<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>
Answer:
they have short life spans and many predators.. hope this works
The statement which best explains how the redox component of this reaction contributes to the reaction's ability to be reversible under cellular conditions is; <em>Choice D: The change in the biochemical standard reduction potential is small.</em>
Discussion:
A reversible process is one in which the system and environment can be restored to exactly the same initial states that they were in prior to when the process occurred, if we go backward along the path of the process.
- However, the necessary condition for a reversible process is therefore the quasi-static requirement.
- The quasi-static requirement in this case is that the change in the biochemical standard reduction potential is small.
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A courtship dance or mating ritual hope this helped
False, only some . Hope this helps !
