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One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .
Correct answer is '1600'. Can you explain this answer?
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One hundred E.Coli cells are each infected by single λ phage part...
Given information:
- 100 E.Coli cells are infected by a single phage particle.
- The ratio of phage particles committing to lysogeny to those committing to lysis is 4:1.
- The average burst size is 80.
Explanation:
Phage particles committing to lysogeny:
- The ratio of phage particles committing to lysogeny to those committing to lysis is 4:1.
- Let's assume that x phage particles commit to lysogeny.
- Since the ratio is 4:1, the number of phage particles committing to lysis will be x/4.
Phage particles committing to lysis:
- The average burst size is 80, which means that each infected E.Coli cell will release 80 phage particles after lysis.
- Since there are 100 infected E.Coli cells, the total number of phage particles released after lysis will be 100 * 80 = 8000.
Total phage particles:
- The total number of phage particles can be calculated as the sum of the particles committing to lysogeny and the particles committing to lysis.
- Total phage particles = Phage particles committing to lysogeny + Phage particles committing to lysis.
Calculating phage particles committing to lysogeny:
- The ratio of phage particles committing to lysogeny to those committing to lysis is 4:1.
- So, the total number of phage particles committing to lysogeny is x.
Calculating total phage particles:
- Total phage particles = Phage particles committing to lysogeny + Phage particles committing to lysis.
- Total phage particles = x + 8000.
Calculating x:
- Since the ratio of phage particles committing to lysogeny to those committing to lysis is 4:1, we can write an equation:
x/(x/4) = 4/1
Solving this equation, we get x = 400.
Substituting the value of x in the equation for total phage particles:
- Total phage particles = x + 8000
- Total phage particles = 400 + 8000
- Total phage particles = 8400.
Therefore, the number of free phage particles released after one round of infection is 8400.
- 100 E.Coli cells are infected by a single phage particle.
- The ratio of phage particles committing to lysogeny to those committing to lysis is 4:1.
- The average burst size is 80.
Explanation:
Phage particles committing to lysogeny:
- The ratio of phage particles committing to lysogeny to those committing to lysis is 4:1.
- Let's assume that x phage particles commit to lysogeny.
- Since the ratio is 4:1, the number of phage particles committing to lysis will be x/4.
Phage particles committing to lysis:
- The average burst size is 80, which means that each infected E.Coli cell will release 80 phage particles after lysis.
- Since there are 100 infected E.Coli cells, the total number of phage particles released after lysis will be 100 * 80 = 8000.
Total phage particles:
- The total number of phage particles can be calculated as the sum of the particles committing to lysogeny and the particles committing to lysis.
- Total phage particles = Phage particles committing to lysogeny + Phage particles committing to lysis.
Calculating phage particles committing to lysogeny:
- The ratio of phage particles committing to lysogeny to those committing to lysis is 4:1.
- So, the total number of phage particles committing to lysogeny is x.
Calculating total phage particles:
- Total phage particles = Phage particles committing to lysogeny + Phage particles committing to lysis.
- Total phage particles = x + 8000.
Calculating x:
- Since the ratio of phage particles committing to lysogeny to those committing to lysis is 4:1, we can write an equation:
x/(x/4) = 4/1
Solving this equation, we get x = 400.
Substituting the value of x in the equation for total phage particles:
- Total phage particles = x + 8000
- Total phage particles = 400 + 8000
- Total phage particles = 8400.
Therefore, the number of free phage particles released after one round of infection is 8400.
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One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer?
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One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer?.
One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer?.
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One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer?, a detailed solution for One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer? has been provided alongside types of One hundred E.Coli cells are each infected by single λ phage particle. The ratio of the number of phage particles committing to lysogeny to those committing to lysis is 4:1. Assuming that the average burst size is 80, the number of free phage particles released after one round of infection is ____ .Correct answer is '1600'. Can you explain this answer? theory, EduRev gives you an
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