( 30 points ) pls help {brainliest}Which of the following DNA sequences is palindromic?

A. GAGTTC
B. GAATTC
C. GAAGAA
D. CTCTCT

Answer:

18 molecules of

Explanation:

In heterotrophic nutrition, organisms ingest or absorb food made by autotrophs to obtain energy. Planta and other photosynthesizing organisms make sugars via photosynthesis, in a form of (partly) light-dependent biosynthesis.

In mitochondria, they break down sugars through aerobic respiration. For the sugar glucose, this results in the production of carbon dioxide and water along with energy in the form of ATP or adenosine triphosphate.  

C6H12O6 (glucose) + 6 O2 → 6 CO2 + 6 H2O + ≅38 ATP

glucose+ oxygen →  carbon dioxide+ water+ energy

For 3 molecules of glucose, multiply the number of reactants and products...

(C6H12O6+ 6 O2 → 6 CO2 + 6 H2O + ≅38 ATP ) x 3

Thus...

3 C6H12O6+ 18 O2 → 18 CO2 + 18 H2O + ≅114 ATP

Answer:

d = 2 g/mL

Explanation:

Given that,

Water displaced by an object in a graduated cylinder, V = 25 mL

Mass of the object, m = 50 g

We need to find the density of this irregular shaped object. It is given by mass per unit volume. So,

So, the density of the object is 2 g/mL.

Our blood contains blood plasma, leukocytes, and erythrocytes but does not contain lymph. Lymph is a separate body fluid, transported by the lymphatic system.

Blood consists of several components: blood plasma, leukocytes (white blood cells), and erythrocytes (red blood cells). Blood plasma is a yellowish liquid component of blood that holds the blood cells of whole blood in suspension. It makes up about 55% of the body's total blood volume. Leukocytes are the cells of the immune system that are involved in protecting the body against both infectious disease and foreign invaders. Erythrocytes, or red blood cells, are cells present in blood in order to transport oxygen. On the other hand, lymph, although a body fluid, is not a part of the blood. It is mainly composed of white blood cells and is transported by the lymphatic system, separate from the circulatory system that transports blood.

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The classification of living things significantly depends on the structure of their cells and based on this, they are divided into two types. They are as follows:

• Prokaryotes.
• Eukaryotes.

What do you mean by Cell structure?

The structure of a cell may be defined as the components through which it has developed. It includes the cell membrane, the nucleus, and the cytoplasm. The cell membrane is a semi-permeable membrane that separates the inner atmosphere of the cell from the outer environment.

The nucleus is a structure inside the cell that contains the nucleolus and most of the cell's DNA. It is also where most RNA is made. When it comes to prokaryotes, they lack a well-defined nucleus and other membrane-bound cell organelles.

Eukaryotic cells generally have a well-defined nucleus and other membrane-bound cell organelles. Apart from this, these types of cells are generally highly compartmentalized as compared to prokaryotic cells.

Therefore, the classification of living things significantly depends on the structure of their cells that has been described above.

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Answer:

The most basic classification of living things is kingdoms. ... Living things are placed into certain kingdoms based on how they obtain their food, the types of cells that make up their body, and the number of cells they contain. Phylum. The phylum is the next level following kingdom in the classification of living things.

Explanation:

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Answer and Explanation:

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