# How many grams of carbohydrate per day are recommended for adults 19 years and older? A. 75 grams B. 100 grams C. 130 grams D. 160 grams

correct option is C. 130 grams

Explanation:

solution

as carbohydrate is a macro nutrient that is found in many foods and beverage

Most of carbohydrate occur in plant base food

some main natural source of carbohydrate are

1. vegetable
2. fruit
3. nuts
4. milk
5. grain
6. seeds

and for adult age of 19 years to 30 years carbohydrate need is 130 grams per day

and adult male age of 19 years to 30 years fiber need is 38 grams per day

and adult female age of 19 years to 30 years fiber need is 25 grams per day

so correct option is C. 130 grams

## Related Questions

If a DNA molecule was a spiral staircase, what would the steps of the staircase be?​

Nitrogenous bases

Explanation:

DNA is a nucleic acid molecule that acts as the genetic material of the organisms. The structure of DNA is made of two strands of nucleotides that are linked together and form a helical structure. The helical structure is known as the double-helical model with a ladder-like appearance.

The backbone of the ladder is composed of the sugar and phosphate groups at alternate positions whereas the staircase of the ladder are made of the different type of nitrogenous bases (A, G, T, C)  joined by the hydrogen bond.

Thus, Nitrogenous bases is the correct answer.

Fine spines (s), smooth fruit (tu), and uniform fruit color (u) are three recessive traits in cucumbers whose genes are linked on the same chromosome. A cucumber plant heterozygous for all three traits is used in a testcross. The progeny from this testcross are:S U Tu 2
s u Tu 70
S u Tu 21
s u tu 4
S U tu 82
s U tu 21
s U Tu 13
S u tu 17__
Total 230
a. Determine the order of these genes on the chromosome.
b. Calculate the map distances between the genes.
c. Determine the coefficient of coincidence and the interference among these genes.
d. Draw the chromosomes of the parents used in the testcross.

We have the number of descendants of each phenotype product of the tri-hybrid cross.

• S U Tu 2
• s u Tu 70
• S u Tu 21
• s u tu 4
• S U tu 82
• s U tu 21
• s U Tu 13
• S u tu 17

The total number, N, of individuals is 230.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the thrid not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

• s u TU (70 individuals)
• S U tu (82 individuals)

Double recombinant)

• S U Tu (2 individuals)
• s u tu (4 individuals)

Comparing them we will realize that between

s u TU (parental)

s u tu (double recombinant)

and

S U tu (Parental)

S U TU (double recombinant)

They only change in the position of the alleles TU/tu. This suggests that the position of the gene TU is in the middle of the other two genes, S and U, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- S ---- TU -----U ----

In a scheme it would be like:

Chromosome 1:

---s---TU---u--- (Parental chromatid)

---s---tu---u--- (Double Recombinant chromatid)

Chromosome 2

---S---tu---U--- (Parental chromatid)

---S---TU---U--- (Double Recombinant chromatid)

Now we will call Region I to the area between S and TU and Region II to the area between TU and U.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between S and TU genes, and P2 to the recombination frequency between TU and U.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.  So:

• P1 = (R + DR) / N

P1 = (21+17+4+2)/230

P1 = 44/230

P1 = 0.191

• P2= = (R + DR) / N

P2 = (21+13+4+2)/230

P1 = 40/230

P1 = 0.174

Now, to calculate the recombination frequency between the two extreme genes, S and U, we can just perform addition or a sum:

P1 + P2= Pt

0.191 + 0.174 = Pt

0.365=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes for which one of every 100 meiotic products results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.191 x 100 = 19.1 MU

GD2= P2 x 100 = 0.174 x 100 = 17.4 MU

GD3=Pt x 100 = 0.365 x 100 = 36.5 MU

To calculate the coefficient of coincidence, CC, we must use the next formula:

CC= observed double recombinant frequency/expected double recombinant frequency

Note:

• observed double recombinant frequency=total number of observed double recombinant individuals/total number of individuals
• expected double recombinant frequency: recombination frequency in region I x recombination frequency in region II.

CC= ((2 + 4)/230)/0.174x0.191

CC=(6/230)/0.0332

CC=0.7857

The coefficient of interference, I, is complementary with CC.

I = 1 - CC

I = 1 - 0.7857

I = 0.2143

What refers to the process by which cells become specialized in structure and function

Hope this will help you... And please mark me as Brilliant

Explanation:

Permineralization. Permineralization is a process of fossilization that occurs when an organism is buried. The empty spaces within an organism (spaces filled with liquid or gas during life) become filled with mineral-rich groundwater. ... Small-scale permineralization can produce very detailed fossils.

Anything written about about an issue/person after the fact is a ________ sourcea.primary
b.secondary

A secondary source is anything written after the fact

Anything written about an issue/person after the fact is a SECONDARY source. A primary source would be the person them self sharing with you,or you witnessing the action or event.

Match each cell type with its description.___natural killer cell
A. stains with basic dye methylene blue, has large amounts of histamine in granules, and facilitates allergic responses and inflammation

___basophil
B. stains with acidic dye eosin, has histamine and major basic protein in granules, and facilitates responses to protozoa and helminths

___macrophage
C. recognizes abnormal cells, binds to them, and releases perforin and granzyme molecules, which induce apoptosis

___eosinophil
D. large agranular phagocyte that resides in tissues such as the brain and lungs

1. A natural killer cell identifies unusual cells, combines with them, and discharges perforin and molecules of granzyme that stimulates apoptosis.

2. A basophil comprises huge concentrations of histamine in granules, gets stain with basic dye methylene blue, and helps in inflammation and allergic reactions.

3. A macrophage refers to a huge agranular phagocyte, which is found within the tissues like lungs and brain.

4. An eosinophil comprises histamine and major basic protein in granules gets stain with acidic dye eosin and helps in reactions against the helminths and protozoa.

The natural killer cell recognizes and destroys abnormal cells, the basophil facilitates allergic responses and inflammation, the macrophage is a large cell that resides in tissues, and the eosinophil facilitates responses to certain parasites.

### Explanation:

Here is the correct match for each cell type with its description:

1. Natural killer cell: C. recognizes abnormal cells, binds to them, and releases perforin and granzyme molecules, which induce apoptosis
2. Basophil: A. stains with basic dye methylene blue, has large amounts of histamine in granules, and facilitates allergic responses and inflammation
3. Macrophage: D. large agranular phagocyte that resides in tissues such as the brain and lungs
4. Eosinophil: B. stains with acidic dye eosin, has histamine and major basic protein in granules, and facilitates responses to protozoa and helminths

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Primary functions of the medusa?

The primary function of medusa is to carry out sexual reproduction and to allow the species to disperse to different locations.

Explanation:

There are two distinct cnidarian body forms: polypoid and medusoid.

The medusa is more of an umbrella or bell shape, with the mouth facing down. The body of the medusa is often called the bell. Medusae are usually free-swimming and either propel themselves using muscle contractions or float along water currents like plankton.