ESSENTIAL CELL BIOLOGY (LL)-TEXT
ESSENTIAL CELL BIOLOGY (LL)-TEXT
5th Edition
ISBN: 9780393680331
Author: ALBERTS
Publisher: NORTON
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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
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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
Check Mark

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:

Fractionof flygenomeshown=44001.8×108×100%=0.002%

Conclusion

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

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