PART A
- Which of the following is true for acetyl-CoA?
1. it is an acetyl group attached to a type of coenzyme
2. it is another name for oxaloacetate
3. it is a protein
4. it is an acetyl group joined with a form of cobalt
Ans: Acetyl-CoA is a molecule composed of an acetyl group attached to coenzyme A. Coenzyme A is a molecule that is involved in many metabolic pathways and acts as a carrier of acetyl groups.
2. The role of DNA ligase in DNA replication is
1. addition of new nucleotides to the leading strand
2. addition of new nucleotides to the lagging strand
3. formation of a phosphodiester bond between the 3’-OH of one Okazaki fragment and the 5’-phosphate of the next on the lagging strand
4. base pairing of the template and the newly formed DNA strand
Ans: DNA ligase is an enzyme that plays a crucial role in DNA replication by joining the Okazaki fragments on the lagging strand. Okazaki fragments are short segments of newly synthesized DNA that are generated during DNA replication of the lagging strand. DNA ligase catalyzes the formation of phosphodiester bonds between adjacent nucleotides, thereby joining the fragments and completing the synthesis of the lagging strand.
3. A buffer contains 10% glucose, 20 mM Tris, and 50 mM HCI. For making 1 litre of buffer from the following
stock solutions – 50% glucose, 1M Tris, and 1 M HCI, the correct combination of volume each of the stock
solutions will be
1. 200 ml, 50 ml, 20 ml
2. 50 ml, 100 ml, 10 ml
3. 50 ml, 50 ml, 50 ml
4. 200 ml, 20 ml, 50 ml
Ans: The question asks for the correct combination of volumes of stock solutions required to make 1 liter of the buffer. The final buffer contains 10% glucose, 20 mM Tris, and 50 mM HCI. To prepare this buffer, we need to calculate the volumes of the 50% glucose, 1M Tris, and 1 M HCI stock solutions that would provide the desired concentrations. By using the formula:
C1V1 = C2V2
where C1 and V1 are the concentration and volume of the stock solution, and C2 and V2 are the desired concentration and volume of the final buffer, we can calculate the volumes of the stock solutions required. The correct combination of volumes of the stock solutions is:
200 ml of 50% glucose solution, 50 ml of 1M Tris solution, and 20 ml of 1 M HCI solution.
4. In enzyme kinetics, if the enzyme concentration in doubled
1. becomes double
2. does not change
3. becomes half
4. increases 4-fold
Ans: In enzyme kinetics, the rate of an enzyme-catalyzed reaction is dependent on the enzyme concentration. If the enzyme concentration is doubled, the rate of the reaction also doubles, as long as other factors such as substrate concentration remain constant. Therefore, the correct answer is that the rate of the reaction becomes double.
5. During growth, the diameter of a Staphylococcus bacterial cell increases by 5%. The specific surface area
(defined as surface area per unit volume)
1. increase approximately by 5%
2. decreases approximately by 5%
3. decreases approximately by 4 %
4. increases approximately by 4 %
Ans: The specific surface area of a bacterial cell is the ratio of its surface area to its volume. During growth, the diameter of the cell increases, and therefore, its volume increases faster than its surface area. As a result, the specific surface area decreases. The approximate decrease in the specific surface area can be calculated as (1 - (1.05)^-1) x 100%, which is approximately 4%.
6. The enzyme used in glucometers to estimate blood glucose levels is
1. glucose isomerase
2. insulin
3. glucose oxidase
4. hexokinase
Ans: Glucose oxidase is an enzyme that catalyzes the oxidation of glucose to produce hydrogen peroxide and gluconic acid. Glucometers use this enzyme to estimate blood glucose levels by measuring the amount of hydrogen peroxide produced, which is proportional to the amount of glucose in the blood sample.
7. Which of the following is a method of investigating the sequence specificity of DNA- binding proteins in vitro?
1. Gene targeting
2. DNA footprinting
3. polymerase chain reaction
4. Southern hybridization
Ans: DNA footprinting is a method used to investigate the sequence specificity of DNA-binding proteins in vitro. It involves labeling the ends of a DNA fragment, incubating the fragment with the protein of interest, and then digesting the DNA with a nuclease that cleaves only at unbound regions of the DNA. The protected regions, where the protein is bound, will remain intact and can be visualized by gel electrophoresis. This method can be used to identify the specific DNA sequences that are recognized and bound by the protein.
