Why is one Of the strands of DNA considered leading and the other is considered lagging?

The correct answer is Ribosomes is the single molecule that could do both cellular work and replicate itself.

Ribosomes have two primary functions: message decoding and peptide bond formation. The ribosomal subunits are made up of two large ribonucleoprotein particles (RNPs) of unequal size. Each subunit is composed of one or more ribosomal RNAs (rRNAs) and a large number of ribosomal proteins (r-proteins). The ribosome is an ancient molecular fossil that can be used to peer into the origins of life. In all living systems, the ribosome, which is composed of RNA and protein, converts mRNA to coded protein. Translation is infused with universality, economy, centrality, and antiquity. Because the ribosome is so important to all aspects of life and reproduction in an organism, rapid change is unlikely to survive. The ribosome components are an excellent resource for studying the evolution of all organisms because all cellular organism have ribosomes .

Based on a number of considerations, it is hypothesized that at some time in the evolution of life, there must have been a single molecule that could do both cellular work and replicate itself. A possible molecule that has been suggested is a catalytic RNA molecule, now called a ?

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If I am understanding correctly, the first thing they are asking you to do is match nitrogen bases.

Adenine (A) always pairs with thymine (T), and guanine (G) always pairs with cytosine (C).

Then, whichever are the double strands, cut them out and put them into you notebook. (I am not sure where the notebook is, but I am assuming you know)

Then in the speech bubble next to the picture of Chargaff- talk about his rule.

What does the rule mean?

What does the rule explain?

How is the rule demonstrated?

Lastly, finding percents. What percent of the pairs has Adenine in them?

For example:

If I look at this pairing below

A →T

T→ A

G →C

C→ G

50% of the pairs include Adenine.

Hope I helped! Comment if you have questions about my answer :)

Answer:

If the sequence of nitrogenous bases of a DNA strand is ATCTAGGCCG, the complementary strand would be TAGATCCGGC, with a percentage of guanine 30%, cytosine 30%, adenine 20% and thymine 20%, according to Chargaff's rule.

Explanation:

DNA is made up of sequences of nitrogenous bases, which are adenine (A), guanine (G), thymine (T) and cytosine (C).  The bases of one strand are matched with the bases of another, according to the complementarity of nitrogenous bases, where:

The image shows a DNA chain whose chain is complementary:

DNA Strand given      Complementary Strand

    Adenine                                Thymine

    Thymine                                Adenine

    Cytosine                                Guanine

    Thymine                                Adenine

    Adenine                                Thymine

    Guanine                                 Cytosine

    Guanine                                 Cytosine

    Cytosine                                Guanine

    Cytosine                                Guanine

    Guanine                                 Cytosine

Therefore, in a DNA molecule there is as much adenine as thymine, and an equal amount of cytosine and guanine. In this example there are 4 molecules of adenine, 4 molecules of thymine, 6 guanines and 6 cytosines.

The proportion or percentages of nitrogenous bases can be calculated according to Chargaff's rule.

Chargaff was able to establish that in a DNA molecule the ratio of purine:pyrimidine of 1:1, so there must be the same amount of thymine as adenine and a similar amount of guanine for the cytosine, according to the complementarity of bases.

Taking into account the law of the base pair, if in a DNA chain there is 30% of Adenine, in the molecule there is:

Guanine 30%

Cytosine 30%

Adenine 20%

Thymine 20%

Total ..... 100%

Answer:

i say that the answer is c

Explanation:

Answer:

Carbon dioxide

Answer:

First box: Carbohydrates

monomer is monosaccarides single sugar molecules

Elements: C, H, O

examples are glucose and cellulos

Box two: Proteins

Monomer: amino acid

Elements: C, H, O, N, P, S

function: a lot of differnt functions but examples include:

– Enzymes that control the rate of biochemical reactions

– Hormones that regulate cell processes (Ex. Insulin)

– Structurally make up bones and muscles (Ex. Collagen)

– Transport substances in and out of cell (Ex. Hemoglobin)

– Antibodies help immune system fight diseases

– Movement (Ex. Contractile proteins)

– Receptors aid in cell signaling

– Energy source in the food we eat (Ex. Casein)

Examples: meats, nuts, and dairy products, but many are made by your body

Box three: Nucleic Acid

Monomer: Nucleotides

Elements: C, H, O, N, P

function: informational molecules that store, transmit, and express our genetic information; contain the instructions for making proteins

example: DNA and RNA

Box Four: Lipids

Monomer: fatty acids

Elements: C, H, O

Function: Long term energy storage

Examples: fats, oils, phospholipids (in the cell membrane), steroids

Explanation:

Answer: E. Magnesium

Explanation:

