Evening Shift CSIR NET Life Sciences Past Paper PDF July 2024

EVENING SHIFT QUESTION PAPER ANSWER KEYS WITH EXPLANATION CBUSINESS SIR NET LIFE SCIENCES PLAN JULY 2024 www.reallygreatsite.com www.Biotecnika.org www.Biotecnika.org ...

EVENING SHIFT QUESTION PAPER ANSWER KEYS WITH EXPLANATION CBUSINESS SIR NET LIFE SCIENCES PLAN JULY 2024 www.reallygreatsite.com www.Biotecnika.org www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 1. Question ID: 703001 2. Question ID: 703002 If Pencils are Erasers, some Erasers are If liars always lie and truthful persons never, and Sharpeners, some Erasers are Crayons, no Crayons in a group of 10 persons everyone calls all are Sharpeners but some Crayons are Pencils then others liars, then the number of liars among the in the given Venn diagram, which of the following is 10 is represented by the shaded area? 1. 10 2. 9 Correct answer: 4 3. 5 4. 1 EXPLANATION If a person is a liar, they lie about everyone, so they will call everyone else liars. If a person is truthful, they will truthfully call all liars liars and 1. Pencils and Sharpeners but not Erasers and Crayons truth-tellers truth-tellers. If everyone in the 2. Pencils and Erasers but not Sharpeners and Crayons group of 10 calls all others liars, this implies 3. Pencils, Erasers, and Sharpeners but not Crayons that: If there are any truthful persons in the 4. Pencils, Erasers and Crayons but not Sharpeners group, they would truthfully call the other liars as liars. The liars would call everyone else liars, but that would be a lie , because each liar would Correct answer: 4 falsely call the 1 truth-teller a liar. The truth-teller would correctly call the 9 liars as liars. Thus there are total 9 liars in the group(option 2 answer) EXPLANATION The biggest circle is Erasers. The circle, that is 3.Question ID:- 703003 completely inside the biggest circle is Pencils as The speed of a car traveling with variable all pencils are erasers. The circle that is not acceleration along a straight line is shown in the overlapping the circle of pencil, is that of figure. sharpners as some erasers are sharpners but their is no overlapping between sharpner and pencils. Now the 4th circle has to be of crayons. So the shaded region is eraser, pencil and crayons but not sharpners. Answer is option 4. Correct answer: 1 www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION If α​1, α​2 ,α​3 are the accelerations at times t1​, t2, EXPLANATION t3 respectively, then: The lowest mortality risk is when prarameter A is 1. α​1 = α​2 = α​3 100 (lowest) and parameter B is 85 (highest), where 2. α​1 > α​3 > α​2 A-B = 15. When we check the other bars we will see 3. α​3 > α​2 >​α​1 that as the value of parameter A-B increases, the α​3 > ​α​1 > α​2 mortality rate increases. So the lowest mortality is at the lowest valoue of A-B. Answer is option 1. Correct answer: 1 4.Question ID:- 703004 5.Question ID:- 703005 The following graph shows the mortality risk of The difference between a three-digit number a disease with respect to parameters A and B. (with non-repeating digits) and the same number in the reverse order is always divisible by: 1. 33 2. 22 3. 13 4. 31 Correct answer: 1 EXPLANATION Which of the following combinations of Let’s assume that one of the three-digit numbers parameters is associated with the lowest (with non-repeating digits) is 100a + 10b +c. The mortality risk? reverse of this number would be 100c – 10b – a. Let’s assume that value of a is greater than c. 1. The lowest value of A−BA - BA−B Now, subtracting the two numbers we get, (100a + 2. The lowest value of B−AB - AB−A 10b + c) – (100c – 10b – a) = 99a – 99c = 99(a-c) 3. The lowest values of both A and B Since, the difference between the two numbers is 4. The highest values of both A and B divisible by 99, it will also be divisible by 33. Ans option 1. Correct answer: 1 www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 6.Question ID:- 703006 7.Question ID:- 703007 A chessboard contains 64 squares of 5 cm size, A ball of molding clay, whose radius is aaa, is in 8 rows and 8 columns, alternately black and remolded into a cube. What is the approximate white. What is the total length of edges (in m) length of the side of the largest cube that can between the squares in the chessboard? be so made? 1. 2.8 1. 0.8a0.8a0.8a 2. 3.2 2. 1.2a1.2a1.2a 3. 5.6 3. 1.6a1.6a1.6a 4. 6.4 4. 2a2a2a Correct answer: 3 Correct answer: 3 EXPLANATION EXPLANATION The length of a side of each square is 5 cm. Clay ball is of radius 'a'. Hence volume of clay Hence length of the entire side of the chess ball = 4/3na³. board is 40 cm (5 cm*8 squares). The internal Clay ball is moulded into cube. Hence the edges are the sum of the lengths of the red lines volume of the sphere and the cube will remain shown in the 2 images below. Hence the total the same. length of the internal edges is (40 cm*7 vertical The volume of cube = 4/3ra². red lines) + (40 cm*7 horizontal red lines) = 280 + Let's say that the length of one of the sides of 280 = 560 cm = 5.6 m. Ans option 3. the cube is x. Then the formula for volume of cube is x'. The volume of cube = 4/3ra². Let's say that the length of one of the sides of the cube is x. BIOTECNIKA’S Then the formula for volume of cube is x'. So, x" 4/3*n*a" Taking cube root on both sides, CERTIFICATION COURSES x= √(4*3.14/3)* a x = 1.6a Ans option 3 STORES.BIOTECNIKA.ORG www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 8.Question ID:- 703008 9.Question ID:- 703009 The following spider diagram shows the marks The graph shows the distribution of lifespan (in obtained (out of 10) by three students in five years) for individuals from species 1 and species tests. If μ\muμ and σ\sigmaσ represent mean and Which one of the following is INCORRECT? standard deviation of the lifespan, respectively, then which of the following statements is true? 1. A scored more than C in total 2. B scored the highest in total 1. μ1>μ2 ; σ1​>σ2​ 3. A never scored 10 marks in a test 2. μ1=μ2 ; σ1​=σ2​ 4. In Test 5, the combined marks of A and C are 3. μ1=μ2 ​; σ1​>σ2​ equal to the marks of B. 4. μ1=μ2 ; σ1​𝜎2 ​Therefore answer should be student has never scored 10 marks Total marks of A option 3.. is 9+5+8+5+1 = 28 Student C (red shade) has scored 3 in Test 1, 4 in Test 2, 10 in Test 3, 7 in Test 4 and 8 in Test 5 Total marks of C is 3+4+10+7+8 = 32Thus option 1 is INCORRECT as C has scored more than A in total Student B (green shade) has scored 6 in Test 1, 10 in Test 2, 7 in Test 3, 9 in Test 4 and 9 in Test 5 Total marks of B is 6+10+7+9+9 = 41 Thus option 2 is correct. In Test 5 , A scored 1 and C scored 8 , their combined score i.e 1+8 is 9 which is equal to the marks scored by B. Hence option 4 is also correct. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 10.Question ID:- 703010 11.Question ID:- 703011 The cost of 2 mangoes, 1 coconut, and 2 How many integers can divide 1184 leaving a bananas is Rs 71, while the cost of 5 mangoes, remainder of 29? 3 coconuts, and 4 bananas is Rs 182. What is 1. 8 the cost of 1 mango and 1 coconut? 2. 5 3. 7 1. It cannot be calculated 4. 