

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.
251. |
Cellulose is a protein.A. hexapolysaccharideB. pentapolysaccharideC. tripolysaccharideD. none of these |
Answer» Correct Answer - D Cellulose is a polysaccharide and consists of long chain of glucose units linked by `beta (1 to 4)` bonds with each other . |
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252. |
Assertion `:` Combinations of progesterone and estrogen are used as antifertility drugs. Reason `:` Progesterone and estrogen control the pregnancy.A. If both the assertion and reason are true and the reason is the correct explanation of the assertionB. If both assertion and reason are true but reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
Answer» Correct Answer - A Both statements are correct and reason explains assertion correctly. |
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253. |
What is animal cellulose ? |
Answer» Cellulose-like compound named tunicin present in the tunic of ascidians. | |
254. |
Where is the starch stored in the plant cells ? |
Answer» In choroplasts and amyloplasts. | |
255. |
What is the genetic meterial of tobacco mosaic virus ? |
Answer» Ribonucleic acid (RNA) | |
256. |
The genetic material of Tobacoo mosaic virus isA. RNAB. DNAC. ProteinD. NADPH |
Answer» Correct Answer - A | |
257. |
What is/are the products(s) formed in the anaerobic degradation of glucose?A. One molecule of pyruvic acidB. Two molecules of pyruvic acidC. Three molecules of pyruvic acidD. Four molecules of pyruvic acid |
Answer» Correct Answer - B Two molecules of pyruvic acid are formed during the anaerobic degradation of glucose. |
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258. |
Which vitamins is incporported into the structure of NAD/NADP?A. RiboflavinB. Vitamin PPC. Nicotinic acidD. All correct |
Answer» Correct Answer - C | |
259. |
On acetylation with acetic anhydride, glucose givesA. DiacetateB. Hexa-acetateC. PentacetateD. Tetra-acetate |
Answer» Correct Answer - C | |
260. |
Glucose reacts wih acetic anhydride to form:A. monoacetateB. tetra-acetateC. penta-acetateD. hexa-acetate |
Answer» Correct Answer - C | |
261. |
Dextrose, grape sugar and blood sugar αcurs in :(a) fructose (b) glucose (c) sucrose (d) starch |
Answer» Option : (b) glucose |
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262. |
Cane sugar on hydrolysis gives :(a) glucose and maltose (b) glucose and lactose (c) glucose and fructose (d) only glucose |
Answer» Option : (c) glucose and fructose. |
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263. |
Which of the following sugars cannot split into further groups by hydrolysis ?A. GlucoseB. SucroseC. LactoseD. Maltose |
Answer» Correct Answer - A (A) Glucose cannot split into furhter groups by hydrolysis. |
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264. |
Major function of mineral magnesium isA. Formation of bonesB. Maintenance of acid-base balanceC. Storage of energyD. Activator of enzymes |
Answer» Correct Answer - D (D) Major function of mineral magnesium is activator of enzymes. |
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265. |
Steroids areA. LipidsB. ProteinsC. VitaminsD. Carbohydrates |
Answer» Correct Answer - A (A) Steroids are lipids. |
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266. |
Haemoglobin isA. an enzymeB. a proteinC. a carbohydrateD. a vitamin |
Answer» Correct Answer - B Haemoglobin is a globular protein of four sub-units , it contains 94% globin. |
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267. |
In the following series of reactions, compound `Z` can be `Z+NH_(3) overset(H_(2)(catayst))rarrCH_(3)CH(NH_(3))COO^(-)`A. `CH_(3)CHO`B. `CH_(3)COCH_(3)`C. `CH_(3)COCOOH`D. None of these |
Answer» Correct Answer - C The resulting compound is an `alpha-` amino acid and the reagent used is `NH_(3)` and `H_(2)` catalyst, so this is anexample of preparation for which `alpha`- keto acids are staring compounds. |
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268. |
Which set of terms correctly identifies the carbohydrate shown ? (1) Pentose , (2) Pentulose, (3) Hexulose , (4) Hexose (5) Aldose, (6) Ketose, (7) Pyranose , (8) FuranoseA. 2,6,8B. 2,6,7C. 1,5,8D. A set of terms other than these |
Answer» Correct Answer - A | |
