For a person with blood type A- (A negative), they can be transfused with either blood type A negative , or blood type O negative.
This is because any other blood type will immediately be rejected by their body.
A person with blood type A negative has got antibodies against any other blood type which has a rhesus positive factor as well as antibodies against any other blood type of a different group such as B and AB, even if they are rhesus negative.
Therefore because of the rhesus negative factor, a person with blood type A negative can only be transfused with rhesus negative .
Answer:
c) The species underwent punctuated equilibrium
e) The species’ environment underwent a dramatic change for which it was poorly suited
Explanation:
Answer:
Topoisomerase and Primase
Explanation:
Topoisomerase — Relaxes the super-coiled DNA
Primase — Provides the starting point for DNA polymerase to begin synthesis of the new strand
Answer:
The answer to this definition is: ACTION POTENTIAL
Explanation:
Action potential is a phenomenon in nervous system used to describe a change in electrical current that occurs when an impulse passes through a neuron or nerve cell. Action potential is induced when sodium (Na+) ions move from the extracellular environment into the intracellular environment.
Impulse travels across the axon of a neuron and thereby causes a difference in the electric potential i.e. - mv to + mv. This process, according to this question, whereby a change in electrical potential occurs due to the passage of an impulse through a nerve cell is termed ACTION POTENTIAL.
Answer:
(a) Frequency of M = 0.64
Frequency of N = 0.04
Frequency of MN= 0.32
(b) Expected frequencies of M = 0.648
Expected frequencies of MN = 0.304
Expected frequencies of N = 0.048
Explanation:
(a) If random mating takes place in the population, then the expected frequencies are
f(L(M)) = p = 0.8
F(L(N)) = q
q= 1 - p
= 1 - 0.8
= 0.2
Frequency of M = p^2 = ( 0.8)^2 = 0.64
Frequency of N = q^2 = (1-p)^2 = (1 - 0.8)^2 = (0.2)^2 = 0.04
Frequency of MN = 2pq = 2 * 0.8 * 0.2 = 0.32
(b)
F = inbreeding coefficient = 0.05
f(L(M)L(M)) = p^2 + Fpq = (0.8)^2 + 0.05 * 0.8 * 0.2 = 0.648
f(L(M)L(N)) = 2 pq - 2Fpq = 2 * 0.8 * 0.2 - ( 2 * 0.05 * 0.8 * 0.2) = 0.304
f(L(N)L(N)) = q^2 + Fpq = (0.2)^2 + ( 0.05 * 0.8 * 0.2) = 0.048
