The Inconsistencies of the data are led to further investigations conclude about data that are inconsistent with the current, scientific understanding of amphibian reproduction
Option B
<h3><u>Explanation:</u></h3>
Class amphibia is one of the classes under the phylum chordata. This class has typical reproductive features. They are the animals that can live in both land and water, and a few are exclusively aquatic. But they essentially need water for the fertilization process. This is because they generally undergo external fertilization and the females lay eggs beside the water body. The males eject the sperms in water which swim through the water into the eggs and fertilize them.
This is the days old theory regarding the amphibia fertilization. But if some new theory comes up someday, that readily don't exclude this theory. The new theory has to pass through several tests and various further investigations where it's seen if the new theory explains most of the amphibia reproduction or not. Then it can be approved along side with the old theory.
Answer:
Myelination.
Explanation:
Brain acts as one of the most important organ of the body that helps in receiving the stimuli and interpret the information in the form of signal. Two main types of neuron are myelinated neuron and unmyelinated neuron.
The myelinated neuron increases the conductivity of the impulse. The myelinated neurons are much more active than the unmyelinated neuron. In case of the quick thinking and instant actions, the myelinated neurons play an important role.
Thus, the answer is myelination.
Answer:
Thus, 50% of the population are are heterozygous carriers for this condition
Explanation:
Researchers have calculated that 22% of the population have Uner Tan Syndrome. This condition follows the recessive mode of inheritance.
22% have the condition = q² (genotypic frequency uu) = 22/100 = 0.22
using the formular p + q = 1, since q² = 0.22, q = √0.22 = 0.4690
Thus p = 1-q where is 0.4690 = 1-0.4690
p = 0.5310
Then p² = 0.5310² = 0.2820 = 28% (genotypic frequency of homozygous normal)
using the formula p² + 2pq + q² where 2pq is the genotypic frequency of the heterozygous carriers. Thus, we have
2 x 0.5310 x 0.4690 = 0.498 = 50%
Thus, 50% of the population are are heterozygous carriers for this condition
The body would not be able to cool itself so excess heat would remain in the body instead of being lost via evaporation of sweat. The rise in temperature would not be brought back to normal which would affect thermo-regulation in the body.
The high body temperature would most definitely affect the activity of enzymes in the body which control most metabolic reactions.