269. |
Classify the following monosaccharides in proper aldoses and ketose. |
Answer» Correct Answer - (i) Aldopentose (ii) Aldotetrose (iii) Ketopentose (iv) Ketohexose. (i) Aldopentose (ii) Aldotetrose (iii) Ketopentose (iv) Ketohexose. |
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270. |
In osazone formation, glucose reacts with three molecules of phenylhydrazine. Which statement is true regarding this ?A. All the three molecules react in similar fashionB. Two molecules react in similar manner, while the third reacts in different wayC. All the three molecules react in different waysD. None of the above is true. |
Answer» Correct Answer - B First molecule of phenylhydrazine undergoes nucleophilic addition on carbonyl (-CHO in glucose and `gt CO` in fructose) group. Second molecule of the reagent oxidizes -CHOH- at position 2(in aldoses) or `-CH_(2)OH` at position 1 (in ketoses) to from -CHO or `gtC-O` respectivley. The third molecule again undergoes nucleophilic addition on the newly developed carbonyl group to from osazone. |
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271. |
The optical rotation of the `alpha`-form of a pyranose is `+150.7^(@)`, that of the `beta`-form is `+52.8^(@)`. In solution an equilibrium mixture of these anomers has an optical rotation of `+80.2^(@)`. The precentage of the `alpha`-form in equilibrium mixture is :A. `28%`B. `32%`C. `68%`D. `72%` |
Answer» Correct Answer - A `150.7 (X) + 52.8 (1-X) = +80.2` `150.7X + 52.8X = +80.2` `97.9X = 27.4` `X = 0.279 = 0.28` |
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272. |
Anomers are those stereoisomers which differ in the configuration atA. `C_(1)`B. `C_(2)`C. BothD. Glycosidic carbon |
Answer» Correct Answer - D In aldoses (e.g. glucose) the two anomers (`alpha-` and `beta-` group), while in ketoses (e.g. fructose) the two anomers differ at `C_(2)` (Carbon having carbonyl, keto group). Such carbon (-CHO in aldoses which is `C_(1)`: and `gtC=O` in ketoses which is `C_(2)`) reacts with the -OH group present at another carbon atom to form glycosides, also known as anomers. Thus `C_(1)` in aldoses and `C_(2)` in ketoses constitute the anomeric or glycoisidic carbon. |
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273. |
The specific rotation of `alpha-`glucose is `+112^(@)` and `beta-`glucose is `+19^(@)` and the specific rotation of the constant equilibrium mixture is `+52.7^(@)`. Calculate the percentage composition of anomers `(alpha and beta)` in the equilibrium mixture. |
Answer» Let `alpha-` and `beta`-glucos have `a` and `b%`, respectively. Equilibrium specific rotaion `=((axxsp.rot.of alpha)+(bxxsp.rot.of beta))/(100)` `+52.7=(axx112+bxx19)/(100)` `=(112a+(100-a)xx19)/(100)` Solving for `a` and `b: a=36.2%,b=63.8%` |
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274. |
The specific rotation of two glucose anomers are `alpha = + 110^(@)` and `beta = 19^@` and for the constant equilibrium mixtures is `+52.7^@`. Calculate the percentage compositions of the anomers in the equilibrium mixture. |
Answer» Correct Answer - `alpha-`anomer `=37.2 %, beta-` anomer `=62.8 %`. Suppose `%` composition of `alpha= x %` Suppose, `%` composition of `beta = 100 -x` `:.` Optical rotation of equimixture `= ((x xx 110)+(110-x)19)/(100)=52.7` `:. x=37.2 %` `:. %` for `alpha= 37.2` `:. %` for `beta =62.8`. |
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275. |
a) a -D-(+) glucose and p -D(+) glucose arei) Metameresii) Anomersiii) Geometrical Isomersiv) Functional group isomersb) What is the denaturation of proteins?c) Differentiate between nucleoside and nucleotide |
Answer» a) ii) Anomers b) When a protein is treated with acid, alkali or heated or subjected to change in pH, the secondary and primary structure of protein gets ruptured. Denaturation does not change the primary structure of proteins. c) The repeating structural units of nucleic acids are called nucleotides. Pentose sugar + Base → nucleoside Nucleoside + Phosphoric acid → nucleotide |
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276. |
In an aqueous solutions of `D-` glucose the percentages of `alpha` and `beta` anomers at the equilibrium condition are respectivelyA. `64%` and `36%`B. `20%` and `80%`C. `36%` and `64%`D. `80%` and `20%` |
