Answer:
a line graph or a bar graph or anything near that to show that you understand
Explanation:
Answer: D. It yields chromosomes that contain both maternal and paternal DNA.
Explanation:
The recombination or crossing over is a genetic process. It takes place during the Prophase I of meiosis. The homologous pair align and held together by synapsis. This results in the mixture of genes combination formed from both parents that can be represented as a single chromosome.
If the recombination do not occur then each chromosome will be either 100% maternal or paternal. Thus recombination assures the recombinant chromosomes which consists of genetic material from both parents. Thus the genetic diversity of offspring increases.
D.many experimental trials with results that support the claims
Answer:
Approximately 6944 glucose residues are added enzymatically per second
Explanation:
Cellulose is the main structural polysaccharides in plants. It is composed of unbranched glucose monomer units linked to each other by beta 1-4 glycosidic bonds.
The cell wall and stem of plants cells are composed of cellulose fibers. They provide rigidity and support to the plant.
In the given bamboo plant, the enzymatic addition of glucose units to the growing cellulose fiber chains results in the phenomenal growth rate of the bamboo stem.
Since each glucose unit contributes ~0.5 nm to the length of a cellulose molecule, number of glucose units required for daily growth is calculated as follows:
0.5 nm = 10⁻⁹
0.3 m/0.5 x 10⁻⁹ m = 600000000 units of glucose per day
Number of seconds in a day = 24 * 60 * 60 = 86400 seconds
Number of glucose residues added per second = 600000000/86400
Number of glucose residues added per second = 6944.4 glucose molecules per second
Therefore, approximately 6944 glucose residues are added per second
