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skelet666 [1.2K]
3 years ago
15

What is the volume of a cylinder if the height is 6 cm and the radius is 4 cm?​

Biology
1 answer:
Sloan [31]3 years ago
8 0

Answer:

v = πr²h

v = 3.142*4²*6

v = 301.71cm³

Explanation:

volume of the cylinder = πr²h

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the answer is Nonmetal

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3 years ago
In which type of exercise would fat reserves be used to create ATP?
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I believe it's aerobic exercise.
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A group of cells is assayed for DNA content immediately following mitosis and is found to have an average of 4 picograms of DNA
damaskus [11]

Answer:

b: 8;8

Explanation:

Mitotic or meiotic cell division constitute the m phase of the cell cycle. At the end of the m phase, the new cells enter the interphase stage of the cell cycle. The interphase is further sub-divided into;

  • <em>the G_0 phase,</em>
  • <em>the G_1 phase,</em>
  • <em>the S phase; and</em>
  • <em>the G_2 phase</em>.

The G_0 phase is essentially a resting phase. Cells that do not need to divide except when necessary move into this phase after exiting the m phase.

Actively dividing cells enter the G_1 phase after exiting the m phase. Cell development and growth takes place. From there, the cells enter the S phase where DNA replication/synthesis takes place. The cells then enter the G_2 phase where proteins are synthesized in preparation for division or m phase.

At the S phase, the amount of DNA a cell carries is doubled but the chromosome number remains the same. For example, if a cell enters the S phase with 2 g of DNA containing 10 chromosomes, at the end of S phase, the amount of DNA would have come 4 g while the number of chromosomes will remain 10.

Hence, if the average amount of DNA in the assayed cells immediately after mitosis is 4 picograms, the amount would be 8 picograms at the end of S phase and will still remains 8 picograms at the end of G_2 phase.

The correct option is b.

8 0
3 years ago
Explain how we know that DNA breaks and rejoins during recombination.
alisha [4.7K]

Answer:

It occurs through homologous recombination

Explanation:

GENERAL RECOMBINATION OR HOMOLOGIST

           Previously we defined its general characteristics. We will now describe a molecular model of this recombination, based on the classic Meselson and Radding, modified with the latest advances. Do not forget that we are facing a model, that is, a hypothetical proposal to explain a set of experimental data. Not all points of this model are fully clarified or demonstrated:

           Suppose we have an exogenote and an endogenote, both consisting of double helices. In recombination models, the exogenote is usually referred to as donor DNA, and the endogenote as recipient DNA.

1) Start of recombination: Homologous recombination begins with an endonucleotide incision in one of the donor double helix chains. Responsible for this process is the nuclease RecBCD (= nuclease V), which acts as follows: it is randomly attached to the donor's DNA, and moves along the double helix until it finds a characteristic sequence called c

Once the sequence is recognized, the RecBCD nuclease cuts to 4-6 bases to the right (3 'side) of the upper chain (as we have written above). Then, this same protein, acting now as a helicase, unrolls the cut chain, causing a zone of single-stranded DNA (c.s. DNA) to move with its 3 ’free end

2) The gap left by the displaced portion of the donor cut chain is filled by reparative DNA synthesis.

3) The displaced single chain zone of the donor DNA is coated by subunits of the RecA protein (at the rate of one RecA monomer per 5-10 bases). Thus, that simple chain adopts an extended helical configuration.

4) Assimilation or synapse: This is the key moment of action of RecA. Somehow, the DNA-bound RecA c.s. The donor facilitates the encounter of the latter with the complementary double helix part of the recipient, so that in principle a triple helix is formed. Then, with the hydrolysis of ATP, RecA facilitates that the donor chain moves to the homologous chain of the receptor, and therefore matches the complementary one of that receptor. In this process, the chain portion of the donor's homologous receptor is displaced, causing the so-called "D-structure".

It is important to highlight that this process promoted by RecA depends on the donor and the recipient having great sequence homology (from 100 to 95%), and that these homology segments are more than 100 bases in length.

Note that this synapse involves the formation of a portion of heteroduplex in the double receptor helix: there is an area where each chain comes from a DNA c.d. different parental (donor and recipient).

5) It is assumed that the newly displaced chain of the recipient DNA (D-structure) is digested by nucleases.

6) Covalent union of the ends originating in the two homologous chains. This results in a simple cross-linking whereby the two double helices are "tied." The resulting global structure is called the Holliday structure or joint.

7) Migration of the branches: a complex formed by the RuvA and RuvB proteins is attached to the crossing point of the Holliday structure, which with ATP hydrolysis achieve the displacement of the Hollyday crossing point: in this way the portion of heteroduplex in both double helices.

8) Isomerization: to easily visualize it, imagine that we rotate the two segments of one of the DNA c.d. 180o with respect to the cross-linking point, to generate a flat structure that is isomeric from the previous one ("X structure").

9) Resolution of this structure: this step is catalyzed by the RuvC protein, which cuts and splices two of the chains cross-linked at the Hollyday junction. The result of the resolution may vary depending on whether the chains that were not previously involved in the cross-linking are cut and spliced, or that they are again involved in this second cutting and sealing operation:

a) If the cuts and splices affect the DNA chains that were not previously involved in the cross-linking, the result will be two reciprocal recombinant molecules, where each of the 4 chains are recombinant (there has been an exchange of markers between donor and recipient)

b) If the cuts and splices affect the same chains that had already participated in the first cross-linking, the result will consist of two double helices that present only two portions of heteroduplex DNA.

8 0
4 years ago
Explain the roles of mrna and trna in protein synthesis.
docker41 [41]

Answer:

mRNA is transcribed from DNA.

The nucleotides in mRNA code for amino acids.

tRNA translates the mRNA sequence into an amino acid sequence so proteins can be made.

Explanation: FOR REAL THE FACTS

6 0
4 years ago
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