9 2. Rs 40 3. Rs 47 4. Rs 53 Correct answer: 2 Correct answer: 4 EXPLANATION EXPLANATION Let's denote the cost of a mango as 𝑀, the cost We need to find how many integers 𝑑 can divide of a coconut as C, and the cost of a banana as 1184 such that the remainder when dividing 𝐵 We are given the following two equations 1184 by 𝑑 is 29. This can be formulated as: 1184 based on the problem statement: 2M + C + 2B = ≡ 29 (mod 𝑑) This implies: 1184−29=1155 So, 𝑑 71(Equation 1) 5M + 3C + 4B = 182(Equation 2) must be a divisor of 1155, and 𝑑 > 29 Next, find We need to find the cost of ( M + C). To solve for the divisors of 1155: The prime factorization of ( M + C ), we can try to eliminate ( B) from the 1155 is: 1155 = 3×5×7×11 The divisors of 1155 equations. Let's multiply Equation 1 by 2 to are: align the banana terms with Equation 2: 4M + 1,3,5,7,11,15,21,33,35,55,77,105,165,231,385,115 2C + 4B = 142(Equation 3) Now, subtract 5 Among these divisors, those greater than 29 Equation 3 from Equation 2: (5M + 3C + 4B) - are: 33,35,55,77,105,165,231,385,1155 Thus, (4M + 2C + 4B) = 182 - 142 This simplifies to: M there are 9 integers that can divide 1184 leaving + C = 40 Thus, the cost of 1 mango and 1 a remainder of 29 coconut is Rs 40. The correct answer is 2. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 12.Question ID:- 703012 13.Question ID:- 703013 In a class, boys secure 69% marks on the Three comparable brands of 1 litre cans of a average while girls secure 72% marks on the liquid detergent are available in a shop with average. If the average marks of the entire different offers as shown in the table: class is 70%, which of the following statements is valid? 1. The total number of students in the class is two times the number of girls. 2. The total number of students in the class is three times the number of boys. 3. The boys are two times the number of girls. If 4 litres of detergent is to be purchased, then 4. The girls are two times the number of boys. the best choice (based on unit price) would be Correct answer: 3 1. A or B 2. A or C 3. B or C Let no. of boys in the class be B and the no. of 4. B girls as G. Then , Total marks for boys = 69B Total marks for girls = 72G The overall average EXPLANATION marks for the class is given by: (69B + 72G) / (B + G) = 70 or 69B + 72G = 70(B+G) or 69B + Correct answer: 2 72G=70B + 70G or 72G − 70G = 70B − 69B or 2G = B So, the number of boys B is twice the number of girls G (option 3 Answer) For brand A, price per can with offer = 320 price for 3 cans with offer = 320 x 4 = 960 1/3rd extra for each can means 1/3rd x 3 = 1 litre extra with same price i.e 960 Since each can is 1 litre, in Rs. 960, 4 litre detergent can be purchased. For brand B, price per can with offer = 332 price for 3 cans with offer = 332 x 3 = 996 1 can comes free with purchase of 3 cans Thus, 4 litre detergent can be purchased in Rs. 996For brand C, price per can with offer = 300 - (20/100 * 300) = 300-60 = 240 price for 4 cans (for 4 litres) without offer = 240 x 4 = 960 Thus best choice would be either A or C Correct answer: 3 www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 14.Question ID:- 703014 15.Question ID:- 703015 A pen, pencil, and an eraser together cost Rs. 21. Human females have two X chromosomes, The pen costs as much more than the pencil as each of which can be passed on to their son or the pencil does than the eraser. How much does daughter with equal probability. Human males the pencil cost? have one X chromosome which is passed on to their daughters and one Y chromosome which 1. 5 is passed on to their sons. Assuming equal 2. 7 numbers of males and females in a population, 3. 9 if an X chromosome is randomly sampled from 4. 11 the population, what is the probability that it was inherited from a female of the previous Correct answer: 2 generation? EXPLANATION 1. 1/3 2. 1/4 Let the cost of the eraser be E rupees. Then, let 3. 2/3 the cost of the pencil be P rupees then 4. 3/4 difference between pencil and eraser cost = P-E Given that the pen costs same amount than the Correct answer: 3 pencil as the pencil does more than the eraser(i.e P-E), we have: Cost of pen=P+(P−E) EXPLANATION The total cost of the pen, pencil, and eraser is Rs. 21: Thus (P+(P−E)) + P + E = 21 Simplify the Females have 2 X- chromosomes, one inherited equation: P+P−E+P+E=21 or 3P=21 Thus, P=7 from mother and one from father. Males have (option 2 Answer) one X chromosome inherited only from mother. So among 1 female and 1 male, only one X- chromosome is inherited from father and 2 X- chromosomes are inherited from mother. So in a population with equal number of males and females, the Probability of X-chromosome from mother = 2/3 and Probability of X-chromosome from father = 1/3. So answer is option 3. AIMNET 2024 CSIRNET STORES.BIOTECNIKA.ORG PRACTICE TEST SERIES Correct answer: 3 www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 16.Question ID:-703016 17.Question ID:- 703017 A cardboard sheet of size 60 cm × 60 cm A car is moving along a bend in a road. is used to make hollow cubes having The bend forms a large quarter circle. If sides of 5 cm. What is the maximum the distance between the left and right number of cubes that can be made? wheels of the car is 2 m, then the difference between the distances 1. 24 travelled by the inner wheels and the 2. 36 outer wheels (in m) as it traverses the 3. 72 bend is: 4. 144 1. 0 Correct answer: 1 2. 2 3. π EXPLANATION 4. 2π Correct answer: 3 As mentioned in the question, a cardboard sheet of size 60 cm x 60 cm, is used to make hollow EXPLANATION cubes each having side length of 5 cm. To claculate the number of cubes that can be made, The inner and outer wheels will trace out two firstly we will calculate the area of the cardboard concentric quarter-circles Let the radius of the sheet: Area of the cardboard sheet =60 cm×60 quarter-circle traced by the inner wheels be 𝑟. cm=3600cm^2 Now, calculate the surface area of The radius of the quarter-circle traced by the one cube: Each cube has 6 faces, and each face outer wheels will then be 𝑟+2 meters (since the is a square with a side length of 5 cm. Area of wheels are 2 meters apart) Since the perimeter of one face of the cube = 5cm×5cm=25cm^2. a circle is 2𝜋𝑟, the length of a quarter-circle arc Surface area of one cube =6×25cm^2 = 150cm^2. is given by: Arc length = 1/4 × 2𝜋𝑟 = 𝜋𝑟/2 To find the maximum number of cubes that can Distance traveled by the inner wheels: If the be made, divide the total area of the cardboard radius is 𝑟, the distance traveled by the inner sheet by the surface area of one cube. Maximum wheels is : Distance(inner) = 𝜋𝑟/2 Distance number of cubes = 3600 cm^2 / 150 cm^2 =24 ​ traveled by the outer wheels: If the radius is 𝑟+2, Therefore, the maximum number of cubes that the distance traveled by the outer wheels is: can be made from the cardboard sheet is 24. Distance(outer) = 𝜋(𝑟+2)/2 =𝜋𝑟/2 + 𝜋×2/2 = 𝜋𝑟/2 + 𝜋 Thus the difference between the distances traveled by the outer and inner wheels is: Difference = Distance(outer) − Distance(inner) = (𝜋𝑟/2 + 𝜋) − 𝜋𝑟/2 = 𝜋 meters (Option 3 Ans) www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 18.Question ID:-703018 19.Question ID:- 703019 Two rings made of metals A and B with ring A In a family of two males and three females, A is having a larger diameter, are placed the daughter of B and sister of C. E is the spouse concentrically leaving an annular gap. The of B and mother of D. C is not the brother of D. thermal expansion coefficients of the two Which of the following statements is NOT metals are CA & CB​. Identify the correct correct? statement(s) from the following. 1. E is the mother of A A. The gap will decrease if CA < CB​. 2. D is the sister of C B. The gap will remain the same if CA = CB​. 3. C is the daughter of B C. The gap will increase if CA > CB​. 4. A is the sister of D 1. 100 m/h Correct answer: 2 2. 120 m/h 3. 220 m/h 4. 