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adelina 88 [10]
3 years ago
9

Which shows the levels of organizational hierarchy listed from least complex to most complex?

Biology
1 answer:
Alex_Xolod [135]3 years ago
4 0
Organism, population, community (first choice)



I hope this helped out, have a nice day! :)
You might be interested in
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
3 years ago
Now, assume the individuals with the filled symbols in the kindred have the O blood type, whereas all the individuals with the o
jek_recluse [69]

Answer:

When we talk about the inheritance of traits, or the passage of traits from parents to future generations down the line, we are not just talking about the visual (phenotypical) expression of those traits, but also, their underlying explanation, which is the genotype. A genotype is basically how the genes of the parents combine in such a way that the children inherit a set of traits from the parents, and express them phenotypically, or not.

In the case of blood types, we have four phenotypic groups: A, B and O. Each one of these types is characterized by the underlying set of genes that are responsible for what is expressed. While the O blood type presents a genotype ii, which is recessive, the A and B types will have the following genetic patterns: Ia Ia, or, Ia i (characteristic of the O genetic material) for the A type and: Ib Ib, or Ib i, for the B type. When there is a genetic conjugation from parents genetic material, regarding blood type, we would have these sets of genes combining. In most of the possible combinations genetically speaking, we have the recessive i gene appearing, including in the A and B dominant blood types. This means that when crossed, there will always be a chance of at least one offspring presenting the O blood type, even if one of the parents is dominant A, or B.

In answer: it is the fact that all three types present the recessive allele i, typical of the O blood type, that when pairings of genes happen between parents, the genetic characteristic of the O type may present itself in a dominant fashion, instead of the usual recessive pattern.

4 0
3 years ago
Which level of organization is the largest?<br> Domain<br> Kingdom<br> Genus<br> Species
sertanlavr [38]
It's kingdom :)))))))
4 0
3 years ago
Read 2 more answers
Which model of replication did the actual results of the Meselson-Stahl experiment support? Explain your answer.
AfilCa [17]

Answer:

Semi-conservative replication

Explanation:

After the double-helix discovery of Watson and Crick, there were three possible models about the DNI replication:

  • The Conservative model stated that the two strands of DNI together were the template of another new molecule. The final product was the original double-stranded molecule and the new molecule.
  • The semi-conservative model stated that the original DNI molecule separated into two strands, and each of them served as a template for the synthesis of a new complementary strand. The replication product would be two double-stranded DNA molecules, each carrying an original strand a new one.
  • The Dispersive moles stated that the replication product would be two molecules made by a mixture of segments of the original and the new molecules.

Meselson and Stahl joined to discover which of the models was the correct one. To do it they used E. coli and Nitrogen isotopes.

  • First, they extracted DNI from bacteria grown in a medium with N¹⁴ and got its density band by centrifugation.
  • Then they grew bacteria in a medium with N¹⁵, extracted their DNI molecules, centrifugated them, and got the density band, which was heavier than the firsts ones.
  • The researchers then transferred bacteria grown in medium with N¹⁵  to a medium with N¹⁴, and they allowed only one replication process to occur. DNI was extracted and centrifugated again, and a new band appeared. This band was an intermediate form between bands of DNI-N¹⁵ and DNI-N¹⁴.

This event <em>eliminated the conservative model</em>. If this model were correct, the expected result would be to get two bands: one corresponding to the density DNI-N¹⁵ and the other corresponding to the density DNI-N¹⁴.

  • Bacteria grown in a medium with N¹⁵ and then transferred to a medium with N¹⁴ were finally allowed to replicate twice. Their DNI was extracted and centrifugated. The result was two bands: one of them coincided with the intermediate band, and the other one with the DNI-N¹⁴.

<u>This result was conclusive</u> because if the dispersal model were correct, these two bands should not appear, as all the DNI strands would have part of the original molecule.

With this experiment, Meselson and Stahl proved that the correct replication model was the semi-conservative one.

5 0
3 years ago
Please help me I will give you 18 points <br> PLZ I need help
Thepotemich [5.8K]

Answer:

I believe it's budding or twinning

Explanation:

Hope i helped

6 0
2 years ago
Read 2 more answers
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