NATIONAL STANDARD EXAMINATION IN BIOLOGY … · 1 / 20 BIOLOGY NSEB 2013-2014 EXAMINATION CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000 Website : …

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BIOLOGY NSEB 2013-2014 EXAMINATION

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.) Ph.: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

CAREER POINT

Association of Teachers in Biological Sciences NATIONAL STANDARD EXAMINATION IN BIOLOGY 2013-2014

Date of Examination : 24th November 2013 Time 15.00 to 17.00 Hrs

[Ques. Paper Code : 3-3-5]

1. Thought the cud chewing animals mostly feed on fibrous foods which are low on fats, sufficient fatty acids are produced and absorbed from the gastrointestinal tract. This is due to the :

(A) interconversion of food in the liver (B) fermentation by bacteria in pregastric region (C) enzyme catalyzed processes in the duodenum (D) processes mediated by commensal bacteria in the rectum Ans. [B] Sol. In cud chewing animals (Ruminant) fibrous food fermented in Rumen and Reticulum. This rumen and

reticulum is filled with fluid containing some protozoa and bacteria. Which helps in fermentation of carbohydrate to produce volatile fatty acid. (Rumen & Reticulum is not gastric part)

2. Ram was studying anatomy of young roots of maize, magnolia, pine and money plant. He forgot to label the

slides. If he finds only sieve cells without sieve tubes and companion cells, this slide has to be of : (A) Pine (B) Money plant (C) Magnolia (D) Maize Ans. [A] Sol Pine (Gymnosperms) have sieve cells only and lacks sieve tubes and companion cells 3. The function of transport of nutrients and respiratory gases within the body is carried out by water in all the

following except : (A) sponges (B) jelly fishes (C) holothurians (D) cuttle fishes Ans. [D] Sol. Cuttle fish is member of class cephalopoda of phylum mollusca & have closed circulatory system. So

nutrient and gases transported by blood. 4. In an experiment, substrate was incubated with an enzyme to obtain a coloured product. When colour

intensity was measured every 5 minutes using a colourimeter, the following graph was obtained.

5 10 15 20 25Time in minutes

Opt

ical

den

sity

In another experiment, the same enzyme substrate mixture was boiled at the end of 15 minutes. The reaction

was allowed to continue after cooling. The product formed was measured using colourimeter. Assuming that the product formed is heat-stable, the graph obtained will look like :

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(A)


Opt

ical

den

sity

(B)


Opt

ical

den

sity

(C)


Opt

ical

den

sity

(D)


Opt

ical

den

sity

Ans. [C] Sol. Initial graph shows gradual increase in product but mixture is boiled at 15 minutes and since product is heat

stable, amount of product will remain same. Enzyme will get denatured at boiling. 5. The type of inheritance represented in the pedigree given below is :

(A) autosomal dominant (B) mitochondrial inheritance (C) X-linked recessive (D) X-linked dominant Ans. [C] Sol. X-linked recessive - Common in males due to hemizygosity - Affected males usually have unaffected parents, but the mother is carrier 6. A researcher clips pinnae of 20 rabbits trapped in a section of forest and releases them. After a fortnight he

traps 28 rabbits in the same section of forest and notices 4 of them with clipped pinnae. His estimate of rabbit population in the section of the forest should be :

(A) 2240 (B) 560 (C) 140 (D) 112 Ans. [C]

Sol 4

20 × 28 = 140

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7. In the accompanying diagrammatic representation of blood vascular system, organs represented by I, II, III and IV respectively are :

(A) Heart, gills, trunk, liver (B) Trunk, liver, gills, heart (C) Heart, liver, gills, trunk (D) Gills, heart, liver, trunk Ans. [D] Sol. I - Gills, II – Heart, III – Liver, IV – Trunk It is single circulation path of blood flow as seen in fishes. 8. Water potential (Ψ) plays important role in water absorption and conduction from soil to leaf. Under which

condition the process will go on smoothly? (A) Ψatmosphere < Ψleaf < Ψroot < Ψsoil (B) Ψatmosphere > Ψleaf > Ψroot > Ψsoil (C) Ψatmosphere = Ψleaf = Ψroot = Ψsoil (D) Ψatmosphere < Ψleaf = Ψroot > Ψsoil Ans. [A] Sol Water moves from less negative ψw to more negative ψw. So, it is least in atmosphere and maximum in soil. 9. The graph below represents the countercurrent exchange mechanism that takes place in fish gills to maximize

gas exchange.

