Genes A and B are neutral. A weakly beneficial mutation arises in the population. This mutation is 100 base pairs away from Gene A and 1000 base pairs away from Gene B. If this mutation were to go to fixation within the population, which gene would be more likely to go to fixation and what is the term for this process? Is there any reason to suspect that one or both of these genes may not go to fixation? Why or why not?

Both genes would be likely to go to fixationThe term for this process is "linked genes"The reason to suspect that both of these genes may not go to fixation is that they are too close to the mutation and the recombination frequency between them is very very low.

Explanation:

Independent assortment law establishes that the alleles from two or more different genes distribute in gametes independently from each other. In other words, a gamete receives an allele from a gene that does not depend nor influence the allele of another gene in the same gamete. This can only be applied to independent genes. These genes segregate independently after crossing-over because they are located far away from each other.

Some other genes, however, are too close to each other and they do not segregate independently. These are the linked genes that do not exhibit an independent distribution, and they inherit together more frequently.  

Crossing-over between linked genes that are very close to each other in the chromosome is not that common. Crossing-over during meiosis occurs randomly in different positions all along the chromosome, and its occurrence frequency in the area between two genes depends on the distance between them. A short distance between genes is a very little target for crossing-over to occur, which means that only a few of them will happen, compared with the number of events between genes that are more separated between each other.  

Two genes that are very close will have a few recombination events and are strongly bounded.  

The more separated two genes are, the more chances of recombination there will be.  The closer they are, the fewer chances of recombination there will be.

Genes that express 50% of recombination frequency or more are not linked genes.  

To analyze the recombination frequency, we have to know that

1% of recombination = 1 map unit = 1centi Morgan = 1,000,000 base pairs.

And that the maximum recombination frequency is always 50%.  

The map unit is the distance between the pair of genes for which every 100 meiotic products one of them results in a recombinant one.  

In the exposed example we know that the distance of gene A from the mutation is 100 base pairs, and the distance of gene B from the mutation is 1000 base pairs.

1,000,000 base pairs 1% recombination frequency

1000 base pairs X = 0.001% recombination frequency

100 base pairs X = 0.0001% recombination frequency

According to the recombination frequency between the mutation and gene A, and between the mutation and gene B, we can assume that both genes are linked to the mutation, as they seem to be too close to it. They are so close, that their recombination frequency is very little.  

                                             

Option D , All of the above

Explanation:

Capillary electrophoresis (CE) is used to separate STRs in a much effective way as compared to the traditional slab-gel electrophoresis method-  

a)CE method is a speedy process

b)It automates the DNA-typing procedures

Being an emerging technology, CE possesses following advantages –  

a)The critical processes such as injection, separation can be fully automated

b)These process can also be made speedy through appropriate and efficient heat dissipation

c)Waste is reduced  

Hence, option D is correct