440 m/h EXPLANATION Family has 2 males and 3 females A - daughter (1 Correct answer: 1 female) B - father/mother C - brother/sister E - mother (1 female) Since E is mother , means her spouse B would be father (1male) Since B and E The coefficient of thermal expansion is used to are a couple, A,C and D are siblings. (hence predict the expansion of materials in response to option 1 is correct, as E would be mother of A) a known temperature change. The larger this (option 4 is also correct as A and D are siblings coefficient is for a material, the more it will and A is a female) Since C is not the brother of D, expand per degree temperature increase. means C is D's sister (1 female) (Hence option 3 is correct as C is a female) That makes D as male, Metal ring A (blue) and Metal ring B (green) are and hence D is brother of A and C. Hence option 2 placed concentrically leaving an annular gap, as is incorrect, since it says sister. shown in the diagram. If C_A < C_B, then after heating, the inner ring B will expand relatively more as compared to the CSIRNET outer ring A. Hence the annular gap would decrease. Statement A is correct. TOUGHNET If C_A = C_B, then after heating, both rings would QUESTION expand equally, but the annular gap would still decrease. Statement B is incorrect. BANK If C_A > C_B, then after heating, the gap will UNIT WISE MCQS decrease. Statement C is incorrect. WITH TOUGHEST QUESTIONS Ans option 1. ASKED IN PAST 10 YEARS www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 20.Question ID:- 703020 EXPLANATION We know , pH = -log[H+] Thus [H+] = antilog (-pH) The largest integer between 1 and 10510^5105 The [H+] of all days would be 3x10^-11 , 5x10^-11, when written in words that does not contain the 8x10^-11, 4x10^-11, 2x10^-11 and 4x10^-11 letter 'N' or 'n' in its name is Average H+ concentration of all days would be = [(3x10^-11)+(5x10^-11)+(8x10^-11)+(4x10^-11)+ 1. 88 (2x10^-11)+(4x10^-11)]/6 = 4.33 x 10^-11 Average 2. 100000 pH = -log(4.33 x 10^-11) = 10.36 Ans 3. 88888 4. 8 Correct answer: 1 22.Question ID:- 703022 EXPLANATION Which one of the following statements regarding the stereoisomers of D-glucose is We have to check the number names of the given INCORRECT? options to see, if it has "N' or 'n'. option 2 is One Hundred Thousand and option 3 is Eighty eight thousand eight hundred eighty eight. They both 1. D-mannose is a C-2 epimer of glucose. contain 'n' and thus can be eliminated. Out of 88 2. D-allose is a C-3 epimer of glucose. and 8 (none of them has 'n' in their number 3. D-galactose is a C-4 epimer of glucose. name), eighty eight is bigger number. so answer 4. D-talose is a C-5 epimer of glucose. is option 1. Correct answer: 4 EXPLANATION 21.Question ID:- 703021 Diastereomers which differ in only one The pH of water in Lonar lake was found to be stereocenter (out of two or more) are called 10.5, 10.3, 10.1, 10.4, 10.7, and 10.4 for epimers. D-mannose differs from D-glucose only measurements taken once daily over six days. at the C-2 carbon, thus they are epimers. D-allose What would be the average pH of the lake water differs from D-glucose only at the C-3 carbon, during this period? they are also epimers D-galactose differs from D- glucose at the C-4 carbon, which makes them epimers But option 4 is incorrect because D-Idose 1. 10.56 is a C-5 epimer of glucose. D-talose differs from 2. 10.26 D-glucose at C-2 as well as C-4, hence they are 3. 10.36 not epimers. 4. 10.46 Correct answer: 3 www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 23.Question ID:- 703023 24.Question ID:- 703024 Which one of the following properties of Which one of the following compounds can serve grooves is a hallmark of the Z-form of DNA? as a direct acceptor of an additional amino group derived from amino acid catabolism? 1. Narrow and deep major groove 2. Wide and deep major groove 1. Fumarate 3. Narrow and shallow major groove 2. Glutamine 4. Flat major groove 3. α-Ketoglutarate 4. Asparagine Correct answer: 4 Correct answer: 3 EXPLANATION EXPLANATION Z-DNA (a left-handed helical structure) are made up of polynucleotides with two repeating units, α-Ketoglutarate is a key compound in the unlike one in A- and B-DNA, hence forming a process of transamination during amino acid zigzag patterned coil. In Z-DNA purines are catabolism, where it acts as a direct acceptor of rotated into the syn conformation while the amino groups. This transamination reaction is pyrimidines maintain the anti-conformation. Z- crucial for the deamination of amino acids, DNA has a flat major groove and a deep minor leading to the formation of glutamate. groove. Major groove in B-DNA is deep and wide, whereas in A-DNA , it is deep and narrow BIOTECNIKA’S CERTIFICATION COURSES STORES.BIOTECNIKA.ORG www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION Correct answer: 4 EXPLANATION The proton motive force (PMF) generated during mitochondrial electron transport is primarily used for ATP synthesis and the transport of ions and molecules across the mitochondrial membrane. The options involving ATP, ADP, and phosphate ion transport (options 1, 2, and 3) directly or indirectly rely on PMF. However, transport of NADH from the cytosol into the mitochondrial matrix does not directly use the proton motive force; it involves shuttles like the malate-aspartate shuttle or glycerol-phosphate shuttle, which operate independently of the proton gradient BExclusively Designed Hand 26.Question ID:- 703526 Crafted Notes For CSIR NET / Which one of the statements about bacterial GATE / DBT BET / ICMR operons is INCORRECT? Entrance Exams. 1. Operons can encode multiple proteins with linked biological activity. 2. An operon expresses multiple proteins from a single STORES.BIOTECNIKA.ORG mRNA. 3. mRNA transcript of an operon has only one Shine- Dalgarno sequence upstream of the first open reading frame. 25.Question ID:- 703025 4. Operon expression is often tightly regulated. In which one of the following, the proton Correct answer: 3 motive force generated in mitochondrial electron transport is NOT used EXPLANATION Operon is the arrangement of multiple genes under 1. Transport of ATP into the cytosol from mitochondrial the same regulatory proteins, whos preotein matrix. products are linked by their biological activity. The 2. Transport of ADP from the cytosol into the transcript of operon genes are transcribed as mitochondrial matrix. polycistronic mRNA, with each cistron coding for 3. Transport of phosphate ions from the cytosol into one of the proteins of operon. Here, each cistron the mitochondrial matrix. will have it's own Shine Dalgarno sequence so that 4. Transport of NADH from the cytosol into the ribosomes can assemble there and translate each mitochondrial matrix. protein. So, statement 3 is incorrect about operon. Answer is option 3. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 27.Question ID:- 703027 Mitochondria import most of their proteins through specific protein import complexes(TOM Mitotic cyclin increases gradually through the G2 and TIM) in their outer and inner membranes. phase of the cell cycle, but the activity of mitotic Unlike endosomes, lysosomes, and the Golgi CDK1 increases suddenly at the onset of M apparatus, which all receive proteins via vesicular phase. This is because: transport from other parts of the cell, mitochondria rely on direct protein translocation 1. Active CDK1 subunit is synthesized in M phase. mechanisms. 2. Mitotic cyclin is sequestered in the cytosol. 3. Activation of CDK1 requires post-translational THE BEST COACHING FOR modifications. 