<span>In Drosophila + indicates wild-type allele for any gene, m is mahogany and e is ebony.
Female parents are m+/m+ and males are +e/+e.
F1 are m+/+e, all wild type. F1 females are crossed with me/me males - the test cross.
Offspring will be : non recombinant m+/me, mahogany wild type or +e/me wild type ebony. OR
recombinant me/me mahogany ebony or ++/++ wild type.
As the two genes are 25 map units apart, the percentage of recombinants will be 25% and therefore percentage parental types will be 75%.
75% 1000 is 750. There are two parental types, so you would expect 375 of each. Therefore, you would expect 375 m+/me and 375 +e/me.
25% of 1000 is 250 split between two recombinants =125 of each. Therefore you would expect 135 me/me and 125 ++/++</span>
<span>The answer to the question is that the behavior is an example of overproduction. A frog needs to lay thousands of eggs because so many of them die in the process of hatching to tadpole stage to frog. A frog lays so many eggs in the hope that a few of them will be able to reach adulthood and be able to reproduce the species in the future.</span>
Answer:
The cells are identical because the DNA is copied during the S phase before cell division and then each resulting cell gets a matching copy of the chromosomes.
<em>hope it helps :)</em>
