C organs because molecules combine to form cells and organs combine to make organ systems
Viral antigens like EBV are usually recognized by T cells involving an antigen fragment present within class I MHC to the T cell receptor.
Ebstein Barr Virus
The Epstein-Barr virus is a human herpes virus with unusual biological characteristics.
- It lives in practically every human person in a dormant condition in resting memory B lymphocytes
- It is also a strong transforming virus in vitro for B cells and is linked to numerous significant lymphomas, including Burkitt's, Hodgkin's disease, and immunoblastic lymphoma.
With the exception of the memory compartment, each stage of the cycle has been shown to be possibly controlled by the immune response.
Epstein-Barr virus (EBV), commonly known as human herpes virus 4, is a double-stranded DNA herpes virus that is extensively spread. It is the cause of infectious mononucleosis.
- EBV is most usually transmitted by body fluids, particularly saliva. EBV, on the other hand, can transmit by blood and sperm during sexual intercourse, blood transfusions, and organ transplants.
EBV can be transferred by utilizing materials that have recently been used by an infected individual, such as a toothbrush or a drinking glass.
- The virus is likely to survive on an object for at least as long as it is wet.
- When you initially get infected with EBV (primary EBV infection), you might spread the virus for weeks or even months before you notice any symptoms.
- Once the virus has entered your body, it remains dormant.
Hence, the correct answer is option A
Learn more about EBV here,
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Answer:
The correct answer is E. none of the above. The population will drops below 100 when t ≥ 38.
Explanation:
Given A= A0 e^kt. The population 10 years ago is A0, the population today is A(10), and we have to find the value of "k" and then the time when population drops below 100.
So, A(t) = 1700 e^kt ⇒ A(10) = 1700 e^k(10) ⇒ 800 = 1700 e^k(10) ⇒
800/1700 = e^k(10) ⇒ln (800/1700) = k(10) ln e ⇒ -0.754/10 = k ⇒
k = -0.0754.
Now you have all the parameters, so you can find the time at which the population drops below 100.
A(t) = 1700 e^kt ⇒ 100 = 1700 e^(-0.0754)t ⇒100/1700 = e^(-0.0754)t ⇒
ln(100/1700) = (-0.0754)t ln e ⇒ [ln(100/1700)]/(-0.0754) = t ⇒
t = 38.
So, the population will drops below 100 when t ≥ 38.