4. The inhibitor of CDK1 is degraded in the M phase. CSIR NET & GATE Correct answer: 3 EXPLANATION Activation and entry of CDK1 into the M phase are EXAM PREPARATION restricted by protein kinases such as Wee1 and Myt1. Once CDK is coupled to a cyclin, these Exam focused Daily classes kinases phosphorylate CDK1 on a tyrosine Printed Hard copy Study residue (Y15) and the nearby threonine residue Materials (T14), thereby activating them at the onset of M Video backups of all phase. Modification of CDK1 by phosphorylation classes is a post-translational modification. Ans Option 3. Unlimited expert assistance and doubt solving via chat Class tests , monthly 28.Question ID:-703028 practice exams All india level test series Among the organelles listed below, which one does NOT obtain proteins via vesicular transport? 1. Endosomes 2. Lysosome 3. Mitochondria 4. Golgi Correct answer: 3 EXPLANATION STORES.BIOTECNIKA.ORG TOLL FREE 1800-1200-1818 or 080-5099-7000 www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 29.Question ID:-703029 30.Question ID:-703030 Topoisomerase activity was measured in terms A yeast strain has accumulated a mutation that of change in the linking number of DNA in the makes it grow slowly. Investigation reveals that presence of Camptothecin (inhibitor of ribosomal RNA levels have dropped drastically Topoisomerase I) or Etoposide (inhibitor of in this strain. Which RNA polymerase is likely Topoisomerase II). Which one of the following to be mutated in this strain? is the correct expected outcome? 1. RNA Pol I 1. In the presence of Camptothecin, Topoisomerase I 2. RNA Pol II will lead to a change in the linking number by ±2. 3. RNA Pol III 2. In the presence of Etoposide, Topoisomerase I will 4. RNA Pol IV lead to a change in the linking number by ±2. 3. In the presence of Camptothecin, Topoisomerase II Correct answer: 1 will lead to a change in the linking number by ±2. 4. In the presence of Etoposide, Topoisomerase II will EXPLANATION lead to a change in the linking number by ±2. In eukaryotes, Pol I codes for all the rRNAs Correct answer: 3 except for 5SrRNA. so if its a defect of ribosome, then the mutation lies in Pol I EXPLANATION Camptothecin is an inhibitor of Topoismerase I, so in its presence only Topoisomerase II will be working. Since Topoisomerase II breaks double stand DNA to release the stress hence here the linking number changes by 2. so option 3 is correct CSIRNET TOUGHNET QUESTION BANK UNIT WISE MCQS WITH TOUGHEST QUESTIONS ASKED IN PAST 10 YEARS www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 31.Question ID:- 703031 Correct answer: 3 The trp operon can be induced by the addition of EXPLANATION indole propionic acid (IPA), which binds to the trp repressor but does not allow the change in M. tuberculosis can inhibit the fusion of conformation. Upon the addition of IPA, what will phagosomes with lysosomes, allowing it to survive be the order of the translation of the enzymes and replicate within the phagosomal compartment, encoded by the operon? which avoids destruction by the lysosomal enzymes. This strategy helps the bacterium evade 1. TrpA, TrpB, TrpC, TrpD, TrpE the host's immune response. So, the correct 2. TrpE, TrpD, TrpC, TrpB, TrpA answer is: Statement 3 -The bacteria will prevent 3. Only TrpE will be translated the maturation of phagosomes. 4. Only TrpA will be translated Correct answer: 2 33.Question ID: 703033 EXPLANATION In the thymus of a normal mouse, positive selection of T cells is based on recognition of IPA is a structural analog of trp. It can bind to trp which of the following? repressor and derepress the Trp operon. Then the genes in the trp operon will be transcribed in the 1. Foreign antigens in association with self-MHC order as they are present in the operon. So, it will molecules be TrpE, TrpD, TrpC, TrpB, TrpA. Answer is option 2. Self-antigens in association with foreign-MHC 2. molecules 3. Self-antigens in association with self-MHC molecules 4. Foreign antigens in association with TLR 32.Question ID:-703032 ligands When E. coli and macrophages are placed in a Correct answer: 3 petri dish with medium, the macrophages internalize the E. coli into cytoplasmic vesicles called phagosomes, which then fuse with EXPLANATION lysosomes where the bacteria are killed. If E. coli Positive selection occurs when double positive T is replaced by M. tuberculosis in the petri dish, cells bind cortical epithelial cells expressing Class which ONE of the following options will happen I or Class II MHC plus self peptides with a high after attachment of the bacteria? enough affinity to get the survival signal. Negative selection occurs when double positive T cells bind 1. The bacteria will be internalized by pinocytosis. to bone-marrow derived APC (macrophages and 2. The bacteria will be internalized by autophagy. dendritic cells) expressing Class I or Class II MHC 3. The bacteria will prevent the maturation of plus self peptides with a high enough affinity to phagosomes. receive an apoptosis signal. Note that selection 4. The bacteria will prevent the generation of occurs on self peptides in the thymus; MHC lysosomes in macrophages. presents self peptides in the absence of pathogen. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 34.Question ID:-703034 35.Question ID:-703035 Normal human fibroblasts, cancer cells (that The immune recognition of "self-molecules" is originated in stem cells), and fibroblasts important for which of the following events? transduced with hTERT (hTERT cells) were passaged for 35 generations. Southern blot 1. Initiation of B cell activation leading to antibody analysis was performed using DNA from above production. cells using radio-labelled probes for telomeric 2. Promoting the differentiation of hematopoietic stem sequences. Which one of the following band cells. patterns would be observed in the 3. Recombination of the T cell receptor. autoradiogram? 4. Activation of natural killer (NK) cells of the innate immune system. 1. 7-9 kb bands in fibroblasts, 18-20 kb bands in cancer and hTERT cells. 2. 18-20 kb bands in fibroblasts, 7-9 kb bands in Correct answer: 4 cancer and hTERT cells. 3. 7-9 kb bands in fibroblasts and hTERT cells, 18-20 EXPLANATION kb bands in cancer cells. 4. 18-20 kb bands in fibroblasts and hTERT cells, 7-9 NK cells undergo a maturation step kb in cancer cells. through the recognition of self molecules that are constitutively expressed in steady- state conditions. A prototypical example of Correct answer: 1 this process is the MHC class I–induced NK cell 'education'.It has been known that EXPLANATION NK cells follow an education process to become both competent to recognize We know in normal cells Telomerase enzyme is 'missing self' and tolerant to self inactive, hence we see a reduction of the Telomeric end but in cancer cells the telomerase is active and in the case where the normal fibroblast cell is substituted with hTERT. hence for the later 2 cases we will find a larger band than in the case where only normal fibroblast is present. STORES.BIOTECNIKA.ORG www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 36.Question ID:-703036 Correct answer: 1 Which one of the following statements best EXPLANATION describes an acrosomal reaction? The Shh signalling cascade plays an essential 1. It is a repulsive interaction between the sperm role in patterning the dorsoventral axis of the and the egg. neural tube. Complete absence of Shh 2. It involves digestion of the acrosome by the signalling, as in the null mutant for Shh itself sperm when