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CSIR NET June 2018 Booklet Code - C (Question Paper)

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Special Edition CSIR NET JUNE 2018 QUESTION PAPER - BOOKLET CODE : C

PART ‘A’ 1. What is the value of (1 + 3 + 5 + 7 + …… + 4033) + 7983 x 2017 1. 20170000 2. 20172017 3. 20171720 4. 20172020

Explanation: The first person makes 10 handshakes around the room. The second person will make 9 exclusive handshakes around the room, as he has already one common handshake with the first person. The third will have 8 unique handshake and so on. Total count = 10+9+8+7+6+5+4+3+2+1 (last person will not shake hand with anyone as he is common in all handshakes) = 55 Reference: http://www.math-only-math.com/sum-of-first-n-natural-numbers.html

Answer: 1 Explanation: Following Arithmetic Progression- for the series given in brackets (1 +3+5+7.........+4033) sum of progression = N (a1 + an)/2 Values of a1 = 1, an = 4033 N = ? = number of terms in the given series N = (an - a1/ d ) +1 = (4033 -1/2) + 1 = 2017 Sum of series = 2017 (4033 + 1) / 2 = 4068289 Rest the sum of series need to be added to 7983 x 2017 so, (4068289 ) + 7983 x 2017 = 4068289 + 16101711 = 20170000 Reference: html

Answer: 3

4. Suppose (i) “A * B” means “A is the father of B”, (ii) “A Δ B” means “A is the husband of B”, (iii) “ A Δ B” means “ A is the wife of B” and (iv) “ A Δ B”, means “ A is the sister of B”. Which of the following represents “ C is the father-in-law of the sister of D”? 1. C E * F ◻️D 2. C * E F ◻️D 3. C Δ E * F ◻️D 4. C * E Δ F ◻️D Answer: 4

Explanation: In all the option F is sister of D. According to option 1 and 3 E is father of F, then C cannot be the father-in-law. In options 2 and 4 E is further shown to be related to F. As F is sister (female) then E has to be husband as given in option 4, then accordingly husbands father C has to be father-in-law https://www.math10.com/en/algebra/arithmetic-progression. of F.

2. Path of a ray of light between two mirrors is shown in the diagram. If the length of each mirrors is ‘ℓ’, what is the total path length of the ray between the mirrors?

5. In a 100 m race A beats B by 10 m, B beats C by 5 m. By how many meters does A beat C? 1. 15.0 m 2. 5.5 m 3. 10.5 m 4. 14. 5 m Answer: 4

1. ¾ ‘ℓ’ 2. 4/3 ‘ℓ’ 3. 3/2 ‘ℓ’ 4. 2 ‘ℓ’

Explanation: In 100m race, When A runs 100, B ran 90 So, A:B = 100:90 When B runs 100, C ran 95 So, B:C = 100:95 So, A:C = A:B * B:C = 100/90 * 100/95 = 200/171 So, when A runs 200 m, C completes 171 So, when A runs 100m, C completes 171/2 = 85.5m So difference = 100-85.5 = 14.5m

Answer: 4

Explanation: As per the laws of reflection, angle of incidence = angle of reflection = 30 degrees. Each of the ray triangles formed is an equilateral triangle with all angles = 60 degrees Reference: https://www.hitbullseye.com/mba/quant/Races-Games-TSD. Now, all incident rays = reflected rays = say 2x php The first incident ray and the last relected ray is outside the mirror length. Total path travelled by ray = 2x+2x+2x+2x+2x+2x = 12x (2 sides of 3 equi- 6. When a farmer was asked as to how many animals he had, he replied lateral triangles) that all but two were cows, all but two were horses and all but two were Total path travelled by ray within the mirror = 2x+2x+2x+2x = 8x pigs. How many animals did he have? Now the mirror length in terms of the equilateral triangle base = x+2x+x = 4x = L (given) 1. 3 So, if 4x = L 2. 6 8x is equal to 2*4x = 2*L 3. 8 4. 12 Reference: http://www.physicsclassroom.com/class/refln/Lesson-1/TheLaw-of-Reflection Answer: 1 3. In a group of 11 persons, each shakes hand with every other once and only once. What is the total number of such handshakes? 1. 110 2. 121 3. 55 4. 66 Phone: 080-5099-7000

Explanation: He had 3 animals - 1 cow, 1 pig and 1 horse. So, all statements were true. Reference: https://english.stackexchange.com/questions/9967/all-but-idiom-has-two-meanings

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7. Nine-eleventh of the members of a parliamentary committee are men. 11. The University needs to appoint a new Vice Chancellor which will be Of the men, two-thirds are from the Rajya Sabha. Further, 7/11 of the based on seniority. Ms. West is less senior to Mr. North but more senior to total committee members are from the Rajya Sabha. What fraction of the Ms. East. Mr. South is senior to Ms. West but junior to Mr. North. If the senior-most declines the assignment, then who will be the new Vice Chantotal number are women from the Lok Sabha/ cellor of the University. 1. 1/11 1. Mr. North 2. 6/11 2. Ms. East 3. 2/11 3. Ms. West 4. 3/11 4. Mr. South Answer: 1 Explanation: According to question Committee consists of Lok Sabha and Answer: 4 Rajya Sabha (7/11). so Lok Sabha fraction is 4/11. In committee 9/11 are men so 2/11 are women. Among men 2/3 are from Rajya Sabha. so male from Explanation: Decreasing order of seniority: North > South > West > East. Loksabha should be 9/11 x 1/3 = 9/33 =3/11 Women from loksabha = Total Since Nort declined, South will be the new VC. fraction of Lok Sabha- Men from Lok sabha = 4/11 - 3/11 = 1/11 Reference: https://www.bankexamstoday.com/2014/12/coded-inequaliReference: https://math.stackexchange.com/questions/1296651/sat-maths- ties-logical-reasoning.html question-about-fractions 12. The prices of diamonds having a articular colour and clarity are tab8. A librarian is arranging a thirteen-volume encyclopaedia on the shelf ulated below: from left to right in the following order of volume numbers : 8, 11, 5, 4, 9, 1, 7, 6, 10, 3, 12, 2. In this pattern, where should the volume 13 be placed? 1. Leftmost 2. Rightmost 3. Between 10 and 3 4. Between 9 and 1 Answer: 3 Explanation: Numbers have been arranged inalphabetical order: eigh, eleven, seven,etc. So, thirteen will come in between ten and three. Reference: https://www.wyzant.com/resources/answers/16653/where_ does_13_go_in_this_sequence_8_11_5_4_9_1_7_6_10_3_12_2 9. Pick the correct statement: 1. The sky is blue because Sir. C. V. Raman gave the correct explanation. 2. Copernicus believed that the Sun, and not the Earth, was at the centre of the solar system. 3. The sky appears blue when seen from the moon. 4. No solar eclipse is visible for an astronaut standing on the moon. Answer: 2 Explanation: Nicolaus Copernicus was a Polish astronomer who put forth the theory that the Sun is at rest near the center of the Universe, and that the Earth, spinning on its axis once daily, revolves annually around the Sun. This is called the heliocentric, or Sun-centered, system. Reference: http://www.nmspacemuseum.org/halloffame/detail.php?id=123 10. What is the last digit of (2017) 2017?

1. 8 2. 16 3. 32 4. 64 Answer: 4 Explanation: Since per carat value of 2carat diamond is 2 lakh, so value of two carat = 2 x 8 = 16 lakh. Per carat value of 0.25 carat is 1 lakh. So the number of 0.25 carat bought by 16 lakh will be 16/0.25 = 64. 13. In a sequence of 24 positive integers, the product of any two consecutive integers is 24. If the 17th member of the sequence is 6, the 7th member is? 1. 24 2. 4 3. 6 4. 17 Answer: 3 Explanation: If any two consecutive number has a product of 24 and 17th number is 6, then 16th and 18th number should be 4. Similarly, if the 16th number is 4, the 15th number should be 6. So, all even placed numbers are 4 and all odd placed numbers are 6. So, 7th place has 6 as the interger

1. 1 2. 3 3. 7 4. 9 Answer: 3 Explanation: (2017)^2017 can also be written as (2017)^2016 x 2017 = (2017^4)504 x 2017 = (...1)^504 x 2017 = .....7 at the unit place. ( 7^4 = 2041) Reference: https://nzmaths.co.nz/resource/powers-seven

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How many 0.25 carat diamonds can be purchased for the price of a 2 carat diamond?

Reference: https://www.talentsprint.com/blog/2017/06/number-series-inumber-series-tricks.html 14. Mohan lent Geeta as much money as she already had. She then spent Rs. 10. Next day, he again lent as much money as Geeta now had, and she spent, Rs. 10 again. On the third day. Mohan again lent as much money as Geeta now had, and she again spent Rs. 10. If Geeta was left with no money at the end of the third day, how much money did she have initially?

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Reference: http://whatis.techtarget.com/definition/prime-number 17. A water tank that is 40% empty holds 40L more water, than when it is 40% full. How much water does it hold when it is full?

Answer: 4

1. 100L 2. 75L 3. 120L 4. 200L

Explanation: Let Mohan have Rs x/- and Geeta have Rs. y/Mohan Geeta Day 1 x y Day 2 x-y 2y-10 Day 3 x-y -(2y-10) 2(2y-10)-10 Day 4 x-y-(2y-10) - [2(2y-10)-10] 2[2(2y-10)-10] - 10

Answer: 4

Given, 2[2(2y-10)-10] - 10 = 0 So, 2[4y - 30] = 10 So, 4y -30 = 5 y = 35/4 = 8.75/-

Explanation: Let the full tank hold x lts When it is 60% full (40% empty), it holds 40 lts more than when it is 40% full So, 0.6x - 0.4x = 40 0.2x = 40 x = 200 lts

Reference: https://simple.wikipedia.org/wiki/Equation

18. How much gold and copper (in g.) respectively, are required to make 15. The distribution of marks of students in a class is given by the follow- a 120 g bar of 22 carat gold? ing chart. 1. 90 and 30 2. 100 and 20 3. 110 and 10 4. 120 and 0 Answer: 3 Explanation: 22 carat gold denote the purity of gold in the final product, i.e. 91.6 grams of pure gold in 100 gram alloy. According to this for 120 g bar of 22 carat, the amount of gold required = (91.6 x 120) /100 = 109.02 Remianing amount of copper in grams = (8.4 x120) / 100 = 10.08 Approximately, 110 g gold and 10 g copper Reference: https://www.bankbazaar.com/gold-rate/hallmark-kdm-916-varieties-of-gold.html?ck=Y%2BziX71XnZjIM9ZwEflsyDYlRL7gaN4W0xhuJSr9Iq7aMYwRm2IPACTQB2XBBtGG&rc=1 If 3.30 marks is the passing score in a 10 mark question paper, which of the following is false?

19. Which should be the correct pattern in the empty square?

1. Majority of the students have scored above the pass mark. 2. Mode of the distribution is 3 3. Average marks of passing students is above 55% 4. Average marks of students who have failed is below 20% Answer: 4 Explanation: From the graph we have, 80 students secured 3 marks; 10 students scored 2 marks and 10 students scored 1 mark each. Total marks secured by these 100 students = (80*3)+(10*2)+(10*1)=270. Therefore, average marks of these students who actually failed to score more than 3.30 = 270/100=2.7 which is greater than 20%. Reference: https://clay6.com/qa/88623/the-average-marks-in-statistics-obtained-by-30-students-is-52-the-average-m 16. If all the angles of a triangle are prime numbers, which of the following could be one such angle? 1. 89o 2. 79o 3. 59o 4. 29o

Answer: 3

Answer: 1

Explanation: Mirror Image

Explanation: The 3 angles can be 89, 89 and 2 Phone: 080-5099-7000

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20. Areas of the three rectangles inside the full rectangle are given in the

the formation of an α helix neutralizes the polar character of the peptide bonds Reference: https://www.ncbi.nlm.nih.gov/books/NBK9928/ 23. Which one of the listed below is a P-type ion transporter?

diagram.

1. Mg+2 and Fe+2 2. Mg+2 and Fe+3 3. Mg+2 and Cl-1 4. Na+-K+, Ca+2 and H+

What is the area of the full rectangle? 1. 36 2. 48 3. 72 4. 96

Answer: 4 Explanation: P type ion transporter forms a phosphorylated (P-) intermediate state during their ion transport cycle. Members of this family generate and maintain crucial (electro-) chemical gradients across cellular membranes, by translocating cations, heavy metals and lipids. The ions bind to the transmembrane domain, coordinated by negatively charged and polar residues, in a region between transmembrane helices M4, M5, M6 and M8. The number of transported ions varies: the fungal H+-ATPase transports one H+, SERCA transports two Ca2+ out versus two to three H+ in, and the NKA transports three Na+ out and two K+ in

Answer: 2 Explanation: l*b = 8 b*y = 4 x*y = 12 So, x*l = ? = (l*b) * (x*y) / (b*y) = 8*12/4 = 24 So, total area = 8+4+12+24 = 48 sq units

Reference: http://jcs.biologists.org/content/joces/124/15/2515.full.pdf 24. The cell maintains a high concentration of protons inside the lysosome because of 1. antiporter in the lysosomal membrane 2. ATP-powered proton pump in the lysosomal membrane 3. facilitated diffusion proton channel in the lysosomal membrane 4. facilitated diffusion proton uniporter in the lysosomal membrane

Reference: https://www.mathsisfun.com/area.html

Answer: 2

PART ‘B’

Explanation: Lysosomes must maintain an acidic luminal pH to activate hydrolytic enzymes and degrade internalized macromolecules. Acidification results from the action of a vacuolar-type H+-ATPase (V-ATPase), which uses the free energy of ATP hydrolysis to pump protons into the lumen of the lysosome.

21. The [OH-] of 0.1 N HCl solution is 1. 10-14M 2. 10-13M 3. 10-12M 4. 10-7M

Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3664703/

25. It is known that there is a large difference in the DNA content between two organisms with similar developmental complexity. This is due Explanation:HCl is a strong acid and that will dissociate in water complete- to large differences in the number of ly. 1 normal HCl would have 1 gm-equivalent of H+ ions in one liter (equivalent to 1 M of HCl). So, 0.1 N =0.1 M. The concentration of H+ = HCl = 0.1 1. transposable elements, repetitive DNA and coding sequences. 2. transposable elements and repetitive DNA M or 10^-1 M. So, OH- = Kw/ H+ => 10^-14 M^2/ 10^-1 M => 10^-13 M 3. introns and coding sequences 4. introns and transposable elements Reference:Lehninger’s Principles of Biochemistry Answer: 2

22. Ability of a membrane protein to span the lipid bilayer strictly depends on the presence of 1. Zinc finger domain 2. α Helices 3. parallel β sheets 4. antiparallel β sheets Answer: 2 Explanation: Many integral membrane proteins (called transmembrane proteins) span the lipid bilayer, with portions exposed on both sides of the membrane. The membrane-spanning portions of these proteins are usually α-helical regions of 20 to 25 nonpolar amino acids. The hydrophobic side chains of these amino acids interact with the fatty acid chains of membrane lipids, and Phone: 080-5099-7000

Answer: 4 Explanation: Genome size/DNA content varies considerably among organisms due to differences in the amplification, deletion, and divergence of various kinds of repetitive sequences, including the transposable elements (from 25 percent to 40 percent of mammalian chromosomes consist of transposable elements that have accumulated in the course of evolution) and introns (the amount of total introns varies in different species) which constitute a large fraction of the genome. Coding sequences will be same for both the species so probably the developmental complexity is similar. Reference: https://academic.oup.com/mbe/article/23/1/162/1193365 https://www.ncbi.nlm.nih.gov/books/NBK21981/

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is close to diffusion-controlled rate of encounter.

26. A uracil containing plasmid was constructed and used in transformation into the wild type (ung+) and uracil-N-glycosylase mutated (ung-) Answer: 3 E.coli cells and scored for transformants in the presence of appropriate antibiotics. Which one of the following statements correctly describes the Explanation: The hallmark of competitive inhibition is that it can be overexperimental outcome? come by a sufficiently high concentration of substrate. Noncompetitive inhibition cannot be overcome by increasing the substrate concentration. The 1. ung+ cells will have fewer transformants compared to ung- cells. Michaelis-Menten and Lineweaver Burk graphs are also different for these 2. ung+ cells will give fewer transformants compared to ung- cells. two. 3. No transformants will be obtained in ung- cells as uracil excision repair will not occur and the plasmid would not replicate. Reference: https://www.ncbi.nlm.nih.gov/books/NBK22530/ 4. Presence of uracil in DNA is unnatural and the plasmid DNA with uracils in it will not produce transformants in either ung+ or ung- cells. 29. Which one of the following pairs of amino acids are glucogenic and ketogenic in nature? Answer: 3 Explanation: The human gene encodes one of several uracil-DNA glycosylases. Alternative promoter usage and splicing of this gene leads to two different isoforms: the mitochondrial UNG1 and the nuclear UNG2.[5] One important function of uracil-DNA glycosylases is to prevent mutagenesis by eliminating uracil from DNA molecules by cleaving the N-glycosylic bond and initiating the base-excision repair (BER) pathway. Uracil bases occur from cytosine deamination or misincorporation of dUMP residues. After a mutation occurs, the mutagenic threat of uracil propagates through any subsequent DNA replication steps.[7] Once unzipped, mismatched guanine and uracil pairs are separated, and DNA polymerase inserts complementary bases to form a guanine-cytosine (GC) pair in one daughter strand and an adenine-uracil (AU) pair in the other.[8] Half of all progeny DNA derived from the mutated template inherit a shift from GC to AU at the mutation site.[8] UDG excises uracil in both AU and GU pairs to prevent propagation of the base mismatch to downstream transcription and translation processes.[8] With high efficiency and specificity, this glycosylase repairs more than 10,000 bases damaged daily in the human cell.[9] Human cells express five to six types of DNA glycosylases, all of which share a common mechanism of base eversion and excision as a means of DNA repair Reference: https://www.ncbi.nlm.nih.gov/pubmed/324994 27. Which one of the following peptides can coexist in both cis- and transconformation? 1. Ala-Ala-CONH2 2. Pro-Gly-CONH2 3. Asn-Gly-CONH2 4. Val-Pro-CONH2 Answer: 4 Explanation: The peptide bond nearly always has the trans configuration since it is more favourable than cis, which is sometimes found to occur with proline residues. For proline residues, the cyclic nature of the side chain means that both cis and trans configurations have more equivalent energies. Thus proline is found in the cis configuration more frequently than other amino acids. The omega torsion angle of proline will be close to zero degrees for the cis configuration, or most often, 180 degrees for the trans configuration. Reference: tide2.html

http://www.cryst.bbk.ac.uk/PPS95/course/3_geometry/pep-

28. Which one of the following statements is NOT correct? 1. Allosteric enzymes do not obey Michaelis-Menten kinetics. 2. The free-energy change provides information about the spontaneity but not rate of a reaction. 3. Competitive and non-competitive inhibitions are kinetically indistinguishable. 4. A kcat/KM(M-1s-1) of ~2 x 108 for an enzyme indicates that the value Phone: 080-5099-7000

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a uniform narrow gap of about 2–4 nm. The gap is spanned by channel-forming proteins (connexins). The channels they form (connexons) allow inorganic ions and other small water-soluble molecules to pass directly from the cytoplasm of one cell to the cytoplasm of the other, thereby coupling the cells both electrically and metabolically.

Answer: 1 Explanation: Depending on their metabolic fates, amino acids are referred to as ‘glucogenic’, ‘ketogenic’ or ‘glucogenic and ketogenic’.Glucogenic amino acids are broken down into one of the following metabolites: pyruvate, α-ketoglutarate, succinyl CoA, fumarate or oxaloacetate which can be oxidized into CO2 and H2O to generate ATP via the tricarboxylic acid cycle (TCA cycle) and oxidative phosphorylation. Alternatively the carbon skeletons from these metabolites can be used to synthesise glucose by the process of gluconeogenesis hence the term glucogenic amino acids.In contrast ketogenic amino acids are broken down into acetyl-CoA and or the ketone body acetoacetate, neither of which can bring about net glucose production in humans. A third group of amino acids are catabolized to both glucogenic and ketogenic intermediates. Thirteen amino acids: glycine, alanine, serine, aspartic acid, asparagine, glutamic acid, glutamine, proline, valine, methionine, cysteine, histidine and arginine, are purely ‘glucogenic’; five amino acids: tryptophan, phenylalanine, tyrosine, isoleucine and threonine are both ‘glucogenic and ketogenic’; while the remaining two amino acids - lysine and leucine - are purely ‘ketogenic.

