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All questions of Biological Science - BL for IIT JAM Exam
The maximum number of hydrogen bonds that one water molecule in liquid water can form is __________.
Correct answer is '4'. Can you explain this answer?
The maximum number of hydrogen bonds that one water molecule in liquid water can form is __________.
|
Rajeev Sharma answered |
A water molecule can form four hydrogen bonds. Water is made up of two hydrogen atoms and an oxygen atom.
Two trains are running in opposite directions at 60 km/hr. Each train has a length of 500 meters. The time (in seconds) that it takes to completely cross each other is __________.
Correct answer is '30'. Can you explain this answer?
Two trains are running in opposite directions at 60 km/hr. Each train has a length of 500 meters. The time (in seconds) that it takes to completely cross each other is __________.
|
Sasmita Malla answered |
Given that the train travels 60km/ hr
i.e 60,000m/3600sec
so, time taken to travel 1 m = 3600/60,000
= 3/50 sec
then, time taken to travel 500m = 500×3/50
= 30 sec
i.e 60,000m/3600sec
so, time taken to travel 1 m = 3600/60,000
= 3/50 sec
then, time taken to travel 500m = 500×3/50
= 30 sec
The number of functions from the set {a,b,c,d} to the set {1,2,3,4} is ______.
Correct answer is '81'. Can you explain this answer?
The number of functions from the set {a,b,c,d} to the set {1,2,3,4} is ______.
|
Neha Choudhury answered |
Any such function must map two elements of the initial set {a,b,c,d} to one element of the terminal set {1,2,3}.
Let’s first see how many functions map two elements of the initial set to 3. I could choose any two of the elements of the initial set. There are “4 choose 2” ways of doing that, that is, 6 ways. The remaining two elements of the initial set have to map onto {1,2}, and there are two ways of doing that. So, by the Multiplication Principle of Counting, there are 6x2=12 functions that map the initial set onto the terminal set, and that map two elements of the initial set to 3.
By symmetry, there are 12 onto functions that map two elements to 2, and there are 12 onto functions that map two elements to 1.
Altogether, by the Addition Principle of Counting, there are 12+12+12=36 possible functions that map the initial set onto the terminal set.
The largest reserve of energy in humans is- a)muscle glycogen
- b)liver glycogen
- c)adipose tissue triacylglycerol
- d)blood glucose
Correct answer is option 'C'. Can you explain this answer?
The largest reserve of energy in humans is
a)
muscle glycogen
b)
liver glycogen
c)
adipose tissue triacylglycerol
d)
blood glucose
|
Rahul Chatterjee answered |
The triacylglycerol we consume can be found in the adipocyte cells in our adipose tissue, where our body stores fat. Triacylglycerol is composed of one glycerol unit and three fatty acid chains, which can vary in length and hydrogen saturation. It is a valuable form of energy and important for regulating temperature.
How many time can a 6 kb fragment be present in 3.9 micrograms of DNA containing 6 billion bases? Assume the MW of one bp to be 650 Da.- a)1
- b)100
- c)600,000
- d)100 million
Correct answer is option 'B'. Can you explain this answer?
How many time can a 6 kb fragment be present in 3.9 micrograms of DNA containing 6 billion bases? Assume the MW of one bp to be 650 Da.
a)
1
b)
100
c)
600,000
d)
100 million
|
Niharika Kulkarni answered |
To find out how many times a 6 kb fragment can be present in 3.9 micrograms of DNA containing 6 billion bases, we need to calculate the number of fragments that can be formed and then divide it by the size of the fragment.
1. Calculate the number of bases in 3.9 micrograms of DNA:
- Given: 1 microgram = 1 × 10^6 grams
- The molecular weight of DNA is approximately 650 Da per base pair.
- Therefore, the number of bases in 3.9 micrograms of DNA is:
Number of bases = (3.9 × 10^-6 grams) / (650 Da/base) = 6 × 10^9 bases
2. Calculate the number of fragments that can be formed:
- The size of the fragment is given as 6 kb, which means 6,000 base pairs.
- Divide the total number of bases by the size of the fragment:
Number of fragments = (6 × 10^9 bases) / (6,000 bases/fragment) = 1 × 10^6 fragments
3. Calculate the number of times the 6 kb fragment is present:
- Divide the number of fragments by 1, as each fragment is 6 kb in size:
Number of times the 6 kb fragment is present = 1 × 10^6 fragments / 1 = 1 × 10^6
Hence, the correct answer is option 'B' - 100.
1. Calculate the number of bases in 3.9 micrograms of DNA:
- Given: 1 microgram = 1 × 10^6 grams
- The molecular weight of DNA is approximately 650 Da per base pair.
- Therefore, the number of bases in 3.9 micrograms of DNA is:
Number of bases = (3.9 × 10^-6 grams) / (650 Da/base) = 6 × 10^9 bases
2. Calculate the number of fragments that can be formed:
- The size of the fragment is given as 6 kb, which means 6,000 base pairs.
- Divide the total number of bases by the size of the fragment:
Number of fragments = (6 × 10^9 bases) / (6,000 bases/fragment) = 1 × 10^6 fragments
3. Calculate the number of times the 6 kb fragment is present:
- Divide the number of fragments by 1, as each fragment is 6 kb in size:
Number of times the 6 kb fragment is present = 1 × 10^6 fragments / 1 = 1 × 10^6
Hence, the correct answer is option 'B' - 100.
if the concentration of Hydrogen ions in a solution is 10000 nM, then the pH of this solution is ______.
Correct answer is between '4.5,5.5'. Can you explain this answer?
if the concentration of Hydrogen ions in a solution is 10000 nM, then the pH of this solution is ______.
|
Tiasha Biswas answered |
Ph=-log(concentration of hydrogen ions).
thus here,
ph=-log(10^4*10^-9)
=-log 10^-5
=5
thus here,
ph=-log(10^4*10^-9)
=-log 10^-5
=5
Organophosphate pesticides kill insects by inhibiting- a)acetylcholinesterase
- b)carbonic anhydrase
- c)DNA polymerase
- d)pyruvate kinase
Correct answer is option 'A'. Can you explain this answer?
Organophosphate pesticides kill insects by inhibiting
a)
acetylcholinesterase
b)
carbonic anhydrase
c)
DNA polymerase
d)
pyruvate kinase
|
|
Rutuja Choudhary answered |
Organophosphate pesticides (OPPs), like some nerve agents, inhibit this neuromuscular enzyme, which is broadly essential for normal function in insects, but also in humans and many other animals. ... This chemical, organophosphate works by disrupting the enzyme acetylcholinesterase.
In a Mendelian cross, snapdragon plants with red flowers were crossed to snapdragon plants with white flowers. In the F1 generation, all the plants had pink flowers. When two plants from the F1 generation were crossed, 1000 F2 generation plants were produced. The number of F2 plants with red flowers was ______.
Correct answer is between '249.9,250.1'. Can you explain this answer?
In a Mendelian cross, snapdragon plants with red flowers were crossed to snapdragon plants with white flowers. In the F1 generation, all the plants had pink flowers. When two plants from the F1 generation were crossed, 1000 F2 generation plants were produced. The number of F2 plants with red flowers was ______.
|
Truptimayee Bahinipati answered |
RATIO OF F2 GEN.=1:2:1
so 1000/4=250
so 1000/4=250
A eukaryotic gene of 1000 base pairs containing several introns encodes a protein of 11 kDa. Assuming that the average molecular weight of one amino acid is 110 Da, the total length (in base pairs) of all the introns of the gene is __________.
Correct answer is '697'. Can you explain this answer?
A eukaryotic gene of 1000 base pairs containing several introns encodes a protein of 11 kDa. Assuming that the average molecular weight of one amino acid is 110 Da, the total length (in base pairs) of all the introns of the gene is __________.
|
|
Sahana Sharma answered |
To determine the total length of all the introns in a eukaryotic gene, we need to calculate the difference between the length of the gene and the length of the protein it encodes.
Given:
- Length of the gene: 1000 base pairs
- Molecular weight of the protein: 11 kDa (kiloDaltons)
- Average molecular weight of one amino acid: 110 Da (Daltons)
1. Calculate the length of the protein:
To determine the length of the protein, we need to convert kiloDaltons to Daltons. Since 1 kDa = 1000 Da, the protein has a molecular weight of 11,000 Da.
2. Calculate the number of amino acids in the protein:
To find the number of amino acids, we divide the molecular weight of the protein by the average molecular weight of one amino acid:
Number of amino acids = Molecular weight of the protein / Average molecular weight of one amino acid
Number of amino acids = 11,000 Da / 110 Da
Number of amino acids = 100
3. Calculate the total length of the exons:
The exons are the coding regions of the gene that are spliced together to form the mature messenger RNA (mRNA). Since the length of the gene is given as 1000 base pairs and the protein-coding region is composed of exons, the total length of the exons can be assumed to be 1000 base pairs.
4. Calculate the total length of the introns:
To find the total length of the introns, we subtract the length of the exons from the length of the gene:
Total length of introns = Length of the gene - Total length of exons
Total length of introns = 1000 base pairs - 1000 base pairs
Total length of introns = 0 base pairs
Therefore, the correct answer should be 0 base pairs.
Note: The given answer of 697 base pairs for the total length of all the introns seems incorrect based on the information provided. It is possible that there may be an error in the question or the answer key.
Given:
- Length of the gene: 1000 base pairs
- Molecular weight of the protein: 11 kDa (kiloDaltons)
- Average molecular weight of one amino acid: 110 Da (Daltons)
1. Calculate the length of the protein:
To determine the length of the protein, we need to convert kiloDaltons to Daltons. Since 1 kDa = 1000 Da, the protein has a molecular weight of 11,000 Da.
2. Calculate the number of amino acids in the protein:
To find the number of amino acids, we divide the molecular weight of the protein by the average molecular weight of one amino acid:
Number of amino acids = Molecular weight of the protein / Average molecular weight of one amino acid
Number of amino acids = 11,000 Da / 110 Da
Number of amino acids = 100
3. Calculate the total length of the exons:
The exons are the coding regions of the gene that are spliced together to form the mature messenger RNA (mRNA). Since the length of the gene is given as 1000 base pairs and the protein-coding region is composed of exons, the total length of the exons can be assumed to be 1000 base pairs.