What would happen if concurrent mechanism replaces countercurrent exchange mechanism?

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(A) (B)

(C)

(D)

Ans. [C] Sol. The graphical representation shows that percentage saturation of O2 will be just half i.e. 50% in concurrent

flow of blood as compared to countercurrent flow of blood. 10. Which of these intercellular junctions in mucosal cells of internal lining of intestine allows direct passage of

solute from one cell to another ? (A) Gap junctions (B) Plasmodesmata (C) Desmosomes (D) Tight junctions Ans. [D] Sol. Tight junction doesn't allow movement of any substance through it. It is a protective mechanism for stomach

against action of HCl. 11. The cross section of pinna of leaf in the accompanying diagram indicates that the plant must be growing in a

habitat with :

(A) abundant light and moisture (B) abundant light but less moisture (C) less light and abundant moisture (D) less light and less moisture Ans. [B] Sol Diagram indicates hypostomatic conditions and thick cuticle on both surface. Thus, habitat will have

abundant light but less moisture.

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12. In an experiment, one population of copepod females was fed with diatoms while another one was fed with dinoflagellates and number of eggs hatched was monitored for 10 days

The graphs indicates that : (A) diatoms are killing female copepods (B) diatoms are useful for sustaining copepod population (C) older female copepods produce less viable eggs (D) the diatoms are toxic for eggs Ans. [D] Sol. Population of copepodes decreases because the diatoms are toxic for eggs.

13. Which of the following tissue can be observed after boiling a piece of pinewood in nitric acid for 15 minutes?

(A) Transfusion tissue (B) Xylem vessels (C) Xylem tracheids (D) Phloem parenchyma Ans. [C] Sol. Pinewood lacks xylem vessels, Phloem parenchyma & transfusion tissue being live cell will not survive after

boiling in nitric acid. So, only xylem tracheids remains

14. Following graphs represents oxyhemoglobin dissociation curve. Point E and G represent venous O2 pressure during exercise and resting state respectively. How much total percentage of oxygen is unloaded in tissue during exercise ?

(A) R (B) Q (C) P + Q (D) P + Q + R Ans. [B] Sol. Q. in graph shows the amount of O2 unloaded in tissue during exercise.

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15. In the accompanying diagram distribution of variables like light intensity (L), nutrient availability (N), primary producers (P) and consumers (C), over 12 months in a year has been shown. The lines marked 1 to 4 correspond to :

(A) N, L, P, C (B) L, P, N, C (C) P, C, L, N (D) C, P, N, L Ans. [A] Sol. P (primary producer) and C (consumers) are correlated lines. L (Light intensity) is represented by 2nd line 16. In a given diagram of cell membrane, which letter represents a molecule of cholesterol ?

(A) P (B) Q (C) R (D) S Ans. [C] Sol. Cholesterol has cyclo-penteno phenantherene ring. 17. Green manure involves growing leguminous plants and plowing them in. It ensures enrichment of soil since : (A) root nodules on them add nitrogen salts (B) legumes grow faster and accumulate phosphate (C) legumes incorporate higher amount of potassium (D) all of the above Ans. [A] Sol Leguminous plants as green manure adds nitrogen salts due to root nodules.

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18. The major limiting factors for phostosynthesis in a plant community shown in the accompanying diagram at levels X, Y, Z respectively are :

(A) temperature, light intensity and carbon dioxide (B) carbon dioxide, temperature and light intensity (C) light intensity, carbon dioxide and temperature (D) carbon dioxide, light intensity and temperature Ans. [D] Sol X – CO2 Y – Light intensity Z – temperature 19. Denitrification is a process carrid out by microbes in which nitratres are reduced to molecular nitrogen. This

process is predominant in : (A) desert (B) bog (C) leached soil (D) tilled farm Ans. [D] Sol. Till farm increases aeration and thus microbes growth, increasing denitrification 20. In the absence of bile, what would be the status of digestion in the duodenum ? (A) Only lipids will not be digested through proteins and carbohydrates will get digested (B) Digestion of all kinds of food will be incomplete (C) All enzymes except pepsin will be denatured due to acidic pH of chyme (D) Nucleic acids will not be digested while other biomolecules will be digested Ans. [B] Sol. In absence of Bile in Intestine all enzyme functions will be hampered because bile makes the medium

alkaline. Without bile pancreatic enzyme may not work efficiently. 21. Which of the accompanying diagrams pertain to the animals with the post gastric symbiotic digestion?

