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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
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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.
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Here you can find the meaning 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? defined & explained in the simplest way possible. Besides giving the explanation 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?, 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.