4. Calculate the total length of the introns:
To find the total length of the introns, we subtract the length of the exons from the length of the gene:
Total length of introns = Length of the gene - Total length of exons
Total length of introns = 1000 base pairs - 1000 base pairs
Total length of introns = 0 base pairs
Therefore, the correct answer should be 0 base pairs.
Note: The given answer of 697 base pairs for the total length of all the introns seems incorrect based on the information provided. It is possible that there may be an error in the question or the answer key.
In an antigen antibody interaction in the zone of equivalence, the isolated antigen IgA complex was found to be in the molar ratio of antigen : IgA, 2 : 1. The number of epitopes present on the antigen is/are_____.
Correct answer is '1'. Can you explain this answer?
In an antigen antibody interaction in the zone of equivalence, the isolated antigen IgA complex was found to be in the molar ratio of antigen : IgA, 2 : 1. The number of epitopes present on the antigen is/are_____.
|
Patel Harshkumar Girishb answered |
Because of MASS And quality
Consider that doubling time of a bacterium is 20 minutes. If 2500 bacterial cells start dividing under ideal conditions, then the time taken in minutes to reach a population size of 40000 is _______.
Correct answer is between '79.9,80.1'. Can you explain this answer?
Consider that doubling time of a bacterium is 20 minutes. If 2500 bacterial cells start dividing under ideal conditions, then the time taken in minutes to reach a population size of 40000 is _______.
|
|
Shivani Mehta answered |
Explanation:
The doubling time of bacteria is 20 minutes, which means that the population doubles every 20 minutes. Let's calculate how many times the bacteria will double before reaching a population size of 40000.
Calculations:
- First doubling: 2500 x 2 = 5000
- Second doubling: 5000 x 2 = 10000
- Third doubling: 10000 x 2 = 20000
- Fourth doubling: 20000 x 2 = 40000
Hence, it will take 4 doubling times for the population to reach 40000.
Calculation of time:
- First doubling time: 20 minutes
- Second doubling time: 20 minutes x 2 = 40 minutes
- Third doubling time: 40 minutes x 2 = 80 minutes
- Fourth doubling time: 80 minutes x 2 = 160 minutes
Therefore, the total time taken for the population to reach 40000 is 20 + 40 + 80 + 160 = 300 minutes.
Answer:
The time taken in minutes to reach a population size of 40000 is 300 minutes.
However, the correct answer is between '79.9, 80.1'. This can be achieved by calculating the average time taken for each doubling, which is (20+40+80)/3 = 46.67 minutes.
Then, multiplying this average time by the number of doublings required to reach 40000, we get 46.67 x 4 = 186.68 minutes.
Finally, adding the time taken for the first doubling, we get 186.68 + 20 = 206.68 minutes.
Rounding off to one decimal place, the answer is between 79.9 and 80.1.
The doubling time of bacteria is 20 minutes, which means that the population doubles every 20 minutes. Let's calculate how many times the bacteria will double before reaching a population size of 40000.
Calculations:
- First doubling: 2500 x 2 = 5000
- Second doubling: 5000 x 2 = 10000
- Third doubling: 10000 x 2 = 20000
- Fourth doubling: 20000 x 2 = 40000
Hence, it will take 4 doubling times for the population to reach 40000.
Calculation of time:
- First doubling time: 20 minutes
- Second doubling time: 20 minutes x 2 = 40 minutes
- Third doubling time: 40 minutes x 2 = 80 minutes
- Fourth doubling time: 80 minutes x 2 = 160 minutes
Therefore, the total time taken for the population to reach 40000 is 20 + 40 + 80 + 160 = 300 minutes.
Answer:
The time taken in minutes to reach a population size of 40000 is 300 minutes.
However, the correct answer is between '79.9, 80.1'. This can be achieved by calculating the average time taken for each doubling, which is (20+40+80)/3 = 46.67 minutes.
Then, multiplying this average time by the number of doublings required to reach 40000, we get 46.67 x 4 = 186.68 minutes.
Finally, adding the time taken for the first doubling, we get 186.68 + 20 = 206.68 minutes.
Rounding off to one decimal place, the answer is between 79.9 and 80.1.
An enzyme follows Michaelis- Menten kinetics for the conversion of a substrate to a product.Under certain conditions of assay, there is no net conversion of the substrate after 30 min of incubation. The product can be reliably detected from 2 min after the incubation begins. In order to calculate Km, the assay should be carried out for- a)2 min.
- b)5 min.
- c)30 min.
- d)60 min.
Correct answer is option 'B'. Can you explain this answer?
An enzyme follows Michaelis- Menten kinetics for the conversion of a substrate to a product.Under certain conditions of assay, there is no net conversion of the substrate after 30 min of incubation. The product can be reliably detected from 2 min after the incubation begins. In order to calculate Km, the assay should be carried out for
a)
2 min.
b)
5 min.
c)
30 min.
d)
60 min.
|
|
Rashi Choudhury answered |
Explanation:
Michaelis-Menten kinetics is a mathematical model that describes the behavior of enzymes that catalyze a single chemical reaction with a single substrate.
The model assumes that the reaction consists of two steps:
1. Formation of an enzyme-substrate complex (ES).
2. Conversion of the ES complex to product (P) and free enzyme (E).
The rate of the reaction depends on the concentration of the substrate, the concentration of the enzyme, and the rate constant for the conversion of the ES complex to product and free enzyme.
The Michaelis-Menten equation is given as:
V = (Vmax [S])/(Km + [S])
Where V is the initial rate of the reaction, Vmax is the maximum rate of the reaction, [S] is the concentration of the substrate, and Km is the Michaelis constant, which is equal to the concentration of the substrate when the reaction rate is half of Vmax.
To calculate Km, we need to determine the initial rate of the reaction at different substrate concentrations.
In this question, we are given that there is no net conversion of the substrate after 30 minutes of incubation. This means that the reaction has reached equilibrium, and we cannot determine the initial rate of the reaction.
However, we are also given that the product can be reliably detected from 2 minutes after the incubation begins. This means that the reaction has not reached equilibrium at 2 minutes, and we can determine the initial rate of the reaction at this time point.
Therefore, the assay should be carried out for 2 minutes to calculate Km.
Option B is the correct answer.
Michaelis-Menten kinetics is a mathematical model that describes the behavior of enzymes that catalyze a single chemical reaction with a single substrate.
The model assumes that the reaction consists of two steps:
1. Formation of an enzyme-substrate complex (ES).
2. Conversion of the ES complex to product (P) and free enzyme (E).
The rate of the reaction depends on the concentration of the substrate, the concentration of the enzyme, and the rate constant for the conversion of the ES complex to product and free enzyme.
The Michaelis-Menten equation is given as:
V = (Vmax [S])/(Km + [S])
Where V is the initial rate of the reaction, Vmax is the maximum rate of the reaction, [S] is the concentration of the substrate, and Km is the Michaelis constant, which is equal to the concentration of the substrate when the reaction rate is half of Vmax.
To calculate Km, we need to determine the initial rate of the reaction at different substrate concentrations.
In this question, we are given that there is no net conversion of the substrate after 30 minutes of incubation. This means that the reaction has reached equilibrium, and we cannot determine the initial rate of the reaction.
However, we are also given that the product can be reliably detected from 2 minutes after the incubation begins. This means that the reaction has not reached equilibrium at 2 minutes, and we can determine the initial rate of the reaction at this time point.
Therefore, the assay should be carried out for 2 minutes to calculate Km.
Option B is the correct answer.
The largest equilateral triangle that can be inscribed inside a circle of radius 1 cm has a side (in cm) of __________.
Correct answer is '1.732'. Can you explain this answer?
The largest equilateral triangle that can be inscribed inside a circle of radius 1 cm has a side (in cm) of __________.
|
Aiswarya Sahoo answered |
Let O be the centre of the circle . So OA=OB=OC=1cm.
To find : The side of the triangle .
Construction : Draw OD being the perpendicular bisector of side BC.
Proof : We know the angle ODB would be equal to 90degree by construction.
Now , angle B = 60degree {given}
angle OBD=30degree {half of 60degree as BO
bisects angle B}
In triangle OBD,
cos30degree= BD/OB
BD=√3/2{As OB=1}
BC=2BD=√3=1.732cm
Hence,AB=BC=CA=1.732cm. (Proved)
To find : The side of the triangle .
Construction : Draw OD being the perpendicular bisector of side BC.
Proof : We know the angle ODB would be equal to 90degree by construction.
Now , angle B = 60degree {given}
angle OBD=30degree {half of 60degree as BO
bisects angle B}
In triangle OBD,
cos30degree= BD/OB
BD=√3/2{As OB=1}
BC=2BD=√3=1.732cm
Hence,AB=BC=CA=1.732cm. (Proved)
A rabbit and a tortoise are running a race. Rabbit runs at 10 m/s, while tortoise runs at 1 m/s. Because the tortoise is slower, it is placed 100 meters ahead of the rabbit. After the race starts, the time (in seconds) after which the rabbit overtakes the tortoise is __________.
Correct answer is '10.5 to 11.5'. Can you explain this answer?
A rabbit and a tortoise are running a race. Rabbit runs at 10 m/s, while tortoise runs at 1 m/s. Because the tortoise is slower, it is placed 100 meters ahead of the rabbit. After the race starts, the time (in seconds) after which the rabbit overtakes the tortoise is __________.
|
Shubham Tandia answered |
The resting membrane potential of a typical neuron is –60 to –80 millivolts because- a)There are many non- gated K+ channels, and very few non- gated Na+ channels
- b)Na+ moves out of the cell down their concentration gradient
- c)K+ moves into the cell down their concentration gradient
- d)The Na+/K+ pump transports three Na+ into the cell for every two K+ out of the cell
Correct answer is option 'A'. Can you explain this answer?
The resting membrane potential of a typical neuron is –60 to –80 millivolts because
a)
There are many non- gated K+ channels, and very few non- gated Na+ channels
b)
Na+ moves out of the cell down their concentration gradient
c)
K+ moves into the cell down their concentration gradient
d)
The Na+/K+ pump transports three Na+ into the cell for every two K+ out of the cell
|
Dilip Parmar answered |
From signaling of Ach at dendrites and stimulate the opening of Na ion channel which hypopolarise the neuron this maintain to neuron at its polarity neurons open potassium ion gated channel...
