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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

Expert verified

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

For each type of offspring of the test-cross in Figure 15.9, explain the relationship between its phenotype and the alleles contributed by the female parent. (It will be useful to draw out the chromosomes of each fly and follow alleles throughout the cross.)

Butterflies have an X-Y sex determination system that is different from that of flies or humans. Female butterflies may be either XY or X0, while butterflies with two or more X chromosomes are males. This photograph shows a tiger swallowtail gynandromorphy, which is half male (left side) and half female (right side). Given that the first division of the zygote divides the embryo into the future right and left halves of the butterfly, propose a hypothesis that explains how nondisjunction during the first mitosis might have produced this unusual looking butterfly.

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\({\chi ^2}\)value means nothing on its own- it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are consistent with the hypothesis, and thus the hypothesis should be rejected, A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the\({\chi ^2}\)value is 0.05or considered statistically significant, the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically significant: the observed data are consistent with the hypothesis.

To find the probability, locate your\({\chi ^2}\)value in the\({\chi ^2}\)Distribution table in Appendix F. The โ€œdegree of freedomโ€ (pdf) of your data set is the number of categories (here,4 phenotypes), minus 1, so df=3.

(a). Determines which values on the df =3 line of the table your calculated\({\chi ^2}\)value lies between.

(b). The column headings for these values show the probability range for your\({\chi ^2}\)number. Based on whether there is non-significant (p\( \le \)0.05) or significant (p>0.05) difference between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

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

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