To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other. Introduction : DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

Question

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Answer 1

Question

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%

Conclusion

The fraction of fly genome shown in the micrograph is 0.002%.

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Question #3: In the KeyGene paper, the authors state that it would be useful if pollen from an apomict would transmit apomixis-inducing genes to the female in the cross (assuming the pollen is viable). Assuming there was just one gene conferring gametophytic obligate apomixis, and that the two parents are inbreds, what would be the consequences of such a cross if: a) The apomixis was a dominant trait? Indicate the genotypes and phenotypes (apomict or non- apomict) of the parents, F1 and F2 generations. Remember to include the expected genotypic and phenotypic ratios (or percentages) in the F1 and F2 generations, and to position the female first (left side) in the parental cross. b) The apomixis was a recessive trait? Indicate the genotypes and phenotypes (apomict or non- apomict) of the parents, F1 and F2 generations. Remember to include the expected genotypic and phenotypic ratios (or percentages) in the F1 and F2 generations, and to position the female first (left side) in the…

Question #5: Assume that two genes are identified that confer gametophytic facultative apomixis in soybean. The genes show independent assortment. Recessive alleles at both loci are required for the facultative apomixis. Facultative apomixis is triggered when the temperature at pollination is above 20 degrees C. At temperatures below 20 degrees C, all reproduction is sexual, independent of genotype. A facultative apomict male, capable of producing viable pollen, was crossed with a sexually reproducing female. Assuming the parents are completely inbred, what are the predicted phenotypic ratios (apomict: non-apomict) for the F1, F2, and DH (F1-derived) generations at each of the following temperatures*: a) 15°C? b) 25°C? *for full credit, show crosses and genotypes where appropriate. Remember to position the female first (left side) in the cross. Type your answer here:

Answer 2

Question

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%

Conclusion

The fraction of fly genome shown in the micrograph is 0.002%.

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Answer 3

Question

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%

Conclusion

The fraction of fly genome shown in the micrograph is 0.002%.

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Answer 4

Question

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%

Conclusion

The fraction of fly genome shown in the micrograph is 0.002%.

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Answer 5

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%

Conclusion

The fraction of fly genome shown in the micrograph is 0.002%.

Answer 6

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

Answer 7

Chapter 6, Problem 1Q

A.

Summary Introduction

To estimate: The length of DNA strands between the replication forks in the given micrograph and the time taken until the forks 4 and 5, and forks 7 and 8 will collide with each other.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

Answer 8

A.

Expert Solution

Explanation of Solution

Refer to Fig. 6-9, “The two replication forks move away in opposite directions at each replication origin” in the textbook. The micrograph shows the scale bar for measurement of DNA strand undergoing replication. Drawing 2 shows the replication process in a fly where replication forks form. As per the scale bar, the distance between replication forks 4 and 5 is approximately 0.28 µm, that is, 280 nm. It corresponds to 824 nucleotides (280/0.34), since the distance between two bases is 0.34 nm. The replication forks in eukaryotes move at about 100 nucleotides per second. Thus, the time taken for forks 4 and 5 to collide would be about 8 seconds. As shown in the drawing, replication forks 7 and 8 would never collide as they move away from each other.

Answer 9

A.

Expert Solution

Answer 10

A.

Expert Solution

Answer 11

Expert Solution

Answer 12

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%

Conclusion

The fraction of fly genome shown in the micrograph is 0.002%.

Answer 13

B.

Summary Introduction

To evaluate: The fraction of genome of fly shown in the given micrograph.

Introduction: DNA is the blueprint of life and contains genetic information. Replication of DNA is semi-conservative. It begins at certain locations on the DNA known as origins of replication having a replication fork. The two replication forks move in opposite directions from origin and separate the two DNA strands.

Answer 14

B.

Expert Solution

Explanation of Solution

The above-mentioned micrograph shown in Fig. 6-9 represents that the length of DNA is about 1.5 µm (1500). It corresponds to 4400 nucleotides (1500/0.34), since the distance between two bases is 0.34 nm. If the fly genome has 1.8×10⁸ nucleotide pairs, then the fraction shown in micrograph is evaluated as follows:

Fraction of fly genome shown=44001.8×108×100% =0.002%