(A)

(B)

(C)

(D)

Ans. [D] Sol. Vermiform appendix have symbiotic micro-organism which helps in digestion of cellulose.

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22. If a branch of day neutral plant is grafted on a long day plant exposed to light for less than the critical period: (A) graft as well as mother plant will flower (B) graft as well as mother plant will remain vegetative (C) graft will bear flowers while mother plant will not (D) graft will not flower while mother plant will Ans. [C] Sol. LDP needs light more than critical period while DNP has no effect of light period. 23. Transpiration is an energy consuming process. Spraying anti-transpirants in flowering stages may prove

advantageous but it is not advantageous in vegetative stages since it will have a direct effect on : (A) photosynthesis resulting in stunted plant (B) translocation of food through phloem (C) absorption and translocation of minerals (D) synthesis and accumulation of florigen Ans. [A] Sol. Spraying of anti-transpirants will have direct effect on photosynthesis 24. Due to leakage, the bags of biennial verity of grain in a warehouse got wet. So as to prevent it from rotting,

the bags were transferred to adjacent cold storage for a fortnight. When inspected, the grains were found to have dried completely. On sowing, the harvest could be obtained in the same year instead of the next year. This must have been due to :

(A) genetic mutation (B) vernalization (C) production of ethylene in wet grains (D) accumulation of dormancy hormones Ans. [B] Sol. Vernalization or springization reduces flowering period and has more effect on biennial variety. 25. Glucose is a high threshold substance and gets reabsorbed from ultra filtrate during urine formation. After

consuming a lot of sweets, urine of the otherwise healthy person also shows presence of glucose. This proves that reabsorption of glucose :

(A) varies from time to time (B) is time independent (C) is concentration dependent (D) is energy intensive Ans. [C] Sol. Glucose diffuses according to its concentration gradient in the nephrons. 26. A nucleic acid extracted from animal liver is loaded and run on agarose gel. After staining it shows following pattern.

If the remaining sample is treated with RNAse and loaded in gel what result would you expect ?

(A) (B)

(C) (D)

Ans. [A] Sol. RNA since lighter than DNA moves faster on agarose gel. After RNAse treatment, only DNA band near

loading well remains.

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27. Evolutionary relationship can be established by studying genetic makeup of organisms. Suppose organisms P, Q and R have similar ancestral origin but the sequence of bases in DNA may differ to some extent at certain place as shown below. How will you put them in a cladogram on the basis of their genetic differences?

P : CAAGCTA Q : ATAGCTT R : AAACCTA

(A)

P Q R

(B)

P R Q

(C)

P R Q

(D)

Q R

P

Ans. [B] Sol. P : CAAGCTA Q : ATAGCTT R : AAACCTA Common between PQR – 3 PQ – 1 QR – 1 PR – 2 This means PR are more closely related while Q is equally related to both P and R 28. To study the ability to secrete a specific protein, cells were homogenized mechanically and organelles were

separated by centrifugation. Which organelle should be used for further investigation? (A) Microsome (B) Peroxisome (C) Lysosome (D) Ribosome Ans. [A] Sol. Upon mechanical homogenization and centrifugation thereafter , ER breaks into fragments called microsome. (ER is part of GERL which is involved in secretion of protein) 29. Which of the following statements about cholesterol in animal cell membrane are correct : (i) It reduces membrane fluidity at moderate temperature. (ii) It prevents membrane gelling at low temperature by disrupting close packing of phospholipids (ii) It prevents kinking of unsaturated hydrocarbon at all temperatures (iv) It promotes kinking of unsaturated tails of phospholipids at low temperature (A) i, ii, and iv (B) i, iii and iv (C) i, ii and iii (D) i and ii Ans. [A] Sol. Cholesterol donot prevents kinking of unsaturated hydrocarbon at all temperature & rest of the statements are

correct. 30. Elaborate courtship behaviour helps in preventing inter-spacific breeding constituting : (A) temporal barrier (B) pre-zygotic reproductive barrier (C) post-zygotic reproductive barrier (D) physiological barrier Ans. [A] Sol. elaborate courtship behaviour shows a type of temporal barrier in preventing interspacific breeding. 31. Tilling is a common agricultural practice. In which of the following ways is it helpful? (A) It improves soil aeration (B) It improves soil micorrhizal content (C) It improves organic matter of the soil (D) All the above Ans. [A] Sol. Tilling improves soil aeration but decreases mycorrhizal content.

