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.
Answer:
True
Explanation:
Animals can be categorized into 3 based on body symmetry
- <em>Those without any body symmetry (asymmetrical)</em>
- <em>Those with bilateral body symmetry (bilateria)</em>
- <em>Those with radial body symmetry (Radiata)</em>
Animals can be categorized into 2 based on number of embryonic germ layer;
- <em>Those with two layers - endoderm and ectoderm (diplobastic)</em>
- <em>Those with three layers - mesoderm in addition to ectoderm and endoderm (triploblastic)</em>
Animals can be categorized based on presence/absence of body cavity or coelom;
- <em>No body cavity - acoelomates</em>
- <em>False body cavity - pseudocoelomates</em>
- <em>True body cavity - coelomates</em>
Animals can be categorized into 2 based on characteristics of embryonic development;
- <em>Deuterostomes</em>
- <em>Protosomes</em>
Answer:
Answer D
Explanation:
improved water habitats for human recreation
Answer:
Transformed into chemical energy
Explanation:
Answer:
The correct answer would be 0.85 liters.
It can be explained with the help of unit rate that is, when the rate is expressed as the quantity of 1 for example, 5 meter per second, 5 liters per day et cetera.
In this case, unit rate (liter per day) can be calculated by dividing the total amount of water used by total number of days.
Amount of water used in 28 days = 23.8 liters
The amount of water used in one day = 
It comes out to be 0.85 liters per day or 850 milliliter per day.
