987 + Interview Questions in KVPY IN GENERAL AWARENESS Page 18 InterviewSolution

851.	A long cylindrical pipe of radius 20 cm is closed at its upper end and has an airtight piston of negligible mass as shown. When a 50 Kg mass is attached to the other end of the pistion, it moves down by a distance`Deltal` before coming to equilibrium. Assuming air to be an ideal gas, `Deltal//L` (see figure) is close to (g = 10 `ms^(2)` , atmospheric pressure is `10^(5)` Pascal), A. `0.01`B. `0.02`C. `0.04`D. `0.09`
Answer» Correct Answer - C

Discussion

852.	The word ‘’KVPY’’ is written on a board and viewed through different lense such that board is at a distance beyond the focal length of the lens. lgnorging magnification effects, consider the following statements (I) Image (i) has been viewed from the planar side of a plano-convex lens and image (ii) from the planar side of a plano-convex lens. (II) Image (i) has been viewed from the concave side of a plano-concave lens and image (ii) from the convex side of a plano-convex lens. (iii) Image (i) has been viewed from the cocave side of a plano-concave lens and image (ii) from the planar side of a plano-convex lens. (iv) Image (i) has been viewed from the planar side of a plano-concave lens and image (ii) from the convex side of a plano-convex lens. Which of the above statements are correct ?A. Only (III)B. Only (IV)C. Only (III) and (IV).D. All four.
Answer» Correct Answer - D

Discussion

853.	The word ‘’KVPY’’ is written on a board and viewed through different lense such that board is at a distance beyond the focal length of the lens.lgnorging magnification effects, consider the following statements(I) Image (i) has been viewed from the planar side of a plano-convex lens and image (ii) from the planar side of a plano-convex lens. (II) Image (i) has been viewed from the concave side of a plano-concave lens and image (ii) from the convex side of a plano-convex lens. (iii) Image (i) has been viewed from the cocave side of a plano-concave lens and image (ii) from the planar side of a plano-convex lens. (iv) Image (i) has been viewed from the planar side of a plano-concave lens and image (ii) from the convex side of a plano-convex lens. Which of the above statements are correct ? (A) Only (III) (B) Only (IV). (C) Only (III) and (IV). (D) All four.
Answer» Correct Option :-(D) All four. Explanation :- (i) For plano-concave lens or concave lens if object is placed beyond focus image is erected (ii) For convex lens If object is placed beyond focus image is inverted.

853.

The word ‘’KVPY’’ is written on a board and viewed through different lense such that board is at a distance beyond the focal length of the lens.lgnorging magnification effects, consider the following statements(I) Image (i) has been viewed from the planar side of a plano-convex lens and image (ii) from the planar side of a plano-convex lens. (II) Image (i) has been viewed from the concave side of a plano-concave lens and image (ii) from the convex side of a plano-convex lens. (iii) Image (i) has been viewed from the cocave side of a plano-concave lens and image (ii) from the planar side of a plano-convex lens. (iv) Image (i) has been viewed from the planar side of a plano-concave lens and image (ii) from the convex side of a plano-convex lens. Which of the above statements are correct ? (A) Only (III) (B) Only (IV). (C) Only (III) and (IV). (D) All four.

Answer»

Correct Option :-(D) All four.

Explanation :-

(i) For plano-concave lens or concave lens if object is placed beyond focus image is erected

(ii) For convex lens If object is placed beyond focus image is inverted.

Discussion

854.	The process of cell death involving DNA cleavage in cells is known as -A. necrosisB. apoptosisC. cytokinesisD. endocytosis
Answer» Correct Answer - B

Discussion

855.	Puffs in the polytene chromosomes of Drosophilia melanogaster salivary glands represent-A. transcriptionally active genesB. transcriptionally inactive genesC. heterochromatinD. housekeeping genes
Answer» Correct Answer - A

Discussion

856.	Puffs in the polytene chromosomes of Drosophila melanogaster salivary glands represent- (A) transcriptionally active genes (B) transcriptionally inactive genes(C) heterochromatin (D) housekeeping genes
Answer» Correct Option :- (A) transcriptionally active genes

Discussion

857.	A couple has two sons and two daughters. Only one son is colour blind and the rest of the siblings are normal. Assuming colour blindness is sex-linked, which ONE of the following would be the phenotype of the parents ? (A) Mother would be colour blind, father would be normal.(B) Father would be colour blind, mother would be normal. (C) Both the parents would be normal. (D) Both the parents would be colour blind.
Answer» Correct Option :- (C) Both the parents would be normal. Explanation :- Male child recieve X-chromosome from mother only. Another normal son indicates that mother is carrier.