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32. In an aquatic ecosystem, a student observed that the lowest tier of the biomass pyramid was much narrower than the rest. The main producers of this ecosystem most likely are :

(A) single-celled protists (B) aquatic plants (C) grasses (D) all the above Ans. [A] Sol. Inverted biomass pyramid - Main producers – Single celled protists 33. A typical growth curve for a bacterium is shown in the diagram

In which of the following phases, the total cell count will increase while the viable count will remain same? (A) P (B) Q (C) R (D) S Ans. [C] Sol. R phase of the graph has same viable count. 34. Epididymis is the part of male reproductive tract and produces glycoproteins that coat the sperm. It also

secretes a variety of substances such as sialic acid. These functions indicate that the tissue is rich in : (A) mitochondria (B) microsomes (C) golgi bodies (D) lysosomes Ans. [C] Sol. Golgi bodies produces varieties of substances like Sialic acid and Glycoprotein etc. required for sperm

maturation in epididymis. 35. Which of the following respiratory parts/surfaces are associated with capillary beds? (A) Spiracles (B) Tracheae of invertebrates (C) Tracheae of vertebrates (D) Skin of earthworms Ans. [D] Sol. In earth-worm capillary beds helps in Cutaneous Respiration.

36. Gut fluid from a tunicate was collected and subjected to biochemical analysis. Results are shown in the brackets following the analysis :

I. 10 c.c. fluid + 10 c.c. 1% starch → Fehling test (+) II. 10 c.c. fluid + 10 c.c. 1% starch → boiling → Fehling test (–) III. 10 c.c. fluid + 10 c.c. 1% glycogen → Fehling test (+) IV. 10 c.c. fluid + 10 c.c. 1% glycogen → boiling → Fehling test (–) V. 10 c.c. fluid + 10 c.c. 1% maltose → Fehling test (–) VI. 10 c.c. fluid + 10 c.c. 1% maltose → boiling → Fehling test (–) Which of the following conclusions is most appropriate ? (A) The gut fluid is rich in carbohydrate content (B) The gut fluid contains disaccharide digesting enzymes (C) The gut fluid contains proteolytic enzymes (D) The gut fluid contains amylolytic enzymes Ans. [D] Sol. The given data shows that the gut fluid contains amylolytic enzymes.

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37. A few major discoveries in cell biology are listed I. Schleiden and Schwann proposed the cell theory II. Leewenhock discovered bacteria III. Golgi stained cells with silver nitrate, discovered golgi apparatus IV. First transmission electron microscope was developed The correct chronological order of these events starting with the earliest event is : (A) I, II, III, IV (B) II, III, I, IV (C) II, I, III, IV (D) II, I, IV, III Ans. [C] Sol. This option shows the correct sequence of discoveries in the field of cell biology. 38. Given the fact that pluripotent stem cells can be induced to differentiate by environmental signals. Which of

the following is true ? (A) Even differentiated cells retain their ability to differentiate into other cell type when given appropriate

chemical signals (B) Only totipotent cells can differentiate into any cells type (C) Only early embryonic cells are totipotent and pluripotent stem cells have limited capability to

differentiate into different cell types (D) All the above Ans. [D] Sol. Pluripotent cell → It can be differentiated only in few specific cell types. Totipotent cell → It can differentiated in all cell types. 39. All of the following statements are true for the typical eukaryotic cell cycle EXCEPT : (A) The S phase is far removed from cell division (B) Gl occurs after the S phase (C) The shortest phase in terms of time is usually M (D) Many of the enzymes necessary for DNA replication are typically produced during Gl Ans. [B] Sol. G1 occurs before S-phase. 40. Which of the following represent the correct order of electron flow during photosynthesis ? (A) Chlorophyll to NADPH to water (B) Water to NADPH to chlorophyll (C) Chlorophyll to water to NADPH (D) Water to chlorophyll to NADPH Ans. [D] Sol. Electron is transported as under. Photolysis →

