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Ch. 15 - Gene Mutation, DNA Repair, and Transposition

Chapter 15, Problem 23

Many of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. The dnaE gene encodes the α subunit of DNA polymerase III. What effect is expected from a mutation in this gene? How could the mutant strain be maintained?

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Hello everyone and welcome to today's video. So today we have that D N. A polymerase three is going to be profiting and correcting DNA replication errors through Exxon nuclear activity. DNA polymerase three consists of three sub units which are the following gene encodes the alpha subunit. So just like the question states, I want you to recall that D N. A preliminary three is going to be made up by three main sub units which are going to be the alpha, which is the one that we're looking for. Then we have the epsilon and the data sub unit. And just in case you forgot the specific genes that encode for these three, we can refer to the stable in order to figure this out. Remember that we're looking at DNA polymerase three which is here we have that the three genes are going to be encoding depending on the sub unit, D N A E. Here, D N A Q and whole E. These are the three genes that are going to be encoding the three sub units of this D N. A polymerase three. Now this data sub unit is going to be encoded by the whole E jean because of this. We're going to cancel out whole E. Which is answer choice A. Then this epsilon sub unit is going to be encoded by DNA que. We're going to be canceling out D N A Q. Because it is going to be encoding for epsilon and not the alpha subunit. So the only one left really is this D N A E and D N A. Is the gene that is going to be encoding for this off a sub unit. Because of this, we're going to be selecting answer choice c d n a e jean as a correct answer to our question. I really hope this video helped you, and I hope to see you on the next video.
Related Practice
Textbook Question
Describe the 'end-replication problem' in eukaryotes. How is it resolved?
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Textbook Question
The SOS repair genes in E. coli (discussed in Chapter 15) are negatively regulated by the lexA gene product, called the LexA repressor. When a cell's DNA sustains extensive damage, the LexA repressor is inactivated by the recA gene product (RecA), and transcription of the SOS genes is increased dramatically. One of the SOS genes is the uvrA gene. You are a student studying the function of the uvrA gene product in DNA repair. You isolate a mutant strain that shows constitutive expression of the UvrA protein. Naming this mutant strain uvrAᶜ, you construct the diagram shown above in the right-hand column showing the lexA and uvrA operons: Outline a series of genetic experiments that would use partial diploid strains to determine which of the two possible mutations you have isolated.
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Textbook Question
The human genome contains approximately 10⁶ copies of an Alu sequence, one of the best-studied classes of short interspersed elements (SINEs), per haploid genome. Individual Alu units share a 282-nucleotide consensus sequence followed by a 3'-adenine-rich tail region [Schmid (1998)]. Given that there are approximately 3 x 10⁹ base pairs per human haploid genome, about how many base pairs are spaced between each Alu sequence?
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Textbook Question
Many of the gene products involved in DNA synthesis were initially defined by studying mutant E. coli strains that could not synthesize DNA. The dnaQ gene encodes the ε subunit of DNA polymerase. What effect is expected from a mutation in this gene?
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Textbook Question
A fellow student considers the issues in Problem 22 and argues that there is a more straightforward, nongenetic experiment that could differentiate between the two types of mutations. The experiment requires no fancy genetics and would allow you to easily assay the products of the other SOS genes. Propose such an experiment.
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Textbook Question
In a bacterial culture in which all cells are unable to synthesize leucine (leu⁻), a potent mutagen is added, and the cells are allowed to undergo one round of replication. At that point, samples are taken, a series of dilutions are made, and the cells are plated on either minimal medium or minimal medium containing leucine. The first culture condition (minimal medium) allows the growth of only leu⁺ cells, while the second culture condition (minimal medium with leucine added) allows growth of all cells. The results of the experiment are as follows: Culture Condition Dilution Colonies Minimal medium 10⁻¹ 18 Minimal medium + leucine 10⁻⁷ 9 What is the rate of mutation at the locus associated with leucine biosynthesis?
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