Answer:
This phenomenon is known by neuroscientists as Neuroplasticity, or brain plasticity
Explanation:
Think of plasticity, Neuroplasticity, or brain plasticity, is the simplest terms, as the ability of parts of the brain to adapt their function or ability to change throughout life.
Structural plasticity - brains ability to change in response to the environment.
Functional plasticity-brains ability to change in response to the activities.
Answer:
One example of a recessive inherited trait is a smooth chin, as opposed to a dominant cleft chin. Let (S) represent the dominant allele, and (s) represent the recessive allele. Only (ss) individuals will express a smooth chin. To determine the probability of inheritance of a smooth chin (or any other recessive trait), the genotypes of the parents must be considered. If one parent is heterozygous (Ss) and the other is homozygous recessive (ss), then half of their offspring will have a smooth chin.
Explanation:
Analysis of a blood sample from a fasting individuals who had not eaten for 24 hours be expected to reveal high levels of glucagon .
Hence the option B is correct .
<h3>What is glucagon ?</h3>
Glucagon is on of the natural hormone of the body .it is peptide hormone.
It is produced by alpha cells of the pancreas .
It controls bodily function which include it controls the glucose level in our body .
It will release in the response to drop in blood sugar due to prolonged fast ,exercise and protein rich meals .
Hence , OPTION B is correct .
Learn more about glucagon here :
brainly.com/question/14697323
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Answer:
2% of the progeny will be double crossovers for the trihybrid test cross
Explanation:
By knowing the positions of genes, we can estimate the distances in MU between them per region.
- Genes A and B are 10 map units apart (Region I)
- Genes B and C are 20 map units apart (Region II)
- Genes A and C are 30 map units apart
----A-------10MU--------B-------------20MU-------------C---
Region I Region II
We can estimate the recombination frequencies by dividing each distance by 100.
• recombination frequency of A-B region = 10MU / 100 = 0.10
• recombination frequency of B-C region = 20MU / 100 = 0.20
Now that we know the recombination frequencies in each region, we can calculate the expected double recombinant frequency, EDRF, like this:
EDRF = recombination frequency in region I x recombination frequency in region II.
EDRF = 0.10 x 0.20 = 0.02
2% of the progeny will be double crossovers for the trihybrid test cross
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