Reference: https://www.ncbi.nlm.nih.gov/books/NBK26857/ 32. Which one of the following is NOT true for the alternative pathway of complement activation? 1. Alternative pathway uses the same membrane-attack complex as the classical pathway. 2. Alternative pathway does not require antigen-antibody interactions. 3. Alternative pathway produces C3 by the same route as the classical pathway. 4. Certain microbial surfaces have physico-chemical properties that may result in activation of alternative pathway. Answer: 4

Explanation: Specific protein:protein or protein:carbohydrate interactions Reference: https://www.diapedia.org/metabolism-insulin-and-other-hor- that characterize classical and lectin pathway activation whereas the alternamones/5105758814/amino-acid-metabolism tive pathway is capable of autoactivation because of a process termed “tickover” of C3. Tickover occurs spontaneously at a rate of ∼1% of total C3 per hour, generating a conformationally altered C3, designated C3(H2O), that is 30. Cervical cancer-causing papilloma virus produces two oncoproteins capable of binding factor B. Once factor B associates with C3(H2O), factor B E6 and E7 which are responsible for interfering with cell cycle regulation itself changes conformation and can then be cleaved by the constitutively acby tive serum protease factor D, generating Ba and Bb. The Bb fragment remains associated with the complex and can then, through its own serine protease 1. Inactivating pRb and p53, respectively domain, cleave additional C3 molecules, generating a form designated C3b. 2. Modulating p53 and pRb, respectively Once C3b is generated, it associates with factor B to generate more C3-con3. Binding to cyclin D1 and CDK4 vertase. This overall series of successive proteolytic steps is enhanced by the 4. Activating expression of p21 serum protein properdin, which stabilizes protein:protein interactions during the process. Answer: 2 Reference: http://www.jimmunol.org/content/176/3/1305 Explanation: The E6 protein is thought to promote cell proliferation by stimulating degradation of the tumor suppressor p53 protein via the formation of a trimeric complex comprising E6, p53 and the cellular ubiquitination enzyme 33. E.coli takes 40 min. to duplicate its genome using a bi-directional E6-AP. E6-stimulated degradation interferes with such biological functions of mode of replication. If E.coli were to use unidirectional mode of replicap53; thus perturbing the control of cell cycle progression, leading finally to tion to synthesize a full copy of DNA complementary to just one of the increased tumor cell growth. strands of the genome, it would take The E7 proteins encoded by the high-risk type HPVs, such as HPV 16 and HPV 18, bind Rb with a much higher affinity compared to those encoded by 1. 40 min the low-risk type HPVs, such as HPV 6 and HPV 11. E7 binds to a region of 2. 80 min the Rb protein commonly referred to as the ‘pocket domains’. The ‘pocket 3. 20 min domain’ sequences of Rb are essential for its tumor suppressor function, with 4. 60 min many naturally occurring loss-of-function mutations of Rb appearing to cluster in these ‘pocket domains’. E7 disrupts the interaction between Rb and E2F, Answer: 1 resulting in the release of E2F factors in their transcriptionally active forms. Explanation: 40 minutes is replicates in a bidirectional mode to completeReference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2785934/ ly synthesized ( copy) the genome (dsDNA), and for synthesizing ssDNA in unidirectional mode it takes same 40 minutes since , the leading strand and lagging strand synthesis happens simultaneously So it will be 40 minutes 31.Which one of the following permits the rapid diffusion of small, water-soluble molecules between the the cytoplasm of the adjacent cells? Reference: https://www.golifescience.com/prokaryotic-dna-replication/ 1. Tight junctions 2. Anchoring junctions 3. Adherens junctions 4. Gap junctions

34. Transcriptional regulation of trp operon by tryptophan involves binding of tryptophan to

1. the repressor protein and inhibition of transcription by its interaction Answer: 4 with the operator region. 2. RNA polymerase and inhibition of transcription. Explanation: With the exception of a few terminally differentiated cells such 3. the repressor protein leading to structural changes and its degradaas skeletal muscle cells and blood cells, most cells in animal tissues are in tion by proteases. communication with their neighbors via gap junctions. Each gap junction ap- 4. the repressor protein leading to its interaction with the sigma subunit pears as a patch where the membranes of two adjacent cells are separated by and inhibition of transcription. Phone: 080-5099-7000

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https://www.sciencedirect.com/topics/agricultural-and-biological-sciences/interferon-gamma

Answer: 1

Explanation: When tryptophan is present, these tryptophan repressor dimers bind to tryptophan, causing a change in the repressor conformation, allowing 37. Which one of the following secondary metabolites is characterized the repressor to bind to the operator. This prevents RNA polymerase from by the presence of a central carbon atom that is bound by a sulphur to a binding to and transcribing the operon, so tryptophan is not produced from its glycone group, and by a nitrogen to a sulphonated oxime group? precursor. When tryptophan is not present, the repressor is in its inactive conformation and cannot bind the operator region, so transcription is not inhibited 1. Alkaloids 2. Terpenes by the repressor. 3. Phenolics Reference: https://www.sciencedirect.com/topics/biochemistry-genet- 4. Glucosinolates ics-and-molecular-biology/trp-operon Answer: 4 Explanation: Glucosinolates are a group of sulphur and nitrogen containing secondary metabolites mostly found in the cruciferae family and it consists of beta-D thioglucose group and a sulfonated oxime group.

35. Phosphorylation of elF2 α subunits (at Ser51) leads to

1. inactivation of Met-tRNA binding activity of elF2B. 2. sequestration of elF2B because of tight binding between elF2 and Reference: https://www.sciencedirect.com/topics/pharmacology-toxicoloelF2B. gy-and-pharmaceutical-science/glucosinolates 3. degradation of elF2B 4. Enhanced guanine exchange activity of elF2B. 38. Dark grown Arabidopsis seedlings show ‘triple response’ when exposed to ethylene hormone. Which one of the following options is characteristic of ‘triple response’?

Answer: 1

Explanation: Phosphorylation of the α subunit of the translation initiation factor eIF2 at serine 51 (eIF2αP) is a master regulator of cell adaptation to var- 1. Reduced shoot elongation, increased shoot thickness and tightening of ious forms of stress with implications in antitumor treatments with chemother- apical hook. apeutic drugs.Phosphorylated eIF2α prevents the recycling of the eIF2-bound 2. Reduced shoot elongation, reduced shoot thickness and loosening of GDP to GTP by the guanine nucleotide exchange factor (GEF) eIF2B (Sonen- apical hook. berg and Hinnebusch, 2009 ). As such, formation of the eIF2-GTP-tRNAMet 3. Increased shoot elongation, increased shoot thickness and loosening of ternary complex is impeded resulting in the inhibition of translation initiation apical hook. 4. Increased shoot elongation, reduced shoot thickness and tightening of (Sonenberg and Hinnebusch, 2009) apical hook. Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2938387/ Answer: https://www.nature.com/articles/cddis2014554 36. If a disease caused by an intracellular pathogen is associated with host anti-inflammatory response, which one of the following may lead to an effective therapeutic approach? 1. Treatment with TGF-β 2. Treatment with macrophage activating agent 3. Depletion of IFN-γ from the system 4. Treatment with IL-4 and IL-10 Answer: 1 Explanation: Bacteria have various strategies to attack the inflammatory process including escape of the host defense , blockage of leukocyte recruitment to an inflamed area , inactivation of antimicrobial peptides, stabilization of the endogenous inflammatory inhibitors , indication of the expression of the anti-inflammatory cytokine, and cleavage of p65/relA of the NF-κB pathway. Overall, microbial pathogens have applied a variety of mechanisms to manipulate host-cell functions, presumably for their own benefit. TGFβ1 was initially identified as a potent chemotactic cytokine to initiate inflammation-treatment with TGF beta can reduce the degree of infection according to the above mentioned anti inflammatory mechanism of pathogen IFN gamma is another important molecule involved in the effector functions of cytotoxic T-cells. IFNγ is a tH1 antiviral and immunomodulatory cytokine that is critically involved in host resistance to multiple pathogens.-but in option it is depletion, unless this can be the best option rather TGF beta Interleukin-4 (IL-4) and interleukin-10 (IL-10) have typically the biologic anti-inflammatory effects on monocytes-so this can’t be the option Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2808140/ https://www.ncbi.nlm.nih.gov/pubmed/11850581 https://www.ncbi.nlm.nih.gov/pubmed/20018551 Phone: 080-5099-7000

Explanation: Reference: 39. Brassinosteroids are a group of steroid hormones that function in a variety of cellular and developmental contexts in plants. Which one of the following acts as an inhibitor of the brassinosteroids receptor? 1. BRI 1 2. BKI 1 3. BAK 1 4. BSK 2 Answer: 2 Explanation: BKI1 is the only reported inhibitor of receptor kinases in Arabidopsis, which negatively regulates BRI1 in the brassinosteroid pathway. BKI1 can also interact with another important LRR-RLK, ERECTA (ER). Phenotypic analysis showed that BKI1 and ER together regulate plant architecture, including pedicel orientation, which is a newly reported phenotype in the BRand ER-mediated developmental processes. Gene expression analysis revealed that BKI1 regulates a subset of ER-responsive genes. Reference: S1674205216302994

https://www.sciencedirect.com/science/article/pii/

40. Which one of the following metabolites moves from mitochondria to peroxisome during the operation of the C2 oxidative photosynthetic cycle? 1. Glycerate 2. Glycolate 3. Glycine

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Special Edition 4. Serine

tional information, established in part by the activity of signaling pathways. Mutations in the Ultrabithorax (Ubx) Hox gene illustrate this assertion. Wings Answer: 4 and halteres are homologous structures located in the second and third thoracic segments, respectively. These appendages greatly differ in size and pattern, Explanation: Glycine leaves the peroxisome and enters the mitochondrion, and derive from imaginal discs, the wing and haltere discs, which also differ where two molecules of glycine are converted to serine and CO2. The newly in size but bear a similar morphology. Ubx, which is expressed in the haltere formed serine diffuses from the mitochondrion back to the peroxisome, where disc but not in the wing disc, determines the difference between these two it is converted to glycerate structures: mutations or deletion of Ubx transform halteres into wings whereas Ubx ectopic expression changes wings into halteres Reference: https://www.tankonyvtar.hu/en/tartalom/tamop425/0010_1A_ Book_angol_01_novenyelettan/ch03s02.html Reference: http://dev.biologists.org/content/133/22/4495 41. Sympathetic postganglionic neurons that are cholinergic, innervate 1. Sweat glands 2. Parotid glands 3. Hair follicles 4. Pancreas Answer: 1 Explanation: The sweat glands are innervated by the sympathetic nervous system and are part of the fight or flight response system. Their innervation consists of two parts, a preganglionic and postganglionic neuron. The preganglionic neuron is short, originates from the thoracolumbar region of the spinal cord, uses acetylcholine as its neurotransmitter, and synapses with the postganglionic neuron via a nicotinic acetylcholine.

44. Which one of the following statements with respect to amphibian development is correct? 1. The organizer is itself induced by the Nieuwkoop Centre located in the dorsal-most mesodermal cells 2. The organizer functions by secreting proteins like Noggin, Chordin and Follistatin that blocks BMP signal that would otherwise dorsalize the mesoderm. 3. In the presence of BMP activators the ectodermal cells form neural tissue. 4. Wnt signalling causes a gradient of β-catenin along the anterior-posterior axis of the neural tube that appears to specify the regionalization of the neural tube. Answer: 4

Explanation: Wnt/β-catenin signalling occurs in a direct and long-range fashion within the ectoderm, and there is an endogenous AP gradient of Wnt/β-catenin signalling in the presumptive neural plate of the Xenopus gastrula. 42. Fertilization in sea urchin eggs involves Ca2+ release from the endo- An activity gradient of Wnt/β-catenin signalling acts as transforming morphoplasmic reticulum for cortical granule reactivation. The major molecule gen to pattern the Xenopus central nervous system. responsible for releasing Ca2+ from intracellular stores is Reference: http://dev.biologists.org/content/128/21/4189 1. Zona pellucida glycoproteins 2. Protamines 45. Which one of the following statements related to components/fea3. Inositol 1, 4, 5-triphosphate tures of senescence in plants is INCORRECT? 4. N-acetylglucosaminidase 1. Programmed cell death in plants may generate functional cells or tisAnswer: 3 sues. 2. Senescence can be induced by application of cytokinins and delayed Explanation: In echinoderms eggs as in many other cell types, , the ER is by overexpression of salicylic acid. the main organelle involved in calcium storage and calcium release events. 3. Plants defective in autophagy demonstrate accelerated plant senesCalcium accumulation in the ER is driven by a SERCA pump (Smooth endo- cence. plasmic reticulum calcium ATPase pump) and elementary release events from 4. Leaf senescence is regulated by NAC and WRKY genes families. the ER either by IP3 or RyR receptors – transmembrane calcium channels that open upon binding their specific ligand enabling calcium to diffuse out along Answer: 1 a concentration gradient Explanation: PCD is largely used to describe the processes of apoptosis and Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4778076/ autophagy (although some use PCD and apoptosis interchangeably) while necrosis is generally described as a chaotic and uncontrolled mode of death. In plants, the term PCD is widely used to describe most instances of death ob43. What is the observed phenotype when the ultrabithorax gene is de- served. leted in Drosophila? Reference: https://academic.oup.com/jxb/article/59/3/435/574378 1. The third thoracic segment is transformed into another second thoracic segment resulting in a fly with four wings. 2. Since it specifies the second thoracic segment, instead of antenna leg 46. In an organism, allele for red eye color is dominant over the allele grows out of the head socket. for white eye colour. A cross is made between a white eyed male and a red 3. Since it specifies the third thoracic segment, a fly with two pairs of eyed female. In the progeny all males are red eyed while the females are halters are developed. white eyed. The reciprocal cross leads to all red eyed progeny . Based on 4. Since the gene fails to be expressed in the second thoracic segment, the the above information which one of the following conclusions is correct? antenna sprout in the leg position. 1. This is a sex-limited trait, and the male is the homomorphic sex. Answer: 1 2. This is a sex-linked trait, with male being the homomorphic sex. 3. This is a sex-linked trait, with female being the homomorphic sex. Explanation: A particular class of selector genes is formed by the Hox 4. This is a case of autosomal inheritance with incomplete penetrance. genes, which specify different structures along the anterior-posterior axis of metazoans . In Drosophila, mutations in these genes frequently transform one Answer: 2 structure into another keeping the coordinates provided by underlying posiReference: https://www.openanesthesia.org/sweat_glands_innervation/

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Special Edition Explanation: it is a sex-linked trait because the trait is not limited to just one gender, but is expressed in both the genders depending upon the genotype of the male and the female parents. in the first cross, the male genotype is ZrZr and the female genotype is ZRW, therefore the offsprings are ZRZR (male:red) and ZrW (female: white). in the reciprocal cross, the male parent was,ZRZR and the female was ZrW, so the offsprings were all red, ZRZr (male:red) and ZRW (female: red). here the male is having homomorphic sex (Homomorphic chromosomes :Morphologically identical members of a homologous pair of chromosomes. Heteromorphic chromosomes :A chromosome pair with some homology but differing in size, shape, or staining properties).

in a specific area of the kidney. Dietary sodium intake increases vasopressin secretion in man and the increased urine flow results from alcohol’s acute inhibition of the release of antidiuretic hormone (ADH) When A consumed a salted diet and alcohol-there will be increase in ADH release and subsequently inhibition in the release, but the inhibition of ADH release is not possible in Individual B, so option 1

Reference: http://www.yourhormones.info/hormones/anti-diuretic-hormone/ https://pubs.niaaa.nih.gov/publications/arh21-1/84.pdf Reference: https://www.sciencedirect.com/topics/biochemistry-genet- https://www.ncbi.nlm.nih.gov/pubmed/3915319 ics-and-molecular-biology/reciprocal-cross 50. An intron was cloned with a transposable element. Absence of the intron following transposition of the element, will indicate that

47. Melanopsin is found in which cell of the retina? 1. Cones 2. Rods 3. Ganglion cells 4. Bipolar cells

1. follows conservative mode of transposition 2. follows replicative mode of transposition 3. is a retrotransposon 4. is an insertion element

Answer: 3

Explanation: Melanopsin is a type of photopigment belonging to a larger Explanation: The conservative mode of transposition its is a “cut & paste” family of light-sensitive retinal proteins called opsins and encoded by the gene mechanism where the whole transposable element is transposed to some other Opn4. In humans, melanopsin is found in intrinsically photosensitive retinal part of the gemone or another locus in the same chromosome. The Replicative ganglion cells (ipRGCs) mode of transposition is a “copy- paste” mechanism of transposition. An Insertion element codes for a protein. In this case, the intron that was cloned in Reference: https://www.ncbi.nlm.nih.gov/pubmed/14500998 the transposable element was absent after transposition of the element. This indicates that the element is a retrotransposon. Because, they are first tran48. Prestin, a membrane protein, is found in which one of the following scribed into RNA, then converted back into identical DNA sequences using cells of the organ of corti? reverse transcription, and these sequences are then inserted into the genome at target sites. Introns cannot be transcribed into mRNA. Hence, are lost. 1. Inner hair cells 2. Inner phalangeal cells Reference: https://www.sciencedirect.com/science?_ob=PdfExcerptURL&_im3. Outer hair cells agekey=3-s2.0-B0122270800011113-main.pdf&_piikey=B0122270800011113&_ 4. Outer phalangeal cells cdi=272999&_orig=PublicationURL&_zone=rslt_list_item&_fmt=abst&_eid=3-s2.0-B0122270800011113&_isbn=9780122270802&_user=12975512&mAnswer: 3 d5=11c5ebd588f491e4505c237b0c0c0c1c&ie=/excerpt.pdf%20target= Explanation: Prestin, a transmembrane protein found in the outer hair cells of the cochlea, represents a new type of molecular motor, which is likely to 51. A male snail homozygous for dextral alleles is crossed with a female be of great interest to molecular cell biologists. In contrast to enzymatic-ac- homozygous for sinistral alleles. All the F1 individuals showed sinistral tivity-based motors, prestin is a direct voltage-to-force converter, which uses phenotype. When F1 progeny snails were self fertilized all individuals of cytoplasmic anions as extrinsic voltage sensors and can operate at microsec- F2 progeny had dextral coiling. This experiment demonstrated ond rates. As prestin mediates changes in outer hair cell length in response to membrane potential variations, it might be responsible for sound amplification 1. dominant epistasis as dextral alleles is dominant over sinistral allele in the mammalian hearing organ. 2. recessive epistasis as in F2 dextral allele appeared in homozygous condition Reference: https://www.ncbi.nlm.nih.gov/pubmed/11836512 3. maternal effect as the nuclear genotype of the F1 mother has governed the phenotype of the F2 individuals 4. maternal inheritance as the mitochondrial genes of the F1 mother has 49. Two individuals A and B, each of 75 kg body weight, have similar vol- governed the phenotype of the F2 individuals. ume of body water. Both of them had high salt snack. However individual A also had a glass of alcoholic drink. Based on this information, which one Answer: 3 of the following statements is true? Explanation: This is a classical example of Genetic Maternal Effect where, 1. A will have lower circulating level of antidiuretic hormone (ADH) than the direction of coiling is determined not by that snail’s own genotype, but by B the genotype of its mother 2. B will have lower circulating level of antidiuretic hormone (ADH) than A Reference: iGenetics: A Molecular Approach 3rd Edition by Peter J. Rus3. The level of ADH will not change in these two individuals sell. Chapter 13 Extensions of and Deviations from Mendelian Genetic 4. The reabsorption of water in kidney will be more in A than B Principles (P-376) Answer: 1 Explanation: Antidiuretic hormone helps to control blood pressure by acting on the kidneys and the blood vessels. Its most important role is to conserve the fluid volume of your body by reducing the amount of water passed out in the urine. It does this by allowing water in the urine to be taken back into the body Phone: 080-5099-7000

52. Bipinnaria and brachiolaria are the larval forms of 1. Crustacea 2. Arthropoda and Mollusca, respectively 3. Ophiuroidea and Holothuroidea, respectively

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Special Edition 4. Asteroidea

Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3883366/

Answer: 4

Explanation: There are three larval stages in Asteroidea in the course of their development to adult stage. 56. A group of paleontologists digging in an area discovers a prehistoric Early Bipinnaria, Bipinnaria and brachiolaria human burial site. The same group, while exploring a nearby area, discovered fossil remains of what appeared to be more than 100 million year Reference: http://www.iaszoology.com/echinoderm-larvae/ old dinosaur bones. Which of the following combinations of modern radiometric dating techniques should they use to calculate the age of these fossils most accurately? 53. In a sample from a population there were 65 individuals with BB genotype, 30 individuals with Bb genotype and 15 individuals with bb 1. 14C dating for human remains and 235U dating for dinosaur remains genotype. The frequency of the ‘b’ allele is 2. 87Rb dating for both human and dinosaur remains 3. 14C dating for both human and dinosaur remains 1. 0.59 4. 129I dating for human remains and 129Xe for dinosaur remains 2. 0.27 3. 0.41 Answer: 1 4. 0.73 Explanation: The most widely known form of radiometric dating is carAnswer: 2 bon-14 dating, used to determine the age of human-made artifacts and remains. But carbon-14 dating won’t work on dinosaur bones. The half-life of Explanation: Given that- BB genotype = 65 individuals; Bb genotype = carbon-14 is only 5,730 years, so carbon-14 dating is only effective on sam30 individuals; bb genotype= 15 individuals. So, The total no. of individu- ples that are less than 50,000 years old. Dinosaur bones, on the other hand, are als = 65+30+15=110. Therefore, total no. of alleles = 110*2=220, since millions of years old -some fossils are billions of years old. To determine the each individual carries 2 alleles for the genotype. Frequency of b allele = ages of these specimens, scientists need an isotope with a very long half-life. (15*2+30)/220=60/220=0.27 Some of the isotopes used for this purpose are uranium-238, uranium-235 and potassium-40, each of which has a half-life of more than a million years. Reference: archives.evergreen.edu/webpages/curricular/2004-2005/generalbiologyecology/evoProblemSetAns.pdf Reference: https://science.howstuffworks.com/environmental/earth/geology/dinosaur-bone-age1.htm; https://www.nature.com/scitable/knowledge/library/dating-rocks-and-fos54. Ruderal species are those which are found in the environment with sils-using-geologic-methods-107924044 1. Low disturbance, high competition 2. High disturbance, low competition 3. Low disturbance, low competition 4. High disturbance, high competition

57. Given below are the statements related to the two competing hypotheses on the origin of modern humans: the Out-of-Africa hypothesis and the multi-regional hypothesis. Which of the following statements is INCORRECT?