−e PS II →−e NADPH

of water (Chlorophyll) 41. Which of the following pairs is mismatched? (A) Spliceosome – removal of introns (B) Transposase – insertion of DNA segments into DNA (C) DNA polymerase – makes a molecule of DNA from a DNA template (D) RNA polymerase – makes a molecule of RNA from an RNA template Ans. [D] Sol. DNA dependent RNA polymerase enzyme (RNA polymerase uses DNA template)

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42. The structure of guanine is given below :

N|

C

CC||C

NN|

H

H N

H

CH

N

O||

H

To which of the following Guanine will have to bind to form RNA nucleotide?

(I) O

–O– P– O–

O–

(II)

O

H

OH

H

OH

OH

H

HOH2C

H

(III)

O

CN

C

C

CN

H

H

H

OH

(IV)

O

H

OH

H

OH

OH

H

HO–C–H

H

H

(V) – OH Options : (A) I, II and III only (B) I and IV only (C) III, IV and V only (D) I or II only Ans. [B] Sol. In RNA molecule Sugar + Base + Phosphate = Nucleotide (Ribose) Guanine −

4PO IV I so, I and IV only 43. To study the changes in the shape of moving amoebae isolated from a pond sample one would use - (A) Transmission electron microscope with unstained specimen (B) Dissecting microscope with fixed specimen (C) Bright field microscope with fixed specimen (D) Phase contrast microscope with unstained specimen Ans. [D] Sol. Unstained living organism can be observed with help of phase contract microscope. 44. A plant growing in marshy shine areas have certain adaptation to help them sustain in such soils. One such

adaptation is that these plants accumulate Na+ and Cl– ions and transport them to the leaves. Which of the following are consequences of this adaptation ?

a. It helps in making the water potential of leaves less negative. b. It helps reduce the transpirational loss of water c. It helps in increasing uptake of water by the roots d. It helps in making the leaves more succulent. (A) (a) and (c) (B) (a) and (b) (C) (b) and (c) (D) (c) and (d) Ans. [A] Sol. A and C Accumulation of Na+ and Cl– ions increases uptake of water by roots as the ψw becomes more negative in

leaves.

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45. The evolutionary strategy is termed as 'r' for those species that live in unstable environments and produce large number of offspring. What other characteristic correlates with this life strategy ?

(A) Large size of organisms (B) A prolonged period of parental care (C) Long life expectancy (D) Early maturity Ans. [D] Sol. r strategy includes - Large number of offspring - Small size of organisms - Early maturity 46. Cystic fibrosis is ian autosomal recessive disorder. Shyam's parents do not suffer from the disorder but he has

a brother who is duffering from cystic fibrosis. What is the probability that shyam is a carrier ? (A) 1/4 (B) 1/2 (C) 2/3 (D) 3/4 Ans. [C] Sol. Cystic fibrosis – autosomal recessive disorder

Shyam's parent Aa Aa

A A a a AA

aa Aa Aa

Shyam's brother

Shyam's brother is suffering from cystic fibrosis but Shyam isn't so, probability of him being carrier is 32

47. The changes in a plant cell when it is transferred from solution P to solution Q are shown in the figure below–

Solutions P and Q most likely are : (A) P : Saturated salt solution Q : Dilute salt solution (B) P : Solution with water potential – 5mPa Q : Solution with water potential – 10mPa (C) P : Solution with water potential – 5mPa Q : Solution with water potential – 0mPa (D) P : Alcohol Q : Water Ans. [B] Sol. Figure shows exosmosis So, Q is having more negative ψw than P 48. If 'K' is the carrying capacity of the habitat, 'N' is the total number of individuals in a population with

intrinsic rate of reproduction as 'r', the growth rate of such a population will be directly proportional to: I. r II. 1/N III. 1/K-N IV. (K-N)/K (A) I and II (B) II and III (C) I and IV (D) Only I Ans. [C]

Sol. dtdN = rN

−

KNK

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49. The brown lizard species (Anolis sagrei) was introduced in a habitat where the green lizard (Anolis carolinensis) was already dwelling. After some years, it was found that low shrubs and grasses were occupied by brown lizards while green lizards lived higher up in the trees and foliage. This is an example of niche separation due to :

(A) Competition (B) Commensalism (C) Ammensalism (D) Symbiosis Ans. [A] Sol. Niche separation i.e. resource partitioning is due to competition.