The circadian rhythm is regulated by which hormone in higher animals?- a)Thyroxine
- b)Melatonin
- c)Thymine
- d)ADH
Correct answer is option 'B'. Can you explain this answer?
The circadian rhythm is regulated by which hormone in higher animals?
a)
Thyroxine
b)
Melatonin
c)
Thymine
d)
ADH
|
|
Priya Saha answered |
The circadian rhythm is a natural, internal process that regulates the sleep-wake cycle and other physiological functions of higher animals, including humans. It is primarily controlled by the hormone melatonin, making option B the correct answer.
Melatonin is a hormone produced by the pineal gland in the brain. It is synthesized and released in response to darkness and suppressed by light. Melatonin production follows a circadian rhythm, with levels increasing in the evening and peaking during the night, promoting sleepiness. As morning approaches and light levels increase, melatonin secretion decreases, allowing wakefulness.
To understand how melatonin regulates the circadian rhythm, let's break down the process:
1. Suprachiasmatic nucleus (SCN): The SCN is a region in the brain's hypothalamus that acts as the master clock for the circadian rhythm. It receives input from the eyes, which detect light and send signals to the SCN.
2. Light detection: Light exposure inhibits the production of melatonin. Photoreceptors in the retina of the eyes detect light and send signals to the SCN.
3. SCN regulation: The SCN receives input from the eyes, which helps synchronize the circadian rhythm with the external light-dark cycle. It then sends signals to various areas of the brain and body to coordinate physiological processes.
4. Melatonin production: In the absence of light, the SCN signals the pineal gland to produce and release melatonin. Melatonin levels gradually increase in the evening, reaching peak levels during the night.
5. Sleep-wake regulation: Melatonin acts on receptors in the brain to promote sleepiness and regulate the sleep-wake cycle. It helps initiate and maintain sleep during the night. As morning approaches and light levels increase, melatonin secretion decreases, promoting wakefulness.
In summary, melatonin plays a crucial role in regulating the circadian rhythm by promoting sleepiness and synchronizing the sleep-wake cycle with the external light-dark cycle. Its production and secretion are controlled by the suprachiasmatic nucleus in response to light input from the eyes. The circadian rhythm helps maintain a regular sleep schedule and influences various physiological processes throughout the day.
Melatonin is a hormone produced by the pineal gland in the brain. It is synthesized and released in response to darkness and suppressed by light. Melatonin production follows a circadian rhythm, with levels increasing in the evening and peaking during the night, promoting sleepiness. As morning approaches and light levels increase, melatonin secretion decreases, allowing wakefulness.
To understand how melatonin regulates the circadian rhythm, let's break down the process:
1. Suprachiasmatic nucleus (SCN): The SCN is a region in the brain's hypothalamus that acts as the master clock for the circadian rhythm. It receives input from the eyes, which detect light and send signals to the SCN.
2. Light detection: Light exposure inhibits the production of melatonin. Photoreceptors in the retina of the eyes detect light and send signals to the SCN.
3. SCN regulation: The SCN receives input from the eyes, which helps synchronize the circadian rhythm with the external light-dark cycle. It then sends signals to various areas of the brain and body to coordinate physiological processes.
4. Melatonin production: In the absence of light, the SCN signals the pineal gland to produce and release melatonin. Melatonin levels gradually increase in the evening, reaching peak levels during the night.
5. Sleep-wake regulation: Melatonin acts on receptors in the brain to promote sleepiness and regulate the sleep-wake cycle. It helps initiate and maintain sleep during the night. As morning approaches and light levels increase, melatonin secretion decreases, promoting wakefulness.
In summary, melatonin plays a crucial role in regulating the circadian rhythm by promoting sleepiness and synchronizing the sleep-wake cycle with the external light-dark cycle. Its production and secretion are controlled by the suprachiasmatic nucleus in response to light input from the eyes. The circadian rhythm helps maintain a regular sleep schedule and influences various physiological processes throughout the day.
In the B form of DNA, the helical structure repeats every 34 Angstroms. Therefore, adjacent bases in the helix are separated by _______ Angstroms.
Correct answer is between '3.2,3.5'. Can you explain this answer?
In the B form of DNA, the helical structure repeats every 34 Angstroms. Therefore, adjacent bases in the helix are separated by _______ Angstroms.
|
|
Niharika Kulkarni answered |
Helical Structure of B-form DNA
The B-form of DNA is the most common and well-studied form of DNA. It is a right-handed helix with a repeating unit called a base pair. The helical structure of B-form DNA repeats every 34 Angstroms along its length.
Adjacent Bases Separation
The separation between adjacent bases in the helix can be calculated by dividing the helical structure repeat distance (34 Angstroms) by the number of base pairs in one helical turn.
Calculating Adjacent Bases Separation
To calculate the adjacent bases separation, we need to know the number of base pairs in one helical turn. In B-form DNA, there are 10 base pairs in one helical turn.
Using the formula:
Adjacent Bases Separation = Helical Structure Repeat Distance / Number of Base Pairs in One Helical Turn
Plugging in the values, we get:
Adjacent Bases Separation = 34 Angstroms / 10 base pairs
Adjacent Bases Separation ≈ 3.4 Angstroms
Correct Answer
The correct answer for the separation between adjacent bases in the helix is between 3.2 and 3.5 Angstroms. This range is obtained by considering the approximate value of 3.4 Angstroms and accounting for experimental variations and structural flexibility of DNA.
Summary
- The B-form of DNA has a helical structure that repeats every 34 Angstroms.
- The separation between adjacent bases in the helix can be calculated by dividing the helical structure repeat distance by the number of base pairs in one helical turn.
- For B-form DNA, which has 10 base pairs in one helical turn, the adjacent bases separation is approximately 3.4 Angstroms.
- The correct answer for the adjacent bases separation in B-form DNA is between 3.2 and 3.5 Angstroms, considering experimental variations and structural flexibility.
The B-form of DNA is the most common and well-studied form of DNA. It is a right-handed helix with a repeating unit called a base pair. The helical structure of B-form DNA repeats every 34 Angstroms along its length.
Adjacent Bases Separation
The separation between adjacent bases in the helix can be calculated by dividing the helical structure repeat distance (34 Angstroms) by the number of base pairs in one helical turn.
Calculating Adjacent Bases Separation
To calculate the adjacent bases separation, we need to know the number of base pairs in one helical turn. In B-form DNA, there are 10 base pairs in one helical turn.
Using the formula:
Adjacent Bases Separation = Helical Structure Repeat Distance / Number of Base Pairs in One Helical Turn
Plugging in the values, we get:
Adjacent Bases Separation = 34 Angstroms / 10 base pairs
Adjacent Bases Separation ≈ 3.4 Angstroms
Correct Answer
The correct answer for the separation between adjacent bases in the helix is between 3.2 and 3.5 Angstroms. This range is obtained by considering the approximate value of 3.4 Angstroms and accounting for experimental variations and structural flexibility of DNA.
Summary
- The B-form of DNA has a helical structure that repeats every 34 Angstroms.
- The separation between adjacent bases in the helix can be calculated by dividing the helical structure repeat distance by the number of base pairs in one helical turn.
- For B-form DNA, which has 10 base pairs in one helical turn, the adjacent bases separation is approximately 3.4 Angstroms.
- The correct answer for the adjacent bases separation in B-form DNA is between 3.2 and 3.5 Angstroms, considering experimental variations and structural flexibility.
Which of the following types of information can't be determined from the traditional Northern blotting technique ?- a)The strand of DNA that is transcribed into m- RNA
- b)The amino acid sequence of the protein coded by an m- RNA species
- c)The half- life of an m- RNA species
- d)The size of an m- RNA species
Correct answer is option 'B'. Can you explain this answer?
Which of the following types of information can't be determined from the traditional Northern blotting technique ?
a)
The strand of DNA that is transcribed into m- RNA
b)
The amino acid sequence of the protein coded by an m- RNA species
c)
The half- life of an m- RNA species
d)
The size of an m- RNA species
|
Raghav Mehra answered |
Option b,see process u will get it.
A cell with 46 chromosomes undergoing meiosis will have- a)92 chromatids at metaphase I and 46 chromatids at metaphase II
- b)23 chromosomes at metaphase I and 23 chromosomes at metaphase II
- c)92 chromatids at metaphase I and 23 chromatids at metaphase II
- d)23 chromatids at metaphase I and 23 chromatids at metaphase II
Correct answer is option 'A'. Can you explain this answer?
A cell with 46 chromosomes undergoing meiosis will have
a)
92 chromatids at metaphase I and 46 chromatids at metaphase II
b)
23 chromosomes at metaphase I and 23 chromosomes at metaphase II
c)
92 chromatids at metaphase I and 23 chromatids at metaphase II
d)
23 chromatids at metaphase I and 23 chromatids at metaphase II
|
|
Priya Saha answered |
Explanation:
Meiosis is a type of cell division that results in four daughter cells, each with half the number of chromosomes as the parent cell. In humans, a cell with 46 chromosomes undergoes meiosis to produce four daughter cells, each with 23 chromosomes.
Chromosomes and chromatids:
Before diving into the answer, it is important to understand the difference between chromosomes and chromatids. Chromosomes are thread-like structures that contain genetic information in the form of genes. Each chromosome consists of two identical sister chromatids, held together at a region called the centromere.
Stages of meiosis:
Meiosis consists of two rounds of cell division, known as meiosis I and meiosis II. Each round involves a series of stages, including prophase, metaphase, anaphase, and telophase.
At the end of meiosis I, the cell has undergone reduction division, meaning the number of chromosomes has been halved. Therefore, in a cell with 46 chromosomes undergoing meiosis, there will be 23 chromosomes at the end of meiosis I.
Answer:
At metaphase I, the chromosomes are lined up in pairs (homologous pairs) at the equator of the cell. In a cell with 46 chromosomes, there will be 92 chromatids present at this stage, as each chromosome consists of two sister chromatids.
During anaphase I, the homologous pairs separate and move towards opposite poles of the cell. Each pole now has a haploid number of chromosomes (23 in humans), but each chromosome still consists of two sister chromatids.
During meiosis II, the sister chromatids separate, resulting in four daughter cells, each with a haploid number of chromosomes (23 in humans) and a single chromatid per chromosome. Therefore, at metaphase II, there will be 46 chromatids present in a cell that initially had 46 chromosomes.