Answer: 2

1. Both the hypotheses support that Homo erectus originated in Africa Explanation: Ruderals exploit high disturbance, low stress and low compe- and expanded to Eurasia tition environments. They are fast growing and generally complete their life 2. Mitochondrial DNA (mtDNA) and Y chromosome DNA evidence supcycle very fast and are also associated with the production of a large number port the ‘Out-of-Africa’ hypothesis. of seeds 3. The principal conflict between the two hypotheses is that multi-regional hypothesis does not support African origin of Homo erectus 4. The multi-regional hypothesis states that independent multiple origins Reference: Page 102, Fundamentals of Biogeography occurred in the million years since Homo erectus came out of Africa. By Richard J. Huggett Answer: 3 55. Scientists discovered two new plant species, “A” and “B” that look similar except that, species “A” bears flowers and leaves that are twice the size of those in species “B”. Which method should the scientists use to appropriately investigate if species “A” is a result of gene duplication in species “B”? 1. Sequence similarity, gene structure and gene size 2. Plant size, physical proximity of gene and genome size 3. Sequence similarity, physical proximity of gene and genome size 4. Sequence length, gene structure and chromosome count Answer: 4 Explanation: Examining the sequence relationships among these genes can deduce the presence of duplication in the chromosome. the gene structure might get altered if there is a duplication. change in the ploidy level is also associated with increase in the copy number of the gene. Sequence studies can be done by PCR techniques while karyotyping will reveal the ploidy level of the cell. Phone: 080-5099-7000

Explanation: Multiregional hypothesis: all living humans derive from the species Homo erectus that left Africa nearly two million-years-ago.Out of Africa Model:after Homo erectus migrated out of Africa the different populations became reproductively isolated, evolving independently, and in some cases like the Neanderthals, into separate species.Hence both the hypothesis supports African origin of Homo erectus which suggest that statement 3 is wrong. Reference: http://www.actionbioscience.org/evolution/johanson.html, https://www.pathwayz.org/Tree/Plain/OUT+OF+AFRICA+HYPOTHESIS 58. Which one of the following statements is TRUE for positive-frequency dependant selection? 1. Fitness of a genotype increases as it becomes less common 2. Fitness of a genotype increases as it becomes more common 3. Fitness of a genotype decreases as it becomes less common 4. Fitness of a genotype decreases as it becomes common and gets fixed Answer: 2

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Special Edition 5a7bdc86835035ea969f41a20580.pdf

Explanation: In positive frequency-dependent selection, the fitness of a phenotype increases as it becomes more common. In “positive frequency-dependent selection,” fitness increases with frequency so that rare genotypes are 62. Given below are biodiversity hotspots in decreasing order of endemic plant species recorded in them. Select the correct order. eliminated and local genetic diversity reduced. Reference: ent+selection

https://genetics-notes.wikispaces.com/Frequency-depend-

59. The animals belonging to the phylum Onychophora 1. have arthropodan characteristics and thus also considered as a class of Arthropoda 2. have both annelidan and arthropodan characteristics 3. have both arthropodan and molluscan characteristics 4. Serve as a connecting link between Annelida and Mollusca Answer: 2 Explanation: Research indicates that they are either modified arthropods or represent a link between annelids and arthropods, most recent molecular phylogenies favor the first hypothesis.Onychophorans were previously grouped under arthropods, but recently they are considered to be a separate phylum that is closely related to the arthropods.They have many similarities with both annelids and arthropods, and have been described as a missing link between the two groups. Reference: http://www.notesonzoology.com/phylum-onychophora/phylum-onychophora-classification-and-features-anthropods/1771

1. Western Ghats and Sri Lanka > Indo Burma > Sundaland > Philippines 2. Philippines > Sundaland > Indo Burma > Western Ghats and Sri Lanka 3. Sundaland > Indo Burma > Philippines > Western Ghats and Sri Lanka 4. Western Ghats and Sri Lanka > Sundaland > Philippines > Indo Burma Answer: 3 Explanation: It is estimated that there are four thousand species of flowering plants known from the Western Ghats and 1,500 (nearly 38 percent) of these are endemic. Indo-Burma- A conservative estimate of total plant diversity in the hotspot reveals about 13,500 vascular plant species, of which about 7,000 (52 percent) are endemic (van Dijk et al. 2004). Sundaland: In terms of vascular plant diversity and endemism, no truly reliable estimates exist. However, by extrapolating from what is known of the principal islands and countries that comprise the hotspot, total vascular plant diversity is estimated at roughly 25,000 species, and the number of endemics at 15,000. At the very least, one-third of the more than 6000 vascular plant species native to the Philippines are endemic.

60. Which one of the following parameters is NOT used in phenetic clas- Reference: https://www.cepf.net/our-work/biodiversity-hotspots/westernghats-and-sri-lanka/species, Key Biodiversity Areas in the Indo-Burma Hotsification of bacteria? spot: Process, Progress and Future Directions 1. trophism A.W. Tordoff 1, M.C. Baltzer 2, J.R. Fellowes 3, J.D. Pilgrim 4 & P.F. Lang2. susceptibility of a bacteria to a particular bacteriophage hammer,http://lntreasures.com/philippines.html 3. reaction to a particular antibody 4. 16S rRNA sequence 63. Which of the following options lists ecosystems in increasing order of plant productivity per day per unit leaf area?

Answer: 4

Explanation: In numerical taxonomy (also called computer or phenetic taxonomy) many (50 to 200) biochemical, morphological, and cultural character- 1. Tropical forests, hot deserts, temperate forests istics, as well as susceptibilities to antibiotics and inorganic compounds, are 2. Hot deserts, temperate forests, tropical forests used to determine the degree of similarity between organisms. The sequencing 3. Hot deserts, temperate grasslands, tropical forests of ribosomal RNA (rRNA) genes, which have been highly conserved through 4. Tropical forests, temperate grasslands, hot deserts evolution, allows phylogenetic comparisons to be made between species whose total DNAs are essentially unrelated. It also allows phylogenetic classi- Answer: 4 fication at the genus, family, and higher taxonomic levels. The rRNA sequence data are usually not used to designate genera or families unless supported by Explanation: Deserts have the highest of the three Net Primary productivity per unit area per day i.e. 2.50 per meter^2 per day followed by Temperate similarities in phenotypic tests. forest= 1.43 per meter^2 per day and lastly Tropical forest = 1.14 per meter^2 per day. Reference: https://www.ncbi.nlm.nih.gov/books/NBK8406/ 61. Which of the following groups represents SAR clade of proteins? 1. Euglenozoans, Red algae, Parabasilids 2. Brown algae, Forams, Radiolarians 3. Slime moulds, Entamoebas, Diplomonads 4. Charophyes, Choanoflagellates, Tubulinids

64. A general increase in the average body mass of animal population within a species with latitude is known as

Answer: 2 Explanation: The SAR include taxa that are very diverse including some of the most important photosynthetic organisms on Earth. This supergroup can be divided into three clades: the Stramenopiles, the Alveolates and the Rhizaria. Brwon Alga is a Stramenopile, While Foram & Radiolarians belong to Rhizaria. Reference:

Reference: Principles of Terrestrial Ecosystem Ecology By F Stuart Chapin III, Pamela A. Matson, Peter (Page Number 179)

1. Allen’s rule 2. Bergmann’s rule 3. Allee effect 4. Hamilton’s rule Answer: 2

Explanation: It is generally defined as a within-species tendency in homeothermic (warm-blooded) animals to have increasing body size with increasing https://pdfs.semanticscholar.org/presentation/99ea/cf8c6efe- latitude and decreasing ambient temperature. That is, Bergmann’s rule states

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Special Edition

that among mammals and birds, individuals of a particular species in colder 4. Fluorescence microscopy areas tend to have greater body mass than individuals in warmer areas. For instance, white-tailed deer are larger in Canada than in the Florida Keys, and Answer: 4 the body size of wood rat populations are inversely correlated with ambient temperature. This principle is named after a nineteenth-century German biol- Explanation: Ion Mobility-MS can be used to gain insights into the conogist, Karl Bergmann, who published observations along these lines in 1847. formational dynamics of a system, offering a unique means of characterizing flexibility and folding mechanisms. Reference:http://www.newworldencyclopedia.org/entry/Bergmann%27s_ NMR spectroscopy plays a major role in the determination of the structures and dynamics of proteins and other biological macromolecules. Chemical rule shifts are the most readily and accurately measurable NMR parameters, and they reflect with great specificity the conformations of native and nonnative 65. Following are statements to depict relationship among measures of states of proteins. Isothermal titration calorimetry (ITC) is particularly useful for determining central tendency in a skewed dataset: dynamic events such as kinetics and binding. Not fluorescent microscopy but FRET from fluorescent spectroscopy is used A. In positively skewed datasets, mean > median > mode in dynamic studies. B. In positively skewed datasets, mode > median > mean C. In negatively skewed datasets, mean > median > mode Reference: https://www.ncbi.nlm.nih.gov/pubmed/24651194, http:// D. In negatively skewed datasets, mode > median > mean www.pnas.org/content/104/23/9615, https://www.news-medical.net/whitepaper/20150624/Characterization-of-Protein-Stability-Using-DifferenWhich of the above statements are TRUE? tial-Scanning-Calorimetry.aspx, 1. A and B 2. A and C 68. Given below are a few statements on Agrobacterium mediated trans3. B and D formation of plants. Which one of the following statements is CORRECT? 4. A and D 1. T-DNA transfer occurs from left border to right border 2. The gfp reporter gene can never be used for selection of transgenic Explanation: A symmetrical distribution is a distribution where the mean, plants median and mode are the same. A skewed distribution, on the other hand, is a 3. Transformation frequencies will decrease an overexpression of virudistribution with extreme values on one side or the other that force the median lence genes. away from the mean in one direction or the other. If the mean is greater than 4. Host plant genes play an important role in influencing transformation the median, the distribution is said to be positively skewed. If the mean is frequencies smaller than the median, the distribution is said to be negatively skewed. Answer: 4 Reference: http://www.abs.gov.au/websitedbs/a3121120.nsf/home/statistiExplanation: TDNA transfer occurs from right border to left border. gfp recal+language+-+measures+of+central+tendency; porter genes are used for selection in transgenic plants. And also Vir genes are https://soc.utah.edu/sociology3112/central-tendency-variability.php reponsible for successful transformation event, hence its overexpression will increase gene transfer. 66. The MALDI mass spectrum of a peptide gave a single peak with M/z of 2000. The ESI mass spectrum of the same peptide gave multiple peaks. Reference: http://www.globalsciencebooks.info/Online/GSBOnline/images/0812/TPJ_2(2)/TPJ_2(2)127-137o.pdf These observations indicate that: Answer: 4

1. Degradation has occurred while acquiring ESI mass spectrum 2. Multiple charged species of the same compound are observed in the ESI spectrum 3. The sample is impure 4. ESI induces polymerization of the peptide Answer: 2 Explanation: Fragmentation of the protonated or de-protonated molecular ions generated from ESI is generally limited, and the mass spectra are relatively simple. However, multiple-protonation of proteins and peptides occurs in the ESI process, and the ESI mass spectra might become more complicated. A single protein will show its characteristic cluster ions containing multiple-charged ions. The number of charges on the protein molecules will depend on the molecular weight of the protein and the number of accessible basic sites.

69. Which one of the following assay systems can specifically detect apoptotic cells? 1. Tetrazolium dye (MTT) based colorimetric assay 2. FITC - annexin V based FACS analysis 3. 51Cr release assay 4. Trypan blue exclusion assay

Answer: 2 Explanation: The MTT assay is a colorimetric assay for assessing cell metabolic activity. Cr51 release assay is used to determine the number of lymphocytes produced in response to infection or drug treatment and Trypan blue exclusion assay is used to determine the number of viable cells present in a cell suspension. Annexin V (or Annexin A5) is a member of the annexin family of intracellular proteins that binds to phosphatidylserine (PS) in a calcium-dependent Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1853331/pdf/ manner. PS is normally only found on the intracellular leaflet of the plasma membrane in healthy cells, but during early apoptosis, membrane asymmetry cbr24_1p003.pdf is lost and PS translocates to the external leaflet. Fluorochrome-labeled Annexin V can then be used to specifically target and identify apoptotic cells. 67. Protein conformational dynamics CANNOT be determined by which Reference: https://www.ncbi.nlm.nih.gov/books/NBK144065/ http://www. one of the following techniques? perkinelmer.com/lab-products-and-services/application-support-knowledgebase/radiometric/chromium-51-release-assay.html https://www.ncbi. 1. NMR spectroscopy nlm.nih.gov/pubmed/18432654 https://www.biolegend.com/fr-lu/products/ 2. Differential scanning calorimetry fitc-annexin-v-5161 3. Mass spectroscopy Phone: 080-5099-7000

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Special Edition lar-biology/exoglycosidase 70. From the various techniques listed below, which one CANNOT be used precisely map the transcription start-site of a gene? 1. S1 Mapping 2. Sequencing the region downstream of promoter 3. 5’ RACE 4. Primer Extension Method Answer: 2 Explanation: S1 nuclease mapping is a nuclease protection assay using nuclease S1. This technique is used to quantify and map RNA transcripts. It is capable of identifying individual RNAs in a mixture of RNA sample of known sequence. It can particularly map introns and 5’ and 3’ ends of transcribed gene regions.Primer extension is a method typically used to map the 5′ end(s) of an RNA, thus defining the transcription start site and providing initial evidence for where the promoter is located within a cloned gene. RACE results in the production of a cDNA copy of the RNA sequence of interest, produced through reverse transcription, followed by PCR amplification of the cDNA copies (RT-PCR). The amplified cDNA copies are then sequenced and, if long enough, should map to a unique genomic region. RACE with high-throughput sequencing was first introduced in 2009 as Deep-RACE to perform mapping of Transcription start sites (TSS) of 17 genes in a single cell-line. However, sequencing the region downstream of promoter cannot be used to map transcription start site.

72. Genes translocated to the heterochromatic regions of chromosomes are silenced. In S. pombe, a translocation event was detected wherein a gene of interest was translocated to the centromere region and is silenced. Mutagenesis leading to loss of function of the following target genes was done to allow expression of the gene of interest from its new locus. A. Mutation in histone deacetylase (Clr3) B. Mutation in histone acetyltransferase (HAT-8) C. Mutation in histone H3 lysine 9 methyl transferase (Clr4) D. Loss of Dicer, an RNA processing enzyme Answer: 1

Explanation: Gene of insert is translocated to Centromere and geen slencing would be prevented by loss of function mutation at histone deacetyl transferease. In unacetylated histones, the N-terminal lysines are positively charged and interact strongly with DNA phosphates. The unacetylated histone tails also interact with neighboring histone octamers, favoring the folding of chromatin into condensed, higher-order structures whose precise conformation is not well understood. The net effect is that general transcription factors cannot assemble into a preinitiation complex on a promoter associated with hypoacetylated histones. In contrast, binding of general transcription factors is repressed much less by histones with hyperacetylated tails in which the positively charged lysines are neutralized and electrostatic interactions with DNA phosphates are eliminated and promote the gene expression. Reference: https://www.biotecharticles.com/Microbiology-Article/Meth- So mutation in deacetyl transferase facilitates the expression rather acetyl transferase. ods-of-Determining-Transcription-Start-Site-3812.html Reference: Molecular Biology of the cell- Lodish, 5th Edition- Chapter 11Trasncriptional control of gene expression, Page no:474

PART ‘C’ 71. Two liposome preparations (“X” and “Y”) are made using basic lipid composition as phosphatidylcholine (PC) and cholesterol (Chol). In “Y” a ganglioside (asialo-GM1) is incorporated during the preparation besides PC and Chol. In an attempt to find out the localization of asialo-GM1 in the membrane bilayer of “Y” (taking “X” as a negative control) and considering liposome as a true depiction of lipid bilayer structure of cellular membrane, following reagents are provided as probes: A. Phospholipase A B. Galactose binding lectin C. Exoglycosidase D. Cyclodextrin Choose the most appropriate reagent(s) from the above list to ascertain the localization of asialo-GM1 1. Only A 2. Both C and D 3. Both B and C 4. Both A and D

73. A group of scientists performed an experiment where they artificially fused mouse cells with monkey cells. The resulting fused cells were labelled with fluorescently tagged antibodies against mouse and monkey surface receptor proteins, X and Y respectively. At the time of 0 minute just after fusion of events, two receptors were confirmed to their own half in in the heterokaryon. However, such surface receptors (X and Y) intermixed on the cell surface after 60 minutes. Which one of the given statements correctly reflects the outcome of the experiment? 1. The proteins in cytoplasm are in a dynamic state 2. The proteins on the membrane surface are in a dynamic state 3. Surface membrane proteins exchange with the cytosolic proteins 4. Membrane surface proteins are in a static phase Answer: 2

