An action potential traveling along the entire length of an axon demonstrates <u>continuous</u> propagation.
Propagation of action potential:
- An action potential is created by the rapid input of Na+ ions and the slightly slower outflow of K+ ions.
- Action potential initiation occurs in the axon's first segment, and as action potentials spread throughout the axon, cells can convey their output to synapses located further distant from the cell.
- The action potential conducts continuously in unmyelinated axons, whereas it conducts rapidly and saltatorily in myelinated axons where sodium channels are concentrated near the nodes of Ranvier.
- The action potential always advances in the same direction whenever a local current begins to form due to the prior axon segment still being in the absolute refractory period. As a result, action potentials travel away from the location of their generation and do not change course.
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Answer:
The correct answer is -
Parental - BB and bb
F1
Genotypes: 100% Bb
F2
Genotypes: ¼ BB, ½ Bb, ¼ bb
Explanation:
It is given that a cross between a brown and a silver-blue mink produces all brown offspring and it is only possible if brown is dominant over silver blue in true breed cross, assume brown trait is represented by B and silver-blue is b then parents will be
BB and bb
B B
b Bb Bb
b Bb Bb
Cross between F1 individuals: Bb cross Bb
gametes : B, b and B, b
B b
B BB Bb
b Bb bb
the genotype of F2 : 1:2:1 = ¼ BB, ½ Bb, ¼ bb
Answer:
A. Respiration
Explanation:
The process in which oxygen atoms are made available to the cell of animals is called cell respiration. In cell respiration the food molecules undergo process which is called food oxidization, that make it available to be absorbed by the animal cell.
Answer:
9 amino acids
Explanation:
A segment of a gene that has 27 nucleotides will code for<u> 9 amino acids.</u>
<em>Each codon on the gene potentially represents an amino acid and a codon is made up of 3 nucleotides each. Hence, in order to find out the total number of amino acids that a gene segment with 27 nucleotides can code for, the number of nucleotides should be divided by 3.</em>
number of amino acids = number of nucleotides/3
27/3 = 9 codons/amino acids
