Chromosome #4, an autosome of the diploid smorglefuger, is rod-shaped and normal
ID: 67309 • Letter: C
Question
Chromosome #4, an autosome of the diploid smorglefuger, is rod-shaped and normally has its centromere positioned at one end (telocentric). A circular variant of this chromosome exists, and except for the lack of physical ends (telomeres) of the chromosome, it is identical in genetic content to the rod-shaped variety; therefore, the two variants will synapse normally. The position of five genes (a-e) on the two forms of this chromosome are shown below; the five genes are all equally spaced along the chromosome.
Here are three different females of the genotype a+b+c+d+e+/a b c d e. The structures of their homologous pair of chromosome #4 are indicated. Each of these types of female is testcrossed to homozygous a b c d e males that carry chromosome #4 only as its normal rod form.
Everything else being equal, which, if any, female type(s) will produce the highest fraction of phenotypically a+b+c+d e progeny?
Select all that apply.
Please explain Thanks
1. female A 2. female B 3. female C 4. none will produce those progenyExplanation / Answer
According to the given data, the possibility of crossing over among genes is high in case of female A when compared to others. This is because the female A is homozygous for rod form, while the other two females are homozygous for circular form and heterozygous for rod form respectively.
The crossing over is not possible between a rod and circular chromosome. Similarly, the crossing over is not possible between two circular chromosomes.
Thus, the female type(s) that will produce the highest fraction of phenotypically a+b+c+d e progeny is 1-female A.