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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 351. |
A Skeleton of four interlocking carbon rings is found inA. steroidsB. waxesC. fatsD. glycerol |
| Answer» Correct Answer - A | |
| 352. |
The most abundant protein in the whole of biosphere isA. CollagenB. RuBisCOC. TrypsinD. Insulin |
| Answer» Correct Answer - B | |
| 353. |
Biomolecules are formed by certain specic linkages between simple monomeric units. Write the names of linkages and monomeric units in the following class of biomolecules. i) Starch ii) Protein iii) Nucleic acid |
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Answer» i) Starch – Monomer: α – D – (+) – Glucose Linkage: Glycosidic linkage ii) Protein – Monomer: α – amino acids Linkage: Peptide linkage or Peptide bond iii) Nucleic acid – Monomer: Nucleotide Linkage: Phosphodiesterlinkage |
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| 354. |
STATEMENT-1 : Sucrose is a reducing sugar STATEMENT-2 : Sucrose has two glycosidic linkages. STATEMENT-3: Sucrose shows mutarolation.A. TTFB. TFTC. FTFD. FTT |
| Answer» Correct Answer - C | |
| 355. |
The number of chiral centres in the cyclic hemiacetal form of glucose isA. 3B. 4C. 5D. 6 |
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Answer» Correct Answer - C An open- chain structure of glucose has 4 chiral centres. The open chain structure exists in equilibrium with two cyclic forms. The cyclic forms of `D-(+)-` glucose are hemiacetals formed by an intramoleculer reaction of the `-OH` group of `C5` with the aldehyde group. Cyclization creates a new stereocenter at `C1` and thus there are 5 chiral centres in the cyclic hemiacetal form of glucose. This new stereocentre explains how two cyclic forms are possible. |
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| 356. |
Glucose forms an oxime but glucose pentaacetate does not. Explain. |
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Answer» Glucose reacts with NH2OH via open-chain form which has the free - CH = O group to form glucose oxime. Glucose pentaacetate, on the other hand, cannot be converted into the open-chain form because its anomeric hydroxyl group (C1 - OH) is acetylated and hence does not form the oxime. |
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| 357. |
What are glycosidic linkages ? In which type of biomolecules, are they present ? |
| Answer» It is present in disaccharides as well as in polysaccharides which are both carbohydrates. | |
| 358. |
How do you explain the presence of an aldehydic group in a glucose molecule? |
| Answer» The presence of an aldehydic group in a glucose molecule can be established by its reactions with hydroxyl with amine, hydrogen cyanide and bromine water. | |
| 359. |
Despite having an aldehyde group,Glucose does not give 2, 4-DNP test. What does this indicate? What is the significance of D and (+) here? |
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Answer» Glucose does not have open chain structure and hence it does not have a free –CHO group. Actually –CHO group combines with C5 –OH to form a hemiacetal. Glucose largely exists in the cyclic hemiacetal form along with a very small amount (0.5%) of the open chain form. Since the concentration of the open chain form is low and its reaction with 2, 4—DNP is reversible, therefore, formation of 2,4—DNP derivative cannot disturb the equilibrium to generate more of the open chain form from the cyclic hemiacetal form and hence it does not react with 2,4—DNP. The capital letter D in the name D-(+)-glucopyranose indicates that the C5—OH group is oriented towards right while the sign (+) indicates that glucopyranose is dextrorotatory. |
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| 360. |
Assertion : Glucose and fructose cannot be distinguished by silver mirror test. Reason : Glucose contains aldehydic group while fructose has ketonic group.A. If both assertion and reason are correct and reason is correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If assertion and reason are both incorrect. |
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Answer» Correct Answer - B Correct explanation. The distinction cannot be made. In the basic medium, fructose rearranges to glucose and therefore, both give the silver mirror test. |
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| 361. |
(a) Despite having an aldehydic group, glucose does not give 2, 4-DNP test. What does it indicate ? (b) Draw Haworth structure of `alpha-D(+)` glucopyranose (c ) What is the significance of D and (+) here ? |
