Anaplasia in Wilm’s tumor is evident by –
**Question:** Anaplasia in Wilm's tumor is evident by -
A. Loss of HLA expression
B. Loss of chromosome 13
C. Loss of chromosome 16
D. Increased Ki-67 proliferation index
**Core Concept:**
Anaplasia refers to the loss of differentiation in neoplastic cells, often associated with an increased risk of progression and poor prognosis. In Wilm's tumor, which is a pediatric kidney cancer, anaplasia can be assessed by specific molecular markers and cytogenetic abnormalities.
**Why the Correct Answer is Right:**
Differentiating between benign and malignant Wilm's tumors is crucial for appropriate treatment planning and patient management. Anaplasia can be detected using specific markers and chromosomal abnormalities. In this case, the correct answer is **D**. Increased Ki-67 proliferation index indicates a higher rate of cell division and is associated with malignant tumors.
**Why Each Wrong Option is Incorrect:**
A. Loss of HLA expression (human leukocyte antigen) is a marker for immune surveillance, not a marker for tumor differentiation or malignancy.
B. Loss of chromosome 13 is not specific to Wilm's tumor and can occur in other malignancies as well.
C. Loss of chromosome 16 is also not specific to Wilm's tumor and can be found in various cancers.
**Clinical Pearl:**
Assessing tumor differentiation and malignancy in Wilm's tumor is essential for appropriate treatment planning. Increased Ki-67 index indicates anaplasia, while loss of HLA expression, chromosome 13, and chromosome 16 are not specific markers for Wilm's tumor and should not be solely relied upon for tumor grading. Instead, consider Ki-67 index as a more reliable marker for assessing tumor malignancy and guiding treatment decisions.