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Alexxandr [17]
3 years ago
8

in an experiment, you mutate the retinoblastoma gene such that its gene product behaves as if hyperphosphorylated. the result wo

uld be a _____________ association between rb and e2f, with __________ transcription of genes necessary for dna replication.
Biology
1 answer:
lakkis [162]3 years ago
3 0

Answer:

1. Weaker

2. Continuous

Explanation:

A tumor suppressor gene codes for retinoblastoma protein (pRb). The retinoblastoma protein binds to transcription factor E2F when DNA damage is detected. The E2F bound to pRb can not stimulate expression of genes that code for proteins required during the process of DNA synthesis. The cell can not enter the S phase.

On the other hand, when the retinoblastoma protein is phosphorylated by cyclin E-CDK2, E2F transcription factor is free to stimulate the expression of genes required for DNA synthesis and the cell proceeds from G1 to S phase.  

When the mutated retinoblastoma gene code for a protein that serves as phosphorylated protein, it will not be able to bind strongly with E2F and there would be continuous expression of genes required for S phase.  

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