Biology, asked by saniyamansoori1691, 6 months ago

Mutations in the genome of e coli are introduced at the rate of 1/10^9 bp per generation if a scientist starts with a colony of 10^6 cells having 1000 bp dna, the number of mutant cells observed after two doubling times will be?

Answers

Answered by Subhanshu83
0

Answer:

Sequencing measurements of fixed mutations over 20,000 generations in E. coli. Because of this long-term experiment, it is possible to compare the full genome sequence at different times to the reference sequence for the genome at the time the experiment started. The labels in the outer ring show the specific mutations that were present after 20,000 generations. Adapted from J. E. Barrick et al. Nature, 461:1243, 2009.

Explanation:

The genomic era has ushered in the ability to read out mutation rates directly. It replaced older methods of inference that were based on indirect evolutionary comparisons or studies of mutations that are visually remarkable such as those resulting in color changes of an organism or changes in pathogenic outcomes. A landmark effort at chasing down mutations in bacteria is a long-term experiment in evolution that has been running for more than two decades in the group of Richard Lenski. In this case it is possible to query the genome directly through sequencing at different time points in the evolutionary process and to examine both where these mutations occur as shown in Figure 1 as well as how they accumulate with time as shown in Figure 2. Sequencing of 19 whole genomes detected 25 synonymous mutation

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