Answer: There is presence of tumor.
Explanation: The adhesion of cells to extracellular matrix (EMC) through integrins ( cell-EMC binding molecules, which are collagens, laminins and fibronectin) causes the activation of kinases in the cytoplasm.
However, kinanes helps in controlling the epithelial cell differentiation and upholding the epithelial tissues. This is done by the addition of phosphate groups to a substrate protein which is termed Protein phosporylation. Then, the kinases direct the affairs of the cell and it's activities. For example, it determines the cell division, anabolic and catabolic activities of the cell, movement of ions between the cell and it's environment (signal transduction), protein functions and etc.
Conclusively, since the activities of the cell like cell division and protein functions is dictated by the kinase, reduction in cell division that gave rise to rapid growth is put on hold. Hence, the tumor is been suppressed.
Note: the binding of cell-EMC is regulated by Transforming Growth Factor (TGF) β.
Yellow-indicates acid production from fermentation of the carbohydrate lower the pH below the neutral range
Answer/ Explanation:
a. The genotype of a homozygous white eyed long winged female would be Vg+Vg+XrXr. We denote the white allele as recessive (r) because the XY male only has one copy and yet has red eyes, so the red eye trait (R) must be dominant. A homozygous red eyed vestigial winged male would have be VgVgXRY. The possible gametes for the female are Vg+Xr only. For the male, the possible gametes are VgXR or VgY
The attached punnett square shows the results of the cross. The females will all be Vg+VgXRXr. The males will all be Vg+VgXRY (must inherit Y from father). That means they will all have normal length wings, the males will have white eyes and the females will have red eyes.
b. The F2 flies arise from intercrossing the F1, so the cross will be Vg+VgXRXr x Vg+VgXRY. The possible gametes for the mother are: Vg+XR, Vg+Xr, VgXR or VgXr. The possible gametes for the father are Vg+Xr
, Vg+Y
, VgXr
, VgY
. The attached punnet square shows this cross. The ratio of the phenotypes will be 6:6:2:2, or 3:3:1:1 (long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye), genotypes shown in the attachment.
c. F1 cross back to the mother would be Vg+VgXRY x Vg+Vg+XrXr. The genotypes are shown in the attached punnet square. The offspring will all be long-winged with white eyes. The F1 to the father would be Vg+VgXRXr x VgVgXRY. The ratio would be 3:3:1:1 long-winged red eye: long-winged white eye: vestigial wing red eye: vestigial wing white eye
A toxic substance is absorbed by a green plant and is completely blocking the production of ATP
It would be called a biome, like a tundra or a desert.