50. DNA was extracted from two bacterial species to find out how genetically different they were. Which of the following methods would help ?

(A) Comparing the A + G contents (B) Comparing the denaturation temperatures (C) Finding the percent complementarity (D) Comparing the genome sizes Ans. [A] Sol. Same size, percent complementarity and denaturation temperature may have different sequence of genes.

51. Repeated freezing and melting of water trapped in crevices of rock causes its fragmentation because ice : (A) is harder than water (B) occupies greater space than water (C) absorbs heat from rock (D) hydrolyses salts more readily than water Ans. [B] Sol. Ice occupies more volume than water. 52. Which of the following process is catalysed by ribozyme ? (A) Delinking of larger and smaller subunits of ribosomes (B) Linking of amino acids into polypeptide chain in a ribosome (C) Making complementary copies of short pieces of RNA (D) Making complmentary DNA from pieces of RNA Ans. [B] Sol. Ribozyme – 23S (prokaryotes) or 28S (Eukaryotes) of ribosome is peptidyl transferase enzyme which links

amino acids into polypeptides.

53. The shaded parts in all the animals exhibit :

(A) Homology (B) Analogy (C) Paralogy (D) Orthology Ans. [B] Sol. All shaded structure helps in same function (Locomotion) but they are different in origin. so they are

analogous organs.

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54. Accompanying pictures show the closely related species inhabiting environments varying mainly in temperatures. Which of them seems to be from the warmest habitat ?

(A) (B) (C) (D)

Ans. [A] Sol. According to Allen's law warm blooded animals living in cold climate have an short extrimities of body &

animals living in warm climate have long extrimities.

55. Average primary production (g/m2/yr) of the following has been depicted in the accompanying diagram. i – Tropical rain forests ii- Tundra iii- Swamps iv- Taiga v-Coral reef Match these with the column from 1 to 5

(A) v,i,iii,iv,ii (B) i,v,iv,iii,ii (C) ii,iv,iii,v,i (D) iv,i,v,iii,ii Ans. [B] Sol. Tropical rain forest has maximum average primary production. Coral reaf too has more productivity but have

more consumers. Tundra has least productivity.

56. Study the diagram of seaweed cover in presence and absence of limpets and urchins. Which of the following statements is appropriate ?

(A) Neither limpets nor urchins influence seaweed cover (B) Limpets and urchins both influence seaweed cover but the influence of limpets is higher (C) Limpets and urchins both influence seaweed cover but the influence of urchins is higher (D) Only urchins seem to influence seaweed cover Ans. [C] Sol. The graph dipicts that both limpets sea urchins influence seaweed cover but the effect of urchins is higher as

compared to limpets.

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57. All Ca2+ in the physiologically balanced salt solution, in which a nerve muscle preparation was kept, has got precipitated. On stimulating the nerve you would expect :

(A) Stronger contraction of skeletal muscle (B) No contraction of skeletal muscle (C) Muscle going into titany (D) Contraction of skeletal muscle after a time lapse Ans. [C] Sol. Precipitation of Ca+2 in muscle cell leads in its contraction for prolonged time. This is tetany. Because

calcium ions are required for both contraction & relaxation of muscle. 58. If all the septa in the body of an earthworm were thick, muscular and non-porous, which of the following

processes would have been impaired ? (A) Excretion and osmoregulation (B) Exchange of respiratory gases (C) Circulation of blood (D) Locomotion Ans. [D] Sol. Due to non-porous septa body fluid can not flow intersegmentally during locomotion. 59. Floating ribs are named so since they are - (A) the only ribs surrounded by body fluid (B) without intercostal muscles (C) not connected to sternum (D) not connected to vertebral column Ans. [C] Sol. 11th and 12th pair of Ribs are not connected with sternum hence called as floating Ribs. 60. The bars in accompanying diagram compare an ectotherm (Alligator) and endotherm (humans) of same body

weight for their maximum potential metabolic rate and source of ATP. What would i, ii, iii respectively be the sources of ATP ?