Answer» Correct Answer - C The `alpha` and `beta - D -` glucose have different specific rotations. When either anomer is dissolved in water their rotation change until the same fixed value results. This rotation change is called mutarotation. The `alpha` and `beta-D-` glucose are each in equilibrium with the open chain aldehyde form and therefore with each other. `alpha-D-"Glucose"hArr underset("form")("aldehyde")hArr beta-D-"Glucose"` As each anomer begins to establish this equilibrium, its specific rotation changes. When equilibrium is reached, the experimentaly determined rotation remains constant. A base such as `NaOH` catalyzes the attainment of the equilibrium. Assuming that the concentration of the open-chain form is negligible, one can, by use of the specific rotations, calculate the precentage of the `alpha` and `beta` anomers, respectively Let `a` and `b` be the mole fractions of the `alpha -` and `beta -` anomers, respectively. Solving the simultaneous equations `a + b = 1` `112^(@) a + 18.7^(@) b = 52.7^(@)` gives `a xx 100% = 36%` `b xx 100% = 64%` Note that these percentages, `36% alpha` anomer and `64% beta -` another are in accord with a greater stability for `beta - D - (+) -` glucopyranose. The preference is (what we might expect) on the basis of its having only equatorial groups. However the `beta`- anomer of a pyranose is not always the more stabel. With `D`- mannose, the equilibrium favours the `alpha-` anomer, and this result is called an anomeric effect. |
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277. |
Study the structures of `alpha-D`-(+) glucopyranose and `beta-D`(+) glucopyranose and mark the correct statement. A. Structures I and II are enantiomersB. Structures I and II are anomersC. The two structures I and II differ in the configuration of `C_(1)` and `C_(4)`D. Both the strucrtures I and II give 2,4-DNP test. |
Answer» Correct Answer - B Such isomers i.e., `alpha` form and `beta`-form which differ only in the configuration of the hydroxyl group of C-1 , are called anomers. |
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278. |
The specific rotation of freshly prepared solution of `alpha-D-` glucose changes from a value of `X^(@)` to a constant value of `Y^(@)`. The values of `X` and `Y` are respectivelyA. `112^(@)` and `52.5^(@)`B. `52.5^(@)` and `112^(@)`C. `19^(@)` and `52.5^(@)`D. `52.5^(@)` and `19^(@)` |
Answer» Correct Answer - A A solution of ordinary `D - (+) -` glucose, the `alpha`- form, `(mp 146^(@)C)` has an initial specific rotation of `112^(@)C`, but, ultimately, the specific rotation of this solution falls to `+ 52.7^(@)`. A solution of the second form of `D - (+) -` glucose `(mp 150^(@)C)` has an initial specific rotation of `+ 18.7^(@)C`, but slowly, the specific rotation of this solution rises to `+ 52.7^(@)`. |
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279. |
Which of the following configurations repersents `beta-D - (+) -` glucopyranose?A. B. C. D. |
Answer» Correct Answer - B The cyclic structures of monosaccharides can be better represented by Haworth projection formula. To get such a formula for any monosaccharide (say `beta -D-` glucose), draw a hexagon with its oxygen atom at the upper right hand corner. The lower edge is usually thickend to indicate that this edge is towards the viewer/observer. Place all the groups (on `C1, C2, C3`, and `C4`) which are present on left hand side in Fischer projection above the plane of the ring and all those groups on the right hand side below the plane of the ring. The terminal `- CH_(2)OH` group is always placed above the plane of the hexagon ring (in D-series.) |
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280. |
`alpha-D-` Glucopyranose and `beta-D-` glucopyranose areA. anomersB. epimersC. diastereomersD. Meso compounds |
Answer» Correct Answer - A | |
281. |
Which of the following structures represnts `alpha`-D-glucopyranose?A. B. C. D. |
Answer» Correct Answer - A | |
282. |
Which one of the following can act as energy carriers? (a) GTN (b) ATP (c) FAD (d) Cyclic AMP |
Answer» ATP can act as energy carriers. |
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283. |
Who invented DNA finger printing? (a) Sir Alec Jeffrey (b) Rosalind Franklin (c) Watson and Crick (d) Maurice Wilkins |
