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Bingel [31]
3 years ago
9

mc004-1.jpg Photo by NIH/Bill Branson Which type of bond is found in many carbon-to-carbon bonds in canola oil, but very few car

bon-to-carbon bonds in butter?
Biology
1 answer:
Xelga [282]3 years ago
7 0
One of the main differences between butter and canola oil is that are room temperature one is a solid and the other is a liquid. This change in phase is due to the bonding between carbon bonds, the more double bonds that there are present the higher the melting point and the fluidity changes. Based on this the correct answer is C=C is found in canola oil while more C-C bonds are found in butter, making option 2 (C=C) the correct answer. In addition option 3 (C=H) can be ignored since hydrogen can only make single bonds. hope this helps:)

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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
A group on the periodic table is best described as a
sdas [7]
It can also be called a column.
7 0
3 years ago
Read 2 more answers
Please help please thank you
Misha Larkins [42]

Answer:

Hi there!

The correct answer is igneous rock.

Igneous rocks they are most likely to experience brittle deformation is termed as breaking of rocks.

5 0
2 years ago
Which statement describes how predation and parasitism are different?
lakkis [162]
The correct answer is answer choice A, 'In predation the predator usually kills its prey, whereas in parasitism the parasite does not usually kill its host'. While parasites can occasionally kill their hosts, it is more beneficial for the parasite to keep its host alive so that it can continue surviving off of the nutrients it gathers. Answer choice B is incorrect because the prey in predator-prey relationships and the host in parasitism do not benefit. Answer choice C is incorrect because the host does not benefit in parasitism, and is often harmed by the parasite. Answer choice D is incorrect because a predator usually kills its prey, while a parasite usually does not kill its host.

Hope this helps!
3 0
2 years ago
Read 2 more answers
PLEASE HELP!!!
diamong [38]
East or west because of the being of north and south which the latitude is on the equator
8 0
3 years ago
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