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marissa [1.9K]
3 years ago
9

The jones family gas had their dog, Sadie since she was a puppy. Sadie was raised in the home, which has a letter carrier visit

at the same time ever day. Though Sadie used to bark and scratch the door when the letter carrier deopped off packages, now she doesnt even wake up from her nap when the letter carrier comes. Sadie is demonstrating a learned behavior that she acquired through what method?
a) habituation
b) trial and error
c) imprinting
d) mimicry

please help?
Biology
2 answers:
bezimeni [28]3 years ago
5 0

a) due to a learned behavior so habituation

Aleks [24]3 years ago
4 0

Answer:

a) habituation

Explanation:

Habituation is the adaptive learning behaviour in which the response strength is reduced when stimulus is given repeatedly. Barking and scratching of door by the dog was the response towards arrival of letter carrier. However, repeated visits of the person at the same time reduced the strength of response through the process of habituation.

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Peter suffers damage to his left frontal lobe and loses the ability to speak, although he can still understand speech. Despite t
ELEN [110]

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.

6 0
2 years ago
How can two parents who show the dominant trait have offspring who shows the recessive trait? Give an example to support your an
natta225 [31]

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:

7 0
3 years ago
Analysis of a blood sample from a fasting individual who had not eaten for 24 hours would be expected to reveal high levels of _
qaws [65]

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 :

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6 0
1 year ago
Genes A and B are 10 map units apart and genes B and C are 20 map units apart, with gene B located between genes A and C. What w
Marizza181 [45]

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

   

7 0
3 years ago
Awnser this quetion<br>please
irinina [24]
Sorry the photo isn't clear
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