Explanation: The fluorescent tags are attached to surface receptors X and Y, on mouse and monkey cells respectively. After fusion both the tags remain on the surface and are not lost with time. Only with time they change thier position. This shows that membrane proteins are not static but dynamic. Since the Answer: 3 tags are not lost with time so, they surface proteins are not exchanged with the Explanation: Option B is correct because galactose binding proteins are ca- cytoplasmic protein. Since the tags are specific so for surface proteins so no pable of binding strongly to the oligosaccharide moiety of ganglioside GM1. conclusion about the cytoplasmic proteins can be drawn for this experiment. Optin C is correct because GM1 gangliosidosis is an autosomal recessive lysosomal storage diseases associated with a neurodegenerative disorder or Reference:https://www.sciencedirect.com/science/article/pii/ dwarfism and skeletal abnormalities, respectively. These diseases are caused S0005273611001933 by deficiencies in the lysosomal enzyme beta-D-galactosidase (beta-Gal), which lead to accumulations of the beta-Gal substrates, GM1 ganglioside, and keratan sulfate. beta-Gal is an exoglycosidase that catalyzes the hydrolysis 74. Eg5 is a well-studied protein in Xenopus. To understand the function of terminal beta-linked galactose residues. So, both galactose binding lectins of Eg5 in mammalian cells, a team of researchers treated mammalian and exoglycosidase have affinity to GM1 and can be utilized for localizing the cells during late G2 phase with Eg5-inhibitor. The following diagrams represent images of mitotic cells. gangliosides located on membrane surface. Reference: http://www.jbc.org/content/287/20/16720.full, http://www.jbc. Based on the above information, what function might be attributed to Eg5 during mitosis? org/content/287/3/1801.full.pdf https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecuPhone: 080-5099-7000

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in the position of hydroxyl group at C4, similarly mannose is the C-2 epimer of glucose,Mannose and galactose are not epimers; they differ from each other in respect to the stereochemistry revolving around two carbons. Reference: https://chemistry.tutorvista.com/organic-chemistry/epimer. html, http://www.chemistryexplained.com/Bo-Ce/Carbohydrates.html

1. Eg5 Inhibits actin dynamics 2. Eg5 can activate GPCR signalling 3. Eg5 has motor activity 4. Eg5 can impact mitochondrial dynamics

76. Following statements are made related to protein structure

A. The hydrogen bonding patterns between the CO and NH groups are n →n + 3 in α helices; n→ n + 4 in 310 helix and n →n + 5 in π helix B. In a β turn, there are 10 atoms between the hydrogen bond donor and Answer: 3 acceptor C. In a γ turn, there are 6 atoms between the hydrogen bond donor and Explanation: Eg5 is a Mitotic motor protein is the homotetrameric, plus-end acceptor directed kinesin, which is involved in separating the two centrosomes dur- D. Parallel sheets have evenly spaced hydrogen bonds, which bridge the ing spindle assembly. Eg5 consists of three domains, a motor head that has strands at an angle. ATP binding capability and a C-terminal tail which has a MT binding domain, linked by a coiled-coil domain. Which one of the following combinations of the above statements is corEg5 remains static in the midzone of spindle appratus and important protein rect? for the assembly but for the maintenance of the bipolar spindle. So Inhibition of Eg5 during spindle formation or at metaphase causes a monopolar spindle 1. A and C phenotype or spindle shortening, respectively 2. A and B Reference: https://scholarworks.umass.edu/cgi/viewcontent.cgi?referer=https://www.google.co.in/&httpsredir=1&article=1583&context=theses

3. C and D 4. B and D

Answer: 4 75. Following are structures of stereoisomers of aldohexoses which differ in the stereochemistry

Explanation: Hydrogen bonds within an alpha-helix also display repeating pattern in which the backbone C=O of residue i hydrogen bonds to the backbone HN of residue i+4.Hydrogen bonds within an 3.10-helix also display a repeating pattern in which the backbone C=O of residue i hydrogen bonds to the backbone HN of residue i+3. Hydrogen bonds within an pi-helix display a repeating pattern in which the backbone C=O of residue i hydrogen bonds to the backbone HN of residue i+5. Hence statement A is incorrect. In the pattern characteristic of parallel β sheet, the hydrogen bonds are evenly spaced but slanted in alternate directions. Hence statement D is correct. One additional sort of tight turn involving only three residues has been described theoretically and also observed at least once in a protein structure . This is the γ turn, which has a very tight hydrogen bond across a seven-atom ring between the CO of the first residue and the NH of the third. Hence statement C is incorrect. Hydrogen-bonded β-turns in proteins occur in four categories: type I (the Based on above structures, following information was given below: most common), type II, type II’, and type I’. Asx-turns resemble β-turns, in that both have an NH. . .OC hydrogen bond forming a ring of 10 atoms. Hence A. D-glucose and D-mannose are epimers because they differ in the ste- statement B is correct. reochemistry at C-2 position. B. D-glucose and D-galactose are epimers because they differ in the ste- Reference: http://www.cryst.bbk.ac.uk/PPS2/course/section8/ss-960531_8. reochemistry at C-4 position. html#HEADING7, http://kinemage.biochem.duke.edu/teaching/anatax/ C. D-mannose and D-glucose are epimers because they differ in the ste- html/anatax.2b.html, reochemistry at C-3 position. http://kinemage.biochem.duke.edu/teaching/anatax/html/anatax.2c.html, D. D-galactose and D-glucose are epimers because they differ in the ste- https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2286581/ reochemistry at C-5 position. Choose one of the correct combinations of above statements:

77. The following statements are made on nucleic acid structure:

1. A and B 2. C and D 3. B, C and D 4. A and D

A. In the B-form of DNA, the sugar pucker is C2’ endo B. In RNA, the sugar pucker is C3’ exo C. The wobble base pair is formed between G and A in RNA D. A change in the sugar pucker from C2’ endo in the B-form of DNA to C3’ endo alters the width and depth of the major groove.

Answer: 1 Explanation: Monosaccharide like glucose, fructose, mannose and galactose can show different isomerism. They also show stereoisomerism due to the presence of chiral carbon atoms. Those stereoisomers which are differing in its configuration at only one chiral carbon atom are called as Epimers. For example, glucose and galactose are Epimers of each other, as they differ in only Phone: 080-5099-7000

Which one of the following combinations of above statements is correct? 1. A and C 2. B and D 3. A and D 4. B and C

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Answer: 3

Answer: 2

Explanation: The G·U wobble base pair is a fundamental unit of RNA secondary structure that is present in nearly every class of RNA from organisms of all three phylogenetic domains. It has comparable thermodynamic stability to Watson–Crick base pairs and is nearly isomorphic to them. Therefore, it often substitutes for G·C or A·U base pairs. Hence statement C is incorrect. B-DNA, the form described by Watson and Crick, adopts 2’-endo (example at right), whereas RNA and the A form of DNA adopt the 3’-endo twist. Hence statement A is correct, whereas, statement 2 is incorrect as it says 3’exo. Simple modification of the sugar puckering of 2’deoxyriboses leads to a reversible change between two stable forms of DNA which resemble very closely the canonical A and B duplex forms. A and B forms differ in the width and depth of the major groove. Hence Statement D is correct.

Explanation: The Gibbs free energy (G) of a system is a measure of the amount of usable energy (energy that can do work) in that system. The change in Gibbs free energy during a reaction provides useful information about the reaction’s energetics and spontaneity . Endergonic - NON-SPONTANEOUS, delta G0 > 0 Exergonic - SPONTANEOUS, delta G0 < 0Oxidative phosphorylation involves the reduction of O2 to H2O with electrons donated by NADH and FADH2, and occurs equally well in light or darkness. Photophosphorylation involves the oxidation of H2O to O2, with NADP+ as electron acceptor, and it is absolutely dependent on light. This electrochemical gradient creates potential energy across the inner mitochondrial membrane known as the proton-motive force

Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1083677/, Reference: http://www.bioinfo.org.cn/book/biochemistry/chapt18/bio1. http://fbio.uh.cu/sites/genmol/adic/na_arch.htm, https://www.ncbi.nlm.nih. htm; http://www.chem.tamu.edu/class/fyp/stone/tutorialnotefiles/thermo/ gov/pubmed/10496424 gibbs.htm 80. Match the enzymes in Column A with their respective biological function in Column B: 78. The Vmax and Km from a Lineweaver-Burk plot of an enzyme reac- Choose the correct combinations of answers from the options given betion where 1/v = 40 μM-1 min at 1/[s] = 0 and 1/[s] = -1.5 x 102 mM-1 at low: 1/v = 0 are 1. A - iii, B - i, C - ii, D - iv 1. 0.025 μM min-1 and 0.67 x 102 mM B. A - i, B - iii, C - iv, D - ii 2. 0.025 μM-1 min and 0.67 x 102 mM-1 C. A - iv, B - ii, C - i, D - iii 3. 0.025 μM min-1 and 1.5 x 102 mM-1 D. A - ii, B - iv, C - iii, D - i 4. 0.038 μM min-1 and 0.67 x 102 mM Answer: 1

Answer: 1

Explanation: Just take the Vmax = 1/ 40 => 0.025 micromol/min and Km = 1/150 => 0.0067 mM

Explanation: Lipases (triacylglycerol acylhydrolases, EC 3.1.1.3) catalyze the hydrolysis of triacylglycerols to glycerol, diacylglycerols, monoglycerols, and free fatty acids Flippases involved in creating membrane asymmetry are type-IV P-type

Reference:https://www.ncbi.nlm.nih.gov/books/NBK22557/ 79. From the following statements : A. Biosynthesis of proteins and nucleic acids from precursors results in production of chemical energy in the form of ATP, NADH, NADPH and FADH2 B. The spontaneity of a reaction in cells does not depend whether ΔG0 for the reaction is positive or negative C. Both oxidative phosphorylation and photophosphorylation involve oxidation of H20 to O2. D. Only chemical potential energy contributes to proton motive force in mitochondria Which one of the following combinations represents all INCORRECT statements? 1. A, B, C 2. B, C, D 3. A, B, D 4. A, C, D Phone: 080-5099-7000

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Phone: 080-5099-7000

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Special Edition ATPases (P4-ATPases) that catalyze the movement of specific phospholipid species from the extracellular leaflet to the cytosolic leaflet, whereas floppases are ABC-transporters that mediate the movement of phospholipids in the reverse direction. In contrast, scramblases are ATP-independent and act to randomize lipid distribution by bidirectionally translocating lipids between leaflets. Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4275391/ https://www.sciencedirect.com/topics/medicine-and-dentistry/lipases http://www.jbc.org/content/282/2/821.full https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/phospholipid-scramblase 81. An integral membrane protein (P) has been identified as a cell surface protein of hepatocytes and assigned to bind to hepatitis B virus (HBV) and promote its entry into cytosol. Upon binding to HBV particles, the C-terminal of P interacts with F-actin in the cytosol and and in turn, helps in the entry of the HBV particles. P was successfully cloned and expressed in animal cells in culture wherein its N-terminal is exposed on the surface while the C-terminal resides in the cytosol. The recombinant protein P so expressed retains its complete structure and function. From the list of experiments given below, which one of the experiments will you perform to show that C- terminal of the protein P via interacting with F-actin helps in HBV entry?

the distance from A to B and E is uniform electric field. uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates, labeled A and B. The potential difference or voltage between the plates is VAB=Ed. Thus E = V(AB)/d. Substituting the values into the equation we get E = -60mV/7nm = 8.571 X 10^6 V/m = 8.6 X 10^6 V/m Reference: tial-in-a-Unifor

https://cnx.org/contents/fl1dytBo@2/Electric-Poten-

83. E.coli DNA ligase catalyses formation of a phosphodiester bond between the adjoining 3’ hydroxyl, and 5’ phosphoryl ends in DNA duplexes. The energetic need for this reaction is met by the hydrolysis of NAD+ to NMN+ and AMP in a three step reaction. Following statements are being made about the mechanism of this reaction.. (i) AMP is linked to the 5’ phosphoryl end of the nicked DNA (ii) Adenylyl group of NAD+ is transferred to the ε-amino group of Lys in DNA ligase to form a phosphoamide adduct. (iii) DNA ligase catalyses the formation of a phosphodiester bond by the nucleophilic attack of the 3’ hydroxyl group onto the phosphate and releases AMP. Based on the statements made above, identify the correct sequence of the reaction steps. 1. (i)-(ii)-(iii) 2. (i)-(iii)-(ii) 3. (ii)-(i)-(iii) 4. (iii)-(i)-(ii)

1. Incubating radiolabelled HBV with hepatocytes in culture and follow up its association with F-actin by immunoprecipitation analysis using anti-F-actin antibody. 2. Incubating radiolabelled HBV with hepatocytes over-expressing the C-terminal mutant of P and repeat the rest of the experiment as in “1” Answer: 3 3. Incubating radiolabelled HBV with hepatocytes over-expressing the N-terminal mutant of P and repeat the rest of the experiment as in “1” Explanation: All DNA ligases catalyse the synthesis of phosphodiester 4. Using wild type P as well as C-terminal mutant of P and their indi- bonds in a very similar manner, by esterification of a 5′-phosphoryl to a 3′ hyvidual over-expression in a heterologous cell line (completely devoid of droxyl group. The reaction mechanism can be split into three distinct catalytic endogenous P protein) and then repeat the experiment as in “1” events. The first involves activation of the ligase through the formation of a covalent Answer: 4 protein–AMP intermediate. The nucleotide has been shown to be linked to the enzyme through a phosphoramidate bond to the ɛ amino group of a conserved Explanation: In host pathogene interaction the C terminal of HBV(P) pro- active site lysine. tein interacts wih F actin of host and viral entry occurs. To confirm the impor- In the second step of the reaction, the AMP moiety is transferred from the tance of C terminal part of (P) protein, Wild type and mutant type (P) protein ( ligase to the 5′-phosphate group at the single-strand break site. defective for C terminal end ) can be compared. The protein protein interaction Finally, DNA ligase catalyses the DNA ligation step with loss of free AMP. can be studied using immunoprecipitation and blotting. This can be done by In spite of these similarities between the two classes of enzymes, the manner incubating radiolabelled HBV(P) along with F actin and F actin autoantibod- by which the eubacterial and ‘eukaryotic’ proteins become activated is rather ies. The (P) with C part will interact with radiolabelled HBV and can be de- different. For eukaryotic ligases, the enzyme–AMP complex is formed after tected. The defective (P) will not interact rather F actin and autoanibodies for reaction of the enzyme and ATP with the release of free pyrophosphate. The F actin will interact. bacterial ligases become adenylated in an unusual reaction which involves cleavage of NAD+ and release of nicotinamide mononucleotide Reference: https://www.journal-of-hepatology.eu/article/S01688278(98)80750-2/abstract Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC113121/ https://www.sciencedirect.com/topics/immunology-and-microbiology/ immunoprecipitation 84. Given below are the types of vaccination (column A), the disease or conditions against which these vaccination types are used (column B) and 82. The cell membrane of neuron maintains intracellular conditions that the advantages or disadvantages for using these vaccination types (coldiffer from those of the extracellular environment. Such difference in in- umn C). tra- and extracellular conditions are critical to the function of the nerve cell as the nerve cell membrane resembles a charged capacitor. Assuming the electric field (E) across a parallel-plate capacitor is uniform and if membrane thickness is 7 nm and potential difference across the membrane is -60 m V, calculate E of the membrane 1. 6 x 105 V m-1 2. 7 x 105 V m-1 3. 8.6 x 106 V m-1 4. 6.6 x 106 V m-1

Which one of the following combinations is the most appropriate match?

Answer: 3 Explanation: The potential difference between points A and B is V(AB). d is Phone: 080-5099-7000

1. a-i-z, b-ii-y, c-iii-x 2. a-ii-x, b-i-y, c-iii-z 3. a-iii-y, b-ii-x, c-i-z

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Special Edition 4. a-ii-z, b-iii-x, c-i-y Answer: 2 Explanation: Live attenuated vaccines-Because of their capacity for transient growth, such vaccines provide prolonged immune-system exposure to the individual epitopes on the attenuated organisms, resulting in increased immunogenicity and production of memory cells. As a consequence, these vaccines often require only a single immunization, eliminating the need for repeated booster. Diphtheria and tetanus vaccines, for example, can be made by purifying the bacterial exotoxin and then inactivating the toxin with formaldehyde to form a toxoid-Not stable and recurrent booster is required since its a protein. and Inactivaetd vaccines are stable.

1. A only 2. A and D 3. A and B 4. C and D Answer: 3

Explanation: In group II introns the reaction pattern is similar except for the nucleophile in the first step, which in this case is the 2’ -hydroxyl group of an A residue within the intron .A branched lariat 2’-5’ structure is formed as an intermediate. Within the spliceosome, a series of RNA–RNA, RNA–protein, and protein– protein interactions is needed to identify and remove intronic regions and join exons, producing a mature transcript. The mature spliceosome carries out splicing through two transesterification reactions. First, the 2′-OH of a branch Reference: Immunology-Janis Kuby-Chapter Vaccines-Pg No: 421 point nucleotide performs a nucleolytic attack on the first nucleotide of the 85. Actinomycin D inhibits the process of transcription in both prokary- intron, forming a lariat intermediate. Second, the 3′-OH from the free exon otic and eukaryotic organisms. The following statements are made about performs a nucleolytic attack on the last nucleotide of the intron, thereby joining exons and releasing the lariat intron. Intron identification relies on certain actinomycin D- mediated inhibition of transcription. sequences, including the 5′ splice site, branch point (and downstream polypyA. Actinomycin D inhibits transcription from a double stranded DNA rimidine tract) and 3′ splice site. template by either E.coli or yeast RNA polymerases. B. Actinomycin D inhibits transcription from a single stranded RNA Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3140690/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2982171/ template by eukaryotic viral RNA polymerases. C. Actinomycin D inhibits transcription from a single stranded Φx174 DNA template by E.coli RNA polymerase immediately after viral DNA 87. A plasmid with a linking number (Lk) of 200, topological winding entry. D. Actinomycin D inhibits transcription from a double stranded RNA (Tw) number of 200 and a writhing number (Wr) of 0 was transformed into E.coli. The plasmid was re-isolated from the culture of the transtemplate by eukaryotic polymerase II. formant. The re-isolated plasmid was found to possess the same molecuWhich of the combinations of the above statements is a true representa- lar weight as the original plasmid, but it possessed a writhing number of tion of the mechanism of actinomycin D mediated inhibition. -5. Following statements are made about this observation. 1. A only 2. A, B and C 3. A, B and D 4. A and D only

A. Lk of the re-isolated plasmid would be 195 B. Lk of the re-isolated plasmid would remain 200 C. Tw of the re-isolated plasmid would remain 200 D. Tw of the re-isolated plasmid would be 195

Answer: 1 Explanation: Actinomycin D is an intercalating compounds with sequence selectivity. The actinomycin chromophore favours guanine-cytosine pairs and is therefore inserted between the G-C step. Hydrogen bonds are established between the guanine 2-amino group and the carbonyl oxygen of threonine, and also between the guanine N-3 atom and the NH group of the same threonine residue, helping to stabilize the actinomycin–DNA complex. The formation of this very stable actinomycin–DNA complex prevents the unwinding of the double helix which leads to inhibition of the DNA-dependent RNA polymerase activity and hence transcription. Reference: https://www.sciencedirect.com/topics/neuroscience/actinomycin; https://pubchem.ncbi.nlm.nih.gov/compound/actinomycin_D#section=Top

Which one of the following combinations of the above statements is the correct representation of the facts. 1. A only 2. A and C 3. A and D 4. D only Answer: 2 Explanation: Linking number= Twist + Writhe. Linking number of the structure does not change.Since L is constant in an intact duplex DNA circle, for every new double helical twist,deltaT, there must be an equal and opposite superhelical twist, that is, deltaW = delta T. Reference: Voet and Voet Chapter, 29, page 1160

86. During maturation process of some RNA molecules, formation of a 2’-5’ phosphodiester bond takes place. Following statements are made 88. Following statements are being made about the orientation of the N-glycosidic bond between the base and the sugar in the following DNA about this phenomenon: duplexes. A. Spliceosome mediated removal of intronic sequences occurs through A. ‘anti’ for B form DNA duplexes the formation of a 2’-5’ phosphodiester bond. B. Removal of group II introns occurs through the formation of 2’-5’ B. ‘syn’ for B form DNA duplexes C. ‘anti’ for A form DNA duplexes phosphodiester bond. C. Enzymatic removal of introns from the yeast tRNA precursors in- D. ‘syn’ for A form DNA duplexes volves 2’-5’ phosphodiester bond formation. D. RNaseP mediated 5’-end maturation of tRNA precursors involves for- Which one of the following combinations of the above statements is correct? mation of a 2’-5’ phosphodiester bond. Which one of the following combinations of the statements is a true representation? Phone: 080-5099-7000

