________________________________
<h3>1. YES - Because progression is shown.</h3><h3 /><h3>2. NO - Because the data is evidence that the athlete can run much faster.</h3>
________________________________
Answer:
I think it is the last one
Answer:
C. Chlamydia. is the correct answer.
Explanation:
Chlamydia is a bacterial sexually transmitted disease, caused by obligate intracellular Chlamydia trachomatis.
In the life of chlamydia, there are two stages of developmental(Reticulate and elementary bodies are two forms of Chlamydia.)
The elementary body is the infectious substance, they are non replicating and have a rigid outer membrane that binds to the receptors present on the host cell and it initiates the infection and due to the rigid outer membrane there, no fusion between the phagosome and the lysosome hence oppose the intracellular killing.
Reticulate bodies are the metabolically active form of a chlamydia and non-infectious.
<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>
