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Chapter 15: Q15.4-3CC (page 309)

The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Review the discussion of cell cycle control in Concept 2.3 and explain how the activation of this gene could contribute to the development of cancer.

Short Answer

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The gene activated in the Philadelphia chromosome codes for a particular tyrosine kinase enzyme. This enzyme does the triggering of cell division. Overexpression of this enzyme can lead to the development of cancer.

Step by step solution

01

Description of cancer

Cancer is the form of tumor that may result due to the uncontrolled proliferation of cells. The loss in the control of the cell cycle can lead to cancer conditions.

Cancer has the ability to metastasis from the primary site of formation to the other secondary sites. There are different types of cancer, such as breast cancer, uterine cancer, lung cancer, etc. The condition of cancer is also known as carcinoma.

02

Description of tyrosine kinase

Tyrosine kinase is the enzyme that acts on cell growth and survival. This enzyme controls many important processes such as cell differentiation, proliferation, and apoptosis. It is an important enzyme in cell cycle regulation.

03

Philadelphia chromosome

Philadelphia chromosome contains the gene that results in the activation of tyrosine kinase. The over activation of this enzyme can deregulate the cell cycle division.

The over-activation of this enzyme can lead to the multiplication of cells in an uncontrolled fashion. It, in turn, results in the accumulation of more cells than the normal level resulting in tumor formation.

Hence, the activation of the gene in the Philadelphia chromosome can result in cancer.

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Most popular questions from this chapter

Crossing over is thought to be evolutionarily advantageous because it continually shuffles genetic alleles into novel combinations. Until recently, it was thought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosome with which to pair up prior to crossing over. However, when the Y chromosome was sequenced, eight large regions were found to be internally homologous to each other, and quite a few of the 78 genes represent duplicates. (Y chromosome researcher David Page has called it a "hall of mirrors.鈥�). Explain what might be a benefit of these regions.

The goodness of fit is measured by\({\chi ^{^2}}\). This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is

\({\chi ^{}} = \sum \frac{{{{\left( {o - e} \right)}^2}}}{e}\)

Where o=observed and e= expected. Calculate the\({\chi ^{^2}}\)value for the data using the table below. Fill out the table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the\({\chi ^{^2}}\)value.

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

220

(aaB-)

210

(A-bb)

231

(aabb)

239

\({\chi ^2}\) =sum

A wild-type fruit fly (heterozygous for the gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for the body color and eye color? Using information for problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

The continuity of life is based on heritable information in the form of DNA. In a short essay (100-150 words), relate the structure and behavior of chromosomes to inheritance in both asexually and sexually reproducing species.

The results in the data table are from a simulated F1 dihybrid test cross. The hypothesis that the two genes are unlinked predicts that the offspring phenotypic ratio will be 1:1:1:1. Using this ratio, calculate the expected number of each phenotype out of the 900 total offspring, and enter the values in that data table.
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