Answer:
In pharmacology, the term mechanism of action (MOA) refers to the specific biochemical interaction through which a drug substance produces its pharmacological effect. A mechanism of action usually includes mention of the specific molecular targets to which the drug binds, such as an enzyme or receptor.
sand dunes form from wind
Answer:
Option 4. Autonomy vs. shame and doubt is correct answer.
Explanation:
Erik Erickson presented theory of personality development which is very popular and acceptable. He divided his theory into 8 sages each having 2 components. I have attached an info graph explaining clearly each stage.
Answer:
First off, the info tells me that bandicoots, sea lions, and zebras could have all shared a common ancestor at one point in time and then developed differences to be better suited to their respective environments. Further, the similarities in their body structures cannot be attributed to convergent evolution (evolution that produces analogous structures) because the three organisms grow in very different environments; thus it had to have come from homology (common ancestry). Having said that, this tells us that the ancestors most likely had the trait of whatever is shared amongst the body structures of bandicoots, sea lions, and zebras, as this was the trait they passed on. Hope this is helpful :)
Answer:
0.9342
Explanation:
The Hardy-Weinberg equation states that p² + 2pq + q² = 1,
where p is the frequency of the dominant 'normal' (n) allele and q is the frequency of the recessive 'albino' (a) allele in the population, while q² represents the frequency of the homo-zygous albino genotype (aa), p² represents the frequency of the homo-zygous normal genotype (nn) and 2pq represents the frequency of the heterozygous genotype (na).
In this case, the frequency of individuals in the population that have the genotype aa (q²) is equal to 26/6000 = 0.004333. In consequence, q is equal to √ 0.004333 = 0.0658.
Moreover, the allele frequency of the normal (n) allele p is equal to 1 - q = 1 - 0.0658 = 0.9342, so p² (nn) = (0.9342)² = 0.8727.
Finally, the frequency of the heterozygous genotype (na) is 2pq = 2 x 0.9342 x 0.0658 = 0.123.