1. A and C 2. B and C

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Special Edition 3. A and D 4. B and D Answer: 1 Explanation: The bases can exist in two distinct orientations about the N-glycosidic bond. These conformations are identified as, syn and anti. It is the anti conformation that predominates in naturally occurring nucleotides. i.e in both B and A forms of DNA duplexes. There is only one known case where a purine adopts a syn conformation. The unusual form is to find in an even more unusual structure namely left handed Z-DNA. Reference: https://themedicalbiochemistrypage.org/nucleic-acids.php

trimeric G proteins, which further transduce these signals intracellular to appropriate downstream effectors and thereby play an important role in various signaling pathways. Hence statement B is correct. Upon activation by a ligand, the GPCR undergoes a conformational change and then activate the G proteins by promoting the exchange of GDP/GTP associated with the Gα subunit. This leads to the dissociation of Gβ/Gγ dimer from Gα. Hence statement D is incorrect as it states that the “on” form binds to β/γ. Both these moieties then become free to act upon their downstream effectors and thereby initiate unique intracellular signaling responses. After the signal propagation, the GTP of Gα-GTP is hydrolyzed to GDP and Gα becomes inactive (Gα-GDP), which leads to its re-association with the Gβ/Gγ dimer to form the inactive heterotrimeric complex. Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2801357/

89. During translation in prokaryotes, when ribosomes reach the termination codon, the termination codon is recognized by the class I release factors (RF1 or RF2) leading to the release of the polypeptide. A second 91. The following intracellular event occurs in a cell that is subjected to class II release factor (RF3) facilitates the termination process. Which of conditions of starvation. the following statements regarding the mechanism of action of the release factors is INCORRECT? 1. Class I release factors decode the stop codons while the RF3 is a GTPase that stimulates recycling of the class I release factors. 2. Free RF3 has a higher affinity for GTP than GDP 3. RF1 and RF2 share a conserved segment of ‘GGQ’ sequence which is essential for the polypeptide release 4. RF1 and RF2, individually possess another stretch of tripeptide sequences which are involved in the recognition of the termination codon Answer: 2 Explanation: Free RF3 binds GDP very strongly, and the spontaneous dissociation of the guanine nucleotide is slow, implying that cytoplasmic RF3 is in the GDP, rather than the GTP, conformation. When RF3·GDP associates with a ribosome in complex with a class 1 RF, the GDP on RF3 can rapidly dissociate and be exchanged for GTP. RF3 in the GTP conformation has a high affinity for the ribosome in the absence of a class 1 RF, and its binding destabilizes that of RF1 or RF2. The formation of RF3·GTP therefore leads to Which one of the following statements correctly represents the event rapid dissociation of RF1 or RF2 from the ribosome. Subsequent hydrolysis of shown above? GTP on RF3 leads to its fast dissociation in the GDP conformation, the normal free state of the factor 1. The cell is undergoing apoptotic cell death with the help of lysosomes (A). Reference: https://www.sciencedirect.com/science/article/pii/ 2. The cell is undergoing autophagy by formation of autophagolysosomes S1097276502006913 (C) 3. The cell is undergoing necroptosis. 90. G protein-coupled receptors (GPCRs) are used to detect and respond 4. The cell is undergoing autophagy and fusion occurs between lysosome to many different types of signals, including neurotransmitters, hormones (B) and autophagolysosome (A) involved in glycogen and fat metabolism and even photons of light. Which one of the following statements regarding GPCR is INCOR- Answer: 2 RECT? Explanation: Autophagy is an evolutionarily conserved, lysosomal pathway 1. GPCRs are a large family with a common structure of seven mem- of engulfment, degradation and recycling of cellular contents including longbrane spanning α helices. lived proteins and organelles. Autophagy promotes cell survival, maintaining 2. GPCRs are coupled to trimeric G proteins comprising three subunits, cellular homeostasis under resting and stress conditions.The process requires α, β and γ formation of a double-membrane structure containing the sequestered cyto3. The Gα subunit is a GTPase switch protein that alternates between an plasmic material, the autophagosome (A), that ultimately fuses with the lysoactive (‘on’) state with bound GTP and an inactive (‘off’) state with GDP some (B) and gives autophagolysosome (C). 4. The ‘on’ form gets bound to β and γ subunits and activates a membrane-bound effector like adenylyl cyclase, phospholipase C or ion chan- Reference: http://docs.abcam.com/pdf/cardiovascular/autophagy_in_ nel heart_disease.pdf https://www.ncbi.nlm.nih.gov/pubmed/28933638 Answer: 4 Explanation: A remarkable property of GPCRs which make it distinguishable from other classes of receptors is the presence of seven TM (7TM) α-helical region. Members of the GPCR superfamily share the same basic architecture i.e., 7TM α-helices, an extracellular amino-terminal segment and an intracellular carboxy-terminal tail. Hence statement A is correct. Heterotrimeric G proteins (Gα, Gβ/Gγ subunits) constitute one of the most important components of cell signaling cascade. G Protein Coupled Receptors (GPCRs) perceive many extracellular signals and transduce them to heteroPhone: 080-5099-7000

92. The extracellular matrix (ECM) is a complex combination of secreted proteins that is involved in holding cells and tissues together. The components of ECM form a network by binding to each other and communicate with cells by binding to adhesion receptors on the cell surface. ECM comprises mainly two classes of macromolecules, proteoglycans and very high molecular weight large proteins. Which one of the following statements regarding statements regarding

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Special Edition ECM constituents is INCORRECT? 1. Proteoglycans are a subset of secreted or cell surface-attached glycoproteins containing covalently linked specialized polysaccharide chains called glycosaminoglycans (GAGs) 2. GAGS are long branched polymers of specific repeating disaccharides of sialic acid and glucose or galactose. 3. Major types of GAGs present in ECM are heparan sulphate, chondroitin sulphate, dermatan sulphate, keratan sulphate and hyaluronan. 4. Major types of large proteins present in ECM are collagen, laminin, elastin and fibronectin.

Reference: https://www.sciencedirect.com/topics/medicine-and-dentistry/ tetrahydrofolatehttps://pubchem.ncbi.nlm.nih.gov/compound/etoposidehttps://www.ncbi.nlm.nih.gov/pmc/articles/PMC4652635 https://www.ncbi. nlm.nih.gov/pmc/articles/PMC4161504/https://www.ncbi.nlm.nih.gov/pmc/ articles/PMC5214563https://www.researchgate.net/.../Mechanism-of-action-of-paclitaxel-Paclitaxel-targets-michttps://www.nature.com â&#x20AC;ş nature reviews cancer â&#x20AC;ş review articleshttps://pubchem.ncbi.nlm.nih.gov/compound/5-Fluorouracil

94. Activation-induced cytidine deaminase (AID) is the key mediator of somatic hypermutation, gene conversion and class-switch recombination. In order to ascertain the role of AID in class-switch recombination, Explanation: Glycosaminoglycans (GAGs) are large linear (UN- immune response against a target antigen was compared between AID BRANCHED) (hence statement 2 is incorrect) polysaccharides constructed knock-out mice and (AID-/-) with that of mice retaining a functional copy of repeating disaccharide units with the primary configurations containing an of the AID gene (AID+/-). Development of IgM and IgG antibodies against amino sugar (either GlcNAc or GalNAc) and an uronic acid (either glucuronic the target antigen was then measured following successive immunization acid and/or iduronic acid). There are five identified glycosaminoglycan chains and plotted graphically. : Hyaluronan, Chondroitin, Dermatan, Heparin/heparan, Keratan (Hence Statement 3 is correct) . Proteoglycans consist of a core protein and one or more Which one of the following is the most appropriate representation of the covalently attached glycosaminoglycan chains. Virtually all mammalian cells experiment? produce proteoglycans and secrete them into the ECM, insert them into the plasma membrane, or store them in secretory granules. Hence statement 1 is correct. The major components of the ECM are fibrillar proteins that provide tensile strength and elasticity (e.g., various collagens and elastins), adhesive glycoproteins (e.g., fibronectin, laminins, and tenascins), and proteoglycans . Hence statement 4 is also correct. Answer: 2

Reference: https://www.ncbi.nlm.nih.gov/books/NBK453033/, https:// www.sigmaaldrich.com/technical-documents/articles/biology/glycobiology/ glycosaminoglycans-and-proteoglycans.html 93. Present day cancer treatment uses many approaches. Beyond surgery and radiation treatment, which are most often employed in cases of larger, more discrete tumors, drug therapies can be used target residual tumor cells and to attack dispersed cancers. Chemotherapies by anti-cancer drugs are mostly aimed at blocking DNA synthesis and cell division. A list of anti-cancer drugs is given in Column A, their chemical nature in Column B and their mechanism of action in Column C

Which one of the following is the most appropriate match? 1. i-a-I, ii-b-II, iii-c-III, iv-d-IV 2. i-b-II, ii-a-III, iii-d-I, iv-c-IV 3. i-c-III, ii-d-IV, iii-a-II, iv-b-I 4. i-d-I, ii-a-IV, iii-b-II, iv-c-III Answer: 4 Explanation: Methotrexate:Methotrexate inhibits dihydrofolic acid reductase, the enzyme that reduces dihydrofolate to tetrahydrofolate, which is utilized in purine nucleotide synthesis. Etoposide:Etoposide derives from podophyllotoxin, etoposide targets DNA topoisomerase II activities thus leading to the production of DNA breaks and eliciting a response that affects several aspects of cell metabolisms. Paclitaxel: Lead alkaloids isolated from the trees are taxol and camptothecins that currently have annual sales in billion dollars. Paclitaxel is also commonly known as taxol. 5-flurouracil: 5-FU acts in several ways, but principally as a thymidylate synthase (TS) inhibitor. Interrupting the action of this enzyme blocks synthesis of the pyrimidine thymidine, which is a nucleoside required for DNA replication Phone: 080-5099-7000

Answer: 1 Explanation: Primary response always shows higher production of IgM whereas IgG expression occurs in secondary response, which is due to class switching mechanism. In the given graphs of IgM and IgG production, AID containing mouse should be able to perform class switching and as a result in secondary response will produce more amount of IgG compared to IgM. In case of AID deficient mouse there should not be any class switching as a result secondary response will not produce IgG. But it can still produce IgM in both primary and secondary response.

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Reference: Immunology by Kuby, chapter 11, page no. 264-265

of KNAT1 and KNAT2, which encode homeobox proteins. Hence statement D is also correct.

95. Following statements were made regarding vulval development in Caenorhabditis elegans:

Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1221894/

Which one of the following options represents a combination of correct statements?

Which combination of the above statements is correct?

97. cAMP signalling plays a very important role in the development of A. The six vulval precursor cells (VPCs) are influenced by the anchor cell Dictyostelium discoideum. Below are few statements related to it: to form an equivalence group. B. In the loss of function lin-12 mutants, both cells become uterine where- A. Every amoeba at the time of aggregation has the potential to make, receive and relay cAMP as in gain of function mutants, both become anchor cell. C. If the anchor cell is destroyed early in development, all the six VPCs B. acb- mutants develop normally but the spores formed appear glassy and are unable to germinate. divide once and contribute towards the formation of hypodermal cells. D. The anchor cell/ventral uterine precursor decision is due to Notch-Del- C. The spores formed by the acg- mutants germinate in the sorus itself. D. RegA is an extracellular phosphodiesterase ta mediated mechanism of restricting adjacent cell fates. E. The paracrine factor secreted by the anchor cell directly activates the E. cAMP is continuously secreted in nanomolar amounts during aggregation Notch-Delta pathway.

1. A, C and D 2. A, B and D 3. C, D and E 4. B, D and E

1. A and D 2. A and B 3. A and E 4. B and D Answer: 2

Explanation: Throughout the aggregation stage cAMP synthesis is oscillatory with a period of about 6 min as the result of control of ACA activity by the PKA circuit. So option A is correct, while E is incorrect. Explanation: RegA is an internal cAMP phosphodiesterase. So option D is incorrect. Reference: Mutants of acrA (ACB gene) show severe defects in terminal differentiation. They aggregate normally but they produce only about 10% of viable spores. 96. Change in leaf morphology is observed during transition from veg- Their fruiting bodies show abnormally long stalks and glassy spore heads due etative to reproductive phase in several plants. The following statements to the reduced number of spores - hence option B is correct. Expression of adenylyl cyclase G, encoded by the acgA gene, is induced after are proposed to explain the above observation: 12 hours of development in prespore cells and greatly increased during spore A. Alteration in the gene content of leaves of reproductive phase from differentiation. AcG activity is regulated by osmolarity of the external media and this enzyme is required to avoid spore germination inside the sorus. - So those of vegetative phase. B. Differential methylation pattern of genes influencing leaf develop- option C is incorrect ment and morphology. https://www.sciencedirect.com/science/article/pii/ C. Mutation in transcription factor that prevents its association with Reference: S0012160614001924 promoter elements of genes regulating leaf development. D. Small RNA mediated inhibition of gene expression of a homeotic gene. https://openaccess.leidenuniv.nl/bitstream/handle/1887/12476/General+introduction.pdf;jsessionid=0DA6957EE6D67E43027411C1FD5DAFWhich one of the following options represents a correct combination of 6D?sequence=2 statements that could explain the observed changes? Answer: 1

98. Torpedo, is known to serve as a receptor for Gurken. Deficiencies of the torpedo gene in Drosophila cause ventralization of the embryo. In an experiment, the germ cell precursors from a wild type embryo were transplanted into embryos whose mother carried the torpedo mutation. Also, the reverse experiment, i.e., transplantation of germ cell precursors from torpedo mutants into wild type embryos was done. The torpedo deficient Answer: 2 germ cells developed in a wild type female showed normal dorso-ventral Explanation: Plant development follows a coordinated and progressive tran- axis, while the wild type germ cells developed in a torpedo deficient fesition from the reproductively incompetent juvenile phase to the mature phase, male showed ventralized egg. which is competent for flowering. The transition from juvenile to mature development involves various physiological and morphological changes, includ- Some of the following statements are drawn from the above experiments ing alterations in leaf structure. Variation in leaf morphology between vege- and some from known facts to understand the functioning of Torpedo. tative phases, also known as heteroblasty, is observed in many plant species, including the model plant Arabidopsis. This leaf morphogenesis is regulated A. Zygotic contribution of Torpedo is essential for the developmental of by post-translational modifications of histones and chromatin remodeling. Ex- dorso-ventral axis pression of genes involved in leaf-axis specification is regulated by acetylation B. Maternal contribution of Torpedo is essential for the development of of histones (hence statement B should be incorrect as it says methylation) and dorso-ventral axis chromatin remodeling. GENERAL TRANSCRIPTION FACTOR GROUP E6 C Since Torpedo is a receptor for Gurken and follicle cells surround the (GTE6), which encodes a bromodomain-containing protein, plays an essen- part of the oocyte where Gurken is expressed, it is likely that Torpedo is tial role in controlling differences in the development of primordia produced expressed in follicle cells. during juvenile and mature phases. GTE6 regulates the development of ma- D. Gurken signalling initially dorsalizes the follicle cells which in turn ture leaf shape through acetylation of histones present on ASYMMETRIC send signal to organize the dorso-ventral polarity in oocyte. LEAVES1 (AS1), a gene involved in leaf-axis specification in Arabidopsis. E. Gurken signalling initially dorsalizes the nurse cells which help in Hence statement A should be correct). AS1 in turn represses the transcription generation of dorso-ventral polarity in oocyte. 1. B and C 2. A and D 3. B and D 4. A and C

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structures within the cytoplasm of vegetal blastomeres. In these cells, Dsh Which one of the following combination of statements is most appropri- stabilizes β-catenin and causes it to accumulate in nuclei, resulting in the acate? tivation of transcriptional gene regulatory networks that drive mesoderm and endoderm formation. So if Beta catenin is not allowed to enter the nucleus, 1. B, C and D all cells become ectodermal. Hence statement E is correct and D is incorrect. 2. A, C and D 3. B, C and E Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3827468/, 4. A, D and E https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2697034/ Answer: 1 Explanation: The oocyte nucleus is originally located at the posterior end of the oocyte, away from the nurse cells. It then moves to an anterior dorsal position and signals the overlying follicle cells to become the more columnar dorsal follicle cells. Hence statement D is correct and Statement E incorrect. The dorsalizing signal from the oocyte nucleus appears to be the product of the gurken gene. The Gurken signal is received by the follicle cells through a receptor encoded by the torpedo gene. Hence Statement C is correct. Maternal deficiency of torpedo causes the ventralization of the embryo. Moreover, the torpedo gene is active in the ovarian follicle cells, not in the embryo. Hence statement B is also correct.

1. A normal limb is regenerated after amputation, irrespective of whether the cut was made below the elbow or through the humerus 2. It occurs by compensatory regeneration and does not include formation of an apical ectodermal cap 3. Regeneration occurs through formation of a blastema, which essentially consists of unspecified multipotential progenitor cells. 4. Proliferation of the blastema cells does not require nerves factors secreted by the nerves. Answer: 1

Reference: https://www.ncbi.nlm.nih.gov/books/NBK10082/ 99. Given below are few statements regarding the role Disheveled (Dsh) and β-catenin (β-cat) in the development of sea urchin. A. Dsh is localized in the vegetal cortex of the oocyte before fertilization and in the region of the 16-cell embryo about to become the micromeres. B. Dsh is localized in the cytosol of the oocyte during oogenesis and in the micromere forming blastomeres of a 16-cell embryo C. β-cat accumulates predominantly in the micromeres and somewhat in the veg2 tier cells. D. Treatment of embryos with lithium chloride does not allow the accumulation of β-cat in the nuclei of all blastula cells, and the animal cells thus become specified as endoderm and mesoderm. E. When β-cat is prevented from entering the nucleus, the embryo develops as a ciliated ectodermal ball Which one of the following options represents a combination of correct statements? 1. B. C and E 2. A, C and D 3. A, C and E 4. B, D and E

Explanation: When an adult salamander limb is amputated, the remaining cells are able to reconstruct a complete limb, with all its differentiated cells arranged in the proper order. In other words, the new cells construct only the missing structures and no more. For example, when a wrist is amputated, the salamander forms a new wrist and not a new elbow. Experiments done with regeneration of a salamander forelimb showed that in the amputation made below the elbow and in the amputation through the humerus, the correct positional information was respecified in both cases. Hence statement A is correct. Salamanders accomplish this feat by dedifferentiation and respecification. Upon limb amputation, a plasma clot forms, and within 6 to 12 hours, epidermal cells from the remaining stump migrate to cover the wound surface, forming the wound epidermis. This single-layered structure is required for the regeneration of the limb, and it proliferates to form the apical ectodermal cap. Hence statement B is incorrect. The formerly well-structured limb region at the cut edge of the stump forms a proliferating mass of indistinguishable, dedifferentiated cells (Hence not progenitor cells, which makes Statement C also incorrect) just beneath the apical ectodermal cap. This dedifferentiated cell mass is called the regeneration blastema. These cells will continue to proliferate, and will eventually redifferentiate to form the new structures of the limb. The proliferation of the salamander limb regeneration blastema is dependent on the presence of nerves. Hence statement D is also incorrect. Reference: https://www.ncbi.nlm.nih.gov/books/NBK9971/#_A4361_

Answer: 3 Explanation: Pattern formation along the animal-vegetal (AV) axis in sea urchin embryos is initiated when canonical Wnt (cWnt) signaling is activated in vegetal blastomeres. The egg’s vegetal cortex plays a critical role in this process by mediating localized “activation” of Disheveled (Dsh). Dsh is broadly expressed during early sea urchin development, but immunolocalization studies revealed that this protein is enriched in a punctate pattern in a novel vegetal cortical domain (VCD) in the egg. Vegetal blastomeres inherit this VCD during embryogenesis, and at the 60-cell stage Dsh puncta are seen in all cells that display nuclear β-catenin. Analysis of Dsh post-translational modification using two-dimensional Western blot analysis revealed that compared to Dsh pools in the bulk cytoplasm, this protein is differentially modified in the VCD and in the 16-cell stage micromeres that partially inherit this domain. Dsh protein begins to accumulate asymmetrically in the cortex of small oocytes and this pole was identified as the vegetal pole. Hence statement A is correct. A critical, early event in the patterning of the animal-vegetal (A-V) axis in the early sea urchin embryo is the maternally regulated entry of β-catenin into the nuclei of vegetal cells. So statement C is correct. Later differentiation of vegetal cell progeny requires augmentation of nuclear β-catenin through zygotic expression of Wnt and activation of the canonical Wnt pathway. Dishevelled (Dsh) is a key signaling molecule in the canonical Wnt pathway. In the early sea urchin embryo, Dsh is concentrated in punctate Phone: 080-5099-7000

100. Which one of the following statements regarding limb regeneration in Salamander is correct?

101. To characterize the mechanism/s by which heat-stress is perceived in Arabidopsis, a team of researchers fused a Heat Shock promoter with luciferase gene. Transgenic plants having promoter:luciferase fusion were raised. Such plants revealed strong luciferase expression under unstressed control condition. Subsequently, these transgenic plants were mutagenized by EMS and seeds from F2 generation were obtained. To analyze the downstream positive regulators of heat-stress, the researchers should analyze seedlings that are. 1. expressing luciferase in the presence of heat-stress 2. not expressing luciferase in the presence of heat-stress 3. expressing luciferase in the absence of heat-stress 4. not expressing luciferase in the absence of heat-stress Answer: 3 Explanation: The question is about the perceiving of heat stress by Arabidopsis. So obviously under heat stress conditions, the heat stress promoter should be recognized and that leads to strong expression of luciferase ( which is the same as option 1, so these seedlings cannot be selected to analyze any downstream expression regulation). And control, does not express luciferase

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Special Edition as the promoter has not been recognized as stress conditions are not there (again same condition as Option 2) So now, the EMS mutagenesis, might mutate the heat stress promoter which renders it not capable of perceiving any heat stress conditions. In this case, if still luciferase is expressed means the expression is clearly controlled by some downstream positive regulators triggered by stress conditions justifying option 3 to be correct. On the other hand, EMS mutating the luciferase enzyme, will not express luciferase in any conditions, so any kind of expressional regualtion cannot be analyzed in seedlings that are not expressing luciferase even in stress conditions. So option 2 cannot be correct. Reference: pdf/2075.pdf

Answer: 2 Explanation: The apoplastic (non-living) pathway provides a route toward the vascular stele through free spaces and cell walls of the epidermis and cortex. The symplastic (living) route to the vascular stele involves cell to cell transport by plasmodesmata. Plasmodesmata are channels of cytoplasm lined by plasma membrane that transverse cell walls. Transcellular transport involves passage through two plasma membranes and can occur by secretion and subsequent endocytosis, diffusion or transporter activity.

