C. A capsid is a protein coat that protects the genetic material of the virus
Hope that helps!!
Answer:
The answer is <em><u>D.
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Explanation:
Viruses can evade the processing and presentation of antigens, for example by interfering with the expression of MHC class I proteins, although this increases their susceptibility to be detected by natural killer cells (NK). However, some members of the herpesvirus, papillomavirus, retrovirus, poxvirus and flavivirus families have also developed strategies to escape the attack of NK cells and promote their survival, inhibiting cell apoptosis. Finally, some pathogens often change surface antigens frequently, as in the case of influenza viruses (orthomyxovirus).
The human immunodeficiency virus (HIV) affects CD4 + T cells and degrades the host's ability to counterattack with a strong cell-mediated immune response. There are so many tactics of immune evasion used by HIV, which have so far hindered the development of an effective vaccine.
Answer:
An animal nutritionist specializes in the dietary needs of animals in captivity, such as pets, farm animals, and zoo animals.
Explanation:
hope this helps
More water means more pee (theoretically). Pee is actually just garbage materials from the body (toxins). These color the urine. The more you drink, the more you have to pee and just pee clean because all toxins are out, or you have more water to take out the toxins, having less of it per let's say 0.5 l
Answer:
6,25%
Explanation:
Considering that the couple has a trait of sickle cell anemia, we know that both are heterozygous for the disease (Aa) and therefore can have children with the following genotypes:
Parents: Aa X Aa
Children: AA(A x A), Aa(A x a), Aa (a x A) and aa(a x a)
Knowing that sickle cell anemia only occurs in homozygous individuals, the probability for children to have the disease according to each crossing is:
A x A = 1/4 = 25%
A x a = 1/4 = 25%
a x A = 1/4 = 25%
a x a = 1/4 = 25%
The probability of forming each homozygous child (aa) is 1/4 or 25%. Since they are two children, the probability of both having sickle cell anemia is calculated by multiplying the probability of each, so:
1/4 × 1/4 = 1/16 = 0.0625 = 6.25%
It is concluded that the probability of a heterozygous couple for sickle cell anemia to have two children with the disease is 6.25%.
