HELP!!!!!

The hyperalgesia that is common in phantom limb pain could potentially be blocked if a new drug was developed that could prevent

gregori [183]

Answer: The answer is B- Diffusion of calcium ions through NMDA receptors.

Explanation:

Phantom limb pain (PLP) is a painful sensation experienced after the loss or amputation of a limb. The limb is no longer there but the individual can still feel real and severe pains in the limb. This occurs soon after surgery.

NMDA is a receptor for Glutamate which is an excitatory neurotransmitter. The activation of NMDA receptors are associated with hyperalgesia which is an abnormally heightened sensitivity to pain.

8 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:

The following four traits are -:

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

Explanation:

Explaination of each pedigree chart-

Pedigree 1 demonstrates the recessive trait 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)
Recessive inheritance is defined by Pedigree 2. This is X-related inheritance as autosomal recessive 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 dominant trait is demonstrated by Pedigree 3 because affected children still have affected parents (remember that all four diseases are rare). The trait must be autosomal dominant 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.
Pedigree 4 is an X-linked dominant function 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

The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61

erastova [34]

C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately 6890 years.

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

Available data:

The half-life of C14 is 5730 years
Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g

C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g

λ ⇒ radioactive decay constant = (Ln2)/T₀,₅

T₀,₅ ⇒ The half-life of carbon 14 = 5730 years

T₀ = Time when the organism was alive

T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

λ = 0.0001209

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 = - 0.833

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ = 6889.99 ≅ 6890 years

The bones are approximately 6890 years.

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

#SPJ1

6 0

2 years ago

The storage form of carbohydrates in animals is __________; and in plants, it is __________.

noname [10]

Glycogen(for animals), Starch(for plants)

7 0

3 years ago

The cell division process that produces daughter cells with half the chromosome of the parent cell is _________.

STatiana [176]

Answer: Mitosis (A)

Explanation:

5 0

3 years ago