(A) Aerobic respiration, glycolysis, existing cellular (B) Existing cellular, glycolysis, aerobic respiration (C) Glycolysis, existing cellular, aerobic respiration, (D) Glycolysis, aerobic respiration, existing cellular Ans. [B] Sol. Initially both alligator & man will use the existing cellular energy, then glycolysis will occur followed by

aerobic respiration.

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61. For sedentary / semi sedentary and free living animals, the most suited type of symmetry has been - (A) Asymmetry and symmetry respectively (B) Radial and bilateral respectively (C) Bilateral and spherical respectively (D) Bilateral both Ans. [B] Sol. Organisms with Radial symmetry are mostly sedentary or semisedentary while with Bilateral symmetry are

free living. 62. When there is sepsis (pus formation), lymph nodes close to region swell. This is due to the fact that - (A) The leucocytes are degraded in lymph nodes (B) Pathogens easily invade lymph nodes (C) Lymph nodes from and store leucocytes (D) Release of histamines leads to inflammation of lymph nodes Ans. [B] Sol. Lymph node traps pathogen. So there is accumulation of pathogen in lymph node leading its swelling. 63. Continuous starting causes profuse secretion of tears. The most appropriate explanation for this is : (A) It protects the cornea from damage due to desiccation (B) Straing on muscles of eyelids leads to contraction of tear glands (C) It causes capillaries to dilate leading to profuse secretion of tears (D) The dust particles entering the eye irritate the tear glands causing their hyperactivity. Ans. [A] Sol. Tear layer prevents cornea from desiccation and infection. 64. Neuroglial cells contain large amount of lipids. Choose the correct functions the cells from those given

below - i. Electric insulation ii. Thermal insulation iii. Mechanical cushioning iv. Effective synapsing v. Effective blood brain barrier (A) i, ii and v (B) ii, iii and iv (C) ii, iv and v (D) i, ii and iii Ans. [C] Sol. Neuroglial cell helps in thermal insulation. Synapsing and forms Blood Brain Barrier.

65. A bacterium growing on an agar surface divides. Most bacteria are not very motile on solid surface. The progeny of bacteria remain very near to the location of original bacterium. If 10 cells are plated and a cell takes 30 minutes to multiply then number of colonies visible next day would be -

(A) 10 (B) 102 (C) 104 (D) 108 Ans. [A] Sol. Since bacteria are not very motile and remain near to location of original spot, So, although bacteria will

divide exponentially but only 10 colonies will be observed next day. 66. An organism with the genotype BbDD is mated to one with the genotype BBDd. Assuming independent

assortment of these two genes. Write the genotype of possible offspring from this cross and calculate the probability of each genotype occurring -

(A) 1/16 BBDD, 1/2 BbDD, 1/4 BBDd, 1/4 BbDd (B) 1/4 BBDD, 1/4 BbDD, 1/4 BBDd, 1/4 BbDd (C) 1/4 BBDD, 1/4 BBDd (D) 1/2 BBDD, 1/2 BbDD, 1/2 BBDd, 1/2 BbDd Ans. [B]

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Sol.

BbDD BBDd ×

BD Bd

BBDD BBDd

BbDD BbDd

BD

bD

So, 1/4 BBDD; 1/4 BbDD; 1/4 BBDd; 1/4 BbDd

67. A standard bacterial growth curve at 37ºC is shown below –

A student inoculated a bacterial culture in a flask containing culture medium and incubated the flask at 37ºC

for 6 hours. He then kept the flask at 4ºC for a couple of hours and then reincubated it at 37ºC for further period. The most probable growth curve that these bacteria would show is :

(A) (B)

(C)

(D)

Ans. [A] Sol. Bacteria becomes dormant at 4ºC but continue to divide thereafter on reincubation at 37º

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68. In Linnaean system, each animal species is placed in a taxonomic hierarchy with seven levels. The level in the middle (4th) is order. It contains -