857.

A couple has two sons and two daughters. Only one son is colour blind and the rest of the siblings are normal. Assuming colour blindness is sex-linked, which ONE of the following would be the phenotype of the parents ? (A) Mother would be colour blind, father would be normal.(B) Father would be colour blind, mother would be normal. (C) Both the parents would be normal. (D) Both the parents would be colour blind.

Answer»

Correct Option :- (C) Both the parents would be normal.

Explanation :-

Male child recieve X-chromosome from mother only. Another normal son indicates that mother is carrier.

Discussion

858.	What is the probability that a human individual would receive the entire haploid set of chromosomes from his/her grandfather ? (A) 1/2 (B) (1/2)23 (C) (1/2)2 (D) (1/2)46
Answer» Correct option (B) (1/2)²³ Explanation: Human have 46 chromosome in diploid cells while haploid cells have 23 chromosomes in gamete, every human receive 1/2 of 46 from each parent.

858.

What is the probability that a human individual would receive the entire haploid set of chromosomes from his/her grandfather ? (A) 1/2 (B) (1/2)23 (C) (1/2)2 (D) (1/2)46

Answer»

Correct option (B) (1/2)²³

Explanation:

Human have 46 chromosome in diploid cells while haploid cells have 23 chromosomes in gamete, every human receive 1/2 of 46 from each parent.

Discussion

859.	There are exactly twelve sundays in the period from january 1 to march 31 in a certain year. Then the day corresponding to february 15 in that year isA. TuesdayB. WednesdayC. ThursdayD. not possible to determine from the given data
Answer» Correct Answer - C Obviously, 1st Jan will be Monday as there will be 90 days from jan. 1 to march 31 (Non leap year) (If year is leap year then days will be 91 =13 weaks not possible `therefore` 15th February will be Thursday

Discussion

860.	Let S be the set of all ordered pairs (x,y) of positive integers, with HCF (x,y) = 16 and LCM (x,y) = 48000. The number of elements in S is (A) 4 (B) 8 (C) 16 (D) 32
Answer» Correct Option :- (B) 8 Explanation :- 48000 = 16 × 3000 = 16 × [3¹× 2³ × 5³ ] As H.C.F. is 16 So 2³ can be selected in 1 way & 3¹ × 5³can be seleted in (1+ 1) (3 +1) = 8 ways No of ordred pairs = 8

860.

Let S be the set of all ordered pairs (x,y) of positive integers, with HCF (x,y) = 16 and LCM (x,y) = 48000. The number of elements in S is (A) 4 (B) 8 (C) 16 (D) 32

Answer»

Correct Option :- (B) 8

Explanation :-

48000 = 16 × 3000 = 16 × [3¹× 2³ × 5³ ]

As H.C.F. is 16 So 2³ can be selected in 1 way & 3¹ × 5³can be seleted in (1+ 1) (3 +1) = 8 ways

No of ordred pairs = 8

Discussion

861.	There are exactly twelve sundays in the period from january 1 to march 31 in a certain year. Then the day corresponding to february 15 in that year is (A) Tuesday (B) Wednesday (C) Thursday (D) not possible to determine from the given data
Answer» Correct Option :- (C) Thursday Explanation :- Obviously, 1st Jan will be monday as there will be 90 days from jan. 1 to march 31 (Non leap year) (If year is leap year then days will be 91 = 13 weaks not possible ) ∴ 15th February will be Thursday

861.

