D. Growing and reproducing
Hope this helps :)
Answer: If this population were in equilibrium and if the sickle-cell allele is recessive, the proportion of the population susceptible to sickle-cell anemia under typical conditions should be 0.20
Explanation: Hardy-Weinberg law provides an equation to relate genotype frequencies and allele frequencies in a randomly mating population. The equation is;
p² + 2pq + q² = 1
For 2 alleles such as A and a, where
p² = homozygous dominant
q² = homozygous recessive and
2pq = heterozygous
From the question, it is said that the sickle-cell allele (SS) constitutes 20% (that is, 20/100) of the hemoglobin alleles in the human gene pool and it is also said to be the homozygous recessive allele.
Therefore, q² = 20/100 = 0.20
During mitosis, the <em />nuclear membrane breaks down and the duplicated chromosomes are separated and evenly distributed to opposite sides of the cell. Cytokinesis involves the division of the cytoplasm and the organelles that are contained within.
This describes the process of cell division. All of these events happen in the M phase of mitosis. The breaking down and fragmentation of the nuclear membrane to expose the genetic material is in the prophase. The next phase is the metaphase when the duplicated chromosomes align in a straight line (metaphase plate). Anaphase involves the separation of chromatids and by the time the chromatids reach the opposite side then the cell enters telophase. Lastly, cytokinesis involves restoration of the nuclear membrane, division of the cytoplasm, and production of two daughter cells.
You need two recessive alleles (xx). If even one dominant allele is present, the dominant trait will show (Xx or XX will both result in the dominant trait showing). That’s why they’re called dominant traits, because they overpower the recessive traits. In order for a recessive trait to show you must have two recessive alleles.