it encounters an egg. results in a dramatic effect on DV patterning 3. It leads to digestion of the zona pellucida. within the neural tube, with absence of ventral 4. It is the fusion of the sperm and egg plasma cell types and failure of floor plate formation. membranes. The eye fields are abnormally close together or fused in the midline, a condition termed cyclopia. Correct answer: 3 EXPLANATION 38.Question ID:- 703038 When a sperm cell makes contact with the zona pellucida it undergoes morphological changes Which one of the following is NOT an example of which are termed the acrosome reaction. The programmed cell death in plants? acrosome reaction also leads to the release and 1. Aerenchyma formation in cortical root cells activation of acrosin, a proteolytic enzyme that 2. Embryonic suspensor cell degeneration is most likely one of the enzymes involved in 3. Tracheary element formation in vasculature the digestion of the zona pellucida necessary to 4. Casparian strip formation in root endodermis facilitate the transport of sperm into the perivitelline space Correct answer: 4 37.Question ID:- 703037 EXPLANATION Mutations in a specific mammalian signaling The formation of the Casparian strip is not pathway result in early defects observed in the associated with programmed cell death. It establishment or maintenance of midline involves the deposition of suberin in the cell structures, such as the notochord and the floor walls of the endodermis to regulate water and plate. Later defects include the absence of distal nutrient uptake, but it does not require the death limb structures, ventral cell types within the of the cells. On the other hand, Aerenchyma neural tube, spinal column and most of the ribs formation in cortical root cells, Embryonic and cyclopia. Mutations in which one of the suspensor cell degeneration and Tracheary following signaling pathways is the most element formation in vasculature involves reported cause for these congenital defects? programmed cell death. 1. Sonic Hedgehog 2. Wingless 3. Notch 4. Epidermal Growth Factor www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 39.Question ID:- 703039 EXPLANATION Water is a weak reductant because the product of Loss of function mutations in snapdragon its oxidation (i.e., removal of electrons), oxygen, (Antirrhinum) genes CYCLOIDEA (CYC) and is a powerful oxidant. Therefore, an oxidant that DICHOTOMA (DICH) will result in the: is more oxidizing (i.e., more positive redox potential) than molecular oxygen is required to 1. Conversion of bilaterally symmetric flower to a drive water oxidation in vivo. This is radially symmetric flower. accomplished by the primary reactions of PSII, 2. Conversion of radially symmetric flower to a which generate the most powerful oxidant known bilaterally symmetric flower. in biology, P680+. Likewise, the function of PSI is 3. Conversion of bisexual flower to a male flower. to generate a reductant that is more reducing 4. Conversion of bisexual flower to a female flower (i.e., more negative redox potential) than ferredoxin and NADP+ Correct answer: 1 41.Question ID:-703041 EXPLANATION Which of the following is a likely consequence of a A wild-type Antirrhinum flower is typically loss of function mutation in the gene encoding the bilateral, with two large dorsal petals, two enzyme phenylalanine ammonia-lyase (PAL) in lateral petals and one small ventral petal. CYC, coffee plants? DICH and RAD have been shown to interact to control dorsal petal identity, crucial for 1. Increased levels of caffeine. producing bilateral flowers. In the CYC;DICH 2. Decreased lignins in cell walls. double mutant, bilateral flower symmetry is 3. Increased lignins in cell walls. completely lost, forming a radially symmetrical FR 4. Decreased levels of caffeine. R flower with only ventral petals Correct answer: 2 40.Question ID:- 703040 EXPLANATION Which one of the following is the strongest Phenylalanine ammonia-lyase (PAL) is an oxidizing agent produced during photosynthesis? enzyme involved in the phenylpropanoid pathway, which is crucial for the synthesis of various secondary metabolites, including lignins. 1. NADPH Lignins are important for the structural integrity 2. P680⁺ and rigidity of plant cell walls. A loss of function 3. Ferredoxin mutation in the PAL gene would impair the 4. P700⁺ production of phenylpropanoid intermediates required for lignin biosynthesis, leading to Correct answer: 2 decreased levels of lignins in cell walls. This is why option 2 is the correct answer. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 42.Question ID:-703042 44.Question ID:- 703044 Which one of the following hormones is NOT Which one of the following statements regarding exclusively synthesized from a single location in the invasion of blast fungus, Magnaporthe the body? oryzae, in rice is INCORRECT? 1. Thyrotropin releasing hormone 1. A biotrophic interfacial complex is formed. 2. Corticotropin releasing hormone 2. Fungal effector proteins are translocated into the 3. Somatostatin host cell cytoplasm. 4. Somatotropin 3. Appressorium is produced to invade the plant. 4. Haustorium is mostly formed to extract nutrients from the host. Correct answer: 3 Correct answer: 4 EXPLANATION Thyrotropin releasing hormone is synthesized EXPLANATION in the hypothalamus. Corticotropin releasing hormone is primarily synthesized in the While many fungal pathogens form a hypothalamus. Somatostatin is produced in haustorium to extract nutrients from the host, many locations, which include the Magnaporthe oryzae does not typically form a gastrointestinal (GI) tract, pancreas, haustorium. Instead, it relies on the formation hypothalamus, and central nervous system of invasive hyphae within the host cells to (CNS). Somatotropin is synthesized in the extract nutrients. The haustorium is more pituitary gland. Ans option 3. commonly associated with obligate biotrophic fungi, such as rusts and powdery mildews. Thus statement 4 is false. 45.Question ID:- 703045 43.Question ID:-703043 Which one of the following adrenoceptors decreases cAMP in the post-synaptic target after How many molecules of acetyl-CoA condense to stimulation with norepinephrine? produce isopentenyl diphosphate, the precursor for the formation of terpenoids by the mevalonate 1. α₁ pathway? 2. α₂ 3. β₁ 1. Two 4. β₂ 2. Three Correct answer: 3 3. Four Correct answer: 2 4. Five EXPLANATION EXPLANATION The α2 adrenoceptor is a G-protein-coupled In the mevalonate pathway, three molecules receptor (GPCR) that, when stimulated by of acetyl-CoA are used to produce one norepinephrine, activates an inhibitory G- molecule of isopentenyl diphosphate (IPP). protein (Gi). This activation decreases the levels of cyclic AMP (cAMP) in the post- synaptic target cells. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 46.Question ID:- 703046 47.Question ID:- 703047 Which one of the following gases diffuses Which one of the following is considered as a renal through the alveolocapillary membrane in the hormone? shortest time at the resting condition? 