9. Using only random VDJ recombination, from 40 V, 30 D, and 6 J gene segments, the number of possible
variable regions of the resulting antibody would be
1. 7,200
2. 76
3. 40
4. 1.4 x
Ans: The number of possible variable regions of the resulting antibody can be calculated by multiplying the number of V, D, and J segments. Therefore, the number of possible variable regions is 40 x 30 x 6 = 7,200. The answer is 1.
10. The fruit of a particular tree species formed the predominant diet of the dodo. After dodo became extinct, that tree species also became extinct. Which of the following is the most likely cause for the tree’s extinction?
1. the dodo habitat was destroyed
2. the seeds of that tree required passage through the digestive system of the dodo for germination
3. by living close to the tree, the dodo protected the tree from other birds
4. other birds ate the fruit of that tree, as well as the fruit of other trees, and dispersed more seeds than the dodo did
Ans: The most likely cause for the extinction of the tree species that formed the predominant diet of the dodo is that the seeds of that tree required passage through the digestive system of the dodo for germination. Therefore, after the dodo became extinct, the tree species also became extinct. The answer is 2.
11. What is the frequency with which a 4bp cutter will cut the DNA, assuming random distribution of base in the genome?
1. 1/254
2. 1/64
3. 1/256
4.1/4096
Ans: The frequency with which a 4bp cutter will cut the DNA, assuming a random distribution of bases in the genome, can be calculated as (1/4)^4, which is equal to 1/256. The answer is 3.
12. Keshav and Kunal are good in Maths and Chemistry. Sumit and Keshav are good in Maths and Biology. Vineet and Kunal are good at Cricket and Chemistry. Sumit, Vineet and Rohit are good in Football and Biology. Who is good in Biology, Cricket, Chemistry and Football?
1. Vineet
2. Keshav
3. Kunal
4. Sumit
Ans: To determine who is good in Biology, Cricket, Chemistry, and Football, we need to find the person who is good in both Biology and Football, as well as either Cricket or Chemistry. From the given information, we can see that Sumit is good in Biology and Football, Vineet is good in Cricket and Chemistry, and Keshav and Kunal are good in Maths and Chemistry. Therefore, the person who is good in Biology, Cricket, Chemistry, and Football is Sumit. The answer is 4.
13. The molecular weight of Val and Ser are 117 Dalton and 105 Dalton, respectively. Val and Ser form a
dipeptide Val-Ser. The molecular weight (in Daltons) of the dipeptide is
1. 204
2. 222
3. 186
4. 240
Ans: The molecular weight of Val and Ser are 117 Dalton and 105 Dalton, respectively. The molecular weight of the dipeptide Val-Ser can be calculated by adding the molecular weights of Val and Ser and subtracting the molecular weight of a water molecule. Therefore, the molecular weight of Val-Ser is (117 + 105) - 18 = 204 Dalton. The answer is 1.
14. Which of the following is NOT a rational grouping of amino acids based on their polarity properties?
1. Val and Leu
2. Met and Leu
3. Asn and Gin
4. Glu and Ile
Ans: The rational grouping of amino acids based on their polarity properties includes grouping Val and Leu together as nonpolar amino acids, grouping Met and Ile together as moderately nonpolar amino acids, grouping Asn and Gln together as polar uncharged amino acids, and grouping Glu and Asp together as negatively charged amino acids. Therefore, the grouping of Glu and Ile is not a rational grouping of amino acids based on their polarity properties. The answer is 4.
15. Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a., respectively. If the total amount of simple interest earned in 2 years was Rs. 3508, what was the amount invested in Scheme B?
1. Rs. 6400
2. Rs. 7200
3. Rs. 7500
4. Rs. 6500
Ans: To find the amount invested in Scheme B, we can use the formula I = P x R x T, where I is the interest, P is the principal amount, R is the rate of interest, and T is the time period. Let x be the amount invested in Scheme A, then the amount invested in Scheme B is (13900 - x). We can then set up the equation:
(x x 14 x 2)/100 + [(13900 - x) x 11 x 2]/100 = 3508
Solving this equation gives x = 7200, which is the amount invested in Scheme A. Therefore, the amount invested in Scheme B is (13900 - 7200) = 6700. The answer is not given in the options.
16. Someone tells you that the pH of a solution is minus 2. Which one of the following is false?
1. concentration of H3O+is 100 m
2. such a value of pH is possible in theory
3. such a value of pH is unlikely to occur in practice
4. such a value of pH is impossible even in theory
Ans: A pH of -2 is impossible even in theory. pH is a measure of the concentration of hydrogen ions in a solution and is defined as the negative logarithm of the hydrogen ion concentration. The pH scale ranges from 0 to 14, where a pH of 7 is neutral, a pH less than 7 is acidic, and a pH greater than 7 is basic. Therefore, the statement that the pH of a solution is minus 2 is false. The answer is 4.