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Answer» (a) Since glucose does not given 2, 4-DNP test, this means that the aldehydic group present in it is not available for chemical reaction. In fact, it is involved in the formation of hemiacetal ring which is a six membered oxide ring. (b) For the structure of `alpha-D(+)` glucopyranose, (c ) The symbol D signifies that the configuration of last chiral carbon `(C_(5))` in glucose is same as that of D-glyceraldehyde. Similarly, (+) sign means that glucose is dextro rotatory in nature. |
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| 362. |
The number of chiral centers in glucopyranose and fructofuranose are:A. 4 and 3B. 5 and 4C. 4 in eachD. 5 in each |
| Answer» Correct Answer - B | |
| 363. |
A saturated fatty acid isA. Oleic acidB. Linoleic acidC. Stearic acidD. All |
| Answer» Correct Answer - C | |
| 364. |
Lipids are translocated through blood byA. glycolipidsB. sulpholpidsC. lipo proteinsD. phospholids |
| Answer» Correct Answer - C | |
| 365. |
in Brain , most common types of lipids areA. glycolipidsB. lipoproteinsC. cholesterolD. steroids |
| Answer» Correct Answer - A | |
| 366. |
Milk tastes sour when kept in the open for sometime due to formation ofA. carbonic acidB. citric acidC. lactic acidD. malic acid |
| Answer» Correct Answer - C | |
| 367. |
D(+) Glucose has melting point `146^(@)C` and specific rotation `[alpha]_(C)6(25)` is +`122^(@)C`. Another D(+) Glucose has melting `150^(@)`C and specific rotation `[alpha]_(D)^(25)` is +`18.7^(@)C`. The two form have significantly different optical rotation but when an aqueous solution of either form is allowed to stand, it rotation changes. The specific rotation of one form decrease and rotation of other increases until both solution show the same value `+52.7^(@)`. The change in rotation towards an equilibrium value is called mutarotation. What percentage of `beta`-D-(+) glucopyrnsone found at equillibrium in the aqueous solution?A. 0.5B. `~~100%`C. 0.38D. 0.64 |
| Answer» Correct Answer - D | |
| 368. |
non-reducing sugars haveA. free CHO group and free CO groupB. neither free CO nor free CHO groupC. free CHO and bound CO groupD. free CO group and bound CHO group. |
| Answer» Correct Answer - B | |
| 369. |
D(+) Glucose has melting point `146^(@)C` and specific rotation `[alpha]_(C)6(25)` is +`122^(@)C`. Another D(+) Glucose has melting `150^(@)`C and specific rotation `[alpha]_(D)^(25)` is +`18.7^(@)C`. The two form have significantly different optical rotation but when an aqueous solution of either form is allowed to stand, it rotation changes. The specific rotation of one form decrease and rotation of other increases until both solution show the same value `+52.7^(@)`. The change in rotation towards an equilibrium value is called mutarotation. Mutarotation is characteristic feature ofA. EpimerB. EnantiomerC. AnomerD. Ring chain isomer |
| Answer» Correct Answer - C | |
| 370. |
How many moles of (CH3CO)2O will be required to form glucose pentaacetate form 2 moles of glucose?(a) 2 (b) 5 (c) 10 (d) 2.5 |
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Answer» Option : (c) 10 |
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| 371. |
Which of the following are the reducing sugars?A. X and YB. X and ZC. Y and ZD. all of these |
| Answer» Correct Answer - D | |
| 372. |
a. How is the mixture of aspartic acid `(A)` histidine `(B)` and threonine `(C )` separated by electrophoresis method? `pI (pH` at isoelectric point) are given. `pI` of `(A)`,(B)`, and `(C )` are `2.77,7.59`, and `5.60` respectively. b. How are they separated by solubility method? |
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Answer» a. Choose the intermediate `pH` of `5.60`. This is the `pI` of threoine, which has a zero net charge and does not migrate in the electric field or in the electrophoresis experiment. Aspartic acid `(pI=2.77)` [Refer to solved Example No.8 `(d)` above] donates an `H^(o+)` and is converted to anion `(III)`, and migrates to the anode. histidine `(pI=7.59)` accepts an `H^(o+)` and is converted to a cation, and migrates to the cathode. b. At isoelectrical point `(pI)` the amino acids have least solubility in water and this property is exploited in the separation of different amino acids obtained from the hydrolysis of protein. To the mixture of three amino acids `(A, B, and C)` set the `pH` of the solution by adding acid upto `2.77`, at which `(A)` will be least soluble in `H_(2)O` and will precipitate out. It is followed by the separation of amino acid `(A)`. ltbRgt Increase the `pH` of the remaining solution by adding base upto `5.60`, at which amino acid `(C )` will precipate out. Similarly. `(B)` will precipate out at the `pH` of `7.59`. |