There are exactly twelve sundays in the period from january 1 to march 31 in a certain year. Then the day corresponding to february 15 in that year is (A) Tuesday (B) Wednesday (C) Thursday (D) not possible to determine from the given data

Answer»

Correct Option :- (C) Thursday

Explanation :-

Obviously, 1st Jan will be monday as there will be 90 days from jan. 1 to march 31 (Non leap year) (If year is leap year then days will be 91 = 13 weaks not possible )

∴ 15th February will be Thursday

Discussion

862.	Consider the set A of natural numbers n whose units digit is nonzero, such that if this units digit is erased, then the resulting number divides n. If K is the number of elements in the set A, thenA. K is infiniteB. `"K is finite but " K gt 100`C. `25leKle100`D. `Klt25`
Answer» Correct Answer - D Such numbers are =9 from 11 to 19 4 i.e. (22,24,26,28) 3 i.e. (33,36,39) 2 i.e. (44,48) 5 i.e. (55,66,77,88,99,) `bar(23)`

Discussion

863.	Consider the set A of natural numbers n whose units digit is nonzero, such that if this units digit is erased, then the resulting number divides n. If K is the number of elements in the set A, then(A) K is infinite (B) K is finite but K > 100 (C) 25 ≤ K ≤ 100 (D) K < 25
Answer» Correct Option :- (D) K < 25 Explanation :- Such numbers are = 9 from 11 to 19 4 i.e. (22, 24, 26, 28) 3 i.e. (33, 36, 39) 2 i.e. (44, 48) 5 i.e. (55, 66, 77, 88, 99) 23

863.

Consider the set A of natural numbers n whose units digit is nonzero, such that if this units digit is erased, then the resulting number divides n. If K is the number of elements in the set A, then(A) K is infinite (B) K is finite but K > 100 (C) 25 ≤ K ≤ 100 (D) K < 25

Answer»

Correct Option :- (D) K < 25

Explanation :-

Such numbers are = 9 from 11 to 19

4 i.e. (22, 24, 26, 28)

3 i.e. (33, 36, 39)

2 i.e. (44, 48)

5 i.e. (55, 66, 77, 88, 99)

23

Discussion

864.	Suppose the quadratic polynomial `p(x) = ax^2 + bx + c` has positive coefficient `a, b, c` such that `b- a=c-b`. If `p(x) = 0` has integer roots `alpha and beta` then what could be the possible value of `alpha+beta+alpha beta` if `0 leq alpha+beta+alpha beta leq 8`A. 3B. 5C. 7D. 14
Answer» Correct Answer - C `P(x)=ax^(2)+bx+c=a(x-alpha)9x-beta)` and `alpha+beta+alphabeta+1-1=(alpha=1)(beta=1)-1` `((a-b+c))/a-1` `Rightarrowalpha+beta+alphabeta=b/a-1=lambda_(1)-1` i.e. ,` b/a` is interger= `lambda_(1)` if b = `alambda_(1)` then `c=a(2lambda_(1)-1)` (because a,b,c are in A.P) `P(x)=ax^(2)=alambda_(1)x+a(2lambda_(1)-1)` `a[x^(2)+lambda_(1)x+(2lambda_(1)-1)]` `D=lambda_(1)^(2)-4 (2lambda_(1)-1)` is perfect square for integral roots `D=lambda_(1)^(2)-8lambda_(1)+4`is perfect square Let `D= (lambda_(1)-4-k) (lambda_(1)-4+k)=12` this gives `lambda_(1)-4-k=2` `(&lambda_(1)-4+k=6)/(lambda_(1)-4 " " 4&k=1)` `lambda_(1)=8` `alpha+beta=alphabeta=8-1=7`

Discussion

865.	Let I, `omega` and `omega^(2)` be the cube roots of unity. The least possible degree of a polynomial, with real coefficients having `2omega^(2), 3 + 4 omega, 3 + 4 omega^(2) ` and `5- omega - omega^(2)` as roots is -A. 4B. 5C. 6D. 8
Answer» Correct Answer - B `{:("roots"rarr, 2omega^(2),3+4omega,3+4omega^(2),5-omega-omega^(2)),(,alpha,beta,gamma,delta):}` `delta = 5-(omega + omega^(2)) = 5 - (-1) = 6` If `alpha = 2omega^(2)` is a root then `2omega` has to be a root too. total `rarr` min. 5 roots, hence min. degree `rarr 5`.

Discussion

866.	Erythropoietin is produced by ?A. HeatB. KidneyC. Bone marrowD. Adrenal gland
Answer» Correct Answer - B

Discussion

867.	Action potential in neurons is generated by a rapid influx of ?A. Chloride ionsB. Potassium ionsC. calcium ionsD. sodium ions
Answer» Correct Answer - D

Discussion

868.	Two students P and Q perform an experiment to verify Ohm’s law for a conductor with resistance R. They use a current source and a voltmeter with least counts of `0.1` mA and `0.1` mV, respectively. The plots of the variation of voltage drop (V) across R with current (I) for both are shown below The statement which is most likely to be correct is:A. P has only random error (s).B. Q has only systematic error (s).C. Q has both random and systematic errors.D. P has both random and systematic errors.
Answer» Correct Answer - D

Discussion

869.	In the circuit shown below, a student performing Ohm’s law experiment accidently puts the voltmeter and the ammeter as shown in the circuit below; the reading in the voltmeter will be close to(A) 0V (B) 4.8 V (C) 6.0 V (D) 1.2 V
Answer» Correct Option :- C) 6.0 V Explanation :- Voltmeter has very high Resistane thus it is put in parallel. If it is put in series maximum of potential difference will be across voltmeter

869.

In the circuit shown below, a student performing Ohm’s law experiment accidently puts the voltmeter and the ammeter as shown in the circuit below; the reading in the voltmeter will be close to(A) 0V (B) 4.8 V (C) 6.0 V (D) 1.2 V

Answer»

Correct Option :- C) 6.0 V

Explanation :-

Voltmeter has very high Resistane thus it is put in parallel. If it is put in series maximum of potential difference will be across voltmeter

Discussion

870.	A cylindrical copper rod has length L and resistance R. If it is melted and formed into another rod of length 2L, the resistance will beA. RB. 2RC. 4RD. 8R
Answer» Correct Answer - C

Discussion

871.	Consider the following vision defects listed in Column I & II and the corrective measures in Column III. Choose the correct combination. Column I Column II Column IIIP. Hypermetropia i. near-sightedness a. convex lens Q. Myopia ii. Far-sightedness b. concave lens (A) P-ii-b (B) Q-i-b (C) P-i-a (D) Q-i-a
Answer» The Correct option is (B) Q-i-b

Discussion

872.	Ice is used in a cooler in order to cool its contents. Which of the following will speed up the cooling process ?A. Wrap the ice in a metal foil.B. Drain the water from the cooler periodically.C. Put the ice as single block.D. Crush the ice.
Answer» Correct Answer - D

Discussion

873.	Ice is used in a cooler in order to cool its contents. Which of the following will speed up the cooling process ?A. Wrap the ice in a metal foil.B. Drain the water from the cooler periodically.C. Put the ice as single block.D. Crush the ice.
Answer» Correct Answer - D

Discussion

874.	Which sequence of events gives rise to flaccid guard cells and stomatal closure at night ?A. log [Glucose] `rArr` low osmotic pressure `rArr` low pH `rArr` high `pCO_(2)`B. low pH `rArr` high `pCO_(2) rArr` low [Glucose] `rArr` low osmotic pressureC. low osmotic pressure `rArr` high `pCO_(2) rArr` low pH `rArr` low [Glucose]D. high `pCO_(2) rArr` low pH `rArr` low [Glucose] `rArr` low osmotic pressure
Answer» Correct Answer - D

Discussion

875.	A line is drawn from the exterior of an animal cell to the centre of the nucleus , crossing through one mitochondrion . What is the minimum of membrane bilyers that the line will cross ?A. 4B. 3C. 8D. 6
Answer» Correct Answer - Bonus There chould be five membrane bilyer that line will cross 1. Cell membrane 2. Mitochdrial membrane 2. Nucleus

Discussion

876.	Assume a spherical mammalian cell has a diameter of 27 microns. If a polypeptide chain with alpha helical conformation has to stretch across the cell, how many amino acids should it be comprised of ?A. 180000B. 1800C. 27000D. 12000
Answer» Correct Answer - A No. of amino acid in one term in `alpha` helix = 3.6 Pitch length for `alpha` Helix `= 5.4 A^(@)` `:.` The No. of Amino acid in polypeptide `(3.6)/(5.4xx10^(-10))xx27xx10^(-6)` `=1.8xx10^(4)` Amino acid = 18000 amino acid

Discussion

877.	Immunosuppressive drugs like cyclosporin delay the rejection of graft post organ transplantation byA. inhibiting T cell infiltrationB. killing B cellsC. killing macrophagesD. killing dendrite cells
Answer» Correct Answer - A Immunosuppressive Drug inhibits T-ell infiltration

Discussion

878.	the element which readily froms an ionic bond has the electronic configurationA. `1s^(2) 2s^(2) 2p^(3)`B. `1s^(2) 2s^(2) 2p^(1) `C. `1s^(2) 2s^(2) 2p^(2)`D. `1s^(2) 2s^(2) 2p^(6) 3s^(1)`
Answer» Correct Answer - D Alkali metals has higest tenency to form ionic bond readily `1s^(2) 2s^(2) 2p^(6)3s^(1) `[ Na metal ]

Discussion

879.	A small child tries to moves a large rubber toy placed on the ground. The toy does not move but gets deformed under her pushing force `(vecF)` which is obliquely upward as shown. Then A. the resultant of the pushing force `(vecF)`, weight of the toy, normal force by the ground on the toy and the frictional force is zeroB. the normal force by the ground is equal and opposite to the weight of the toyC. the pushing force `(vecF)` of the child is balanced by the equal and opposite frictional forceD. the pushing force `(vacF)` of the child is balanced by the total internal force in the toy generated due to deformation
Answer» Correct Answer - A

Discussion

880.	Molar conductivities `(Lambda_(m)^(@))` at infinite dilution of `NaCl, HCl` and `CH_(3)COONa` arc `126.4, 425.9` and `91.0 S cm^(2) mol^(-1)` respectively. `Lambda_(m)^(@)` for `CH_(3)COOH` will beA. 390.5B. 299.5C. 208.5D. 217.4
Answer» Correct Answer - A `Lambda (HOAc)= Lambda^(@) (NaO Ac)+Lambda^(@)(HCl)- Lambda^(@) (NaCl)` `=91+425.9-126.4=390.5`

Discussion

881.	Selection of lysine auxotroph (bacteria which requires lysine for growth) from a mixed population of bacteria can be done by growing the bacterial population in the presence of -A. lysineB. penicillinC. lysine and penicillinD. glucose
Answer» Correct Answer - D

Discussion

882.	Selection of lysine auxotroph (bacteria which requires lysine for growth) from a mixed population of bacteria can be done by growing the bacterial population in the presence of- (A) lysine (B) penicillin (C) lysine and penicillin (D) glucose
Answer» Correct Option :- (D) glucose

882.

Selection of lysine auxotroph (bacteria which requires lysine for growth) from a mixed population of bacteria can be done by growing the bacterial population in the presence of- (A) lysine (B) penicillin (C) lysine and penicillin (D) glucose

Answer»

Correct Option :- (D) glucose

Discussion

883.	Enzyme X catalyzes hydrolysis of GTP into GDP. The GTP-bound form of X transmits a signal that leads to cell proliferation. The GDP-bound form does transmit any such signal. Mutations in X are found in many cancers. Which of the following alterations of X are most likely to contribute to cancer ?A. Mutations that increase the of X for GDPB. Mutations that decrease the affinity of X for GTPC. Mutations that decrease the rate of GTP hydrolysisD. Mutations that prevent expression of enzyme X.
Answer» Correct Answer - C `"GTP-X Complex"to"Active cell division"` `"GDP-X Complex"to"No cell division"` `and GTPunderset("X enzyme")overset("Hydrolysis")toGDP` Mutation cause the decrease of rate of hydrolysis of GTP. Thus GTP-X complex remain present for long time cell division become uncontrolled.

Discussion

884.	E.coli about to replicate was pulsed with tritiated thymidine for 5 min and then transferred to normal medium. After one cell division which one of the following observations would be correct ? (A) both the strands of DNA will be radioactive (B) one strand of DNA will be radioactive (C) none of the strands will be radioactive (D) half of one strand of DNA will be radioactive
Answer» Correct Option :- (B) one strand of DNA will be radioactive

884.

E.coli about to replicate was pulsed with tritiated thymidine for 5 min and then transferred to normal medium. After one cell division which one of the following observations would be correct ? (A) both the strands of DNA will be radioactive (B) one strand of DNA will be radioactive (C) none of the strands will be radioactive (D) half of one strand of DNA will be radioactive

Answer»

Correct Option :- (B) one strand of DNA will be radioactive

Discussion

885.	Mutation in a single gene can lead to changes in multiple traits. This is an example ofA. HeterotrophyB. Co-dominanceC. PenetranceD. Pleiotrophy
Answer» Correct Answer - D Pleiotrophy : A single gene affects more than one phenotype.

Discussion

886.	A mixture of equal numbers of fast and slow dividing cells is cultures in a medium containing a trace amount of radioactively labeled thymidine for one hour. The cells are then transferred to regular (unlabelled) medium. After 24 hrs of growth in regular mediaA. fast dividing cells will have maximum radioactivityB. slow dividing cells will have maximum radioactivityC. both will have same amount of radioactivityD. there will be no radioactivity in either type of cells
Answer» Correct Answer - A Radioactivity is mostly incorporated in rapidly dividing cell during S phase. Thus after 24 hour it will be mainly present in few rapidly dividing cell

Discussion

887.	The activity of a certain protein is dependent on its phosphorylation. A mutation in its gene changed a single amino acid which affected the function of the molecule. Which amino acid change is most likely to account for this observation ?A. Tyrosine to TryptophanB. Lysine to valineC. Leucine to isoleucineD. Valine to alanine
Answer» Correct Answer - A Serine, tyrosine, threonine and histidine are most common target for phosphorylation in eukaryotes. If Tyrosine get mutated in tryptophan than phosphorylation of gene does not occur and gene can not ne activated

Discussion

888.	If a double stranded DNA has `15%` cytosine, what is the `%` of adenine in the DNA ?A. `15%`B. `70%`C. `35%`D. `30%`
Answer» Correct Answer - C Cytosine `=15%` According to chargaff principle A = T & G = C A+T+G+C = 100 Thus A will be `35%`

Discussion

889.	Let S = {1, 2, 3, ….., 40} and let A be a subset of S such that no two elements in A have their sum divisible by 5. What is the maximum number of elements possible in A ? (A) 10 (B) 13 (C) 17 (D) 20
Answer» Correct option (C) 17 Explanation: A {1, 2,6,7,11,12,16,17, 21,22, 26,27,31,32,36,37} = & One of the element which is multiple of 5 B {3, 4,8,9,13,14,18,19,23, 24,28,29,33,34,38,39} = & One of the element which is multiple of 5.

889.

Let S = {1, 2, 3, ….., 40} and let A be a subset of S such that no two elements in A have their sum divisible by 5. What is the maximum number of elements possible in A ? (A) 10 (B) 13 (C) 17 (D) 20

Answer»

Correct option (C) 17
Explanation:

A {1, 2,6,7,11,12,16,17, 21,22, 26,27,31,32,36,37} = & One of the element which is multiple of 5 B {3, 4,8,9,13,14,18,19,23, 24,28,29,33,34,38,39} = & One of the element which is multiple of 5.

Discussion

890.	Let `S = {1,2,3,..., 40}` and let A be a subset of S such that notwo elements in A have their sum divisible by 5. What is themaximum number of elements possible in A?A. 10B. 13C. 17D. 20
Answer» Correct Answer - C `A={1,2,6,7,11,12,16,17,21,22,26,27,31,32,36,37}` & One of the element which is multiple of 5 `B={3,4,8,9,13,14,18,19,23,24,28,29,33,34,38,39}` & One of the element which is multiple of 5

Discussion

891.	How many ordered pairs of (m,n) integers satisfy `(m)/(12)=(12)/(n)`?A. 30B. 15C. 12D. 10
Answer» Correct Answer - A `(m)/(12)=(12)/(n)` `mn=144impliesn=(144)/(3)` we want to find here total number of divisors of 144 `144=2^(4).3^(2)` Total divisors are `(4+1)(2+1)=15` But negative paris are also posible hence=30.

Discussion

892.	Two workers A and B are engaged to do a piece of work. Working alone, A takes 8 hours more to complete the work than if both worked together. On the other hand, working alone, B would need `4(1)/(2)` hours more to complete the work than if both worked together. How much time would they take to complete the job working together?A. 4HoursB. 5HoursC. 6 HoursD. 7 Hours
Answer» Correct Answer - C If A & B work together take time t hr. If A work alone takes t+8hr.to complete the work Work /hr by A is `(1)/(t+8)` Similarly for B work/hr is `(1)/(t+9//2)` When A & B work together there work/hr `(1)/(t+8)+(1)/(t+9//2)` Work will be done in t hr. `t((1)/(t+8)+(1)/(t+9//2))=1` By solving t=6

Discussion

893.	Definciency of which one of the following vitamins can cause impaired blod clotting ?A. vitamin BB. Vitamin CC. VitaminD. Vitamin K
Answer» Correct Answer - D Vitamin K help in synthesis of blood clotting factor in liver .

