The letters ‘D’ or ‘L’ before the name of a stereoisomer
Question: The letters D or L before the name of a stereoisomer of a compound indicates the correlation of configuration of that particular stereoisomer. This refers to their relationship with one of the isomers of glyceraldehyde. Predict whether the following compound has D or L configuration. Solution: The OH group is linked on the left side of the C5 carbon atom. Hence, the given compound has L configuration....
Read More →Monosaccharides contain carbonyl group
Question: Monosaccharides contain carbonyl group hence are classified, as aldose or ketose. The number of carbon atoms present in the monosaccharide molecule is also considered for classification. In which class of monosaccharide will you place fructose? Solution: The molecular formula of fructose is C6H12O6. It also contains a ketone group, hence it is placed in the class of ketohexoses. Monosaccharides are classified depending on the number of carbon atoms the molecule contains....
Read More →Under what conditions glucose is converted
Question: Under what conditions glucose is converted to gluconic and saccharic acid? Solution: Glucose is converted to gluconic acid which is a six-carbon carboxylic acid, on treatment with a mild oxidizing agent like Br2 water. Glucose is converted to saccharic acid, which is a dicarboxylic acid, on treatment with nitric acid....
Read More →Name the linkage connecting monosaccharide
Question: Name the linkage connecting monosaccharide units in polysaccharides. Solution: The monosaccharide units in polysaccharides are linked by glycosidic bonds. A glycosidic linkage is when an oxide linkage is formed between two monosaccharide units with the loss of a water molecule....
Read More →In nucleoside, a base is attached at 1C position of the sugar moiety.
Question: In nucleoside, a base is attached at 1C position of the sugar moiety. A nucleotide is formed by linking the phosphoric acid unit to the sugar unit of a nucleoside. At which position of sugar unit is the phosphoric acid linked in a nucleoside to give a nucleotide? Solution: A nucleoside is formed when a nitrogenous base is attached to a 1 position of a five-carbon sugar. Phosphoric acid is linked to the 5 carbon of the sugar in a nucleoside molecule to give a nucleotide molecule....
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{\mathrm{x}^{2}+\mathrm{x}+1}{\left(\mathrm{x}^{2}+1\right)(\mathrm{x}+2)} \mathrm{dx}$ Solution: Denominator is factorised, so let separate the fraction through partial fraction, hence let $\frac{\mathrm{x}^{2}+\mathrm{x}+1}{\left(\mathrm{x}^{2}+1\right)(\mathrm{x}+2)}=\frac{\mathrm{Ax}+\mathrm{B}}{\mathrm{x}^{2}+1}+\frac{\mathrm{Cx}+\mathrm{D}}{\mathrm{x}+2}$......(i) $\Rightarrow \frac{\mathrm{x}^{2}+\mathrm{x}+1}{\left(\mathrm{x}^{2}+1\ri...
Read More →How do you explain the presence of
Question: How do you explain the presence of all the six carbon atoms in glucose in a straight chain? Solution: When glucose is heated for a prolonged time with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight chain....
Read More →Name the sugar present in milk.
Question: Name the sugar present in milk. How many monosaccharide units are present in it? What are such oligosaccharides called? Solution: The sugar present in milk is lactose. Lactose contains two monosaccharides, glucose and galactose. Oligosaccharides containing two monosaccharide units are called disaccharides....
Read More →Which of the following terms are correct
Question: Which of the following terms are correct about enzyme? (i) Proteins (ii) Dinucleotides (iii) Nucleic acids (iv) Biocatalysts Solution: Option (i) Proteins and (iv)Biocatalystsare the answers....
Read More →Let Q0 be the set of all nonzero rational numbers
Question: Let $Q_{0}$ be the set of all nonzero rational numbers. Let $*$ be a binary operation on $Q_{0}$, defined by $a^{*} b=\frac{a b}{4}$ for $a l l a, b \in Q_{0}$ (i) Show that $*$ is commutative and associative. (ii) Find the identity element in $\mathrm{Q}_{0}$. (iii) Find the inverse of an element $\mathrm{a}$ in $\mathrm{Q}_{0}$. Solution: (i) For commutative binary operation, $a^{*} b=b^{*} a$. $a^{*} b=\frac{a b}{4}$ and $b^{*} a=\frac{b a}{4}$ as multiplication is commutative ab = ...
Read More →Which of the following are purine bases?
Question: Which of the following are purine bases? (i) Guanine (ii) Adenine (iii) Thymine (iv) Uracil Solution: Option (i)Guanine and (ii)Adenine are the answers....
Read More →In fibrous proteins, polypeptide chains
Question: In fibrous proteins, polypeptide chains are held together by ___________. (i) van der Waals forces (ii) disulphide linkage (iii) electrostatic forces of attraction (iv) hydrogen bonds Solution: Option (ii)disulphide linkage and (iv)hydrogen bondsare the answers....
Read More →Which of the following monosaccharides
Question: Which of the following monosaccharides are present as five-membered cyclic structure (furanose structure)? (i) Ribose (ii) Glucose (iii) Fructose (iv) Galactose Solution: Option (i) and (iii) are the answers....
Read More →Lysine, is _______________.
