose. It is an anaerobic process and involves the degradation of glucose into two molecules of pyruvate with the generation of two molecules of ATP.

(c) The aqueous solution of sucrose is dextrorotatory having ${\left[\mathrm{\alpha }\right]}_D=+66.4°$. On hydrolysis with dilute acids or enzyme invertase, cane sugar(sucrose) gives equimolar mixture of D(+)-glucose and D-(-)-fructose

$$\begin{gathered} {\mathrm{C}}_{12}{\mathrm{H}}_{22}{\mathrm{O}}_{11} + {\mathrm{H}}_2\mathrm{O} \stackrel{ \mathrm{HCl} }{\to } {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6 + {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6 \\ \mathrm{cane} \mathrm{sugar} \mathrm{D}-\mathrm{glucose} \mathrm{D}-\mathrm{fructose} \\ [\mathrm{\alpha }{]}_D=66.4° [\mathrm{\alpha }{]}_D=+52.5 [\mathrm{\alpha }{]}_D=-92.4° \end{gathered}$$

So, sucrose is dextrorotatory but after hydrolysis gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)-glucose. Therefore, the resultant solution is laevorotatory in nature with specific rotation of -39.9$°$.

(a) The number of molecules of ATP (Adenosine triphosphate) produced in lipid metabolism of a molecule of palmitic acid is 130. Monosacchride ribose is an ATP. which acts as energy currency of the cells during metabolism of carbohydrates, protein and fats.

(b) The organic compounds other than carbohydrates, proteins which maintain normal growth and nutrition in the human body (but not produced in human body) are called vitamins.

(c) Digestion is basically hydrolysis reaction in which large molecules are hydrolysed to give smaller molecules. For example when we eat proteins, it will hydrolyse and form amino acids.