1. Megalin 1. CO 2. Cubilin 2. O₂ 3. Renalase 3. CO₂ 4. Uroguanylin 4. N₂O Correct answer: 4 Correct answer: 3 Explanation Explanation The gas that diffuses through the Renalase is a relatively recently discovered alveolocapillary membrane in the shortest time kidney protein connected to chronic kidney at resting conditions is N2O (nitrous oxide). This disease and blood pressure disorders. It is because the rate of gas diffusion across the functions by metabolizing catecholamines alveolocapillary membrane depends on the and similar substances, such as adrenaline gas's solubility in the blood and its molecular and dopamine. weight. N2O has a high diffusion rate due to its low molecular weight and relatively high solubility in the blood, which allows it to diffuse quickly across the alveolar membrane. ENROLL IN BIOTECNIKA’S CERTIFICATION COURSES STORES.BIOTECNIKA.ORG www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 48.Question ID:- 703048 1. Transcripts of the target gene are paternally Which one of the following statements regarding contributed. genetics of quantitative traits in plants is 2. Transcripts of the target gene are maternally INCORRECT? contributed. 3. The transcripts are observed due to 1. Loci responsible for a quantitative trait can show mitochondrial inheritance. variations in their individual contributions to the 4. The transcripts are being detected from yeast trait. that larvae eat. 2. Quantitative trait loci (QTL) always have identical effects on a phenotypic trait in different Correct answer: 2 environments. 3. Recombinant Inbred Lines (RIL) populations EXPLANATION used for QTL mapping are immortal. 4. F₂:₃ families can measure both additive and dominant effects at specific QTL. 25% of the embryos did not show the gene in genomic DNA PCR. But northern blot of all the embryos showed the gene transcript. This is Correct answer: 2 only possible if the trait is under maternal effect. This means the mothers genome codes for the EXPLANATION transcript and that transcript is passed from The effect of QTL on a phenotypic trait can vary mother to the embryos through the cytoplasm. depending on environmental conditions, due to So, the transcript of the gene are maternally genotype-environment interactions. Thus contributed. Answer is option 2. statement 2 is incorrect. 49.Question ID:-703049 50.Question ID:-703050 A Drosophila stock that is heterozygous null for The pedigree in Panel (i) represents the a unique nuclear target gene was sib-mated. The inheritance pattern of a given trait. The trait is target gene is essential for the development of NOT 100% penetrant. Panel (ii) represents PCR Drosophila. The embryos from the cross were amplification profile of each member of the analyzed and the following results were family using a specific primer pair. (M: mother, obtained: F: father, C: child) PCR analysis of the genomic DNA isolated from embryos showed that 25% of the embryos did not have the target gene. Northern analysis of the RNA isolated from the above embryos showed the presence of transcript corresponding to the target gene. No lethality was observed in the progeny. Which one of the following options can best explain the above observations? www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION EXPLANATION What is the mode of inheritance of this trait? Bacteriophage Mu represents a hybrid gene Autosomal recessive creature that exists as both a transposable Autosomal dominant element and a bacteriophage (a virus that infects X-linked recessive bacteria). Hence statemnt 1 is correct. The X-linked dominan bacteriophage P1 encodes a site-specific recombination system which is composed of Correct answer: 2 two components: the loxP DNA site at which recombination takes place and the Cre protein EXPLANATION which carries out that recombination. Hence statemnt 2 is correct. M13 is an example of a In all the three families, all the affected filamentous phage,6407 nucleotides in length. It individuals have at least one affected parent. So is circular and is unusual in that it consists the trait has to be a dominant trait. From the entirely of single-stranded DNA and not ssRNA band patterns obtained in PCR amplification, we genome. Hence statement 3 is incorrect. Phage can associated one band from one allele and P1 is a temperate bacteriophage. The other two bands from the other allele. In family bacteriophage ΦX174 has a [+] sense circular 2, father is male and still is heterozygous as he single-stranded DNA genome of 5,386 has 3 bands representing 2 different alleles. In nucleotides. Hence statement 4 is correct. Since the same way, in family 3 the child is male but is incorrect stament has to be choosen, answer is still heterozygous. This means the trait is option 3. autosomal as males cannot be heterozygous for X-linked traits. So answer is option 2. 52.Question ID:-703052 The proponents of sustainable development argue for a switch to a predominantly plant-based diet, in 51.Question ID:-703051 order to reduce the human footprint of food production. The statements given below present Which one of the following statements is some of the arguments put forward by them. INCORRECT? A. Animal-based diets involve greater thermodynamic energy loss. 1. Phage Mu is used to create insertion mutations. B. The production of animal-based foods involves high 2. Phage P1 is a source of Cre-LoxP recombination carbon burn-off. system. C. Animal tissues have high C 3. Phage M13 has single-stranded circular RNA ratios. genome. D. Animal tissues have high water content. 4. Phage ΦX174 has single-stranded circular DNA genome. Select the option that constitutes the basis of their argument. 1. A and B Correct answer: 3 2. A and C Correct answer: 1 3. B and C 4. B and D www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 54.Question ID:- 703054 EXPLANATION Which one of the following is most commonly According to sustainable development, used for barcoding-based identification of switching to plant based alternatives (PBA) for animal species? food can be beneficial for our environment. By reducing the demand for meat and increasing 1. Cytochrome oxidase I the normalization of vegan foods, we can see 2. Microsatellites some dramatic changes in the trajectory of our 3. 28S planet. Two major benefits: 1. Reduces the 4. MatK emission of green house gases, thus reducing carbon footprint. 2. Reduces energy Correct answer: 1 consumption. So statements A and B are correct. Answer is option 1. EXPLANATION Microsatellites are used in genetic studies and population genetics but are not as commonly used for species identification through barcoding. 28S rRNA is 53.Question ID:- 703053 used in phylogenetic studies and taxonomy but is not typically used for species barcoding in animals. MatK Which one of the following organisms is NOT is used primarily for plant barcoding, not animal paedomorphic? species. Cytochrome oxidase I (COI) is the most commonly used gene for barcoding-based 1. Oikopleura identification of animal species. It is part of the 2. Branchiostoma mitochondrial genome and is used in the DNA 3. Ambystoma barcoding technique to differentiate species due to its 4. Triturus high variability among species and relatively conserved regions within species. This method has become a standard approach for identifying and Correct answer: 2 classifying animal species. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 55.Question ID:- 703055 donesian islands of Bali and Lombok to the south, and extends north through the Makassar Strait The term gynodioecious species refers to between Kalimantan (Borneo) and Sulawesi. plants with: Wallace’s hypothesis was based on observations from islands in the Indonesian archipelago. He 1. Female flowers and hermaphrodite flowers on noted a distinct divide between the fauna of the separate individuals. western islands, with animals largely of Asian 2. Female flowers and male flowers on separate origin, and that of the eastern portion where individuals. animals were Australasian. So Indomalaya and 3. Female flowers and hermaphrodite flowers on the Australasia are divided by Wallace line. Answer is same individual. option 2. 