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Chapter 20: Problem 9

Why do cells exposed to visible light following irradiation with ultraviolet light have a greater survival rate than cells kept in the dark after irradiation with ultraviolet light?

Short Answer

Expert verified
Cells exposed to visible light following irradiation with ultraviolet light have a greater survival rate because the visible light facilitates photoreactivation. During this process, the enzyme photolyase absorbs the visible light and uses it to repair the UV-induced DNA damage, specifically thymine dimers, restoring normal DNA functioning and resulting in improved cell survival.

Step by step solution

01

Understand the Effects of Ultraviolet Light on Cells

Ultraviolet (UV) light can cause damage to cells, particularly by causing the formation of thymine dimers in DNA. These dimers distort the DNA helix, leading to mistakes in DNA replication and transcription, resulting in mutations and cell death.
02

Role of Visible Light in Photoreactivation

Visible light can stimulate a process known as photoreactivation, which is a light-dependent DNA repair process. In this process, an enzyme called photolyase absorbs visible light and uses this energy to break the bonds in the thymine dimer, essentially reversing the damage and allowing the DNA to function normally.
03

Compare Survival Rates in Visible Light and Dark Following UV Exposure

If cells are kept in the dark after UV exposure, they do not have the opportunity to undergo photoreactivation since photolyase requires visible light to function. Thus, they may suffer from higher rates of mutations and cell death. In contrast, cells exposed to visible light after UV irradiation can harness this light for photoreactivation, which repairs the UV-induced damage and restores the normal functioning of the DNA. This results in a greater survival rate of these cells.

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Most popular questions from this chapter

The chromosome of a certain bacterium is a circular, double-stranded DNA molecule of \(5.2 \times 10^{6}\) base pairs. The chromosome contains one origin of replication, and the rate of replication-fork movement is 1000 nucleotides per second. (a) Calculate the time required to replicate the chromosome. (b) Explain how the bacterial generation time can be as short as 25 minutes under extremely favorable conditions.

The entire genome of the fruit fly \(D\). melanogaster consists of \(1.65 \times 10^{8} \mathrm{bp}\). If replication at a single replication fork occurs at the rate of \(30 \mathrm{bp}\) per sec, calculate the minimum time required to replicate the entire genome if replication were initiated (a) at a single bidirectional origin (b) at 2000 bidirectional origins (c) In the early embryo, replication can require as few as 5 minutes. What is the minimum number of origins necessary to account for this replication time?

Will DNA repair in \(E\). coli be dependent on the enzymatic cofactor \(\mathrm{NAD}^{\oplus}\) ?

Why are two different DNA polymerase enzymes required to replicate the E. coli chromosome?

Ethyl methane sulfonate (EMS) is a reactive alkylating agent that ethylates the O-6 residue of guanine in DNA. If this modified \(\mathrm{G}\) is not excised and replaced with a normal \(\mathrm{G}\), what would be the outcome of one round of DNA replication?
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