<span>Below are useful in creating a phylogenetic tree of a vertebrate taxon:</span>
<span>1. </span>DNA sequence from living organism or also known as genetic sequence, it is used to determine strand of DNA
<span>2. </span>Behavioral data from living species, a method to identify the origin of individual in behavioral aspect
<span>3. </span><span>Morphological data from fossils, study the arrangement and physical aspect of organism </span>
I believe it's the long arm of chromosome 17.
<span>To find the number of trees per hectare, simply divide the two figures together. 11,025 (the number of trees) divided by 3,150 (the number of overall hectares in the plot of land) gives a value of 3.5 trees per hectare.</span>
Answer:
a) Photosystem I requires the use of light converting an excited NADP + electron, reducing it to NADPH. In photosystems I and II, the passage of electrons to a series of redox reactions will be shown due to the absorption of light energy by the chlorophyll molecules.
(b) The control must be RuBisco, because for most photosynthesis responses to light, temperature and carbon dioxide, it produces oxygenation and carboxylation reactions, thus reflecting kinetic properties.
(c) plants can increase the ability in a variable ecosystem thanks to the flexibility of photosynthetic pigments, through these the necessary solar energy for photosynthesis is captured, according to each light spectrum there will be a specific pigment.
Answer:
D. 0.60
Explanation:
If the population is in Hardy-Weinberg equilibrium, the genotypic frequencies are:
- freq AA = p²
- freq Aa = 2pq
- freq aa = q²
<em>p</em> is the frequency of the Rh positive allele (A) and <em>q</em> is the frequency of the Rh-negative allele (a).
If 84% of the population is Rh-positive, then 16% is Rh-negative and has the genotype <em>aa</em>. Therefore:
q² = 0.16
q = √0.16
q=0.4
And because p+q=1,
p = 1 - 0.4
p = 0.6
The frequency of the Rh-positive allele is 0.6
