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2 years ago

Increased concentrations of alpha-fetoprotein (AFP) in adults are most characteristically associated with:

Chemistry

1 answer:

Vlad1618 [11]2 years ago

8 0

Concentrations of alpha-fetoprotein (AFP):

Increased concentrations of alpha-fetoprotein (AFP) in adults are most characteristically associated with Hepatocellular carcinoma.

Alpha Fetoprotein (AFP):

It is a protein produced in a developing baby's liver. When a newborn is born, their AFP levels are typically high, but by the time they turn one, they are quite low. AFP levels in healthy persons should be extremely low. For adults, a typical AFP level ranges from 10 ng/mL to 20 ng/mL.

In reaction to cancer in the body, normal or cancerous cells produce chemicals known as tumor markers. High levels of AFP can indicate noncancerous liver illnesses such as cirrhosis and hepatitis as well as cancers of the testicles, ovaries, or liver.

Thus, one of the risk factors for Hepatocellular carcinoma has been proven to be a persistent increase in AFP serum levels.

Learn more about the carcinoma here,

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Which of these elements has the highest first ionization energy?

julia-pushkina [17]

Answer:

Cesium (Cs)

Explanation:

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3 years ago

What volume of chlorine gas (Cl2), measured at STP, can be produced by the decomposition of 73.0g of hydrogen chloride gas (HCl)

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Answer:

Explanation:

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3 years ago

What is the sum of 16+ -20?

mr Goodwill [35]

Answer:

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3 years ago

Read 2 more answers

A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

$k=9.98\times 10^{-2}hr^{-1}$

Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

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3 years ago

1: The atomic number of an atom is equivalent to which of the following?

IrinaVladis [17]

I think the correct answer among the choices presented above is option C. The <span>atomic number of an atom is equivalent to the number of protons in the nucleus. For a neutral atom, it is also the number of electrons since in a neutral atom protons and electrons are present in equal number.</span>

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3 years ago