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| 373. |
Find the odd one out and give the reasons.Valine, phenyl alanine, histidine, lysine, alanine. |
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Answer» Alanine, It is non essential amino acid whereas others are essential amino acids. |
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| 374. |
In `alpha`-D-Glucose, the anomeric carbon is at: A. 1B. 2C. 4D. 5 |
| Answer» Correct Answer - A | |
| 375. |
Assertion: Reducing sugars undergo mutarotion. Reason: during mutarotaiton, one anomeric form is converted into an equilibrium mixture of two anomers.A. Both A and R are true and R is the correct explanation of AB. both A and R true but R is not a correct explanation of AC. A is true but R is falseD. A is false but R is true |
| Answer» Correct Answer - B | |
| 376. |
Statement I: There is a releatikonship between the ability of a sugar to mutarotate and to reduce Tollens reafent. Statement II: The reduction of Tollens reafent and mutarotation both depend on the presence of free carbonyl form.A. Statement `I` is true, Statement `II` is true, Statements `II` is the correct explanation of Statement `I`.B. Statement `I` is true, Statement `II` is true, Statement `II` is not the correct explanation of Statement `I`.C. Statement `I` is true, Statement `II` is false.D. Statement `I` is false, Statement `II` is true. |
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Answer» Correct Answer - A Both the statement are correct and statement`II` the is correct explanation of statement `I`. |
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| 377. |
Statement-1 : Glucose and fractose cannot give similar osazone on reaction with `Ph-NH-NH_(2)`. Statement-2 : Glucose and fructose have similar configuration on `C_(3),C_(4),C_(5)` carbon.A. STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1B. STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-1C. STATEMENT-1 is true, STATEMENT-2 is falseD. STATEMENT-1 is false, STATEMENT-2 is true |
| Answer» Correct Answer - D | |
| 378. |
`D`-Glucose and `D`-fructose both form the same osazone, Which statements are correct about the above reaciton?A. Glucose and fructose are epimers.B. Glucose and fruc tose are anomers.C. The c onfigurations of the OH group at C-3 and c-4 in glucose and fruc tose are same.D. The configurations of the OH group at C -4 and C-5 in glucose and fructose are same. |
| Answer» Correct Answer - C::D | |
| 379. |
Statement I: Glycosides mutarotate. Statement II: The anomeric `OH` is etherified and the equilibrium with the free carbonyl form is destroyed.A. Statement `I` is true, Statement `II` is true, Statements `II` is the correct explanation of Statement `I`.B. Statement `I` is true, Statement `II` is true, Statement `II` is not the correct explanation of Statement `I`.C. Statement `I` is true, Statement `II` is false.D. Statement `I` is false, Statement `II` is true. |
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Answer» Correct Answer - D Statements `I` is wrong and statement `II` is the correct explanation of why glycosides do not mutarotate. |
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| 380. |
Explain, in neutral or basic aqueous solution, glycosides do not show mutarotation, but in acidic medium they show mutarotation. |
| Answer» Since glycosides are acetals, they undergo hydrolysis in aqueous acid to form cyclic hemiacetals which undergo mutarotation. | |
| 381. |
`0.89 g` of an `alpha-` amino acid `(A)` gave `0.224` lit. `N_(2)` gas at `NTP` on reaction with `HNO_(2)`. In this process an optically active `(B)` is formed. A as well as `B`, gave cyclic compounds `C` and `D` on intermolecular dehydration Identify `A` to `D`. |
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Answer» Correct Answer - (A) `CH_(3)-underset(underset(NH_(2))(|))(CH)-overset(overset(O)(||))(C)-OH` (B) `CH_(3)-underset(underset(OH)(|))(CH)-overset(overset(O)(||))(C)-OH` `n_(N_(2))=(0.224)/(22.4)=10^(-2)` `because 1` mole of amino = mole of `N_(2)` `:. 10^(-2)` mole of `N_(2)` is liberated `=10^(-2)` moles of amino acid. `:. 10^(-2)` mole of amino acid weigh `= 0.89 g` `:. 1` mole of amino acid weigh `= 89 g`. |
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| 382. |
A compound [X] discharges bromine water in `CCL_(4)`.The compound neither gives any colour with `FeCl_(3)` nor effervescences with aq. `NaHCO_(3)`solution .However ,its hyldrolysate with conc.KOH followed by acidification gives another compound Y which gives colour with `FeCl_(3)`solution as well as effervescences of `CO_(2)` with `NaHCO_(3)` solution .Compounds X and Y respectively are:A. B. C. D. |
