Concepts of Biology
Concepts of Biology
1st Edition
ISBN: 9781938168116
Author: Samantha Fowler, Rebecca Roush, James Wise
Publisher: OpenStax College
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Chapter 9, Problem 1ACQ

Figure 9.10 You isolate a cell strain in which the joining together of Okazaki fragments is impaired and suspect that a mutation has occurred in an enzyme found at the replication fork. Which enzyme is most likely to be mutated?

Chapter 9, Problem 1ACQ, Figure 9.10 You isolate a cell strain in which the joining together of Okazaki fragments is impaired

Expert Solution & Answer
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Summary Introduction

To write:

The enzyme which is most likely mutated when joining of Okazaki fragments is impaired in the replication.

Introduction:

DNA replication is the process in which the copy of DNA is made. This is done in 3 steps in a eukaryotic cell. They are initiation, elongation, and termination. It is a semiconservative type of replication. Enzymes involved in this process are helicase, DNA polymerase, and DNA ligase.

Explanation of Solution

DNA replication is started at the origin of replication where a Y shaped fork known as replication is formed when both strands are separated by helicase enzyme. Both the strands of DNA act as a template. One strand is known as a leading strand at which replication is continuous and other strand is known as a lagging strand at which replication occurs in fragments. This is because DNA polymerase can synthesize DNA only in 5' to 3' direction. These fragments are known as Okazaki fragments. These fragments at the lagging strands are then sealed by the enzyme DNA ligase. The joining of Okazaki fragments is impaired when the DNA ligase is not able to join the strands. This can occur due to the mutation in enzyme DNA ligase.

Conclusion

When joining of DNA fragments on the lagging strand is impaired in DNA replication, the mutation is most likely to occur in DNA ligase as it is an important enzyme which joins the strands.

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DNA polymerase occasionally incorporates the wrong nucleotide during DNA replication. If left unrepaired, the base-pair mismatch that results will lead to mutation in the next replication. As part of a template strand, the incorporated wrong base will direct the incorporation of a base complementary to itself, so the bases on both strands of the DNA at that position will now be different from what they were before the mismatch event. The MER-minus strain of yeast does not have a functional  mismatch excision repair system, but it has normal base excision repair and nucleotide excision repair systems. Which of the following statements is correct about differences in the mutation spectrum between MER-minus and wildtype yeast? More than one answer is correct. Options:   More point mutations will arise in MER-minus yeast.   Fewer point mutations will arise in MER-minus yeast as compared with wildtype.   Of the total point mutations that…
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