The correct answer is antigens and epitope.
An antigen is a substance which triggers an immune response in an organism. This can be either a foreign substance, such as a pathogen, or a specific part of the host organism (in this case this is an autoimmune response). The presence of an antigen activates a specific part of the immune system, called antibody. Each antigen has a specific antibody, which is tailored by the immune system accordingly. More specifically, the antibody has a specific structure, called paratope, which is complementary and binds like a key to a specific structure of the antigen, called epitope.
Flora:Calochortus flexuosus, Calochortus kennedyi, Opuntia echinocarpa.
Fauna:bighorn, coyote, bobcat, mountain lion, and mule deer. Mule deer
The person would be unable to fully digest dairy products
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 .
Answer: The stoichiometry products of the citric acid cycle assuming one molecule of glucose is oxidized are as follows:
2 mol acetyl CoA, 2 mol ATP equivalent, 2 mol OAA, 6 mol NADH, 2 mol FADH2, 4 mol CO2
Explanation:
Citric acid cycle is a pathway designed to burn away carboxylic acids as two molecules. The citric acid cycle accepts 2 carbon molecules and oxidizes it to water and carbon dioxide. After glycolysis, pyruvate is converted to acetyl CoA which enters the citric acid cycle.
Considering the oxidation of sugar, one molecule of glucose generates 4 NADH while being converted to acetyl CoA. One molecule of glucose undergoes two round of citric acid cycle . 2 mol of CO2 are released in each turn of the cycle.
Each turn of the cycle yields 3 NADH molecules and one FADH2 molecule. One ATP molecule is formed in each cycle.
Through the two rounds of citric acid cycle, one molcule of glucose generates 6 NADH, 2 FADH2, 2 ATP and 4 CO2 in total.
