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djverab [1.8K]
3 years ago
14

The activation of beta2 receptors in the bronchi causes bronchodilation (an increase in diameter of the bronchi), making it easi

er to breathe. Which of the following neurotransmitters would be useful for a person suffering from an episode of bronchoconstriction (a reduction in the diameter of the bronchi)?
a. acetylcholine
b. epinephrine
c. vasopressin (or ADH)
d. oxytocin
Biology
1 answer:
UNO [17]3 years ago
8 0

Answer:

The correct answer is (b) epinephrine.

Explanation:

Epinephrine, also known as <em>Adrenaline</em>, is a hormone secreted by the medulla of the adrenal glands.

In medicine, epinephrine is used as a stimulant in cardiac arrest, as a vasoconstrictor in shock and as a bronchodilator and antispasmodic in bronchial asthma.

When epinephrine is inhaled in small doses, it causes short-term relief from the symptoms by widening the bronchial tubes allowing air to pass through.

I hope it helps!

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What are the main digestive enzymes secreted by the pancreas? (Anatomy and Physiology)​
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The enzymes made by the pancreas include:
Pancreatic proteases (such as trypsin and chymotrypsin) - which help to digest proteins.
Pancreatic amylase - which helps to digest sugars (carbohydrates).
Pancreatic lipase - which helps to digest fat.
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3 years ago
How does the conservation of matter relate to the process of digestion?
pshichka [43]

Answer: According to the the law of Conservation of Matter, matter is neither created nor destroyed. ... When matter changes from one form to another, e.g., a portion of beef steak (beef protein) is eaten and digested, the macro-structure of the beef is broken by our digestive enzymes into its component amino acids.

Explanation:

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Overtime a species can change - due to environmental factors and geographic changes. Sometimes, these changes can lead to new sp
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Answer:

geographic isolation

Explanation:

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3 years ago
In a lab about the activity series, a solution of copper(II)chloride is placed in a test tube with zinc metal. If zinc is higher
Tcecarenko [31]

Answer:

Option D, Zinc is higher on the activity series because there is clear evidence of a reaction.

Explanation:

It is very clear that a metal which is higher in activity will replace a metal having lower activity.  

Since, Zinc has higher activity than copper, it is very sure that it in a copper chloride solution it will replace copper and produce zinc chloride solution leaving behind copper as residue.  

Residues of copper is a clear evidence of reaction  

The chemical equation for this reaction would be -

CuCl_2 + 2Zn^+ -----> 2ZnCl + Cu^{2+}

Hence, option D is correct

5 0
3 years ago
Each of the four pedigrees that follow represents a human family within which a genetic disease is segregating. Affected individ
Ne4ueva [31]

 Answer:

<u> The following four traits are -: </u>

  • <u>Pedigree 1 -</u> A recessive trait (autosomal recessive)  is expressed by pedigree 1.
  • <u>Pedigree 2- Recessive inheritance is defined by Pedigree 2. </u>
  • <u>Pedigree 3</u> - The inheritance of the dominant trait (autosomal dominant) is illustrated by Pedigree 3.
  • <u>Pedigree 4-</u> An X-like dominant trait is expressed by Pedigree 4.    

Explanation:

<u>Explaination of each pedigree chart</u>-

  • Pedigree 1 demonstrates the <u>recessive trait </u>since their children have been affected by two unaffected individuals. If the characteristics were X-linked, in order to have an affected daughter, I-1 would have to be affected. X^A In this, both parents are autosomal recessive trait carriers, so the child will be affected by a 1/4 (aa)
  • <u> Recessive inheritance</u> is defined by <u>Pedigree 2</u>. This is<u> X-related inheritance as autosomal recessive</u> inheritance has already been accounted for in part 1. This inference is confirmed by evidence showing that the father (I-1) is unaffected and that only the sons exhibit the characteristic in generation II, suggesting that the mother must be the carrier. The individual I-2 is a carrier for this X-linked trait. A typical  Xa chromosome is attached to the unaffected father (I-1), so the chance of carrier II-5 is 1/2. Probability of an affected son = 1/2 (probability II-5 is a carrier) x 1/2 (probability II -5 contributes (X^A) x 1/2 (probability of Y from father II-6) = 1/8. An affected daughter's likelihood is 0 because a typical X^A must be contributed by II-6.
  • The inheritance of the<u> dominant trait</u> is demonstrated by <u>Pedigree 3 </u>because affected children still have affected parents (remember that all four diseases are rare). The trait must be <u>autosomal dominant</u> because it is passed down to the son by the affected father. There is a 1/2 risk that the heterozygous mother (II-5) would pass on mutant alleles to a child of either sex for an autosomal dominant feature.
  • <u>Pedigree 4</u> is an <u>X-linked dominant function</u> characterized by the transmission to all of his daughters from the affected father but none of his son. On the mutant X chromosome, the father (I-1) passes on to all his daughters and none of his sons. As seen by his normal phenotype, II-6 therefore does not bear the mutation. An affected child's likelihood is 0.    

In the question the pedigree chart was missing ,hence it is given below.

     

7 0
3 years ago
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