4. Female flowers and male flowers on the same individual. 57.Question ID:-703057 Correct answer: 1 Ian Pavlov conducted experiments to demonstrate EXPLANATION that a dog that associates the sound of a bell with food, would salivate on hearing the bell even when Gynodioecy is a rare breeding system that is the food was not presented. This is an example of found in certain flowering plant species in which female and hermaphroditic plants coexist within 1. Operant conditioning. a population. 2. Classical conditioning. 3. Sensitization. 4. Habituation. 56.Question ID:- 703056 Correct answer: 2 Which of the following biogeographic realms are divided by the Wallace Line? EXPLANATION Classical conditioning theory says that 1. Indomalaya and Neotropical behaviors are learned by connecting a neutral 2. Indomalaya and Australasia stimulus with a positive one, such as when 3. Nearctic and Palearctic Pavlov's dogs heard a bell (neutral) and 4. Palearctic and Afrotropical expected food (positive). In classical conditioning, the subject firmly associates the Correct answer: 2 neutral stimulus with the unconditioned response. This creates a new behavior, or what's known as the conditioned response. If the link EXPLANATION between the two weakens or breaks, this leads to what's called extinction. When Pavlov's dogs no The Wallace Line does not actually exist in reality. longer got food after hearing the bell, they It is an imaginary line that intersects the Lombok eventually stopped associating the bell with Strait between the In food. Answer is option 2. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 58.Question ID:- 703058 59.Question ID:- 703059 Runaway selection was proposed by R. A. Fisher Greenhouse gas emissions are considered the to explain the evolution of extravagant secondary primary driver of global warming through their sexual characters. The model is based on the influence on the radiative forcing of the exaggeration of characters in males and female atmosphere. This radiative forcing occurs choice for these exaggerated characters. Which because: one of the following statements is considered an assumption of this model? 1. Longwave radiation emitted by the earth's surface is absorbed and scattered. 1. Exaggeration of characters in males, and female 2. Shortwave radiation received from the sun is choosiness for exaggeration are both heritable. absorbed and scattered. 2. Neither exaggeration of characters in males, nor 3. Shortwave radiation emitted by the earth's surface is female choosiness for exaggeration are heritable. absorbed and scattered. 3. Exaggeration of characters in males is heritable but 4. Longwave radiation arriving from the sun is selectively female choosiness for exaggeration is not heritable. transmitted. 4. Exaggeration of characters in males is not heritable but female choosiness for exaggeration is heritable. Correct answer: 1 Correct answer: 1 EXPLANATION EXPLANATION The correct understanding of how greenhouse Runaway selection hypothesis, in biology, an gases influence radiative forcing and global explanation first proposed by English statistician warming is based on the interaction with R.A. Fisher in the 1930s to account for the rapid radiation emitted by the Earth's surface. evolution of specific physical traits in male Greenhouse gases primarily trap heat by animals of certain species. Some traits—such as absorbing and re-emitting longwave radiation prominent plumage, elaborate courtship (infrared radiation) that is emitted by the Earth's behaviours, or extreme body ornamentation— surface. This process prevents the heat from are so strongly preferred by females of certain escaping into space, thereby warming the species that they will mate only with those males atmosphere. Thus the correct statement is 1. possessing the strongest expression of the trait. In subsequent generations, male offspring are more likely to possess that physical trait, whereas female offspring are more likely to possess a preference for that trait in males. Over time, the species may be characterized by extreme sexual dimorphism. Both the trait in male and preference of trait in female are genetically correlated. Answer is option 1. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 60.Question ID:- 703060 61.Question ID:-703061 The Burgess Shale in the Canadian Rocky Natural selection can maintain genetic Mountains is known for its Cambrian fossils. This polymorphisms. Which one of the following CAN site is abundant in which one of the following fossil NOT contribute to the maintenance of assemblages? polymorphisms? 1. Arthropods 1. When the direction of selective forces is different in 2. Dinosaurs different environments 3. Woody plants 2. When heterozygotes have superior fitness over 4. Fishes homozygotes 3. When gradients of selective forces favor different Correct answer: 1 morphs 4. When frequency-dependent selection confers an advantage to a morph which is common EXPLANATION The Burgess Shale is found in an area of the Correct answer: 4 Canadian Rocky Mountains known as the Burgess Pass, and is located in British EXPLANATION Columbia's Yoho National Park. Part of the ancient landmass called Laurentia, centered in direction of selective forces in different Hudson Bay, the Burgess Shale represents one environments will definately lead to of the most diverse and well-preserved fossil polymorphism as variations will form. localities in the world. The dominant fossils heterozygotes always increases polymorphism found are arthropods, but other fossils are found as two different allels are involved. in the 3rd in great abundance, including worms, crinoids, case also we will find variation. but when it sea cucumbers, chordates, and other organisms comes to frequency dependent selection then a with no mineralized shell. balanced polymorphismis maintained. In this selection model, a single genotype will never replace its competitors, because increasing its frequency will lead to lowering its fitness and a subsequent decrease of its frequency. Thus, rare genes will tend to increase and common genes Correct answer: 2 will decrease in frequency, leading to a balanced polymorphism THE BEST COACHING FOR CSIR NET & GATE EXAM PREPARATION TOLL FREE 1800-1200-1818 or 080-5099-7000 STORES.BIOTECNIKA.ORG www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 62.Question ID:-703062 63.Question ID:- 703063 Which one of the following molecular phylogenetic Which one of the statements about homoplasy is trees depicts the correct relationship among NOT true? invertebrates? 1. It represents an independent acquisition of traits in unrelated lineages. 2. It refers to a character shared by a set of species but not present in their common ancestor. 3. It refers to a character state that evolved because of convergent evolution. 