Answer:
Q(0) = 0C, Q(1) = 264nC, Q(2) = 952C Q(3) = 2088nC, Q(4) = 3696C Q(5) = 5800nC
Explanation:
I = 4t³ + 200t² + 60t
But charge of an object =》 Q = IT
Charge of an object is the product of the current and the time in which the current passes through the membrane.
When t = 0
Q = 4(0)³ + 200(0)² + 60(0) = 0C
When t = 1
Q = 4(1)³ + 200(1)² + 60(1) = 264nC
When t = 2
Q = 4(2)³ + 200(2)² + 60(2) = 952nC
When t= 3
Q = 4(3)³ + 200(3)² + 60(3) = 2088nC
When t= 4
Q = 4(4)³ + 200(4)² + 60(4) = 3696nC
When t = 5
Q = 4(5)³ + 200(5)² + 60(5) = 5800nC
A plant cell placed in a hypotonic solution will not lyse because the plant cells tend not to bust..
Answer:
- The lac operon can be activated by the binding of allolactose to the repressor protein, releasing it from DNA and thereby allowing for transcription to occur.
- In response to low glucose levels, cAMP is upregulated; the binding of cAMP to the cAMP receptor protein triggers the activation of the operon.
Explanation:
Lactose operon or lac operon (includes lacZ, lacY and lacA genes) is found in some bacteria and the products of its genes are involved in lactose metabolism. So, this operon is active (genes are transcribed) when lactose is present and glucose is absent (or at low level). The operon is regulated by the lac repressor which acts as a lactose sensor and catabolite activator protein (CAP) which acts as a glucose sensor.
When there is lactose (in the form of allolactose) lac repressor detects it and stops being repressor. This enables transcription.
CAP detects glucose (via cAMP) and activates transcription when glucose levels are low.
Answer: it would be natural selection
Explanation:
