The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 10 10 base pairs. The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified. The fractional abundance of this target sequence after 10 , 15 , 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process. Concept introduction: The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

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Answer 1

Question

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6

The sequence is 10−6 of the original genome.

The PCR amplification doubles the number of copies each time. Amplification is 2n , where, n is number of cycles.

So, after 10 cycles, this is calculated as:

F=210×1( 2 10 + 10 −6 )=0.00102

Thus, after 10 cycles the product is 0.00102 of the genome.

After 15 cycles, this is calculated as:

F=215×1( 2 15 + 10 −6 )=0.0317

Thus, after 15 cycles the product is 0.0317 of the genome.

After 20 cycles, this is calculated as:

F=220×1( 2 20 + 10 −6 )=0.511

Thus, after 20 cycles the product is 0.511 of the genome.

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Answer 2

Question

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6

The sequence is 10−6 of the original genome.

The PCR amplification doubles the number of copies each time. Amplification is 2n , where, n is number of cycles.

So, after 10 cycles, this is calculated as:

F=210×1( 2 10 + 10 −6 )=0.00102

Thus, after 10 cycles the product is 0.00102 of the genome.

After 15 cycles, this is calculated as:

F=215×1( 2 15 + 10 −6 )=0.0317

Thus, after 15 cycles the product is 0.0317 of the genome.

After 20 cycles, this is calculated as:

F=220×1( 2 20 + 10 −6 )=0.511

Thus, after 20 cycles the product is 0.511 of the genome.

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Answer 3

Question

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6

The sequence is 10−6 of the original genome.

The PCR amplification doubles the number of copies each time. Amplification is 2n , where, n is number of cycles.

So, after 10 cycles, this is calculated as:

F=210×1( 2 10 + 10 −6 )=0.00102

Thus, after 10 cycles the product is 0.00102 of the genome.

After 15 cycles, this is calculated as:

F=215×1( 2 15 + 10 −6 )=0.0317

Thus, after 15 cycles the product is 0.0317 of the genome.

After 20 cycles, this is calculated as:

F=220×1( 2 20 + 10 −6 )=0.511

Thus, after 20 cycles the product is 0.511 of the genome.

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Answer 4

Question

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6

The sequence is 10−6 of the original genome.

The PCR amplification doubles the number of copies each time. Amplification is 2n , where, n is number of cycles.

So, after 10 cycles, this is calculated as:

F=210×1( 2 10 + 10 −6 )=0.00102

Thus, after 10 cycles the product is 0.00102 of the genome.

After 15 cycles, this is calculated as:

F=215×1( 2 15 + 10 −6 )=0.0317

Thus, after 15 cycles the product is 0.0317 of the genome.

After 20 cycles, this is calculated as:

F=220×1( 2 20 + 10 −6 )=0.511

Thus, after 20 cycles the product is 0.511 of the genome.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6

The sequence is 10−6 of the original genome.

The PCR amplification doubles the number of copies each time. Amplification is 2n , where, n is number of cycles.

So, after 10 cycles, this is calculated as:

F=210×1( 2 10 + 10 −6 )=0.00102

Thus, after 10 cycles the product is 0.00102 of the genome.

After 15 cycles, this is calculated as:

F=215×1( 2 15 + 10 −6 )=0.0317

Thus, after 15 cycles the product is 0.0317 of the genome.

After 20 cycles, this is calculated as:

F=220×1( 2 20 + 10 −6 )=0.511

Thus, after 20 cycles the product is 0.511 of the genome.

Answer 6

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6

The sequence is 10−6 of the original genome.

The PCR amplification doubles the number of copies each time. Amplification is 2n , where, n is number of cycles.

So, after 10 cycles, this is calculated as:

F=210×1( 2 10 + 10 −6 )=0.00102

Thus, after 10 cycles the product is 0.00102 of the genome.

After 15 cycles, this is calculated as:

F=215×1( 2 15 + 10 −6 )=0.0317

Thus, after 15 cycles the product is 0.0317 of the genome.

After 20 cycles, this is calculated as:

F=220×1( 2 20 + 10 −6 )=0.511

Thus, after 20 cycles the product is 0.511 of the genome.

Answer 7

Chapter 21, Problem 1P

Interpretation Introduction

Interpretation: The exponential nature of PCR allows spectacular increases in the abundance of a DNA sequence being amplified. Consider a 10 kbp DNA sequence in a genome of 1010 base pairs.

The fraction of the genome represented by this sequence should be determined. The fractional abundance of this sequence in this genome should be identified.

The fractional abundance of this target sequence after 10, 15, 20 cycles of PCR, should be calculated starting with DNA representing the whole genome and if no other sequences in the genome undergo amplification in the process.

Concept introduction:

The method of making of copies of specific region of DNA uses polymerase chain reaction or PCR. The amount of target sequence in the reaction is doubled with each PCR cycle. Example: ten cycles will theoretically multiply the DNA segment by a factor of about one thousand and so on.

Answer 8

Expert Solution & Answer

Explanation of Solution

A 10 kbp DNA sequence in a genome of 1010 base pairs. The fraction of the genome is represented by this sequence. That is, the fractional abundance of this sequence can be calculated as:

F=10 kbp× 103 bp/kbp 10 10 bp=10−6