To prevent cancer, over exposure to radiation
If a species cannot reproduce then it will die out. This depends on the climatic conditions of the region. If there is a natural disaster, for example, it will affect a species' ability to reproduce. This reduces the number of diverse species on earth. Biodiversity is the recognised variety of different types of life found on the earth. It also measures the variety of number and variety of plants, animals and other organisms found in different ecosystems.
Explanation:
Since the 2 species co-inhabit a similar home ground, home ground isolation will be forgotten as an element keeping the 2species separate. There are but another broad class of mechanisms for reproductive isolation. One is behaviour. several birds specifically have terribly specific coupling rituals, together with coupling dances, specific birdsong, the event of specific coupling animal material, the discharge of specific pheromones and the building of a selected nest structure. Another broad class is mechanical isolation. you'll notice that the reproductive organ organs of specific species are terribly specific, and preclude different shut species from with success coupling. Another broad class is gametic isolation. for many species combos, it's inconceivable for the gametes to fuse with success to make a hybrid. If hybrid games do fuse, in several cases, the hybrids sterile, like mules as an example.
Answer:
Because mRNA tells you what type of veins
Explanation:
Answer:
underwater ecosystems formed in shallow water by the dense growth of several different species known as kelp's. ... Like those systems, though, kelp forests provide important three-dimensional, underwater habitat that is home to hundreds or thousands of species of invertebrates, fishes, and other algae.
Answer:
(a) 0.16
(b) 1
Explanation:
Let Probability that ticks in the Midwest carried Lyme disease, P(L) = 0.16
Probability that ticks in the Midwest carried HGE disease, P(H) = 0.10
Probability that ticks in the Midwest carried either Lyme disease or HGE disease, P(
) = 0.10
(a) Probability that a tick carries both Lyme disease (L) and HE (H) is given by
P(L 
H);
As we know that P(A 
B) = P(A) + P(B) - P(A B)
So, in our question;
P(L H) = P(L) + P(H) - P(L H)
0.10 = 0.16 + 0.10 - P(L H)
P(L H) = 0.16 + 0.10 - 0.10 = 0.16
Therefore, the probability that a tick carries both Lyme disease (L) and HE (H) is 0.16 or 16% .
(b) <em>Conditional Probability P(A/B) is given by</em> = 
So, the conditional probability that a tick has HE given that it has Lyme disease is given by = P(H/L)
P(H/L) = 
= 
= 1 .