4. It represents characters that are similar due to parsimony. Correct answer: 4 Correct answer: 2 EXPLANATION EXPLANATION Homoplasy is a fascinating and unusual occurrence in evolution. It is the independent acquisition of the same trait in unrelated lineages. Hence statement 1 is true. Unlike homology, if the character shared by two organisms is not traced to a common ancestor, the similarity may be the result of homoplasy (sometimes considered synonym of analogy). Hence statement 2 is true. Homoplasy can result in three different ways. One, the organisms have a common ancestor but the character-state was not present in their common ancestor (parallelism). It could also result from two different characters in different ancestors evolving into identical character-states (convergence). Hence statement 3 is true. Similarity could also arise from loss of a particular character (reversal), thus reverting to ancestral condition (loss of perianth in some families). Parsimony, in the context of phylogenetics, is a principle used to select the simplest scientific explanation that fits the evidence, which often involves choosing the tree with the least number of evolutionary changes. Since, homoplasy represents characters that are similar not due to shared ancestry but due to convergent evolution, parallel evolution, or evolutionary reversals. Homoplasy can complicate phylogenetic analysis because it can lead to similarities that do not reflect true evolutionary relationships. Hence, statement 4 is "NOT" true and is the answer. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 64.Question ID:- 703064 65.Question ID:- 703065 The Ti plasmid from Agrobacterium Which of the following describes 'Empty Forest'? tumefaciens has genes for auxin, cytokinin, and opine synthesis, while genes for opine 1. Absence of large trees catabolism and Vir genes lie outside the T- 2. Less species diversity due to natural reasons DNA region. Which one of the following 3. Habitat void of large mammals due to genes are involved in providing a carbon anthropogenic impacts source to Agrobacterium in their ecological 4. Loss of habitat niche? Correct answer: 3 1. Genes for auxin synthesis only. 2. Genes for auxin as well as cytokinin synthesis. EXPLANATION 3. Genes for opine synthesis and opine catabolism. 4. Genes for auxin synthesis as well as opine An "empty forest" refers to an ecosystem that synthesis. is void of large mammals. Correct answer: 3 EXPLANATION In Agrobacterium tumefaciens, opines are a unique class of compounds that serve as a carbon and nitrogen source for the bacterium. These compounds are synthesized in plant cells following infection and are not typically found in plants. They are produced as a result of the action of the T-DNA (transfer DNA) from the Ti plasmid (tumor-inducing plasmid) of the bacterium. KONCEPTIKA LITE STORES.BIOTECNIKA.ORG Flowcharts on 100 Important topics of CSIR NET Exam www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 66.Question ID:-703066 67.Question ID:- 703067 The following graphs represent plots for the A 160 kDa complex of four protein molecules volume (dotted lines) and bacterial viable cell consists of a dimer formed by a 25 kDa protein count curves (solid line) for a fermenter connected by two disulfide bonds, and three culture. other proteins of 10, 30, and 70 kDa, respectively. It was isolated and analyzed on an SDS-PAGE gel without DTT in the gel loading dye. Which one of the following options would represent the SDS-PAGE profile? 1. Four bands corresponding to 10, 30, 50, and 70 kDa 2. Four bands corresponding to 10, 25, 30, and 70 kDa 3. One band corresponding to 160 kDa 4. Two bands corresponding to 50 kDa and 110 (Image with graphs labeled A, B, C, and D) kDa Which one of the following corresponds to Correct answer: 1 the features applicable to a fed-batch mode of fermenter culture 1. A EXPLANATION 2. B 3. C In SDS-PAGE, proteins are typically denatured 4. D and separated based on their molecular weight. DTT is a dithiol-reducing agent that breaks disulfide bonds and also inhibits their formation. Correct answer: 4 Without DTT (a reducing agent), disulfide bonds remain intact, preventing the separation of subunits connected by these bonds. As mentioned in the question that 160 KDa protein was isolated and analyzed on SDS-PAGE gel without DTT in gel laoding dye, so the dimer formed by two 25 kDa proteins connected by two disulfide bonds, will appear as a single 50 kDa band. Three other proteins of 10 kDa, 30 kDa, and 70 kDa, which are likely not covalently bound to the 50 kDa dimer, will appear as separate bands. Thus, the SDS-PAGE profile will show four bands corresponding to 10, 30, 50 and 70 KDa. Therefore, answer is option 1. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 69.Question ID:- 703069 JULY 2024 Two students performed an ELISA to determine the amount of anti-Spike antibody PUBLISH in serum of a Covid-19 patient. They used the same ELISA plates, the same reagents for RESEARCH coating, blocking and detection, and the same ELISA reader. Both generated PAPERS IN independent standard curves of absorbance vs concentration using the same Spike AI ML IN BIOLOGY protein. Student 'A' correctly reported a concentration of 100 µg/ml, but student 'B' WORK ON REAL-TIME PROJECTS reported 450 µg/ml. Which one of the following could most likely explain the wrong GET WORK result of student 'B'? EXPERIENCE 1. The ELISA plate was not washed properly 2024 STARTS 26th Aug between coating with antigen and blocking. 2. The ELISA plate was not washed properly after addition of the sample. 3. The slope of the standard curve generated by student B was lower than optimal. 4. The negative control showed very little 68.Question ID:- 703068 absorbance. Which one of the following is an example of Correct answer: 3 parametric statistical test? EXPLANATION 1. Kruskal Wallis test 2. Fisher's exact test The most likely explanation for student 'B' 3. Unpaired t-test reporting a much higher concentration (450 4. Wilcoxon signed rank test µg/ml) compared to student 'A' (100 µg/ml) is option 3: The slope of the standard curve Correct answer: 3 generated by student B was lower than optimal. If the standard curve has a lower slope, it means that the relationship between EXPLANATION absorbance and concentration is not as steep, If the data are normally distributed, parametric which can result in higher calculated tests such as the paired or unpaired sample t- concentrations for given absorbance values. test, ANOVA or Pearson correlation are used. If This is because a flatter standard curve can the data are not normally distributed, the lead to an overestimation of the concentration nonparametric tests such as Mann-Whitney U- of the analyte in the samples. Test, Wilcoxon-signed rank test, Kruskal Wallis test, or Fisher's extract test are used. www.Biotecnika.org  CSIR NET JULY 2024 QUESTION PAPER + ANSWER KEYS WITH EXPLANATION 70.Question ID:- 703070 71.Question ID:- 703071 Which one of the following statements is correct A decapeptide composed of MFTGYPCPRW was with respect to the 95% confidence interval of the dissolved in 20 mM HEPES (pH 7.0), 50 mM NaCl, estimated mean from a set of observations? 50 mM Na₂SO₄, 5 mM DTT, and 4 mM EDTA. Which one of the following statements about the peptide 1. They are limits b.etween which, in the long run, in the given buffer conditions is correct? 95% of observations fall. 2. They are a way of measuring the precision of the 1. The peptide forms a dimer through disulfide estimate of the mean. bonds. 3. They are limits within which the sample mean 2. The peptide has a net positive charge. falls with probability 0.95. 3. The peptide has a net negative charge. 4. They are a way of measuring the variability of a 4. The peptide is neutral set of observations Correct answer: 2 Correct answer: 2 EXPLANATION EXPLANATION In the decapeptide (MFTGPYCPRW) , the amino acid residues with ionizable side chains are : A confidence interval is a statistical tool used to Tyrosine (OH group) with pkr 10.1 Cysteine (SH estimate the range of values within which a group) w