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harkovskaia [24]
3 years ago
10

PLEASE HELP!! (I will mark brainiest) (REAL ANSWERS ONLY PLEASE!)

Chemistry
2 answers:
Inessa [10]3 years ago
8 0

The phosphate group of one nucleotide bonds covalently with the sugar molecule of the next nucleotide, and so on, forming a long polymer of nucleotide monomers. The sugar–phosphate groups line up in a “backbone” for each single strand of DNA, and the nucleotide bases stick out from this backbone. The carbon atoms of the five-carbon sugar are numbered clockwise from the oxygen as 1′, 2′, 3′, 4′, and 5′ (1′ is read as “one prime”). The phosphate group is attached to the 5′ carbon of one nucleotide and the 3′ carbon of the next nucleotide. In its natural state, each DNA molecule is actually composed of two single strands held together along their length with hydrogen bonds between the bases.

Stells [14]3 years ago
7 0

Answer:

Explanation:

The discovery that DNA is the prime genetic molecule, carrying all the hereditary information within chromosomes, immediately focused attention on its structure. It was hoped that knowledge

of the structure would reveal how DNA carries the genetic messages that are replicated when chromosomes divide to produce two identical copies of themselves. During the late 1940s and early 1950s, several research groups in the United States and in Europe engaged in serious efforts—both cooperative and rival—to understand how the atoms of DNA are linked together by covalent bonds and how the resulting molecules are arranged in three-dimensional space. Not surprisingly, there initially were fears that DNA might have very complicated and perhaps bizarre structures that differed radically from one gene to another. Great relief, if not general elation, was thus expressed when the fundamental DNA structure was found to be the double helix. It told us that all genes have roughly the same three-dimensional form and that the differences between two genes reside in the order and number of their four nucleotide building blocks along the complementary strands.

Now, some 50 years after the discovery of the double helix, this simple description of the genetic material remains true and has not had to be ap- preciably altered to accommodate new findings. Nevertheless, we have come to realize that the structure of DNA is not quite as uniform as was first thought. For example, the chromosome of some small viruses have single-stranded, not double-stranded, molecules. Moreover, the precise orientation of the base pairs varies slightly from base pair to base pair in a manner that is influenced by the local DNA sequence. Some DNA se- quences even permit the double helix to twist in the left-handed sense, as opposed to the right-handed sense originally formulated for DNA’s general structure. And while some DNA molecules are linear, others are circular. Still additional complexity comes from the supercoiling (further twisting) of the double helix, often around cores of DNA-binding proteins.

Likewise, we now realize that RNA, which at first glance appears to be very similar to DNA, has its own distinctive structural features. It is principally found as a single-stranded molecule. Yet by means of intra-strand base pairing, RNA exhibits extensive double-helical character and is capable of folding into a wealth of diverse tertiary structures. These structures are full of surprises, such as non-classical base pairs, base-backbone interactions, and knot-like configurations. Most remarkable of all, and of profound evolutionary significance, some RNA molecules are enzymes that carry out reactions that are at the core of information transfer from nucleic acid to protein.

Clearly, the structures of DNA and RNA are richer and more intricate than was at first appreciated. Indeed, there is no one generic structure for DNA and RNA. As we shall see in this chapter, there are in fact vari- ations on common themes of structure that arise from the unique physi- cal, chemical, and topological properties of the polynucleotide chain

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Copper carbonate (CuCO3) reacts with hydrochloric acid (HCl) according to this equation:
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Answer:

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Explanation:

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What is the representative particle for copper metal, Cu
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A representative particle is the smallest unit of a substance that can be broken down without altering the composition. Matter is composed of three types of representative particles: atoms, molecules and formula units.

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Atoms are the smallest particle that can be split. Substances that contain only one kind of atom are called elements.

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Determinar los números cuánticos del electrón desapareado en el átomo del cloro (Z = 17) y diga si es paramagnético
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Answer:

n = 3

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ml = +1

ms = +1/2

Es paramagnético

Explanation:

Siguiendo las reglas de llenado de orbitales, los 17 electrones del cloro se llenan así:

1S = <u>⇅</u>

2S = <u>⇅</u>

2P = <u>⇅</u> <u>⇅</u> <u>⇅</u>

3S = <u>⇅</u>

3P = <u>⇅</u> <u>⇅</u> <u>↑</u>

<u />

El número cuántico principal n, es el nivel energético donde se encuentra este electrón:

n = 3 (Porque está en el orbital 3P

El número cuántico secundario, l, para el orbital 3P es  = 1:

l = 1

El número cuántico magnético, ml, es determinado por la posición del electrón. Como está en el tercer orbital 3P:

ml = +1

Y el número cuántico de spin, ms (↑ = +1/2; ↓ = -1/2)=

ms = +1/2

Dado que el último electrón se encuentra desapareado, el cloro es paramagnético dado que el espín de el último electrón no tiene su electrón complementario haciendo que este compuesto pueda interactuar con un campo magnético.

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The protein catalase catalyzes the reaction The Malcolm Baldrige National Quality Award aims to:
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The question is missing a part, so the complete question is as follows:

The protein catalase catalyzes the reaction The Malcolm Bladrigde National Quality Awards aims to: 2H2O2 (aq) ⟶ 2H2O (l) + O2 (g) and has a Michaelis-Menten constant of KM = 25mM and a turnover number of 4.0 × 10 7 s -1. The total enzyme concentration is 0.012 μM and the intial substrate concentration is 5.14 μM. Catalase has a single active site. Calculate the value of Rmax (often written as Vmax) for this enzyme. Calculate the initial rate, R (often written as V0), of this reaction.

1) Calculate Rmax

The turnover number (Kcat) is a ratio of how many molecules of substrate can be converted into product per catalytic site of a given concentration of enzyme per unit of time:

Kcat = \frac{Vmax}{Et},

where:

Vmax is maximum rate of reaction when all the enzyme sites are saturated with substrate

Et is total enzyme concentration or concentration of total enzyme catalytic sites.

Calculating:

Kcat = \frac{Vmax}{Et}

Vmax = Kcat · Et

Vmax = 4×10^{7} · 1.2 × 10^{-8}

Vmax = 4.8 × 10^{-1} M

2) Calculate the initial rate of this reaction (R):

The Michaelis-Menten equation studies the dynamics of an enzymatic reaction. This model can explain how an enzyme enhances the rate of a reaction and how the reaction rate depends on the concentration of the enzyme and its substrate. The equation is:

V0 = \frac{[S].(Vmax)}{KM + [S]}, where:

[S] is the substrate's concentration

KM is the Michaelis-Menten constant

Substituting [S] = 5.14 × 10^{-6}, KM = 2.5 × 10^{-4} and Vmax = 4.8 × 10^{-1}, the result is V0 = 0.478 M.

The answers are Vmax = 4.8 × 10^{-1} M and V0 = 0.478 M.

 

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