Discussion

894.	which one of the following cell types is a part of innatae immunity ?A. skin epithelial cellsB. B cellsC. T lymphocytesD. Liver cells
Answer» Correct Answer - A innate immunity is general defense of body eq. 1. phagocyosis of invanders by macrophage 2. Restisance of skin to invading microphage 3. Destruction of micro- organisms by HCl in digestive juice etc.

Discussion

895.	the process of transfer of electrons from glucose to molecular oxygen in baceria and m mitochdira is known as-A. TCA cycleB. Oxidative phosphorylationC. fermentationD. Glycolysis
Answer» Correct Answer - B the process of electron from glucose to molecular oxygen in bacteria and mitochondrion is occue by electron tranport system which leads to oxidaive phoorylation .

Discussion

896.	Glycolysis is the breakdown of glucose to pyruvic acid ,How many molecules of pyruiv acid are formed form one molecule of glucose ?A. 1B. 2C. 3D. 4
Answer» Correct Answer - B 2 pyruvic cid molecule are formed from one glucose molecule glycolysis.

Discussion

897.	A scientist has a house just beside a busy highway. He collects leaves from some plants growing in his garden to do radio-carbon dating ( to estimate the age of the plant by estimating the amount of a radioisotope of carbon in its tissues ). Surprisingly the radio-carbon dating shows that the plant is a few thousand years old. (a) Was the result of the radio-carbon dating wrong or can you propose a reason for such an observation ? (b) What simple experiment can be done to test the reason that you have proposed ?
Answer» (a) The result of the radio-carbon dating was correct. Reason : Vehicles running on the highway beside the house emitted carbon dioxide from the combustion of petrol or diesel, which are fossil fuels. The carbon in this carbon dioxide, coming from living material that has been converted into petroleum millions of years ago, would get assimilated into the tissues of the plant as it uses carbon dioxide from the surrounding atmosphere for photosynthesis. Therefore tissues of the plant, when used for radio-carbon dating, would show the age of the plant to be many thousands of years old. (b) A simple experiment to test the validity of this explanation would be to collect seeds from the plant and grow them in a plot of land away from the highway or other sources of carbon dioxide coming from the burning of fossil fuels. Radio-carbon dating of plants growing from these seeds should show them as young plants.

Discussion

898.	The break-down of glucose in a cell occurs in any of the following pathways : Three experiments (A, B, C) have been set up. In each experiment, a flask contains the organism in growth medium, glucose and a brown dye that changes its colour to yellow when the pH decreases. The mouth of the flask is attached to a test tube containing lime water (Calcium hydroxide , as shown in the figure) . In C, but not in A and B , air is removed from the flask before beginning the experiment. After a period of growth, the following observation were made : A : Lime water turns milky , the dye colour remains the same B : The dye colour changes , lime water does not turn milky C : Lime water turns milky , the dye colour remains the same (a) Question : Identify which of the reactions is the pathways depicted above is taking place in each experiment . Give reasons for your answer . (b) Question : Identify which of the reactions in the pathways depicted above is expected to occur in Red Blood Cells (RBCs)
Answer» (a) In experiment A, ethanol fermentation occurs producing `CO_(2)`, turning lime water milky. Since acid is not produced the dye colour does not change. In experiment B, lactic acid fermentation takes place, which produces acid but does not produce `CO_(2)`. Hence dye colour changes to yellow but the lime water does not turn milky. In Experiment C, since the lime water turns milky, ethanol fermentation is occurring. In addition, since removal of air did not affect the reaction, the fermentation is anaerobic and yeast must be the organism in the flask. (b) In RBCs, lactic acid fermentation occurs.

