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The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________
Correct answer is '6000'. Can you explain this answer?
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The stems of bamboo, a tropical grass, can grow at the phenomenal rate...
Fust, calculate the growth per second:
0.3 m/day = 3 x 10-6 m/s
(24 h/day)(60 min/h)(60 s/min)
Given that each glucose residue increases the length of the cellulose chain by 0.5 nm(5 x 10-10 m), the number of residues added per second is 3 x 10-6 ms = 6,000 residues/s
5 x 10-10 m/residue
0.3 m/day = 3 x 10-6 m/s
(24 h/day)(60 min/h)(60 s/min)
Given that each glucose residue increases the length of the cellulose chain by 0.5 nm(5 x 10-10 m), the number of residues added per second is 3 x 10-6 ms = 6,000 residues/s
5 x 10-10 m/residue
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The stems of bamboo, a tropical grass, can grow at the phenomenal rate...
The growth rate of bamboo stems can be determined by considering the rate at which cellulose molecules, the primary component of the stem, are being elongated. Cellulose is a polymer made up of repeating units of D-glucose, and each glucose unit contributes approximately 0.5 nm to the length of a cellulose molecule.
To calculate the number of sugar residues per second that must be added to account for the growth rate of 0.3 m/day, we can follow these steps:
1. Convert the growth rate from meters per day to meters per second:
- There are 24 hours in a day and 60 minutes in an hour, so there are 24 * 60 = 1440 minutes in a day.
- Since there are 60 seconds in a minute, there are 1440 * 60 = 86400 seconds in a day.
- To calculate the growth rate in meters per second, divide 0.3 m/day by 86400 seconds/day:
0.3 m/day / 86400 s/day ≈ 3.47 x 10^(-6) m/s
2. Determine the increase in length of a cellulose molecule per second:
- Since each D-glucose unit contributes approximately 0.5 nm to the length of a cellulose molecule, we need to convert the growth rate from meters to nanometers.
- Multiply the growth rate in meters per second by 10^9 to convert it to nanometers per second:
3.47 x 10^(-6) m/s * 10^9 nm/m = 3.47 nm/s
3. Calculate the number of sugar residues added per second:
- Since each D-glucose unit contributes 0.5 nm to the length of a cellulose molecule, and the increase in length of a cellulose molecule per second is 3.47 nm/s, we can divide the increase in length by the contribution of each glucose unit to get the number of sugar residues added per second:
3.47 nm/s / 0.5 nm = 6.94 sugar residues/s
Rounding up to the nearest whole number, the number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is approximately 7 sugar residues/s. However, the correct answer provided is '6000', which suggests that there may be additional factors or assumptions involved in this calculation that are not mentioned in the given information.
To calculate the number of sugar residues per second that must be added to account for the growth rate of 0.3 m/day, we can follow these steps:
1. Convert the growth rate from meters per day to meters per second:
- There are 24 hours in a day and 60 minutes in an hour, so there are 24 * 60 = 1440 minutes in a day.
- Since there are 60 seconds in a minute, there are 1440 * 60 = 86400 seconds in a day.
- To calculate the growth rate in meters per second, divide 0.3 m/day by 86400 seconds/day:
0.3 m/day / 86400 s/day ≈ 3.47 x 10^(-6) m/s
2. Determine the increase in length of a cellulose molecule per second:
- Since each D-glucose unit contributes approximately 0.5 nm to the length of a cellulose molecule, we need to convert the growth rate from meters to nanometers.
- Multiply the growth rate in meters per second by 10^9 to convert it to nanometers per second:
3.47 x 10^(-6) m/s * 10^9 nm/m = 3.47 nm/s
3. Calculate the number of sugar residues added per second:
- Since each D-glucose unit contributes 0.5 nm to the length of a cellulose molecule, and the increase in length of a cellulose molecule per second is 3.47 nm/s, we can divide the increase in length by the contribution of each glucose unit to get the number of sugar residues added per second:
3.47 nm/s / 0.5 nm = 6.94 sugar residues/s
Rounding up to the nearest whole number, the number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is approximately 7 sugar residues/s. However, the correct answer provided is '6000', which suggests that there may be additional factors or assumptions involved in this calculation that are not mentioned in the given information.
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The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer?
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The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer?.
The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The stems of bamboo, a tropical grass, can grow at the phenomenal rate of 0.3 m/day under optimal conditions. It is given that the stems are composed almost entirely of cellulose fibers oriented in the direction of growth. Each D-glucose unit contributes ~0.5 nm to the length of a cellulose molecule. The number of sugar residues per second that must be added enzymatically to growing cellulose chains to account for the growth rate is ______________Correct answer is '6000'. Can you explain this answer?.
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