17. How much sodium hydroxide will you weigh to prepare 0.25 L of 3 M solution?
1. 40 g
2. 80 g
3. 40 Kg
4. 30 g
Ans: The answer can be found using the formula:
moles = concentration × volume
where concentration is given as 3 M and volume is given as 0.25 L.
moles = 3 M × 0.25 L = 0.75 mol
The molar mass of sodium hydroxide (NaOH) is 40 g/mol. Therefore, the mass of NaOH required can be calculated as:
mass = moles × molar mass = 0.75 mol × 40 g/mol = 30 g
Therefore, the correct answer is 4. 30 g.
18. Which of the following is/are critical for genome replication?
1. all of the given options are correct
2. polymerase
3. ligase
4. helicase
Ans: The correct answer is 1. all of the given options are correct. All of the options (polymerase, ligase, and helicase) are critical for genome replication. The polymerase is required for the synthesis of new DNA strands, ligase is required for joining Okazaki fragments on the lagging strand, and helicase is required for unwinding the double-stranded DNA molecule.
19. A student made 0.15 M solution of copper sulphate. The absorbance of the solution was found to be 0.3 when using a cuvette with a path length of 1 cm. Copper sulphate solution made by a second student gave an absorbance of 0.45 using the same cuvette at the same wavelength. What is the concentration of the copper sulphate solution made by the second student?
1.0.425 M
2. 0.125 M
3. 0.225 M
4. 0.325 M
Ans: The absorbance of a solution is related to its concentration by the Beer-Lambert law, which states that: A = εlc where A is the absorbance, ε is the molar absorptivity (a constant for a given substance and wavelength), l is the path length, and c is the concentration.
We can rearrange this equation to solve for the concentration: c = A/(εl)
For the first student's solution, c = 0.15 M, A = 0.3, and l = 1 cm. Therefore: ε = A/(cl) = 0.3/(0.15 × 1) = 2
For the second student's solution, A = 0.45 and l = 1 cm. Plugging in the values, we get: c = A/(εl) = 0.45/(2 × 1) = 0.225 M
Therefore, the correct answer is 3. 0.225 M.
20. Bacteria protect themselves from phages by producing the following enzymes which fragment the phage genome
1. endonucleases
2. methylases
3. topoisomerases
4. exonucleases
Ans: The correct answer is 1. endonucleases. Endonucleases are enzymes that cleave the phosphodiester bonds within the DNA molecule. Bacteria produce endonucleases that are specific to certain DNA sequences that are found in phages but not in their own genome. These enzymes fragment the phage DNA, preventing it from replicating and killing the host cell.
22. Two particles are moving back and forth in a 10 m-long tube. Particle ‘P’ is moving at a speed of 5 m/s and particle ‘Q’ at a speed of 2 m/s. Consider that both P and Q start at the same time in the same direction. How much times will ‘P’ cross ‘Q” by the time ‘Q’ reaches the end of the tube?
1. 5
2. 0
3. 2
4. 1
Ans: The time it takes for 'Q' to reach the end of the tube can be calculated as:
time = distance/speed = 10 m/2 m/s = 5 s
During this time, particle 'P' will cover a distance of:
distance = speed × time = 5 m/s × 5 s = 25 m
Since particle 'P' is moving back and forth in the tube, it will cross particle 'Q' each time it reaches one end of the tube and turns around. The total number of times it will cross 'Q' can be calculated by dividing the distance it covers (25 m) by the distance between each crossing (10 m):
crossings = distance/distance per crossing = 25 m/10 m = 2.5
Since the number of crossings must be a whole number, particle 'P' will cross 'Q' 2 times by the time 'Q' reaches the end of the tube. Therefore, the correct answer is 3. 2.
23. The most important step of an automated DNA sequencing reaction is
1. specific and systematic termination of the amplified DNA
2. ligation of DNA template
3. addition of calcium chloride
4. cleavage of template DNA
Ans: The correct answer is 1. specific and systematic termination of the amplified DNA. Automated DNA sequencing relies on a specific and systematic termination of DNA synthesis. This is achieved by adding a small amount of dideoxynucleotides (ddNTPs) to the reaction mixture, which are incorporated into the growing DNA strands but prevent further extension. The ddNTPs are labeled with different fluorescent dyes, allowing the sequence of the DNA to be read based on the color of the dye at the end of each fragment.
