The mRNA sequence formed from DNA template 3 ′ -TGCCTAACG- 5 ′ has to be identified. Concept Introduction: DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5 ′ end to 3 ′ end then other strand starts from 3 ′ end to 5 ′ end of nucleotides. In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template. RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C .

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Answer 1

Question

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

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3’TCG-CAG-GGA-GAA-GTC5’ 1)What are the mRNA codons? 2) What are the tRNA anticodons? 3) What is the amino acid sequence that coupled be translated from the mRNA sequence?

Answer 2

Question

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

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Answer 3

Question

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

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Answer 4

Question

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

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Answer 5

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 6

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

Answer 7

Chapter 17.6, Problem 17.13P

(a)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TGCCTAACG-5′ has to be identified.

Concept Introduction:

DNA is a molecule that is stranded with two polynucleotides into a double helix structure. The two polynucleotide strands are placed antiparallel to each other. It means that one strand starts from 5′ end to 3′ end then other strand starts from 3′ end to 5′ end of nucleotides.

In transcription, DNA strand separates and mRNA with DNA template is formed by the catalyzation of RNA polymerase. The mRNA molecule formed has base pairs opposite to DNA template.

RNA does not contain thymine base instead it has uracil base. Therefore, in this purine base adenine (A) is held with pyrimidine base uracil (U) by two hydrogen bonds. Whereas purine base guanine (G) is held with pyrimidine base cytosine (C) by three hydrogen bonds. These base pairs are denoted as A-U and G-C.

Answer 8

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 9

(b)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 10

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 11

(c)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 12

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 13

(d)

Interpretation Introduction

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Answer 14

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Answer 15

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Answer 16

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Answer 17

Ch. 17.1 - Prob. 17.1P Ch. 17.1 - Draw the structure of guanosine. Classify the...Ch. 17.1 - Prob. 17.3P Ch. 17.1 - Give the name that corresponds to each...Ch. 17.1 - Which nucleic acid (DNA or RNA) contains each of...Ch. 17.1 - Label each statement about the compound...Ch. 17.1 - Identify each component as a base, nucleoside, or...Ch. 17.2 - Draw the structure of a dinucleotide formed by...Ch. 17.2 - Label the 5' end and the 3' end in each...Ch. 17.2 - Label each statement about the polynucleotide...

Solution Summary: The author explains the mRNA sequence formed from DNA template 3 ′ -TGCCTAACG- 5′ has to be identified.

Concept explainers

Videos

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

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Chapter 17 Solutions

Solution Summary: The author explains the mRNA sequence formed from DNA template 3 ′ -TGCCTAACG- 5′ has to be identified.

Concept explainers

Videos

Interpretation: The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified. Concept Introduction: Refer to part (a).

Interpretation: The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified. Concept Introduction: Refer to part (a).

Interpretation: The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified. Concept Introduction: Refer to part (a).

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Chapter 17 Solutions

Interpretation:

The mRNA sequence formed from DNA template 3′-GACTCC-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Interpretation:

The mRNA sequence formed from DNA template 3′-TTAACGCGA-5′ has to be identified.

Concept Introduction:

Refer to part (a).

Interpretation:

The mRNA sequence formed from DNA template 3′-CAGTGACCGTAC-5′ has to be identified.

Concept Introduction:

Refer to part (a).