The cell cycle is an ordered series of events involving cell growth and cell division that produces two new daughter cells. Cells on the path to cell division proceed through a series of precisely timed and carefully regulated stages of growth, DNA replication, and division that produces two identical (clone) cells.
sorry if it is wrong
Answer:
The right answer is D. The benthic zone
Explanation:
The limnetic and the littoral zone all get a lot of sunlight so there are a lot of plant life. The pelagic zone hosts a variety of plant species so this cannot be the correct answer. That leaves us with the benthic zone.
Benthic meaning inhabiting bottom areas or substrates. Plant life at the bottom is very scarce but there are a lot of insect species.
Answer:
DNA plasmid that contains both "old" and "new" gene segments and confers new characteristics to the organism in which it is placed.
Explanation:
During the genetic engineering procedure, which aims at producing an organism with better and desirable characteristics, a DNA called PLASMID is usually used to convey the gene of interest into the organism.
A plasmid is a self-replicating extra-chromosomal DNA found in the bacterial genome. The plasmid becomes a RECOMBINANT PLASMID when a foreign DNA of interest is inserted into it, in order to act as a vector (carrier). Therefore, a recombinant plasmid is a DNA plasmid that contains both "old" i.e bacterial genome and "new" i.e foreign gene segments and confers new characteristics to the organism in which it is placed.
Organelle,
Cells,
Tissues,
Organs,
Organ System,
Organisms,
Populations,
Communities,
Ecosystems,
And Biosphere
Answer:
Frequency of B allele is 0.6681
Explanation:
If p represents the frequency of dominant allele and q represents the frequency of recessive allele, according to Hardy-Weinberg equilibrium:
p + q = 1
p² + 2pq + q² = 1
where p² = frequency of homozygous dominant genotype
q² = frequency of homozygous recessive genotype
2pq = frequency of heterozygous genotype
Given that number of recessive chestnut horse = 28
Total horses = 226 + 28 = 254
frequency of b² genotype = 28/254 = 0.1102
frequency of recessive b allele = √0.1102 = 0.3319
So, frequency of B allele =
1 - 0.3319 = 0.6681
Hence frequency of B allele is 0.6681