Reference: http://passel.unl.edu/pages/informationmodule.php?idinforhttps://www.ncbi.nlm.nih.gov/pmc/articles/PMC3613477/ mationmodule=1057703469&topicorder=3&maxto=6,

104. Stomata from detached epidermis of common dayflower (Commelina communis) were treated with saturating photon fluxes of red light. 102. Two near inbred parental lines P1 and P2 of an angiosperm are In a parallel treatment, stomata treated with red light were also illumicrossed to produce F1 seeds in which, the ploidy of the endosperm is 6N. If nated with blue light (indicated by arrow). From the graphs shown below, plants generated from these F1 seeds are backcrossed with P1, what will select the correct pattern of stomata opening (solid lines and dotted lines represent stomatal aperture under red and blue lights, respectively) be the ploidy of the somatic cells in the next generation? 1. 2N 2. 4N 3. 5N 4. 6N Answer: 3 Explanation: In angiosperms, spermatozoa go by pair in each pollen grain and fertilize, in addition to the egg cell, one of its sister cells, called the central cell. This “double fertilization” leads to the embryo on the one hand and to its nutritive tissue, the endosperm, on the other hand. In addition, in most flowering plants, the endosperm is triploid 2N from one parent and N from another) because of a doubled maternal genetic contribution in the central cell. Here the ploidy is 6N means it has underdone nondisjunctio. tthe parents are however 2N. so when F1 (6N) plant is crossed with P1 (2N), a 5N plant will be produced- 3N (half the number from F1 plant) + 2N (from P1). Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC2828724/ 103. The table given below represents the types of intercellular transport in “Column I” in land plants and their transport pathways in “Column II”

Answer: 1 Explanation: Both red and blue light stimulate stomatal opening. Because chlorophyll also absorbs these wavelengths, sensitivity to red and blue light is consistent with a role of guard cells in opening stomata under conditions conducive for photosynthesis. Indeed, in most species guard cells are the only epidermal cells that contain chloroplasts, and guard cell chlorophyll is implicated as a photoreceptor in the light responses of stomata. However, the greater quantum efficiency of blue light over red light in stimulating stomatal opening suggests that guard cells also possess a specific blue-light photoreceptor. So most appropriate response is seen in graph 1. Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1054730/ http://www.plantphysiol.org/content/119/3/809 105. Following are certain statements regarding Rubsico. The predominant protein in plant leaves that catalyses the initial reaction of Calvin-Benson cycle.

Which one of the following combinations matches column I correctly with column II 1. A-i, B-ii, C-iii 2. A-ii, B-i, C-iii 3. A-iii, B-ii, C-i 4. A-i, B-iii, C-ii Phone: 080-5099-7000

A. During the oxygenase activity of Rubisco, O2 is used as substrate to produce three-carbon molecule, 3-phosphoglycerate and two-carbon molecule, 2-phosphoglycolate B. In red and brown algae, the large subunit of Rubisco is localized in the chloroplast while small subunit is localized in the nucleus. C. The bound sugar phosphates in Rubsico are specifically removed by an ATP dependent enzymes, Rubisco activase

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Special Edition D. The active form of Rubisco catalyzes carboxylation or oxygenation reactions in five steps Which one of the following combinations of above statements is correct? 1. A, B and C 2. A, B and D 3. B, C and D 4. A, C and D Answer: 4

Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4451336/ https://www.ncbi.nlm.nih.gov/pmc/articles/PMC102221/ 107. A chemist synthesizes three new chemical compounds in the laboratory and names them as X, Y and Z. After analysing mutagenic potential of all these compounds, the geneticist observed that all are highly mutagenic. The geneticist also tested the potential of mutations induced these compounds to be reversed by other known mutagens and obtained the following results.

Explanation: Rubisco is one of the slowest and largest enzymes, with a molecular mass of 560 kDa . In land plants and green algae, the chloroplast rbcL gene encodes the 55-kDa large subunit, whereas a family of rbcS nuclear genes encodes nearly identical 15-kDa small subunits ; in nongreen algae both the rbcL and rbcS genes are chloroplast encoded. So statement B is wrong. Correct combination is A,C and D. Reference: https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1884142/ 106. Given below are certain statements regarding plant-pathogen interactions:

Assuming that X, Y and Z caused any of the three types of mutations, transition, transversion or single base deletion, what conclusions can you make about the nature of mutations produced by these compounds?

1. X causes transversion; Y causes transition; Z causes single base deleA. The pattern recognition receptor (PRR), upon perceiving pathogen or tion microbe-associated patterns (PAMPs/MAMPs), activates plant defences 2. X causes transition; Y causes transversion; Z causes single base deleresulting in pattern triggered immunity (PTI) tion B. A vrPto is a resistance gene in tomato that acts against pathogenic 3. X causes transition; Y causes single base deletion; Z causes transverattack by the bacterium Pseudomonas syringae pv. Tomato. sion C. The effector molecules produced by pathogen is recognized by resist- 4. X causes transversion; Y causes single base deletion; Z causes transiance ( R ) gene present in plants resulting into a defence strategy known tion as effector triggered immunity (ETI) D. Defence mechanisms triggered in plants during PTI are usually Answer: 2 stronger than those during ETI Explanation: Treatment of cytosine with nitrous acid produces uracil which Which one of the following combinations of above statements is correct? pairs with adenine to produce a CG-to-TA transition mutation during replication. Likewise, nitrous acid modifies adenine to produce hypoxanthine, a base 1. A and B that pairs with cytosine rather than thymine, which results in an AT-to-GC 2. C and D transition mutation. So transition mutations can be reverted by Nitrous acid. 3. A and C Mutations induced by hydroxylamine can only be CG-to-TA transitions. So 4. B and D only AT-GC transitions can be revrted by hydroxylamine. Intercalating agents such as proflavin, acridine, Answer: 3 and ethidium bromide can cause either additions or deletions, frameshift mutations induced by intercalating agents can be reverted by a second treatment Explanation: The first interaction between the plants and microbes take with those same agents. So single base deletion induced by Z can be reverted place in apoplast and is mediated by the recognition of microbial elicitors by by acridine orange. the receptor proteins of the plants. These microbial elicitors, also known as pathogen-associated molecular patterns (PAMPs), are recognized by the mem- Reference: iGenetics, A Molecular Approach brane-localized pattern recognition receptors (PRRs) of plants. The bacterial Third Edition, Peter J. Russell, Page-142 flagellin and elongation factor (EF)-Tu peptide surrogates, flg22 and elf18, and chitin, are common examples of PAMPs, which are recognized by the plant PRRs that include the three receptor-like kinases, flagellin-sensitive22 108. An individual is having an inversion in heterozygous condition. The (FLS2), EF-Tu receptor (EFR), and chitin elicitor receptor kinase1 (CERK1). regions on normal chromosome are marked as A, B, C, D, E, F, G while The successful recognition of microbial derived PAMPs by PRRs of the plants the chromosome having inversion has the regions as a,b,e,d,c,f,g. The diactivates a first line of defense which is known as PAMP-triggered immunity agram given below shows pairing of these two homologous chromosomes (PTI).So A is correct. during meiosis and the site of a crossing over is indicated. However, resistance (R) proteins of plants recognize these effector proteins of pathogens and can induce a second line of defense which is known as the effector-triggered immunity (ETI) . ETI is quantitatively stronger and faster than PTI and can result in a localized cell death (hypersensitive response) to kill both pathogen and pathogen infected plant cells. C is correct and D is wrong. The avrPto gene of Pseudomonas syringae pv tomato triggers race-specific resistance in tomato plants carrying Pto, a resistance gene encoding a protein kinase. When introduced into P. s. tabaci, avrPto triggers resistance in tobacco W38 plants that carry the corresponding R gene. The AvrPto protein is believed to be secreted into host cells through the bacterial type III secretion pathway, The following statements are given to describe the inversion and the where it activates disease resistance in tomato by interacting with Pto.So the consequence of crossing over shown in the above diagram: AvrPto gene is from the bacteria not from the plant. So B is also wrong. A. This is a pericentric inversion So correct combination should be A and C. B. This will generate a dicentric and an acentric chromosome following Phone: 080-5099-7000

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separation of chromosome after crossing over. ing the time graph for both Hfr strains the order can be deduced. The marker C. This will generate monocentric recombinant chromosomes following which has less time enters early and vice versa. separation of chromosomes after crossing over D. All the gametes thus formed will have deletion and/or duplication and Reference: An Introduction to Genetic Analysis. 7th edition. will be non-viable. E. 50% of the gametes having recombinant chromatid will be non-viable, while 50% gametes having non-recombinant chromatid will survive 110. The Cl- content of red blood cells (RBC’s) in the venous blood was E. This is paracentric inversion found to be higher than that in arterial blood in a human subject. Following proposals were made to explain these observations. Which combination of the above statements describe the inversion and meiotic consequences correctly? A. The high pCO2 in venous plasma leads to increased diffusion of CO2 into RBC and the formation of H2CO3. 1. A, B and C B. HCO3- content in the RBC of venous blood becomes much greater 2. A, C and E than that in plasma. 3. B, E and F C. The excess of HCO3- leaves the RBC of venous blood along with Na+ 4. C, D and F to plasma by a Na+ - HCO3-symporter D. The increased Na+ in the venous plasma is transported to the RBC Answer: 3 along with ClExplanation: A paracentric inversion does not include the centromere. In- Select the combination with INCORRECT statements from the followversion heterozygotes during the first meiotic anaphase, the two centromeres ing options migrate to opposite poles of the cell. Because of the crossover, one recombinant chromatid becomes stretched across the cell as the two centromeres begin 1. A and B to migrate in anaphase, forming a dicentric bridge—that is, a chromosome 2. B and C with two centromeres (a dicentric chromosome) 3. A and D . With continued migration, the dicentric bridge breaks due to tension. The 4. C and D other recombinant product of the crossover event is a chromosome without a centromere (an acentric fragment). This acentric fragment is unable to con- Answer: 4 tinue through meiosis and is usually lost. Thus, the only gametes that can give rise to viable progeny are those containing the chromosomes that did not Explanation: Venous blood carries blood from the respiring tissues back to involve crossing-over and gametes or zygotes derived from recombined chro- the heart and therefore has high pCO2 which diffuses into the RBC.The reacmatids are inviable. tion H2O + CO2 ↔ H+ + HCO–3 is accelerated tremendously due to presence of a specific enzyme carbonic anhydrase within the RBC.Concentration of Reference: “iGenetics, A Molecular Approach HCO–3 within the RBC, therefore, becomes very high within a short time and Third Edition, Peter J. Russell, Page-470 attains a level much higher than that in the plasma. the red cell membrane is impermeable to positively charged ions. some HCO–3 from within the RBC migrates into plasma and some Cl– from the 109. Two Hfr strains, Hfr-1 (arg+ leu+ gal+ strs) and Hfr-2 (arg+ his+ plasma migrates within the RBC in exchange of HCO–3.The most abundant gal+ purr strs) were mated with a F- strain (arg+ leu+ gal+ his+ purr strs). available base in plasma being Na+ the HCO–3 in the plasma combines with The results of the interrupted mating experiments are shown as plots ‘a’ Na+ to form NaHCO3. and ‘b’, respectively. Reference: http://www.biologydiscussion.com/human-physiology/respiratory-system/mechanism-of-formation-of-co2-respiration-humans-biology/81318 111. The different waves of a normal electrocardiogram (ECG) of a human subject are shown below:

Based on these results, identify which of the options accurately reflects the order of loci?

The relationship of the events of cardiac cycle to these ECG waves are proposed in the following statements.

Answer: 2

A. The P wave occurs due to the depolarization of atria B. The atrial repolarization is responsible for the T wave C. The QRS complex occurs during ventricular depolarization D. Q - T interval indicates plateau portion of auricular action potential

Explanation: the two-step approach : (1) deter-mine the underlying princi- Select the combination with INCORRECT statements from the followple, and (2) draw a diagram. Here the principle is clearly that each Hfr strain ing options. donates genetic markers from a fixed point on the circular chromosome and that the earliest markers are donated with the highest frequency. By visualiz- 1. A and B Phone: 080-5099-7000

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increasing pressure on the nerve. Some probable changes of compound action are stated below: A. ‘A’ peak was inhibited by lower intensity of pressure B. ‘C’ peak was inhibited by higher of pressure C. ‘B’ peak was inhibited by lower intensity of pressure D. ‘C’ peak was inhibited by lower higher of pressure E. ‘A’ peak was inhibited by higher of pressure

Answer: 4

Explanation: Components of a normal ECG tracings consist of waveform components which indicate electrical events during one heart beat. These waveforms are labeled P, Q, R, S, T and U. The first deflection is the P wave associated with right and left atrial depolar- Select the option with the combination of CORRECT statements ization. P wave is the first short upward movement of the ECG tracing. It indicates that the atria are contracting, pumping blood into the ventricles. Hence 1. A and B statement A is correct. 2. B and C The second wave is the QRS complex. Typically this complex has a series 3. C and D of 3 deflections that reflect the current associated with right and left ventricu- 4. D and E lar depolarization.The QRS complex, normally beginning with a downward deflection, Q; a larger upwards deflection, a peak (R); and then a downwards Answer: S wave. The QRS complex represents ventricular depolarization and contraction. Hence statement C is also correct. Explanation: T wave is normally a modest upwards waveform representing ventricular repolarization. Hence statement B is incorrect. Reference: The QT interval is measured from the beginning of the QRS to the end of the T wave. It represents the time in which the ventricles depolarize and repolarize and is a measure of ventricular action potential (AP) duration. Hence state- 114. The different segments of renal tubule (column A) and the mechament D is also incorrect. nism of Na+ transport in the apical membrane of tubular cells (column B) are tabulated below: Reference: https://www.practicalclinicalskills.com/ecg-interpretation, http://www.medicine-on-line.com/html/ecg/e0001en_files/05.htm 112. The excitation of auditory hair cells by the displacement of stereocilia has been explained in the following proposed statements: A. The gradual increased height of stereocilia is required for the transduction process B. The changes of membrane potential of auditory hair cells are proportional to the direction and magnitude of stereocilia C. The higher concentration of K+ in endolymph and higher concentration of Na+ in perilymph are not required for the excitation of hair cells D. The mechanically sensitive cation channels on the top of stereocilia are not adapted to maintain displacement of stereocilia Select the combination with INCORRECT statements from the following options

Select the option with the correct matches:

1. A and B 2. B and C 3. C and D 4. A and C

1. a - (iii), b - (iv), c - (i), d - (ii) 2. a - (iv), b - (iii), c - (ii), d - (i) 3. a - (i), b - (ii), c - (iii), d - (iv) 4. a - (ii), b - (i), c - (iv), d - (iii)

Answer: 4

Answer: 1

Explanation: The hair cell is a flask-shaped epithelial cell named for the bundle of hairlike processes that protrude from its apical end into the scala media. Each hair bundle contains anywhere from 30 to a few hundred hexagonally arranged stereocilia, with one taller kinocilium.The stereocilia are simpler, containing only an actin cytoskeleton. Each stereocilium tapers where it inserts into the apical membrane, forming a hinge about which each stereocilium pivots.. The stereocilia are graded in height and are arranged in a bilaterally symmetric fashion (in vestibular hair cells, this plane runs through the kinocilium). Displacement of the hair bundle parallel to this plane toward the tallest stereocilia depolarizes the hair cell, while movements parallel to this plane toward the shortest stereocilia cause hyperpolarization. In contrast, displacements perpendicular to the plane of symmetry do not alter the hair cell’s membrane potential. Reference: https://www.ncbi.nlm.nih.gov/books/NBK10867/

Explanation: In proximal tubule, sodium/proton exchanger enables reabsorption of bicarbonate. Glucose is transported alongwith Na+ via Na/glucose symporters. In thick ascending loop of henle, Three-ion cotransporter (sodium/potassium/ chloride) and the sodium/potassium ATPase, which as before maintains the sodium concentration gradient. Sodium is actively pumped out, while potassium and chloride diffuse down their electrochemical gradients through channels in the tubule wall and into the bloodstream. Approximately 5%of sodium filtered load is reabsorbed in the early distal tubule via sodium-chloride co transporter sodium reabsorption in late distal tubule takes place through special channels ENacs Reference: Principals of anatomy and physiology Gerard J. Tortora and grabowski

115. The figure below represents normal sex determination, differentia113. The peaks of the compound action potential (i.e., A, B and C) re- tion and development in humans corded from a mammalia mixed nerve were affected after application of Phone: 080-5099-7000

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females are expected to be wild type- since the male parent is wildtype and the mutations are X-linked recessive (given) all the female progeny must be wildtype. F. F1 wildtype males appeared due to crossing over. This is possible when there occurs a crossing over within the XX female chromosome. Option 1 seems to be the best option because only in this option both statements E & F fall together along. Reference: iGenetics, A Molecular Approach Third Edition, Peter J. Russell, Chapter 14, Page-407 117. The locations of five overlapping deletions have been mapped to a Drosophila chromosome as shown below 1. A = WT1 (Wilm’s Tumor 1), B = MIS (Mullerian Inhibitory Substance), C = SRY, D = Testosterone 2. A = GnRH, B = FSH, C = Testosterone, D = 5α Reductase 3. A = SRY, B = MIS, C = Testosterone, D = DHT (Dihydrotestosterone) 4. A = WT1, B = LH, C = ABP (Androgen Binding Protein), D = Inhibin Answer: 1 Explanation: The product of the Wilms’ tumor gene, WT1, is essential for male sex determination and differentiation in mammals. WT1 regulates SRY in the initial sex determination process in humans and activates a cascade of genes ultimately leading to the complete organogenesis of the testis. SRY (sex-determining region of the Y chromosome), and there is extensive evidence that it is indeed the gene that encodes the human testis-determining factor. SRY regulates SF1 for testis development and AMH for regerssion of mullerian duct. Testosterone appears to be responsible for promoting the formation of the male reproductive structures (the epididymis, seminal vesicles, and vas deferens) that develop from the Wolffian duct primordium. However, it does not directly masculinize the male urethra, prostate, penis, or scrotum. These latter functions are controlled by 5a-dihydrotestosterone

(Horizontal lines in the above figure indicate the deleted regions) Recessive mutations a, b, c, d and e are known to be located within this region, but the order of mutations on the chromosome is not known. When the flies homozygous for the recessive mutations are crossed with flies homozygous for the deletions, the following results are obtained (letter “m” represents mutant phenotype and “+” represents the wild type)

Reference: https://www.ncbi.nlm.nih.gov/books/NBK9967/

116. A virgin Drosophila female was crossed with a wild type male. The On the basis of the above data, the relative order of the five mutant F1 progeny obtained had four types of males as shown below. genes on the chromosome is 1. b c d c a 2. a b c d c 3. b c c a d 4. c d b c a Assuming that white eye and crossveinless mutations are X-linked and recessive , the following statements were made: A. F1 females were also of sour types as that of males B. The white eyed crossveinless male flies appeared due to independent assortment C. The map distance between the genes for white eye and crossveinless is estimated to be 12 cM. D. The map distance between white eye and crossveinless is estimated to be 6 cM. E. All F1 females are expected to be wild type F. The F1 wild type males appeared due to crossing over The combination with correct statements is:

Answer: 1 Explanation: Looking carefully into the length of deletion and the loss of function of genes it can be seen that deletion 1 is affecting function of b, c and d. similarly deletion 2 is affecting function of c and d, 3 is affecting function of d and e, deletion 4 and 5 affecting function of e and a. so this gives the sequencial order of genes bcdea. Reference: iGenetics, A Molecular Approach Third Edition, Peter J. Russell, Page-449

118. The pedigree given below follows the the inheritance pattern of a late onset (after age of 30 years) genetic disease that is 100% penetrant. Affected individuals are indicated by a solid circle (woman) or solid square (males). RFLP analysis of DNA from each individual is shown below in the pedigree.