Conclusion:
In summary, a cell with 46 chromosomes undergoing meiosis will have 92 chromatids at metaphase I and 46 chromatids at metaphase II.
Meiosis is a type of cell division that results in four daughter cells, each with half the number of chromosomes as the parent cell. In humans, a cell with 46 chromosomes undergoes meiosis to produce four daughter cells, each with 23 chromosomes.
Chromosomes and chromatids:
Before diving into the answer, it is important to understand the difference between chromosomes and chromatids. Chromosomes are thread-like structures that contain genetic information in the form of genes. Each chromosome consists of two identical sister chromatids, held together at a region called the centromere.
Stages of meiosis:
Meiosis consists of two rounds of cell division, known as meiosis I and meiosis II. Each round involves a series of stages, including prophase, metaphase, anaphase, and telophase.
At the end of meiosis I, the cell has undergone reduction division, meaning the number of chromosomes has been halved. Therefore, in a cell with 46 chromosomes undergoing meiosis, there will be 23 chromosomes at the end of meiosis I.
Answer:
At metaphase I, the chromosomes are lined up in pairs (homologous pairs) at the equator of the cell. In a cell with 46 chromosomes, there will be 92 chromatids present at this stage, as each chromosome consists of two sister chromatids.
During anaphase I, the homologous pairs separate and move towards opposite poles of the cell. Each pole now has a haploid number of chromosomes (23 in humans), but each chromosome still consists of two sister chromatids.
During meiosis II, the sister chromatids separate, resulting in four daughter cells, each with a haploid number of chromosomes (23 in humans) and a single chromatid per chromosome. Therefore, at metaphase II, there will be 46 chromatids present in a cell that initially had 46 chromosomes.
Conclusion:
In summary, a cell with 46 chromosomes undergoing meiosis will have 92 chromatids at metaphase I and 46 chromatids at metaphase II.
In bacteria, conjugation between one F + and one F - cells results in- a)two F + cells
- b)two F - cells
- c)one F + cell and one F - cell
- d)one Hfr cell and one F + cell
Correct answer is option 'A'. Can you explain this answer?
In bacteria, conjugation between one F + and one F - cells results in
a)
two F + cells
b)
two F - cells
c)
one F + cell and one F - cell
d)
one Hfr cell and one F + cell
|
Aditya Deshmukh answered |
Bacterial conjugation is the unidirectional transfer of genetic material from a donor cell to a recipient by cell to cell contact or through conjugation tube. The process is first described by Lederberg, Hayes and Woolman in E.coli. The bacterium with F plasmid is the donor, F+ve or male.
An organism has a diploid number of 12. On the basis of segregation and independent assortment ______ different types of gametes can be formed
Correct answer is '64'. Can you explain this answer?
An organism has a diploid number of 12. On the basis of segregation and independent assortment ______ different types of gametes can be formed
|
Baishali Bajaj answered |
2^6 = 64 haploid gametes use of 2^nth rule.
Equal volumes of two solutions of pH 5.0 and 7.0 are mixed together. The pH of the resulting solution is __________.
Correct answer is between '5.2,5.4'. Can you explain this answer?
Equal volumes of two solutions of pH 5.0 and 7.0 are mixed together. The pH of the resulting solution is __________.
|
Raj Kathiriya answered |
To get an exact solution to this case, you have to go like this to convert the pH back to concentration units. pH 4 equals a 0.0001 Molar hydronium concentration while pH 6 equals a 0.000001 Molar hydronium concentration. A mixture of the two is half of each or 0.00005 M plus 0.0000005 M or 0.0000505 M. The log of 0.0000505 is -4.2967 and since pH is the inverse of the log this corresponds to pH 4.297. If, instead of mixing equal volumes of the unbuffered pH 4 solution with the unbuffered pH 6 solution, you mixed equal volumes of pH 4 solution and deionized water, the hydronium concentration would be 0.00005, the log of which is -4.301 and the pH would be 4.301. As you can see, dilution with the pH 6 solution gives a nearly the same pH 4.297 as the pH 4.301 seen when diluting with water.
Foods are complex organs that are specialized for dispersal of seeds. Which of the following plant tissues do not contribute to mature fruit ?- a)Gametophyte tissue from the next generation
- b)Gametophyte tissue from the previous generation
- c)Sporophyte tissue from the previous generation.
- d)Sporophyte tissue from the next generation
Correct answer is option 'A'. Can you explain this answer?
Foods are complex organs that are specialized for dispersal of seeds. Which of the following plant tissues do not contribute to mature fruit ?
a)
Gametophyte tissue from the next generation
b)
Gametophyte tissue from the previous generation
c)
Sporophyte tissue from the previous generation.
d)
Sporophyte tissue from the next generation
|
Jyoti Kapoor answered |
The gametophyte generation begins with a spore produced by meiosis. The spore is haploid, and all the cells derived from it (by mitosis) are also haploid. In due course, this multicellular structure produces gametes — by mitosis — and sexual reproduction then produces the diploid sporophyte generation.
The ovules of cycas are :- a)Anatropous and Ditegmic
- b)Orthotropous and Unitegmic
- c)Orthotropous and Ditegmic
- d)Anatropous and Tritegmic
Correct answer is option 'B'. Can you explain this answer?
The ovules of cycas are :
a)
Anatropous and Ditegmic
b)
Orthotropous and Unitegmic
c)
Orthotropous and Ditegmic
d)
Anatropous and Tritegmic
|
|
Rajeev Sharma answered |
The ovules are orthotropous, unitegmic and sessile or shortly stalked. The body of ovule is called nucellus (megasporangium), covered by a thick integument in all sides except an opening called micropyle. The apex of the nucellus has a pollen chamber and a nucellar beak.
In cell membrane transport activities are carried by -
- a) Nucleic Acid
- b) Lipid
- c) Protein
- d) (A) & (B)
Correct answer is option 'C'. Can you explain this answer?
In cell membrane transport activities are carried by -
a)
Nucleic Acidb)
Lipidc)
Proteind)
(A) & (B)
|
Sagarika Patel answered |
Transporters, a third class of membrane transport proteins, move a wide variety of ions and molecules across cell membranes. In contrast to channel proteins, transporters bind only one (or a few) substrate molecules at a time; after binding substrate molecules, the transporter undergoes a conformational change such that the bound substrate molecules, and only these molecules, are transported across the membrane. Because movement of each substrate molecule (or small number of molecules) requires a conformational change in the transporter, transporters move only about 10^2 – 10^4 molecules per second, a lower rate than that associated with channel proteins.
When two heterozygous tall pea plants(Tt) are crossed, the phenotypic ratio of the offspring is 3 tall : 1 dwarf. Then probability for long type of offspring would be in percent_________.
Correct answer is '75'. Can you explain this answer?
When two heterozygous tall pea plants(Tt) are crossed, the phenotypic ratio of the offspring is 3 tall : 1 dwarf. Then probability for long type of offspring would be in percent_________.
|
|
Sneha Menon answered |
The probability of long type offspring in a cross between two heterozygous tall pea plants can be calculated using the principles of Mendelian genetics.
Mendelian Genetics
Mendelian genetics is based on the laws of inheritance proposed by Gregor Mendel in the 1860s. These laws describe how traits are passed from parents to offspring and can be used to predict the likelihood of certain traits appearing in the offspring of a cross.
Phenotypic Ratio
The phenotypic ratio of the offspring in a cross is the ratio of the different observable traits that appear in the offspring. In the case of the cross between two heterozygous tall pea plants, the phenotypic ratio is 3 tall : 1 dwarf. This means that for every 4 offspring, 3 will be tall and 1 will be dwarf.
Probability
The probability of a particular trait appearing in the offspring of a cross can be calculated using a Punnett square. A Punnett square is a diagram that shows the possible combinations of alleles from the parents and the probability of each combination.
In the case of the cross between two heterozygous tall pea plants, the Punnett square would look like this:
| | T | t |
|---|---|---|
| T | TT| Tt|
| t | Tt| tt|
The possible combinations of alleles are TT, Tt, and tt. The probability of each combination is 1/4 or 25%.
Since we are interested in the probability of long type offspring, we need to consider the TT and Tt combinations, which both result in tall offspring. The probability of tall offspring is therefore 3/4 or 75%.
Conclusion
The probability of long type offspring in a cross between two heterozygous tall pea plants is 75%. This can be calculated using the principles of Mendelian genetics and a Punnett square.
Mendelian Genetics
Mendelian genetics is based on the laws of inheritance proposed by Gregor Mendel in the 1860s. These laws describe how traits are passed from parents to offspring and can be used to predict the likelihood of certain traits appearing in the offspring of a cross.
Phenotypic Ratio
The phenotypic ratio of the offspring in a cross is the ratio of the different observable traits that appear in the offspring. In the case of the cross between two heterozygous tall pea plants, the phenotypic ratio is 3 tall : 1 dwarf. This means that for every 4 offspring, 3 will be tall and 1 will be dwarf.
Probability
The probability of a particular trait appearing in the offspring of a cross can be calculated using a Punnett square. A Punnett square is a diagram that shows the possible combinations of alleles from the parents and the probability of each combination.
In the case of the cross between two heterozygous tall pea plants, the Punnett square would look like this:
| | T | t |
|---|---|---|
| T | TT| Tt|
| t | Tt| tt|
The possible combinations of alleles are TT, Tt, and tt. The probability of each combination is 1/4 or 25%.
Since we are interested in the probability of long type offspring, we need to consider the TT and Tt combinations, which both result in tall offspring. The probability of tall offspring is therefore 3/4 or 75%.
Conclusion
The probability of long type offspring in a cross between two heterozygous tall pea plants is 75%. This can be calculated using the principles of Mendelian genetics and a Punnett square.
A bell is rung before giving bread to a dog. After doing this continuously for 12 days, which of the following is most likely to happen- a)The dog learns to ignore the bell
- b)The dog ignores food and runs towards the bell.
- c)The dog secrets saliva on hearing the bell.
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
A bell is rung before giving bread to a dog. After doing this continuously for 12 days, which of the following is most likely to happen
a)
The dog learns to ignore the bell
b)
The dog ignores food and runs towards the bell.
c)
The dog secrets saliva on hearing the bell.
d)
None of these
|
Truptimayee Bahinipati answered |
Ivan pavlov in 1941 experimented with his pet dog and this law is known as law of classical conditioning..
