Answer:
For example, addition of an orange chromic acid reagent to some compounds causes the chromium reagent to change to a blue-green color (Figure 6.37a). This is considered a "positive" test result, and in this case indicates the presence of a functional group that can be oxidized (alcohol or aldehyde).
The answer is <span>root hairs</span>
The
correct answer is they filter urea from blood and produce urine.
<span>
Nitrogenous
wastes tend to form toxic ammonia which needs to be removed. Terrestrial
animals must detoxify ammonia by converting it into a relatively nontoxic molecule-urea
(it occurs in the urea cycle). The urea cycle mainly occurs in the liver and the
produced urea is then released into the blood. It travels to the kidneys where
is filtrated and excreted in urine.</span>
Answer:
D. 6
Explanation:
carbon atom has 6 protons
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