1. C, E, F 2. A, B, D 3. A, D, F 4. B, D, E Answer: 1 Explanation:Statement E and F are true because it states that: E. All F1 Phone: 080-5099-7000

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to a former state, even if it finds itself placed in conditions of existence identical to those in which it has previously lived ... it always keeps some trace of the intermediate stages through which it has passed. Van Valen’s law states that the probability of extinction for species and higher taxa (such as families and orders) is constant for each group over time; groups grow neither more resistant nor more vulnerable to extinction, however old their lineage is. Occam’s razor (or Ockham’s razor) is a principle from philosophy and science. Suppose there exist two explanations for an occurrence. In this case the simpler one is usually better- also called the Principle of Parsimony. Reference: https://en.wikipedia.org/wiki/Biological_rules https://simple.wikipedia.org/wiki/Occam%27s_razor Which grandchildren (IIIb to IIId) will be affected by the disease after attaining the age of 30 years? 1. Only IIIb 2. Only IIIb and IIIc 3. Both IIIc and IIId 4. Both IIIb and IIId

120. Three anatomical characteristics (A, B and C) of invertebrate nervous system are used to build a generalized cladogram given below. Presence of the anatomical character is indicated by ‘+’

Answer: 4 Explanation: Careful onservation of the pedigree RFLP pattern reveals diseased individuals are homozygous having two bands and heterozygus individuals have 3 bands who are phenotypically normal. So among III b IIIc and IIId, since both b and d have same two band RFLP pattern so they should show the diseased condition. Reference: iGenetics, A Molecular Approach Third Edition, Peter J. Russell, Page-245 119. Following table contains some of the generalizations of evolutionary biology:

Based on the pattern of character distribution, pick the correct combination that are represented by A, B and C 1. A - unpaired nerve cord, B - paired nerve cord, C - cephalic ganglia 2. A - cephalic ganglia, B - unpaired nerve cord, C - paired nerve cord 3. A - cephalic ganglia, B - paired nerve cord, C - unpaired nerve cord 4. A - unpaired nerve cord, B - cephalic ganglia, C - paired nerve cord Answer: 2

Explanation: Loriciferans contains a paired ventral nerve cord whereas nematodes and nematomorpha contain unpaired ventral nerve cord The blueprint of the arthropod CNS consists of a dorsal cephalic ganglion, the ‘brain’, followed by a chain of ventral ganglia, the ventral cord . In the Spiralia(e.g. Gnathifera, Trochozoa) the main neural tracts found in theventro-lateral body region are paired. Within Spiralia, Gnathifera may represent thedeepest branching lineage comprising the jaw wormsGnathostomulida and their sister group Micrognathozoa +Syndermata. Their intra-epidermal, unsegmented nervous systems comprisean anterior brain and three to five ventral and two to four dorsallongitudinal nerves, connected by few transverse commissures.Neurites of the stomatogastric nervous system were found lin-ing the pharynx and connecting to a prominent buccal gangli-on. The entire body of Onychophora, including the stub feet, is littered with numerous papillae: warty protrusions that carry a mechanoreceptive bristle (responsive to mechanical stimuli) at the tip, each of which is also connected to further sensory nerve cells lying beneath. The mouth papillae, the exits of the slime glands, probably also have a function in sensory perception. Sensory cells known as “sensills” on the “lips” or labrum respond to chemical stimuli and are known as chemoreceptors. These are also found on the two antennae, which can be regarded as the velvet worm’s most Which of the following is correct match between Column I and Column important sensory organs. Except in a few (typically subterranean) species, one simply constructed eye (ocellus) lies laterally, just underneath the head, II behind each antenna. 1. A - (i), B - (ii), C - (iv), D - (iii) 2. A - (i), B - (iii), C - (iv), D - (ii) 3. A - (ii), B - (iii), C - (i), D - (iv) 4. A - (iv), B - (iii), C - (i), D - (ii) Answer: 2

Reference: The Evolution of Organ Systems By Andreas Schmidt-Rhaesa https://www.cell.com/current-biology/pdf/S0960-9822(16)30857-0.pdf, https://www.researchgate.net/publication/284423720_The_larval_nervous_ system_of_the_penis_worm_Priapulus_caudatus_Ecdysozoa

Explanation: Cope’s rule states that animal population lineages tend to in121. In order to survive in a non-aquatic environment , plants acquired crease in body size over evolutionary time. several adaptations with specialized functions. Given below is a list of Dollo’s law of irreversibility, states that “an organism never returns exactly Phone: 080-5099-7000

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3. A - Anthocerotophyta, B - Marchantiophyta, C - Bryophyta 4. A - Bryophyta, B - Anthocerotophyta, C - Marchantiophyta Answer: 2 Explanation: Hornworts ( Anthocerotophyta ) have only one giant chloroplast per cell and sporophyte body with smotatal opening. Since Marchantiophyta have predominant gametophytic stage which is photosynthetic in nature, they contain numerous chloroplast per cell. Bryophytes have photosynthetic gametophytes and multicellular rhizoid as well as sporophyte body with smotata. So accordingly A should be Bryophyta, B should be Marchantiophyta and C should be Anthocerotophyta.

Which one of the following options represents a correct match between the adaptations and their functions? 1. A - (iv), B - (ii), C - (i), D - (iii) 2. A - (iii), B - (i), C - (iv), D - (ii) 3. A - (ii), B - (iii), C - (ii), D - (i) 4. A - (i), B - (iv), C - (iii), D - (iv)

Reference: pdf

http://www-plb.ucdavis.edu/courses/bis/1c/text/Chapter22nf.

123. Following table shows a list of clades and plants:

Answer: 2 Explanation: Waxy cuticle: A waxy layer known as the cuticle covers the leaves of all plant species. The cuticle reduces the rate of water loss from the leaf surface; Thickened or lignified cell walls: Sclerenchyma cells are characterized by relatively thick, lignified secondary cell walls. All plant cells initially have only a primary wall made predominantly of cellulose. As sclerenchyma develops, a secondary wall with a high proportion of lignin is deposited inside the primary wall. The lignified wall gives sclerenchyma cells their rigidity, and they function primarily in mechanical support and water conduction. They also make up most rigid parts of the plant (e.g., seed coats and some fruit walls) and are often positioned so that they provide mechanical protection for softer plant parts.; Homoiohydry: The ability of an organism to maintain a relatively stable water content independently of short-term fluctuations in water supply, Hydroids are imperforate water-conducting cells specific to advanced mosses ; Pigmentation: photoprotection of photosysyem II (PS II) is essential to avoid the liht induced damage of the photosynthetic spparatus due to the formation of reactive oxygen species (ROS) which might induce photooxidative stress under excess light

Which one of the following is a correct match for the above? 1. A - (iii), B - (iv), C - (ii), D - (i) 2. A - (ii), B - (i), C - (iii), D - (iv) 3. A - (ii), B - (iv), C - (iii), D - (i) 4. A - (iii), B - (i), C - (ii), D - (iv) Answer: 4

Explanation: The basal angiosperms are the flowering plants which diverged from the lineage leading to ... Nymphaeales (water lilies, together with some other aquatic plants) and Austrobaileyales (woody aromatic plants including star anise). Eudicots includes-The rose family (Rosaceae) provides fruits such as strawberries, raspberries, apples, cherries, peaches, and plums and others. Orchids are the largest family of monocots among the angiosperms (flowering plants), with between twenty-five thousand and thirty thousand species, and new species are continually being described. Reference: https://www.sciencedirect.com/science/article/pii/ Piperales, order of flowering plants comprising 3 families, 17 genera, and S0005272811000971, https://www.sciencedirect.com/topics/agricultur- 4,170 species. Along with the orders Laurales, Magnoliales, and Canellales, al-and-biological-sciences/ground-tissue, https://www.ncbi.nlm.nih.gov/ Piperales forms the magnoliid clade, which is an early evolutionary branch in pmc/articles/PMC3310499/, Page 29 The Evolution of Plant Physiology the angiosperm tree; the clade corresponds to part of the subclass Magnoliidae edited by Alan R. Hemsley, Imogen Poole, https://academic.oup.com/ under the old Cronquist botanical classification system. pcp/article-pdf/37/3/395/5441967/37-3-395.pdf, http://www.pnas.org/conReference: http://science.kennesaw.edu/jmcneal7/plantsys/basals.html tent/94/25/14162 http://lifeofplant.blogspot.com/2011/04/eudicots.html http://lifeofplant.blogspot.com/2011/03/orchids.html 122. Following table presents bryophyte phyla with their selected char- https://www.britannica.com/plant/Piperales acteristics 124. The following table shows names of bones (Column A) and specific features (Column B)

In the above table, phyla A, B and C represent 1. A - Marchantiophyta, B - Bryophyta, C - Anthocerotophyta 2. A - Bryophyta, B - Marchantiophyta, C - Anthocerotophyta Phone: 080-5099-7000

Which one of the following options gives the correct match of the bones with their specific features? 1. A - (iii), B - (i), C - (iv), D - (ii)

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Which one of the following represents the combination of correct statements? 1. A and B 2. B and C 3. C and D 4. A and C

Answer: 1

Explanation: The odontoid process (also dens or odontoid peg) is a protuberance (process or projection) of the Axis (second cervical vertebra). It exhibits a slight constriction or neck, where it joins the main body of the vertebra. Th deltoid tuberosity is located on the lateral surface of the humerus a little Answer: 2 more than a third of the way along its length and is the attachment site for the deltoid muscle Explanation: Energy efficiencies were calculated for individuals of half the Acromian process is a bony process of the scapula which forms a part of the adult mass, both ectotherms and endotherms convert about the same proporpectoral girdle tion of food energy into somatic growth although for a given body mass the The articular surface of the ulna is called sigmoid cavity/notch latter expend about 10 times more aerobic power than the former. On the other hand, early in life, during the period of maximum growth, ectotherms channel Reference: Principals of anatomy and physiology Gerard J. Tortora and a 2-3 times greater percentage of metabolic energy into growth than endograbowski therms. Even greater becomes the difference between these two groups if we consider the relative cost of reproduction. It can be shown that, weight by 125. The figure belows shows the nervous system of Mollusca with gan- weight, nematodes, fish, birds and mammals require almost the same amount glia and the connecting nerves. The connecting nerves are labelled as A, of energy for the production of offspring-, roughly 250 kJ per day and kg of B, C and D eggs, hatchlings or litter. However, whereas the cost of producing offspring represents only 2%-6% of the total metabolizable energy of an endotherm, a fish has to spend 35%, a nematode nearly everything it has for this purpose. Reference: https://www.ncbi.nlm.nih.gov/pubmed/28310790

Which one of the following options has correct labelling of A, B, C and D? 1. A - Cerebral commissure; B - Left Cerebro-pedal connective; C - Pedal -commissure; D - Left Pedal-visceral connective 2. A - Cerebral connective; B - Left Cerebro-pedal commissure; C - Pedal -connective; D - Left Pedal-visceral commissure 3. A - Occipital commissure; B - Occipito-pedal connective; C - Pedal -commissure; D - Left Pedo-caudal connective 4. A - Cerebral connective; B - Left Cerebro-pedal commissure; C - Pedal -commissure; D - Pedal-caudal connective Answer: 1 Explanation: The mollusc nervous system is referred to as a tetraneural nervous system, because there are four main neural strands: Two pairs of connectives link the cerebral ganglia to the pedal ganglia on the ventral side. another to the visceral ganglia and parietal ganglia passing the pleural ganglia on the dorsal side. See the diagram in the link. Therefore A. Cco: Cerebral commissura. C. Pdco: Pedal commissure Cco: B. left Cerebral pedal commissura. D. Left pedal Visceral commisure Reference: http://www.molluscs.at/gastropoda/index.html?/gastropoda/ morphology/nervous_system.html, http://www.molluscs.at/images/weichtiere/schnecken/nervensystem.png 126. Following are certain statements regarding energy efficiencies of ectotherms and endotherms: A. Ectotherms have high assimilation efficiency but low production efficiency. B. Ectotherms have low assimilation efficiency but high production efficiency. C. Endotherms have high assimilation efficiency but low production efficiency. D. Endotherms have low assimilation efficiency but high production efficiency. Phone: 080-5099-7000

127. Given below are some properties related to botanical and zoological nomenclature. A. Absence of tautonyms B. Presence of genus and species ranks only C. Absence of principle of coordination D. Presence of only holotype and neotype Select the correct combination that distinguishes botanical nomenclature from zoological nomenclature system. 1. A, B and D 2. A, B and C 3. A and C only 4. A, C and D Answer: 3 Explanation: Correct combination that distinguishes botanical nomenclature from zoological nomenclature system is as follows: In plant nomenclature (ICBN), tautonyms are not valid i.e. generic name and specific name should not be same in plants. ego Mangifera mangifera But tautonyms are valid for animal nomenclature (ICZN-International Code of Zoological Nomenclature) e & Naja naja (Indian cobra), Raitus rattus (Rat). Hence statement A is correct. Article 4.1 of the ICBN states that ‘A plant may be assigned to taxa of the following ranks (in descending sequence) regnum, subregnum, divisio, subdivisio, classis, subclassis, ordo, subordo, familia, subfamilia, tribus, subtribus, genus, subgenus, sectio, subsectio, series, subseries, species, subspecies, varietas, subvarietas, forma, subforma. ARTICLE 24.1: The name of an infraspecific taxon is a combination of the name of a species and an infraspecific epithet connected by a term denoting its rank. An item often overlooked in this respect is that the infraspecific epithet should be preceded by a term denoting its rank. This applies not only when introducing a new taxon, but also when quoting one, since the article clearly states that this term is part of the name. Therefore Discoaster tanii nodifer, which is commonly used, should correctly be cited as Discoaster tanii subsp. nodifer. Although the term is abbreviated as ‘subsp.’ in the ICBN, the abbreviation ‘ssp.’ is also very common, and can be considered correct. Hence statement B is wrong as there are other ranks considered too. Statement of the Principle of Coordination. -A name established for a taxon at any rank in the family group, the phytoflagellates excepted, is deemed to be established with the same author and date for taxa based upon the same

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name-bearing type (type genus) at other ranks in the family group, with appro- 129. The following figure is a “risk-graph” that illustrates the percent priate mandatory change of suffix. A name generated in zoological nomencla- risk a species faces towards extinction. ture in accordance with the Principle of Coordination is not considered validly published under the botanical Code unless it appears in print and is applied to an accepted taxon. Hence statement C is right. Statement D is wrong as there are other type specimen terms like isotype, lectotype etc. Therefore the correct combination is A and C. Reference: http://ib.berkeley.edu/courses/ib200a/lect/ib200a_lect20_Will_ nomenclature.pdf, https://www.floridamuseum.ufl.edu/herbarium/types/ abouttypes.htm, https://www.emedicalprep.com/study-material/biology/diversity-in-the-living-world/the-living-world/main-rules-icbn/, http://ina.tmsoc.org/nannos/icbn.htm 128. Following table gives a list of international environmental agreements and areas covered:

The following are ranks assigned according to IUCN’s red list category: (i) Critically endangered (ii) Near threatened (iii) Vulnerable (iv) Least concern Which one of the following is the most appropriate match between the percent-risk and their assigned rank? 1. a - (i), b - (iii), c - (iv), d - (ii) 2. a - (i), b - (iv), c - (iii), d - (ii) 3. a - (iii), b - (ii), c - (i), d - (iv) 4. a - (iv), b - (iii), c - (ii), d - (i) Answer: 1 Explanation: Species are classified by the IUCN Red List into nine groups, specified through criteria such as rate of decline, population size, area of geographic distribution, and degree of population and distribution fragmentation. Extinct : No known individuals remaining, Extinct in the wild: Known only to survive in captivity, or as a naturalized population outside its historic range,Critically Endangered Extremely high risk of extinction in the wild Endangered: High risk of extinction in the wild, Vulnerable:High risk of endangerment in the wild, Nearly Threatened: Likely to become endangered in the near future, Least Concern: Lowest risk (Does not qualify for a more atrisk category; widespread and abundant taxa are included in this category.) Data Deficinet: Not enough data to make an assessment of its risk of extinction. Not Evaluated:Has not yet been evaluated against the criteria

Which of the following is correct combination? 1. A - (i), B - (ii), C - (iv), D - (iii) 2. A - (ii), B - (i), C - (iii), D - (iv) 3. A - (iv), B - (i), C - (iii), D - (ii) 4. A - (ii), B - (iv), C - (iii), D - (ii

Reference: Guidelines for Using the IUCN Red List Categories and Criteria (PDF), Version 13, IUCN, March 2017, retrieved 2018-01-04. http:// cmsdocs.s3.amazonaws.com/RedListGuidelines.pdf

Answer: 2

Explanation: The Basel Convention on the Control of Transboundary Movements of Hazardous Wastes and their Disposal was adopted on 22 March 1989 130. The complexity of a food web in a community is quantified using by the Conference of Plenipotentiaries in Basel, Switzerland, in response to a certain parameters which are defined below. Which of the following is an public outcry following the discovery, in the 1980s, in Africa and other parts INCORRECT representation? of the developing world of deposits of toxic wastes imported from abroad. The Cartagena Protocol on Biosafety to the Convention on Biological Diversity is an international agreement which aims to ensure the safe handling, transport and use of living modified organisms (LMOs) resulting from modern biotechnology that may have adverse effects on biological diversity, taking also into account risks to human health. It was adopted on 29 January 2000 and entered into force on 11 September 2003. Kyoto Protocol, in full Kyoto Protocol to the United Nations Framework Convention on Climate Change, international treaty, named for the Japanese city in which it was adopted in December 1997, that aimed to reduce the emission of gases that contribute to global warming. The Stockholm Convention is a global treaty to protect human health and the environment from persistent organic pollutants (POPs). Reference: http://www.basel.int/TheConvention/Overview/tabid/1271/Default.aspx https://bch.cbd.int/protocol/ https://www.britannica.com/event/Kyoto-Protocol https://www.unido.org/our-focus/safeguarding-environment/implementation-multilateral-environmental-agreements/stockholm-convention

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Answer: 3 Explanation: All are correct except, Potential links which is calculated as N=n(n-1)/2.