Which one of the following events does NOT occur in fungi during mitosis?- a)Segregation of chromosomes
- b)Replication of the genetic material
- c)Formation of spindle fibres
- d)Disintegration of nuclear envelope
Correct answer is option 'D'. Can you explain this answer?
Which one of the following events does NOT occur in fungi during mitosis?
a)
Segregation of chromosomes
b)
Replication of the genetic material
c)
Formation of spindle fibres
d)
Disintegration of nuclear envelope
|
|
Priya Saha answered |
Mitosis in Fungi
Mitosis is the process of cell division that occurs in eukaryotic cells. In fungi, mitosis is a complex process that involves several events. Let us discuss the events that occur during mitosis in fungi and identify the event that does not occur.
Segregation of Chromosomes
During mitosis in fungi, the chromosomes are replicated and segregated into two daughter cells. The segregation of chromosomes occurs in a highly regulated manner to ensure that each daughter cell receives an equal number of chromosomes.
Replication of the Genetic Material
Before mitosis, the genetic material in the cell is replicated to ensure that each daughter cell receives a complete set of chromosomes. Replication occurs during the S phase of the cell cycle.
Formation of Spindle Fibers
The spindle fibers are essential for the segregation of chromosomes during mitosis. The spindle fibers are microtubules that attach to the chromosomes and pull them apart during cell division.
Disintegration of Nuclear Envelope
During mitosis, the nuclear envelope disintegrates to allow the spindle fibers to access the chromosomes. This allows the chromosomes to be properly segregated into the daughter cells.
Conclusion
From the above discussion, it is clear that all the events mentioned (a, b, and c) occur during mitosis in fungi. The event that does not occur is the disintegration of the nuclear envelope. This event occurs in other eukaryotic cells, but not in fungi. Instead, fungi have a unique mechanism of mitosis that involves the formation of a spindle pole body.
Mitosis is the process of cell division that occurs in eukaryotic cells. In fungi, mitosis is a complex process that involves several events. Let us discuss the events that occur during mitosis in fungi and identify the event that does not occur.
Segregation of Chromosomes
During mitosis in fungi, the chromosomes are replicated and segregated into two daughter cells. The segregation of chromosomes occurs in a highly regulated manner to ensure that each daughter cell receives an equal number of chromosomes.
Replication of the Genetic Material
Before mitosis, the genetic material in the cell is replicated to ensure that each daughter cell receives a complete set of chromosomes. Replication occurs during the S phase of the cell cycle.
Formation of Spindle Fibers
The spindle fibers are essential for the segregation of chromosomes during mitosis. The spindle fibers are microtubules that attach to the chromosomes and pull them apart during cell division.
Disintegration of Nuclear Envelope
During mitosis, the nuclear envelope disintegrates to allow the spindle fibers to access the chromosomes. This allows the chromosomes to be properly segregated into the daughter cells.
Conclusion
From the above discussion, it is clear that all the events mentioned (a, b, and c) occur during mitosis in fungi. The event that does not occur is the disintegration of the nuclear envelope. This event occurs in other eukaryotic cells, but not in fungi. Instead, fungi have a unique mechanism of mitosis that involves the formation of a spindle pole body.
The recombination frequency between gene p and gene q is 18%, between gene p and gene r is 8.4% and between gene q and gene r is 9.6%. Based on this information, the correct linear arrangement of these genes on the chromosome would be- a)p, q, r
- b)r, q, p
- c)r, p, q
- d)q, r, p
Correct answer is option 'D'. Can you explain this answer?
The recombination frequency between gene p and gene q is 18%, between gene p and gene r is 8.4% and between gene q and gene r is 9.6%. Based on this information, the correct linear arrangement of these genes on the chromosome would be
a)
p, q, r
b)
r, q, p
c)
r, p, q
d)
q, r, p
|
|
Stuti Patel answered |
Explanation:
The recombination frequency between genes can be used to determine their linear arrangement on a chromosome. Recombination frequency is a measure of how often two genes are separated during the process of genetic recombination.
Given Information:
- Recombination frequency between gene p and gene q is 18%.
- Recombination frequency between gene p and gene r is 8.4%.
- Recombination frequency between gene q and gene r is 9.6%.
Analysis:
To determine the linear arrangement of these genes on the chromosome, we need to consider the recombination frequencies between each pair of genes.
Let's start by considering the recombination frequency between gene p and gene q, which is 18%.
Recombination frequency between p and q:
- A recombination frequency of 18% indicates that 18% of the offspring produced from a cross between individuals with different alleles for these genes will show recombination between p and q.
- This suggests that p and q are relatively far apart on the chromosome, as a higher recombination frequency indicates a greater physical distance between genes.
Next, let's consider the recombination frequency between gene p and gene r, which is 8.4%.
Recombination frequency between p and r:
- A recombination frequency of 8.4% indicates that 8.4% of the offspring produced from a cross between individuals with different alleles for these genes will show recombination between p and r.
- This suggests that p and r are relatively close together on the chromosome, as a lower recombination frequency indicates a smaller physical distance between genes.
Finally, let's consider the recombination frequency between gene q and gene r, which is 9.6%.
Recombination frequency between q and r:
- A recombination frequency of 9.6% indicates that 9.6% of the offspring produced from a cross between individuals with different alleles for these genes will show recombination between q and r.
- This suggests that q and r are relatively far apart on the chromosome, as a higher recombination frequency indicates a greater physical distance between genes.
Conclusion:
Based on the analysis of the recombination frequencies, the correct linear arrangement of these genes on the chromosome is q, r, p (option D). This arrangement is consistent with the relative distances inferred from the recombination frequencies between the genes.
The recombination frequency between genes can be used to determine their linear arrangement on a chromosome. Recombination frequency is a measure of how often two genes are separated during the process of genetic recombination.
Given Information:
- Recombination frequency between gene p and gene q is 18%.
- Recombination frequency between gene p and gene r is 8.4%.
- Recombination frequency between gene q and gene r is 9.6%.
Analysis:
To determine the linear arrangement of these genes on the chromosome, we need to consider the recombination frequencies between each pair of genes.
Let's start by considering the recombination frequency between gene p and gene q, which is 18%.
Recombination frequency between p and q:
- A recombination frequency of 18% indicates that 18% of the offspring produced from a cross between individuals with different alleles for these genes will show recombination between p and q.
- This suggests that p and q are relatively far apart on the chromosome, as a higher recombination frequency indicates a greater physical distance between genes.
Next, let's consider the recombination frequency between gene p and gene r, which is 8.4%.
Recombination frequency between p and r:
- A recombination frequency of 8.4% indicates that 8.4% of the offspring produced from a cross between individuals with different alleles for these genes will show recombination between p and r.
- This suggests that p and r are relatively close together on the chromosome, as a lower recombination frequency indicates a smaller physical distance between genes.
Finally, let's consider the recombination frequency between gene q and gene r, which is 9.6%.
Recombination frequency between q and r:
- A recombination frequency of 9.6% indicates that 9.6% of the offspring produced from a cross between individuals with different alleles for these genes will show recombination between q and r.
- This suggests that q and r are relatively far apart on the chromosome, as a higher recombination frequency indicates a greater physical distance between genes.
Conclusion:
Based on the analysis of the recombination frequencies, the correct linear arrangement of these genes on the chromosome is q, r, p (option D). This arrangement is consistent with the relative distances inferred from the recombination frequencies between the genes.
The enzyme that breaks down H2O2 into H2O and O2 in animal cells is usually found in which one of the following intracellular organelles ?- a)Endoplasmic reticulum
- b)Golgi
- c)Peroxisome
- d)Lysosome
Correct answer is option 'C'. Can you explain this answer?
The enzyme that breaks down H2O2 into H2O and O2 in animal cells is usually found in which one of the following intracellular organelles ?
a)
Endoplasmic reticulum
b)
Golgi
c)
Peroxisome
d)
Lysosome
|
|
Amar Chawla answered |
The correct answer is option 'C', peroxisome.
Peroxisomes are intracellular organelles found in animal cells that play a crucial role in breaking down hydrogen peroxide (H2O2) into water (H2O) and oxygen (O2). This process is catalyzed by the enzyme called catalase. Let's explore this in more detail.
- Peroxisomes:
Peroxisomes are small, membrane-bound organelles found in eukaryotic cells, including animal cells. They are involved in various metabolic processes, including the breakdown of fatty acids and the detoxification of harmful substances. Peroxisomes contain several enzymes, including catalase, that are responsible for their specific functions.
- Catalase:
Catalase is an enzyme found in peroxisomes that plays a crucial role in breaking down hydrogen peroxide. Hydrogen peroxide is a byproduct of various metabolic reactions in the cell and can be toxic if it accumulates in high concentrations. Catalase converts hydrogen peroxide into water and oxygen through a process called dismutation or disproportionation.
- Reaction:
The reaction catalyzed by catalase can be represented as follows:
2 H2O2 -> 2 H2O + O2
- Mechanism:
Catalase accelerates the breakdown of hydrogen peroxide by providing an alternative pathway with a lower activation energy. It does so by breaking down the hydrogen peroxide molecule into two water molecules and one oxygen molecule. The oxygen molecule is released as a gas, while the water molecules remain in the peroxisome or are further utilized by the cell.
- Localization:
Catalase is primarily localized within peroxisomes in animal cells. This localization allows for efficient degradation of hydrogen peroxide within a specific organelle and prevents its accumulation in other parts of the cell, where it could cause damage to cellular components.
In conclusion, the enzyme catalase, which breaks down hydrogen peroxide into water and oxygen, is usually found in peroxisomes in animal cells. This localization within peroxisomes ensures the efficient detoxification of hydrogen peroxide and protects the cell from its harmful effects.
Peroxisomes are intracellular organelles found in animal cells that play a crucial role in breaking down hydrogen peroxide (H2O2) into water (H2O) and oxygen (O2). This process is catalyzed by the enzyme called catalase. Let's explore this in more detail.
- Peroxisomes:
Peroxisomes are small, membrane-bound organelles found in eukaryotic cells, including animal cells. They are involved in various metabolic processes, including the breakdown of fatty acids and the detoxification of harmful substances. Peroxisomes contain several enzymes, including catalase, that are responsible for their specific functions.