Question: Lysine, is _______________. (i) -Amino acid (ii) Basic amino acid (iii) Amino acid synthesised in the body (iv) -Amino acid Solution: Option (i) -Amino acid , (ii)Basic amino acid and (iii)Amino acid synthesised in the body are the answers....
Read More →Amino acids are classified as acidic,
Question: Amino acids are classified as acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. Which of the following is acidic? Solution: Option (ii) and (iv) are the answers....
Read More →Consider a binary operation on
Question: Consider a binary operation on $\mathrm{Q}-\{1\}$, defined by $\mathrm{a} * \mathrm{~b}=\mathrm{a}+\mathrm{b}-\mathrm{ab}$. (i) Find the identity element in $\mathrm{Q}-\{1\}$. (ii) Show that each $a \in Q-\{1\}$ has its inverse. Solution: (i) For a binary operation $*$, e identity element exists if $a * e=e^{*} a=a .$ As $a * b=a+b-a b$ $a * e=a+e-a e(1)$ $e^{*} a=e+a-e a(2)$ using $a^{*} e=a$ $a+e-a e=a$ $e-a e=0$ $e(1-a)=0$ either $e=0$ or $a=1$ as operation is on $Q$ excluding 1 so...
Read More →Which of the following carbohydrates
Question: Which of the following carbohydrates are branched polymer of glucose? (i) Amylose (ii) Amylopectin (iii) Cellulose (iv) Glycogen Solution: Option (i)Amylose and (iv) Glycogenare the answers....
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{1}{x \log x(2+\log x)} d x$ Solution: Let substitute $u=\log x \Rightarrow d u=\frac{1}{x} d x$, so the given equation becomes $\int \frac{1}{x \log x(2+\log x)} d x=\int \frac{1}{u(2+u)} d u \ldots .(i)$ Denominator is factorised, so let separate the fraction through partial fraction, hence let $\frac{1}{u(2+u)}=\frac{A}{u}+\frac{B}{(2+u)} \ldots$(ii) $\Rightarrow \frac{1}{u(2+u)}=\frac{A(2+u)+B u}{u(2+u)}$ $\Rightarrow 1=\mathrm{A}(2+\math...
Read More →Proteins can be classified into two types
Question: Proteins can be classified into two types on the basis of their molecular shape i.e., fibrous proteins and globular proteins. Examples of globular proteins are : (i) Insulin (ii) Keratin (iii) Albumin (iv) Myosin Solution: Option (i)Insulin and (iii)Albumin are the answers...
Read More →Define * on Z by a * b = a + b - ab. Show that
Question: Define $*$ on $Z$ by $a * b=a+b-a b$. Show that $*$ is a binary operation on $Z$ which is commutative as well as associative. Solution: $*$ is an operation as $a * b=a+b-a b$ where $a, b \in Z$. Let $a=\frac{1}{2}$ and $b=2$ two integers. $\mathrm{a} * \mathrm{~b}=\frac{1}{2} * 2=\frac{1}{2}+2-\frac{1}{2} \cdot 2 \Rightarrow \frac{1+4}{2}-1=\frac{5-2}{2} \Rightarrow \frac{3}{2} \in \mathrm{Z}$ So, $*$ is a binary operation from $Z \times Z \rightarrow Z$. For commutative, $\mathrm{b}^{...
Read More →Carbohydrates are classified on the basis of
Question: Carbohydrates are classified on the basis of their behaviour on hydrolysis and also as reducing or non-reducing sugar. Sucrose is a __________. (i) monosaccharide (ii) disaccharide (iii) reducing sugar (iv) non-reducing sugar Solution: Option (ii)disaccharideand (iv)non-reducing sugar are the answers....
Read More →Three structures are given below in
Question: Three structures are given below in which two glucose units are linked. Which of these linkages between glucose, units are between C1 and C4 and which linkages are between C1 and C6? (i) (A) is between C1 and C4, (B) and (C) is between C1 and C6 (ii) (A) and (B) are between C1 and C4, (C) is between C1 and C6 (iii) (A) and (C) is between C1 and C4, (B) is between C1 and C6 (iv) (A) and (C) is between C1 and C6, (B) is between C1 and C4 Solution: Option (iii)(A) and (C) is between C1 an...
Read More →Structure of a disaccharide formed by glucose
Question: Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units. (i) a carbon of glucose and a carbon of fructose. (ii) a carbon of glucose and e carbon of fructose. (iii) a carbon of glucose and b carbon of fructose. (iv) f carbon of glucose and f carbon of fructose. Solution: Option (iii)a carbon of glucose and b carbon of fructose. is the answer....
Read More →Evaluate the following integral:
Question: Evaluate the following integral: $\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$ Solution: Denominator is factorized, so let separate the fraction through partial fraction, hence let $\frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}=\frac{A x+B}{\left(x^{2}+1\right)}+\frac{C x+D}{x^{2}+3} \ldots \ldots$ (i) $\Rightarrow \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}=\frac{(A x+B)\left(x^{2}+3\right)+(C x+D)\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}+...
Read More →Optical rotations of some compounds along
Question: Optical rotations of some compounds along with their structures are given below which of them have D configuration. (i) I, II, III (ii) II, III (iii) I, II (iv) III Solution: Option (i) I, II, III is the answer....
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