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Reference: https://en.wikibooks.org/wiki/Ecology/Ecosystems; www. pnas.org/content/103/30/11211; https://www.researchgate.net/publica- 1. It is the net gain of direct fitness when sociality is facultative tion/227671258_The_structure_of_aphid-parasitoid_community; https:// 2. It is under positive selection via indirect fitness benefits that exceed en.wikibooks.org/wiki/Ecology/Ecosystems direct fitness costs. 3. It generates indirect benefit by enhancing survivorship of kin 4. It is favored when rb - c > 0 where c is fitness cost to altruist, b is fitness 131. benefit to recipient, and r is genetic relatedness. Answer: 1 Explanation: Hamilton’s rule that altruism is favoured when rb–c>0; where c is the fitness cost to the altruist, b is the fitness benefit to the recipient and r is their genetic relatedness. This predicts that altruism is favoured when r or b are higher and c lower. Hamilton’s theory is referred to in many ways. Hamilton called it ‘inclusive fitness theory’, but it is more often referred to as ‘kin selection’, a term coined by John Maynard Smith. Jerram Brown pointed out that the inclusive fitness of an individual is divided into two components: ‘direct fitness’ and ‘indirect fitness’. Direct fitness is gained through the production of offspring, and indirect fitness through aiding the reproduction of nondescendent relatives. A behaviour is only altruistic if it leads to a decrease in direct fitness, so altruism can only be favoured when an indirect benefit outweighs this direct cost, as shown by Hamilton’s rule.

The above graph illustrates two lines that represent the immigration and extinction rates for an island based on its distance from mainland (solid Reference: line) and its size (dotted line). Which of the following is true for this is- 9822(06)01695-2.pdf land? 1. It is close to the mainland and is very small 2. It is far from the mainland and is very large 3. It is close to the mainland and is very large 4. It is far from the mainland and is very small

https://www.cell.com/current-biology/pdf/S0960-

134. Given are some statements with reference to the use of genes in plant molecular systematics.

Answer: 3 Explanation: Closer and larger island have higher immigration rate and lower extinction rate. Arriving colonists help sustain the presence of a species on a near island and prevent its extinction. Hence if you observe the graph the equilibrium point is reached, it has higher immigration rate as compared to extinction rate. Therefore option 3 is correct. Reference: Biology by Campbell and Reece 132. Inclusive fitness of an animal can be measured as a sum of direct fitness and indirect fitness. Imagine you have 10 offsprings. Through diligent parental care. 5 survive to reproduce. You give your life in a heroic deed to save a total of 5 of your nieces and nephews . What is your inclusive fitness?

A. mtDNA are not preferred over cpDNA or rDNA because they generally show slow rate of sequence evolution and fast rate of structural evolution. B. cpDNA are not preferred because of their haploidy, uniparental inheritance, and absence of recombination among cpDNA molecules. C. rDNA such as ITS are preferred for their higher evolutionary rates as well as shorter sequence length. D. rDNA and cpDNA cannot be used simultaneously in molecular systematics since they represent conflicting patterns of inheritance. Which of the above statements are INCORRECT? 1. A, C and D 2. A, B and C 3. A and C only 4. B and D only Answer: 4

Explanation: Chloroplast DNA (cpDNA) is a traditional workhorse for re1. 15 constructing evolutionary relationships among angiosperms. The frequent use 2. 12.5 of cpDNA in such analyses is predicated on the apparent simplicity of its in3. 7.5 heritance: uniparental through the maternal line, and lacking biparental re4. 3.75 combination.Hence Statement B is wrong. Since the cpDNA molecule is highly conserved, there is very little intra-specific variation, and cpDNA-based Answer: 4 RFLP studies have therefore mostly been conducted on an inter-specific level. By contrast, plant mitochondria have never been much used in molecular analExplanation: Inclusive Fitness = Direct Fitness + Indirect Fitness yses. The major reason is that while the plant mitochondrial DNA (mtDNA) = (Survival of offspring) x (r for parent-offspring) + sequence is usually highly conserved, the size and structure of mtDNA mole(Survival of non-descendant kin) x (the proper r for each cules may vary widely even within individual plants . Moreover, recent studies type of relationship) In this question Offspring survived= 5 and coeffient of indicated that substitution rates of mtDNA genes can vary enormously even relatedness(r) =0.5, Niece+nephews= 5 and r= 0.25, hence inclusive fitness = among closely related plant species.Hence statement A is correct. The 16S–23S (5*0.5)+ (5*0.25)= 2.5+1.25= 3.75 internally transcribed spacer (ITS) regions of the rRNA operon might be under minimal selective pressure during evolution and therefore have more variation Reference: http://ib.berkeley.edu/labs/slatkin/eriq/classes/biol472/lectin the sequences than that of the coding regions of 16S and 23S rRNAs.While notes/lect7_08_Altruism.pdf the sequence of the rRNA genes evolves slowly, the internal transcribed spacers (ITS1 and ITS2), the external transcribed spacer (ETS), and the intergenic spacer (IGS) evolve rapidly. Hence statement C is correct. 133. Altruism describes a behaviour performed by animals that may be disadvantageous to self while benefiting others. Which one of the follow- Reference: Recombination in the Chloroplasts of the Florally ing statements is INCORRECT about altruism? Diverse Andean Subtribe Iochrominae Phone: 080-5099-7000

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Special Edition (Solanaceae) Erin Collier-Zans, https://www.ncbi.nlm.nih.gov/pmc/articles/ PMC2099596/,https://www.sciencedirect.com/science/article/pii/ S0944501310000261, https://www.ncbi.nlm.nih.gov/pmc/articles/ PMC3880010/ 135. Following are key points about the effect of genetic drift:

tracellular dynamics of live cells using microscopic based techniques,so all the microscopic techniques except Dark field microscopy and epifluroscence microscopy can be appointed as both the microscopic techniques cannot determine live cells as well as epifluroscence microscopy Reference: 137. A certain protein has been assumed to play an indispensable role in the survival of an intracellular parasite inside the host cells. Which one of the following techniques will best prove the assumption to be correct?

A. Genetic drift is significant in small populations B. Genetic drift can cause allele frequencies to change in a pre-directed way C. Genetic drift can lead to a loss of genetic variation within populations 1. Treat the parasite-infected host cells with an inhibitor of the protein D. Genetic drift can cause harmful alleles to be fixed and check the number of parasites per host cell under the microscope 2. Check the expression of the protein in parasite-infected host cells Which one of the following combination of the above statements are 3. Check the activity of the protein in parasite-infected host cells true? 4. Treat the parasite-infected host cells with an activator of the protein and check the number of parasites per host cell under the microscope 1. A and B only Answer: 3 2. A and C only 3. A, B and C Explanation: In order to prove the hypothesis, that a certain protein is re4. A, C and D sponsible for the survival of the parasite in the host cell, the best way to do so is by inhibiting the protein and then check for the survival of the parasites in Answer: 4 the host cell. If the parasites die in the absence of the protein, it implies that the protein is required for the survival of the parasites which is no longer availaExplanation: Changes in relative allele frequency due only to random sam- ble. If not, then the assumption is not correct. So, by counting the number of pling error are known as genetic drift, a stochastic process. Hence Statement parasites under a phase‐contrast and UV‐light microscope before and after B is wrong. The smaller the population, the smaller the subset of potentially using an inhibitor for the protein we can prove the assumption. successful gametes, and the more likely that genetic drift will occur. Hence Even expression studies can be performed to check the expression level of Statement A is right. Genetic drift can happen in two ways: protein at different conditions like different phases of infection. Founder Effect in which a small sample of breeding individuals from a large population colonizes a new area. Reference: https://onlinelibrary.wiley.com/doi/full/10.1002/%28SICI Bottleneck Effect: most members of a large population are removed (perhaps %291097-0320%2820000301%2939%3A3%3C235%3A%3AAID-CYdue to some natural disaster such as a hurricane, volcanic eruption, patho- TO10%3E3.0.CO%3B2-L gen invasion or other catastrophe that doesn’t favor any particular genotype www.mdpi.com/1422-0067/17/8/1270/pdf over another) leaving only a few survivors.Through sampling error, genetic drift can cause populations to lose genetic variation. Hence statement C is right.Alleles which are not harmful nor beneficial can become fixed by chance 138. Given below is a table with information on isotopes, their half-life through genetic drift.In very small population genetic drift can cause alleles and type of particle(s) they emit. that are slightly harmful to become fixed. When this occurs the population’s survival can be threatened. Hence option D is right. Reference: http://www.bio.miami.edu/dana/160/160S13_7.html, https://evolution. berkeley.edu/evolibrary/article/side_0_0/genesdrift_01, Campbell Biology, 11th Ed By Pearson (Page number 494) 136. In order to visualize the intracellular organization of a cell, one can utilize various microscopy-based techniques. These include: A. Differential interference contrast (DIC) microscopy B. Phase contrast microscopy C. Dark field microscopy D. Epifluorescence microscopy E. Scanning electron microscopy F. Transmission electron microscopy G. Confocal microscopy

Choose the correct combination from the options given below: 1. (a) - (iii) - (y); (b) - (ii) - (x), (y); ( c ) - (i) - (y) 2. (a) - (iii) - (x); (b) - (i) - (x); ( c ) - (ii) - (x), (y) 3. (a) - (ii) - (x), (y); (b) - (iii) - (x); ( c ) - (i) - (x) 4. (a) - (i) - (x); (b) - (ii) - (x); ( c ) - (iii) - (x), (y) Answer: 2

Explanation: C11: half life 20 min and they decay by emitting positrons Which of the above mentioned microscopes can be used to study the in- which are positively charged beta particles; C14: half life is 5730 years, they decay by emitting negatrons which are negatively charged beta particles; Na tracellular dynamics using live cell imaging? 24: half life 15 hours, decay by beta and gamma emission 1. A, B, E, F, G Reference https://www.ncbi.nlm.nih.gov/pubmed/20388115, http://math2. A, B, C, D, G central.uregina.ca/beyond/articles/ExpDecay/Carbon14.html, https://www. 3. A, D, E, F, G britannica.com/science/sodium-24, 4. C, D, E, F, G Answer: 1

139. Which one of the following set of essential components are required Explanation: As the question has asked about the visualization of the in- for Sanger method of DNA sequencing in a required buffer containing Phone: 080-5099-7000

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Special Edition MgCl2 and Tris-HCl? 1. DNA template, a primer, 4 deoxyribonucleotides, 4 labelled deoxyribonucleotides, DNA polymerase 2. DNA template, a primer, 4 labelled deoxyribonucleotides, DNA polymerase, DNA ligase 3. DNA template, 4 deoxyribonucleotides, 4 labelled deoxyribonucleotides, DNA polymerase, DNA ligase 4. DNA template, a primer, 4 labelled deoxyribonucleotides, DNA polymerase, telomerase Answer: 1 Explanation: The DNA sample is divided into four separate sequencing reactions, containing all four of the standard deoxynucleotides (dATP, dGTP, dCTP and dTTP) and the DNA polymerase. To each reaction is added only one of the four dideoxynucleotides (ddATP, ddGTP, ddCTP, or ddTTP), while the other added nucleotides are ordinary ones. The dideoxynucleotide concentration should be approximately 100-fold lower than that of the corresponding deoxynucleotide (e.g. 0.005mM ddTTP : 0.5mM dTTP) to allow enough fragments to be produced while still transcribing the complete sequence. Reference https://en.wikipedia.org/wiki/Sanger_sequencing 140. A 1257 bp genomic DNA sequence of a prokaryotic gene was cloned under a strong constitutive promoter along with a suitable polyA signal and used for development of transgenic tobacco plants. Molecular analysis revealed the presence of three types/lengths of transgene derived mRNAs: 555kbp, 981 bp and 1257 bp - in the leaves of transgenic plants. The following statements were proposed to explain the above results. A. The three mRNAs represent alternatively spliced transcripts due to the presence of putative intronic sequence in the gene. B. The gene sequence was characterized by the presence of potential polyadenylation signals that resulted in premature termination of transcription. C. Expression of full-length transcripts (1257 bases) was lethal to the transformed cells D. the transgenic plants were chimeric in nature and comprised a mix of transformed and untransformed cells. Which of the following combinations of the above statements would correctly explain the obtained results? 1. A and C 2. B and D 3. A and B 4. C and D Answer: 3 Explanation: According to statement 2 if expression of full length 1257 bp transcripts were lethal to the transformed cells, then it is not possible for the molecular anaylsis to reveal the presence of 1257 bp transgene-derived mRNAs. So Statement C cannot be correct. Statement D has nothing to do with obtained results(as asked in the question), as even if the transgenic plants were chimeric in nature, the transformed cells would still generate different lengths of transgene derived transcripts, which can be detected by molecular analysis of the leaves of the transgenic plants. Reference 141. In order to detect minor variations in antigen concentration, the following procedures were suggested. Which one will likely be the best option? 1. Antigen coated microtitre well → add antibody → add enzyme conjugated secondary antibody → add substrate and measure colour. 2. Antigen coated microtitre well → add antigen → add enzyme conjuPhone: 080-5099-7000

gated secondary antibody → add substrate and measure colour. 3. Preincubate antigen with fixed amount of antibody → add to antigen coated well→ add enzyme conjugated secondary antibody → add substrate and measure colour. 4. Preincubate antigen with fixed amount of antibody → add to antibody coated well→ add enzyme conjugated secondary antibody → add substrate and measure colour. Answer: 3 Explanation: To detect minor variations in antigen out of the options given competitive ELISA can be performed. In which the primary antibody (unlabeled) is incubated with sample antigen. Antibody-antigen complexes are then added to 96-well plates which are precoated with the same antigen. Unbound antibody is removed by washing the plate. (The more antigen in the sample, the less antibody will be able to bind to the antigen in the well, hence "competition.")The secondary antibody that is specific to the primary antibody and conjugated with an enzyme is added. A substrate is added, and remaining enzymes elicit a chromogenic or fluorescent signal. The strength of the signal inversely proportional to antigen concentration

Reference

http://www.microbiologynotes.com/wp-content/uploads/2015/06/lectur2-2.jpg, http:// www.sinobiological.com/competitive-elisa-principle.html

142. Three students (P, Q and R) in a research lab were trying to identify proteins that interact with a transcription factor X. P performed gel filtration experiments and identified that X was found along with proteins A, B, C and D. Q performed co-immunoprecipitation experiments using antibodies to X and identified A, B and C. R did a yeast-2-hybrid screen and identified only B. The following are likely conclusions that may explain the results: (i) A, B, C and D are in a complex with X (ii) X directly interacts with B (iii) Only A, B and C are in complex with X (iv) D is probably weakly associated with X Which of the above conclusions best explains all the results? 1. (i), (ii) and (iii) 2. (i), (ii) and (iv) 3. (i), (iii) and (iv) 4. (ii), (iii) and (iv) Answer: 4 Explanation: Co-IP is a classic technology widely used for protein-protein interaction identification and validation.The principle of the co-IP technique is as follows: Many intracellular protein-protein interactions are retained when cells are lysed under non-denaturing conditions. The bait protein, for example, TF X, can be captured by its specific antibody stabilized to agarose beads. If there is another protein, the prey protein, for example, protein A, binds to protein X in vivo, the TF X - protein A complex can then be precipitated together by the antibody. Subsequently, through the investigation of protein A, we can confirm the TF X - protein A interaction, or discover new interactors of TF X. Hence if protein A, B and C were identified along with protein X means it should be forming a complex. The limitation of this technology lies in: Low affinity or transient interaction between proteins may not be detected. Hence D was not identified as the interaction was weak. The result of CoIP could not determine whether the interaction is direct or indirect, since the possibility of involvement of additional proteins could not be ruled out. To confirm this, yeast two hybrid screen was performed, Yeast two-hybrid screening represents a sensitive in vivo method for the identification and analysis of protein-protein interactions. The principle is based on the ability of a separate DNA-binding domain (DNA-BD) and activation domain (AD) to reconstitute a functional transactivator when brought into proximity. When a bait and a prey protein interact, the DNA-BD and AD form a functional transactivator, resulting in activation of reporter gene expression in yeast reporter strains. This technique can be used to identify novel protein interactions, analyze protein-protein interactions between two known proteins, as well as dissect interacting protein domains. Hence if this screen confirmed interaction between X and B , means B is directly bound with X.

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https://www.profacgen.com/Co-Immunoprecipitation-Co-IP. statements were proposed to explain the above results.

143. Sub-cellular fractionation-based assays have been used to identify various organelles in the mammalian cells. In order to characterize such organelles in a living mammalian cell, which of the following microscopy-based method would be the most accurate? 1. use of fluorescent probes specific for organelles 2. use of organelle specific fluorescent probes followed by microinjection of fluorescent antibodies against organelle-specific protein 3. use of fluorescent probes in permeabilized cells 4. use of organelle specific fluorescent probes followed by cryo-electron microscopy Answer: 2 Explanation: Both of these fluorescent labels have different binding sites e.g. that respond to changes in microenvironments (e.g., pH, viscosity, and polarity) or quantities of biomolecules of interest (e.g., ions, reactive oxygen species, and enzymes). So their use in combination will help locate structure in a better way. Further, unlike antibodies, these fluorescent probes can be used to investigate organelle structure and activity in live cells with minimal disruption of cellular function Reference https://www.thermofisher.com/content/dam/LifeTech/global/ technical-reference-library/Molecular%20Probes%20Handbook/chapter-pdfs/Ch-12-Probes-for-Organelles.pdf?icid=WE216841 144. From the following statements, A. Surface plasmon resonance can be used to determine binding constants only on the range of 102 - 103 M B. de novo sequencing is not possible by mass spectral methods C. The position of hydrogen atoms in proteins is not directly determined by X-ray diffraction D. Circular dichroism and nuclear magnetic resonance spectroscopy do not give the same information on protein structure. Choose the option with all correct statements.

A. Protein encoded by gene ‘Y’ is not lethal to the cell B. Gene ‘X’ has introns, which prevents its expression in E. coli C. Expression of ‘X’ protein is lethal to the cell D. The ‘Y’ gene product inhibits the activity of ‘X’ protein Which one of the following options represents a combination of correct statements to explain the observations? 1. Only A and B 2. B, C and D 3. Only A and D 4. A, C and D Answer: 3 Explanation: Researcher attempted to clone two genes (X and Y) independently in plasmids-Y was easily cloned but not the X, and when the X gene was cloned in Y clone plasmid, gave him a successful experiment, might be because, the Y is not lethal to cell but inhibits the expression of X, when insertional inactivation happens which affects the Y expression and while X will be expressed. We can express the eukaryoric gene in prokaryotic vector like E.coli by adding the cDNA so option B will not be included, and will be able to synthesise a toxic protein too without any harm to the cell ( ES). 1. X can be clones only in Y, and according to the answer Y inhibits X, only when isertional inactivation happens-Y gene will be disrupted and X can be expressed 2. At this point we do not have any protein to inhibit X, and if X is lethl, again the clone will not be a successful clone. So Y is not lethal but inhits X would be tha correct option. Reference http://www.microbiologyresearch.org/docserver/fulltext/micro/138/1/ mic-138-1-205.pdf?expires=1529238008&id=id&accname=guest&checksum=D2E38B3D66054E938A422AF796EBCC76 https://www.ncbi.nlm.nih.gov/books/NBK22390/ https://www.ncbi.nlm.nih.gov/pubmed/2404273

1. A, B, C 2. A, C, D 3. B, D 4. C, D Answer: 4 Explanation: Option C and D are correct. Surface Plasmon resonance has association/dissociation constants ranging even from nM to fM. So, option A is wrong In proteomics, de novo sequencing is the process of deriving peptide sequences from tandem mass spectra without the assistance of a sequence database. So, B is again wrong. Hydrogen atoms can be located accurately and precisely by x-ray crystallography CD gives information about secondary and tertiary conformations of the molecule. However, it does not give the residue-specific information that can be obtained by X-ray crystallography or NMR. Reference Wilson and walker Page 29, https://www.ncbi.nlm.nih.gov/pmc/ articles/PMC2728378/, http://www.bioinfor.com/denovo-tutorial/ 145. A researcher attempted to clone two genes (X and Y) independently in a plasmid vector for over expression and purification in E.coli. All attempts to clone gene X were unsuccessful whereas gene ‘Y’ could be cloned easily. When the researcher attempted to clone gene ‘X’ in the plasmid clone containing gene ‘Y’, gene ‘X’ could be cloned. The following Phone: 080-5099-7000

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