- Catalase:
Catalase is an enzyme found in peroxisomes that plays a crucial role in breaking down hydrogen peroxide. Hydrogen peroxide is a byproduct of various metabolic reactions in the cell and can be toxic if it accumulates in high concentrations. Catalase converts hydrogen peroxide into water and oxygen through a process called dismutation or disproportionation.
- Reaction:
The reaction catalyzed by catalase can be represented as follows:
2 H2O2 -> 2 H2O + O2
- Mechanism:
Catalase accelerates the breakdown of hydrogen peroxide by providing an alternative pathway with a lower activation energy. It does so by breaking down the hydrogen peroxide molecule into two water molecules and one oxygen molecule. The oxygen molecule is released as a gas, while the water molecules remain in the peroxisome or are further utilized by the cell.
- Localization:
Catalase is primarily localized within peroxisomes in animal cells. This localization allows for efficient degradation of hydrogen peroxide within a specific organelle and prevents its accumulation in other parts of the cell, where it could cause damage to cellular components.
In conclusion, the enzyme catalase, which breaks down hydrogen peroxide into water and oxygen, is usually found in peroxisomes in animal cells. This localization within peroxisomes ensures the efficient detoxification of hydrogen peroxide and protects the cell from its harmful effects.
A culture of mycobacterium leprae was subjected to alkaline ethanol extraction prior to acid fast staining. The colour of culture of following staining will be ______- a)Orange
- b)Blue
- c)Light Green
- d)Red
Correct answer is option 'D'. Can you explain this answer?
A culture of mycobacterium leprae was subjected to alkaline ethanol extraction prior to acid fast staining. The colour of culture of following staining will be ______
a)
Orange
b)
Blue
c)
Light Green
d)
Red
|
Vishnu Priya Mohan Balaj answered |
Mycobacterium is a gram negative bacteria so it has thick peptidoglycan wall in outer membrane so it looks red colour
The number of 5- Sylow subgroup(s) in a group of order 4S is ______.
Correct answer is '1'. Can you explain this answer?
The number of 5- Sylow subgroup(s) in a group of order 4S is ______.
|
|
Avinash Mehta answered |
60=22.3.5 No. of Sylow -5 subgroups =1+5k divides 12.So 1+5k=1,6⟹n5=1,6⟹n5=6 as G is a simple group.
Consider n3=1+3k divides 20⟹1+3k=1,4,10⟹1+3k=4,10. If n3=4 then we have 8 elements of order 3 and A5 has 20 elements of order 3 which is a contradiction.Hence n3=10.
Since A5 has no element of order 6.So 3 is false.
A5 has many elements of order 2 viz. (12)(34),(13)(24),,. Hence 1 is correct only
Angle A is larger than angle C and smaller than angle B by the same amount. In ΔABC If angle B = 67º, angle C is _____- a)64º
- b)47º
- c)70º
- d)53º
Correct answer is option 'D'. Can you explain this answer?
Angle A is larger than angle C and smaller than angle B by the same amount. In ΔABC If angle B = 67º, angle C is _____
a)
64º
b)
47º
c)
70º
d)
53º
|
|
Tejas Goyal answered |
Given information:
- Angle A is larger than angle C and smaller than angle B by the same amount.
- Angle B = 67
To find: Angle C
Solution:
Let x be the amount by which angle A is larger than angle C and smaller than angle B.
Then, we can write the following equations based on the given information:
- A = C + x (Angle A is larger than angle C by x)
- A = B - x (Angle A is smaller than angle B by x)
- B = 67 (Given)
Substituting B = 67 in the second equation, we get:
A = 67 - x
Substituting this value of A in the first equation, we get:
67 - x = C + x
Simplifying this equation, we get:
2x = 67 - C
Dividing both sides by 2, we get:
x = (67 - C) / 2
Substituting this value of x in the first equation (A = C + x), we get:
A = C + (67 - C) / 2
Simplifying this equation, we get:
A = (C + 67) / 2
Since the sum of angles in a triangle is 180 degrees, we have:
A + B + C = 180
Substituting the values of A and B from the above equations, we get:
(C + 67) / 2 + 67 + C = 180
Simplifying this equation, we get:
3C = 113
Dividing both sides by 3, we get:
C = 37.67
Rounding off to the nearest whole number, we get:
C = 38
Therefore, angle C is 38 degrees.
- Angle A is larger than angle C and smaller than angle B by the same amount.
- Angle B = 67
To find: Angle C
Solution:
Let x be the amount by which angle A is larger than angle C and smaller than angle B.
Then, we can write the following equations based on the given information:
- A = C + x (Angle A is larger than angle C by x)
- A = B - x (Angle A is smaller than angle B by x)
- B = 67 (Given)
Substituting B = 67 in the second equation, we get:
A = 67 - x
Substituting this value of A in the first equation, we get:
67 - x = C + x
Simplifying this equation, we get:
2x = 67 - C
Dividing both sides by 2, we get:
x = (67 - C) / 2
Substituting this value of x in the first equation (A = C + x), we get:
A = C + (67 - C) / 2
Simplifying this equation, we get:
A = (C + 67) / 2
Since the sum of angles in a triangle is 180 degrees, we have:
A + B + C = 180
Substituting the values of A and B from the above equations, we get:
(C + 67) / 2 + 67 + C = 180
Simplifying this equation, we get:
3C = 113
Dividing both sides by 3, we get:
C = 37.67
Rounding off to the nearest whole number, we get:
C = 38
Therefore, angle C is 38 degrees.
the value of the determinant of adjoint of A is ______.
In a Eukaryotic gene. One of the introns is mutated at the 3’ end as shown :
Normal : AGGGCATTAGC
Mutant : AGGGCTTTGCWhich of the following processes in this mutation most likely to affect ?- a)Frame Shift Mutation
- b)Capping
- c)Splicing
- d)Hybridization
Correct answer is option 'C'. Can you explain this answer?
In a Eukaryotic gene. One of the introns is mutated at the 3’ end as shown :
Normal : AGGGCATTAGC
Mutant : AGGGCTTTGC
Normal : AGGGCATTAGC
Mutant : AGGGCTTTGC
Which of the following processes in this mutation most likely to affect ?
a)
Frame Shift Mutation
b)
Capping
c)
Splicing
d)
Hybridization
|
|
Siddharth Banerjee answered |
Explanation:
Splicing is the process involved in removing introns from pre-mRNA and joining the exons together to form mature mRNA. The mutation in the intron at the 3' end of a eukaryotic gene can affect the splicing process.
Effects of Mutation on Splicing:
The effect of a mutation on splicing depends on the location of the mutation in the intron. Mutations at the splice site (exon-intron junction) can affect the recognition of the splice site by the spliceosome, which may lead to exon skipping or intron retention. Mutations within the intron can also affect splicing by creating new splice sites or disrupting existing ones.
In the given mutation, the nucleotide sequence at the 3' end of the intron has changed from TAGC to TTGC. This change creates a new potential splice site (AGG GCT TTA GC) within the intron. The spliceosome may recognize this new site and use it instead of the normal splice site, which can lead to the retention of a part of the intron or the inclusion of a part of the exon in the mature mRNA.
Conclusion:
Therefore, the mutation in the intron at the 3' end of a eukaryotic gene is most likely to affect the splicing process. The mutation can create new splice sites or disrupt existing ones, leading to abnormal splicing and the production of aberrant mRNA.
Splicing is the process involved in removing introns from pre-mRNA and joining the exons together to form mature mRNA. The mutation in the intron at the 3' end of a eukaryotic gene can affect the splicing process.
Effects of Mutation on Splicing:
The effect of a mutation on splicing depends on the location of the mutation in the intron. Mutations at the splice site (exon-intron junction) can affect the recognition of the splice site by the spliceosome, which may lead to exon skipping or intron retention. Mutations within the intron can also affect splicing by creating new splice sites or disrupting existing ones.
In the given mutation, the nucleotide sequence at the 3' end of the intron has changed from TAGC to TTGC. This change creates a new potential splice site (AGG GCT TTA GC) within the intron. The spliceosome may recognize this new site and use it instead of the normal splice site, which can lead to the retention of a part of the intron or the inclusion of a part of the exon in the mature mRNA.
Conclusion:
Therefore, the mutation in the intron at the 3' end of a eukaryotic gene is most likely to affect the splicing process. The mutation can create new splice sites or disrupt existing ones, leading to abnormal splicing and the production of aberrant mRNA.
Which of the following statement(s) is/are TRUE for oogenesis in humans ?- a)There is unequal cytokinesis during meiosis
- b)All four products of meiosis develop into mature gametes
- c)Mitotic divisions are completed before birth
- d)The primary oocytes are arrested at prophase I of meiosis
Correct answer is option 'A,C,D'. Can you explain this answer?
Which of the following statement(s) is/are TRUE for oogenesis in humans ?
a)
There is unequal cytokinesis during meiosis
b)
All four products of meiosis develop into mature gametes
c)
Mitotic divisions are completed before birth
d)
The primary oocytes are arrested at prophase I of meiosis
|
Chirag Verma answered |
ANSWER
'A,C,D'
, a similar unequal cytokinesis takes place. Most of the cytoplasm is retained by the mature egg (ovum), and a second polar body receives little more than a haploid nucleus.
c)In oogenesis, diploid oogonium go through mitosis until one develops into a primary oocyte, which will begin the first meiotic division, but then arrest; it will finish this division as it develops in the follicle, giving rise to a haploid secondary oocyte and a smaller polar body.
d)Mammalian oocytes are stored in
the ovary arrested at prophase I of meiosis. ... Oocyte growth occurs concomitantly with follicle growth, but the oocyte remains arrested at prophase I until a preovulatory surge of luteinizing hormone (LH) from the pituitary stimulates meiotic resumption.
Insulin action on a target cell causes an increase in- a)acetyl CoA carboxylase
- b)glucokinase
- c)glucose-6-phosphate dehydrogenase
- d)GLUT4 transporter
Correct answer is option 'A,B,C,D'. Can you explain this answer?
Insulin action on a target cell causes an increase in
a)
acetyl CoA carboxylase
b)
glucokinase
c)
glucose-6-phosphate dehydrogenase
d)
GLUT4 transporter
|
|
Vandana Gupta answered |
Insulin Action on Target Cell
Insulin is a hormone that regulates glucose homeostasis in the body. It is synthesized and secreted by the beta cells of the pancreas. Insulin binds to insulin receptors on the surface of target cells, which triggers a cascade of intracellular events, leading to the uptake and utilization of glucose by the cells.