Answer» (a) Sir Alec Jeffrey |
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284. |
What are different types of RNA which are found in cell? |
Answer» RNA molecules are classified according to their structure and function into three major types. 1. Ribosomal RNA (r – RNA) 2. Messenger RNA (m – RNA) 3. Transfer RNA (t – RNA) r-RNA: rRNA is mainly found in cytoplasm and in ribosomes, which contain 60% RNA and 40% protein. Ribosomes are the sites at which protein synthesis takes place. t – RNA: tRNA molecules have lowest molecular weight of all nucleic acids. They consist of 73 – 94 nucleotides in a single chain. The function of tRNA is to carry amino acids to the sites of protein synthesis on ribosomes. m – RNA: mRNA is present in small quantity and very short lived. They are single stranded, and their synthesis takes place on DNA. The synthesis of mRNA from DNA strand is called transcription. mRNA carries genetic information from DNA to the ribosomes for protein synthesis. |
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285. |
Explain about DNA finger printing process. |
Answer» 1. DNA finger printing is one of the most accurate methods for placing an individual at the scene of a crime has been a finger print. 2. The DNA finger print is unique for every person and can be extracted from traces of sample from blood, saliva, hair etc. By using this method, we can detect the individual specific variation in human DNA. 3. In this method, the extracted DNA is cut at specific points of varying lengths in the formation of DNA fragments of varying lengths which were analysed by technique called gel electrophoresis. This method separates the fragments based on their size. The gel containing the DNA fragments are then transferred to a nylon sheet using a technique called blotting. Then the fragments will undergo autoradiography in which they were exposed to DNA probes. 4. A piece of X-ray film was then exposed to the fragments, and a dark mark was produced at any point where a radioactive probe had become attached. The resultant pattern of marks could then be compared with other samples. 5. DNA finger printing is based on slight sequence differences between individuals. These methods are providing decisive in court cases world wide. |
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286. |
Explain about the types of RNA molecules. |
Answer» 1. Ribonucleic acids are similar to DNA. Cells contain upto eight times high quantity of RNA than DNA. RNA is found in large amount in the cytoplasm and a lesser amount in the nucleus. 2. RNA molecules are classified according to their structure and function into three major types. 1. Ribosomal RNA (r – RNA) 2. Messenger RNA (m – RNA) 3. Transfer RNA (t – RNA 3. r – RNA: r – RNA is mainly found in cytoplasm and in ribosomes, which contain 60% RNA and 40% protein. Ribosomes are the sites at which protein synthesis takes place. 4. t – RNA: t – RNA molecules have lowest molecular weight of all nucleic acids. They consist of 73 – 94 nucleotides in a single chain. The function of t – RNA is to carry amino acids to the sites of protein synthesis on ribosomes. 5. m – RNA: m – RNA is present in small quantity and very short lived. They are single stranded and their synthesis take place on DNA. The synthesis m-RNA from DNA strand is called transcription. m – RNA carries genetic information from DNA to the ribosomes for protein synthesis. |
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287. |
Which one of the following bases is not present in DNA?A. OuinolineB. AdenineC. CytosineD. Thymine |
Answer» Correct Answer - A Quinoline is an alkaloid, it is not present in DNA. DNA has four nitrogen bases as adenine, guanine, cytosine and thymine. |
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288. |
Which of the following form pentacetate with acetic anhydide? (a) Glucose (b) Fructose (c) Lactose (d) Both a & b |
Answer» (d) Both a & b |
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289. |
What la an aminoacid sequence? |
Answer» It is a particular order in which amino acids are arranged to form a specific protein. The aminoacids sequence in a protein molecule is of paramount importance. |
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290. |