24. Chlorine is assigned and atomic weight of 35.5. This is due to
1. presence of half of proton
2. none of the given options
3. presence of isotopes
4. presence of half a neutron
Ans; The correct answer is 3. presence of isotopes. Chlorine has two stable isotopes, chlorine-35, and chlorine-37, with relative abundances of 75.77% and 24.23%, respectively. The atomic weight of chlorine is therefore a weighted average of the atomic masses of these two isotopes:
(0.7577 × 35 amu) + (0.2423 × 37 amu) = 35.5 amu
Therefore, the atomic weight of chlorine is not due to the presence of half a proton or half a neutron, but rather to the presence of isotopes with different atomic masses.
26. Histones
1. contain a high amount of basic amino acids
2. contain both sheet
3. have molecular weights in excess of 100,000 Da
4. are negatively-charged globular proteins
Ans: Histones contain high amounts of basic amino acids, particularly lysine and arginine, which gives them a positive charge and allows them to interact with the negatively charged DNA molecule.
27. A mixture of homotetramer ‘X’ and heterodimer ‘Y’ with identical molecular weight were resolved on SDS- PAGE. It gives three bands on a gel with molecular weights 40 kDa, 60 kDa, and 100 kDa, The native molecular wight (in kDa) of the homotetramer ‘X’ is
1. 160
2. 240
3. 320
4. 100
Ans: The native molecular weight of the homotetramer 'X' can be calculated by adding the molecular weights of the three bands on the gel that contain only 'X' subunits, which are 40 kDa, 80 kDa, and 120 kDa. These add up to 240 kDa, so the native molecular weight of 'X' is 240 kDa.
28. Find the next number in the series 23, 30, 38, 47, 57
1. 65
2. 69
3. 67
4. 68
Ans: The next number in the series is 68. The series is obtained by adding consecutive odd numbers starting from 23: 23 + 7 = 30, 30 + 8 = 38, 38 + 9 = 47, 47 + 10 = 57, 57 + 11 = 68.
29. What is the probability of getting 53 Sundays in a ‘Leap’ year?
1. 2/7
2. 1/7
3. 3/7
4. 4/7
Ans: The probability of any given day being a Sunday is 1/7. In a leap year, there are 366 days, of which 52 weeks and 2 days are not Sundays. Therefore, there are 52 Sundays in a leap year. The probability of getting an additional Sunday is the probability that February 1st falls on a Sunday, which is 1/7. Therefore, the probability of getting 53 Sundays in a leap year is (1/7) x 1 = 1/7.
30. Synthesis of the majority of lipids in a cellular system occurs in the
1. mitochondria
2. lysosomes
3. nucleus
4. endoplasmic reticulum
Ans: The majority of lipids are synthesized in the endoplasmic reticulum (ER) of the cell. The ER has two distinct regions: the smooth ER, which is involved in lipid synthesis and metabolism, and the rough ER, which is involved in protein synthesis and modification. The smooth ER contains enzymes that catalyze the synthesis of various types of lipids, including phospholipids, cholesterol, and triglycerides.
31. What will be the generation time of a culture with a specific growth rate constant of 0.01?
1. 1.155 h
2. 11.55 h
3. 0.693 min
4. 6.93 min
Ans: The generation time can be calculated using the formula: generation time = ln(2)/specific growth rate constant. Plugging in the given value of specific growth rate constant, we get: generation time = ln(2)/0.01 = 69.3 minutes ≈ 1.155 hours. Therefore, the answer is 1.155 h.
32. Ligands ‘A’ and ‘B’ bind to protein ‘P’ with dissociation constants of 1 nM and 100 nM, respectively. Which of the following true?
1. ‘A’ binds ‘P’ with more affinity
2. Dissociation constant is not related to affinity
3. ‘B’ binds ‘P’ with more affinity
4. Both ‘A’ and ‘B’ bind ‘P’ with equal affinity
Ans: The dissociation constant (Kd) is a measure of the affinity of a ligand for a protein. A smaller Kd indicates a higher affinity. Here, the Kd of ‘A’ is 1 nM and the Kd of ‘B’ is 100 nM. Therefore, ‘A’ binds ‘P’ with more affinity. The answer is option 1.
33. The Freund’s complete adjuvant is a mixture of
1. oil, water, dried bacterial spores
2. oil, water and dried Mycobacterium cells
3. amino acids, detergent and dried S. aureus cells
4. glucose, oil and dried E. coli cells
Ans: Freund’s complete adjuvant is a mixture of oil, water, and dried Mycobacterium cells. The answer is option 2.