Increase in Acetyl CoA Carboxylase
Acetyl CoA carboxylase (ACC) is an enzyme that plays a key role in fatty acid synthesis. Insulin activates ACC by promoting its dephosphorylation, which leads to an increase in the production of malonyl CoA. Malonyl CoA is a precursor for fatty acid synthesis, and its increased production under insulin stimulation provides the necessary building blocks for lipid synthesis.
Increase in Glucokinase
Glucokinase is an enzyme that catalyzes the conversion of glucose to glucose-6-phosphate. It is primarily expressed in the liver and pancreatic beta cells. Insulin increases the expression and activity of glucokinase, which promotes glucose uptake and utilization by the liver and beta cells.
Increase in Glucose-6-Phosphate Dehydrogenase
Glucose-6-phosphate dehydrogenase (G6PD) is an enzyme that plays a key role in the pentose phosphate pathway, which generates NADPH for biosynthetic processes. Insulin stimulates the activity of G6PD, leading to an increased production of NADPH, which is required for fatty acid synthesis, cholesterol synthesis, and antioxidant defense.
Increase in GLUT4 Transporter
GLUT4 is a glucose transporter that is primarily expressed in adipose tissue and skeletal muscle. Insulin promotes the translocation of GLUT4 from intracellular vesicles to the plasma membrane, which increases glucose uptake by the cells. This process is mediated by the insulin signaling pathway, which activates the PI3K-Akt pathway, leading to the translocation of GLUT4 to the cell surface.
In summary, insulin action on target cells causes an increase in acetyl CoA carboxylase, glucokinase, glucose-6-phosphate dehydrogenase, and GLUT4 transporter, which promotes glucose uptake, utilization, and storage by the cells.
Insulin is a hormone that regulates glucose homeostasis in the body. It is synthesized and secreted by the beta cells of the pancreas. Insulin binds to insulin receptors on the surface of target cells, which triggers a cascade of intracellular events, leading to the uptake and utilization of glucose by the cells.
Increase in Acetyl CoA Carboxylase
Acetyl CoA carboxylase (ACC) is an enzyme that plays a key role in fatty acid synthesis. Insulin activates ACC by promoting its dephosphorylation, which leads to an increase in the production of malonyl CoA. Malonyl CoA is a precursor for fatty acid synthesis, and its increased production under insulin stimulation provides the necessary building blocks for lipid synthesis.
Increase in Glucokinase
Glucokinase is an enzyme that catalyzes the conversion of glucose to glucose-6-phosphate. It is primarily expressed in the liver and pancreatic beta cells. Insulin increases the expression and activity of glucokinase, which promotes glucose uptake and utilization by the liver and beta cells.
Increase in Glucose-6-Phosphate Dehydrogenase
Glucose-6-phosphate dehydrogenase (G6PD) is an enzyme that plays a key role in the pentose phosphate pathway, which generates NADPH for biosynthetic processes. Insulin stimulates the activity of G6PD, leading to an increased production of NADPH, which is required for fatty acid synthesis, cholesterol synthesis, and antioxidant defense.
Increase in GLUT4 Transporter
GLUT4 is a glucose transporter that is primarily expressed in adipose tissue and skeletal muscle. Insulin promotes the translocation of GLUT4 from intracellular vesicles to the plasma membrane, which increases glucose uptake by the cells. This process is mediated by the insulin signaling pathway, which activates the PI3K-Akt pathway, leading to the translocation of GLUT4 to the cell surface.
In summary, insulin action on target cells causes an increase in acetyl CoA carboxylase, glucokinase, glucose-6-phosphate dehydrogenase, and GLUT4 transporter, which promotes glucose uptake, utilization, and storage by the cells.
The area of inner surface of bronchiole is _____ m2.
Correct answer is '10'. Can you explain this answer?
The area of inner surface of bronchiole is _____ m2.
|
|
Stuti Patel answered |
The inner surface of a bronchiolitis 10m^2.
To produce 12 Pollen grains of cyperaceal, required meiotic divisions are _____.
Correct answer is '12'. Can you explain this answer?
To produce 12 Pollen grains of cyperaceal, required meiotic divisions are _____.
|
|
Avinash Mehta answered |
1 meosis produces 4 pollengrains
so to get 12 pollengrains,3 meosis is required.
In a survey of 25 owners of dogs and/or cats, 20 owned dogs and 10 owned cats. The number of people who owned both dogs and cats was _____.
Correct answer is between '4.9,5.1'. Can you explain this answer?
In a survey of 25 owners of dogs and/or cats, 20 owned dogs and 10 owned cats. The number of people who owned both dogs and cats was _____.
|
Crazy Mess answered |
5
20 have dog remain 5
10 have cat
5 how have both
20 have dog remain 5
10 have cat
5 how have both
Slime molds belong to- a)bacteria
- b)fungi
- c)protists
- d)plants
Correct answer is option 'C'. Can you explain this answer?
Slime molds belong to
a)
bacteria
b)
fungi
c)
protists
d)
plants
|
Mrinalini Sen answered |
Slime molds are classified under the Kingdom Protista because, like other protists, they really don't fit in with other kingdoms! They are motile like animals but some are unicellular so they can not be classified as animals.
A polypeptide chain is made up of 101 amino acid residues. The polypeptide has 200 bonds about which rotation can occur. Assume that three orientations are possible about each of these bonds.Based on these assumptions, how many random coil conformations are possible for the polypeptide chain ?- a)3200
- b)2003
- c)101 x 3200
- d)101 x 2003
Correct answer is option 'A'. Can you explain this answer?
A polypeptide chain is made up of 101 amino acid residues. The polypeptide has 200 bonds about which rotation can occur. Assume that three orientations are possible about each of these bonds.Based on these assumptions, how many random coil conformations are possible for the polypeptide chain ?
a)
3200
b)
2003
c)
101 x 3200
d)
101 x 2003
|
|
Anshika Chavan answered |
Calculation of possible random coil conformations for a polypeptide chain
Given:
- Polypeptide chain contains 101 amino acid residues
- 200 bonds can rotate
- Three orientations are possible about each bond
To calculate the number of possible random coil conformations for the polypeptide chain, we need to consider the following:
1. Number of possible orientations for each bond
- Three orientations are possible about each bond
- Therefore, the number of possible orientations for 200 bonds would be 3^200
2. Total number of conformations
- The total number of conformations would be the product of the number of possible orientations for each bond
- Total number of conformations = (3^200)
3. Number of unique conformations
- However, not all of these conformations would be unique, as some may be identical due to rotations that result in the same overall shape
- The number of unique conformations is difficult to calculate accurately, but it is estimated to be around 10^30
4. Calculation of possible random coil conformations
- To calculate the number of possible random coil conformations, we can divide the estimated number of unique conformations by the number of possible orientations for each bond
- Possible random coil conformations = (10^30) / (3^200)
- This value is too large to be practical, and it is unlikely that all possible conformations could ever be sampled or studied.
Therefore, the answer is option A, 3200, which is the number of possible orientations for each bond raised to the power of the number of bonds that can rotate.
Given:
- Polypeptide chain contains 101 amino acid residues
- 200 bonds can rotate
- Three orientations are possible about each bond
To calculate the number of possible random coil conformations for the polypeptide chain, we need to consider the following:
1. Number of possible orientations for each bond
- Three orientations are possible about each bond
- Therefore, the number of possible orientations for 200 bonds would be 3^200
2. Total number of conformations
- The total number of conformations would be the product of the number of possible orientations for each bond
- Total number of conformations = (3^200)
3. Number of unique conformations
- However, not all of these conformations would be unique, as some may be identical due to rotations that result in the same overall shape
- The number of unique conformations is difficult to calculate accurately, but it is estimated to be around 10^30
4. Calculation of possible random coil conformations
- To calculate the number of possible random coil conformations, we can divide the estimated number of unique conformations by the number of possible orientations for each bond
- Possible random coil conformations = (10^30) / (3^200)
- This value is too large to be practical, and it is unlikely that all possible conformations could ever be sampled or studied.
Therefore, the answer is option A, 3200, which is the number of possible orientations for each bond raised to the power of the number of bonds that can rotate.
Let G be a group of order 231. The number of elements of order 11 in G is ______.
Correct answer is '10'. Can you explain this answer?
Let G be a group of order 231. The number of elements of order 11 in G is ______.
|
|
Rajeev Sharma answered |
Well, if g has order 11 in G, then <g> is an order-11 subgroup of G, and so g is an element of a subgroup of G that has order 11. Since G has only one of these --- namely H --- we learn that every element of order 11 in G has to be in H. Since H only has 11 elements, this already tells us something: that there are at most 11 elements of order 11 in G. But we can do better than this. Since H is a subgroup, one of its elements is identity element of H (which has order 1, not 11). So in fact there are at most 10 elements of G that have order 11 --- namely, the ten non-identity elements of H.
But wait --- is it possible to prove that all of the non-identity elements of H have order 11? Yes it is --- by Lagrange's theorem. If g is in H, the order of g has to divide the order of H, which is 11, and since 11 is prime, this means that the order of g has to be either 1 or 11. The order of a nonidentity element of H can't be 1, so it has to be 11. Conclusion: every nonidentity element of H has order 11. Conclusion: the elements of order 11 in G are precisely the non-identity elements of the subgroup H, of which there are 10.
So G has 10 elements of order 11.
If one arginine has a molecular weight of 174 Daltons, the molecular weight (in Dalton) of a circular polymer of 38 arginines would be ______.
Correct answer is '5928'. Can you explain this answer?
If one arginine has a molecular weight of 174 Daltons, the molecular weight (in Dalton) of a circular polymer of 38 arginines would be ______.
|
|
Mrinalini Sen answered |
So we have 38 arginines, and each one weighs 174 daltons. That means we have 6612 daltons of combined weight. However, each arginine is linked, so to form a peptide bond we need to remove 18 daltons (the molecular weight of water) so 38x18 = 684 daltons that are removed for forming the links. If we subtract 6612-684, we will get 5928 daltons.
A single coin- tossing experiment has two possible outcomes, namely head and tail. Suppose three coins are tossed simultaneously, the probability of getting two heads and one tail is ______.
