Mutations in genes encoding cell signaling proteins contribute to many cancers. for example, a chromosomal translocation fuses the bcr gene to the abl gene, leading to a constitutively active bcr–abl kinase and chronic myeloid leukemia (cml). cml is successfully treated with the drug imatinib (trade name gleevec), which mimics an abl kinase substrate and thus inhibits kinase activity. what normal cell molecule, and kinase substrate, does gleevec mimic?

Imatinib is a small molecule kinase inhibitor. The BCR-ABL kinase can phosphorylate a series of downstream substrates, leading to proliferation of mature granulocytes. Bcr-Abl kinase substrate is the tyrosine. The Protein Tyrosine Kinase activity is an important requirement for malignant transformation, and that it cannot be complemented by any downstream effector, though not all interactions of BCR-ABL with other proteins are phosphotyrosine dependent.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years