(A) a single species (B) all the genera of a family (C) only certain families (D) all the classes of a phylum Ans. [C] Sol. Order is formed by group of related families 69. Fish gills are - (A) only respiratory in function (B) only excretory in function (C) osmoregullatory in function (D) All the three are correct Ans. [D] Sol. Gills helps in Respiration, excretion as well as osmoregulation. 70. The blood pressure inside capillaries connected to the arterioles and venules is 32 mm Hg and 7mm Hg

respectively. What will be the pressure of interstitial fluid ? (A) 22 mm Hg (B) 0 mm Hg (C) 7 mm Hg (D) 32 mm Hg Ans. [A] Sol. (I) Blood pressure in interstitial fluid is less than B.P. in capillaries connected with arteriole (II) Blood Pressure in interstitial fluid is greater than B.P. in capillaries connected with venules 71. Which of the ecological parameters, when represented graphically, show inverted pyramid ? (A) Pyramid of number in pond ecosystem (B) Pyramid of number in parasitic ecosystem (C) Pyramid of biomass in pond ecosystem (D) Both b and c Ans. [D] Sol. Inverted pyramid is shown by both - Number pyramid in parasitic ecosystem - Biomass pyramid in pond ecosystem 72. Which of the following are biomagnified at different level of food chain ? (i) Heavy metal (ii) Aerosol (iii) DDT (iv) Green house gases (A) i and ii (B) i and iv (C) i and iii (D) ii and iii Ans. [C] Heavy metal and DDT shows biomagnification at different levels of food chain. 73. In a given food chain suppose the amount of energy at the fourth tropic level is 6 KJ, what will be the energy

available at producer level ? (A) 0.6 KJ (B) 60 KJ (C) 600 KJ (D) 6000 KJ Ans. [D] Producer level → 6000 kJ Following 10% law Energy at 4th trophic level remains 6kJ

74. A few statements are made about the characteristics of various life forms. Mark the correct statement - (A) All chordates are vertebrates but not all vertebrates are chordates (B) All prokaryots lack nuclear envelope but nuclear envelope is present in mitochondria of eukaryotes (C) All eukaryotic chromosomes are associated with histone proteins but the genome of chloroplast lacks

histones (D) All bryophytes show predominant sporophytic generation but all pteridophytes show predominant

gametophytic generation Ans. [C] Sol. Chloroplast is of prokaryotic origin and so lacks histones

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75. Rough endoplasmic reticulum (ER) is associated with the ribosomes, while smooth ER is devoid of these. Ribosomes are associated with protein synthesis. Some cells show presence of RER, others smooth ER. Which of the following cells show predominance of RER ?

(A) Pancreatic cell (B) Bone marrow cell (C) Cells of mammalian testis (D) Both a and b Ans. [D] Sol. Pancreatic and Bone marrow cell both have RER predominantly.

76. The hypothesis that chloroplasts are the descendants of blue-green bacteria is evident from the fact that (Choose the most appropriate answer)-

(A) both are green in color (B) blue green bacteria were the first organisms to contain chloroplasts (C) the blue green bacterial cell is similar to a chloroplast and is approximately the same size (D) Blue green bacteria were the first organisms to contain chlorophyll. Ans. [D] Sol. BGA had chromatophores instead of chloroplast but were first organism to contain chlorophyll.

77. The RQ for a resting human adult is approximetely 0.85. If he/she undertakes violent exercise for 3 to 5 minutes -

(A) RQ will rise (B) RQ will fall (C) RQ will remain same (D) RQ will fall and then rise Ans. [C] Sol. RQ remain same. 78. A drug 'Placlitaxel' interferes with the breakdown of microtubules and stabilizes them. This drug is also used

in cancer therapy. Which of the following stages will be severely affected when animal cell is treated with this drug ?

(A) Prophase (B) G0 (C) Anaphase (D) Cytokinesis Ans. [C] Sol. Cell will get arrested at anaphase due to non-functional spindle fibers

79. Consider a cat population of 100, in which 84 are black and 16 are white. Black trait is dominant over white. What will be number of heterozygote in this cat population -

(A) 16 (B) 36 (C) 48 (D) 84 Ans. [C] Sol. q2 = 0.16 So, q = 0.4 and p = 0.6 Frequency of heterozygotes = 2pq = 2 × 0.4 × 0.6 = 0.48 So, Population of heterozygotes = 0.48 × 100 = 48

80. A short fragment of DNA was analyzed for various mutations. When comparison between normal and mutated types was done, following gel pattern was observed for two individuals A and B. Most probable type of mutation in these individuals is -

Control A B

(A) deletion in A and Inversion in B (B) deletion in A and duplication in B (C) inversion in A and deletion in B (D) duplication in A and deletion in B Ans. [D] Sol. Duplication in A increase its size while deletion in B decreases its size. So, B moves farther than usual

control band.

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