In the heterozygous state (w +/w), a variegated eye is produced, with white and red patches. The following statements are true in relation to this experiment:
b. When heterochromatin spreading does not reach the new location of the w + allele, the gene will be transcribed, producing red eye patches.
c. When heterochromatin spreading encompasses the new location of w + allele, the gene is transcribed, producing white eye patches.
d. Even though eye color phenotypic variegation exists within the eye, all cells have the same genotype.
Explanation:
Heterochromatin in droshophila has genes which gets expressed by position effect variegation. In some normally active cells genes are silenced in some due to transposition of genes which causes change or variegating in phenotype.
This change is due to change in position of the gene in the genome and chromosome but no change in gene.
The variegation in drsophilla melanogaster shows that white gene was not damaged.
The rearrrangement of gene takees place at pericentric heterochromatin.
Heterochromatin is required for maintaining genome integrity and regulating gene expression. When spread in neighboring region it can inactivate the genes present there.
The genes taking part in variegation remains same hence no change in genotype.
Answer:
an individual animal, plant, or single-celled life form.
Answer:
The correct answer would be - CDK inhibitors.
Explanation:
Removing the CDK inhibitors in nonproliferating cells can reactivate the cell cycle that can be useful in various cell and tissue repair and also in cell replacement therapies.
These cells are depends on the parent cells that are able to divide so that the daughter cells are limited to the same cell lineage. If CDK inhibitors are removed from the cells can help in proliferating the cells as the cyclin B will be binding to the CDK1 that helps in cell cycle by rising Cyclin B - CDK complex.
Thus, the correct answer is - CDK.