Correct answer is between '0.37,0.38'. Can you explain this answer?
A single coin- tossing experiment has two possible outcomes, namely head and tail. Suppose three coins are tossed simultaneously, the probability of getting two heads and one tail is ______.
|
|
Jay Nambiar answered |
**Solution:**
To find the probability of getting two heads and one tail when three coins are tossed simultaneously, we need to consider all the possible outcomes and determine the favorable outcomes.
**Possible Outcomes:**
When three coins are tossed simultaneously, there are 2^3 = 8 possible outcomes since each coin can either be a head or a tail.
The 8 possible outcomes are:
1. HHH
2. HHT
3. HTH
4. THH
5. HTT
6. THT
7. TTH
8. TTT
**Favorable Outcomes:**
We are interested in the outcome where we get two heads and one tail. From the possible outcomes, we can see that there are three favorable outcomes:
1. HHT
2. HTH
3. THH
**Probability Calculation:**
To calculate the probability, we need to divide the number of favorable outcomes by the number of possible outcomes.
Number of favorable outcomes = 3
Number of possible outcomes = 8
Probability = Number of favorable outcomes / Number of possible outcomes = 3/8 ≈ 0.375
Therefore, the probability of getting two heads and one tail when three coins are tossed simultaneously is approximately 0.375.
**Answer:**
The correct answer lies between 0.37 and 0.38.
To find the probability of getting two heads and one tail when three coins are tossed simultaneously, we need to consider all the possible outcomes and determine the favorable outcomes.
**Possible Outcomes:**
When three coins are tossed simultaneously, there are 2^3 = 8 possible outcomes since each coin can either be a head or a tail.
The 8 possible outcomes are:
1. HHH
2. HHT
3. HTH
4. THH
5. HTT
6. THT
7. TTH
8. TTT
**Favorable Outcomes:**
We are interested in the outcome where we get two heads and one tail. From the possible outcomes, we can see that there are three favorable outcomes:
1. HHT
2. HTH
3. THH
**Probability Calculation:**
To calculate the probability, we need to divide the number of favorable outcomes by the number of possible outcomes.
Number of favorable outcomes = 3
Number of possible outcomes = 8
Probability = Number of favorable outcomes / Number of possible outcomes = 3/8 ≈ 0.375
Therefore, the probability of getting two heads and one tail when three coins are tossed simultaneously is approximately 0.375.
**Answer:**
The correct answer lies between 0.37 and 0.38.
The body sizes of individuals of a population of frogs on a remote island were measured and found to be normally distributed (Gaussian). After several years with severe winters, the mean body size was found to have significantly increased, although the distribution remained normal (Gaussian).These results can be interpreted to mean that body size is under- a)Directional selection
- b)Stabilizing selection
- c)Disruptive selection
- d)Group selection
Correct answer is option 'A'. Can you explain this answer?
The body sizes of individuals of a population of frogs on a remote island were measured and found to be normally distributed (Gaussian). After several years with severe winters, the mean body size was found to have significantly increased, although the distribution remained normal (Gaussian).
These results can be interpreted to mean that body size is under
a)
Directional selection
b)
Stabilizing selection
c)
Disruptive selection
d)
Group selection
|
|
Sahana Roy answered |
Explanation:
Hence, the correct answer is option A, directional selection.
- The change in mean body size towards larger individuals suggests directional selection.
- In directional selection, individuals with traits that are further away from the population average are favored by the environment. This leads to a shift in the mean value of the trait towards the favored direction over time.
- In this case, the severe winters may have favored larger-bodied individuals, as they were better able to survive and reproduce in the harsh conditions.
- The fact that the distribution remained normal suggests that the selection pressure was not strong enough to cause a significant deviation from normality.
- If stabilizing selection were occurring, the mean body size would remain stable, as individuals with intermediate body sizes would be favored over those with extreme sizes.
- If disruptive selection were occurring, individuals with extreme body sizes would be favored over those with intermediate sizes, leading to a bimodal distribution.
- Group selection refers to the idea that selection can occur at the level of the group, rather than just the individual. This is not relevant to this scenario.
Hence, the correct answer is option A, directional selection.
Which of the following changes would cause a shift in the membrane potential of a neuronal cell from –70 mV to –50 mV?- a)A decrease in K+ permeability
- b)An increase in Na+ permeability
- c)An increase in K+ permeability
- d)A decrease in Na+ permeability
Correct answer is option 'A,B'. Can you explain this answer?
Which of the following changes would cause a shift in the membrane potential of a neuronal cell from –70 mV to –50 mV?
a)
A decrease in K+ permeability
b)
An increase in Na+ permeability
c)
An increase in K+ permeability
d)
A decrease in Na+ permeability
|
Bandana Chauhan answered |
And:b,c
A microscopic unicellular is observed to have the following characteristics a flagellum, chloroplasts, a food gullet, a nucleus, mitochondria. This organism belongs to which kingdom ?- a)Monera
- b)Algae
- c)Protista
- d)Fungi
Correct answer is option 'C'. Can you explain this answer?
A microscopic unicellular is observed to have the following characteristics a flagellum, chloroplasts, a food gullet, a nucleus, mitochondria. This organism belongs to which kingdom ?
a)
Monera
b)
Algae
c)
Protista
d)
Fungi
|
|
Neha Choudhury answered |
The Protista, or Protoctista, are a kingdom of simple eukaryotic organisms, usually composed of a single cell or a colony of similar cells. Protists live in water, in moist terrestrial habitats, and as parasites and other symbionts in the bodies of multicellular eukaroytes.
In the C4 pathway, carbon dioxide is concentrated in - a) pith
- b)mesophyll cells
- c) endodermis
- d)bundle sheath cells
Correct answer is option 'D'. Can you explain this answer?
In the C4 pathway, carbon dioxide is concentrated in
a)
pith
b)
mesophyll cells
c)
endodermis
d)
bundle sheath cells
|
|
Rajeev Sharma answered |
Pyruvate then returns to the mesophyll cells, where a phosphate from ATP is used to regenerate PEP. Thus in C4 plants, C4 carbon fixation has a net added cost of 1 ATP for every CO2 delivered to rubisco; however, C4 plants are less likely to die of dehydration compared to C3 plants in dry conditions.
For the conversion of one N2 to 2NH4+ during biological nitrogen fixation required ATP are _____.
Correct answer is '16'. Can you explain this answer?
For the conversion of one N2 to 2NH4+ during biological nitrogen fixation required ATP are _____.
|
|
Bhavana Dasgupta answered |
Biological nitrogen fixation is the process by which certain bacteria and archaea convert atmospheric nitrogen gas (N2) into ammonia (NH3), which can be used by plants and other organisms. This process is vital for the nitrogen cycle and plays a crucial role in maintaining the availability of nitrogen for living organisms.
The conversion of one molecule of N2 to two molecules of NH4 requires a series of enzymatic reactions involving nitrogenase, an enzyme complex that catalyzes the conversion. This process requires a significant amount of energy in the form of ATP (adenosine triphosphate) to drive the reactions.
Here's a detailed explanation of why 16 ATP molecules are required for the conversion:
1. Nitrogenase Complex Activation:
- The nitrogenase complex consists of two main components: the MoFe protein and the Fe protein.
- In order to activate the nitrogenase complex, the Fe protein binds to ATP and transfers an electron to the MoFe protein.
- This ATP hydrolysis step requires one ATP molecule.
2. Reduction of Nitrogen:
- The MoFe protein uses the energy from the activated nitrogenase complex to reduce N2 to NH3.
- This process involves the transfer of electrons and protons to the nitrogen molecule.
- Each reduction reaction requires 8 electrons and 8 protons.
- The transfer of each electron requires one ATP molecule, resulting in a total of 8 ATP molecules for the reduction of one N2 molecule.
3. Hydrogenation of Nitrogen:
- After the reduction of N2 to NH3, the nitrogenase complex further catalyzes the hydrogenation of NH3 to NH4.
- This process also requires the transfer of electrons and protons.
- Each hydrogenation reaction requires 8 electrons and 8 protons.
- The transfer of each electron requires one ATP molecule, resulting in a total of 8 ATP molecules for the hydrogenation of one N2 molecule.
Therefore, the total number of ATP molecules required for the conversion of one N2 to 2 NH4 is 1 ATP (activation) + 8 ATP (reduction) + 8 ATP (hydrogenation) = 17 ATP molecules.
However, it is worth noting that during the process, one ATP molecule is generated through the electron transport chain, resulting in a net requirement of 16 ATP molecules.
In conclusion, the correct answer is '16' ATP molecules.
The conversion of one molecule of N2 to two molecules of NH4 requires a series of enzymatic reactions involving nitrogenase, an enzyme complex that catalyzes the conversion. This process requires a significant amount of energy in the form of ATP (adenosine triphosphate) to drive the reactions.
Here's a detailed explanation of why 16 ATP molecules are required for the conversion:
1. Nitrogenase Complex Activation:
- The nitrogenase complex consists of two main components: the MoFe protein and the Fe protein.
- In order to activate the nitrogenase complex, the Fe protein binds to ATP and transfers an electron to the MoFe protein.
- This ATP hydrolysis step requires one ATP molecule.
2. Reduction of Nitrogen:
- The MoFe protein uses the energy from the activated nitrogenase complex to reduce N2 to NH3.
- This process involves the transfer of electrons and protons to the nitrogen molecule.
- Each reduction reaction requires 8 electrons and 8 protons.
- The transfer of each electron requires one ATP molecule, resulting in a total of 8 ATP molecules for the reduction of one N2 molecule.
3. Hydrogenation of Nitrogen:
- After the reduction of N2 to NH3, the nitrogenase complex further catalyzes the hydrogenation of NH3 to NH4.
- This process also requires the transfer of electrons and protons.
- Each hydrogenation reaction requires 8 electrons and 8 protons.
- The transfer of each electron requires one ATP molecule, resulting in a total of 8 ATP molecules for the hydrogenation of one N2 molecule.
Therefore, the total number of ATP molecules required for the conversion of one N2 to 2 NH4 is 1 ATP (activation) + 8 ATP (reduction) + 8 ATP (hydrogenation) = 17 ATP molecules.
However, it is worth noting that during the process, one ATP molecule is generated through the electron transport chain, resulting in a net requirement of 16 ATP molecules.
In conclusion, the correct answer is '16' ATP molecules.
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