Trust me bro

Answer:

benedicts solutuin(c)

Organism represents an individual form of life such as an animal, plant, or single-celled life form. A prokaryote or a eukaryote can be a unicellular organism.  Eukaryotic organisms have a membrane-bound cell nucleus as well as extra membrane-bound compartments known as organelles (such as mitochondria in animals and plants and plastids in plants and algae, all generally considered to be derived from endosymbiotic bacteria)

Fungi, mammals, and plants are examples of eukaryotic organism kingdoms. Organisms are classified as producers, consumers, herbivores, carnivores, omnivores, scavengers, parasites, predators, and decomposers. Everything you need to know about the 39 trillion microbes that live in our bodies, according to the human microbiome.

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Answer:

Small populations tend to lose genetic diversity more quickly than large populations due to stochastic sampling error (i.e., genetic drift). This is because some versions of a gene can be lost due to random chance, and this is more likely to occur when populations are small.

Explanation:

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A pulmonary (lung) artery obstruction is referred to as an acute pulmonary embolism or embolus.

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The statement that is not true related to the protein structure is option b.  Proteins' tertiary structure is dictated by R group bonds.

It is classified as the  three dimensional shape of the protein.

In the tertiary structure it does contain the  single polypeptide chain that we called as the backbone or there more than one protein secondary structures

Also, it should not be dictated by R group bonds.

Therefore, we can conclude that the option b is correct.

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Answer:  A)

Explanation:

Answer:

c) deletion is the type of mutation in the following

Answer : Because you can't make something out of thin air.

Explanation:

It has to come from somewhere. And that somewhere is the cell.

Answer:

They will most likely be immune to the treatment.

Explanation:

This is the same as a viruse evolving to survive the vaccine for it.

Answer: biomarker: A substance used as an indicator of a biological state, most commonly disease.

trace fossil: A type of fossil reflecting the reworking of sediments and hard substrates by organisms including structures like burrows, trails, and impressions.

fossil record: All discovered and undiscovered fossils and their placement in rock formations and sedimentary layers.

strata: Layers of sedimentary rock.

fossiliferous: Containing fossils.

Explanation: hope this is what you mean if not ill try again

Answer:

5) summer

6)winter

7)the equator

8)south pole

Answer:

D

Explanation:

Hope it helps

Answer:

Asexual Reproduction

Explanation:

Amoeba is an example of Asexual Reproduction.

Hope it helps!

Answer: Lipids

Explanation:

Answer:

all of them duh even trees do and flowers and many more all plants even humans do it and animals

Answer:

full

Explanation:

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Alkali metals are the ones that belong to the 1st group of the periodic table, they are the most reactive metals having less ionization energy.

And one more thing is common between them, that is :

All alkali metals are soft metals, that can easily be cut through knife.

So, required choice is : B. Softer

Answer: wave B has a higher pitch

Explanation:

I mean its obvious...

Answer:

A. Wave B has a higher pitch.

Explanation:

I took the quiz apix

Answer:

A

Explanation:

Answer:

The correct answer is D, "there are approximately 83 cells in interphase. Disruption of interphase will result in the cell being unable to synthesize proteins and organelles required to divide, which will result in the cell not passing key checkpoints and ceasing to divide."

Hope this helps!

Monosaccharide profiling of glycoproteins by capillary electrophoresis with contactless conductivity detection is saccharides.

Saccharides form one of the major constituents of the biological macromolecules in living organisms. Many biological processes including the protein folding, stability, immune response and receptor activation are regulated by glycosylation.

In this work, we optimized a capillary electrophoresis method with the capacitively coupled contactless conductivity detection for separation of eight monosaccharides commonly found in glycoproteins, namely D-glucose, D-galactose, D-mannose, N-acetyl-D-glucosamine, N-acetyl-D-galactosamine, D-fucose, N-acetylneuraminic acid, and D-xylose. A highly alkaline solution of 50 mM sodium hydroxide, 22.5 mM disodium phosphate, and 0.2 mM CTAB (pH 12.4) was used as background electrolyte in a 10 µm id capillary. To achieve baseline separation of all the analytes, a counter-directional pressure of –270 kPa was applied during the separation. The limits of the detection of our method were below 7 µg/ml (i.e., 1.5 pg or 1 mg/g protein) and the limits of quantification were below 22 µg/ml (i.e., 5 pg or 3 mg/g protein). As a proof of the concept of our methodology, we performed an analysis of monosaccharides released from fetuin glycoprotein by acid hydrolysis. The results show that, when combined with an appropriate pre-concentration technique, developed method can be used as a monosaccharide profiling tool in glyco-proteomics and complement the routinely used LC-MS/MS analysis.

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