Vitamin `B_6` is known asA. PyridoxineB. ThiamineC. TocopherolD. Riboflavin |
Answer» Correct Answer - A Vitatmin `B_6` is called pyridoxine. It is found in fruits, green-vegetables, milk, etc. Due to its deficiency, anaemia disease is caused. |
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291. |
Glucose reacts with methyl alcohol to giveA. `alpha` - methyl glucosideB. `beta` - methyl glucosideC. Both a and bD. None of these |
Answer» Correct Answer - C Glucose reacts with `CH_3OH` in presence of dry HCI gas to give `alpha` and `beta` methyl glucoside. `underset"Glucose"(C_6H_11O_5.OH)+HOCH_3 overset"Dry HCl"to underset(alpha "and" beta "-methyl glucoside")(C_6H_11O_5.OCH_3+H_2O)` |
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292. |
Which of the following is a simplest aminoacid?A. GlycineB. AlanineC. LeucineD. Valine |
Answer» Correct Answer - A `H_2NCH_2COOH` (glycine) is the simplest amino acid. |
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293. |
In Ketohexose the possible optical isomers areA. 12B. 4C. 16D. 8 |
Answer» Correct Answer - D In Ketohexose total no. of chiral centres `=3`. Hence total no. of stereo isomers `= 2^(3) = 8`. |
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294. |
Which of the following is an example of ketohexose?A. mannoseB. galactoseC. maltoseD. Fructose |
Answer» Correct Answer - D Fructose contains six carbon atoms and one keto group. |
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295. |
Which of the following represent a peptide chain?A. `-overset(H)overset(|)(N)-underset(O)underset(||)(C)-underset(H)underset(|)(N)-overset(|)underset(|)(C)-NH-underset(O)underset(||)(C)-NH-`B. `-overset(H)overset(|)(N)-underset(O)underset(||)(C)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-overset(H)overset(|)(H)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-`C. `-overset(H)overset(|)(N)-overset(|)underset(|)(C)-underset(O)underset(||)(C)-overset(H)overset(|)(N)-overset(|)underset(|)(C)--underset(O)underset(||)(C)-overset(H)overset(|)(N)-overset(|)underset(|)(C)-overset(O)overset(||)(C)-`D. `-overset(H)overset(|)(N)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-underset(O)underset(||)(C)-underset(H)underset(|)(N)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-underset(H)underset(|)(N)-overset(|)underset(O)underset(||)(C)-overset(|)underset(|)(C)-overset(|)underset(|)(C)-` |
Answer» Correct Answer - C See polypeptides. |
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296. |
Which of the following is thermosetting polymer?A. PolytheneB. `PVC`C. TeflonD. Bakelite |
Answer» Correct Answer - D Phenol-formaldehyle (Baelite), urea-formaldehyde and meltamine-formaldehyde are thermosettin polymers. |
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297. |
Which of the following is thermosetting polymer?A. Nylon-6B. Nylon-6,6C. BakeliteD. `SBR` |
Answer» Correct Answer - C Baelite is a thermosetting polymer and have three dimensional cross-linked structures |
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298. |
Which of the following is thermosetting polymer?A. PVCB. PVAC. BakeliteD. Perspex |
Answer» Correct Answer - C Bakelit is thermosetting polymer. It becomes infusible on heating and cannot be remoulded. |
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299. |
Assertion `:` Sulpha durgs contain sulphonamide group. Reason `:` Salvarsan is a sulpha drug.A. If both the assertion and reacons are true and reason is a true explanation of the assertion.B. If both the assertion and reasons are true but the reason is not the correct explanation of assertion.C. If the assertion is ture but reason is false.D. If assertion is false but reasion is true. |
Answer» Correct Answer - C Salvarsan is arsenic based drug. |
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300. |
Silph drugs/sulphanilamide kill bacteria by inhibting of which of the following ?A. Para-aminobenzoic acidB. Felic acidC. PhenylalanineD. Methionin |
Answer» Correct Answer - B Sulpha durgs compete P-aminobenzoic acid (PABA) which is essential for the synthesis of folic acid in bacteria . The Sulpha drugs act as competetive inhibitor and combine with enzyme and do not allow enzyme to act with RABA. |
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