34. The aluminium bronze alloy consists of copper and aluminium in the ratio of 10:1 by weight. If an object made of this alloy weighs 77 Kilograms (Kg), how many Kg of aluminium does it contain?
1. 70.7
2. 7.7
3. 7.0
4. 0.7
Ans: The ratio of copper to aluminium in the aluminium bronze alloy is 10:1 by weight. This means that for every 10 kg of copper, there is 1 kg of aluminium. If the total weight of the object made of this alloy is 77 kg, then the weight of aluminium in it is (1/10) x 77 = 7.7 kg. Therefore, the answer is 7.7.
35. Primary cilia biogenesis typically starts at the
1. S phase of the cell cycle
2. S and G2 phase of the cell cycle
3. G2 phase of the cell cycle
4. G1/G0 phase of the cell cycle
Ans: Primary cilia biogenesis typically starts at the G1/G0 phase of the cell cycle, when the centrosome migrates to the cell surface and nucleates the formation of the primary cilium. The answer is option 4.
36. Which of the following is NOT true ?
1. prokaryotes are unicellular organisms
2. eukaryotes can be either multicellular or unicellular organisms
3. prokaryotic cells lack nucleus whereas eukaryotic cells have a nucleus
4. eukaryotic cells are evolutionarily more ancient than prokaryotic cells
Ans: Eukaryotic cells are NOT evolutionarily more ancient than prokaryotic cells. Prokaryotic cells are considered to be more ancient than eukaryotic cells, as they are believed to have been the first form of life on Earth. The answer is option 4.
37. Trypsin cleaves a protein at the
1. C – terminus side of Arg/Lys residues
2. N – terminus side of Arg/Lys residues
3. C – terminus side of Val/lle residues
4. N- terminus side of Val/lle residues
Ans: Trypsin cleaves a protein at the C-terminus side of Arg/Lys residues. The answer is option 1.
38. A, B, C, and D are to be seated in a row, But C and D cannot be together. Also B cannot be at the third place.
Which of the following must be false?
1. A is at the second place
2. A is at the first place
3. A is at the third place
4. A is at the fourth place
Ans: If A is at the second place, then the possible seating arrangements are ACBD or ADCB, which violate the given conditions. If A is at the first place, then the possible seating arrangements are BCDA or BDCA, which satisfy the conditions. If A is at the third place, then the possible seating arrangements are BACD, BDAC, CDAB, or CDBA, of which only BACD and BDAC satisfy the conditions. If A is at the fourth place, then the possible seating arrangements are BCAD or BDAC, which satisfy the conditions. Therefore, the only option that must be false is option 1. The answer is option 1.
39. Single-stranded DNA can be separated from double-stranded DNA efficiently using
1. hydrophobic interaction chromatography
2. RP-HPLC
3. hydroxyapatite chromatography
4. urea PAGE
Ans: Single-stranded DNA can be separated from double-stranded DNA efficiently using denaturing gel electrophoreses, such as urea PAGE (polyacrylamide gel electrophoresis) or agarose gel electrophoresis in the presence of denaturants. The answer is option 4.
40. Eight 3rd year students can finish an experiment in 15 days and eighteen 1st year students can complete the same experiment in 10 days. If all these students work together, in how many days will the experiment get completed?
1.6.67
2.7.67
3.6.33
4.6.00
Ans: Let's assume that the experiment requires a total of "work units" to be completed. We can use the following formula:
(number of students) × (number of days) = (total work units)
Let E be the number of work units required to complete the experiment. Then we have:
(8 × 15) = E (for 3rd year students)
(18 × 10) = E (for 1st year students)
Simplifying these equations, we get:
E = 120 (for 3rd year students)
E = 180 (for 1st year students)
Now, let's assume that x is the number of days required to complete the experiment when both groups work together. We can use the following equation, based on the idea that the total work done by both groups together equals the total work required to complete the experiment:
(8 × 15 + 18 × 10) × x = E
Substituting the values of E, we get:
(8 × 15 + 18 × 10) × x = 180
Simplifying, we get:
x = 6.67
Therefore, the experiment will be completed in approximately 6.67 days when both groups work together. So the answer is option 1: 6.67.
