Answer:
I think the answer us sulfuric acid
Answer:
The correct option is - (a) the activation and inactivation gates must both be open.
Explanation:
Option (a) is correct because of the following reason -
The axonal membrane is at its natural resting potential until an action potential occurs, and channels are deactivated and blocked on the extracellular side by their activation gates. The activation gates open in response to an electric current, allowing positively charged ions to flow into the neuron through the channels. When enough has penetrated the neuron and the membrane potential has reached a certain level, the channels inactivate themselves by closing their inactivation gates at the height of the action potential. The inactivation gate can be thought of as a "plug" tethered to the intracellular alpha subunit's domains III and IV. When the inactivation gate is closed, the flow of through the channel is stopped. As a result, both the activation and inactivation gates must be open to enable sodium ions to enter the cell.
<u>Hence , the correct option is (a).</u>
Answer and Explanation:
<u>Available data:</u>
- New breeding occurred between two different cats.
- Both cats were heterozygous hairy cats,
- One parent loved sinks and the other parent disliked sinks.
Let us say that:
- the gene for the cat hair trait is H, being "H" the dominant allele expressing hair, and "h" the recessive allele expressing hairless.
- the gene for the sinks-liking trait is S, being "S" the dominant allele expressing sinks-liking, and "s" the recessive allele expressing sinks-desliking
If both cats are heterozygous for the hair trait, then their genotypes would be Hh in both animals.
One of the cats likes sinks, so its genotype would be S-. The "-" symbol means that we do not know if this is a homozygous or heterozygous animal. The other cat disliked sinks, so its genotype for the trait would be ss.
Cross: HhS- x Hhss
Gametes) HS hS H- h-
Hs hs Hs hs
Punnet square: HS hS H- h-
Hs HHSs HhSs HHs- Hhs-
Hs HHSs HhSs HHs- Hhs-
hs HhSs hhSs Hhs- hhs-
hs HhSs hhSs Hhs- hhs-
a) List the genotype of each parent based on the information above.
Parent 1: HhS- ---> there are two possibilities: HhSs or HhSS
Parent 2: Hhss
b) Determine how many of the potential offspring would be hairless and love sinks
If parent 1 is HhSs, the potential offspring that would be hairless and sinks-loving would be 2/16=1/8. Its genotype would be hhss.
HS hS Hs hs
Hs HHSs HhSs HHss Hhss
Hs HHSs HhSs HHss Hhss
hs HhSs hhSs Hhss hhss
hs HhSs hhSs Hhss hhss
But if parent 1 is HhSS, there would not be any individual hairless and sinks-loving. All of them would be heterozygous for sinks-liking.
HS hS HS hS
Hs HHSs HhSs HHSs HhSs
Hs HHSs HhSs HHSs HhSs
hs HhSs hhSs HhSs hhSs
hs HhSs hhSs HhSs hhSs
Answer: the answer should be a.
Explanation: