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λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______
Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1 - 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61
Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1 - 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61
Correct answer is '688000'. Can you explain this answer?
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λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer?
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λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. 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 λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. 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 λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer?.
λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. 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 λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. 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 λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer?.
Solutions for λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer? in English & in Hindi are available as part of our courses for IIT JAM.
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λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer?, a detailed solution for λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer? has been provided alongside types of λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice λ EMBL3 is used as a vector to clone 2000 bp fragments generated from a partial Sau3Al digest of the human genome (3 x 109 bp). We wish to isolate a gene contained completely on a 20,000 bp fragment. To have a 99% chance of isolating this gene in the λ EMBL3 recombinant genomic library, the number of independent clones to be examined are _______Hint: (2 x 104 / 3 x 109) = 6.7 x 10-6 and ln(1- 6.7 x 10-6) = In 0.9999 and In (0.01) = -4.61Correct answer is '688000'. Can you explain this answer? tests, examples and also practice IIT JAM tests.
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