Discussion

899.	The break-down of glucose in a cell occurs in any of the following pathways : Glucose → Pyruvic Acid Three experiments (A, B, C) have been set up. In each experiment, a flask contains the organism in growth medium, glucose and a brown dye that changes its colour to yellow when the pH decreases. The mouth of the flask is attached to a test tube containing lime water (Calcium hydroxide ; as shown in the figure). In C, but not in A and B, air is removed from the flask before beginning the experiment. After a period of growth, the following observation were made : A : Lime water turns milky; the dye colour remains the same B : The dye colour changes ; lime water does not turn milky C : Lime water turns milky ; the dye colour remains the same (a) Question : Identify which of the reactions is the pathways depicted above is taking place in each experiment. Give reasons for your answer. (b) Question : Identify which of the reactions in the pathways depicted above is expected to occur in Red Blood Cells (RBCs)
Answer» Sol. (a) In experiment A, ethanol fermentation occurs producing CO₂, turning lime water milky. Since acid is not produced the dye colour does not change. In experiment B, lactic acid fermentation takes place, which produces acid but does not produce CO₂. Hence dye colour changes to yellow but the lime water does not turn milky. In experiment C, since the lime water turns milky, ethanol fermentation is occurring. In addition, since removal of air did not affect the reaction, the fermentation is anaerobic and yeast must be the organism in the flask. (b) In RBCs, lactic acid fermentation occurs.

899.

The break-down of glucose in a cell occurs in any of the following pathways : Glucose → Pyruvic Acid Three experiments (A, B, C) have been set up. In each experiment, a flask contains the organism in growth medium, glucose and a brown dye that changes its colour to yellow when the pH decreases. The mouth of the flask is attached to a test tube containing lime water (Calcium hydroxide ; as shown in the figure). In C, but not in A and B, air is removed from the flask before beginning the experiment. After a period of growth, the following observation were made : A : Lime water turns milky; the dye colour remains the same B : The dye colour changes ; lime water does not turn milky C : Lime water turns milky ; the dye colour remains the same (a) Question : Identify which of the reactions is the pathways depicted above is taking place in each experiment. Give reasons for your answer. (b) Question : Identify which of the reactions in the pathways depicted above is expected to occur in Red Blood Cells (RBCs)

Answer»

Sol.

(a) In experiment A, ethanol fermentation occurs producing CO₂, turning lime water milky. Since acid is not produced the dye colour does not change. In experiment B, lactic acid fermentation takes place, which produces acid but does not produce CO₂. Hence dye colour changes to yellow but the lime water does not turn milky. In experiment C, since the lime water turns milky, ethanol fermentation is occurring. In addition, since removal of air did not affect the reaction, the fermentation is anaerobic and yeast must be the organism in the flask.

(b) In RBCs, lactic acid fermentation occurs.

Discussion

900.	Mohini, a resident of Chandigarh went to Shimla with her parents. There she found the same plant that they have in their backyard, at home. However, she observed that while the plants in their backyard bore white flowers, those in Shimla had pink flowers. She brought home some seeds of the plant from Shimla and planted them in Chandigarh. Upon performing self breeding for several generations she found that the plant from Shimla produced only white flowers. (a) According to you what might be the reason for this observation-genetic or eavironmental factors ? (b) Suggest a simple experiment to determine whether this variation is genetic in nature (c) Suggest another experiment to check whether this variation in flower color is due to environmental fctors.
Answer» Sol. (a) Difference in flower color is most likely due to environmental factors (b) Perform cross breeding between the plants from Chandigarh and those from Shimla to find out whether we get any pink flowers or flowers with any shade of color between pink and white in the F1 generation (c) Grow the plants from Chandigarh in Shimla and check whether they still produce white flowers or bear pink flowers

900.

Mohini, a resident of Chandigarh went to Shimla with her parents. There she found the same plant that they have in their backyard, at home. However, she observed that while the plants in their backyard bore white flowers, those in Shimla had pink flowers. She brought home some seeds of the plant from Shimla and planted them in Chandigarh. Upon performing self breeding for several generations she found that the plant from Shimla produced only white flowers. (a) According to you what might be the reason for this observation-genetic or eavironmental factors ? (b) Suggest a simple experiment to determine whether this variation is genetic in nature (c) Suggest another experiment to check whether this variation in flower color is due to environmental fctors.

Answer»

Sol.

(a) Difference in flower color is most likely due to environmental factors

(b) Perform cross breeding between the plants from Chandigarh and those from Shimla to find out whether we get any pink flowers or flowers with any shade of color between pink and white in the F1 generation

(c) Grow the plants from Chandigarh in Shimla and check whether they still produce white flowers or bear pink flowers

Discussion

Explore topic-wise InterviewSolutions in .

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