Answer:
B) oaks and other sexually reproducing, extant (currently living) trees
Explanation:
The biological species concept defines the species on the basis of their reproductive isolation. It states that when individuals are able to interbreed to produce fertile and viable progeny, they belong to the same species. The members of different biological species cannot interbreed. If they interbreed, either pre-zygotic or post-zygotic isolation mechanisms do not allow the production of fertile progeny.
Therefore, the biological species concept can be applied to the organisms that are able to reproduce sexually. The asexually reproducing organisms would not exhibit any reproductive isolation which is a key criterion to group organisms under different species. Among the given examples, biological species concept can be applied to the sexually reproducing extant trees such as oak.
Since we cannot deduce the reproductive isolation in sexually reproducing extinct species, the concept is not useful for dinosaurs which are extinct now.
Answer: A.Independent assortment
Explanation:
The other three can be introduced into sexually reproducing organisms.
Answer:
Biston betularia f. typica, the white-bodied peppered moth.
Typica and carbonaria morphs on the same tree. ...
Creationists have disputed the occurrence or significance of the melanic carbonaria morph increasing in frequency
Answer:
The new plants are identical to the mother strawberry plant because they are the result of asexual reproduction, that is, from successive mitosis clones of the mother plant are made. The runners do vegetative reproduction, they make the vegetative reproduction by means of which roots and stems are formed with identical cells to the parent.
Answer:
When the pKa is 6.0, we can determine the fraction of protonated H is by:
pH = pKa + log [A]/[HA]
Where
A = Deprotonated imidazole side
HA = Protonated side
Given, pH = 5.0
5 = 6 + log [A]/[HA]
log [A]/[HA] = -1 (take antilog of both side)
[A]/[HA] = 0.1
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 0.1
Given, pH = 7.5
7.5 = 6 + log [A]/[HA]
log [A]/[HA] = 1.5 (take antilog of both sides)
[A]/[HA] = 31.62
The ratio of the deprotonated imidazole side chain to the protonated side chain at pH 5.0 = 31.62