41. The temperature of media post sterilization drops from 100°C to 60°C in 40 min by simply keeping it on the lab bench and allowing slow atmospheric cooling to take place at an ambient temperature of 20°C. In the next 40 min, the approximate temperature (°C) of the media would be around
1.40°C
2. 20°C
3.50°C
4. 30°C
Ans: The media cools down from 100°C to 60°C in 40 minutes on the lab bench, and then it continues to cool for another 40 minutes. Using Newton's Law of Cooling, we can estimate that the approximate temperature of the media after 80 minutes of cooling is around 30°C.
42. Which of the following is a non-reducing sugar?
1. fructose
2. galactose
3. ribose
4. sucrose
Ans: sucrose is a non-reducing sugar because it does not have a free aldehyde or ketone group that can be oxidized.
43. Denaturation of DNA is a
1. temperature-independent process
2. linear process
3. cooperative phenomenon
4. neither linear nor a cooperative process
Ans: Denaturation of DNA is neither linear nor a cooperative process. It is a complex process that involves the disruption of the hydrogen bonds, base stacking interactions, and hydrophobic interactions that hold the double helix structure together.
44. The egg white protein, ovalbumin, is denatured in a hard-boiled egg. Which of the following i least
affected?
1. quaternary structure of ovalbumin
2. primary structure of ovalbumin
3. secondary structure of ovalbumin
4. tertiary structure of ovalbumin
Ans: The quaternary structure of ovalbumin is least affected by denaturation in a hard-boiled egg. This is because the quaternary structure is formed by the association of multiple protein subunits and is stabilized by non-covalent interactions, such as hydrogen bonds, ionic bonds, and hydrophobic interactions.
45. Enzymes bind their substrates via
1. hydrogen bonds
2. hydrophobic interactions
3. all of the given options are correct
4. shape complementarity
Ans: Enzymes bind their substrates via shape complementarity. The active site of an enzyme is a specific 3-dimensional shape that fits the shape of the substrate molecule, allowing for precise binding and catalysis.
46. Nucleic acid structures are stabilized by
1. hydrophilic interactions
2. covalent interactions
3. covalent and hydrophilic interactions
4. hydrophobic interactions and hydrogen bonding
Ans: Nucleic acid structures are stabilized by a combination of hydrophobic interactions and hydrogen bonding. Hydrogen bonding is responsible for base pairing between complementary nucleotides, while hydrophobic interactions stabilize the interior of the double helix.
47. How many peptide fragments can be generated from the complete digestion of the polypeptide
AGRCDKCQANRSLMNF with trypsin?
1.6
2.4
3.3
4.2
Ans: Four peptide fragments can be generated from the complete digestion of the polypeptide AGRCDKCQANRSLMNF with trypsin. Trypsin cleaves peptide bonds on the C-terminal side of arginine (R) and lysine (K) residues, so the peptide will be cleaved into the following fragments: AGR, CDK, CQANR, and SLMNF.
48. While working in the lab, you forgot to keep enzymes back in the fridge. Which of the following enzyme will be least affected on being left outside at room temperature?
1. Topoisomerase
2. Taq DNA polymerase
3. DNA ligase
4. BamHI restriction enzyme
Ans: Topoisomerase is least affected by being left outside at room temperature. This is because topoisomerase is a stable enzyme that is able to function at a wide range of temperatures and pH values.
49. The genetic codon is a triplet and there are 64 codons. How many codons would be possible if the codon is a doublet?
1. 24
2. 16
3. 8
4. 64
Ans: If the genetic codon is a doublet (i.e., consists of two nucleotides), there would be 16 possible codons. This is because there are 4 nucleotides (A, T, C, and G) in DNA, and a doublet can combine any two of these nucleotides, resulting in 4 x 4 = 16 possible combinations.
50. In the first-semester course work at the Biotech Institute, 50 students signed up for both Genetics and Statistics and 90 students signed up for either Genetics or Statistics. If 25 students are taking Genetics but are not taking Statistics, how many students are taking Statistics but not taking Genetics?
1.65
2.25
3.15
4.35
Ans: 35 students are taking Statistics but not taking Genetics. To see why, we can use the principle of inclusion-exclusion. First, we know that a total of 140 students signed up for Genetics and/or Statistics (50 students signed up for both + 25 students taking Genetics only + 90 students taking at least one of the two courses). We also know that 25 students are taking Genetics but not taking Statistics. So, the number of students taking Statistics (with or without Genetics) is 140 - 25 = 115. Finally, we can subtract the number of students taking both courses (50) and the number of students taking Genetics only (25) to get the number of students taking Statistics only: 115 - 50 - 25 = 40. Therefore, 35 students are taking Statistics but not taking Genetics.
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