| Answer» Correct Answer - b | |
| 383. |
Vitamin `B_(2)` is :A. ascorbic acidB. riboflavinC. thiamineD. pyridoxine. |
| Answer» Correct Answer - B | |
| 384. |
Count number of compounds which can decolurize Bromine water out of following list : |
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Answer» Correct Answer - 5 (b),(c ), (e ),(g) and (h) |
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| 385. |
Vitamin `B_(6)` is known asA. PyridoxineB. ThiamineC. TocophercolD. Riboflavin |
| Answer» Correct Answer - A | |
| 386. |
Count water insoluble vatamins out of following : (a) Vitamin A (b) Vitamin B (c )Vitamin `B_(2)` (d) Vitamin `B_(6)` (e ) Vitamin B_(12) (f ) Vitamin C (g) Vitamin D (h) Vitamin E (i) Vitamin K |
| Answer» Correct Answer - 4 | |
| 387. |
The term anomer of glucose refers toA. isomers of glucose that differ in configuration at carbons one and four (C-1 and C-4)B. A mixture of (D) glucose and (L)-glucoseC. enantiomers of glucoseD. isomers of glucose that differ in configuration at carbon one (C-1). |
| Answer» Correct Answer - D | |
| 388. |
The term anomer of glucose refers toA. isomer of glucose that differs in configuration at carbon one and four (C-1 and C-4)B. a mixture of D-glucose and L-glucoseC. enantiomers of glucoseD. isomers of glucose that differ in configuration at carbon on (C-1) |
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Answer» Correct Answer - D |
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| 389. |
The term anomer of glucose refers to:A. Isomers of glucose that differ in configuration at carbons one and four `(C-1 and C-4)`.B. A mixture of `D-`glucose and `L-`glucose.C. Enantiomers of glucose.D. Isomers of glucose that differ in configuration at carbon one `(C-1)` |
| Answer» Correct Answer - D | |
| 390. |
Which of the following is an example of aldopentose?A. ErythroseB. RiboseC. FructoseD. Dihydroxyacetone |
| Answer» Correct Answer - b | |
| 391. |
The term anomer of glucose refers to:A. isomers of glucose that differ in configurations at carbons one and four (C-1 and C-4)B. a mixture of (D)-glucose and (L) -glucoseC. enantiomers of glucoseD. cyclic isomers of glucose that differ in configuration at carbon one (C-1) |
| Answer» Correct Answer - d | |
| 392. |
Correct structure of glycine at pH = 7 is :A. `H_(3)overset(o+)N-CH_(2)COO^(Θ)`B. `Noverset(o+)H_(3)-CH_(2)-COOH`C. `NH_(2)-CH_(2)-COOH`D. `NH_(2)CH_(2)COO^(Θ)` |
| Answer» Correct Answer - d | |
| 393. |
The commonest disaccharide has the molecular formula?A. `C_(12)H_(22)O_(11)`B. `C_(10)H_(20)O_(10)`C. `C_(10)H_(18)O_(9)`D. `C_(10)H_(32)O_(11)` |
| Answer» Correct Answer - A | |
| 394. |
The commonest disaccharide has the molecular formula?A. `C_(10)H_(18)O_(9)`B. `C_(10)H_(20)O_(10)`C. `C_(18)H_(22)O_(11)`D. `C_(12)H_(22)O_(11)` |
| Answer» Correct Answer - d | |
| 395. |
The commonest disaccharide has the molecular formulaA. `C_(12)H_(22)O_(11)`B. `C_(10)H_(18)O_(9)`C. `C_(10)H_(20)O_(10)`D. `C_(18)H_(32)O_(11)` |
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Answer» Correct Answer - A Sucrose, the comonest disaccharides has the molecular formula `C_(12)H_(22)O_(11)`. |
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| 396. |
The commonest disaccharide has the molecular formula?A. `C_(10)H_(8)O_(9)`B. `C_(12)H_(24)O_(12)`C. `C_(18)H_(22)O_(12)`D. `C_(12)H_(22)O_(11)` |
| Answer» Correct Answer - D | |
| 397. |
Which of the following is a disaccharide?(a) Starch(b) Fructose(c) Lactose(d) Cellulose |
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Answer» The answer is (c) Lactose |
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| 398. |
Lactose is a disaccharide ofA. glucose onlyB. glucose and fructoseC. glucose and galactoseD. all of the above |
| Answer» Correct Answer - C | |
| 399. |
Maltose is hydrolysed in the presence of mal- tase toA. glucoseB. glucose & fructoseC. fructoseD. glucose & galactose |
| Answer» Correct Answer - A | |
| 400. |
Consider the following statements: I. Glucose occur freely in nature. II. Glucose occur in the combined form. lll. Glucose present in sweet fruits and honey. IV. Ripe grapes contain glucose in large amounts. Select the correct statement(s) and choose the most appropriate option.A. I and IIIB. II and IIIC. IV and ID. I, II, III and IV |
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Answer» Correct Answer - D Glucose occurs freely in nature as well as in the combined form. It is present . in sweet fruits and honey. Ripe grapes also contain glucose in large amounts |
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